FATE-H.jsonl

{"informal_statement": "Prove that if $H$ is a subgroup of $G$ of index $n$, then there is a normal subgroup $K$ of $G$ such that $K\\leq H$ and $[G:K]\\leq n!$", "formal_statement": "theorem subgroup_normal_index_le_factorial {G : Type} [Group G] {n : ℕ} (hn : n ≠ 0) \n    (H : Subgroup G) (hH : H.index = n) :\n    ∃ K : Subgroup G, K.Normal ∧ K ≤ H ∧ K.index ≠ 0 ∧ K.index ≤ n.factorial := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Group Actions and Sylow theorems"], "name": "Problem_1", "informal_proof": "\nLet $X$ be the set of left cosets of $H$. Consider $\\phi \\colon G \\to \\mathrm{Sym}(X)$ given by $\\phi(x)(aH) = (xa)H$. Then $\\phi$ is a homomorphism. \n\nConsider now $K = \\ker \\phi$. Then $K$ is a normal subgroup of $G$ contained in $H$. \n\nFinally, $G/K$ is isomorphic to a subgroup of $\\mathrm{Sym}(X)$, which has order $n!$, where $n = [G:H]$. Thus, $[G:K]$ is finite and divides $n!$.\n", "header": "import Mathlib\n\n", "informal_prefix": "/- Prove that if $H$ is a subgroup of $G$ of index $n$, then there is a normal subgroup $K$ of $G$\nsuch that $K\\leq H$ and $[G:K]\\leq n!$-/"}
{"informal_statement": "Prove that if $\\#G = 56$ then $G$ is not simple.", "formal_statement": "theorem not_isSimpleGroup_of_card_eq_56 {G : Type} [Group G] (hG : Nat.card G = 56) :\n    ¬ IsSimpleGroup G := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Group Actions and Sylow theorems"], "name": "Problem_2", "informal_proof": "\nWe have that \\( 56 = 2^3 \\cdot 7 \\). Then the number of Sylow 7-subgroups is \\( 1 + 7k \\), for some \\( k \\geq 0 \\) and divides 8. So the number of Sylow 7-subgroups is either 1 or 8. If it is 1, we are done. So let us assume that it is 8. Each Sylow 7-subgroup has order 7. In a group of prime order every nonidentity element has order 7. Thus each Sylow 7-subgroup has 6 elements of order 7. Also, they intersect trivially. It follows that there are \\( 8 \\cdot 6 = 48 \\) elements of order 7. This leaves only \\( 56 - 48 = 8 \\) elements, which is the size of one Sylow 2-subgroup. Thus, there is only room for one such subgroup. That is, either the Sylow 7-subgroup or the Sylow 2-subgroup must be normal.\n", "header": "import Mathlib\n\n", "informal_prefix": "/- Prove that if $\\#G = 56$ then $G$ is not simple.-/"}
{"informal_statement": "Prove that if $\\#G = 3393$ then $G$ is not simple.", "formal_statement": "theorem not_isSimpleGroup_of_card_eq_3393 {G : Type} [Group G] (h : Nat.card G = 3393) : ¬ IsSimpleGroup G := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Group Actions and Sylow theorems"], "name": "Problem_3", "informal_proof": "\nFirst, factorize the order of $G$: $3393=3^2\\times 13\\times 29$. If $G$ is a simple group, let $n$ be the number of Sylow-13 subgroups, we know that $n\\mid 3^2\\times 29$ and $13\\mid n-1$ which shows that $n=261$. These cyclic subgroups shows that there are at least $261\\times (13-1)= 3132$ elements with order $13$. In the same way, we know that the number of Sylow-29 subgroups is 117 and there are at least $117\\times (29-1)=3276$ elements with order 29. These can not hold at the same time, which means either the Sylow-13 subgroup or the Sylow-29 subgroup is normal.\n", "header": "import Mathlib\n\n", "informal_prefix": "/- Prove that if $\\#G = 3393$ then $G$ is not simple. -/"}
{"informal_statement": "Prove that if $p$ is a prime and $P$ is a non-abelian group of order $p^3$, then $|Z(P)| = p$ and $P/Z(P) \\cong \\mathbb{Z}/p\\mathbb{Z} \\times \\mathbb{Z}/p\\mathbb{Z}$.", "formal_statement": "theorem nonempty_mulEquiv_zMod_prod_zMod {p : ℕ} [Fact p.Prime] {P : Type} [Group P] (hp : Nat.card P = p ^ 3)\n    (h : ∃ (a b : P), a * b ≠ b * a) : Nat.card (Subgroup.center P) = p ∧\n    Nonempty ((P ⧸ Subgroup.center P) ≃* Multiplicative ((ZMod p) × (ZMod p))) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "p-Groups, Nilpotent Groups and Solvable Groups"], "name": "Problem_4", "informal_proof": "\nConsider $Z(G)$ the center of the group $G$.\n\n\\textbf{Claim 1.} $Z(G)$ is of order $p$.\n\n\\textbf{Proof of the Claim 1.} We know that $Z(G) \\leq G$. By Lagrange's Theorem $|Z(G)|$ must divide $|G|$. Since $|G| = p^3$ the only possibilities are $1, p, p^2, p^3$. \n\\begin{itemize}\n    \\item $|Z(G)| \\neq p^3$ because otherwise we will have $Z(G) = G$ but $G$ is non-abelian.\n    \n    \\item $|Z(G)| \\neq p^2$ also because otherwise we will have the order of the factor group by the center as $|G/Z(G)| = |G|/|Z(G)| = p^3/p^2 = p$.\n    \n    We have: $|G/Z(G)| = p \\implies G/Z(G)$ is cyclic $\\implies G$ is abelian. But $G$ is non-abelian.\n\n    \\item Now $|Z(G)| \\neq 1$ also because $G$ is a $p$-group and $p$-groups have non-trivial center.\n\\end{itemize}\n\nThus the only possible order for $Z(G)$ is $p$. \\qed\n\n\\textbf{Claim 2.} Any group $H$ of order $p^2$ is abelian, thus either isomorphic to $\\mathbb{Z}/p^2\\mathbb{Z}$ or $\\mathbb{Z}/p\\mathbb{Z} \\times \\mathbb{Z}/p\\mathbb{Z}$.\n\n\\textbf{Proof of the Claim 2.} The center of $H$ is nontrivial. Thus either $H$ is abelian or $H/Z(H)$ is of order $p$. However, in the latter case, $H/Z(H)$ must be cyclic, which implies $H$ is abelian, contradiction. \\qed\n\nCombining Claim 1 and Claim 2, we know that $G/Z(G)$ is a group of order $p^2$ and must be isomorphic to $\\mathbb{Z}/p^2\\mathbb{Z}$ or $\\mathbb{Z}/p\\mathbb{Z} \\times \\mathbb{Z}/p\\mathbb{Z}$. If $G/Z(G)$ is isomorphic to the cyclic group $\\mathbb{Z}/p^2\\mathbb{Z}$, it contradicts to $G$ is non-abelian. So $G/Z(G)$ must be $\\mathbb{Z}/p\\mathbb{Z} \\times \\mathbb{Z}/p\\mathbb{Z}$.\n\n", "header": "import Mathlib\n\n", "informal_prefix": "/- Prove that if $p$ is a prime and $P$ is a non-abelian group of order $p^3$, then $|Z(P)| = p$\nand $P/Z(P) \\cong \\mathbb{Z}/p\\mathbb{Z} \\times \\mathbb{Z}/p\\mathbb{Z}$.-/"}
{"informal_statement": "Let $R$ be a ring with $1 \\neq 0$.\nFor two elements $a,b \\in R$, if $1-ab$ is a unit, then $1-ba$ is a unit.", "formal_statement": "theorem isUnit_one_sub_mul {R : Type} [Ring R] [Nontrivial R] {a b : R} (h : IsUnit (1 - a * b)) :\n    IsUnit (1 - b * a) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Ring Theory", "Basic Definitions and Examples"], "name": "Problem_5", "informal_proof": "\nThere is an element $s\\in R$ such that $s(1-ab)=(1-ab)s=1$. Consider the element $t=1+bsa$, the computation $t(1-ba)= 1+bsa-ba-bsaba=1-ba+b(s-sab)a=1; (1-ba)t= 1+bsa-ba-babsa= 1-ba+b(s-abs)a=1$ shows that t is the inverse of $1-ba$.\n", "header": "import Mathlib\n\n", "informal_prefix": "/- Let $R$ be a ring with $1 \\neq 0$.\\nFor two elements $a,b \\in R$, if $1-ab$ is a unit,\nthen $1-ba$ is a unit. -/"}
{"informal_statement": "Show that a prime $p$ can be written as $p = a^2+ab+b^2$ with $a,b \\in \\mathbb{Z}$ if and only if $p=3$ or $p \\equiv 1 \\pmod 3$.", "formal_statement": "theorem exists_sum_two_squares_iff_mod_three_eq_one (p : ℕ) (hp : p.Prime) :\n    (∃ a b : ℤ, a ^ 2 + a * b + b ^ 2 = p) ↔ p = 3 ∨ p % 3 = 1 := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Other Topics", "Algebraic Number Theory and Valuation Theory"], "name": "Problem_6", "informal_proof": "\nFor `only if' part, given \\( p = x^2 + xy + y^2 = (x - y)^2 + 3xy \\), this expression relates to properties of perfect squares modulo 3.  If \\( p \\neq 3 \\), then \\( x - y \\) is not divisible by 3. Hence, \\( (x - y)^2 \\equiv 1 \\pmod{3} \\) and it follows that \\( p \\equiv 1 \\pmod{3} \\).\n\nFor `if' part, the case when $p=3$ is trivial. We assume \\( p \\equiv 1 \\pmod{3} \\). Let \\( K = \\mathbb{Q}(\\sqrt{-3}) \\). Consider the integer ring \\( \\mathscr{O}_K = \\mathbb{Z}[\\alpha] \\), where \\( \\alpha = \\frac{1+\\sqrt{-3}}{2} \\). Recall that the field norm of an element \\( x + y\\alpha \\in \\mathscr{O}_K \\) is\n\\[\nN_{K/\\mathbb{Q}}(x + y\\alpha) = (x + y\\alpha)(x + y\\overline{\\alpha}) = x^2 + xy + y^2.\n\\]\nSince \\( \\mathscr{O}_K \\) is a PID, it follows that a prime \\( p \\) can be expressed as \\( p = x^2 + xy + y^2 \\) if and only if \\( p \\) is the norm of some element of \\( \\mathscr{O}_K \\).\n\nIf \\( p = \\mathfrak{p}_1 \\mathfrak{p}_2 \\), then since \\( \\mathscr{O}_K \\) is a PID, \\( \\mathfrak{p}_1 \\) is generated by an element of \\( \\mathscr{O}_K \\), which must have norm \\( p \\). Hence,\n$p$ be expressed as \\( p = x^2 + xy + y^2 \\).\n\nTherefore, it suffices to show that \\( p \\) splits completely in \\( \\mathscr{O}_K \\). We use the following version of the \\emph{Kummer-Dedekind Theorem}:\n\n{\\it Theorem: [Kummer-Dedekind]\nLet \\( p \\) be a prime, and let \\( \\beta \\in \\mathscr{O}_K \\) be such that \\( K = \\mathbb{Q}(\\beta) \\) and \\( p \\nmid (\\mathscr{O}_K : \\mathbb{Z}[\\beta]) \\). Let \\( f(X) \\) be the minimal polynomial of \\( \\beta \\) over \\( \\mathbb{Q} \\). Suppose that\n\\[\nf(X) \\equiv f_1(X)^{e_1} \\cdots f_m(X)^{e_m} \\pmod{p}.\n\\]\nThen \\( p \\) splits as \\( p\\mathscr{O}_K = \\mathfrak{p}_1^{e_1} \\cdots \\mathfrak{p}_m^{e_m} \\) in \\( \\mathscr{O}_K \\).}\n\n\nApplying this to \\( K = \\mathbb{Q}(\\sqrt{-3}) \\) with \\( \\beta = \\sqrt{-3} \\), since \\( (\\mathscr{O}_K : \\mathbb{Z}[\\sqrt{-3}]) = 2 \\), the theorem applies to all primes \\( p > 2 \\). We have\n\\begin{align*}\np \\equiv 1 \\pmod{3} \n& \\Rightarrow \\left( \\frac{p}{3} \\right) = 1 \\\\\n& \\Rightarrow (-1)^{\\frac{p-1}{2}} (-1)^{\\frac{p-1}{2} \\cdot \\frac{3-1}{2}} \\left( \\frac{p}{3} \\right) = 1  \\\\\n& \\Rightarrow (-1)^{(p-1)/2} \\left( \\frac{3}{p} \\right) = 1  \\quad \\text{(by quadratic reciprocity)} \\\\\n& \\Rightarrow  \\left( \\frac{-3}{p} \\right) = 1 \\\\\n& \\Rightarrow  X^2 + 3 \\text{ is reducible mod } p \\\\\n& \\Rightarrow X^2 + 3 \\text{ splits into distinct linear factors mod } p \\\\\n& \\Rightarrow p \\text{ splits completely in } \\mathscr{O}_K.\n\\end{align*}\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Show that a prime $p$ can be written as $p = a^2+ab+b^2$ with $a,b \\in \\mathbb{Z}$ if and only\nif $p=3$ or $p \\equiv 1 \\pmod 3$. -/"}
{"informal_statement": "Let $G$ be a group and let $K\\subseteq H$ be subgroups of $G$ with $K \\lhd H$.\n\tIf $H \\lhd G$ and $C_H(K)=1$, prove that $H$ centralizes $C_G(K)$.", "formal_statement": "theorem le_centralizer_centralizer_of_centralizer_eq_bot {G : Type} [Group G] (H K : Subgroup G)\n    [H.Normal] (h1 : K ≤ H) [(K.subgroupOf H).Normal]\n    (h2 : Subgroup.centralizer (K.subgroupOf H) = (⊥ : Subgroup H)) :\n    H ≤ Subgroup.centralizer (Subgroup.centralizer (K : Set G) : Set G) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Basic Axioms and Examples"], "name": "Problem_7", "informal_proof": "\nWe need to show $hgh^{-1}g^{-1}=1$ for any $h\\in H$ and $g\\in C_G(K)$, and it is sufficient to show that $hgh^{-1}g^{-1}\\in C_H(K)$. $H \\lhd G$ leads to the result that $gh^{-1}g^{-1}$ and $hgh^{-1}g^{-1}$ is in the group $H$. Let $k\\in K$ and $h^{-1}kh=k'\\in K$, then $hgh^{-1}g^{-1}k= hgh^{-1}kg^{-1}= hgk'h^{-1}g^{-1}= hk'gh^{-1}g^{-1}=khgh^{-1}g^{-1}$ and it is done.\n", "header": "import Mathlib\n\nopen Subgroup\n\n", "informal_prefix": "/- Let $G$ be a group and let $K\\subseteq H$ be subgroups of $G$ with $K \\lhd H$.\n\tIf $H \\lhd G$ and $C_H(K)=1$, prove that $H$ centralizes $C_G(K)$.-/"}
{"informal_statement": "Show that if a Sylow $2$-subgroup of $G$ is nontrivial and cyclic, then $G$ has a subgroup $H$ with $[G:H] =2$.", "formal_statement": "theorem exists_index_two_of_sylow_two_isCyclic {G : Type} [Group G] [Finite G] (P : Sylow 2 G)\n    (h1 : P.toSubgroup ≠ ⊥) [IsCyclic P] : ∃ H : Subgroup G, H.index = 2 := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Group Actions and Sylow theorems"], "name": "Problem_8", "informal_proof": "\n\n\\textbf{Claim.} For every $g$ in $G$, let $\\pi : G \\to \\text{Sym } G $ be the left translation action of $G$. Let $\\pi_g$ denote $\\pi(g)$ in $\\text{Sym } G$. Then $\\pi_g$ is an odd permutation if and only if $g$ has even order and $[G:\\langle g \\rangle ]$ is odd.\n\n\\textbf{Proof of the Claim. } Let $r$ be the order of $g$. Then for any $x \\in G$, the permutation $\\pi_g$ takes\n$$\n\\begin{tikzcd}\nx \\ar[r, mapsto, \"\\pi_g\"] & gx \\ar[r, mapsto, \"\\pi_g\"] & g^2 x \\ar[r, mapsto, \"\\pi_g\"] & \\cdots  \\ar[r, mapsto, \"\\pi_g\"] & g^rx = x,\n\\end{tikzcd}\n$$\nand it is clear that for any $i \\in \\{1, \\dots, r-1\\}$, $g^ix \\neq x$. \nThus $\\pi_g$ breaks up $G$ into disjoint union of $[G:\\langle g\\rangle ]$ cycles, each is a $r$-cycle.\n\nBut we know that $r$-cycle is the product of $r-1$ transpositions; so for $\\pi_g$ to be an odd permutation, we need and only need $(r-1) \\cdot [G:\\langle g\\rangle]$ to be an odd number, i.e. $r$ is even and $[G:\\langle g \\rangle]$ is odd. \\qed\n\nNow consider the homomorphism $\\pi: G \\to S_n$ given by left translation action of $G$ on itself.  There is a natural homomorphism $\\mathrm{sgn}: S_n \\to \\{\\pm 1\\}$ sending even permutations to $1$ and odd permutation to $-1$. The composition is\n\\begin{equation}\n\\tag{$\\star$}\nG \\xrightarrow{\\ \\pi \\ } S_n \\xrightarrow{ \\mathrm{sgn}} \\{\\pm 1\\}.\n\\end{equation}\nWe need to show that the composition is surjective, then the kernel would give a subgroup of $G$ of index $2$.\n\nFor this, we need to show that for some $g \\in G$, $\\pi_g$ is an odd permutation.  Since the Sylow $2$-subgroup $P_2$ of $G$ is nontrivial and cyclic; let $\\sigma_2$ be its generator. Then $\\sigma_2$ has even order and $[G:\\langle \\sigma_2\\rangle ] = [G:P_2]$ is odd. By (1), $\\pi_\\sigma$ is an odd permutation and hence $(\\star)$ is surjective. We are done.\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Show that if a Sylow $2$-subgroup of $G$ is nontrivial and cyclic, then $G$ has a subgroup $H$\nwith $[G:H] =2$. -/"}
{"informal_statement": "If \\( n \\) is odd and \\( n \\geq 3 \\), show that the identity is the only element of \\( D_{2n} \\) which commutes with all elements of \\( D_{2n} \\).", "formal_statement": "theorem DihedralGroup.centralizer_eq_bot {n : ℕ} (hn : Odd n) (h : n ≥ 3) :\n    Subgroup.centralizer ⊤ = (⊥ : Subgroup (DihedralGroup n)) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Basic Axioms and Examples"], "name": "Problem_9", "informal_proof": "\nGeneral method to find the centre of $D_{2n}$\n\nIf $n=1$ or $n=2$, then $D_{2n}$ is abelian and hence $Z(D_{2n}) = D_{2n}$.\n\nNow suppose $n \\geq 3$. By definition we have\n\\[\nD_{2n} = \\{ a^i b^j : i = 0,1, \\ j = 0,1,\\dots,n-1 \\}\n\\]\nwhere $a$ is an element of order $2$, $b$ is an element of order $n$, and $a,b$ are given by the relation $ba = ab^{-1}$.\n\nThis implies \n\\[\nb^2 a = b(ba) = b(ab^{-1}) = (ba)b^{-1} = (ab^{-1})b^{-1} = ab^{-2}.\n\\]\nIn general, we can write\n\\[\nb^r a = a b^{-r} \\quad \\text{(1)}\n\\]\nfor all integers $r \\geq 0$.\n\nSince $a$ and $b$ together generate $D_{2n}$, an element of $D_{2n}$ is in the center if and only if it commutes with both $a$ and $b$.\n\nTherefore, $x = a^i b^j \\in Z(D_{2n})$ if and only if $xa = ax$ and $xb = bx$.\n\nConsider $xb = a^i b^{j+1} = bx = b a^i b^j$, which simplifies to\n\\[\na^i b = b a^i \\quad \\text{(2)}.\n\\]\n\nClearly, $i = 0$ satisfies (2) because $a^0 b = b a^0$ implies $b = b$.\n\nHowever, $i \\neq 1$ because if $i = 1$, then $ab = ba$, but from (1), for $n \\geq 3$, we have $ab = ba^{-1} \\neq ba$, leading to a contradiction.\n\nThus, if $x = a^i b^j$ is central, then $i = 0$ and $x = b^j$.\n\nNow, the condition $xa = ax$ becomes $b^j a = a b^j$. From (1), we have $a b^{-j} = a b^j$, which implies\n\\[\nb^{-j} = b^j \\quad \\Rightarrow \\quad b^{2j} = 1 \\quad \\text{(3)}.\n\\]\nSince $\\text{ord}(b) = n$, (3) gives $n \\mid 2j$. Therefore, if $n$ divides $2j$, then either $j = 0$ or $2j = n$.\n\nIf $j = 0$, then $x = b^0 = 1$.\n\nIf $2j = n$, then $n$ must be even, and $j = \\frac{n}{2}$, so $x = b^{n/2}$.\n\nWe have thus proved that\n\\[\nZ(D_{2n}) = \n\\begin{cases}\nD_{2n} & \\text{if } n = 1, 2, \\\\\n\\{1\\} & \\text{if } n > 2 \\text{ is odd}, \\\\\n\\{1, b^{n/2}\\} & \\text{if } n > 2 \\text{ is even}.\n\\end{cases}\n\\]\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- If \\( n \\) is odd and \\( n \\geq 3 \\), show that the identity is the only element of\n\\( D_{2n} \\) which commutes with all elements of \\( D_{2n} \\). -/"}
{"informal_statement": "Determine the last two digits of $3^{3^{100}}$.", "formal_statement": "theorem three_pow_three_pow_mod_100 : 3 ^ (3 ^ 100) % 100 = 3 := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Other Topics", "Elementary Number Theory"], "name": "Problem_10", "informal_proof": "\nObserve that $3^{20}\\equiv (-1)^{20} \\equiv 1\\pmod{4}$ and $3^{20}= 3^{\\phi(25)} \\equiv 1 \\pmod{25}$. Hence, $3^{20}\\equiv 1 \\pmod{100}$. On the other hand, $3^{100}= 81^{25} \\equiv 1 \\pmod{20}$. Hence, $3^{3^{100}}= 3^{20k+1}\\equiv 3 \\pmod{100}$.\n", "header": "import Mathlib\n\n", "informal_prefix": "/- prove that the last two digits of $3^{3^{100}} is 03-/"}
{"informal_statement": "Let \\( G \\) be a group of order \\( 3825 \\). Prove that if \\( H \\) is a normal subgroup of order \\( 17 \\) in \\( G \\), then \\( H \\leq Z(G) \\).", "formal_statement": "theorem le_center_of_card_eq_17_of_card_eq_3825 {G : Type} [Group G] (h : Nat.card G = 3825)\n    (H : Subgroup G) [H.Normal] (hH : Nat.card H = 17) : H ≤ Subgroup.center G := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Subgroups and Quotient groups"], "name": "Problem_11", "informal_proof": "\nIf $H \\trianglelefteq G$, then $G$ acts by conjugation on $H$ as automorphisms of $H$, with kernel $C_G(H)$. Hence, $G/C_G(H) \\leq \\mathrm{Aut}(H)$. Since $H$ is of prime order ($17$), $H$ is cyclic. So, $|\\mathrm{Aut}(H)| = \\varphi(17) = 17 - 1 = 16$.\n\nSince $G/C_G(H) \\leq \\mathrm{Aut}(H)$, by Lagrange's theorem $|G/C_G(H)|$ divides $|\\mathrm{Aut}(H)|$. This implies possible orders for $|G/C_G(H)|$ are $1, 2, 4, 8$. \n\\quad \\text{------} (*)\n\nAlso, $C_G(H) \\leq G$, so again by Lagrange's theorem $|C_G(H)|$ divides $|G|$.\n\nFrom (*), $|C_G(H)| = 3825/2$ or $3825/4$ or $3825/8$. But none of $3825/2$, $3825/4$, $3825/8$ are integers. Then the only possibility left is $|C_G(H)| = 3825/1 = 3825 = |G|$, implying $G = C_G(H)$.\n\nSince $H$ is abelian and $C_G(H) = G$, $H \\leq Z(G)$.\n", "header": "import Mathlib\n\n", "informal_prefix": "/- Let $G$ be a group of order $3825$. Prove that if $H$ is a normal subgroup of order $17$ in $G$,\nthen $H \\leq Z(G)$.-/"}
{"informal_statement": "Prove that \\( SL_2(\\mathbb{F}_3) / Z(SL_2(\\mathbb{F}_3)) < A_4 \\).", "formal_statement": "theorem exists_sl_quot_center_monoidHom_alternatingGroup :\n    ∃ φ : SL(2,ZMod 3) ⧸ Subgroup.center SL(2,ZMod 3) →* alternatingGroup (Fin 4),\n    Function.Injective φ := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Group Actions and Sylow theorems"], "name": "Problem_12", "informal_proof": "\nThe number of \\( SL_2(\\mathbb{F}_3) \\) is 24 and the number of Sylow-3 subgroups is 4. (It can not be 1, because the unipotent subgroup is not normal) The conjugation action of  \\( SL_2(\\mathbb{F}_3) \\) on the set of Sylow-3 subgroups is transitive and  we get that the center should be of order 2 or 6. But all those elements of order 3 is not in the center, because everyone is conjugate to some element in the unipotent subgroup. So the center should be the group consisting of \\( \\pm I \\).\n\nThen consider the natural and faithful action of \\( SL_2(\\mathbb{F}_3) / Z(SL_2(\\mathbb{F}_3))  \\) on the projective line over $\\mathbb{F}_3$, which contains 4 points. So \\( SL_2(\\mathbb{F}_3) / Z(SL_2(\\mathbb{F}_3))  \\) is isomorphic to some subgroup of order 12 of $S_4$. The only choice is $A_4$, because it is of index 2 and normal. But the conjugacy classes of $S_4$ are of size 1,3,6,6,8. There is only one way to get 12 from these numbers.\n", "header": "import Mathlib\n\nopen MatrixGroups\n\n", "informal_prefix": "/--\nProve that \\( SL_2(\\mathbb{F}_3) / Z(SL_2(\\mathbb{F}_3)) < A_4 \\).\n-/"}
{"informal_statement": "Prove that the number of Sylow $p$-subgroups of $\\operatorname{GL}_2(\\mathbb{F}_p)$ is $p + 1$.", "formal_statement": "theorem card_sylow_gl_two_eq_add_one (p : ℕ) [Fact p.Prime] :\n    Nat.card (Sylow p <| GL (Fin 2) (ZMod p)) = p + 1 := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Group Actions and Sylow theorems"], "name": "Problem_13", "informal_proof": "\nThe number of the linear group is $p(p-1)^2(p+1)$ and the number of Sylow-p subgroups is one of the following numbers: $(p-1)^2$; $(p+1)$; $(p-1)^2(p+1)$ by Sylow theorem. Consider the conjugation action of $\\operatorname{GL}_2(\\mathbb{F}_p)$ on the set of Sylow-p subgroups, the action is transitive and trivial for the center $C(\\operatorname{GL}_2(\\mathbb{F}_p))$, which is of order $p-1$. So the number of Sylow-p subgroups is not divided by $(p-1)^2$ and the only choice is $p+1$.\n", "header": "import Mathlib\n\nopen Matrix\n\n", "informal_prefix": "/-- Prove that the number of Sylow $p$-subgroups of $\\operatorname{GL}_2(\\mathbb{F}_p)$ is $p + 1$.\n-/"}
{"informal_statement": "Let $S$ be any ring and let $n>2$ be an integer.\n    Prove that if \\( A \\) is any strictly upper triangular matrix in \\( M_n(S) \\), then \\( A^n = 0 \\). (A strictly upper triangular matrix is one whose entries on and below the main diagonal are all zero.)", "formal_statement": "theorem pow_eq_zero_of_strictly_upper_triangle {S : Type} [Ring S] (n : ℕ) (hn : 2 < n)\n    (A : Matrix (Fin n) (Fin n) S) (hA : ∀ (i : Fin n), ∀ (j : Fin n), i ≥ j → A i j = 0) :\n    A ^ n = 0 := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Other Topics", "Linear Algebra"], "name": "Problem_14", "informal_proof": "\nLet $A$ be an $n \\times n$ upper triangular matrix. Then its characteristic polynomial is $X^n$. By the Cayley-Hamilton theorem, it follows that $A^n = 0$.\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Let $S$ be any ring and let $n>2$ be an integer.\nPropose a proof that if $A$ is any strictly upper triangular matrix in $M_n(S)$, then $A^n = 0$.\n(A strictly upper triangular matrix is one whose entries on and below the main diagonal are all\nzero.) -/"}
{"informal_statement": "Prove that the ring $\\mathbb{Z}[\\sqrt{-5}]$ is not a principal ideal domain.", "formal_statement": "theorem not_isPrincipalIdealRing_Zsqrtd_neg_five : ¬IsPrincipalIdealRing (Zsqrtd (-5)) := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Other Topics", "Algebraic Number Theory and Valuation Theory"], "name": "Problem_15", "informal_proof": "\nSpecifically, the ideal $I = (2, 1+\\sqrt{-5})$ is not principal.\n\nThe easiest way to see this is to use the norm map $N \\colon \\mathbb{Z}[\\sqrt{-5}] \\to \\mathbb{Z}$ defined by\n\\[\nN(x + y\\sqrt{-5}) = (x + y\\sqrt{-5})(x - y\\sqrt{-5}) = x^2 + 5y^2.\n\\]\nThis map is well-defined and multiplicative ($N(rs) = N(r)N(s)$ for $r,s \\in \\mathbb{Z}[\\sqrt{-5}]$). Multiplicativity implies that if $z = xy$ then $N(z) = N(x)N(y)$ and so $N(x), N(y) \\mid N(z)$ (since $N(x)$, $N(y)$, and $N(z)$ are all integers).\n\nNow onto the proof. First, notice that $1 \\notin I$, and so $I$ is a proper ideal of $R$. By way of contradiction, assume that $I = (\\alpha)$ for some $\\alpha \\in R$. Then $\\alpha \\mid 2$, $\\alpha \\mid (1+\\sqrt{-5})$. Hence by the above observation about the norm map we know $N(\\alpha) \\mid N(2) = 4$ and $N(\\alpha) \\mid N(1+\\sqrt{-5}) = 6$. Hence $N(\\alpha) \\mid \\gcd(4,6) = 2$, which (since $N(\\alpha) \\geq 0$) makes $N(\\alpha) \\in \\{1,2\\}$.\n\nBut we know $\\alpha = a + b\\sqrt{-5}$ for integers $a,b$ and $N(\\alpha) = a^2 + 5b^2$. The equation $a^2 + 5b^2 = 2$ has no solutions $(a,b)$ in $\\mathbb{Z}^2$, and the only solutions $(a,b)$ in $\\mathbb{Z}^2$ to $a^2 + 5b^2 = 1$ are $(a,b) = (\\pm 1, 0)$. Thus $\\alpha \\in \\{\\pm 1\\}$ is a unit, which means $I = (\\alpha) = R$. But we've already noted above (and it's easy to verify directly) that $1 \\notin I$, and so this is a contradiction.\n", "header": "import Mathlib\n\n", "informal_prefix": "/- Prove that the ring $\\mathbb{Z}[\\sqrt{-5}]$ is not a principal ideal domain. -/"}
{"informal_statement": "Let \\( R \\) be an integral domain and let \\( i, j \\) be relatively prime integers. Prove that the ideal \\( (x^i - y^j) \\) is a prime ideal in \\( R[x, y] \\).", "formal_statement": "theorem span_pow_sub_pow_isPrime_of_coprime {R : Type} [CommRing R] [IsDomain R] {i j : ℕ}\n    (hi : i > 0) (hj : j > 0) (h : Nat.Coprime i j) :\n    (span {(X 0 ^ i - X 1 ^ j : MvPolynomial (Fin 2) R)}).IsPrime := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Ring Theory", "Polynomials"], "name": "Problem_16", "informal_proof": "\nAssume $f(X,Y) = X^n - Y^m = g(X,Y)h(X,Y)$. Then $f(Z^m, Z^n) = 0$ implies that one of $g(Z^m, Z^n)$ or $h(Z^m, Z^n)$, without loss of generality $g(Z^m, Z^n)$, is the zero polynomial. That means that for all $k$, the monomials cancel, i.e., if\n\\[\ng(X,Y) = \\sum a_{i,j}X^iY^j\n\\]\nthen\n\\[\n\\sum_{mi+nj=k} a_{i,j} = 0.\n\\]\n\nCan we ever have $mi + nj = mi' + nj'$? That would mean $m(i-i') = n(j'-j)$. It follows that $i \\equiv i' \\pmod{n}$ and $j \\equiv j' \\pmod{m}$. But $i,i' \\in \\{0,1,\\dots,n-1\\}$ and $j,j' \\in \\{0,1,\\dots,m-1\\}$. This implies $i = i'$ and $j = j'$.\n\nOverall, $mi + nj = mi' + nj'$ is equivalent with $i = i'$ and $j = j'$. So distinct terms from $g$ and $h$ correspond to distinct terms in $g(Z^m, Z^n)$ and $h(Z^m, Z^n)$. \n\nWhen we substitute $X = Z^m$ and $Y = Z^n$, none of the terms in $g(Z^m, Z^n)$ and $h(Z^m, Z^n)$ will cancel out, so that $g(Z^m, Z^n) = 0$ and $h(Z^m, Z^n) = 0$ would require all coefficients to be zero. This means $g(Z^m, Z^n) \\neq 0 \\neq h(Z^m, Z^n)$, but then we have the contradiction:\n\\[\n0 = f(Z^m, Z^n) = g(Z^m, Z^n)h(Z^m, Z^n) \\neq 0.\n\\]\n", "header": "import Mathlib\n\nopen MvPolynomial Ideal\n", "informal_prefix": "/- Let \\( R \\) be an integral domain and let \\( i, j \\) be relatively prime integers. Prove that the ideal \\( (x^i - y^j) \\) is a prime ideal in \\( R[x, y] \\).-/"}
{"informal_statement": "Prove that $\\frac{x^{p}-1}{x-1}$ is irreducible in $\\mathbb{Z}[x].$", "formal_statement": "theorem irreducible_X_pow_p_sub_one_div_X_sub_one (p : ℕ) [hp : Fact (Nat.Prime p)] :\n    Irreducible (((Polynomial.X : ℤ[X]) ^ p - 1) /ₘ (Polynomial.X - 1)) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Ring Theory", "Polynomials"], "name": "Problem_17", "informal_proof": "\nLet $x=1+y$ and use the Eisenstein's irreducibility criterion.\n", "header": "import Mathlib\n\nopen Polynomial\n\n", "informal_prefix": "/-- Prove that $\\frac{x^{p}-1}{x-1}$ is irreducible in $\\mathbb{Z}[x].$ -/"}
{"informal_statement": "Prove that \\( x^2 + y^2 - 1 \\) is irreducible in \\( \\mathbb{Q}[x, y] \\).", "formal_statement": "theorem irreducible_pow_two_add_pow_two_sub_one :\n    Irreducible ((X 0) ^ 2 + (X 1) ^ 2 - 1 : MvPolynomial (Fin 2) ℚ) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Ring Theory", "Polynomials"], "name": "Problem_18", "informal_proof": "\nIf not then it should be a product of 2 degree 1 polynomials. WLOG we suppose that $x^2+y^2-1=(x+ay+b)(x+cy+d)$, then $a+c=0$ and $ac=1$, which is impossible.\n", "header": "import Mathlib\n\nopen MvPolynomial\n\n", "informal_prefix": "/- Prove that \\( x^2 + y^2 - 1 \\) is irreducible in \\( \\mathbb{Q}[x, y] \\).-/"}
{"informal_statement": "Prove that any finite group is isomorphic to a subgroup of $A_n$ for some $n$.", "formal_statement": "theorem exists_subgroup_alternatingGroup_mulEquiv {G : Type} [Group G] [Finite G] :\n    ∃ (n : ℕ) (H : Subgroup (alternatingGroup (Fin n))), Nonempty (G ≃* H) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Group Actions and Sylow theorems"], "name": "Problem_19", "informal_proof": "\nBy Cayley's theorem, every group $G$ is isomorphic to a subgroup of $S_n$ for some $n$.\n\nMoreover, $S_n$ embeds into $A_{n+2}$ via the following construction: \n\\begin{itemize}\n    \\item For even permutations $\\sigma \\in S_n$, map $\\sigma \\mapsto \\sigma$\n    \\item For odd permutations $\\sigma \\in S_n$, map $\\sigma \\mapsto \\sigma \\cdot (n+1\\ n+2)$\n\\end{itemize}\nThis mapping defines an injective homomorphism, as verified by straightforward computation.\n\nComposing these embeddings gives:\n\\[\nG \\hookrightarrow S_n \\hookrightarrow A_{n+2}\n\\]\nwhere $n = |G|$. Thus, any finite group $G$ embeds in the alternating group $A_{|G|+2}$.\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Prove that any finite group is isomorphic to a subgroup of $A_n$ for some $n$. -/"}
{"informal_statement": "Prove that\n    if \\( G \\) is a nonabelian group of order \\( p^3 \\) (\\( p \\) prime), then the center of \\( G \\) is the subgroup generated by all elements of the form \\( aba^{-1}b^{-1} \\) (\\( a, b \\in G \\)).", "formal_statement": "theorem center_eq_commutator_of_pow_three_eq_card {G : Type} [Group G] {p : ℕ} [hp : Fact p.Prime]\n    (hG : p ^ 3 = Nat.card G) (h : ∃ x y : G, x * y ≠ y * x) :\n    Subgroup.center G = commutator G := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "p-Groups, Nilpotent Groups and Solvable Groups"], "name": "Problem_20", "informal_proof": "\nLet \\( G \\) be a non-abelian group of order \\( p^3 \\) where \\( p \\) is prime. The center \\( Z(G) \\) of a \\( p \\)-group is nontrivial, so \\( |Z(G)| = p \\) or \\( p^2 \\).\n\nIf \\( |Z(G)| = p^2 \\), then \\( |G/Z(G)| = p \\). This implies \\( G/Z(G) \\) is cyclic, which forces \\( G \\) to be abelian, contradicting our assumption.\n\nThus, \\( |Z(G)| = p \\) and \\( |G/Z(G)| = p^2 \\). Since groups of order \\( p^2 \\) are abelian, \\( G/Z(G) \\) is abelian. For all \\( x, y \\in G \\):\n\\[\n(xZ(G))(yZ(G)) = (yZ(G))(xZ(G))\n\\]\nEquivalently:\n\\[\n1 \\cdot Z(G) = (xZ(G))^{-1}(yZ(G))^{-1}(xZ(G))(yZ(G)) = x^{-1}y^{-1}xy Z(G) = [x,y] Z(G)\n\\]\nThus, \\( [x,y] \\in Z(G) \\) for all \\( x, y \\in G \\), so the commutator subgroup \\( [G, G] \\leq Z(G) \\).\n\nSince \\( Z(G) \\) has order \\( p \\), its only subgroups are \\( \\{e\\} \\) and \\( Z(G) \\). As \\( G \\) is non-abelian, \\( [G, G] \\) is nontrivial, so necessarily \\( [G, G] = Z(G) \\).\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Prove that if $G$ is a nonabelian group of order $p^3$ ($p$ prime),\n    then the center of $G$ is the subgroup generated by all elements of\n    the form $aba^{-1}b^{-1}$ ($a, b \\in G$). -/"}
{"informal_statement": "Prove that the order of $\\mathrm{Aut}(\\mathbb{Z}_3\t\\times \\mathbb{Z}_9)$ is 108.", "formal_statement": "theorem card_addAut_eq_108 : Nat.card (AddAut <| ZMod 3 × ZMod 9) = 108 := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Basic Axioms and Examples"], "name": "Problem_21", "informal_proof": "\nIt suffices to choose the image of $(0,1)$ and $(1,0)$. $(0,1)$ must be sent to an order 9 element $(a,b), b \\in 1,2,4,5,7,8$, which has 18 choices. $(1,0)$ must be sent to an order 3 element that is not a multiple of $(a,b)$, which has $9-3 = 6$ choices. In total, there are 108 choices.\n", "header": "import Mathlib\n\n", "informal_prefix": "/- Prove that the order of Aut(Z_3 × Z_9) is 108.-/"}
{"informal_statement": "Let $D_8$ be the dihedral group with $8$ elements. Prove that $\\mathrm{Aut}(D_8) \\cong D_8$.", "formal_statement": "theorem nonempty_mulAut_dihedralGroup_four : Nonempty (MulAut (DihedralGroup 4) ≃* DihedralGroup 4) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Basic Axioms and Examples"], "name": "Problem_22", "informal_proof": "\nLet $D_8 \\cong H \\leq D_{16} = G$, where\n\\[ D_{16} = \\{1, a, a^2, a^3, a^4, a^5, a^6, a^7, b, ba, ba^2, ba^3, ba^4, ba^5, ba^6, ba^7\\} \\]\nand\n\\[ H = \\{1, a^2, a^4, a^6, b, ba^2, ba^4, ba^6\\}. \\]\nLet $G$ act on $H$ by conjugation. That is, $g \\cdot h = ghg^{-1}$. Since $H$ is normal in $G$, this action is well-defined.\n\nConsider the permutation representation $\\theta: G \\to S_H$. Recall that $\\ker \\theta = C_G(H)$. In this case, $\\theta(g)$ is a group homomorphism on $H$, the image of $\\theta$ is contained in $\\operatorname{Aut} H$. Then\n\\[ G / \\ker \\theta \\cong \\operatorname{Im} \\theta \\leq \\operatorname{Aut} H. \\]\n\nIt is easy to show that $\\ker \\theta = C_G(H) = Z(G)$. Hence, $G / \\ker \\theta = G / Z(G) \\cong D_8$. Now we show that $\\theta$ is onto. Then $D_8 \\cong \\operatorname{Aut} D_8$ follows from the First Isomorphism Theorem.\n\nFor any $\\varphi \\in \\operatorname{Aut} H$, we want to find $g \\in G$ such that $\\theta(g) = \\varphi$. Note that $a^2$ and $b$ are independent generators of $H$. $\\varphi$ is completely determined by $\\varphi(a^2)$ and $\\varphi(b)$. Since $\\varphi$ is an automorphism, it preserves the order of the element. Thus, $|\\varphi(a^2)| = |a^2| = 4$ and $|\\varphi(b)| = |b| = 2$. Comparing the order, we have $\\varphi(a^2) \\in \\{a^2, a^6\\}$ and $\\varphi(b) \\in \\{b, ba^2, ba^4, ba^6\\}$. ($\\varphi(b) \\neq a^4$, otherwise, $\\theta$ is not onto.) We have the following $8$ possibilities.\n\n\\[\n\\begin{cases}\n\\varphi(a^2) = a^2 \\\\\n\\varphi(b) = b\n\\end{cases}\n\\Rightarrow \\varphi = \\theta(a^4)|_H \\text{ or } \\theta(a^8)|_H,\n\\quad\n\\begin{cases}\n\\varphi(a^2) = a^6 \\\\\n\\varphi(b) = b\n\\end{cases}\n\\Rightarrow \\varphi = \\theta(ba^4)|_H \\text{ or } \\theta(b)|_H\n\\]\n\\[\n\\begin{cases}\n\\varphi(a^2) = a^2 \\\\\n\\varphi(b) = ba^2\n\\end{cases}\n\\Rightarrow \\varphi = \\theta(a^3)|_H \\text{ or } \\theta(a^7)|_H,\n\\quad\n\\begin{cases}\n\\varphi(a^2) = a^6 \\\\\n\\varphi(b) = ba^2\n\\end{cases}\n\\Rightarrow \\varphi = \\theta(ba^5)|_H \\text{ or } \\theta(ba)|_H\n\\]\n\\[\n\\begin{cases}\n\\varphi(a^2) = a^2 \\\\\n\\varphi(b) = ba^4\n\\end{cases}\n\\Rightarrow \\varphi = \\theta(a^2)|_H \\text{ or } \\theta(a^6)|_H,\n\\quad\n\\begin{cases}\n\\varphi(a^2) = a^6 \\\\\n\\varphi(b) = ba^4\n\\end{cases}\n\\Rightarrow \\varphi = \\theta(ba^6)|_H \\text{ or } \\theta(ba^2)|_H\n\\]\n\\[\n\\begin{cases}\n\\varphi(a^2) = a^2 \\\\\n\\varphi(b) = ba^6\n\\end{cases}\n\\Rightarrow \\varphi = \\theta(a)|_H \\text{ or } \\theta(a^5)|_H,\n\\quad\n\\begin{cases}\n\\varphi(a^2) = a^6 \\\\\n\\varphi(b) = ba^6\n\\end{cases}\n\\Rightarrow \\varphi = \\theta(ba^7)|_H \\text{ or } \\theta(ba^3)|_H\n\\]\nTherefore, $\\theta$ is onto. (For any $\\varphi \\in \\operatorname{Aut} H$, if $\\varphi(a^2) = a^{2+4i}$ and $\\varphi(b) = ba^{2j} \\in G = D_{16}$, where $i \\in \\{0,1\\}$ and $j \\in \\{0,1,2,3\\}$, then choose $g = a^{4-j}b^i$. We have $\\theta(g) = \\varphi$.)\n", "header": "import Mathlib\n\n", "informal_prefix": "/--Let $D_8$ be the dihedral group with $8$ elements. Prove that $\\mathrm{Aut}(D_8) \\cong D_8$.-/"}
{"informal_statement": "Prove that if $\\#G = 1053$ then $G$ is not simple.", "formal_statement": "theorem not_isSimpleGroup_of_card_eq_1053 (G : Type) [Group G]\n    [Finite G] (h_card : Nat.card G = 1053) : ¬ IsSimpleGroup G := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Group Actions and Sylow theorems"], "name": "Problem_23", "informal_proof": "\nObserve that $1053=3^4 \\times 13$. Let $n_{13}$ be the number of Sylow-$13$ subgroups of $G$. By Sylow Theorem, we have $n_{13}\\mid 81$ and $n_{13} \\cong 1 (\\mod 13)$. Thus, $n_{13}=1$ or $27$. \n\nCase 1, $n_13 = 27$. Let $|P \\cap Q|=3$, then $|PQ|=37$ which exceeds the order of the group and is not possible.\n\nLet $|P \\cap Q|=32$, then $|PQ|=36$. But then what can be the order of $N_G(P \\cap Q)$? The only possibility is $|N_G(P \\cap Q)|=34 \\cdot 13$. Hence $P \\cap Q$ is normal in $G$ and we are done.\n\nLet $|P \\cap Q|=33$, then $|PQ|=35$. Hence the choices for $|N_G(P \\cap Q)|$ are $33 \\cdot 13$, $1053$. For the latter case again $P \\cap Q$ is normal.\n\nBut if $|N_G(P \\cap Q)|=33 \\cdot 13$, we have $[G:N_G(P \\cap Q)]=3$. Since $1053$ does not divide $3!$, by index theorem we conclude that $G$ has a non-trivial normal subgroup.\n\nHence all the cases have been exhausted and we conclude $G$ is not simple.\n\nCase $2$, $n_13 = 1$. $G$ has a unique a Sylow-$13$ subgroup, denoted by $P$. By Sylow Theorem, $P$ is normal. Therefore, $G$ is not simple.\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Prove that if $\\#G = 1053$ then $G$ is not simple. -/"}
{"informal_statement": "Prove that $\\mathbb{Q}/\\mathbb{Z}$ has no proper subgroups of finite index.", "formal_statement": "theorem eq_top_of_finiteIndex (H : AddSubgroup (ℚ ⧸ (Int.castAddHom ℚ).range)) (h_fin : H.FiniteIndex) :\n    H = ⊤ := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Subgroups and Quotient groups"], "name": "Problem_24", "informal_proof": "\nFirst part: We prove that $\\mathbb{Q}$ has no proper subgroups of finite index by contradiction. Assume that $\\mathbb{Q}$ has a proper subgroup $H$ such that $[\\mathbb{Q}:H]=n$ is finite. Since $\\mathbb{Q}$ is Abelian, $H$ is normal. Then, $\\mathbb{Q}/H$ is a group of order $n$. For any $x\\in \\mathbb{Q}$, we have $n(x+H)=0+H\\in \\mathbb{Q}/H$. Hence, $nx\\in H$. Thus, $n \\mathbb{Q}\\subseteq H$. Clearly, $n\\mathbb{Q}=\\mathbb{Q}$. It follows that $\\mathbb{Q} \\subseteq H$. Thus, $\\mathbb{Q} = H$. This contradicts with our assumption that $H$ is proper. Therefore, $\\mathbb{Q}$ has no proper subgroups of finite index. \n\nSecond part: We prove that $\\mathbb{Q}/\\mathbb{Z}$ has no proper subgroups of finite index by contradiction. Assume that $\\mathbb{Q}/\\mathbb{Z}$ has a proper subgroup $S$ such that $[\\mathbb{Q}/\\mathbb{Z}:S]=k$ is finite. Consider the quotient morphism $\\phi:\\mathbb{Q}\\to \\mathbb{Q}/\\mathbb{Z}$ of groups. The preimage $\\phi^{-1}(S)$ is a subgroup of $\\mathbb{Q}$. Clearly,  $\\phi^{-1}(S)$ is proper and $[\\mathbb{Q}: \\phi^{-1}(S)]=k$ is finite. This contradicts with first part. Therefore, $\\mathbb{Q}/\\mathbb{Z}$ has no proper subgroups of finite index.\n", "header": "import Mathlib\n\n", "informal_prefix": "/--\nProve that $\\mathbb{Q}/\\mathbb{Z}$ has no proper subgroups of finite index.\n-/"}
{"informal_statement": "Let $R=\\mathbb{Z}+x\\mathbb{Q}[x]\\subset \\mathbb{Q}[x]$be the set of polynomials in x with rational coefficients whose constant term is an integer. Prove that $R$ is not a U.F.D.", "formal_statement": "theorem not_uniqueFactorizationMonoid_adjoin_int : ¬ UniqueFactorizationMonoid\n    (Algebra.adjoin ℤ ({f | ∃ h : ℚ[X], f = X * h} : Set ℚ[X])) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Ring Theory", "Polynomials"], "name": "Problem_25", "informal_proof": "\n$R$ is not a unique factorization domain because not every element has a factorization.\n\n\\begin{itemize}\n    \\item Explicitly, the element \\( x \\) is not irreducible since  \n          \\[\n          x = 2 \\cdot \\frac{1}{2}x\n          \\]  \n          and neither \\( 2 \\) nor \\( \\frac{1}{2}x \\) is a unit.\n    \\item However, \\( x \\) cannot be written as a finite product of irreducible elements: any such factorization would necessarily consist of a product of constants times a rational multiple of \\( x \\), but no rational multiple of \\( x \\) is irreducible in \\( \\mathbb{Z} + x\\mathbb{Q}[x] \\).\n    \\item So, no matter how much we attempt to factor \\( x \\), we can never finish.\n\\end{itemize}\n\n", "header": "import Mathlib\n\nopen Polynomial\n", "informal_prefix": "/--Let $R=\\mathbb{Z}+x\\mathbb{Q}[x]\\subset \\mathbb{Q}[x]$be the set of polynomials in x with rational coefficients\nwhose constant term is an integer. Prove that $R$ is not a U.F.D.-/"}
{"informal_statement": "Suppose \\( G \\) is a group and \\( H \\) is a maximal subgroup of \\( G \\). Show that either \\( Z(G) \\leq H \\) or \\( [G,G] \\leq H \\). (A maximal subgroup contains either the center or the commutator subgroup.)", "formal_statement": "theorem center_le_or_commutator_le_of_isCoatom {G : Type} [Group G] (H : Subgroup G)\n    (h : IsCoatom H) : Subgroup.center G ≤ H ∨ commutator G ≤ H := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Subgroups and Quotient groups"], "name": "Problem_26", "informal_proof": "We can suppose that $Z=Z(G)$ is not contained in $H$, and we have to prove that $G' \\le H$. So $HZ$ properly contains $H$, hence by maximality equals $G$. Then we are going to prove $H$ is normal subgroup of G. If not, there exist $g$ such that $gHg^{-1}\neq H $. By $HZ=G$ we know $g=hz$ for $h\\in H,z\\in Z$. But $z$ is in center so $hzH(hz)^{-1}=hHh^{-1}=H$ is contradict. So H is normal subgroup of $G$. Consider $G/H$, if it is not commutative, then $Z(G/H)\neq G/H$, so there is a bigger group contaning $H$ which is not $G$, which is contradict to condition. So $G/H$ commutative, $[G,G]\\subset H$.", "header": "import Mathlib\n\n", "informal_prefix": "/--Suppose \\( G \\) is a group and \\( H \\) is a maximal subgroup of \\( G \\). Show that either \\( Z(G)\n\\leq H \\) or \\( [G,G] \\leq H \\). (A maximal subgroup contains either the center or the commutator\n subgroup.)-/"}
{"informal_statement": "Let $F$ be a field contained in the ring of $n \\times n$ matrices over $\\mathbb{Q}$. Prove that $[F:\\mathbb{Q}] \\leq n$.", "formal_statement": "theorem rank_le_of_subalgebra_matrix {n : ℕ} (F : Subalgebra ℚ (Matrix (Fin n) (Fin n) ℚ))\n    (h : IsField F) : Module.rank ℚ F ≤ n := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Other Topics", "Linear Algebra"], "name": "Problem_27", "informal_proof": "\nNote that $F/\\mathbb{Q}$ is a finite extension, thus $F=\\mathbb{Q}(\\alpha)$ for some $\\alpha\\in M_n(\\mathbb{Q})$ by the primitive element theorem. Then $[F:\\mathbb{Q}]$ is equal to the degree of $\\alpha$, and bounded by $n$ by Cayley-Hamilton's theorem.\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Let $F$ be a field contained in the ring of $n \\times n$ matrices over $\\mathbb{Q}$.\nProve that $[F:\\mathbb{Q}] \\leq n$. -/"}
{"informal_statement": "Let $k$ be a perfect field of characteristic $p > 0$. Let $F = k(t)$ be the field of rational functions in one variable over $k$. Show that every finite extension $E$ of $F$ can be generated by one element, that is, there exists $\\alpha \\in E$ such that $E = F(\\alpha)$.", "formal_statement": "theorem exists_ratFunc_adjoin_eq_top {k : Type} [Field k] [PerfectField k] {p : ℕ} [Fact p.Prime] [CharP k p]\n    {E : Type} [Field E] [Algebra (RatFunc k) E] [FiniteDimensional (RatFunc k) E] :\n    ∃ α : E, (RatFunc k)⟮α⟯ = ⊤ := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Field Extensions"], "name": "Problem_28", "informal_proof": "\n需要再精细一点，如果（1）（2）之一成立的（2）。\nWe first remark that by the proof of primitive element theorem, a finite separable extension can be generated by one element.\n\nLet $E$ be a finite extension of $F = k(t)$, we need to show that $E$ is generated by one element over $F$. We prove by induction on $[E:F]$.\nWe will show that\n\\begin{enumerate}\n\\item either $E/F$ is separable,\n\\item or $k(t^{1/p}) \\subseteq E$.\n\\end{enumerate}\nSuppose that we have proved this. Then in case (1), the separable extension $E/F$ is generated by one element. In case (2), $E$ is a finite extension of $k(t^{1/p})$. Write $u = t^{1/p}$; then $E$ is a finite extension of $k(u)$. By inductive hypothesis, $E = k(u)(\\alpha)$. Let $f(x) \\in k(u)[x]$ be the minimal polynomial of $\\alpha$ over $k(u)$; write $f(x) = x^n + a_{n-1}(u)x^{n-1} + \\cdots + a_0(u)$.\nThen $(f(x))^p = x^{pn} + a_{n-1}(u)^px^{p(n-1)} + \\cdots + a_0(u)^p \\in F[x]$ is the minimal polynomial of $\\alpha$ over $k(t)$. (This can be seen as follows: clearly, $f(\\alpha)^p = 0$; if $(f(x))^p$ factors nontrivially as $g(x)h(x) \\in F[x]$, then viewing $g(x), h(x) \\in k(u)[x]$, we see that they must be powers of $f(x)$. Contradiction.)\n\nNow we prove that one of (1) and (2) happens. Suppose that (2) does not hold; we prove (1).\nIt suffices to show that each $\\alpha \\in E$ is separable over $F$. Suppose some $\\alpha$ is not separable over $F$; let $f(x)$ be the minimal polynomial. By a theorem from the class, $f(x) = g(x^p)$ for some $g(x) = x^{n} + a_{n-1}(t)x^{n-1} + \\cdots + a_0(t) \\in F[x]$. As $k$ is perfect, there exists $b_i(t^{1/p}) \\in k(t^{1/p})$ such that $b_i(t^{1/p})^p = a_i(t)$. So the minimal polynomial of $\\alpha$ over $k(t^{1/p})$ divides $x^n+b_{n-1}(t^{1/p})x^{n-1}+ \\cdots + b_0(t^{1/p})$.  Note that\n$$p\\cdot \\deg g \\geq \n[k(t^{1/p})(\\alpha): k(t^{1/p})] [k(t^{1/p}):k(t)] = [k(t^{1/p})(\\alpha):F(\\alpha)] \\cdot [F(\\alpha):F] = [k(t^{1/p})(\\alpha):F(\\alpha)]\\cdot \\deg f.\n$$\nIt follows that $k(t^{1/p})(\\alpha)= F(\\alpha)$; thus $F(\\alpha)$ contains $k(t^{1/p})$, proving (2). \n\n", "header": "import Mathlib\n\nopen IntermediateField\n\n", "informal_prefix": "/-- Let $k$ be a perfect field of characteristic $p > 0$.\nLet $F = k(t)$ be the field of rational functions in one variable over $k$.\nShow that every finite extension $E$ of $F$ can be generated by one element, that is,\nthere exists $\\alpha \\in E$ such that $E = F(\\alpha)$. -/"}
{"informal_statement": "Show that if $F$ has characteristic $p$, then all degree $p$ Galois extension of $F$ are obtained by adjoining a zero of $x^p-x-a$ for some $a \\in F$.", "formal_statement": "theorem exists_nonempty_adjoin_root_X_pow_p_sub_X_sub_C\n    {F E : Type} [Field F] {p : ℕ} [Fact p.Prime] [CharP F p] [Field E]\n    [Algebra F E] [IsGalois F E] (h_deg : Module.finrank F E = p) :\n    ∃ a : F, Nonempty (AdjoinRoot (X ^ p - X - C a : F[X]) ≃+* E) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Galois theory"], "name": "Problem_29", "informal_proof": "Let K be a deg $p$ Galois extension of $F$. Then $\\mathrm{Gal}(K/F)=\\mathbb{Z}/p=<\tau>$ is cyclic group. Because $Tr_{K/L}(1)=0$, by additive Hilbert 90 theorem, there exist $c\\in K $ such that $\tau(c)-c=1$. So $c,\tau(c),...,\tau^{p-1}(c)$ are distinct number in $K$, so degree of minimal polynomial of $c$ is p. So $K=F(c)$. $\tau(c^p-c)=c^p-c$, so $c^p-c-a=0$ for $a \\in F$. So $a$ is root of $x^p-x-a$.", "header": "import Mathlib\n\nopen IntermediateField Polynomial\n\n", "informal_prefix": "/- Show that if $F$ has characteristic $p$, then all degree $p$ Galois extension of $F$ is to\nadjoin a zero of $x^p-x-a$ for some $a \\in F$.-/"}
{"informal_statement": "Let \\( E \\) be the splitting field of\n\\[\nf(x) = \\frac{x^7 - 1}{x - 1} = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1\n\\]\nover \\( \\mathbb{Q} \\). Let \\( \\zeta \\) be a zero of \\( f(x) \\), i.e., a primitive seventh root of \\( 1 \\). Let \\( \\beta = \\zeta + \\zeta^2 + \\zeta^4 \\). Show that the intermediate field \\( \\mathbb{Q}(\\beta) \\) is actually \\( \\mathbb{Q}(\\sqrt{-7}) \\).", "formal_statement": "theorem nonempty_ringEquiv_adjoin_pow_two_add_seven {E : Type} [Field E] [Algebra ℚ E]\n    [IsCyclotomicExtension {7} ℚ E] (ζ : E)\n    (h : IsPrimitiveRoot ζ 7) (β : E) (hb : β = ζ + ζ ^ 2 + ζ ^ 4) :\n    Nonempty (ℚ⟮β⟯ ≃+* AdjoinRoot (X ^ 2 + 7 : ℚ[X])) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Explicit Computations"], "name": "Problem_30", "informal_proof": "\nWe have $\\beta^2+\\beta=2(\\zeta+\\zeta^2+\\cdots\\zeta^6)=-2$, so $\\beta=(-1\\pm\\sqrt{-7})/2$ and $\\mathbb{Q}(\\beta)=\\mathbb{Q}(\\sqrt{-7})$\n", "header": "import Mathlib\n\nopen Polynomial\nopen scoped IntermediateField\n\n", "informal_prefix": "/-- Let $E$ be the splitting field of\n\\[\nf(x) = \\frac{x^7 - 1}{x - 1} = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1\n\\]\nover $\\mathbb{Q}$. Let $\\zeta$ be a zero of $f(x)$, i.e., a primitive seventh root of $1$.\nLet $\\beta = \\zeta + \\zeta^2 + \\zeta^4$. Show that the intermediate field $\\mathbb{Q}(\\beta)$\nis actually $\\mathbb{Q}(\\sqrt{-7})$. -/"}
{"informal_statement": "Prove that the primitive $n^{\\textrm{th}}$ roots of unity form a basis over $\\mathbb{Q}$ for the cyclotomic field of $n^{\\textrm{th}}$ roots of unity if and only if $n$ is squarefree.", "formal_statement": "theorem exists_basis_primitiveRoots_iff_squarefree {n : ℕ+} :\n    (∃ b : Basis (primitiveRoots n (CyclotomicField n ℚ)) ℚ (CyclotomicField n ℚ),\n    (b : _ → _) = (↑)) ↔ Squarefree n := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Field Extensions"], "name": "Problem_31", "informal_proof": "\n\n(1) $n$ is not squarefree $\\implies \\exists$ prime $p$ s.t. $p^2 \\mid n$.  \nFor $\\eta_n$ is a primitive $n$th root of unity, consider $\\beta := \\eta_n^{n/p}$.  \nNow, $\\beta^p = \\left(\\eta_n^{n/p}\\right)^p = 1 \\implies \\beta$ is a primitive $p$th root of unity.  \nHence, $\\beta$ satisfies the equation\n\\[\n1 + \\beta + \\beta^2 + \\dots + \\beta^{p-1} = 0.\n\\]\n\nObserve also that when $j$ satisfies $1 \\leq j \\leq p-1$, it follows that\n\\[\n\\eta_n \\beta^j = \\eta_n \\cdot \\eta_n^{j n / p} = \\eta_n^{j n / p + 1}.\n\\]\nSince $p^2$ divides $n$, every prime $q$ dividing $n$ will also divide $\\frac{n j}{p}$.  \nBut we can be sure that $\\gcd\\left(\\frac{n j}{p} + 1, n\\right) = 1$.  \nThus, $\\eta_n^{j n / p + 1}$ is a primitive $n$th root of unity.  \nThen the $\\eta_n \\beta^j$'s satisfy the equation\n\\[\n1 + \\eta_n \\beta + \\eta_n \\beta^2 + \\dots + \\eta_n \\beta^{p-1} = 0.\n\\]\n\nBut this shows that $\\{1, \\alpha, \\alpha^2, \\dots, \\alpha^{p-1}\\}$ is linearly dependent and hence not a basis for the cyclotomic field of $n$th roots of unity.\n\n(2) Let us show that in case of an $n$ of the shape\n\\[\nn = \\prod_{p \\in P} p,\n\\]\nwith a product of primes from a finite set of primes $P$, the primitive $n$th roots of unity are generating $F$, the cyclotomic field of $n$th roots of unity, as a vector space over $\\mathbb{Q}$. The field extension $F:\\mathbb{Q}$ has degree $\\phi(n)$, exactly as many as primitive $n$th roots of unity.\n\nFix for each $p$ a primitive $p$th root of unity $\\zeta_p$. Then a primitive $n$th root of unity is of the shape:\n\\[\n\\zeta(k) := \\prod_{p \\in P} \\zeta_p^{k(p)},\n\\]\nwhere $k = (k(p))_{p \\in P}$ is a multiindex, $0 < k(p) < p$. \n(Their number is thus $\\prod (p-1) = \\prod p \\prod \\left(1 - \\frac{1}{p}\\right) = n \\prod_{p \\mid n} \\left(1 - \\frac{1}{p}\\right) = \\varphi(n)$.)\n\nConsider an element $x$ of $F$, and write it as a linear combination with rational scalars over all roots of order $n$, primitive or not (seen as \"vectors\"). This is possible, since $F$ is by definition generated by all powers of $\\zeta_n := \\exp \\left( \\frac{2\\pi i}{n} \\right)$. Using the (multiplicative version of the) Chinese Remainder Theorem, each \"vector\" above is of the form\n\\[\n\\zeta(l) := \\prod_{p \\in P} \\zeta_p^{l(p)},\n\\]\nwhere $l = (l(p))_{p \\in P}$ is a multiindex, $0 \\leq l(p) < p$.\n\nFor each $\\zeta(l)$ which has one or more $l$-components equal to zero, replace in the $\\zeta(l)$-product the corresponding factor $1$ by the obvious linear combination of $\\zeta_p$-powers, extracted from\n\\[\n0 = 1 + \\left( \\zeta_p + \\zeta_p^2 + \\dots + \\zeta_p^{p-1} \\right).\n\\]\nThis leads to a wanted representation of $x$ in terms of $\\zeta(k)$ primitive roots, $k$ having no component equal to zero.\n\n", "header": "import Mathlib\n\n", "informal_prefix": "/- Prove that the primitive $n^{\\textrm{th}}$ roots of unity form a basis over $\\mathbb{Q}$ for\nthe cyclotomic field of $n^{\\textrm{th}}$ roots of unity if and only if $n$ is squarefree.-/"}
{"informal_statement": "Prove that the splitting field of $x^4 - 2x^2 - 2$ over $\\mathbb{Q}$ is of degree $8$ with dihedral Galois group", "formal_statement": "theorem nonempty_galois_mulEquiv_dihedralGroup_four :\n    Nonempty (Gal (X ^ 4 - 2 * X ^ 2 - 2 : ℚ[X]) ≃* DihedralGroup 4) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Explicit Computations"], "name": "Problem_32", "informal_proof": "\n\n\\begin{enumerate}\n    \\item[(a)] Prove that $x^4 - 2x^2 - 2$ is irreducible over $\\mathbb{Q}$.\n\n    \\item[(b)] Show that the roots of this quartic are \n    $\\alpha_1 = \\sqrt{1 + \\sqrt{3}}$, \n    $\\alpha_2 = \\sqrt{1 - \\sqrt{3}}$, \n    $\\alpha_3 = -\\sqrt{1 + \\sqrt{3}}$,\n    $\\alpha_4 = -\\sqrt{1 - \\sqrt{3}}$.\n\n    \\item[(c)] Let $K_1 = \\mathbb{Q}(\\alpha_1)$ and $K_2 = \\mathbb{Q}(\\alpha_2)$. Show that $K_1 \\neq K_2$ and $K_1 \\cap K_2 = \\mathbb{Q}(\\sqrt{3}) = F$.\n\n    \\item[(d)] Prove that $K_1$, $K_2$ and $K_1K_2$ are Galois over $F$ with $\\operatorname{Gal}(K_1K_2/F)$ the Klein 4-group. Write out the elements of $\\operatorname{Gal}(K_1K_2/F)$ explicitly. Determine all the subgroups of the Galois group and give their corresponding fixed subfields of $K_1K_2$ containing $F$.\n\n    \\item[(e)] Prove that the splitting field of $x^4 - 2x^2 - 2$ over $\\mathbb{Q}$ is of degree 8 with dihedral Galois group.\n\\end{enumerate}\n\n\\begin{enumerate}\n        \\item[(a)] The polynomial $x^4 - 2x^2 - 2$ is irreducible by Eisenstein's criterion for $p = 2$.\n\n        \\item[(b)] Note that \n        \\begin{align*}\n            (\\pm\\sqrt{1 \\pm \\sqrt{3}})^4 - 2(\\pm\\sqrt{1 \\pm \\sqrt{3}})^2 - 2 \n            &= (4 \\pm 2\\sqrt{3}) - 2(1 \\pm \\sqrt{3}) - 2 \\\\\n            &= 0.\n        \\end{align*}\n\n        \\item[(c)] Observe that $\\alpha_1$ is real, while $\\alpha_2$ is complex, so $K_1 \\neq K_2$. Now $F \\subseteq K_1 \\cap K_2$. $K_1$, $K_2$ are each of degree 4, and they're not equal, so $2 \\leq [K_1 \\cap K_2 : \\mathbb{Q}] < 4$. Therefore $K_1 \\cap K_2 = F$.\n\n        \\item[(d)] We have the following factorization\n        \\[\n        x^4 - 2x^2 - 2 = (x^2 - 1 - \\sqrt{3})(x^2 - 1 + \\sqrt{3}) \\in F[x],\n        \\]\n        and clearly $K_1$ is the splitting field of $x^2 - 1 - \\sqrt{3} \\in F[x]$ so $K_1/F$ is Galois. Similarly, $K_2/F$ is also Galois.\n\n        Now $K_1K_2$ is the splitting field of the polynomial $x^4 - 2x^2 - 2$ over $F$ and $\\operatorname{Gal}(K_1K_2/F)$ is generated by\n        \\[\n        \\tau : \\begin{cases}\n            \\alpha_1 \\mapsto \\alpha_1 \\\\\n            \\alpha_2 \\mapsto \\alpha_4\n        \\end{cases}, \\quad\n        \\sigma : \\begin{cases}\n            \\alpha_1 \\mapsto \\alpha_3 \\\\\n            \\alpha_2 \\mapsto \\alpha_2\n        \\end{cases}\n        \\]\n        so it has the structure of the Klein 4-group. The subgroup $\\{1, \\tau\\}$ corresponds to the fixed field $K_1$, $\\{1, \\sigma\\}$ corresponds to $K_2$, $\\{1, \\sigma\\tau\\}$ corresponds to $F(\\sqrt{-2})$, the identity subgroup corresponds to $K_1K_2$, and $\\{1, \\sigma, \\tau, \\sigma\\tau\\}$ corresponds to $F$.\n\n        \\item[(e)] Since $K_1K_2$ is the splitting field of $x^4 - 2x^2 - 2$ over $\\mathbb{Q}$ we obtain $[K_1K_2 : \\mathbb{Q}] = [K_1K_2 : F][F : \\mathbb{Q}] = 4 \\cdot 2 = 8$ so $G = \\operatorname{Gal}(K_1K_2/\\mathbb{Q})$ is of order 8. From the previous part, we see that $G$ has at least 3 subgroups of order 2. Also, $G$ is not abelian. Since the only nonabelian subgroups of order 8 are $D_8$ and $Q_8$, we conclude that $G$ must be the dihedral group.\n\\end{enumerate}\n\n", "header": "import Mathlib\n\nopen Polynomial\n\n", "informal_prefix": "/- The Galois group of the splitting field of $x^4 - 2x^2 - 2$ over $\\mathbb{Q}$ is the\ndihedral group with eight elements-/"}
{"informal_statement": "Let $f(x) \\in \\mathbb{Q}[x]$ be a polynomial of degree $n$ ($n>4$) and the splitting field $E$ of $f(x)$ has Galois group $S_n$ over $\\mathbb{Q}$. Let $\\alpha$ be a zero of $f(x)$ in $E$. Prove that for any other root $\\beta$ of $f(x)$, there are precisely $(n-1)!$ elements in $\\mathrm{Gal}(E/\\mathbb{Q})$ that takes $\\alpha $ to $\\beta$.", "formal_statement": "theorem card_gal_map_eq_eq_factorial {n : Nat} (hn : n > 4) (f : ℚ[X]) (hf : f.degree = n)\n    (hf' : Irreducible f) (h : f.Gal ≃* (Equiv.Perm <| Fin n))\n    {a b : SplittingField f} (ha : f.aeval a = 0) (hb : f.aeval b = 0) (neq : a ≠ b) :\n    Nat.card {h : f.Gal // h a = b} = Nat.factorial (n - 1) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Group Theory", "Group Actions and Sylow theorems"], "name": "Problem_33", "informal_proof": "\nAny Galois action permutes the $n$ roots of $f$ in $E$, and if two Galois action take same images on these roots then they are same in the whole field $E$ since $E$ is generated by these roots. So we have an isomorphism between the Galois group $S_n$ and the permutation group of roots of $f$, so the number of elements which take a root to another is $(n-1)!$.\n", "header": "import Mathlib\n\nopen Polynomial\n\n", "informal_prefix": "/- Let $f(x) \\in \\mathbb{Q}[x]$ be a polynomial of degree $n$ ($n>4$) and the splitting field $E$\nof $f(x)$ has Galois group $S_n$ over $\\mathbb{Q}$. Let $\\alpha$ be a zero of $f(x)$ in $E$.\nProve that for any other root $\\beta$ of $f(x)$, there are precisely $(n-1)!$ elements in\n$\\mathrm{Gal}(E/\\mathbb{Q})$ that takes $\\alpha $ to $\\beta$ -/"}
{"informal_statement": "Let \\( E \\) be a field of characteristic zero. Consider a prime \\( q \\) and an element \\( b \\in E^\\times \\) that isn’t a \\( q \\)-th power. Let \\( E' = E(a) \\) with \\( a^q = b \\) and \\( E' \\neq E \\). Show that \\( X^q - b \\) is reducible over \\( E \\) if and only if \\( [E' : E] < q \\).", "formal_statement": "theorem not_irreducible_iff_finrank_lt {E E': Type} [Field E] [CharZero E] [Field E'] [Algebra E E'] {q : ℕ}\n    [Fact q.Prime] {b : E} (hb : b ≠ 0) (not_pow : ∀ c : E, c ^ q ≠ b) {a : E'}\n    (ha : a ^ q = algebraMap E E' b) (haE : ⊤ = IntermediateField.adjoin E {a}) :\n    ¬ Irreducible (X ^ q - C b) ↔ Module.finrank E E' < q := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Field Extensions"], "name": "Problem_34", "informal_proof": "\nIf $f(x):=x^q-b$ is reducible, we write $f(x)=f_1(x)f_2(x)$. Since $f(a)=0$, $f_1(a)$ or $f_2(a)$ is $0$. The degrees of $f_1$ and $f_2$ are strictly smaller than $q$. So $[E':E]<q$, as we can write $E'$ as $\\sum_{i=0}^{deg(f_1)-1} Ea^i$ assuming that $f_1(a)=0$. If $[E':E]<q$, it follows that $a$ satisfies a polynomial $g(x)$ of degree $<q$. If $f$ is irreducible, it is the minimal polynomial of $a$, but this gives a contradiction since $a$ satisfies a polynomial of degree $<q$.\n", "header": "import Mathlib\n\nopen Polynomial\n", "informal_prefix": "/- Let \\( E \\) be a field of characteristic zero. Consider a prime \\( q \\) and an element \\( b\n\\in E^\\times \\) that isn’t a \\( q \\)-th power. Let \\( E' = E(a) \\) with \\( a^q = b \\). Show that\n\\( X^q - b \\) is reducible over \\( E \\) if and only if \\( [E' : E] < q \\)-/"}
{"informal_statement": "Let $D \\in \\mathbb{Z}$ be a squarefree integer and let $a \\in \\mathbb{Q}$ be a nonzero rational number. Show that $\\mathbb{Q}(\\sqrt{a\\sqrt D})$ cannot be a cyclic extension of degree $4$ over $\\mathbb{Q}$.", "formal_statement": "theorem isEmpty_adjoinRoot_zMod_four {d : ℤ} (hd : Squarefree d) {a : ℚ} (ha : a ≠ 0) :\n    IsEmpty ((AdjoinRoot ((a⁻¹ • X ^ 2) ^ 2 - C (d : ℚ)) ≃ₐ[ℚ]\n      AdjoinRoot ((a⁻¹ • X ^ 2) ^ 2 - C (d : ℚ))) ≃* Multiplicative (ZMod 4)) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Explicit Computations"], "name": "Problem_35", "informal_proof": "\nThis extension is not Galois, as $[\\mathbb{Q}(\\sqrt{a\\sqrt{D}}):\\mathbb{Q}]=4$, but $\\sqrt{-a\\sqrt{D}}\\notin \\mathbb{Q}(\\sqrt{a\\sqrt{D}})$ but \\textbf{is} a root of $x^4=a^2D$. (If this were a Galois extension of degree $4$, adjoining $\\sqrt{a\\sqrt{D}}$ could give us the splitting field.)\n", "header": "import Mathlib\n\nopen Polynomial IntermediateField\n\n", "informal_prefix": "/- Let $D \\in \\mathbb{Z}$ be a squarefree integer and let $a \\in \\mathbb{Q}$ be a nonzero rational number.\nShow that $\\mathbb{Q}(\\sqrt{a\\sqrt D})$ cannot be a cyclic extension of degree $4$ over $\\mathbb{Q}$.-/"}
{"informal_statement": "Let $f(X)=X^4-X^2-1\\in \\mathbb{Q}[X]$, $K$ is the splitting field of $f$ over $\\mathbb{Q}$, prove that the number of intermediate fields of $K/\\mathbb{Q}$ is 10.", "formal_statement": "theorem card_intermediateField_splittingField_eq_ten :\n    Nat.card (IntermediateField ℚ (X ^ 4 - X ^ 2 - 1 : ℚ[X]).SplittingField) = 10 := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Galois theory"], "name": "Problem_36", "informal_proof": "\nThe roots of $f(X)$ are:\n\\[\nX = \\pm \\alpha, \\quad \\pm i\\beta \\quad \\text{where} \\quad \\alpha = \\sqrt{\\frac{1+\\sqrt{5}}{2}}, \\quad \\beta = \\sqrt{\\frac{\\sqrt{5}-1}{2}} = \\frac{1}{\\alpha}.\n\\]\n\nThe splitting field \\( K \\) over \\( \\mathbb{Q} \\) is generated by these roots, so $K = \\mathbb{Q}(\\alpha, i)$, since:\n\\begin{itemize}\n    \\item \\( \\alpha \\) has minimal polynomial \\( f(X) \\), irreducible over \\( \\mathbb{Q} \\), so \\( [\\mathbb{Q}(\\alpha) : \\mathbb{Q}] = 4 \\).\n    \\item Adjoining \\( i \\) to \\( \\mathbb{Q}(\\alpha) \\) gives \\( K \\), a degree-2 extension, so \\( [K : \\mathbb{Q}] = 8 \\).\n\\end{itemize}\n\nThe Galois group \\( G = \\text{Gal}(K/\\mathbb{Q}) \\) has order 8. Define automorphisms:\n\\begin{itemize}\n    \\item \\( \\sigma \\): Rotation generator with \\( \\sigma(\\alpha) = i/\\alpha \\) and \\( \\sigma(i) = -i \\). This satisfies \\( \\sigma^4 = \\text{id} \\).\n    \\item \\( \\tau \\): Reflection generator (complex conjugation) with \\( \\tau(\\alpha) = \\alpha \\) and \\( \\tau(i) = -i \\), satisfying \\( \\tau^2 = \\text{id} \\).\n\\end{itemize}\n\nThese generators obey \\( \\tau\\sigma = \\sigma^{-1}\\tau \\), characteristic of the dihedral group \\( D_4 \\). Thus: $G \\cong D_4$, which has exactly 10 subgroups. By the fundamental theorem of Galois theory, intermediate fields correspond bijectively to subgroups of \\( G \\). Hence, the number of intermediate fields is 10.\n", "header": "import Mathlib\n\nopen Polynomial IntermediateField\n\n", "informal_prefix": "/-- Let $f(X)=X^4-X^2-1\\in \\mathbb{Q}[X]$, $K$ is the splitting field of $f$ over $\\mathbb{Q}$,\nprove that the number of intermediate fields of $K/\\mathbb{Q}$ is 10. -/"}
{"informal_statement": "Let $L/K$ be a Galois extension of fields such that $\\mathrm{Gal}(L/K)$ is cyclic of order $n$, generated by $\\sigma$. Write $n = ab$ with $\\gcd(a,b) =1$. Let $F_1$ be the fixed field of $\\sigma^a$ and $F_2$ be the fixed field of $\\sigma^b$. Suppose that $F_1 = K(\\alpha)$ and $F_2 = K(\\beta)$.  Prove that $L = K(\\alpha+\\beta)$.", "formal_statement": "theorem adjoin_add_eq_top_of_fixedField_zpowers {K L : Type} [Field K]\n    [Field L] [Algebra K L] [IsGalois K L] (n a b : ℕ)\n    (σ : L ≃ₐ[K] L) (hσ : orderOf σ = n) (cycle : Subgroup.zpowers σ = ⊤) (hn : n > 0)\n    (hn' : n = a * b) (hab : Nat.Coprime a b) {α β : L}\n    (hα : K⟮α⟯ = IntermediateField.fixedField (Subgroup.zpowers (σ ^ a)))\n    (hβ : K⟮β⟯ = IntermediateField.fixedField (Subgroup.zpowers (σ ^ b))) : K⟮α + β⟯ = ⊤ := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Galois theory"], "name": "Problem_37", "informal_proof": "\nAs each subextension of $L/K$ cyclic, assume $\\mathrm{Gal}(L/K(\\alpha+\\beta))=\\langle\\sigma^d\\rangle$ for some $d|n$. Note that $\\alpha+\\beta=\\sigma^{\\text{lcm}(d,b)}(\\alpha+\\beta)=\\sigma^{\\text{lcm}(d,b)}\\alpha+\\beta$, we get $a|{\\text{lcm}(d,b)}$. Hence $a|d$ as $\\gcd(a,b)=1$, and similarly $b|d$, then ${\\text{lcm}(a,b)}=n|d$. Therefore $d=n$, thus $L=K(\\alpha+\\beta)$.\n", "header": "import Mathlib\n\nopen Polynomial IntermediateField AdjoinRoot\n\n", "informal_prefix": "/-- Let $L/K$ be a Galois extension of fields such that $\\mathrm{Gal}(L/K)$ is cyclic of order $n$,\ngenerated by $\\sigma$. Write $n = ab$ with $\\gcd(a,b) =1$. Let $F_1$ be the fixed field of\n$\\sigma^a$ and $F_2$ be the fixed field of $\\sigma^b$. Suppose that $F_1 = K(\\alpha)$ and\n$F_2 = K(\\beta)$.  Prove that $L = K(\\alpha+\\beta)$. -/"}
{"informal_statement": "Let $p$ be a prime number and let $F$ be a field containing $p$-th roots of unity. Let $K$ be a Galois extension of $F$ with Galois group $\\mathbb{Z}/p \\times \\mathbb{Z}/p$. Show that there exist two elements $\\alpha, \\beta \\in K^\\times$ such that $K= F(\\alpha, \\beta)$ and $a = \\alpha^p, b = \\beta^p \\in F$.", "formal_statement": "theorem exists_pow_p_mem_algebraMap_and_adjoin_eq_top {p : Nat} [Fact p.Prime]\n    {F K : Type} [Field F] (hF : (primitiveRoots p F).Nonempty) [Field K] [Algebra F K]\n    [IsGalois F K] (hK : (K ≃ₐ[F] K) ≃* Multiplicative (ZMod p × ZMod p)) :\n    ∃ (α β : K), α ≠ 0 ∧ β ≠ 0 ∧ α ^ p ∈ (algebraMap F K).range ∧ β ^ p ∈ (algebraMap F K).range ∧\n    IntermediateField.adjoin F {α, β} = ⊤ := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Galois theory"], "name": "Problem_38", "informal_proof": "\nDenote by $K_1,K_2/F$ the subextensions fixed by the two factors $\\mathbb{Z}/p$, then $K=K_1K_2$. By Kummer theory each $K_i$ is a radical extension of deg $p$.\n", "header": "import Mathlib\n\nopen IntermediateField\n\n", "informal_prefix": "/-Let $p$ be a prime number and let $F$ be a field containing $p$-th roots of unity.\nLet $K$ be a Galois extension of $F$ with Galois group $\\mathbb{Z}_p \\times \\mathbb{Z}_p$.\nShow that there exist two elements $\\alpha, \\beta \\in K^\\times$ such that\n$K= F(\\alpha, \\beta)$ and $a = \\alpha^p, b = \\beta^p \\in F$.-/"}
{"informal_statement": "Prove that a splitting field of \\( X^{15} - 2 \\) over \\( \\mathbb{F}_7 \\) is generated by a primitive \\( 45 \\)-th root of unity.", "formal_statement": "theorem exists_isPrimitiveRoot_and_adjoin_eq_top :\n    letI : Fact (Nat.Prime 7) := by decide\n    ∃ ζ : (X ^ 15 - 2 : (ZMod 7)[X]).SplittingField, IsPrimitiveRoot ζ 45 ∧\n    IntermediateField.adjoin (ZMod 7) {ζ} = ⊤ := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Explicit Computations"], "name": "Problem_39", "informal_proof": "\nLet $\\alpha$ be a root. Then $\\alpha^{45}\\equiv 1$ mod $7$. Then all roots of $X^{15}-2$ are $45$-th roots of unity. So this polynomial splits over $\\mathbb{F}_{7^{n}}$ where $45|7^n-1$. We compute and find $n=12$. We show that the splitting field cannot be smaller. When $n=4$, $7^4-1=2400$, then $\\alpha^{2400}\\equiv2^{160}\\equiv 2$ mod $7$, which is impossible to be contained in $\\mathbb{F}_{7^4}$. \nWhen $n=6$, $7^6-1=117648=2^43^2\\cdot 19\\cdot 43$, then $\\alpha^{117648}\\equiv(2\\alpha)^{9\\cdot19\\cdot 43}\\equiv \\alpha^{9\\cdot 19\\cdot 43}\\equiv\\alpha^{15\\times490+3}\\equiv \\alpha^3\\cdot 2^{490}\\equiv 2\\alpha^3$ mod $7$. But $\\alpha$ cannot always satisfy $2\\alpha^3\\equiv 1$. (In fact, suppose $\\alpha$ is also a root of $2X^3-1\\equiv X^3+3$. One can verify that $X^3+3|X^{15}-2$, but $X^{15}-2\\neq (X^3+3)^5$ mod $7$.) So $\\mathbb{F}_{7^{12}}$ is the (smallest) splitting field.\\par\nMoreover, $\\mathbb{F}_{7^{12}}$ is the smallest field extension over $\\mathbb{F}_7$ that contains a primitive $45$-th root of unity. So it is generated by $45$-th roots.\n", "header": "import Mathlib\n\nopen Polynomial\n", "informal_prefix": "/- Prove that a splitting field of \\( X^{15} - 2 \\) over \\( \\mathbb{F}_7 \\) is generated by a\n primitive \\( 45 \\)-th root of unity.-/"}
{"informal_statement": "Let $K$ be the splitting field of an irreducible quintic polynomial $f(x) \\in \\mathbb{Q} [x]$ and let $\\{\\alpha_1, \\dots, \\alpha_5\\}$ be zeros of $f(x)$ in $K$. Show that if $\\mathbb{Q}(\\alpha_1, \\alpha_2, \\alpha_3) \\neq K$, then $\\mathrm{Gal}(K/\\mathbb{Q}) \\cong S_5$.", "formal_statement": "theorem nonempty_gal_mulEquiv_perm_fin_five {f : ℚ[X]} {K : Type} [Field K] [Algebra ℚ K]\n    [IsSplittingField ℚ K f] (hf1 : Irreducible f) (hf2 : f.natDegree = 5) (a₁ a₂ a₃ : K)\n    (ha1 : a₁ ∈ rootSet f K ∧ a₂ ∈ rootSet f K ∧ a₃ ∈ rootSet f K)\n    (ha2 : a₁ ≠ a₂ ∧ a₂ ≠ a₃ ∧ a₃ ≠ a₁)\n    (h : IntermediateField.adjoin ℚ {a₁, a₂, a₃} ≠ ⊤) :\n    Nonempty (f.Gal ≃* (Equiv.Perm <| Fin 5)) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Galois theory"], "name": "Problem_40", "informal_proof": "\nIf $H:=\\mathbb{Q}(\\alpha_1,\\alpha_2,\\alpha_3)\\neq K$, then $K/\\mathbb{Q}(\\alpha_1,\\alpha_2,\\alpha_3)$ is quadratic. Indeed, $f$ splits to $(x-\\alpha_1)(x-\\alpha_2)(x-\\alpha_3)f_2(x)$ in $H$, and $f_2$ is irreducible because otherwise $H\\supset K$. Then $K/H$ is quadratic and is Galois by the fundamental theorem of Galois theory. Then $\\mathrm{Gal}(K/H)$ is a group of order $2$.\\par\nThe group $G:=\\mathrm{Gal}(K/\\mathbb{Q})$ permutes the roots $\\{\\alpha_1,\\alpha_2,\\alpha_3,\\alpha_4,\\alpha_5\\}$. By the definition of splitting fields, there is an injective group homomorphism\n$$G\\hookrightarrow S_5.$$\nBy the last paragraph, there is a permutation $(4\\ 5)\\in G$. Since the order of $G$, $|G|=[K:\\mathbb{Q}]$ is divisible by $5=[\\mathbb{Q}(\\alpha_1:\\mathbb{Q})]$, and since $<(1\\ 2\\ 3\\ 4\\ 5)>$ is the only subgroup of order $5$ in $S_5$, we have that $\\sigma=(1\\ 2\\ 3\\ 4\\ 5)$ is in $G$. By doing conjugates of $\\sigma^i$ for $(4\\ 5)$, we know that $(n\\ n+1)$ are in $G$. By multiplying with $\\sigma$, $\\tau_5=(1\\ 2\\ 3\\ 4)$, $\\tau_4=(1\\ 2\\ 3\\ 5)$, $\\tau_3=(1\\ 2\\ 4\\ 5)$, $\\tau_2=(1\\ 3\\ 4\\ 5)$ and $\\tau_1=(2\\ 3\\ 4\\ 5)$ are in $G$. By conjugating $\\tau_i$ to $(4\\ 5)$, we see that $G$ contains all permutations of the form $(a\\ b)$. So $G=S_5$.\n", "header": "import Mathlib\n\nopen Polynomial\n\n", "informal_prefix": "/- Let $K$ be the splitting field of a irreducible quintic polynomial $f(x) \\in \\mathbb{Q} [x]$\nand let $\\{\\alpha_1, \\dots, \\alpha_5\\}$ be zeros of $f(x)$ in $K$. Show that if $\\mathbb{Q}\n(\\alpha_1, \\alpha_2, \\alpha_3) \\neq K$, then $\\\\mathrm{Gal}(K/\\mathbb{Q}) \\cong S_5$.-/"}
{"informal_statement": "Let $p$ be a prime integer. Suppose that the degree of every finite extension of a field $F$ is divisible by $p$. Prove that the degree of every finite extension of $F$ is a power of $p$.", "formal_statement": "theorem exists_finrank_eq_pow_of_dvd_finrank {F : Type} [Field F] (p : ℕ) [Fact (Nat.Prime p)]\n    (h : ∀ (E : Type) [Field E] [Algebra F E] [FiniteDimensional F E],\n    Module.finrank F E ≠ 1 → p ∣ Module.finrank F E)\n    (E : Type) [Field E] [Algebra F E] [FiniteDimensional F E] :\n    ∃ k, Module.finrank F E = p ^ k := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Galois theory"], "name": "Problem_41", "informal_proof": "\nLet $E$ be a finite separable extension of $F$ and $K$ its Galois closure; write $G: = \\mathrm{Gal}(K/F)$. Take $H$ to be the $p$-Sylow subgroup of $G$, then $K^H$ is an extension of $F$ of degree $[G:H]$, which is prime to $p$, unless $G=H$.  But we have assumed that the degree of every finite extension of $F$ is divisible by $p$. So $G=H$, and thus any finite separable extension of $F$ has degree a power of $p$.\n\nIf $F$ is perfect, the above already proved the problem. Otherwise, $F$ has an extension of degree $ \\ell = \\mathrm{char}(k)$ given by $F(\\sqrt[\\ell]{a})$ for some $a \\in F \\backslash F^\\ell$. Thus $\\ell = p$. Then all finite extensions of $F$ have degree a power of $p$.\n\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Let $p$ be a prime integer. Suppose that the degree of every finite extension of a field $F$\n is divisible by $p$. Prove that the degree of every finite extension of $F$ is a power of $p$. -/"}
{"informal_statement": "Let \\( K \\) be a field with \\( \\operatorname{char}(K) \\neq 2 \\). Consider Galois extensions \\( L/K \\) with \\( \\operatorname{Gal}(L/K) \\cong (\\mathbb{Z}/2\\mathbb{Z})^2 \\). Let \\( c \\in L^\\times \\) be a nonsquare, and let \\( E = L(\\sqrt{c}) \\). Prove that \\( E \\) is Galois over \\( K \\) if and only if for each \\( \\sigma \\in \\operatorname{Gal}(L/K) \\), the ratio \\( \\sigma(c)/c \\) is a square in \\( L \\).", "formal_statement": "theorem isGalois_adjoin_iff_isSquare (K L : Type) [Field K] [Field L] (h : ¬ CharP K 2) [Algebra K L] [IsGalois K L]\n    (f : (L ≃ₐ[K] L) ≃* Multiplicative (ZMod 2 × ZMod 2)) (c : Lˣ) (hc : c.1 ≠ 0)\n    (hcs : ¬ IsSquare c.1) [Fact (Irreducible (X ^ 2 - C c.1))] :\n    IsGalois K (AdjoinRoot (X ^ 2 - C c.1)) ↔ ∀ σ : (L ≃ₐ[K] L), IsSquare (σ c / c) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Galois theory"], "name": "Problem_42", "informal_proof": "\n``if'': This case $\\sqrt c$ is a root of $\\prod_{\\mathrm{Gal}(L/K)}(T^2-\\sigma c)\\in K[T]$ with all roots in $E$.\\\\\n``only if'': else each $\\sigma\\in\\mathrm{Gal}(L/K)$ extends to some $\\sigma'\\in\\mathrm{Gal}(E/K)$. Note that $\\sigma'(\\sqrt c)/\\sqrt c\\in E$ is a square root of $\\sigma(c)/c$. Referring to Exercise 131, if the ratio is a nonsquare in $L$, then $[E:L]>2$, a contradiction. \n", "header": "import Mathlib\n\nopen IntermediateField AdjoinRoot Polynomial\n\n", "informal_prefix": "/--Let \\( K \\) be a field with \\( \\operatorname{char}(K) \\neq 2 \\). Consider Galois extensions\n\\( L/K \\) with \\( \\operatorname{Gal}(L/K) \\cong (\\mathbb{Z}/2\\mathbb{Z})^2 \\). Let \\( c \\in L^\\times\n\\) be a nonsquare, and let \\( E = K(\\sqrt{c}) \\). Prove that \\( E \\) is Galois over \\( K \\) if and\nonly if for each \\( \\sigma \\in \\operatorname{Gal}(L/K) \\), the ratio \\( \\sigma(c)/c \\) is a square\nin \\( L \\).-/"}
{"informal_statement": "Let $F$ be a field with $\\mathbb{Q} \\subseteq F \\subseteq \\mathbb{C}$, where $F/\\mathbb{Q}$ is a finite \\emph{abelian} Galois extension. Let $\\alpha \\in F$ and let $f(x) \\in \\mathbb{Q}[x]$ be its minimal monic polynomial. Assume that $|\\alpha | =1$. Prove that $|\\beta| = 1$ for every complex root $\\beta$ of $f(x)$.", "formal_statement": "theorem norm_eq_one_of_mem_rootSet (F : IntermediateField ℚ ℂ) [FiniteDimensional ℚ F] [IsGalois ℚ F]\n    (h : ∀ f g : F ≃ₐ[ℚ] F, f * g = g * f) (α : F) (f : ℚ[X])\n    (h_min : f = minpoly ℚ α) (ha : ‖α‖ = 1)\n    (β : ℂ) (hb : β ∈ f.rootSet ℂ) : ‖β‖ = 1 := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Galois theory"], "name": "Problem_43", "informal_proof": "\nWe assume that there are complex roots for $f$, otherwise the statement is empty. So $\\alpha$ is a complex root, since otherwise $f(x)=x\\pm1$ which makes the statement empty. Then there is $\\sigma\\in Gal(F/\\mathbb{Q})$ such that $\\sigma(\\alpha)=\\beta$. Since the extension is abelian and the complex conjugation is a nontrivial element in the Galois group, we have that $\\sigma|\\alpha|^2=\\sigma(\\alpha)\\sigma(\\overline{\\alpha})=|\\sigma(\\alpha)|^2=|\\beta|^2=1$.\n", "header": "import Mathlib\n\nopen Polynomial\n\n", "informal_prefix": "/-- Let $F$ be a field with $\\mathbb{Q} \\subseteq F \\subseteq \\mathbb{C}$, where $F/\\mathbb{Q}$ is a finite \\emph\n{abelian} Galois extension. Let $\\alpha \\in F$ and let $f(x) \\in \\mathbb{Q}[x]$ be its minimal monic\npolynomial. Assume that $|\\alpha | =1$. Prove that $|\\beta| = 1$ for every complex root $\\beta$ of\n$f(x)$. -/"}
{"informal_statement": "Let \\( k \\) be a finite field of size \\( q \\). Show that the number of degree-\\( 19 \\) monic irreducible polynomials over \\( k \\) is \\( \\frac{q^{19} - q}{19} \\).", "formal_statement": "theorem card_monic_and_irreducible_and_natDegree_eq_19 {F : Type} (q : ℕ) [Field F]\n    [Fintype F] (hF : Fintype.card F = q) :\n    Nat.card { P : F[X] | P.Monic ∧ Irreducible P ∧ P.natDegree = 19 } =\n    (q ^ 19 - q) / 19 := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Field Extensions"], "name": "Problem_44", "informal_proof": "\nLet $f$ be such a polynomial. Then any root $\\alpha$ of $f$ generates a field extension of $k$ of degree $19$. But such a field extension over $k$ is unique in $\\overline{k}$, whose elements are exactly the roots of $x^{q^{19}}-x$. We denote it by $k_{q^{19}}$. Moreover, there is no intermediate field between $k$ and $k_{q^{19}}$ since $19$ is a prime number. So every element in $k_{q^{19}}$ is either of degree $19$ or of degree $1$ over $k$. Then $\\#\\{\\text{elements in }k_{q^{19}}\\text{ of degree }19\\}=q^{19}-q$. Then the number of monic irreducible polynomials is $\\frac{q^{19}-q}{19}.$\n", "header": "import Mathlib\n\nopen Polynomial\n\n", "informal_prefix": "/-- Let $k$ be a finite field of size $q$. Show that the number of degree-$19$ monic irreducible\npolynomials over $k$ is $\\frac{q^{19} - q}{19}$. -/"}
{"informal_statement": "Let \\( F \\) be a field, and let \\( f, g \\in F[x] \\setminus \\{0\\} \\) be relatively prime and not both constant. Show that \\( F(x) \\) has finite degree \\( d = \\max(\\deg(f), \\deg(g)) \\) over its subfield \\( F\\left(\\frac{f}{g}\\right) \\).", "formal_statement": "theorem finrank_adjoin_dvd_eq_max_natDegree (F : Type) [Field F] (f g : F[X]) (hf : f ≠ 0)\n    (hg : g ≠ 0) (hcoprime : IsCoprime f g) (hfg : ¬(f.natDegree = 0 ∧ g.natDegree = 0)) :\n    Module.finrank F⟮(f : RatFunc F) / g⟯ (RatFunc F) = max f.natDegree g.natDegree := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Ring Theory", "Polynomials"], "name": "Problem_45", "informal_proof": "\nLet $y = \\frac{f(x)}{g(x)}$, making $F(y)$ a subfield of $F(x)$. Consider the polynomial:\n\\[ P(T) = yg(T) - f(T) \\in F(y)[T]. \\]\nWe will show that $[F(x):F(y)] = \\max(\\deg f, \\deg g)$ by proving that $P(T)$ is the minimal polynomial of $x$ over $F(y)$.\n\n\\medskip\n\n\\noindent \\textbf{Step 1: $P(T)$ is the minimal polynomial if irreducible}\\\\\nSince $P(x) = yg(x) - f(x) = 0$, it suffices to prove that $P(T)$ is irreducible over $F(y)$.\n\n\\medskip\n\n\\noindent \\textbf{Step 2: Viewing $P(T)$ in $F[y,T]$}\\\\\nWe can consider $P(T) = g(T)y - f(T)$ as an element of $F[y,T]$. Note that:\n\\begin{itemize}\n    \\item $y$ is transcendental over $F$ because $x$ is transcendental and $y = f(x)/g(x)$ is a non-constant rational function.\n    \\item $F[y,T]$ is a UFD since $F[y]$ is a polynomial ring over a field.\n\\end{itemize}\n\n\\medskip\n\n\\noindent \\textbf{Step 3: Irreducibility in $F[y,T]$}\\\\\nSuppose $P(T)$ were reducible in $F[y,T]$. Then it would factor as:\n\\[ P(T) = (a(T)y + b(T)) \\cdot c(T) \\]\nwhere $a(T), b(T), c(T) \\in F[T]$ with $\\deg a, \\deg b, \\deg c \\geq 1$. \n\nComparing coefficients:\n\\begin{align*}\ng(T) &= a(T)c(T) \\\\\n-f(T) &= b(T)c(T)\n\\end{align*}\nThis implies $c(T)$ is a common divisor of both $f(T)$ and $g(T)$. But since $\\gcd(f,g) = 1$ in $F[T]$, we must have $c(T) \\in F^\\times$, making the factorization trivial. \n\nTherefore, $P(T)$ is irreducible in $F[y,T]$.\n\n\\medskip\n\n\\noindent \\textbf{Step 4: Application of Gauss's Lemma}\\\\\nBy Gauss's Lemma, since $P(T)$ is irreducible in $F[y][T] = F[y,T]$ and primitive (as the coefficients $g(T)$ and $-f(T)$ are coprime), it remains irreducible in $F(y)[T]$.\n\n\\medskip\n\n\\noindent \\textbf{Step 5: Degree calculation}\\\\\nThe degree of $P(T)$ in $T$ is clearly $\\max(\\deg f, \\deg g)$ because:\n\\begin{itemize}\n    \\item When $\\deg f > \\deg g$, the leading term comes from $-f(T)$\n    \\item When $\\deg g > \\deg f$, the leading term comes from $yg(T)$\n    \\item When $\\deg f = \\deg g$, the leading terms cannot fully cancel because $f$ and $g$ are coprime (any cancellation would imply a common root, contradicting $\\gcd(f,g)=1$)\n\\end{itemize}\n\n\\medskip\n\n\\noindent \\textbf{Conclusion}\\\\\nSince $P(T)$ is the minimal polynomial of $x$ over $F(y)$ and has degree $\\max(\\deg f, \\deg g)$, we conclude:\n\\[ [F(x):F(y)] = \\max(\\deg f, \\deg g). \\]\n", "header": "import Mathlib\n\nopen IntermediateField Polynomial\n\n", "informal_prefix": "/-- Let $F$ be a field, and let $f, g \\in F[x] \\setminus \\{0\\}$ be relatively prime and not both\nconstant. Show that $F(x)$ has finite degree $d = \\max(\\deg(f), \\deg(g))$ over its subfield $F\\left\n(\\frac{f}{g}\\right)$. -/"}
{"informal_statement": "Let \\( \\alpha \\) be a non-zero complex number such that \\( \\alpha + \\alpha^{-1} \\) is contained in a quadratic number field. Let \\( L \\) be the normal closure of \\( \\mathbb{Q}(\\alpha) \\). Show that \\( [L : \\mathbb{Q}] \\) divides \\( 8 \\).", "formal_statement": "theorem finrank_dvd_eight_of_add_inv_eq_algebraMap {α : ℂ} {K L : Type} [Field K] [NumberField K]\n    [Algebra K ℂ] (hk : Module.finrank ℚ K = 2) {a : K} (h : α + α⁻¹ = algebraMap K ℂ a)\n    [Field L] [Algebra ℚ L] [IsNormalClosure ℚ ℚ⟮a⟯ L] : Module.finrank ℚ L ∣ 8 := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Explicit Computations"], "name": "Problem_46", "informal_proof": "\nSet $\\alpha+\\alpha^{-1}=\\beta$. Then $\\alpha^2-\\beta\\alpha+1=0$. Denote $f(x):=(x^2-\\beta x+1)(x^2-\\sigma(\\beta)x+1)\\in \\mathbb{Q}[x]$. To split this polynomial, we can adjoin $\\beta$, $\\alpha$, $\\alpha'$ to $\\mathbb{Q}$ one-by-one, where $\\alpha'$ is a root of $x^2-\\sigma(\\beta)x+1$. Then $[\\mathbb{Q}(\\beta):\\mathbb{Q}]=[\\mathbb{Q}(\\beta,\\alpha):\\mathbb{Q}(\\beta)]=[\\mathbb{Q}(\\beta,\\alpha'):\\mathbb{Q}(\\beta))]=2$, so $[L:\\mathbb{Q}]|[\\mathbb{Q}(\\beta,\\alpha,\\alpha'):\\mathbb{Q}]|8$.\n", "header": "import Mathlib\n\nopen IntermediateField\n", "informal_prefix": "/--Let \\( \\alpha \\) be a non-zero complex number such that \\( \\alpha + \\alpha^{-1} \\) is contained\nin a quadratic number field. Let \\( L \\) be the normal closure of \\( \\mathbb{Q}(\\alpha) \\). Show\nthat \\( [L : \\mathbb{Q}] \\) divides \\( 8 \\).-/"}
{"informal_statement": "Let \\( \\zeta \\) be a primitive \\( 9 \\)-th root of unity in \\( \\mathbb{C} \\), so \\( \\zeta^9 = 1 \\), and let \\( \\omega = \\zeta^3 \\) be a primitive \\( 3 \\)-rd root of unity, so \\( \\omega^3 = 1 \\). If \\( \\alpha^3 = 3 \\), show that \\( \\alpha \\) is not a cube in \\( \\mathbb{Q}(\\zeta) \\).", "formal_statement": "theorem not_exists_eq_pow_three_of_pow_three_eq_three (α : ℂ) (hα : α ^ 3 = 3) (ζ : ℂ)\n    (hζ : IsPrimitiveRoot ζ 9) : ¬ ∃ β : ℚ⟮ζ⟯, α = β ^ 3 := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Other Topics", "Algebraic Number Theory and Valuation Theory"], "name": "Problem_47", "informal_proof": "\nIf $\\alpha$ is a cube in the field \\( \\mathbb{Q}(\\zeta) \\), it is also a cube in the field $\\mathbb{Q}_3(\\zeta)$. The field $\\mathbb{Q}_3(\\zeta)$ is a totally ramified extension of degree 6 over $\\mathbb{Q}_3$. The normalized 3-valuation of $\\alpha$ is $\\frac{1}{3}$, and the valuations can be took by elements in $\\mathbb{Q}_3(\\zeta)$ is of the form $\\frac{1}{6}\\mathbb{Z}$. There is no $\\frac{1}{9}$, so $\\alpha$ can not be a cube.\n", "header": "import Mathlib\n\nopen Algebra\n\nopen scoped IntermediateField\n\n", "informal_prefix": "/-- Let $\\zeta$ be a primitive $9$-th root of unity in $\\mathbb{C}$, so $\\zeta^9 = 1$, and let\n$\\omega = \\zeta^3$ be a primitive $3$-rd root of unity, so $\\omega^3 = 1$. If $\\alpha^3 = 3$,\nshow that $\\alpha$ is not a cube in $\\mathbb{Q}(\\zeta)$. -/"}
{"informal_statement": "Prove that the cardinality $\\mathrm{Aut}(\\mathbb{C})$ (i.e. the group of field automorphism of $\\mathbb{C}$) is infinite.", "formal_statement": "theorem infinite_complex_algEquiv : Infinite (ℂ ≃ₐ[ℚ] ℂ) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Explicit Computations"], "name": "Problem_48", "informal_proof": "\nLet $\\overline{\\mathbb{Q}}$ denote the algebraic closure of $\\mathbb{Q}$ in $\\mathbb{C}$. The complex numbers $\\mathbb{C}$ can be viewed as a transcendental extension of $\\overline{\\mathbb{Q}}$. By Zorn's Lemma, there exists a transcendence basis $S \\subset \\mathbb{C}$ over $\\overline{\\mathbb{Q}}$ with $|S| = 2^{\\aleph_0}$. Notably, this basis contains infinitely many algebraically independent transcendental elements.\n\nFor any finite permutation $\\sigma$ of distinct elements in the transcendence basis $S$, we may construct a unique automorphism as follows:\n\n\\begin{itemize}\n    \\item Fix an enumeration $\\{s_1, s_2, s_3, ...\\} \\subset S$ (using a countable subset of $S$);\n    \\item For each $n \\in \\mathbb{N}$, define $\\sigma_n$ as the transposition swapping $s_n$ and $s_{n+1}$ while fixing all other elements.\n\\end{itemize}\n\nEach $\\sigma_n$ induces a unique automorphism of $\\overline{\\mathbb{Q}}(S)$ by:\n\\begin{align*}\n    \\tilde{\\sigma}_n: \\overline{\\mathbb{Q}}(S) &\\to \\overline{\\mathbb{Q}}(S) \\\\\n    f(s_1, ..., s_n, s_{n+1}, ...) &\\mapsto f(s_1, ..., s_{n+1}, s_n, ...)\n\\end{align*}\nwhere $f$ is a rational function with coefficients in $\\overline{\\mathbb{Q}}$.\n\nSince $\\mathbb{C}$ is the algebraic closure of $\\overline{\\mathbb{Q}}(S)$, each automorphism $\\tilde{\\sigma}_n$ extends uniquely to an automorphism $\\hat{\\sigma}_n$ of $\\mathbb{C}$ by algebraic continuation. Specifically:\n\\begin{itemize}\n    \\item For $\\alpha \\in \\mathbb{C}$ algebraic over $\\overline{\\mathbb{Q}}(S)$, map it to the conjugate root of its minimal polynomial under $\\tilde{\\sigma}_n$;\n    \\item The automorphism preserves field operations by construction.\n\\end{itemize}\n\nThe infinite collection $\\{\\hat{\\sigma}_n\\}_{n\\in\\mathbb{N}}$ consists of distinct automorphisms because:\n\\begin{itemize}\n    \\item $\\hat{\\sigma}_m(s_n) \\neq \\hat{\\sigma}_k(s_n)$ for $m \\neq k$ when $n = \\min(m,k)$;\n    \\item Each transposition generates a unique permutation pattern.\n\\end{itemize}\n\nThus $\\operatorname{Aut}(\\mathbb{C})$ contains infinitely many distinct automorphisms. Moreover, using Cantor's diagonal argument, one can show $|\\operatorname{Aut}(\\mathbb{C})| = 2^{2^{\\aleph_0}}$, but our construction already suffices to prove infiniteness.\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Prove that the cardinality $\\mathrm{Aut}(\\CC)$ (i.e. the group of field automorphism of $\\CC$)\nis infinite. -/"}
{"informal_statement": "Prove that every finitely generated extension of $\\mathbb{Q}$ can be embeded into $\\mathbb{C}$.", "formal_statement": "theorem exists_algHom_complex_injective {F : Type} [Field F] [Algebra ℚ F]\n    (h : (⊤ : IntermediateField ℚ F).FG) : ∃ f : F →ₐ[ℚ] ℂ, Function.Injective f := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Explicit Computations"], "name": "Problem_49", "informal_proof": "\nAssume the extension $F/\\mathbb{Q}$ is finitely generated, so its transcendence degree $\\ell$ is finite and $\\ell \\le n$. Let us fix a transcendence basis $\\{v_1, v_2, \\dots, v_\\ell\\}$. For its part $\\mathbb{C}/\\mathbb{Q}$ is known to have infinite transcendence degree. This follows from a cardinality argument, but using the \\href{https://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem}{Lindemann-Weierstrass theorem} we can more explicitly state that the numbers $z_1 = e^{\\sqrt{2}}$, $z_2 = e^{\\sqrt{3}}$, \\dots, $z_\\ell = e^{\\sqrt{p_\\ell}}$ are all transcendental and algebraically independent. Here the exponents are the square roots of the $\\ell$ smallest primes.\n\nThis means that the subfield $K = \\mathbb{Q}(v_1, v_2, \\dots, v_\\ell) \\subseteq F$ is isomorphic to the subfield $\\mathbb{Q}(z_1, z_2, \\dots, z_\\ell) \\subseteq \\mathbb{C}$ by an embedding of fields $\\phi' : K \\to \\mathbb{C}$ fully determined by $v_i \\mapsto z_i, i = 1, 2, \\dots, \\ell$.\n\nBy the properties of transcendence bases we know that $F/K$ is algebraic and finitely generated. Therefore actually $[F : K]$ is finite. As $\\mathbb{C}$ is algebraically closed, a textbook lemma I described \\href{https://math.stackexchange.com/a/898059/11619}{here} shows that we can then extend $\\phi'$ to an embedding $\\phi : F \\to \\mathbb{C}$ --- adding one generator to the domain at a time. First extending it to an embedding of $F_1 = K(u_1) \\to \\mathbb{C}$, then to an embedding of $F_2 = F_1(u_2)$ etc. Observe that all the extensions $F_{i+1}/F_i$ are simple and algebraic.\n", "header": "import Mathlib\n\n", "informal_prefix": "/- Prove that every finitely generated extension of $\\mathbb{Q}$ can be embeded into $\\mathbb{C}$-/"}
{"informal_statement": "Let \\( m \\) be a maximal ideal of \\( \\mathbb{Z}[x_1, \\dots, x_n] \\) and \\( F = \\mathbb{Z}[x_1, \\dots, x_n]/m \\). Show that \\( F \\) cannot have characteristic \\( 0 \\).", "formal_statement": "theorem not_charZero_mvPolynomial_quot {n : ℕ} (m : Ideal (MvPolynomial (Fin n) ℤ))\n    [hm : m.IsMaximal] : ¬ CharZero (MvPolynomial (Fin n) ℤ ⧸ m) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Ring Theory", "Polynomials"], "name": "Problem_50", "informal_proof": "\n$m \\cap \\mathbb{Z}$ can not be $(0)$, otherwise it will be in $\\mathrm{Spec}\\ \\mathbb{Q}[x_1, \\dots, x_n]$. But the height of $m$ is $n + 1$ and there is no ideal in $\\mathbb{Q}[x_1, \\dots, x_n]$ whose height $> n$.\n", "header": "import Mathlib\n\nopen MvPolynomial\n\n", "informal_prefix": "/-- Let $m$ be a maximal ideal of $\\mathbb{Z}[x_1, \\dots, x_n]$ and $F = \\mathbb{Z}[x_1, \\dots, x_n]\n/m$. Show that $F$ cannot have characteristic $0$. -/"}
{"informal_statement": "Let $K/F$ be a simple algebraic extension. Let $K = F(\\theta)$. Let $L$ be an intermediate field of $K/F$. Show that the minimal polynomial of $\\theta$ over $L$: $g(x) = x^r+\\alpha_1x^{r-1} + \\cdots +\\alpha_r$, satisfies that $F(\\alpha_1, \\dots, \\alpha_r) = L$.", "formal_statement": "theorem adjoin_minpoly_coeffs_toSet_eq {F K : Type} [Field F] [Field K]\n    [Algebra F K] {θ : K} (L : IntermediateField F K) (h : F⟮θ⟯ = ⊤) :\n    IntermediateField.adjoin F (minpoly L θ).coeffs.toSet = L := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Field Extensions"], "name": "Problem_51", "informal_proof": "\nBy the definition, we have $L'= F(\\alpha_1,...\\alpha_r)\\subset L$. If this is not an equality, $r\\geq[K:L']=[F(\\theta): L']>[K: L]=r$, which leads to a contradiction. \n", "header": "import Mathlib\n\nopen IntermediateField\n\n", "informal_prefix": "/-- Let $K/F$ be a simple algebraic extension. Let $K = F(\\theta)$. Let $L$ be an intermediate field\nof $K/F$. Show that the minimal polynomial of $\\theta$ over $L$: $g(x) = x^r+\\alpha_1x^{r-1} +\n\\cdots +\\alpha_r$, satisfies that $F(\\alpha_1, \\dots, \\alpha_r) = L$. -/"}
{"informal_statement": "Let $q$ denote a power of a prime $p$. Show that the extension $\\mathbb{F}_q(t^{1/n})$ over $\\mathbb{F}_q(t)$ is Galois if and only if $q \\equiv 1 \\bmod n$.", "formal_statement": "theorem isGalois_galoisField_X_pow_iff_modEq_one (p m n : ℕ) (hn : 1 ≤ n) (hm : 1 ≤ m)\n    [Fact p.Prime] :  IsGalois (GaloisField p m)⟮(RatFunc.X ^ n : RatFunc (GaloisField p m))⟯\n    (RatFunc (GaloisField p m)) ↔ p ^ m ≡ 1 [MOD n] := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Field Extensions"], "name": "Problem_52", "informal_proof": "\n$\\impliedby$: \\begin{enumerate}\n    \\item \\textbf{Separability:}\n    \\begin{itemize}\n        \\item Let $q = p^m$ where $p$ is prime. If $q \\equiv 1 \\pmod{n}$ then $q-1$ is divisible by $n$, hence $\\gcd(n,p) = 1$.\n        \\item The polynomial $f(x) = x^n - t$ is irreducible over $\\mathbb{F}_q(t)$ by Eisenstein's criterion.\n        \\item The derivative $f'(x) = nx^{n-1} \\neq 0$ in $\\mathbb{F}_q(t)$ since $p \\nmid n$, proving $f(x)$ is separable.\n    \\end{itemize}\n    \n    \\item \\textbf{Normality:}\n    \\begin{itemize}\n        \\item From $q \\equiv 1 \\pmod{n}$, $\\mathbb{F}_q^\\times$ contains all $n^\\text{th}$ roots of unity.\n        \\item The splitting field of $x^n - t$ is $\\mathbb{F}_q(t^{1/n})$, as roots are $\\zeta \\cdot t^{1/n}$ ($\\zeta \\in \\mathbb{F}_q$).\n    \\end{itemize}\n\\end{enumerate}\n\n$\\implies$: \\begin{enumerate}\n    \\item \\textbf{Separability requirement:}\n    \\begin{itemize}\n        \\item If the extension is Galois, $x^n - t$ must be separable. This forces $\\gcd(n,p) = 1$, as $f'(x) = 0$ would otherwise imply $p \\mid n$.\n    \\end{itemize}\n    \\item \\textbf{Normality requirement:}\n    \\begin{itemize}\n        \\item For $\\mathbb{F}_q(t^{1/n})$ to contain all roots of $x^n - t$, $\\mathbb{F}_q$ must contain all $n^\\text{th}$ roots of unity.\n        \\item This requires $n \\mid (q-1) \\implies q \\equiv 1 \\pmod{n}$.\n    \\end{itemize}\n\\end{enumerate}\n", "header": "import Mathlib\n\nopen IntermediateField\n\n", "informal_prefix": "/-- Let $q$ denote a power of a prime $p$. Show that the extension $\\mathbb{F}_q(t^{1/n})$ over\n$\\mathbb{F}_q(t)$ is Galois if and only if $q \\equiv 1 \\bmod n$. -/"}
{"informal_statement": "Prove that every field automorphism of $\\mathbb{R}$ that fixes $\\mathbb{Q}$ is trivial.", "formal_statement": "theorem real_algEquiv_eq_one (f : ℝ ≃ₐ[ℚ] ℝ) : f = 1 := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Explicit Computations"], "name": "Problem_53", "informal_proof": "\nLet \\(\\phi\\) be an automorphism of the field \\(\\mathbb{R}\\). Let \\(x > 0\\) be a positive real number. Then there exists \\(y \\in \\mathbb{R}\\) such that \\(x = y^2\\). Hence\n\\[\n\\phi(x) = \\phi(y^2) = \\phi(y)^2 > 0.\n\\]\n\nIf \\(a < b\\), then \\(b - a > 0\\). By the previous result,\n\\[\n\\phi(b) - \\phi(a) = \\phi(b - a) > 0.\n\\]\nThis implies \\(\\phi(a) < \\phi(b)\\), meaning \\(\\phi\\) is strictly increasing.\n\nFor any natural number \\(n\\), we can write \\(n = 1 + \\cdots + 1\\) (with \\(n\\) terms). Therefore,\n\\[\n\\phi(n) = \\phi(1) + \\cdots + \\phi(1) = 1 + \\cdots + 1 = n.\n\\]\nAny rational number \\(r\\) can be expressed as \\(r = (a - b)c^{-1}\\) for natural numbers \\(a, b, c\\). Thus,\n\\[\n\\phi(r) = \\phi(a - b)\\phi(c)^{-1} = (a - b)c^{-1} = r.\n\\]\n\nLet \\(x \\in \\mathbb{R}\\). For any rational numbers \\(r, s\\) satisfying \\(r < x < s\\), the monotonicity of \\(\\phi\\) gives\n\\[\nr = \\phi(r) < \\phi(x) < \\phi(s) = s.\n\\]\nSince \\(s - r\\) can be made arbitrarily small, it follows that \\(\\phi(x) = x\\). Therefore, \\(\\phi\\) is the identity automorphism.\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Prove that every field automorphism of $\\mathbb{R}$ that fixes $\\mathbb{Q}$ is trivial. -/"}
{"informal_statement": "Let $\\mathbb{F}_4$ be the field with $4$ elements, $t$ a transcendental over $\\mathbb{F}_4$, and $F =\\mathbb{F}_4(t^4 + t)$ and $K =\\mathbb{F}_4(t)$. Show that $K$ is Galois over $F$.", "formal_statement": "theorem isGalois_galoisField_adjoin_X_pow_four_add_X :\n    IsGalois (GaloisField 2 2)⟮(X ^ 4 + X : RatFunc (GaloisField 2 2))⟯\n    (RatFunc (GaloisField 2 2)) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Field Extensions"], "name": "Problem_54", "informal_proof": "\nFirst observe that \\( t \\) is algebraic over \\( F \\) as it satisfies the polynomial \\( f(x) = x^4 + x - s \\in F[x] \\), where \\( s = t^4 + t \\), implying \\( [K:F] \\leq 4 \\). Irreducibility of \\( f(x) \\) over \\( F \\) follows since if it split into lower-degree factors with coefficients in \\( F \\), terms like \\( t^2 \\) would need to lie in \\( F \\), which is impossible as \\( F \\) contains only rational functions in \\( s \\). The polynomial \\( f(x) \\) has all roots \\( \\{t + u \\mid u \\in \\mathbb{F}_4\\} \\subset K \\), verified by substituting \\( t + u \\) into \\( f(x) \\): \\((t + u)^4 + (t + u) = t^4 + u^4 + t + u = s + (u^4 + u) = s\\) because \\( u^4 = u \\) in \\( \\mathbb{F}_4 \\), so \\( f(x) \\) splits completely in \\( K \\). Thus \\( K \\) is the splitting field of \\( f(x) \\), making \\( K/F \\) normal. Separability holds because the derivative \\( f'(x) = 1 \\neq 0 \\) (in characteristic 2), ensuring distinct roots, and \\( \\mathbb{F}_4 \\) being perfect guarantees all algebraic extensions are separable. Hence \\( K/F \\) is both normal and separable, and therefore Galois.\n", "header": "import Mathlib\n\nopen IntermediateField RatFunc\n\n", "informal_prefix": "/-- Let $\\mathbb{F}_4$ be the field with $4$ elements, $t$ a transcendental over $\\mathbb{F}_4$,\nand $F =\\mathbb{F}_4(t^4 + t)$ and $K =\\mathbb{F}_4(t)$. Show that $K$ is Galois over $F$. -/"}
{"informal_statement": "Let $K$ be a field with $\\operatorname{char}(K) \\neq 2$. Consider a Galois extension $L/K$.\nShow that $\\operatorname{Gal}(L/K) \\cong (\\mathbb{Z}/2\\mathbb{Z})^2$ if and only if\nthe extensions $L/K$ has the form $L = K(\\sqrt{a}, \\sqrt{b})$ for $a, b \\in K^\\times$ such that\n$a$, $b$, and $a/b$ are nonsquares in $K^\\times$.", "formal_statement": "theorem exists_pow_two_eq_algebraMap_and_adjoin_eq_top {K L : Type} [Field K] [Field L]\n    [Algebra K L] [IsGalois K L] (hK : ¬ CharP K 2) : IsKleinFour (L ≃ₐ[K] L) ↔\n    ∃ a b : Kˣ, ∃ x y : L, x ^ 2 = algebraMap K L a.1 ∧ y ^ 2 = algebraMap K L b.1 ∧\n    K⟮x, y⟯ = ⊤ ∧ ¬IsSquare a ∧ ¬IsSquare b ∧ ¬IsSquare (a / b) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Explicit Computations"], "name": "Problem_55", "informal_proof": "\n$\\impliedby$: Suppose \\( L = K(\\sqrt{a}, \\sqrt{b}) \\) where \\( a, b \\in K^\\times \\) such that \\( a \\), \\( b \\), and \\( a/b \\) are nonsquares in \\( K^\\times \\). The extensions \\( K(\\sqrt{a})/K \\) and \\( K(\\sqrt{b})/K \\) are both quadratic. If \\( K(\\sqrt{a}) = K(\\sqrt{b}) \\), there exists \\( c, d \\in K \\) such that \\( \\sqrt{b} = c + d\\sqrt{a} \\). Squaring both sides gives \\( b = c^2 + 2cd\\sqrt{a} + d^2a \\). Since \\( \\sqrt{a} \\notin K \\), we must have \\( 2cd = 0 \\). As \\( \\operatorname{char}(K) \\neq 2 \\), this implies \\( cd = 0 \\). If \\( c = 0 \\), then \\( b = d^2a \\), contradicting \\( a/b \\in (K^\\times)^2 \\). If \\( d = 0 \\), then \\( \\sqrt{b} \\in K \\), contradicting \\( b \\notin (K^\\times)^2 \\). Thus \\( K(\\sqrt{a}) \\neq K(\\sqrt{b}) \\), and \\( L/K \\) is a degree 4 extension. The Galois group is generated by automorphisms \\( \\sigma \\) and \\( \\tau \\), where \\( \\sigma(\\sqrt{a}) = -\\sqrt{a} \\), \\( \\sigma(\\sqrt{b}) = \\sqrt{b} \\), and \\( \\tau(\\sqrt{a}) = \\sqrt{a} \\), \\( \\tau(\\sqrt{b}) = -\\sqrt{b} \\), hence \\( \\operatorname{Gal}(L/K) \\cong (\\mathbb{Z}/2\\mathbb{Z})^2 \\).\n\n$\\implies$: Conversely, suppose \\( L/K \\) is a Galois extension with \\( \\operatorname{Gal}(L/K) \\cong (\\mathbb{Z}/2\\mathbb{Z})^2 \\). The group has three subgroups of order 2, corresponding to three quadratic intermediate fields \\( K(\\sqrt{a}) \\), \\( K(\\sqrt{b}) \\), and \\( K(\\sqrt{c}) \\) where \\( a, b, c \\in K^\\times \\setminus (K^\\times)^2 \\). Let \\( c = ab \\). If \\( a/b \\in (K^\\times)^2 \\), then \\( a = d^2b \\) for some \\( d \\in K \\), implying \\( ab = d^2b^2 \\in (K^\\times)^2 \\), a contradiction. Thus \\( a \\), \\( b \\), and \\( a/b \\) must all be nonsquares. Therefore, \\( L = K(\\sqrt{a}, \\sqrt{b}) \\) where \\( a, b \\in K^\\times \\) satisfy the required nonsquare conditions.\n\n", "header": "import Mathlib\n\nopen IntermediateField\n\n", "informal_prefix": "/--\nLet $K$ be a field with $\\operatorname{char}(K) \\neq 2$. Consider a Galois extension $L/K$.\nShow that $\\operatorname{Gal}(L/K) \\cong (\\mathbb{Z}/2\\mathbb{Z})^2$ if and only if\nthe extensions $L/K$ has the form $L = K(\\sqrt{a}, \\sqrt{b})$ for $a, b \\in K^\\times$ such that\n$a$, $b$, and $a/b$ are nonsquares in $K^\\times$.\n-/"}
{"informal_statement": "Prove that for $n$ odd, $n>1$, $\\Phi_{2n}(x) = \\Phi_n(-x)$, where $\\Phi_n$ is the $n$th cyclotomic polynomial over $\\mathbb{Q}$.", "formal_statement": "theorem cyclotomic_two_mul_eq_cyclotomic_comp_neg {n : ℕ} (hn : Odd n) (hn' : 1 < n) :\n    Polynomial.cyclotomic (2 * n) ℚ = (Polynomial.cyclotomic n ℚ).comp (-X) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Field Extensions"], "name": "Problem_56", "informal_proof": "\nThe roots of \\( \\Phi_n(x) \\) are the primitive \\( n \\)-th roots of unity \\(\\zeta = e^{2\\pi i k/n}\\) where \\(\\gcd(k, n) = 1\\).  \nThe roots of \\( \\Phi_{2n}(x) \\) are the primitive \\( 2n \\)-th roots of unity \\(\\eta = e^{\\pi i (2m+1)/n}\\) with \\(\\gcd(2m+1, 2n) = 1\\).\n \nReplace \\( x \\) with \\(-x\\) in \\( \\Phi_n(x) \\). The roots of \\( \\Phi_n(-x) \\) become \\(-\\zeta\\), which are primitive \\( 2n \\)-th roots of unity because \\( n \\) is odd (their order is \\(\\operatorname{lcm}(n, 2) = 2n\\)).\n\nBoth \\( \\Phi_{2n}(x) \\) and \\( \\Phi_n(-x) \\) are monic, irreducible over \\( \\mathbb{Q} \\), and share the same roots. Their degrees match:  \n\\[\n\\deg \\Phi_{2n}(x) = \\phi(2n) = \\phi(2)\\phi(n) = \\phi(n) = \\deg \\Phi_n(-x).\n\\]\nThus, \\( \\Phi_{2n}(x) = \\Phi_n(-x) \\).\n", "header": "import Mathlib\n\nopen Polynomial\n\n", "informal_prefix": "/- Prove that for $n$ odd, $n>1$, $\\Phi_{2n}(x) = \\Phi_n(-x)$, where $\\Phi_n$ is the $n$th\ncyclotomic polynomial over \\mathbb{Q}.-/"}
{"informal_statement": "If $F$ is a field that is not perfect, show that $F$ has a nontrivial purely inseparable extension.", "formal_statement": "theorem exists_isPurelyInseparable_and_bot_lt {F : Type} [Field F] (h : ¬ PerfectField F) :\n    ∃ E : IntermediateField F (AlgebraicClosure F), IsPurelyInseparable F E ∧ ⊥ < E := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Field Extensions"], "name": "Problem_57", "informal_proof": "\nSince $F$ is imperfect, its character is some prime $p$, and there is some $\\alpha\\in F^\\times-F^{\\times p}$. Take $E=F(\\alpha^{1/p})$ over $F$, it is of degree $p$ as $T^p-\\alpha$ irreducible over $F$, with $\\alpha$ an inseparable element. Then $E/F$ is purely inseparable as it is prime and inseparable.\n", "header": "import Mathlib\n\n", "informal_prefix": "/- If $F$ is a field that is not perfect, show that $F$ has a nontrivial purely inseparable\nextension.-/"}
{"informal_statement": "Show that there is at most one extension \\( F(\\alpha) \\) of a field \\( F \\) such that \\( \\alpha^4 \\in F \\), \\( \\alpha^2 \\notin F \\), and \\( F(\\alpha) = F(\\alpha^2) \\).", "formal_statement": "theorem exists_eq_adjoin_and_pow_four_mem_bot_and_not_pow_two_mem_bot (F : Type) [Field F] :\n    Subsingleton {M : IntermediateField F (AlgebraicClosure F) //\n    ∃ α : AlgebraicClosure F, M = F⟮α⟯ ∧ α ^ 4 ∈ (⊥ : IntermediateField F (AlgebraicClosure F)) ∧\n    ¬ α ^ 2 ∈ (⊥ : IntermediateField F (AlgebraicClosure F)) ∧ F⟮α⟯ = F⟮α ^ 2⟯} := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Explicit Computations"], "name": "Problem_58", "informal_proof": "\nNote that $F(\\alpha^2)$ is a quadratic radical extension. Write $\\alpha=a+b\\alpha^2$, $a,b\\in F^\\times$ by condition, may assume $a=1$. Taking square then $1+b^2\\alpha^4=0$, $2b\\alpha^2=1$, then $F$ is of character 5 and $F(\\alpha)=F(b\\alpha^2)=F(\\sqrt{-1})$, unique over $F$ up to isom. \n", "header": "import Mathlib\n\nopen IntermediateField\n", "informal_prefix": "/--Show that there is at most one extension \\( F(\\alpha) \\) of a field \\( F \\) such that \\( \\alpha^4\n\\in F \\), \\( \\alpha^2 \\notin F \\), and \\( F(\\alpha) = F(\\alpha^2) \\).-/"}
{"informal_statement": "Show that \\( x^7 - 11 \\) has no root in the splitting field of \\(x^7 - 12\\) over \\(\\mathbb{Q}\\).", "formal_statement": "theorem rootSet_isEmpty_in_splittingField :\n    (.X ^ 7 - 11 : Polynomial ℚ).rootSet ((.X ^ 7 - 12 : Polynomial ℚ).SplittingField) = ⊥ := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Other Topics", "Algebraic Number Theory and Valuation Theory"], "name": "Problem_59", "informal_proof": "\nLet \\(K = \\mathbb{Q}(\\alpha)\\). Consider the decomposition of (11) in \\(O_K\\). By the decomposition theorem of prime ideals, the ideal $(11)$ is ramified if \\(x^7-11\\) has a root. But the polynomial $x^7-12=x^7-1 (\\text{mod 11})$ is separable in $\\mathbb{F}_{11}$, thus the $(11)$ is not ramified, contradiction.\n", "header": "import Mathlib\n\n", "informal_prefix": "/--\nShow that \\( x^7 - 11 \\) has no root in the splitting field of \\(x^7 - 12\\) over \\(\\mathbb{Q}\\).\n-/"}
{"informal_statement": "For a positive integer $a$, consider the polynomial $$ f_a = x^6 + 3ax^4 + 3x^3 + 3ax^2 + 1. $$ Let $F$ be the splitting field of $f_a$. Show that its Galois group is solvable.", "formal_statement": "theorem isSolvable_X_pow_six_add_gal {a : ℤ} (ha : a > 0) : IsSolvable\n    (X ^ 6 + C (3 * a : ℚ) * X ^ 4 + C 3 * X ^ 3 + C (3 * a : ℚ) * X ^ 2 + C 1 : ℚ[X]).Gal:= by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Galois theory"], "name": "Problem_60", "informal_proof": "\nWe note that $\\frac{f_a}{x^3}= x^3+\\frac{1}{x^3}+3a(x+\\frac{1}{x})+3= (x+\\frac{1}{x})^3+(3a-3)(x+\\frac{1}{x})+3$. Let $F_1$ be the splitting field of the polynomial $y^3+(3a-3)y+3$ whose Galois group is of order less than 7 and solvable. $F/F_1$ is the composition of finite degree 2 extensions, so the Galois group $Gal(F/F_1)$ is solvable. As a result, the Galois group of $F$ is solvable.  \n", "header": "import Mathlib\n\nopen Polynomial\n", "informal_prefix": "/--For a positive integer $a$, consider the polynomial $$ f_a = x^6 + 3ax^4 + 3x^3 + 3ax^2 + 1.\n $$ Let $F$ be the splitting field of $f_a$. Show that its Galois group is solvable.-/"}
{"informal_statement": "Prove that the polynomial $x^4+1$ is not irreducible over any field of positive characteristic.", "formal_statement": "theorem not_irreducible_X_pow_four_add_one {F : Type} [Field F] {p : ℕ} [Fact p.Prime]\n    [CharP F p] : ¬ Irreducible (X ^ 4 + C 1 : F[X]) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Ring Theory", "Polynomials"], "name": "Problem_61", "informal_proof": "First if charateristic is $2$, then $x^4+1=(x+1)^4$ then we are done. So we may assmue that characteristic is odd. $X^4 + 1 = (X^2 + i)(X^2 - i) = (X^2 - \\sqrt{2}X + 1)(X^2 + \\sqrt{2}X + 1) = (X^2 - i\\sqrt{2}X - 1)(X^2 + i\\sqrt{2}X - 1)$. By product of two non-squares is a square, we are done.", "header": "import Mathlib\n\nopen Polynomial\n", "informal_prefix": "/--Prove that the polynomial $x^4+1$ is not irreducible over any field of positive characteristic.-/"}
{"informal_statement": "Let $F$ be a field and let $f(x) \\in F[x]$ be an irreducible polynomial with splitting field $E$ over $F$. Choose $\\alpha \\in E$ with $f(\\alpha) =0$. Furthermore, for some fixed integer $n \\geq 1$, let $g(x)$ be an irreducible polynomial in $F[x]$ with $g(\\alpha^n)=0$. If $\\deg(f) / \\deg(g) = n$ and if the characteristic of $F$ does not divide $n$, prove that $E$ contains a primitive $n$th root of unity.", "formal_statement": "theorem primitiveRoots_not_empty (E F : Type) [Field E] [Field F] [Algebra F E]\n    (f : Polynomial F) (h_f_irr : Irreducible f) (h_splitting_field : f.IsSplittingField F E)\n    (a : E) (ha : f.aeval a = 0) (n : ℕ) (hn : n ≥ 1)\n    (g : Polynomial F) (h_g_irr : Irreducible g) (h_ga : g.aeval (a ^ n) = 0)\n    (h_deg : f.degree = g.degree * n) (h_char : (n : F) ≠ 0) : primitiveRoots n E ≠ ∅ := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Field Extensions"], "name": "Problem_62", "informal_proof": "\nLet $d_f = \\deg(f)$ and $d_g = \\deg(g)$. By the definition of an irreducible polynomial, we have $[F(\\alpha):F] = d_f$ and $[F(\\alpha^n):F] = d_g$. The problem states that $d_f / d_g = n$.\n\nLet $K = F(\\alpha^n)$. We have a tower of fields $F \\subseteq K \\subseteq F(\\alpha)$. By the tower law for field extensions:\n\\[ [F(\\alpha):F] = [F(\\alpha):K] \\cdot [K:F] \\]\nSubstituting the known degrees, we get:\n\\[ d_f = [F(\\alpha):K] \\cdot d_g \\]\nUsing the given condition $d_f = n \\cdot d_g$, this becomes:\n\\[ n \\cdot d_g = [F(\\alpha):K] \\cdot d_g .\\]\nwhich implies that $[F(\\alpha):K] = n$.\n\nNow, consider the polynomial $h(x) = x^n - \\alpha^n$. The coefficients of $h(x)$ belong to $K=F(\\alpha^n)$, so $h(x) \\in K[x]$. Furthermore, $\\alpha$ is a root of $h(x)$, since $h(\\alpha) = \\alpha^n - \\alpha^n = 0$.\n\nSince the degree of the minimal polynomial of $\\alpha$ over $K$ is $[F(\\alpha):K]=n$, and $h(x)$ is a monic polynomial in $K[x]$ of degree $n$ which has $\\alpha$ as a root, it follows that $h(x)$ must be the minimal polynomial of $\\alpha$ over $K$. This means $h(x)$ is irreducible over $K$.\n\nThe polynomial $f(x) \\in F[x]$ can also be viewed as a polynomial in $K[x]$, since $F \\subseteq K$. As $f(\\alpha)=0$, the minimal polynomial of $\\alpha$ over $K$ must divide $f(x)$ in the ring $K[x]$. Thus, $h(x)$ divides $f(x)$.\n\nThis divisibility implies that every root of $h(x)$ is also a root of $f(x)$. By definition, $E$ is the splitting field of $f(x)$, so $E$ contains all the roots of $f(x)$. Consequently, $E$ must contain all the roots of $h(x)$.\n\nThe roots of $h(x) = x^n - \\alpha^n$ are the elements $\\{\\zeta_n^k \\alpha \\mid k=0, 1, \\dots, n-1\\}$, where $\\zeta_n$ is a primitive $n$-th root of unity. Since the characteristic of $F$ does not divide $n$, these $n$ roots are distinct.\n\nWe can assume $\\alpha \\neq 0$. If $\\alpha=0$, then $f(0)=0$ and $g(0)=0$. As $f$ and $g$ are irreducible, this forces $f(x)=c_1x$ and $g(x)=c_2x$, so $d_f=d_g=1$. The given condition $d_f/d_g = n$ implies $n=1$. The theorem holds trivially, as the primitive 1st root of unity is $1$, which is in $F \\subseteq E$.\n\nFor $n > 1$, we have $\\alpha \\neq 0$. Since all roots of $h(x)$ are in $E$, we have both $\\alpha \\in E$ and $\\zeta_n\\alpha \\in E$. As $E$ is a field and $\\alpha \\neq 0$, the quotient of these two elements must also be in $E$:\n\\[ \\frac{\\zeta_n\\alpha}{\\alpha} = \\zeta_n \\in E. \\]\nTherefore, $E$ contains a primitive $n$-th root of unity.\n\n", "header": "import Mathlib\n\n", "informal_prefix": "/--\nLet $F$ be a field and let $f(x) \\in F[x]$ be an irreducible polynomial with splitting field $E$ over $F$.\nChoose $\\alpha \\in E$ with $f(\\alpha) =0$. Furthermore, for some fixed integer $n \\geq 1$,\nlet $g(x)$ be an irreducible polynomial in $F[x]$ with $g(\\alpha^n)=0$. If $\\deg(f) / \\deg(g) = n$\nand if the characteristic of $F$ does not divide $n$, prove that $E$ contains a primitive $n$th root of unity.-/"}
{"informal_statement": "Let \\( f \\in \\mathbb{Q}[X] \\) and \\( \\xi \\in \\mathbb{C} \\) be a root of unity. Show that \\( f(\\xi) \\neq 2^{\\frac{1}{4}} \\).", "formal_statement": "theorem eval₂_algebraMap_ne_two_pow_one_dvd_four {n : ℕ} (hn : 1 ≤ n) (f : ℚ[X]) (ξ : ℂˣ)\n    (h : ξ ∈ rootsOfUnity n ℂ) : f.eval₂ (algebraMap ℚ ℂ) ξ ≠ (2 : ℂ) ^ ((1 : ℂ) / 4) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Galois theory"], "name": "Problem_63", "informal_proof": "\nIt suffices to show that $a=2^{\\frac{1}{4}}\\notin \\mathbb{Q}(\\xi)$. The minimal polynomial of $a$ is $x^4-2$, and the splitting field is $\\mathbb{Q}(i,a)$. If $a\\in\\mathbb{Q}(\\xi)$ then $$D_4\\cong\\mathbb{Q}(i,a)\\subset\\mathbb{Q}(\\xi)\\cong(\\mathbb{Z}/n\\mathbb{Z})^\\times$$ for some $n\\in \\mathbb{Z}$. This is impossible since the left hand side is non-abelian but the right hand side is abelian.\n", "header": "import Mathlib\n\nopen Polynomial\n", "informal_prefix": "/--Let \\( f \\in \\mathbb{Q}[X] \\) and \\( \\xi \\in \\mathbb{C} \\) be a root of unity. Show that \\( f(\\xi) \\neq 2^{\\frac{1}{4}} \\).-/"}
{"informal_statement": "Let $K$ be a finite extension of a field $F$, and let $f(x) \\in K[x]$. Prove that there exists a nonzero polynomial $g(x) \\in K[x]$ such that $f(x)g(x) \\in F[x]$.", "formal_statement": "theorem exists_mul_eq_map_algebraMap {F K : Type} [Field F] [Field K] [Algebra F K] [FiniteDimensional F K]\n    (f : K[X]) : ∃ g ≠ 0, ∃ h : F[X], f * g = h.map (algebraMap F K) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Ring Theory", "Polynomials"], "name": "Problem_64", "informal_proof": "\nSince $K$ is a finite extension of $F$, every element $k \\in K$ is integral over $F$. Therefore:\n\\begin{itemize}\n    \\item For every $k \\in K$, the element $k \\in K[x]$ is integral over $F[x]$\n    \\item For every $k \\in K$ and $i \\in \\mathbb{N}$, the element $kx^i \\in K[x]$ is integral over $F[x]$ (since both $k$ and $x^i$ are integral over $F[x]$, and products of integral elements remain integral)\n    \\item Any polynomial $f(x) \\in K[x]$ is integral over $F[x]$, as it can be expressed as a sum of integral elements $kx^i$\n\\end{itemize}\n\nWe now prove the key fact:\n\\begin{enumerate}\n    \\item[(1)] If $A \\subseteq B$ is a ring extension and $u \\in B$ is integral over $A$, then there exists nonzero $v \\in B$ such that $uv \\in A$.\n\\end{enumerate}\n\n\\begin{proof}[Proof of (1)]\nLet $n$ be the minimal positive integer such that there exists a monic polynomial:\n\\[ P(Y) = Y^n + \\sum_{i=0}^{n-1} a_iY^i \\in A[Y] \\]\nsatisfying $P(u) = 0$. Define:\n\\[ v = \\sum_{i=1}^{n-1} a_iu^{i-1} + u^{n-1} \\]\n\nThen calculate:\n\\begin{align*}\nuv &= u\\left(\\sum_{i=1}^{n-1} a_iu^{i-1} + u^{n-1}\\right) \\\\\n&= \\sum_{i=1}^{n-1} a_iu^i + u^n \\\\\n&= \\sum_{i=0}^{n-1} a_iu^i + u^n - a_0 \\\\\n&= P(u) - a_0 \\\\\n&= -a_0 \\in A\n\\end{align*}\n\nThe element $v \\neq 0$ by minimality of $n$, completing the proof of (1).\\qedhere\n", "header": "import Mathlib\n\nopen Polynomial\n", "informal_prefix": "/--Let $K$ be a finite extension of a field $F$, and let $f(x) \\in K[x]$. Prove that there exists a\nnonzero polynomial $g(x) \\in K[x]$ such that $f(x)g(x) \\in F[x]$.-/"}
{"informal_statement": "Prove that for any $a,b \\in \\mathbb{F}_{p^n}$ that if $x^3+ax+b$ is irreducible then $-4a^3-27b^2$ is a square in $\\mathbb{F}_{p^n}$.", "formal_statement": "theorem isSquare_discriminant_of_irreducible {p n : ℕ} [Fact p.Prime] (a b : GaloisField p n)\n    (h_irr : Irreducible (X ^ 3 + C a * X + C b)) :\n    IsSquare (- 4 * a ^ 3 - 27 * b ^ 2) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Galois theory"], "name": "Problem_65", "informal_proof": "Let $G$ be Galois group of $x^3+ax+b$. Actually, as \\( p(x) = x^3 + ax + b \\) is irreducible, \\( G \\) can be viewed as a subgroup of \\( S_3 \\), not \\( S_n \\), and \\[G \\subset A_3 \\iff \\sqrt{D} \\in \\mathbb{F}_p.\\] By the theory of finite fields, if \\( p \\) is irreducible, then the splitting field of \\( p \\) over \\( \\mathbb{F}_p \\) is isomorphic to \\( \\mathbb{F}_{p^n} \\), thus \\( |G| = [\\mathbb{F}_{p^n} : \\mathbb{F}_p] = 3 \\). Notice that \\( S_3 \\) has only one subgroup of order 3, by Sylow's theorem, therefore \\( G = A_3 \\).", "header": "import Mathlib\n\nopen Polynomial\n", "informal_prefix": "/--Prove that for any $a,b \\in \\mathbb{F}_{p^n}$ that if $x^3+ax+b$ is irreducible then $-4a^3-27b^2$ is a\n square in $\\mathbb{F}_{p^n}$.-/"}
{"informal_statement": "Prove that, if $n \\geq 3$, then $x^{2^n}+x+1$ is \\emph{reducible} in $\\mathbb{F}_2$.", "formal_statement": "theorem not_irreducible_X_pow_two_pow_add_X_add_C_one {n : ℕ} (hn : n ≥ 3) :\n    ¬ Irreducible (X ^ 2 ^ n + X + C 1 : (ZMod 2)[X]) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Field Extensions"], "name": "Problem_66", "informal_proof": "Let \\( p(x) \\mid x^{2n} + x + 1 \\), \\( p \\) irreducible, and \\( a \\in \\mathbb{F}_2 \\) a root of \\( f \\). Then \\( a^{2n} + a + 1 = 0 \\), equivalently \\( a^{2n} = a + 1 \\). It follows that \\( a^{2^{n}} = a \\), so \\( \\mathbb{F}_2(a) \\subset \\mathbb{F}_{2^{2n}} \\). This shows that \\( \\deg p \\mid 2n \\). Because $2^n> 2n$ for $n\\geq 3$, so $x^{2^n}+x+1$ cannot be irreducible.", "header": "import Mathlib\n\nopen Polynomial\n", "informal_prefix": "/--Prove that, if $n \\geq 3$, then $x^{2^n}+x+1$ is \\emph{reducible} in $\\mathbb{F}_2$.-/"}
{"informal_statement": "Prove that the prime ideals of \\( \\mathbb{F}_7[\\alpha] \\otimes_{\\mathbb{F}_7} \\mathbb{F}_7[\\alpha] \\) are principal, where \\( \\alpha^3 = 2 \\).", "formal_statement": "theorem isPrincipal_of_ideal_tensor_zMod_seven\n    (p : Ideal (TensorProduct (ZMod 7) (AdjoinRoot (X ^ 3 - 2 : Polynomial (ZMod 7)))\n    (AdjoinRoot (X ^ 3 - 2 : Polynomial (ZMod 7))))) [p.IsPrime] : p.IsPrincipal := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Ideals and Modules"], "name": "Problem_67", "informal_proof": "\nLet $K=\\mathbb{F}_7[\\alpha]$, then the ring is same as $K[x]/(x^3-2)$, which is a quotient of the principal ideal domain $K[x]$. \n", "header": "import Mathlib\n\nopen Polynomial\n\n", "informal_prefix": "/-\nProve that the prime ideals of $\\mathbb{F}_7[\\alpha] \\otimes_{\\mathbb{F}_7} \\mathbb{F}_7[\\alpha]$ are principal, where $\\alpha^3 = 2$.\n-/"}
{"informal_statement": "Let \\( A \\) be a Noetherian ring and let \\( x \\in A \\) be an element which is neither a unit nor a zero-divisor. Prove that the ideals \\( xA \\) and \\( x^nA \\) for \\( n = 1, 2, \\dots \\) have the same prime divisors:\n\\[\n\\operatorname{Ass}_A(A/xA) = \\operatorname{Ass}_A(A/x^nA).\n\\]", "formal_statement": "theorem associatedPrimes_quot_span_eq_quot_span_pow {A : Type} [CommRing A] [IsNoetherianRing A]\n    (x : A) (hx : x ∈ nonZeroDivisors A) (hx' : ¬ IsUnit x) (n : ℕ) (h : n ≥ 1) :\n    associatedPrimes A (A ⧸ span {x}) = associatedPrimes A (A ⧸ span {x ^ n}) := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Localization and Decomposition of Ideals"], "name": "Problem_68", "informal_proof": "Note that $\text{Supp}_A(A/x)=V(x)=V(x^n)=\text{Supp}_A(A/x^n)$.  As associated primes of $A$ modules are the minimal primes which appears in the longest filtration of $M$. Come to $A/x^n$, because $x$ is not zero divisor or unit $x^k/x^{k+1}\\cong A/x$, so the minimal primes appears in the longest filtration of $A/x^n$ will be same as $A/x$.", "header": "import Mathlib\n\nopen Ideal\n", "informal_prefix": "/-\nLet $A$ be a Noetherian ring and let $x \\in A$ be an element which is\nneither a unit nor a zero-divisor. Prove that the ideals $xA$ and $x^nA$\nfor $n = 1, 2, \\dots$ have the same prime divisors:\n\\[\n\\operatorname{Ass}_A(A/xA) = \\operatorname{Ass}_A(A/x^nA).\n\\]\n-/"}
{"informal_statement": "Prove that if \\( K \\) is a field of finite degree over \\( \\mathbb{Q} \\) and \\( x_1, \\dots, x_n \\) are finitely many elements of \\( K \\), then the subring \\( \\mathbb{Z}[x_1, \\dots, x_n] \\) they generate over \\( \\mathbb{Z} \\) is not equal to \\( K \\).", "formal_statement": "theorem int_adjoin_range_ne_top {K : Type} [Field K] [NumberField K] {n : ℕ} (x : Fin n → K) :\n    Algebra.adjoin ℤ (Set.range x) ≠ ⊤ := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Other Topics", "Algebraic Number Theory and Valuation Theory"], "name": "Problem_69", "informal_proof": "\nAll $x_i$ are algebraic over $\\mathbb{Q}$, hence there exists an integer $c$ such that $cx_i$ are algebraic integers. Fix a prime $p$ which is prime to $c$, and a $p$-adic valuation $v:K\\to\\mathbb{Q}$, we have $\\mathbb{Z} \\subset \\text{Im}(v) $. But for any element $y\\in\\mathbb{Z}[x_1, \\dots, x_n]$, we have $v(y)\\geq 0$. So the ring is not equal to $K$.\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Prove that if $K$ is a field of finite degree over $\\mathbb{Q}$ and $x_1, \\dots, x_n$ are\nfinitely many elements of $K$, then the subring $\\mathbb{Z}[x_1, \\dots, x_n]$ they generate over\n$\\mathbb{Z}$ is not equal to $K$. -/"}
{"informal_statement": "Show that $\\mathbb{Z}[x]/(x^2+4)$ is not integrally closed.", "formal_statement": "theorem not_isIntegrallyClosed_adjoinRoot_pow_two_add_four :\n    ¬ IsIntegrallyClosed (AdjoinRoot (X ^ 2 + C 4 : ℤ[X])) := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Integral Dependence and the Nullstellensatz"], "name": "Problem_70", "informal_proof": "\nThe ring $\\mathbb{Z}[x]/(x^2+4)$ is same as $\\mathbb{Z}[2i]$, its fractional fields is $\\mathbb{Q}[i]$, whose ring of algebraic integers is $\\mathbb{Z}[i]$.\n", "header": "import Mathlib\n\nopen Polynomial\n\n", "informal_prefix": "/-- Show that $\\mathbb{Z}[X]/(x^2+4)$ is not integrally closed. -/"}
{"informal_statement": "Prove that $k[x,y]$ is not a Dedekind ring.", "formal_statement": "theorem not_isDedekindRing_mvPolynomial_fin_two {k : Type} [Field k] :\n    ¬ IsDedekindRing (MvPolynomial (Fin 2) k) := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Dedekind Domains and DVRs"], "name": "Problem_71", "informal_proof": "\n$\\dim k[x, y] = 2$ so it is not a Dedekind ring.\n", "header": "import Mathlib\n\nopen Polynomial\n\n", "informal_prefix": "/-- Prove that $k[x,y]$ is not a Dedekind ring. -/"}
{"informal_statement": "Let \\( A \\) be a ring such that for each maximal ideal \\( \\mathfrak{m} \\) of \\( A \\), the local ring \\( A_{\\mathfrak{m}} \\) is Noetherian; and for each \\( x \\neq 0 \\) in \\( A \\), the set of maximal ideals of \\( A \\) which contain \\( x \\) is finite. Show that \\( A \\) is Noetherian.", "formal_statement": "theorem isNoetherianRing_of_finite_isMaximal_and_mem {A : Type} [CommRing A]\n    (h_local : ∀ (m : Ideal A), [m.IsMaximal] → IsNoetherianRing (Localization.AtPrime m))\n    (h_finite : ∀ x : A, x ≠ 0 → Set.Finite {m : Ideal A | m.IsMaximal ∧ x ∈ m}) :\n    IsNoetherianRing A := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Noetherian rings and Chain Conditions"], "name": "Problem_72", "informal_proof": "\nOnly need to show that any ascending chain of ideals $I_1\\subset I_2\\subset\\cdots$ of $A$ is eventually stable. Assume $I_1\\neq 0$ and pick $x\\in I_1-0$. By conditions there are only finitely many maximal ideals with $x\\in\\mathfrak m$, and there is some $n_\\mathfrak m$ such that $I_{n_\\mathfrak m,\\mathfrak m}=I_{n_\\mathfrak m+1,\\mathfrak m}=\\cdots$. For $x\\notin\\mathfrak m$, we get $A_\\mathfrak m=(x)_\\mathfrak m=I_{1,\\mathfrak m}=I_{2,\\mathfrak m}=\\cdots$. Hence for $N=\\max_{x\\in\\mathfrak m}n_\\mathfrak m\\in \\mathbb{N}$ we have $I_{N,\\mathfrak m}=I_{N+1,\\mathfrak m}=\\cdots$ for any maximal ideal $\\mathfrak m$. Therefore $I_N=I_{N+1}=\\cdots$, thus $A$ noetherian.\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Let $A$ be a ring such that for each maximal ideal $\\mathfrak{m}$ of $A$, the local ring\n$A_{\\mathfrak{m}}$ is Noetherian; and for each $x \\neq 0$ in $A$, the set of maximal ideals of $A$\nwhich contain $x$ is finite. Show that $A$ is Noetherian.-/"}
{"informal_statement": "If \\( R \\) is a valuation ring, then an \\( R \\)-module \\( A \\) is flat if it is torsion-free.", "formal_statement": "theorem flat_of_noZeroSMulDivisor {R A : Type} [CommRing R] [IsDomain R] [ValuationRing R] [AddCommGroup A]\n    [Module R A] [NoZeroSMulDivisors R A] : Module.Flat R A := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Integral Dependence and the Nullstellensatz"], "name": "Problem_73", "informal_proof": "\nWrite $A$ as a direct limit of its finitely generated submodules, it is sufficient to prove for finitely generated $A$ because the direct limit of flat modules is flat. \nThe valuation ring $R$ is a local ring with maximal ideal $\\mathfrak{m}$. Choose $x_1,\\dots,x_n\\in A$ such that their image $[x_1],\\dots ,[x_n]\\in A/\\mathfrak{m}A$ form an $R/\\mathfrak{m}$ basis. Then we have $A=\\Sigma Rx_i$. Suppose that $\\Sigma a_ix_i=0$ for some $a_i$, then there exists a $j$ that $a_j$ divides all other $a_i$, so we have $\\Sigma (a_i/a_j)\\cdot x_i=0$ by torsion-freeness of $A$. Now we have $a_i/a_j\\in R$ and $\\Sigma[a_i/a_j]x_i=0$, it implies $[a_i/a_j]=0$, which is impossible when $i=j$.\n", "header": "import Mathlib\n\n", "informal_prefix": "/- If $R$ is a valuation ring, then an $R$-module $A$ is flat if it is torsion-free.-/"}
{"informal_statement": "If \\( R \\) is a valuation ring, then an \\( R \\)-module \\( A \\) is torision-free if it is flat.", "formal_statement": "theorem noZeroSMulDivisors_of_flat {R A : Type} [CommRing R] [IsDomain R] [ValuationRing R] [AddCommGroup A]\n    [Module R A] [Module.Flat R A] : NoZeroSMulDivisors R A := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Tensor Product and Flatness"], "name": "Problem_74", "informal_proof": "\nSuppose that $A[a]$ is nonzero. Then $R\\xrightarrow[]{a}R$ is injective but $A\\otimes R\\xrightarrow[]{a}A\\otimes R$ isn't, hence $M$ is not flat. \n", "header": "import Mathlib\n\n", "informal_prefix": "/- If $R$ is a valuation ring, then an $R$-module $A$ is torsion-free if it is flat.-/"}
{"informal_statement": "Suppose \\( A \\) and \\( B \\) are commutative rings containing a field \\( k \\), with \\( B \\) finitely generated over \\( k \\) as a ring. If \\( \\varphi : A \\to B \\) is a ring homomorphism with \\( \\varphi|_k = \\text{Id} \\) and if \\( Q \\subset B \\) is a maximal ideal, prove that \\( \\varphi^{-1}(Q) \\subset A \\) is a maximal ideal.", "formal_statement": "theorem comap_isMaximal_of_finiteType {A B k : Type} [CommRing A] [CommRing B] [Field k] [Algebra k A] [Algebra k B]\n    [Algebra.FiniteType k B] (φ : A →ₐ[k] B) (Q : Ideal B) [hQ : Q.IsMaximal] :\n    (Ideal.comap φ Q).IsMaximal := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Integral Dependence and the Nullstellensatz"], "name": "Problem_75", "informal_proof": "\n$B/Q$ is a field which is finite generated over $k$, so it is finite over $k$ (Atiyah P82 Prop 7.9). Then $A / \\varphi^{-1}(Q)$ is a $k$-linear subspace of $B/Q$, so it is finite over $k$. As a result, $A / \\varphi^{-1}(Q)$ is a field.\n", "header": "import Mathlib\n\n", "informal_prefix": "/--Suppose \\( A \\) and \\( B \\) are commutative rings containing a field \\( k \\), with \\( B \\)\nfinitely generated over \\( k \\) as a ring. If \\( \\varphi : A \\to B \\) is a ring homomorphism with\n\\( \\varphi|_k = \\text{Id} \\) and if \\( Q \\subset B \\) is a maximal ideal, prove that \\( \\varphi^{-1}\n(Q) \\subset A \\) is a maximal ideal.-/"}
{"informal_statement": "Show that a finite torsion-free module over a Dedekind domain is projective.", "formal_statement": "theorem projective_of_noZeroSMulDivisor {R M : Type} [CommRing R] [AddCommGroup M] [Module R M]\n    [IsDedekindDomain R] [Module.Finite R M] [NoZeroSMulDivisors R M] : Module.Projective R M := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Localization and Decomposition of Ideals"], "name": "Problem_76", "informal_proof": "\n设 $A$ 是一个戴德金整环, $M$ 是一个有限的 $A$ 模. 注意对 $M$ 而言 \\href{https://github.com/mbkybky/module_localProperties/blob/4e1c42d05da4ee7f2ef0016247734450d9dd2f07/ModuleLocalProperties/NoZeroSMulDivisors.lean#L264}{torsion-free} 和 \\href{https://leanprover-community.github.io/mathlib4_docs/Mathlib/RingTheory/LocalProperties/Projective.html#Module.projective_of_localization_maximal}{projective} 都是局部性质, 故不妨设 $A$ 是 DVR. 此时 $M$ 是自由的, 进而是投射的.\n", "header": "import Mathlib\n\n", "informal_prefix": "/- Show that a finite torsion-free module over a Dedekind domain is projective. -/"}
{"informal_statement": "Let \\( k \\) be a field, \\( A \\) a local \\( k \\)-algebra with maximal ideal \\( \\mathfrak{m} \\). Assume that \\( A \\) is a localization of a \\( k \\)-algebra \\( R \\) and that \\( A / \\mathfrak{m} = k \\). Prove that there exists maximal ideal \\( \\mathfrak{n} \\) of \\( R \\) with \\( R_{\\mathfrak{n}} = A \\).", "formal_statement": "theorem exists_isMaximal_atPrime_of_bijective {k R A : Type} [Field k] [CommRing R] [CommRing A] [Algebra k R]\n    [Algebra R A] [Algebra k A] [IsScalarTower k R A] [IsLocalRing A]\n    (h : Function.Bijective <| (IsLocalRing.residue A).comp (algebraMap k A))\n    (S : Submonoid R) [IsLocalization S A] :\n    ∃ n : Ideal R, ∃ hn : n.IsMaximal, IsLocalization.AtPrime A n := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Localization and Decomposition of Ideals"], "name": "Problem_77", "informal_proof": "\nSuppose that $A=S^{-1}R$ and let $\\mathfrak{n}$ be the prime image of $\\mathfrak{m}$ under $R\\to A$. Since $\\mathfrak{n}\\cap S=\\emptyset$ we have $k$-algebra morphisms $R_{\\mathfrak{n}}\\to A$. Viewing $A$ as an $R_{\\mathfrak{m}}$-module, by Nakayama Lemma it's sufficient to show $R_{\\mathfrak{n}}/\\mathfrak{n}\\to A/\\mathfrak{m}$ is an isomorphism. And its a $k$-algebra morphism form a field $R_{\\mathfrak{n}}/\\mathfrak{n}$ to $k$ so it is an isomorphism.\n", "header": "import Mathlib\n\n", "informal_prefix": "/--Let \\( k \\) be a field, \\( A \\) a local \\( k \\)-algebra with maximal ideal \\( \\mathfrak{m} \\).\nAssume that \\( A \\) is a localization of a \\( k \\)-algebra \\( R \\) and that \\( A / \\mathfrak{m} = k\n\\). Prove that there exists maximal ideal \\( \\mathfrak{n} \\) of \\( R \\) with \\( R_{\\mathfrak{n}} = A\n\\).-/"}
{"informal_statement": "Let \\( R' / R \\) be an integral extension of rings. Show that \\( \\text{rad}(R) = \\text{rad}(R') \\cap R \\), where $\\text{rad}(R)$ denotes the nilpotent radical of $R$.", "formal_statement": "theorem nilpotent_eq_contraction_nilpotent_of_integral (R R' : Type) [CommRing R] [CommRing R']\n    [Algebra R R'] (int : Algebra.IsIntegral R R') :\n    nilradical R = Ideal.comap (algebraMap R R') (nilradical R') := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Integral Dependence and the Nullstellensatz"], "name": "Problem_78", "informal_proof": "\nBy going up Theorem, for any prime ideal $\\mathfrak{p}\\subset R$, there exists a prime ideal $\\mathfrak{P}\\subset R'$ such that $\\mathfrak{P}\\cap R=\\mathfrak{p}$. Thus the prime ideals of form $\\mathfrak{P}\\cap R$ run over the prime ideals of $R$, then we have $rad(R')\\cap R =\\bigcap\\limits_{\\mathfrak{P}\\subset R'}\\mathfrak{P}\\cap R=\\bigcap\\limits_{\\mathfrak{p}\\subset R}\\mathfrak{p}\\cap R=rad(R)$.\n", "header": "import Mathlib\n\n", "informal_prefix": "/--\nLet \\( R' / R \\) be an integral extension of rings. Show that \\( \\text{rad}(R) = \\text{rad}(R') \\cap R \\),\nwhere $\\text{rad}(R)$ denotes the nilpotent radical of $R$.-/"}
{"informal_statement": "Let \\( R \\) be a commutative ring. If all submodules of finitely generated free modules over \\( R \\) are free over \\( R \\), then \\( R \\) is a PID.", "formal_statement": "theorem isPrincipalIdealRing_of_forall_free {R : Type} [CommRing R]\n    (h : ∀ (M : Type) [AddCommGroup M] [Module R M] [Module.Finite R M] [Module.Free R M],\n      ∀ (N : Submodule R M), Module.Free R N) :\n    IsPrincipalIdealRing R := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Ideals and Modules"], "name": "Problem_79", "informal_proof": "\nIf \\( a \\in R \\), \\( a \\neq 0 \\), then \\( aR \\) is free (as a submodule of the free \\( R \\)-module \\( R \\)), so \\( a \\) is a non-zero divisor. To prove this, let \\( ax \\) be an element of an \\( R \\)-basis of \\( aR \\). If \\( ba = 0 \\) for some \\( b \\in R \\), then \\( b(ax) = 0 \\) and therefore \\( b = 0 \\).  This shows that \\( R \\) is necessarily an integral domain.\n\nNow let \\( I \\) be a non-zero ideal of \\( R \\). Since \\( I \\) is free (as a submodule of the free \\( R \\)-module \\( R \\)) it must have a basis with a single element (base change to fraction field), so \\( I \\) is principal.\n", "header": "import Mathlib\n\n", "informal_prefix": "/--Let \\( R \\) be a commutative ring. If all submodules of finitely generated free modules over\n\\( R \\) are free over \\( R \\), then \\( R \\) is a PID.-/"}
{"informal_statement": "Let \\( R \\subset R' \\) be an extension of integral domains, and let \\( \\overline{R} \\) be the integral closure of \\( R \\) in \\( R' \\). Show that for any two monic polynomials \\( f, g \\in R'[t] \\) with \\( f \\cdot g \\in R[t] \\), we have \\( f, g \\in \\overline{R}[t] \\).", "formal_statement": "theorem mem_polynomial_integral_closure_of_prod_mem_polynomial (R S : Type) [CommRing R]\n    [IsDomain R] [CommRing S] [IsDomain S] [Algebra R S] [NoZeroSMulDivisors R S] (f g : S[X])\n    (mem : f * g ∈ lifts (algebraMap R S)) (hf : f.Monic) (hg : g.Monic) :\n    f ∈ lifts (integralClosure R S).subtype ∧ g ∈ lifts (integralClosure R S).subtype := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Integral Dependence and the Nullstellensatz"], "name": "Problem_80", "informal_proof": "\n\nSince the coefficients of a monic polynomial can be expressed as elementary symmetric polynomials in its roots, and the roots of \\( f \\) and \\( g \\) are integral over \\( R \\), it follows that the coefficients of \\( f \\) and \\( g \\) must also be integral over \\( R \\). Therefore, \\( f, g \\in \\overline{R}[t] \\).\n\n", "header": "import Mathlib\n\nopen Polynomial\n\n", "informal_prefix": "/--\nLet \\( R \\subset R' \\) be an extension of integral domains, and let \\( \\overline{R} \\) be the integral closure of \\( R \\) in \\( R' \\).\nShow that for any two monic polynomials \\( f, g \\in R'[t] \\) with \\( f \\cdot g \\in R[t] \\),\nwe have \\( f, g \\in \\overline{R}[t] \\).-/"}
{"informal_statement": "Let \\( R = \\mathbb{C}[x_1, \\dots, x_n]/(x_1^2 + x_2^2 + \\dots + x_n^2) \\). Then \\( R \\) is not a unique factorization domain for \\( n = 3 \\) or \\( 4 \\).", "formal_statement": "/--\nLet \\( R = \\mathbb{C}[x_1, \\dots, x_n]/(x_1^2 + x_2^2 + \\dots + x_n^2) \\).\n-/\nabbrev R (n : ℕ) : Type :=\n  MvPolynomial (Fin n) ℂ ⧸ Ideal.span {(∑ i : Fin n, X i ^ 2 : MvPolynomial (Fin n) ℂ)}\n\n\ntheorem not_UFD_of_3_or_4 (n : ℕ) (h : n = 3 ∨ n = 4) [IsDomain (R n)] :\n    ¬ UniqueFactorizationMonoid (R n) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Ring Theory", "Basic Definitions and Examples"], "name": "Problem_81", "informal_proof": "\nIn case $n=3$, we have $-x_1^2=(x_2+ix_3)(x_2-ix_3)$, and $x_1,(x_2+ix_3),(x_2-ix_3)$ are not units in $R$. \nIn case $n=4$, we have $-(x_1+ix_2)(x_1-ix_2)=(x_3+ix_4)(x_3-ix_4)$, and $(x_1+ix_2),(x_1-ix_2),(x_2+ix_3),(x_2-ix_3)$ are not units in $R$. \n", "header": "import Mathlib\n\nopen MvPolynomial", "informal_prefix": "/--\nLet \\( R = \\mathbb{C}[x_1, \\dots, x_n]/(x_1^2 + x_2^2 + \\dots + x_n^2) \\).\nThen \\( R \\) is not a unique factorization domain for \\( n = 3 \\) or \\( 4 \\).-/"}
{"informal_statement": "Let \\( D \\) be a unique factorization domain. Prove that if every nonzero prime ideal of \\( D \\) is maximal, then \\( D \\) is a principal ideal domain.", "formal_statement": "theorem isPrincipalIdealRing_of_isPrime_ne_bot_isMaximal {D : Type} [CommRing D] [IsDomain D]\n    [UniqueFactorizationMonoid D] (h : ∀ P : Ideal D, [P.IsPrime] → P ≠ ⊥ → P.IsMaximal) : IsPrincipalIdealRing D := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Ring Theory", "PID, ED and UFD"], "name": "Problem_82", "informal_proof": "\n\n{\\it Fact 1:\nIf \\( p \\) is a nonzero irreducible element of \\( D \\), then \\( (p) \\) is a prime ideal of \\( D \\).\n\n\nProof:\nLet \\( a \\) and \\( b \\) be nonzero elements of \\( D \\) such that \\( ab \\in (p) \\), i.e., \\( ab = pf \\) for some nonzero element \\( f \\in D \\). If \\( a \\) and \\( b \\) are both non-units, express their unique factorizations:\n\\[\na = u p_1 \\cdots p_r, \\quad b = v q_1 \\cdots q_s\n\\]\nwhere \\( p_i \\) and \\( q_j \\) are irreducible elements, and \\( u, v \\in D^\\times \\) are units. Then:\n\\[\nab = (uv) p_1 \\cdots p_r q_1 \\cdots q_s = pf\n\\]\nBy unique factorization, \\( p \\) must be associate to one of the \\( p_i \\) or \\( q_j \\). Thus either \\( a \\in (p) \\) or \\( b \\in (p) \\) (or both). \n\nIf one of \\( a \\) or \\( b \\) is a unit (say \\( a \\in D^\\times \\)), then \\( ab \\in (p) \\) implies \\( b = a^{-1}pf \\in (p) \\). Similarly if \\( b \\) is a unit. Therefore, \\( (p) \\) is prime.\n}\n\n{\\it Fact 2:\nIf \\( I \\) is a nonzero prime ideal of \\( D \\), then \\( I \\) is principal.\n\nProof:\nLet \\( I \\) be a nonzero prime ideal and take \\( r \\in I \\setminus \\{0\\} \\). Factorize \\( r \\) into irreducibles:\n\\[\nr = p_1 \\cdots p_n \\quad \\text{(each \\( p_i \\) irreducible)}\n\\]\nSince \\( I \\) is prime, some \\( p_i \\in I \\). By Fact 1, \\( (p_i) \\) is a prime ideal. As \\( (p_i) \\) is maximal among proper principal ideals and \\( (p_i) \\subseteq I \\), it follows that \\( I = (p_i) \\).\n}\n\nWe are now ready to finish the proof. Let \\( \\mathscr{N} \\) be the set of non-principal ideals in \\( D \\). Suppose that \\( \\mathscr{N} \\) is non-empty.  \n\\( \\mathscr{N} \\) is a partially ordered set (with inclusion of ideals as the ordering relation). Any totally ordered subset of ideals \\( J_1 \\subseteq J_2 \\subseteq \\cdots \\) in \\( \\mathscr{N} \\) has an upper bound, namely \\( J = \\bigcup_j J_j \\). \\( J \\) is an ideal of \\( D \\) because the subset is totally ordered. If \\( J = (x) \\) for some \\( x \\in D \\), then \\( x \\in J_j \\) for some \\( j \\), and hence \\( (x) = J_j \\). This contradiction shows \\( J \\in \\mathscr{N} \\). By Zorn's Lemma, \\( \\mathscr{N} \\) has a maximal element, call it \\( \\mathfrak{n} \\).\n\nBy Fact 2, \\( \\mathfrak{n} \\) is not a prime ideal. Thus, there exist \\( a, b \\in D \\) such that \\( a, b \\notin \\mathfrak{n} \\), but \\( ab \\in \\mathfrak{n} \\). Then \\( \\mathfrak{n} + (a) \\) and \\( \\mathfrak{n} + (b) \\) are ideals strictly larger than \\( \\mathfrak{n} \\), so they must be principal:  \n\\[\n\\mathfrak{n} + (a) = (u), \\quad \\mathfrak{n} + (b) = (v) \\quad \\text{for some} \\quad u, v \\in D.\n\\]\nThen we have:\n\\[\n\\mathfrak{n} = \\mathfrak{n} + (ab) = (\\mathfrak{n} + (a))(\\mathfrak{n} + (b)) = (u)(v) = (uv).\n\\]\nThis contradicts \\( \\mathfrak{n} \\) being non-principal.\n\nThus, \\( \\mathscr{N} \\) must be empty, i.e., all ideals in \\( D \\) are principal.\n", "header": "import Mathlib\n\n", "informal_prefix": "/--\nLet \\( D \\) be a unique factorization domain.\nProve that if every nonzero prime ideal of \\( D \\) is maximal, then \\( D \\) is a principal ideal domain.\n-/"}
{"informal_statement": "Let \\( M \\) be a finitely-generated module over a Dedekind domain. Prove that \\( M \\) is flat if and only if $M$ is torsion-free.", "formal_statement": "theorem flat_iff_noZeroSMulDivisor {R M : Type} [CommRing R] [AddCommGroup M]\n    [Module R M] [IsDedekindDomain R] [Module.Finite R M] :\n    Module.Flat R M ↔ NoZeroSMulDivisors R M := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Tensor Product and Flatness"], "name": "Problem_83", "informal_proof": "\nSince flat and torsion-free are local properties, we can assume $R$ is local. In this case they are both equivalent to free.\n", "header": "import Mathlib\n\n\n", "informal_prefix": "/-- Let $M$ be a finitely-generated module over a Dedekind domain. Prove that $M$ is flat if and\nonly if $M$ is torsion-free. -/"}
{"informal_statement": "Let \\( A \\) be a Dedekind domain and \\( \\mathfrak{a} \\neq 0 \\) an ideal in \\( A \\). Show that every ideal in \\( A/\\mathfrak{a} \\) is principal.", "formal_statement": "theorem isPrincipalIdealRing_quot_of_isDedekind {A : Type} [CommRing A]\n    [IsDedekindDomain A] (a : Ideal A) (ha : a ≠ ⊥) :\n    IsPrincipalIdealRing (A ⧸ a) := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Dedekind Domains and DVRs"], "name": "Problem_84", "informal_proof": "\nGiven an ideal \\( I \\) in a Dedekind domain \\( A \\), factor \n\\[\nI = \\prod_{i=1}^n \\mathfrak{p}_i^{e_i}\n\\]\nwhere \\( \\mathfrak{p}_i \\) are prime ideals and \\( e_i \\) are positive integers. By the Chinese Remainder Theorem,\n\\[\nA/I \\cong \\bigoplus_{i=1}^n A/\\mathfrak{p}_i^{e_i}.\n\\]\nTo show \\( A/I \\) is principal, it suffices to show each component \\( A/\\mathfrak{p}_i^{e_i} \\) is principal.\n\nThe ideals of \\( A/\\mathfrak{p}_i^{e_i} \\) correspond to ideals of \\( A \\) containing \\( \\mathfrak{p}_i^{e_i} \\), which are exactly \\( \\mathfrak{p}_i^k \\) for \\( 0 \\leq k \\leq e_i \\), under the projection map \n\\[\n\\pi : A \\to A/\\mathfrak{p}_i^{e_i}.\n\\]\n\nConsider two cases:\n\\begin{enumerate}\n    \\item If \\( \\pi(\\mathfrak{p}_i) = \\pi(\\mathfrak{p}_i^2) \\), then \\( \\pi(\\mathfrak{p}_i) = 0 \\), so \\( A/\\mathfrak{p}_i^{e_i} \\) is a field (hence principal).\n    \n    \\item Otherwise, select \n    \\[\n    \\alpha \\in \\pi(\\mathfrak{p}_i) \\setminus \\pi(\\mathfrak{p}_i^2).\n    \\]\n    Then \\( (\\alpha) \\) is a proper ideal with \\( (\\alpha) \\not\\subseteq \\pi(\\mathfrak{p}_i^n) \\) for any \\( n \\geq 2 \\). It follows that \\( (\\alpha) = \\pi(\\mathfrak{p}_i) \\), and consequently,\n    \\[\n    \\pi(\\mathfrak{p}_i^n) = (\\alpha^n).\n    \\]\n    Thus all ideals are principal.\n\\end{enumerate}\n\nWe conclude that \\( A/\\mathfrak{p}_i^{e_i} \\) is principal for each \\( i \\), so \\( A/I \\) is principal.\n", "header": "import Mathlib\n\n\n", "informal_prefix": "/-- Let $A$ be a Dedekind domain and $\\mathfrak{a} \\neq 0$ an ideal in $A$. Show that every ideal in\n $A/\\mathfrak{a}$ is principal. -/"}
{"informal_statement": "Let \\( A \\) be a Noetherian ring and let \\( \\mathfrak{a}, \\mathfrak{b} \\) be ideals in \\( A \\). If \\( M \\) is any \\( A \\)-module, let \\( M_{\\mathfrak{a}} \\), \\( M_{\\mathfrak{b}} \\) denote its \\( \\mathfrak{a} \\)-adic and \\( \\mathfrak{b} \\)-adic completions respectively. If \\( M \\) is finitely generated, prove that \\( (M_{\\mathfrak{a}})_{\\mathfrak{b}} \\cong M_{\\mathfrak{a} + \\mathfrak{b}} \\).", "formal_statement": "theorem nonempty_adicCompletion_adicCompletion_linearEquiv {A M : Type} [CommRing A]\n    [IsNoetherianRing A] (𝒜 : Ideal A) [AddCommGroup M]\n    [Module A M] [Module.Finite A M] (ℬ : Ideal A) :\n    Nonempty (AdicCompletion ℬ (AdicCompletion 𝒜 M) ≃ₗ[A] AdicCompletion (𝒜 + ℬ) M) := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Other Topics", "Completions and Hensel's lemma"], "name": "Problem_85", "informal_proof": "\n(Atiyah P114 Exercise 10.5) \n\nTake the $\\mathfrak{a}$-adic completion of the exact sequence\n$$ 0 \\to \\mathfrak{b}^m M \\to M \\to M/\\mathfrak{b}^m M \\to 0 $$\nand apply (10.13). Then use the isomorphism\n$$ \\lim_{\\leftarrow m} \\left( \\lim_{\\leftarrow n} M/(\\mathfrak{a}^n M + \\mathfrak{b}^m M) \\right) \\cong \\lim_{\\leftarrow n} M/(\\mathfrak{a}^n M + \\mathfrak{b}^n M) $$\nand the inclusions $(\\mathfrak{a} + \\mathfrak{b})^{2n} \\subseteq \\mathfrak{a}^n + \\mathfrak{b}^n \\subseteq (\\mathfrak{a} + \\mathfrak{b})^n$.\n", "header": "import Mathlib\n\n\nopen Submodule\nopen Finset\nopen Submodule\n\n", "informal_prefix": "/-- Let $A$ be a Noetherian ring and let $\\mathfrak{a}, \\mathfrak{b}$ be ideals in $A$. If $M$ is\nany $A$-module, let $M_{\\mathfrak{a}}$, $M_{\\mathfrak{b}}$ denote its $\\mathfrak{a}$-adic and\n$\\mathfrak{b}$-adic completions respectively. If $M$ is finitely generated, prove that\n$(M_{\\mathfrak{a}})_{\\mathfrak{b}} \\cong M_{\\mathfrak{a} + \\mathfrak{b}}$. -/"}
{"informal_statement": "Let \\( M \\) be an \\( R \\)-module. The following are equivalent:\n\n\\begin{enumerate}\n    \\item \\( M \\) is finitely generated.\n    \\item For every family \\( (Q_{\\alpha})_{\\alpha \\in A} \\) of \\( R \\)-modules, the canonical map \\( M \\otimes_{R} \\left( \\prod_{\\alpha} Q_{\\alpha} \\right) \\to \\prod_{\\alpha} (M \\otimes_{R} Q_{\\alpha}) \\) is surjective.\n\\end{enumerate}", "formal_statement": "theorem finite_iff_surjective_linearMap {R : Type} [CommRing R]\n    (M : Type) [AddCommGroup M] [Module R M] :\n    Module.Finite R M ↔ ∀ {α : Type} (Q : α → Type),\n    ∀ [(a : α) → AddCommGroup (Q a)] [(a : α) → Module R (Q a)],\n        Function.Surjective (LinearMap.pi (\n            fun i => LinearMap.lTensor M (\n                LinearMap.proj i (φ := Q) (R := R)))) := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Ideals and Modules"], "name": "Problem_86", "informal_proof": "\nFor $M$ finitely generated, there exists some surjective $p:R^N\\to M$ from a f.g. free module. Consider the diagram\n\\begin{center}{\\begin{tikzcd}\n    R^N\\otimes_R\\prod_\\alpha Q_\\alpha\\arrow[r,\"\\iota_{R^N}\"]\\arrow[d,\"p\\otimes\\text{id}\"]&\\prod_\\alpha(R^N\\otimes_R Q_\\alpha)\\arrow[d,\"\\prod p\\otimes\\text{id}\"]\\\\M\\otimes_R\\prod_\\alpha Q_\\alpha\\arrow[r,\"\\iota_{M}\"]&\\prod_\\alpha(M\\otimes_R Q_\\alpha)\n\\end{tikzcd}}\\end{center}\nNote that the upper map decomposites as $R^N\\otimes_R\\prod_\\alpha Q_\\alpha\\cong(\\prod_\\alpha Q_\\alpha)^N\\cong\\prod_\\alpha Q_\\alpha^N\\cong\\prod_\\alpha (R^N\\otimes_RQ_\\alpha)$. The right maps are surjective as each term is. We get that the lower is also surjective.\n\nFor $M$ infinitely generated, take $\\alpha=\\{N:N\\subset M\\ \\text{f.g.}\\}$, pick $(m_N)_{N\\in\\alpha}\\in\\prod_\\alpha M$ such that $m_N\\notin N$. As $M\\otimes_R\\prod_{N\\in\\alpha}R\\cong\\text{colim}_{N\\in\\alpha}(N\\otimes_R\\prod_\\alpha R)\\cong\\text{colim}_{N\\in\\alpha}\\prod_\\alpha N$ is a filtered colimit, $(m_N)_{N\\in\\alpha}$ cannot lie in its image in $\\prod_\\alpha M$. Hence 2. is false for the family $(R)_\\alpha$.\n\nWe repeatedly use that $\\otimes$ preserves colim, and that finite $\\prod$ coincides with $\\oplus$, and that limit diagrams commute.\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Let $M$ be an $R$-module. The following are equivalent:\n\n\\begin{enumerate}\n    \\item $M$ is finitely generated.\n    \\item For every family $(Q_{\\alpha})_{\\alpha \\in A}$ of $R$-modules, the canonical map\n    $M \\otimes_{R} \\left( \\prod_{\\alpha} Q_{\\alpha} \\right) \\to \\prod_{\\alpha} (M \\otimes_{R}\n    Q_{\\alpha})$ is surjective.\n\\end{enumerate} -/"}
{"informal_statement": "If \\( R \\) is a valuation ring of Krull dimension 1 and \\( K \\) its field of fractions, then there does not exist any rings intermediate between \\( R \\) and \\( K \\).", "formal_statement": "theorem eq_bot_or_eq_top_of_ringKrullDim_eq_one {R K : Type} [CommRing R] [IsDomain R]\n    [Field K] [Algebra R K] [IsFractionRing R K] [ValuationRing R] (hD : ringKrullDim R = 1)\n    (S : Subalgebra R K) : S = ⊥ ∨ S = ⊤ := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Ideals and Modules"], "name": "Problem_87", "informal_proof": "\nValuation rings are local domains, and every finitely generated ideal is principal. If $A$ is of dimension $1$, then the maximal ideal $\\mathfrak{m}$ is the only nontrivial prime ideal. By the definition of $A^*$ and $\\mathfrak{m}$ of a valuation ring, if $R\\subset R'\\subset K$, $R'$ inverts an element $x\\in \\mathfrak{m}$, which makes the localization $\\mathfrak{m}R'=K$.\\par \nLet $I$ be a prime ideal of $R'$, then $I\\cap R$ is a prime ideal of $R$. It can not be $\\mathfrak{m}$ by the last paragraph. So $I\\cap R=0$, but then $I\\subset 0\\cup R^*$, which must be $0$. So $R'=K$.\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- If $R$ is a valuation ring of Krull dimension 1 and $K$ its field of fractions, then there do\nnot exist any rings intermediate between $R$ and $K$. -/"}
{"informal_statement": "Let \\((R, \\mathfrak{m})\\) be a Noetherian local ring. Let \\(x, y \\in \\mathfrak{m}\\) be a regular sequence of length \\(2\\). For any \\(n \\geq 2\\) show that there do not exist \\(a, b \\in R\\) with\n\n\\[\nx^{n-1}y^{n-1} = a x^{n} + b y^{n}\n\\]", "formal_statement": "theorem not_exists_pow_sub_one_mul_pow_sub_one_eq {R : Type} [CommRing R] [IsNoetherianRing R]\n    [IsLocalRing R] {n : ℕ} {x y : R} (hn : n ≥ 2) (hx : x ∈ IsLocalRing.maximalIdeal R)\n    (hy : y ∈ IsLocalRing.maximalIdeal R) (h : Sequence.IsRegular R [x, y]) :\n    ¬ ∃ a b : R, x ^ (n - 1) * y ^ (n - 1) = a * x ^ n + b * y ^ n := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Dimension Theory and Smoothness", "Regular Sequences and Regular Local Rings"], "name": "Problem_88", "informal_proof": "\nElse if $b\\in(x^{n-1})$, then $y^{n-1}=ax+x^{1-n}by^n$ as $x\\in R$ nonzerodivisor. If $b\\in(x^n)$ then $y^{n-1}=0$ in $R/x$, contradicts; else $1=x^{1-n}by\\in\\mathfrak m$ in $R/x$ as $y$ nonzerodivisor, contradicts. Else $b\\in (x^j)-(x^{j+1})$ for some $j<n-1$, then $x^{-j}by^{n-1}=0$ in $R/x$, contradicts.\n", "header": "import Mathlib\n\nopen RingTheory\n", "informal_prefix": "/--Let \\((R, \\mathfrak{m})\\) be a Noetherian local ring. Let \\(x, y \\in \\mathfrak{m}\\) be a regular\nsequence of length \\(2\\). For any \\(n \\geq 2\\) show that there do not exist \\(a, b \\in R\\) with\n\\[\nx^{n-1}y^{n-1} = a x^{n} + b y^{n}\n\\]-/"}
{"informal_statement": "Let \\( K \\) be a field and \\( L \\) an extension field of \\( K \\). If \\( P \\) is a prime ideal of \\( L[X_1, \\dots, X_n] \\) and \\( \\mathfrak{p} = P \\cap K[X_1, \\dots, X_n] \\), then \\( \\operatorname{ht}(P) \\geq \\operatorname{ht}(\\mathfrak{p}) \\).", "formal_statement": "theorem primeHeight_le_of_comap_eq {K L : Type} (n : ℕ) [Field K] [Field L] [Algebra K L]\n    (P : Ideal (MvPolynomial (Fin n) L)) (p : Ideal (MvPolynomial (Fin n) K)) [P.IsPrime]\n    [p.IsPrime] (h : P.comap (MvPolynomial.map (algebraMap K L)) = p) :\n    p.primeHeight ≤ P.primeHeight := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Integral Dependence and the Nullstellensatz"], "name": "Problem_89", "informal_proof": "\n$K[X_1, \\dots, X_n] \\to L[X_1, \\dots, X_n]$ is flat so $\\operatorname{ht}(P) \\geq \\operatorname{ht}(\\mathfrak{p})$ by virtue of \"Going down\" lemma.\n", "header": "import Mathlib\n\nopen MvPolynomial\n", "informal_prefix": "/--Let \\( K \\) be a field and \\( L \\) an extension field of \\( K \\). If \\( P \\) is a prime ideal of\n\\( L[X_1, \\dots, X_n] \\) and \\( \\mathfrak{p} = P \\cap K[X_1, \\dots, X_n] \\), then \\(\n\\operatorname{ht}(P) \\geq \\operatorname{ht}(\\mathfrak{p}) \\).-/"}
{"informal_statement": "Suppose that \\( R \\) is a Noetherian local ring with maximal ideal \\( \\mathfrak{m} \\) and residue field \\( \\kappa \\). In this case the projective dimension of \\( \\kappa \\) is \\( \\geq \\dim_{\\kappa} \\mathfrak{m} / \\mathfrak{m}^{2} \\).", "formal_statement": "theorem not_hasProjectiveDimensionLT_finrank_cotangentSpace {R : Type} [CommRing R] [IsLocalRing R]\n    [IsNoetherianRing R] :\n      ¬ CategoryTheory.HasProjectiveDimensionLT\n        (ModuleCat.of R (IsLocalRing.ResidueField R))\n          (Module.finrank (IsLocalRing.ResidueField R) (IsLocalRing.CotangentSpace R)) := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Dimension Theory and Smoothness", "Regular Sequences and Regular Local Rings"], "name": "Problem_90", "informal_proof": "\nIf $R$ is regular, then the projective dimension of \\( \\kappa = \\dim_{\\kappa} \\mathfrak{m} / \\mathfrak{m}^{2}\\). Otherwise the projective dimension of \\( \\kappa = +\\infty\\).  \n", "header": "import Mathlib\n\n", "informal_prefix": "/--\nSuppose that \\( R \\) is a Noetherian local ring with maximal ideal \\( \\mathfrak{m} \\) and residue field \\( \\kappa \\).\nIn this case the projective dimension of \\( \\kappa \\) is \\( \\geq \\dim_{\\kappa} \\mathfrak{m} / \\mathfrak{m}^{2} \\).-/"}
{"informal_statement": "If a ring \\( R \\), not a field, is a maximal proper subring of a field \\( K \\), show \\( R \\) is a valuation ring of Krull dimension 1.", "formal_statement": "theorem exists_toSubring_eq_and_ringKrullDim_eq_one {K : Type} [Field K] (R : Subring K)\n    (h : IsCoatom R) (hR : ¬ IsField R) :\n    (∃ V : ValuationSubring K, V.toSubring = R) ∧ ringKrullDim R = 1 := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Dimension Theory and Smoothness", "Heights and Dimensions"], "name": "Problem_91", "informal_proof": "\nA maximal subring must be a local ring (otherwise, its localization is strictly bigger). Maximal local subrings must be valuation rings. It cannot be Krull dimension greater then or equal to 2 because one can do further localization w.r.t smallest prime ideal to enlarge the ring. It cannot be Krull dimension 0 because every Krull dimension 0 domain is a field.\n", "header": "import Mathlib\n\n", "informal_prefix": "/--If a ring \\( R \\), not a field, is a maximal proper subring of a field \\( K \\), show \\( R \\) is\na valuation ring of Krull dimension 1.-/"}
{"informal_statement": "Let \\( R \\) be a Dedekind domain. Show the following:\n\nIf \\( P_1, \\dots, P_n \\in \\operatorname{Spec}(R) \\) are pairwise distinct, non-zero prime ideals and \\( e_1, \\dots, e_n \\) are non-negative integers, there exists \\( a \\in R \\setminus \\{0\\} \\) such that\n\\[\n(a) = P_1^{e_1} \\cdots P_n^{e_n} \\cdot J,\n\\]\nwith \\( J \\subseteq R \\) an ideal in whose factorization none of the \\( P_i \\) appear.", "formal_statement": "theorem exists_factor_principal_ideal (R : Type) [CommRing R] [IsDedekindDomain R]\n    (n : ℕ) (p : (Fin n) → PrimeSpectrum R) (h_nonzero : ∀ i, (p i).asIdeal ≠ ⊥)\n    (h_inj : Function.Injective p) (e : (Fin n) → ℕ) :\n    ∃ (a : R) (J : Ideal R), a ≠ 0 ∧ Ideal.span {a} = J * ∏ (i : Fin n), (p i).1 ^ (e i) ∧\n    (∀ (i : Fin n), ¬ ∃ (K : Ideal R), J = K * (p i).1) := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Dedekind Domains and DVRs"], "name": "Problem_92", "informal_proof": "\nPick elements $a_i\\in P_i^{e_i}-P_i^{e_i+1}$ as $R$ is Dedekind. By the Chinese Remainder Theorem, there exists some $a\\in R$ such that $a\\equiv a_i\\mod{P^{e_i+1}}$ for each $i\\leq n$. Each $P_i$ in the factorization of $a$ has multiplicity $e_i$ determined by the congruences.\n", "header": "import Mathlib\n\n", "informal_prefix": "/--\nLet \\( R \\) be a Dedekind domain. Show the following:\n\nIf \\( P_1, \\dots, P_n \\in \\operatorname{Spec}(R) \\) are pairwise distinct, non-zero prime ideals\nand \\( e_1, \\dots, e_n \\) are non-negative integers, there exists \\( a \\in R \\setminus \\{0\\} \\) such that\n\\[\n(a) = P_1^{e_1} \\cdots P_n^{e_n} \\cdot J,\n\\]\nwith \\( J \\subseteq R \\) an ideal in whose factorization none of the \\( P_i \\) appear.\n-/"}
{"informal_statement": "Let $\\mathcal{O}$ be an integral domain in which all nonzero ideals admit a unique factorization into prime ideals. Show that $\\mathcal{O}$ is a Dedekind domain.", "formal_statement": "theorem isDedekindDomain_of_ideal_UFD (O : Type) [CommRing O] [IsDomain O]\n    [CancelCommMonoidWithZero (Ideal O)] [UniqueFactorizationMonoid (Ideal O)] :\n    IsDedekindDomain O := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Dedekind Domains and DVRs"], "name": "Problem_93", "informal_proof": "\nFirst we show that every Nonzero Prime Ideal is Maximal. In fact, let $P$ be a nonzero prime ideal of $\\mathcal{O}$. To show $P$ is maximal, we must show that if $M$ is a maximal ideal such that $P \\subseteq M$, then $P = M$.\n\nSince $P \\subseteq M$, $M$ must divide $P$. The ideal $P$ is prime, so by the definition of unique factorization, its only factorization is $P$ itself. The ideal $M$ is also prime, so its factorization is $M$.\n\t\t\nThe division relation $P = M \\cdot I$ for some ideal $I$ is a factorization of $P$. By the uniqueness of prime ideal factorization, this factorization can only have one prime factor, which must be $P$. Therefore, we must have $M = P$ (and $I = \\mathcal{O}$).\n\t\t\nThus, any maximal ideal containing $P$ is $P$ itself, which proves that $P$ is a maximal ideal.\n\n只需证明对任意理想 $I \\ne (0)$, $I * I^{-1} = (1)$, 其中 $I^{-1} := \\{a \\in \\mathrm{Frac} A\\ |\\ aI \\subset A\\}$ (\\href{https://leanprover-community.github.io/mathlib4_docs/Mathlib/RingTheory/DedekindDomain/Ideal.html#isDedekindDomain_iff_isDedekindDomainInv}{isDedekindDomain\\_iff\\_isDedekindDomainInv}). 由唯一分解性，只需证明 $I = \\mathfrak{p}$ 是素理想的情形.\n\n先证明 $\\mathfrak{p}^{-1} \\ne (1)$. Let $a \\in \\mathfrak{p}$, $a \\neq 0$, and $\\mathfrak{p}_1\\mathfrak{p}_2\\cdots\\mathfrak{p}_r = (a) \\subseteq \\mathfrak{p}$. Then one of the $\\mathfrak{p}_i$, say $\\mathfrak{p}_1$, is contained in $\\mathfrak{p}$, and so $\\mathfrak{p}_1 = \\mathfrak{p}$ because $\\mathfrak{p}_1$ is a maximal ideal. (Indeed, if none of the $\\mathfrak{p}_i$ were contained in $\\mathfrak{p}$, then for every $i$ there would exist $a_i \\in \\mathfrak{p}_i \\setminus \\mathfrak{p}$ such that $a_1 \\cdots a_r \\in \\mathfrak{p}$. But $\\mathfrak{p}$ is prime.) Since $\\mathfrak{p}_2 \\cdots \\mathfrak{p}_r \\not\\subseteq (a)$, there exists $b \\in \\mathfrak{p}_2 \\cdots \\mathfrak{p}_r$ such that $b \\notin (1)$, i.e., $a^{-1}b \\notin (1)$. On the other hand we have $b\\mathfrak{p} \\subseteq (a)$, i.e., $a^{-1}b\\mathfrak{p} \\subseteq (1)$, and thus $a^{-1}b \\in \\mathfrak{p}^{-1}$. It follows that $\\mathfrak{p}^{-1} \\neq (1)$.\n\n假设 $\\mathfrak{p}\\mathfrak{p}^{-1} = \\mathfrak{p}$, 取 $a \\in \\mathfrak{p}$, 则 $(a\\mathfrak{p}^{-1})$ 是 $A$ 的理想, 故$ \\mathfrak{p} (a\\mathfrak{p}^{-1}) = (a) \\mathfrak{p}$. 由于理想唯一分解知其乘法可消去, 故 $(a\\mathfrak{p}^{-1}) = (a)$, 进而 $\\mathfrak{p}^{-1} \\subset A$, 矛盾! 因此 $\\mathfrak{p} < \\mathfrak{p}\\mathfrak{p}^{-1} \\subset A$, 结合 $\\mathfrak{p}$ 是极大理想知 $\\mathfrak{p}\\mathfrak{p}^{-1} = (1)$.\n        \n", "header": "import Mathlib\n\n", "informal_prefix": "/--\nLet $\\mathcal{O}$ be an integral domain in which all nonzero ideals admit a unique factorization into prime ideals.\nShow that $\\mathcal{O}$ is a Dedekind domain. -/"}
{"informal_statement": "Suppose that a ring \\( S \\) is integral over the image of a ring homomorphism \\( R \\to S \\). Show that the Krull dimension of \\( M \\) as an \\( S \\)-module is the same as the Krull dimension of \\( M \\) as an \\( R \\)-module.", "formal_statement": "theorem ringKrullDim_quot_annihilator_eq {R S M : Type} [CommRing R] [CommRing S] [AddCommGroup M] [Algebra R S]\n    [Module S M] [Module R M] [IsScalarTower R S M] [Algebra.IsIntegral R S] :\n    ringKrullDim (S ⧸ (Module.annihilator S M)) = ringKrullDim (R ⧸ (Module.annihilator R M)) := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Integral Dependence and the Nullstellensatz"], "name": "Problem_94", "informal_proof": "\n$\\dim_R M = \\dim R / \\mathrm{Ann}_R(M)$, $\\dim_S M = \\dim S / \\mathrm{Ann}_S(M)$. 注意 $R / \\mathrm{Ann}_R(M) \\to S / \\mathrm{Ann}_S(M)$ 也是整的, 故 $ \\dim R / \\mathrm{Ann}_R(M) = \\dim S / \\mathrm{Ann}_S(M)$.\n", "header": "import Mathlib\n\n", "informal_prefix": "/-- Suppose that a ring $S$ is integral over the image of a ring homomorphism $R \\to S$. Show that\nthe Krull dimension of $M$ as an $S$-module is the same as the Krull dimension of $M$ as an\n$R$-module. -/"}
{"informal_statement": "Show that if $x_1, \\dots, x_r$ is a regular sequence in $R$, then so is $x_1^{a_1}, \\dots, x_r^{a_r}$ for any positive integers $a_1, \\dots, a_r$.", "formal_statement": "theorem isRegular_ofFn_pow (R M : Type) [CommRing R] [AddCommGroup M] [Module R M]\n    (rs : List R) (a : Fin rs.length → ℕ+) (h : Sequence.IsRegular M rs) :\n    Sequence.IsRegular M (List.ofFn (fun i => rs[i] ^ (a i).1)) := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Dimension Theory and Smoothness", "Regular Sequences and Regular Local Rings"], "name": "Problem_95", "informal_proof": "\n由归纳法, 只需证明:\nLet $x_1, \\dots, x_i, \\dots, x_n$ and $x_1, \\dots, x'_i, \\dots, x_n$ be an $M$-regular sequence, then $x_1, \\dots, x_ix'_i, \\dots, x_n$ is an $M$-Regular sequence.\n\nIn fact, we can assume $ i = 1 $ after replacing $ M $ by $ M / (x_1, \\dots, x_{i -1})M $. Note that $ x_1 $ is not a zero-divisor of $ M $, so we have a short exact sequence\n    \\begin{center}\n    \t\\begin{tikzcd}\n    \t\t0 \\arrow[r] & M/x_1'M \\arrow[r, \"\\cdot\\ x_1\"] & M/x_1x_1'M \\arrow[r] & M/x_1M \\arrow[r] & 0.\n    \t\\end{tikzcd}\n    \\end{center}\n    Since $ x_2, \\dots, x_n $ are $ M/x_1M $ and $ M/x_1'M $-regular sequences, so it is an $ M / x_1x_1'M $ -regular sequence by \\href{https://stacks.math.columbia.edu/tag/0F1T}{Stacks Project, Lemma 0F1T}.\n", "header": "import Mathlib\n\nopen RingTheory\n\n", "informal_prefix": "/-- Show that if $x_1, \\dots, x_r$ is a regular sequence in $R$,\nthen so is $x_1^{a_1}, \\dots, x_r^{a_r}$ for any positive integers $a_1, \\dots, a_r$. -/"}
{"informal_statement": "Let \\( I_1, I_2 \\subseteq K[x_1, \\dots, x_n] \\) be two ideals. With \\( y \\) an additional indeterminate, form the ideal\n\\[\nJ := \\big( y \\cdot I_1 \\cup (1 - y) \\cdot I_2 \\big) K[x_1, \\dots, x_n, y] \\subseteq K[x_1, \\dots, x_n, y].\n\\]Show that\n\\[\nI_1 \\cap I_2 = K[x_1, \\dots, x_n] \\cap J.\n\\]", "formal_statement": "theorem inf_eq_span_X_smul_sup_one_sub_X_smul_comap_C {K : Type} [Field K] (n : ℕ) \n    (I1 I2 : Ideal (MvPolynomial (Fin n) K)) :\n    I1 ⊓ I2 = (Ideal.span {(X : (MvPolynomial (Fin n) K)[X])} • (I1.map C) ⊔\n    Ideal.span {(1 - X : (MvPolynomial (Fin n) K)[X])} • (I2.map C) ).comap C := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Ideal Theory", "Ideals and Modules"], "name": "Problem_96", "informal_proof": "\n记 $A := K[x_1, \\dots, x_n]$. 对任意 $x \\in I_1 \\cap I_2$, $x = xy + (1-y)y \\in J$, 故 $I_1 \\cap I_2 \\subseteq K[x_1, \\dots, x_n] \\cap J$. 另一方面, 对 $c \\in K[x_1, \\dots, x_n] \\cap J$, 只需证 $c \\in I_1 \\cap I_2$. 事实上, 设 $c = f(y)ya + g(y)(1 - y)b$, $f(y), g(y) \\in A[y]$, $a \\in I_1, b \\in I_2$. 取 $y = 0$ 知 $c = g(0)b \\in I_2$. 同理 $c \\in I_1$ 故 $c \\in I_1 \\cap I_2$.\n", "header": "import Mathlib\n\nopen Polynomial\n", "informal_prefix": "/--Let \\( I_1, I_2 \\subseteq K[x_1, \\dots, x_n] \\) be two ideals. With \\( y \\) an additional\nindeterminate, form the ideal\n\\[\nJ := \\big( y \\cdot I_1 \\cup (1 - y) \\cdot I_2 \\big) K[x_1, \\dots, x_n, y] \\subseteq K[x_1, \\dots,\nx_n, y].\n\\]Show that\n\\[\nI_1 \\cap I_2 = K[x_1, \\dots, x_n] \\cap J.\n\\]-/"}
{"informal_statement": "Prove that $\\sin 1^{\\circ}$ is algebraic over $\\mathbb{Q}$.", "formal_statement": "theorem isAlgebraic_sin_pi_div_180 : IsAlgebraic ℚ (sin (π / 180)) := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Field Theory", "Explicit Computations"], "name": "Problem_97", "informal_proof": "\nLet $\\theta = 1^\\circ = \\frac{\\pi}{180}$ radians. Consider the complex number:\n\\[\n\\zeta = e^{i\\theta} = \\cos\\theta + i\\sin\\theta\n\\]\nThis is a primitive $360$\\textsuperscript{th} root of unity, satisfying:\n\\[\n\\zeta^{360} = e^{i\\cdot 360\\theta} = e^{i2\\pi} = 1\n\\]\nThus $\\zeta$ is algebraic over $\\mathbb{Q}$ as it roots $x^{360} - 1$.\n\nThe complex conjugate $\\overline{\\zeta} = e^{-i\\theta} = \\cos\\theta - i\\sin\\theta$ is also algebraic. Using algebraic closure properties:\n\\[\n\\zeta - \\overline{\\zeta} = 2i\\sin\\theta \\implies \\sin\\theta = \\frac{\\zeta - \\overline{\\zeta}}{2i}\n\\]\nSince algebraic numbers form a field closed under addition, subtraction, multiplication, and division (with non-zero denominators), and $i$ is algebraic (root of $x^2 + 1$), the expression $\\frac{\\zeta - \\overline{\\zeta}}{2i}$ must be algebraic.\n", "header": "import Mathlib\n\nopen Real\n\n", "informal_prefix": "/- Prove that $\\sin 1^{\\circ}$ is algebraic over $\\mathbb{Q}$.-/"}
{"informal_statement": "Let \\( A \\) be a Noetherian ring, and suppose that \\( P \\) is a height \\( r > 0 \\) prime ideal generated by \\( r \\) elements, which may \\textbf{not} be a regular sequence. Show that \\( P \\) can be generated by an \\( A \\)-sequence.", "formal_statement": "theorem exists_isRegular_and_eq_ofList {A : Type} [CommRing A] [IsNoetherianRing A] (P : Ideal A) [P.IsPrime] (r : ℕ)\n    (hr : 0 < r) (h : r = P.primeHeight) (a : Fin r → A) (hP : P = Ideal.span (Set.range a)) :\n    ∃ b : List A, Sequence.IsRegular A b ∧ P = Ideal.ofList b := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Dimension Theory and Smoothness", "Regular Sequences and Regular Local Rings"], "name": "Problem_98", "informal_proof": "\n我们来证明若 $P$ 被 $r$ 个元素 $x_1 \\dots x_r$ 生成, 则这 $r$ 个元素一定是一个 regular sequence. 对 $P$ 做局部化, 不妨 $A$ 是局部环, $P$ 是其极大理想, 则 $\\dim A = r$. 对 $r$ 归纳. $r = 1$ 时, 假设 $x_1$ 是零因子, 则 $\\dim R / (x_1) = 1$. 但 $R / (x_1) = R / P$ 是域, 矛盾! 假设对 $r - 1$ 成立, 考虑 $r$ 的情形. 此时 $P / (x_1)$ 可以被 $r - 1$ 个元素生成, 故 $\\dim R / (x_1) \\le r - 1$. 因此 $\\dim R / (x_1) = r - 1$ 且 $x_1$ 不是零因子. 对 $A / (x_1)$ 用归纳假设即知结论成立.\n", "header": "import Mathlib\n\nopen RingTheory\n", "informal_prefix": "/--Let \\( A \\) be a Noetherian ring, and suppose that \\( P \\) is a height \\( r > 0 \\) prime ideal\ngenerated by \\( r \\) elements, which may **not** be a regular sequence. Show that \\( P \\) can be\ngenerated by an \\( A \\)-sequence.-/"}
{"informal_statement": "For a commutative ring \\(R\\) and \\(n \\in \\mathbb{N}\\) such that \\(2\\) is invertible in \\(R\\). If \\(A \\in SO(2n + 1,  R)\\), then \\(det(I - A) = 0\\).", "formal_statement": "theorem determinant_eq_zero (R : Type) [CommRing R] (n : ℕ) (h2_inv : IsUnit (2 : R))\n    (A : Matrix.specialOrthogonalGroup (Fin (2 * n + 1)) R) : (1 - A.1).det = 0 := by", "source": "FATE-H", "tag": ["Abstract Algebra", "Other Topics", "Linear Algebra"], "name": "Problem_99", "informal_proof": "\nLet $A \\in SO(2n+1, R)$. The dimension $k=2n+1$ is odd, and we have $A^TA = I$ with $\\det(A) = 1$.\nA chain of equalities gives:\n\\begin{align*}\n    \\det(I - A) &= \\det(I - A^T) \\\\\n    &= \\det(I - A^{-1}) \\\\\n    &= \\det(A^{-1})\\det(A - I) \\\\\n    &= \\det(A-I) \\\\\n    &= (-1)^k \\det(I - A) \\\\\n    &= -\\det(I-A)\n\\end{align*}\nThis implies $2\\det(I-A) = 0$. Since $2$ is invertible in $R$, it follows that $\\det(I-A) = 0$.\n", "header": "import Mathlib\n\n", "informal_prefix": "/--\nFor a commutative ring \\(R\\) and \\(n \\in \\mathbb{N}\\) such that \\(2\\) is invertible in \\(R\\).\nIf \\(A \\in SO(2n + 1,  R)\\), then \\(det(I - A) = 0\\).-/"}
{"informal_statement": "There exists a commutative ring with finite prime spectrum but is not Noetherian.", "formal_statement": "theorem exists_finite_primeSpectrum_not_isNoetherianRing :\n    ∃ (R : Type) (_ : CommRing R), Finite (PrimeSpectrum R) ∧ ¬ IsNoetherianRing R := by", "source": "FATE-H", "tag": ["Commutative Algebra", "Dimension Theory and Smoothness", "Heights and Dimensions"], "name": "Problem_100", "informal_proof": "\nThe ring $A=k[x_i,i\\in \\mathbb{N}]/(x_i^2,i\\in\\mathbb{N})$ has exactly one prime ideal $(x_i,i\\in\\mathbb{N})$, which is not finitely generated.\n", "header": "import Mathlib\n\n", "informal_prefix": "/--\nThere exists a commutative ring with finite prime spectrum but is not Noetherian.\n-/"}