fixed a few errors

proofnet.py

@@ -53,7 +53,7 @@ class ProofNet(datasets.GeneratorBasedBuilder):
     BUILDER_CONFIGS = [
         ProofNetConfig(
             name="plain_text",
-            version=datasets.Version("2.0.0", ""),
+            version=datasets.Version("2.1.0", ""),
             description="Plain text",
         ),
     ]

test.jsonl

@@ -67,7 +67,7 @@
 {"id": "Axler|exercise_5_13", "formal_statement": "theorem exercise_5_13 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] [finite_dimensional F V] {T : End F V}\n  (hS : \u2200 U : submodule F V, finrank F U = finrank F V - 1 \u2192\n  map T U = U) : \u2203 c : F, T = c \u2022 id :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $T \\in \\mathcal{L}(V)$ is such that every subspace of $V$ with dimension $\\operatorname{dim} V-1$ is invariant under $T$. Prove that $T$ is a scalar multiple of the identity operator.", "nl_proof": "\\begin{proof}\n\n    First off, let $T$ isn't a scalar multiple of the identity operator. So, there does exists that $v \\in V$ such that $u$ isn't an eigenvector of $T$. Therefore, $(u, T u)$ is linearly independent.\n\n\n\nNext, you should extend $(u, T u)$ to a basis of $\\left(u, T u, v_1, \\ldots, v_n\\right)$ of $V$. So, let $U=\\operatorname{span}\\left(u, v_1, \\ldots, v_n\\right)$. Then, $U$ is a subspace of $V$ and $\\operatorname{dim} U=\\operatorname{dim} V-1$. However, $U$ isn't invariant under $T$ since both $u \\in U$ and $T u \\in U$. This given contradiction to our hypothesis about $T$ actually shows us that our guess that $T$ is not a scalar multiple of the identity must have been false.\n\n\\end{proof}"}
 {"id": "Axler|exercise_5_24", "formal_statement": "theorem exercise_5_24 {V : Type*} [add_comm_group V]\n  [module \u211d V] [finite_dimensional \u211d V] {T : End \u211d V}\n  (hT : \u2200 c : \u211d, eigenspace T c = \u22a5) {U : submodule \u211d V}\n  (hU : map T U = U) : even (finrank U) :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $V$ is a real vector space and $T \\in \\mathcal{L}(V)$ has no eigenvalues. Prove that every subspace of $V$ invariant under $T$ has even dimension.", "nl_proof": "\\begin{proof}\n\n    First off, let us assume that $U$ is a subspace of $V$ that is invariant under $T$. Therefore, $\\left.T\\right|_U \\in \\mathcal{L}(U)$. If $\\operatorname{dim}$ $U$ were odd, then $\\left.T\\right|_U$ would have an eigenvalue $\\lambda \\in \\mathbb{R}$, so there would exist a nonzero vector $u \\in U$ such that\n\n$$\n\n\\left.T\\right|_U u=\\lambda u .\n\n$$\n\nSo, this would imply that $T_u=\\lambda u$, which would imply that $\\lambda$ is an eigenvalue of $T$. But $T$ has no eigenvalues, so $\\operatorname{dim} U$ must be even.\n\n\\end{proof}"}
 {"id": "Axler|exercise_6_3", "formal_statement": "theorem exercise_6_3 {n : \u2115} (a b : fin n \u2192 \u211d) :\n  (\u2211 i, a i * b i) ^ 2 \u2264 (\u2211 i : fin n, i * a i ^ 2) * (\u2211 i, b i ^ 2 / i) :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that $\\left(\\sum_{j=1}^{n} a_{j} b_{j}\\right)^{2} \\leq\\left(\\sum_{j=1}^{n} j a_{j}{ }^{2}\\right)\\left(\\sum_{j=1}^{n} \\frac{b_{j}{ }^{2}}{j}\\right)$ for all real numbers $a_{1}, \\ldots, a_{n}$ and $b_{1}, \\ldots, b_{n}$.", "nl_proof": "\\begin{proof}\n\n    Let $a_1, a_2, \\ldots, a_n, b_1, b_2, \\ldots, b_n \\in R$.\n\nWe have that\n\n$$\n\n\\left(\\sum_{j=1}^n a_j b_j\\right)^2\n\n$$\n\nis equal to the\n\n$$\n\n\\left(\\sum_{j=1}^n a_j b_j \\frac{\\sqrt{j}}{\\sqrt{j}}\\right)^2=\\left(\\sum_{j=1}^n\\left(\\sqrt{j} a_j\\right)\\left(b_j \\frac{1}{\\sqrt{j}}\\right)\\right)^2\n\n$$\n\nThis can be observed as an inner product, and using the Cauchy-Schwarz Inequality, we get\n\n$$\n\n\\begin{aligned}\n\n&\\left(\\sum_{j=1}^n a_j b_j\\right)^2=\\left(\\sum_{j=1}^n\\left(\\sqrt{j} a_j\\right)\\left(b_j \\frac{1}{\\sqrt{j}}\\right)\\right)^2 \\\\\n\n&=\\left\\langle\\left(a, \\sqrt{2} a_2, \\ldots, \\sqrt{n} a_n\\right),\\left(b_1, \\frac{b_2}{\\sqrt{2}}, \\ldots, \\frac{b_n}{\\sqrt{n}}\\right)\\right\\rangle \\\\\n\n& \\leq\\left\\|\\left(a, \\sqrt{2} a_2, \\ldots, \\sqrt{n} a_n\\right)\\right\\|^2\\left\\|\\left(b_1, \\frac{b_2}{\\sqrt{2}}, \\ldots, \\frac{b_n}{\\sqrt{n}}\\right)\\right\\|^2 \\\\\n\n&=\\left(\\sum_{j=1}^n j a_j^2\\right)\\left(\\sum_{j=1}^n \\frac{b_j^2}{j}\\right) \\\\\n\n& \\text { Hence, }\\left(\\sum_{j=1}^n a_j b_j\\right)^2=\\left(\\sum_{j=1}^n j a_j^2\\right)\\left(\\sum_{j=1}^n \\frac{b_j^2}{j}\\right) .\n\n\\end{aligned}\n\n$$\n\n\\end{proof}"}
-{"id": "Axler|exercise_6_13", "formal_statement": "theorem exercise_6_13 {V : Type*} [inner_product_space \u2102 V] {n : \u2115}\n  {e : fin n \u2192 V} (he : orthonormal \u2102 e) (v : V) :\n  \u2225v\u2225^2 = \u2211 i : fin n, \u2225\u27eav, e i\u27eb_\u2102\u2225^2 \u2194 v \u2208 span \u2102 (e '' univ) :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $\\left(e_{1}, \\ldots, e_{m}\\right)$ is an or thonormal list of vectors in $V$. Let $v \\in V$. Prove that $\\|v\\|^{2}=\\left|\\left\\langle v, e_{1}\\right\\rangle\\right|^{2}+\\cdots+\\left|\\left\\langle v, e_{m}\\right\\rangle\\right|^{2}$ if and only if $v \\in \\operatorname{span}\\left(e_{1}, \\ldots, e_{m}\\right)$.", "nl_proof": "\\begin{proof}\n\nIf $v \\in \\operatorname{span}\\left(e_1, \\ldots, e_m\\right)$, it means that\n\n$$\n\nv=\\alpha_1 e_1+\\ldots+\\alpha_m e_m .\n\n$$\n\nfor some scalars $\\alpha_i$. We know that $\\alpha_k=\\left\\langle v, e_k\\right\\rangle, \\forall k \\in\\{1, \\ldots, m\\}$. Therefore,\n\n$$\n\n\\begin{aligned}\n\n\\|v\\|^2 & =\\langle v, v\\rangle \\\\\n\n& =\\left\\langle\\alpha_1 e_1+\\ldots+\\alpha_m e_m, \\alpha_1 e_1+\\ldots+\\alpha_m e_m\\right\\rangle \\\\\n\n& =\\left|\\alpha_1\\right|^2\\left\\langle e_1, e_1\\right\\rangle+\\ldots+\\left|\\alpha_m\\right|^2\\left\\langle e_m, e_m\\right\\rangle \\\\\n\n& =\\left|\\alpha_1\\right|^2+\\ldots+\\left|\\alpha_m\\right|^2 \\\\\n\n& =\\left|\\left\\langle v, e_1\\right\\rangle\\right|^2+\\ldots+\\left|\\left\\langle v, e_m\\right\\rangle\\right|^2 .\n\n\\end{aligned}\n\n$$\n\n$\\Rightarrow$ Assume that $v \\notin \\operatorname{span}\\left(e_1, \\ldots, e_m\\right)$. Then, we must have\n\n$$\n\nv=v_{m+1}+\\frac{\\left\\langle v, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2} v_0,\n\n$$\n\nwhere $v_0=\\alpha_1 e_1+\\ldots+\\alpha_m e_m, \\alpha_k=\\left\\langle v, e_k\\right\\rangle, \\forall k \\in\\{1, \\ldots, m\\}$, and $v_{m+1}=v-$ $\\frac{\\left\\langle v, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2} v_0 \\neq 0$.\n\n\n\nWe have $\\left\\langle v_0, v_{m+1}\\right\\rangle=0$ (from which we get $\\left\\langle v, v_0\\right\\rangle=\\left\\langle v_0, v_0\\right\\rangle$ and $\\left\\langle v, v_{m+1}\\right\\rangle=$ $\\left.\\left\\langle v_{m+1}, v_{m+1}\\right\\rangle\\right)$. Now,\n\n$$\n\n\\begin{aligned}\n\n\\|v\\|^2 & =\\langle v, v\\rangle \\\\\n\n& =\\left\\langle v, v_{m+1}+\\frac{\\left\\langle v, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2} v_0\\right\\rangle \\\\\n\n& =\\left\\langle v, v_{m+1}\\right\\rangle+\\left\\langle v, \\frac{\\left\\langle v, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2} v_0\\right\\rangle \\\\\n\n& =\\left\\langle v_{m+1}, v_{m+1}\\right\\rangle+\\frac{\\left\\langle v_0, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2}\\left\\langle v_0, v_0\\right\\rangle \\\\\n\n& =\\left\\|v_{m+1}\\right\\|^2+\\left\\|v_0\\right\\|^2 \\\\\n\n& >\\left\\|v_0\\right\\|^2 \\\\\n\n& =\\left|\\alpha_1\\right|^2+\\ldots+\\left|\\alpha_m\\right|^2 \\\\\n\n& =\\left|\\left\\langle v, e_1\\right\\rangle\\right|^2+\\ldots+\\left|\\left\\langle v, e_m\\right\\rangle\\right|^2 .\n\n\\end{aligned}\n\n$$\n\nBy contrapositive, if $\\left\\|v_1\\right\\|^2=\\left|\\left\\langle v, e_1\\right\\rangle\\right|^2+\\ldots+\\left|\\left\\langle v, e_m\\right\\rangle\\right|^2$, then $v \\in \\operatorname{span}\\left(e_1, \\ldots, e_m\\right)$.\n\n\\end{proof}"}
+{"id": "Axler|exercise_6_13", "formal_statement": "theorem exercise_6_13 {V : Type*} [inner_product_space \u2102 V] {n : \u2115}\n  {e : fin n \u2192 V} (he : orthonormal \u2102 e) (v : V) :\n  \u2016v\u2016^2 = \u2211 i : fin n, \u2016\u27eav, e i\u27eb_\u2102\u2016^2 \u2194 v \u2208 span \u2102 (e '' univ) :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $\\left(e_{1}, \\ldots, e_{m}\\right)$ is an or thonormal list of vectors in $V$. Let $v \\in V$. Prove that $\\|v\\|^{2}=\\left|\\left\\langle v, e_{1}\\right\\rangle\\right|^{2}+\\cdots+\\left|\\left\\langle v, e_{m}\\right\\rangle\\right|^{2}$ if and only if $v \\in \\operatorname{span}\\left(e_{1}, \\ldots, e_{m}\\right)$.", "nl_proof": "\\begin{proof}\n\nIf $v \\in \\operatorname{span}\\left(e_1, \\ldots, e_m\\right)$, it means that\n\n$$\n\nv=\\alpha_1 e_1+\\ldots+\\alpha_m e_m .\n\n$$\n\nfor some scalars $\\alpha_i$. We know that $\\alpha_k=\\left\\langle v, e_k\\right\\rangle, \\forall k \\in\\{1, \\ldots, m\\}$. Therefore,\n\n$$\n\n\\begin{aligned}\n\n\\|v\\|^2 & =\\langle v, v\\rangle \\\\\n\n& =\\left\\langle\\alpha_1 e_1+\\ldots+\\alpha_m e_m, \\alpha_1 e_1+\\ldots+\\alpha_m e_m\\right\\rangle \\\\\n\n& =\\left|\\alpha_1\\right|^2\\left\\langle e_1, e_1\\right\\rangle+\\ldots+\\left|\\alpha_m\\right|^2\\left\\langle e_m, e_m\\right\\rangle \\\\\n\n& =\\left|\\alpha_1\\right|^2+\\ldots+\\left|\\alpha_m\\right|^2 \\\\\n\n& =\\left|\\left\\langle v, e_1\\right\\rangle\\right|^2+\\ldots+\\left|\\left\\langle v, e_m\\right\\rangle\\right|^2 .\n\n\\end{aligned}\n\n$$\n\n$\\Rightarrow$ Assume that $v \\notin \\operatorname{span}\\left(e_1, \\ldots, e_m\\right)$. Then, we must have\n\n$$\n\nv=v_{m+1}+\\frac{\\left\\langle v, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2} v_0,\n\n$$\n\nwhere $v_0=\\alpha_1 e_1+\\ldots+\\alpha_m e_m, \\alpha_k=\\left\\langle v, e_k\\right\\rangle, \\forall k \\in\\{1, \\ldots, m\\}$, and $v_{m+1}=v-$ $\\frac{\\left\\langle v, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2} v_0 \\neq 0$.\n\n\n\nWe have $\\left\\langle v_0, v_{m+1}\\right\\rangle=0$ (from which we get $\\left\\langle v, v_0\\right\\rangle=\\left\\langle v_0, v_0\\right\\rangle$ and $\\left\\langle v, v_{m+1}\\right\\rangle=$ $\\left.\\left\\langle v_{m+1}, v_{m+1}\\right\\rangle\\right)$. Now,\n\n$$\n\n\\begin{aligned}\n\n\\|v\\|^2 & =\\langle v, v\\rangle \\\\\n\n& =\\left\\langle v, v_{m+1}+\\frac{\\left\\langle v, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2} v_0\\right\\rangle \\\\\n\n& =\\left\\langle v, v_{m+1}\\right\\rangle+\\left\\langle v, \\frac{\\left\\langle v, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2} v_0\\right\\rangle \\\\\n\n& =\\left\\langle v_{m+1}, v_{m+1}\\right\\rangle+\\frac{\\left\\langle v_0, v_0\\right\\rangle}{\\left\\|v_0\\right\\|^2}\\left\\langle v_0, v_0\\right\\rangle \\\\\n\n& =\\left\\|v_{m+1}\\right\\|^2+\\left\\|v_0\\right\\|^2 \\\\\n\n& >\\left\\|v_0\\right\\|^2 \\\\\n\n& =\\left|\\alpha_1\\right|^2+\\ldots+\\left|\\alpha_m\\right|^2 \\\\\n\n& =\\left|\\left\\langle v, e_1\\right\\rangle\\right|^2+\\ldots+\\left|\\left\\langle v, e_m\\right\\rangle\\right|^2 .\n\n\\end{aligned}\n\n$$\n\nBy contrapositive, if $\\left\\|v_1\\right\\|^2=\\left|\\left\\langle v, e_1\\right\\rangle\\right|^2+\\ldots+\\left|\\left\\langle v, e_m\\right\\rangle\\right|^2$, then $v \\in \\operatorname{span}\\left(e_1, \\ldots, e_m\\right)$.\n\n\\end{proof}"}
 {"id": "Axler|exercise_7_5", "formal_statement": "theorem exercise_7_5 {V : Type*} [inner_product_space \u2102 V] \n  [finite_dimensional \u2102 V] (hV : finrank V \u2265 2) :\n  \u2200 U : submodule \u2102 (End \u2102 V), U.carrier \u2260\n  {T | T * T.adjoint = T.adjoint * T} :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Show that if $\\operatorname{dim} V \\geq 2$, then the set of normal operators on $V$ is not a subspace of $\\mathcal{L}(V)$.", "nl_proof": "\\begin{proof}\n\n    First off, suppose that $\\operatorname{dim} V \\geq 2$. Next let $\\left(e_1, \\ldots, e_n\\right)$ be an orthonormal basis of $V$. Now, define $S, T \\in L(V)$ by both $S\\left(a_1 e_1+\\ldots+a_n e_n\\right)=a_2 e_1-a_1 e_2$ and $T\\left(a_1 e_1+\\ldots+\\right.$ $\\left.a_n e_n\\right)=a_2 e_1+a_1 e_2$. So, just by now doing a simple calculation verifies that $S^*\\left(a_1 e_1+\\right.$ $\\left.\\ldots+a_n e_n\\right)=-a_2 e_1+a_1 e_2$\n\n\n\nNow, based on this formula, another calculation would show that $S S^*=S^* S$. Another simple calculation would that that $T$ is self-adjoint. Therefore, both $S$ and $T$ are normal. However, $S+T$ is given by the formula of $(S+T)\\left(a_1 e_1+\\ldots+a_n e_n\\right)=2 a_2 e_1$. In this case, a simple calculator verifies that $(S+T)^*\\left(a_1 e_1+\\ldots+a_n e_n\\right)=2 a_1 e_2$.\n\n\n\nTherefore, there is a final simple calculation that shows that $(S+T)(S+T)^* \\neq(S+$ $T)^*(S+T)$. So, in other words, $S+T$ isn't normal. Thereofre, the set of normal operators on $V$ isn't closed under addition and hence isn't a subspace of $L(V)$.\n\n\\end{proof}"}
 {"id": "Axler|exercise_7_9", "formal_statement": "theorem exercise_7_9 {V : Type*} [inner_product_space \u2102 V]\n  [finite_dimensional \u2102 V] (T : End \u2102 V)\n  (hT : T * T.adjoint = T.adjoint * T) :\n  is_self_adjoint T \u2194 \u2200 e : T.eigenvalues, (e : \u2102).im = 0 :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that a normal operator on a complex inner-product space is self-adjoint if and only if all its eigenvalues are real.", "nl_proof": "\\begin{proof}\n\n    First off, suppose $V$ is a complex inner product space and $T \\in L(V)$ is normal. If $T$ is self-adjoint, then all its eigenvalues are real. So, conversely, let all of the eigenvalues of $T$ be real. By the complex spectral theorem, there's an orthonormal basis $\\left(e_1, \\ldots, e_n\\right)$ of $V$ consisting of eigenvectors of $T$. Thus, there exists real numbers $\\lambda_1, \\ldots, \\lambda_n$ such that $T e_j=\\lambda_j e_j$ for $j=$ $1, \\ldots, n$.\n\nThe matrix of $T$ with respect to the basis of $\\left(e_1, \\ldots, e_n\\right)$ is the diagonal matrix with $\\lambda_1, \\ldots, \\lambda_n$ on the diagonal. So, the matrix equals its conjugate transpose. Therefore, $T=T^*$. In other words, $T$ s self-adjoint.\n\n\\end{proof}"}
 {"id": "Axler|exercise_7_11", "formal_statement": "theorem exercise_7_11 {V : Type*} [inner_product_space \u2102 V]\n  [finite_dimensional \u2102 V] {T : End \u2102 V} (hT : T*T.adjoint = T.adjoint*T) :\n  \u2203 (S : End \u2102 V), S ^ 2 = T :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $V$ is a complex inner-product space. Prove that every normal operator on $V$ has a square root. (An operator $S \\in \\mathcal{L}(V)$ is called a square root of $T \\in \\mathcal{L}(V)$ if $S^{2}=T$.)", "nl_proof": "\\begin{proof}\n\n    Let $V$ be a complex inner product space.\n\nIt is known that an operator $S \\in \\mathcal{L}(V)$ is called a square root of $T \\in \\mathcal{L}(V)$ if\n\n$$\n\nS^2=T\n\n$$\n\nNow, suppose that $T$ is a normal operator on $V$.\n\nBy the Complex Spectral Theorem, there is $e_1, \\ldots, e_n$ an orthonormal basis of $V$ consisting of eigenvalues of $T$ and let $\\lambda_1, \\ldots, \\lambda_n$ denote their corresponding eigenvalues.\n\nDefine $S$ by\n\n$$\n\nS e_j=\\sqrt{\\lambda_j} e_j,\n\n$$\n\nfor each $j=1, \\ldots, n$.\n\nObviously, $S^2 e_j=\\lambda_j e_j=T e_j$.\n\nHence, $S^2=T$ so there exist a square root of $T$.\n\n\\end{proof}"}
@@ -110,7 +110,7 @@
 {"id": "Herstein|exercise_2_5_23", "formal_statement": "theorem exercise_2_5_23 {G : Type*} [group G] \n  (hG : \u2200 (H : subgroup G), H.normal) (a b : G) :\n  \u2203 (j : \u2124) , b*a = a^j * b:=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a group such that all subgroups of $G$ are normal in $G$. If $a, b \\in G$, prove that $ba = a^jb$ for some $j$.", "nl_proof": "\\begin{proof}\n\n    Let $G$ be a group where each subgroup is normal in $G$. let $a, b \\in G$.\n\n$$\n\n\\begin{aligned}\n\n    \\langle a\\rangle\\triangleright  G  &\\Rightarrow b \\cdot\\langle a\\rangle=\\langle a\\rangle \\cdot b . \\\\\n\n& \\Rightarrow \\quad b \\cdot a=a^j \\cdot b \\text { for some } j \\in \\mathbb{Z}.\n\n\\end{aligned}\n\n$$\n\n(hence for $a_1 b \\in G \\quad a^j b=b \\cdot a$ ).\n\n\\end{proof}"}
 {"id": "Herstein|exercise_2_5_31", "formal_statement": "theorem exercise_2_5_31 {G : Type*} [comm_group G] [fintype G]\n  {p m n : \u2115} (hp : nat.prime p) (hp1 : \u00ac p \u2223 m) (hG : card G = p^n*m)\n  {H : subgroup G} [fintype H] (hH : card H = p^n) : \n  characteristic H :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Suppose that $G$ is an abelian group of order $p^nm$ where $p \\nmid m$ is a prime.  If $H$ is a subgroup of $G$ of order $p^n$, prove that $H$ is a characteristic subgroup of $G$.", "nl_proof": "\\begin{proof}\n\n    Let $G$ be an abelian group of order $p^n m$, such that $p \\nmid m$. Now, Given that $H$ is a subgroup of order $p^n$. Since $G$ is abelian $H$ is normal. Now we want to prove that $H$ is a characterestic subgroup, that is $\\phi(H)=H$ for any automorphism $\\phi$ of $G$. Now consider $\\phi(H)$. Clearly $|\\phi(H)|=p^n$. Suppose $\\phi(H) \\neq H$, then $|H \\cap \\phi(H)|=p^s$, where $s<n$. Consider $H \\phi(H)$, this is a subgroup of $G$ as $H$ is normal. Also $|H \\phi(H)|=\\frac{|H||\\phi(H)|}{|H \\cap \\phi(H)|}=\\frac{p^{2 n}}{p^s}=p^{2 n-s}$, where $2 n-s>n$. By lagrange's theorem then $p^{2 n-s}\\left|p^n m \\Longrightarrow p^{n-s}\\right| m \\Longrightarrow p \\mid m$-contradiction. So $\\phi(H)=H$, and $H$ is characterestic subgroup of $G$.\n\n\\end{proof}"}
 {"id": "Herstein|exercise_2_5_43", "formal_statement": "theorem exercise_2_5_43 (G : Type*) [group G] [fintype G]\n  (hG : card G = 9) :\n  comm_group G :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that a group of order 9 must be abelian.", "nl_proof": "\\begin{proof}\n\n    We use the result from problem 40 which is as follows: Suppose $G$ is a group, $H$ is a subgroup and $|G|=n$ and $n \\nmid\\left(i_G(H)\\right) !$. Then there exists a normal subgroup $\\$ K \\backslash$ neq $\\{$ e $\\} \\$$ and $K \\subseteq H$.\n\nSo, we have now a group $G$ of order 9. Suppose that $G$ is cyclic, then $G$ is abelian and there is nothing more to prove. Suppose that $G$ s not cyclic,then there exists an element $a$ of order 3 , and $A=\\langle a\\rangle$. Now $i_G(A)=3$, now $9 \\nmid 3$ !, hence by the above result there is a normal subgroup $K$, non-trivial and $K \\subseteq A$. But $|A|=3$, a prime order subgroup, hence has no non-trivial subgroup, so $K=A$. So $A$ is normal subgroup. Now since $G$ is not cyclic any non-identity element is of order 3.So Let $a(\\neq$ $e) \\in G$.Consider $A=\\langle a\\rangle$. As shown before $A$ is normal. $a$ commutes with any if its powers. Now Let $b \\in G$ such that $b \\notin A$. Then $b a b^{-1} \\in A$ and hence $b a b^{-1}=a^i$.This implies $a=b^3 a b^{-3}=a^{i^3} \\Longrightarrow a^{i^3-1}=e$. So, 3 divides $i^3-1$. Also by fermat's little theorem 3 divides $i^2-1$.So 3 divides $i-1$. But $0 \\leq i \\leq 2$. So $i=1$, is the only possibility and hence $a b=b a$. So $a \\in Z(G)$ as $b$ was arbitrary. Since $a$ was arbitrary $G=Z(G)$. Hence $G$ is abelian.\n\n\\end{proof}"}
-{"id": "Herstein|exercise_2_5_52", "formal_statement": "theorem exercise_2_5_52 {G : Type*} [group G] [fintype G]\n  (\u03c6 : G \u2243* G) {I : finset G} (hI : \u2200 x \u2208 I, \u03c6 x = x\u207b\u00b9)\n  (hI1 : 0.75 * card G \u2264 card I) : \n  \u2200 x : G, \u03c6 x = x\u207b\u00b9 \u2227 \u2200 x y : G, x*y = y*x :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a finite group and $\\varphi$ an automorphism of $G$ such that $\\varphi(x) = x^{-1}$ for more than three-fourths of the elements of $G$. Prove that $\\varphi(y) = y^{-1}$ for all $y \\in G$, and so $G$ is abelian.", "nl_proof": "\\begin{proof}\n\nLet us start with considering $b$ to be an arbitrary element in $A$. \n\n\n\n1. Show that $\\left|A \\cap\\left(b^{-1} A\\right)\\right|>\\frac{|G|}{2}$, where\n\n$$\n\nb^{-1} A=\\left\\{b^{-1} a \\mid a \\in A\\right\\}\n\n$$\n\nFirst notice that if we consider a map $f: A \\rightarrow b^{-1} A$ defined by $f(a)=b^{-1} a$, for all $a \\in A$, then $f$ is a 1-1 map and so $\\left|b^{-1} A\\right| \\geq|A|>\\frac{3}{4}|G|$. Now using inclusion-exclusion principle we have\n\n$$\n\n\\left|A \\cap\\left(b^{-1} A\\right)\\right|=|A|+\\left|b^{-1} A\\right|-\\left|A \\cup\\left(b^{-1} A\\right)\\right|>\\frac{3}{4}|G|+\\frac{3}{4}|G|-|G|=\\frac{1}{2}|G|\n\n$$\n\n2. Argue that $A \\cap\\left(b^{-1} A\\right) \\subseteq C(b)$, where $C(b)$ is the centralizer of $b$ in $G$.\n\n\n\nSuppose $x \\in A \\cap\\left(b^{-1} A\\right)$, that means, $x \\in A$ and $x \\in b^{-1} A$. Thus there exist an element $a \\in A$ such that $x=$ $b^{-1} a$, which gives us $x b=a \\in A$. Now notice that $x, b \\in A$ and $x b \\in A$, therefore we get\n\n$$\n\n\\phi(x b)=(x b)^{-1} \\Longrightarrow \\phi(x) \\phi(b)=(x b)^{-1} \\Longrightarrow x^{-1} b^{-1}=b^{-1} x^{-1} \\Longrightarrow x b=b x\n\n$$\n\nTherefore, we get $x b=b x$, for any $x \\in A \\cap\\left(b^{-1} A\\right)$, that means, $x \\in C(b)$.\n\n\n\n3. Argue that $C(b)=G$.\n\nWe know that centralizer of an element in a group $G$ is a subgroup (See Page 53). Therefore $C(b)$ is a subgroup of $G$. From statements $\\mathbf{1}$ and $\\mathbf{2}$, we have\n\n$$\n\n|C(b)| \\geq\\left|A \\cap\\left(b^{-1} A\\right)\\right|>\\frac{|G|}{2}\n\n$$\n\nWe need to use the following remark to argue $C(b)=G$ from the above step.\n\nRemark. Let $G$ be a finite group and $H$ be a subgroup with more then $|G| / 2$ elements then $H=G$.\n\n\n\nProof of Remark. Suppose $|H|=p$ Then by Lagrange Theorem, there exist an $n \\in \\mathbb{N}$, such that $|G|=n p$, as $|H|$ divide $|G|$. Now by hypothesis $p>\\frac{G]}{2}$ gives us,\n\n$$\n\np>\\frac{|G|}{2} \\Longrightarrow n p>\\frac{n|G|}{2} \\Longrightarrow n<2 \\Longrightarrow n=1\n\n$$\n\nTherefore we get $H=G$.\n\n\n\nNow notice that $C(b)$ is a subgroup of $G$ with $C(b)$ having more than $|G| / 2$ elements. Therefore, $C(b)=G$.\n\n\n\n4. Show that $A \\in Z(G)$.\n\n\n\nWe know that $x \\in Z(G)$ if and only if $C(a)=G$. Now notice that, for any $b \\in A$ we have $C(b)=G$. Therefore, every element of $A$ is in the center of $G$, that means, $A \\subseteq Z(G)$.\n\n\n\n5. 5how that $Z(G)=G$.\n\n\n\nAs it is given that $|A|>\\frac{3|G|}{4}$ and $A \\leq|Z(G)|$, therefore we get\n\n$$\n\n|Z(G)|>\\frac{3}{4}|G|>\\frac{1}{2}|G| .\n\n$$\n\nAs $Z(G)$ is a subgroup of $G$, so by the above Remark we have $Z(G)=G$. Hence $G$ is abelian.\n\n\n\n6. Finally show that $A=G$.\n\n\n\nFirst notice that $A$ is a subgroup of $G$. To show this let $p, q \\in A$. Then we have\n\n$$\n\n\\phi(p q)=\\phi(p) \\phi(q)=p^{-1} q^{-1}=(q p)^{-1}=(p q)^{-1}, \\quad \\text { As } G \\text { is abelian. }\n\n$$\n\nTherefore, $p q \\in A$ and so we have $A$ is a subgroup of $G$. Again by applying the above remark. we get $A=G$. Therefore we have\n\n$$\n\n\\phi(y)=y^{-1}, \\quad \\text { for all } y \\in G\n\n$$\n\n\n\n\\end{proof}"}
+{"id": "Herstein|exercise_2_5_52", "formal_statement": "theorem exercise_2_5_52 {G : Type*} [group G] [fintype G]\n  (\u03c6 : G \u2243* G) {I : finset G} (hI : \u2200 x \u2208 I, \u03c6 x = x\u207b\u00b9)\n  (hI1 : (0.75 : \u211a) * card G \u2264 card I) : \n  \u2200 x : G, \u03c6 x = x\u207b\u00b9 \u2227 \u2200 x y : G, x*y = y*x :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $G$ be a finite group and $\\varphi$ an automorphism of $G$ such that $\\varphi(x) = x^{-1}$ for more than three-fourths of the elements of $G$. Prove that $\\varphi(y) = y^{-1}$ for all $y \\in G$, and so $G$ is abelian.", "nl_proof": "\\begin{proof}\n\nLet us start with considering $b$ to be an arbitrary element in $A$. \n\n\n\n1. Show that $\\left|A \\cap\\left(b^{-1} A\\right)\\right|>\\frac{|G|}{2}$, where\n\n$$\n\nb^{-1} A=\\left\\{b^{-1} a \\mid a \\in A\\right\\}\n\n$$\n\nFirst notice that if we consider a map $f: A \\rightarrow b^{-1} A$ defined by $f(a)=b^{-1} a$, for all $a \\in A$, then $f$ is a 1-1 map and so $\\left|b^{-1} A\\right| \\geq|A|>\\frac{3}{4}|G|$. Now using inclusion-exclusion principle we have\n\n$$\n\n\\left|A \\cap\\left(b^{-1} A\\right)\\right|=|A|+\\left|b^{-1} A\\right|-\\left|A \\cup\\left(b^{-1} A\\right)\\right|>\\frac{3}{4}|G|+\\frac{3}{4}|G|-|G|=\\frac{1}{2}|G|\n\n$$\n\n2. Argue that $A \\cap\\left(b^{-1} A\\right) \\subseteq C(b)$, where $C(b)$ is the centralizer of $b$ in $G$.\n\n\n\nSuppose $x \\in A \\cap\\left(b^{-1} A\\right)$, that means, $x \\in A$ and $x \\in b^{-1} A$. Thus there exist an element $a \\in A$ such that $x=$ $b^{-1} a$, which gives us $x b=a \\in A$. Now notice that $x, b \\in A$ and $x b \\in A$, therefore we get\n\n$$\n\n\\phi(x b)=(x b)^{-1} \\Longrightarrow \\phi(x) \\phi(b)=(x b)^{-1} \\Longrightarrow x^{-1} b^{-1}=b^{-1} x^{-1} \\Longrightarrow x b=b x\n\n$$\n\nTherefore, we get $x b=b x$, for any $x \\in A \\cap\\left(b^{-1} A\\right)$, that means, $x \\in C(b)$.\n\n\n\n3. Argue that $C(b)=G$.\n\nWe know that centralizer of an element in a group $G$ is a subgroup (See Page 53). Therefore $C(b)$ is a subgroup of $G$. From statements $\\mathbf{1}$ and $\\mathbf{2}$, we have\n\n$$\n\n|C(b)| \\geq\\left|A \\cap\\left(b^{-1} A\\right)\\right|>\\frac{|G|}{2}\n\n$$\n\nWe need to use the following remark to argue $C(b)=G$ from the above step.\n\nRemark. Let $G$ be a finite group and $H$ be a subgroup with more then $|G| / 2$ elements then $H=G$.\n\n\n\nProof of Remark. Suppose $|H|=p$ Then by Lagrange Theorem, there exist an $n \\in \\mathbb{N}$, such that $|G|=n p$, as $|H|$ divide $|G|$. Now by hypothesis $p>\\frac{G]}{2}$ gives us,\n\n$$\n\np>\\frac{|G|}{2} \\Longrightarrow n p>\\frac{n|G|}{2} \\Longrightarrow n<2 \\Longrightarrow n=1\n\n$$\n\nTherefore we get $H=G$.\n\n\n\nNow notice that $C(b)$ is a subgroup of $G$ with $C(b)$ having more than $|G| / 2$ elements. Therefore, $C(b)=G$.\n\n\n\n4. Show that $A \\in Z(G)$.\n\n\n\nWe know that $x \\in Z(G)$ if and only if $C(a)=G$. Now notice that, for any $b \\in A$ we have $C(b)=G$. Therefore, every element of $A$ is in the center of $G$, that means, $A \\subseteq Z(G)$.\n\n\n\n5. 5how that $Z(G)=G$.\n\n\n\nAs it is given that $|A|>\\frac{3|G|}{4}$ and $A \\leq|Z(G)|$, therefore we get\n\n$$\n\n|Z(G)|>\\frac{3}{4}|G|>\\frac{1}{2}|G| .\n\n$$\n\nAs $Z(G)$ is a subgroup of $G$, so by the above Remark we have $Z(G)=G$. Hence $G$ is abelian.\n\n\n\n6. Finally show that $A=G$.\n\n\n\nFirst notice that $A$ is a subgroup of $G$. To show this let $p, q \\in A$. Then we have\n\n$$\n\n\\phi(p q)=\\phi(p) \\phi(q)=p^{-1} q^{-1}=(q p)^{-1}=(p q)^{-1}, \\quad \\text { As } G \\text { is abelian. }\n\n$$\n\nTherefore, $p q \\in A$ and so we have $A$ is a subgroup of $G$. Again by applying the above remark. we get $A=G$. Therefore we have\n\n$$\n\n\\phi(y)=y^{-1}, \\quad \\text { for all } y \\in G\n\n$$\n\n\n\n\\end{proof}"}
 {"id": "Herstein|exercise_2_7_7", "formal_statement": "theorem exercise_2_7_7 {G : Type*} [group G] {G' : Type*} [group G']\n  (\u03c6 : G \u2192* G') (N : subgroup G) [N.normal] : \n  (map \u03c6 N).normal  :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $\\varphi$ is a homomorphism of $G$ onto $G'$ and $N \\triangleleft G$, show that $\\varphi(N) \\triangleleft G'$.", "nl_proof": "\\begin{proof}\n\nWe first claim that $\\varphi(N)$ is a subgroup of $G'$. To see this, note that since $N$ is a subgroup of $G$, the identity element $e_G$ of $G$ belongs to $N$. Therefore, the element $\\varphi(e_G) \\in \\varphi(N)$, so $\\varphi(N)$ is a non-empty subset of $G'$.\n\n\n\nNow, let $a', b' \\in \\varphi(N)$. Then there exist elements $a, b \\in N$ such that $\\varphi(a) = a'$ and $\\varphi(b) = b'$. Since $N$ is a subgroup of $G$, we have $a, b \\in N$, so $ab^{-1} \\in N$. Thus, we have\n\n$$\\varphi(ab^{-1}) = \\varphi(a) \\varphi(b^{-1}) = a'b'^{-1} \\in \\varphi(N),$$\n\nwhich shows that $a', b' \\in \\varphi(N)$ implies $a'b'^{-1} \\in \\varphi(N)$. Therefore, $\\varphi(N)$ is a subgroup of $G'$.\n\n\n\nNext, we will show that $\\varphi(N)$ is a normal subgroup of $G'$. Let $\\varphi(N) = N'$, a subgroup of $G'$. Let $x' \\in G'$ and $h' \\in N'$. Since $\\varphi$ is onto, there exist elements $x \\in G$ and $h \\in N$ such that $\\varphi(x) = x'$ and $\\varphi(h) = h'$.\n\n\n\nSince $N$ is a normal subgroup of $G$, we have $xhx^{-1} \\in N$. Thus,\n\n$$\\varphi(xhx^{-1}) = \\varphi(x)\\varphi(h)\\varphi(x^{-1}) = x'h'x'^{-1} \\in \\varphi(N),$$\n\nwhich shows that $x' \\in G'$ and $h' \\in N'$ implies $x'h'x'^{-1} \\in \\varphi(N)$. Therefore, $\\varphi(N)$ is a normal subgroup of $G'$. This completes the proof.\n\n\\end{proof}"}
 {"id": "Herstein|exercise_2_8_15", "formal_statement": "theorem exercise_2_8_15 {G H: Type*} [fintype G] [group G] [fintype H]\n  [group H] {p q : \u2115} (hp : nat.prime p) (hq : nat.prime q) \n  (h : p > q) (h1 : q \u2223 p - 1) (hG : card G = p*q) (hH : card G = p*q) :\n  G \u2243* H :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $p > q$ are two primes such that $q \\mid p - 1$, then any two nonabelian groups of order $pq$ are isomorphic.", "nl_proof": "\\begin{proof}\n\n    For a nonabelian group of order $p q$, the structure of the group $G$ is set by determining the relation $a b a^{-1}=b^{k^{\\frac{p-1}{q}}}$ for some generator $k$ of the cyclic group. Here we are using the fact that $k^{\\frac{p-1}{q}}$ is a generator for the unique subgroup of order $q$ in $U_p$ (a cyclic group of order $m$ has a unique subgroup of order $d$ for each divisor $d$ of $m$ ). The other possible generators of this subgroup are $k^{\\frac{l(p-1)}{q}}$ for each $1 \\leq l \\leq q-1$, so these give potentially new group structures. Let $G^{\\prime}$ be a group with an element $c$ of order $q$, an element $d$ of order $p$ with structure defined by the relation $c d c^{-1}=d^{k^{\\frac{l(p-1)}{q}}}$. We may then define\n\n$$\n\n\\begin{aligned}\n\n\\phi: G^{\\prime} & \\rightarrow G \\\\\n\nc & \\mapsto a^l \\\\\n\nd & \\mapsto b\n\n\\end{aligned}\n\n$$\n\nsince $c$ and $a^l$ have the same order and $b$ and $d$ have the same order this is a well defined function.\n\nSince\n\n$$\n\n\\begin{aligned}\n\n\\phi(c) \\phi(d) \\phi(c)^{-1} & =a^l b a^{-l} \\\\\n\n& =b^{\\left(k^{\\frac{p-1}{q}}\\right)^l} \\\\\n\n& =b^{k^{\\frac{l(p-1)}{q}}} \\\\\n\n& =\\phi(d)^{k^{\\frac{l(p-1)}{q}}}\n\n\\end{aligned}\n\n$$\n\n$\\phi\\left(c^i d^j\\right)=a^{l i} b^j=e$ only if $i=j=0$, so $\\phi$ is 1-to-l. Therefore $G$ and $G^{\\prime}$ are isomorphic and so up to isomorphism there is only one nonabelian group of order $p q$.\n\n\\end{proof}"}
 {"id": "Herstein|exercise_2_10_1", "formal_statement": "theorem exercise_2_10_1 {G : Type*} [group G] (A : subgroup G) \n  [A.normal] {b : G} (hp : nat.prime (order_of b)) :\n  A \u2293 (closure {b}) = \u22a5 :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $A$ be a normal subgroup of a group $G$, and suppose that $b \\in G$ is an element of prime order $p$, and that $b \\not\\in A$. Show that $A \\cap (b) = (e)$.", "nl_proof": "\\begin{proof}\n\nIf $b \\in G$ has order $p$, then $(b)$ is a cyclic group of order $p$. Since $A$ is a subgroup of $G$, we have $A \\cap (b)$ is a subgroup of $G$. Also, $A \\cap (b) \\subseteq (b)$. So $A \\cap (b)$ is a subgroup of $(b)$. Since $(b)$ is a cyclic group of order $p$, the only subgroups of $(b)$ are $(e)$ and $(b)$ itself.\n\n\n\nTherefore, either $A \\cap (b) = (e)$ or $A \\cap (b) = (b)$. If $A \\cap (b) = (e)$, then we are done. Otherwise, if $A \\cap (b) = (b)$, then $A \\subseteq (b)$. Since $A$ is a subgroup of $G$ and $A \\subseteq (b)$, it follows that $A$ is a subgroup of $(b)$.\n\n\n\nSince the only subgroups of $(b)$ are $(e)$ and $(b)$ itself, we have either $A = (e)$ or $A = (b)$. If $A = (e)$, then $A \\cap (b) = (e)$ and we are done. But if $A = (b)$, then $b \\in A$ as $b \\in (b)$, which contradicts our hypothesis that $b \\notin A$. So $A \\neq (b)$.\n\n\n\nHence $A \\cap (b) \\neq (b)$. Therefore, $A \\cap (b) = (e)$. This completes our proof.\n\n\\end{proof}"}
@@ -129,7 +129,7 @@
 {"id": "Artin|exercise_2_3_2", "formal_statement": "theorem exercise_2_3_2 {G : Type*} [group G] (a b : G) :\n  \u2203 g : G, b* a = g * a * b * g\u207b\u00b9 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the products $a b$ and $b a$ are conjugate elements in a group.", "nl_proof": "\\begin{proof}\n\n    We have that $(a^{-1})ab(a^{-1})^{-1} = ba$. \n\n\\end{proof}"}
 {"id": "Artin|exercise_2_8_6", "formal_statement": "theorem exercise_2_8_6 {G H : Type*} [group G] [group H] :\n  center (G \u00d7 H) \u2243* (center G) \u00d7 (center H) :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the center of the product of two groups is the product of their centers.", "nl_proof": "\\begin{proof}\n\n    We have that $(g_1, g_2)\\cdot (h_1, h_2) = (h_1, h_2)\\cdot (g_1, g_2)$ if and only if $g_1h_1 = h_1g_1$ and $g_2h_2 = h_2g_2$. \n\n\\end{proof}"}
 {"id": "Artin|exercise_3_2_7", "formal_statement": "theorem exercise_3_2_7 {F : Type*} [field F] {G : Type*} [field G]\n  (\u03c6 : F \u2192+* G) : injective \u03c6 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that every homomorphism of fields is injective.", "nl_proof": "\\begin{proof}\n\n    Suppose $f(a)=f(b)$, then $f(a-b)=0=f(0)$. If $u=(a-b) \\neq 0$, then $f(u) f\\left(u^{-1}\\right)=f(1)=1$, but that means that $0 f\\left(u^{-1}\\right)=1$, which is impossible. Hence $a-b=0$ and $a=b$.\n\n\\end{proof}"}
-{"id": "Artin|exercise_3_7_2", "formal_statement": "theorem exercise_3_7_2 {K V : Type*} [field K] [add_comm_group V]\n  [module K V] {\u03b9 : Type*} [fintype \u03b9] (\u03b3 : \u03b9 \u2192 submodule K V)\n  (h : \u2200 i : \u03b9, \u03b3 \u03b9 \u2260 \u22a4) :\n  (\u22c2 (i : \u03b9), (\u03b3 i : set V)) \u2260 \u22a4 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $V$ be a vector space over an infinite field $F$. Prove that $V$ is not the union of finitely many proper subspaces.", "nl_proof": "\\begin{proof}\n\n    If $V$ is the set-theoretic union of $n$ proper subspaces $W_i$ ( $1 \\leq i \\leq n$ ), then $|F| \\leq n-1$.\n\nProof. We may suppose no $W_i$ is contained in the union of the other subspaces. Let $u \\in W_i, \\quad u \\notin \\bigcup_{j \\neq i} W_j$ and $v \\notin W_i$.\n\nThen $(v+F u) \\cap W_i=\\varnothing$ and $(v+F u) \\cap W_j(j \\neq i)$ contains at most one vector since otherwise $W_j$ would contain $u$. Hence\n\n$$\n\n|v+F u|=|F| \\leq n-1 .\n\n$$\n\nCorollary: Avoidance lemma for vector spaces.\n\nLet $E$ be a vector space over an infinite field. If a subspace is contained in a finite union of subspaces, it is contained in one of them.\n\n\\end{proof}"}
+{"id": "Artin|exercise_3_7_2", "formal_statement": "theorem exercise_3_7_2 {K V : Type*} [field K] [add_comm_group V]\n  [module K V] {\u03b9 : Type*} [fintype \u03b9] (\u03b3 : \u03b9 \u2192 submodule K V)\n  (h : \u2200 i : \u03b9, \u03b3 i \u2260 \u22a4) :\n  (\u22c2 (i : \u03b9), (\u03b3 i : set V)) \u2260 \u22a4 :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $V$ be a vector space over an infinite field $F$. Prove that $V$ is not the union of finitely many proper subspaces.", "nl_proof": "\\begin{proof}\n\n    If $V$ is the set-theoretic union of $n$ proper subspaces $W_i$ ( $1 \\leq i \\leq n$ ), then $|F| \\leq n-1$.\n\nProof. We may suppose no $W_i$ is contained in the union of the other subspaces. Let $u \\in W_i, \\quad u \\notin \\bigcup_{j \\neq i} W_j$ and $v \\notin W_i$.\n\nThen $(v+F u) \\cap W_i=\\varnothing$ and $(v+F u) \\cap W_j(j \\neq i)$ contains at most one vector since otherwise $W_j$ would contain $u$. Hence\n\n$$\n\n|v+F u|=|F| \\leq n-1 .\n\n$$\n\nCorollary: Avoidance lemma for vector spaces.\n\nLet $E$ be a vector space over an infinite field. If a subspace is contained in a finite union of subspaces, it is contained in one of them.\n\n\\end{proof}"}
 {"id": "Artin|exercise_6_4_2", "formal_statement": "theorem exercise_6_4_2 {G : Type*} [group G] [fintype G] {p q : \u2115}\n  (hp : prime p) (hq : prime q) (hG : card G = p*q) :\n  is_simple_group G \u2192 false :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that no group of order $p q$, where $p$ and $q$ are prime, is simple.", "nl_proof": "\\begin{proof}\n\n    If $|G|=n=p q$ then the only two Sylow subgroups are of order $p$ and $q$.\n\nFrom Sylow's third theorem we know that $n_p \\mid q$ which means that $n_p=1$ or $n_p=q$.\n\nIf $n_p=1$ then we are done (by a corollary of Sylow's theorem)\n\nIf $n_p=q$ then we have accounted for $q(p-1)=p q-q$ elements of $G$ and so there is only one group of order $q$ and again we are done.\n\n\\end{proof}"}
 {"id": "Artin|exercise_6_4_12", "formal_statement": "theorem exercise_6_4_12 {G : Type*} [group G] [fintype G]\n  (hG : card G = 224) :\n  is_simple_group G \u2192 false :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that no group of order 224 is simple.", "nl_proof": "\\begin{proof}\n\n    The following proves there must exist a normal Sylow 2 -subgroup of order 32 ,\n\nSuppose there are $n_2=7$ Sylow 2 -subgroups in $G$. Making $G$ act on the set of these Sylow subgroups by conjugation (Mitt wrote about this but on the set of the other Sylow subgroups, which gives no contradiction), we get a homomorphism $G \\rightarrow S_7$ which must be injective if $G$ is simple (why?).\n\n\n\nBut this cannot be since then we would embed $G$ into $S_7$, which is impossible since $|G| \\nmid 7 !=\\left|S_7\\right|$ (why?)\n\n\\end{proof}"}
 {"id": "Artin|exercise_10_1_13", "formal_statement": "theorem exercise_10_1_13 {R : Type*} [ring R] {x : R}\n  (hx : is_nilpotent x) : is_unit (1 + x) :=", "src_header": "import .common \n\nopen function fintype subgroup ideal polynomial submodule zsqrtd char_p\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "An element $x$ of a ring $R$ is called nilpotent if some power of $x$ is zero. Prove that if $x$ is nilpotent, then $1+x$ is a unit in $R$.", "nl_proof": "\\begin{proof}\n\n    If $x^n=0$, then\n\n$$\n\n(1+x)\\left(\\sum_{k=0}^{n-1}(-1)^k x^k\\right)=1+(-1)^{n-1} x^n=1 .\n\n$$\n\n\\end{proof}"}
@@ -143,7 +143,7 @@
 {"id": "Dummit-Foote|exercise_1_1_2a", "formal_statement": "theorem exercise_1_1_2a : \u2203 a b : \u2124, a - b \u2260 b - a :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove the the operation $\\star$ on $\\mathbb{Z}$ defined by $a\\star b=a-b$ is not commutative.", "nl_proof": "\\begin{proof}\n\n    Not commutative since\n\n$$\n\n1 \\star(-1)=1-(-1)=2\n\n$$\n\n$$\n\n(-1) \\star 1=-1-1=-2 .\n\n$$\n\n\\end{proof}"}
 {"id": "Dummit-Foote|exercise_1_1_4", "formal_statement": "theorem exercise_1_1_4 (n : \u2115) : \n  \u2200 (a b c : \u2115), (a * b) * c \u2261 a * (b * c) [ZMOD n] :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that the multiplication of residue class $\\mathbb{Z}/n\\mathbb{Z}$ is associative.", "nl_proof": "\\begin{proof}\n\n    We have\n\n$$\n\n\\begin{aligned}\n\n(\\bar{a} \\cdot \\bar{b}) \\cdot \\bar{c} &=\\overline{a \\cdot b} \\cdot \\bar{c} \\\\\n\n&=\\overline{(a \\cdot b) \\cdot c} \\\\\n\n&=\\overline{a \\cdot(b \\cdot c)} \\\\\n\n&=\\bar{a} \\cdot \\overline{b \\cdot c} \\\\\n\n&=\\bar{a} \\cdot(\\bar{b} \\cdot \\bar{c})\n\n\\end{aligned}\n\n$$\n\nsince integer multiplication is associative.\n\n\\end{proof}"}
 {"id": "Dummit-Foote|exercise_1_1_15", "formal_statement": "theorem exercise_1_1_15 {G : Type*} [group G] (as : list G) :\n  as.prod\u207b\u00b9 = (as.reverse.map (\u03bb x, x\u207b\u00b9)).prod :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $(a_1a_2\\dots a_n)^{-1} = a_n^{-1}a_{n-1}^{-1}\\dots a_1^{-1}$ for all $a_1, a_2, \\dots, a_n\\in G$.", "nl_proof": "\\begin{proof}\n\n    For $n=1$, note that for all $a_1 \\in G$ we have $a_1^{-1}=a_1^{-1}$.\n\nNow for $n \\geq 2$ we proceed by induction on $n$. For the base case, note that for all $a_1, a_2 \\in G$ we have\n\n$$\n\n\\left(a_1 \\cdot a_2\\right)^{-1}=a_2^{-1} \\cdot a_1^{-1}\n\n$$\n\nsince\n\n$$\n\na_1 \\cdot a_2 \\cdot a_2^{-1} a_1^{-1}=1 .\n\n$$\n\nFor the inductive step, suppose that for some $n \\geq 2$, for all $a_i \\in G$ we have\n\n$$\n\n\\left(a_1 \\cdot \\ldots \\cdot a_n\\right)^{-1}=a_n^{-1} \\cdot \\ldots \\cdot a_1^{-1} .\n\n$$\n\nThen given some $a_{n+1} \\in G$, we have\n\n$$\n\n\\begin{aligned}\n\n\\left(a_1 \\cdot \\ldots \\cdot a_n \\cdot a_{n+1}\\right)^{-1} &=\\left(\\left(a_1 \\cdot \\ldots \\cdot a_n\\right) \\cdot a_{n+1}\\right)^{-1} \\\\\n\n&=a_{n+1}^{-1} \\cdot\\left(a_1 \\cdot \\ldots \\cdot a_n\\right)^{-1} \\\\\n\n&=a_{n+1}^{-1} \\cdot a_n^{-1} \\cdot \\ldots \\cdot a_1^{-1},\n\n\\end{aligned}\n\n$$\n\nusing associativity and the base case where necessary.\n\n\\end{proof}"}
-{"id": "Dummit-Foote|exercise_1_1_17", "formal_statement": "theorem exercise_1_1_17 {G : Type*} [group G] {x : G} {n : \u2115}\n  (hxn: order_of x = n) :\n  x\u207b\u00b9 = x ^ (n-1) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $x$ be an element of $G$. Prove that if $|x|=n$ for some positive integer $n$ then $x^{-1}=x^{n-1}$.", "nl_proof": "\\begin{proof}\n\n    We have $x \\cdot x^{n-1}=x^n=1$, so by the uniqueness of inverses $x^{-1}=x^{n-1}$.\n\n\\end{proof}"}
+{"id": "Dummit-Foote|exercise_1_1_17", "formal_statement": "theorem exercise_1_1_17 {G : Type*} [group G] {x : G} {n : \u2115}\n  (hxn: order_of x = n) :\n  x\u207b\u00b9 = x ^ (n - 1 : \u2124) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $x$ be an element of $G$. Prove that if $|x|=n$ for some positive integer $n$ then $x^{-1}=x^{n-1}$.", "nl_proof": "\\begin{proof}\n\n    We have $x \\cdot x^{n-1}=x^n=1$, so by the uniqueness of inverses $x^{-1}=x^{n-1}$.\n\n\\end{proof}"}
 {"id": "Dummit-Foote|exercise_1_1_20", "formal_statement": "theorem exercise_1_1_20 {G : Type*} [group G] {x : G} :\n  order_of x = order_of x\u207b\u00b9 :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "For $x$ an element in $G$ show that $x$ and $x^{-1}$ have the same order.", "nl_proof": "\\begin{proof}\n\n    Recall that the order of a group element is either a positive integer or infinity.\n\nSuppose $|x|$ is infinite and that $\\left|x^{-1}\\right|=n$ for some $n$. Then\n\n$$\n\nx^n=x^{(-1) \\cdot n \\cdot(-1)}=\\left(\\left(x^{-1}\\right)^n\\right)^{-1}=1^{-1}=1,\n\n$$\n\na contradiction. So if $|x|$ is infinite, $\\left|x^{-1}\\right|$ must also be infinite. Likewise, if $\\left|x^{-1}\\right|$ is infinite, then $\\left|\\left(x^{-1}\\right)^{-1}\\right|=|x|$ is also infinite.\n\nSuppose now that $|x|=n$ and $\\left|x^{-1}\\right|=m$ are both finite. Then we have\n\n$$\n\n\\left(x^{-1}\\right)^n=\\left(x^n\\right)^{-1}=1^{-1}=1,\n\n$$\n\nso that $m \\leq n$. Likewise, $n \\leq m$. Hence $m=n$ and $x$ and $x^{-1}$ have the same order.\n\n\\end{proof}"}
 {"id": "Dummit-Foote|exercise_1_1_22b", "formal_statement": "theorem exercise_1_1_22b {G: Type*} [group G] (a b : G) : \n  order_of (a * b) = order_of (b * a) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Deduce that $|a b|=|b a|$ for all $a, b \\in G$.", "nl_proof": "\\begin{proof}\n\n    Let $a$ and $b$ be arbitrary group elements. Letting $x=a b$ and $g=a$, we see that\n\n$$\n\n|a b|=\\left|a^{-1} a b a\\right|=|b a| .\n\n$$\n\n\\end{proof}"}
 {"id": "Dummit-Foote|exercise_1_1_29", "formal_statement": "theorem exercise_1_1_29 {A B : Type*} [group A] [group B] :\n  \u2200 x y : A \u00d7 B, x*y = y*x \u2194 (\u2200 x y : A, x*y = y*x) \u2227 \n  (\u2200 x y : B, x*y = y*x) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $A \\times B$ is an abelian group if and only if both $A$ and $B$ are abelian.", "nl_proof": "\\begin{proof}\n\n    $(\\Rightarrow)$ Suppose $a_1, a_2 \\in A$ and $b_1, b_2 \\in B$. Then\n\n$$\n\n\\left(a_1 a_2, b_1 b_2\\right)=\\left(a_1, b_1\\right) \\cdot\\left(a_2, b_2\\right)=\\left(a_2, b_2\\right) \\cdot\\left(a_1, b_1\\right)=\\left(a_2 a_1, b_2 b_1\\right) .\n\n$$\n\nSince two pairs are equal precisely when their corresponding entries are equal, we have $a_1 a_2=a_2 a_1$ and $b_1 b_2=b_2 b_1$. Hence $A$ and $B$ are abelian.\n\n$(\\Leftarrow)$ Suppose $\\left(a_1, b_1\\right),\\left(a_2, b_2\\right) \\in A \\times B$. Then we have\n\n$$\n\n\\left(a_1, b_1\\right) \\cdot\\left(a_2, b_2\\right)=\\left(a_1 a_2, b_1 b_2\\right)=\\left(a_2 a_1, b_2 b_1\\right)=\\left(a_2, b_2\\right) \\cdot\\left(a_1, b_1\\right) .\n\n$$\n\nHence $A \\times B$ is abelian.\n\n\\end{proof}"}

valid.jsonl

@@ -65,8 +65,8 @@
 {"id": "Axler|exercise_5_4", "formal_statement": "theorem exercise_5_4 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] (S T : V \u2192\u2097[F] V) (hST : S \u2218 T = T \u2218 S) (c : F):\n  map S (T - c \u2022 id).ker = (T - c \u2022 id).ker :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose that $S, T \\in \\mathcal{L}(V)$ are such that $S T=T S$. Prove that $\\operatorname{null} (T-\\lambda I)$ is invariant under $S$ for every $\\lambda \\in \\mathbf{F}$.", "nl_proof": "\\begin{proof}\n\n    First off, fix $\\lambda \\in F$. Secondly, let $v \\in \\operatorname{null}(T-\\lambda I)$. If so, then $(T-\\lambda I)(S v)=T S v-\\lambda S v=$ $S T v-\\lambda S v=S(T v-\\lambda v)=0$. Therefore, $S v \\in \\operatorname{null}(T-\\lambda I)$ since $n u l l(T-\\lambda I)$ is actually invariant under $S$.\n\n\\end{proof}"}
 {"id": "Axler|exercise_5_12", "formal_statement": "theorem exercise_5_12 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] {S : End F V}\n  (hS : \u2200 v : V, \u2203 c : F, v \u2208 eigenspace S c) :\n  \u2203 c : F, S = c \u2022 id :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $T \\in \\mathcal{L}(V)$ is such that every vector in $V$ is an eigenvector of $T$. Prove that $T$ is a scalar multiple of the identity operator.", "nl_proof": "\\begin{proof}\n\n    For every single $v \\in V$, there does exist $a_v \\in F$ such that $T v=a_v v$. Since $T 0=0$, then we have to make $a_0$ be the any number in F. However, for every single $v \\in V\\{0\\}$, then the value of $a_V$ is uniquely determined by the previous equation of $T v=a_v v$.\n\n\n\nNow, to show that $T$ is a scalar multiple of the identity, then me must show that $a_v$ is independent of $v$ for $v \\in V\\{0\\}$. We would now want to show that $a_v=a_w$.\n\n\n\nFirst, just make the case of where $(v, w)$ is linearly dependent. Then, there does exist $b \\in F$ such that $w=b v$. Now, you would have the following: $a_W w=T w=T(b v)=b T v=b\\left(a_v v\\right)=a_v w$. This is showing that $a_v=a_w$.\n\nFinally, make the consideration to make $(v, w)$ be linearly independent. Now, we would have the following: $\\left.a_{(} v+w\\right)(v+w)=T(v+w)=T v+T w=a_v v+a_w w$.\n\n\n\nThat previous equation implies the following: $\\left.\\left.\\left(a_{(} v+w\\right)-a_v\\right) v+\\left(a_{(} v+w\\right)-a_w\\right) w=0$. Since $(v, w)$ is linearly independent, this would imply that both $\\left.a_{(} v+w\\right)=a_v$ and $\\left.a_{(} v+w\\right)=a_w$. Therefore, $a_v=a_w$.\n\n\\end{proof}"}
 {"id": "Axler|exercise_5_20", "formal_statement": "theorem exercise_5_20 {F V : Type*} [add_comm_group V] [field F]\n  [module F V] [finite_dimensional F V] {S T : End F V}\n  (h1 : @card T.eigenvalues (eigenvalues.fintype T) = finrank F V)\n  (h2 : \u2200 v : V, \u2203 c : F, v \u2208 eigenspace S c \u2194 \u2203 c : F, v \u2208 eigenspace T c) :\n  S * T = T * S :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose that $T \\in \\mathcal{L}(V)$ has $\\operatorname{dim} V$ distinct eigenvalues and that $S \\in \\mathcal{L}(V)$ has the same eigenvectors as $T$ (not necessarily with the same eigenvalues). Prove that $S T=T S$.", "nl_proof": "\\begin{proof}\n\n    First off, let $n=\\operatorname{dim} V$. so, there is a basis of $\\left(v_1, \\ldots, v_j\\right)$ of $V$ that consist of eigenvectors of $T$. Now, let $\\lambda_1, \\ldots, \\lambda_n$ be the corresponding eigenvalues, then we would have $T v_j=\\lambda_1 v_j$ for every single $j$.\n\n\n\nNow, for every $v_j$ is also an eigenvector of S, so $S v_j=a_j v_j$ for some $a_j \\in F$. For each $j$, we would then have $(S T) v_j=S\\left(T v_j\\right)=\\lambda_j S v_j=a_j \\lambda_j v_j$ and $(T S) v_j=T\\left(S v_j\\right)=a_j T v_j=a_j \\lambda_j v_j$. Since both operators, which are $S T$ and $T S$, agree on a basis, then both are equal.\n\n\\end{proof}"}
-{"id": "Axler|exercise_6_2", "formal_statement": "theorem exercise_6_2 {V : Type*} [add_comm_group V] [module \u2102 V]\n  [inner_product_space \u2102 V] (u v : V) :\n  \u27eau, v\u27eb_\u2102 = 0 \u2194 \u2200 (a : \u2102), \u2225u\u2225 \u2264 \u2225u + a \u2022 v\u2225 :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $u, v \\in V$. Prove that $\\langle u, v\\rangle=0$ if and only if $\\|u\\| \\leq\\|u+a v\\|$ for all $a \\in \\mathbf{F}$.", "nl_proof": "\\begin{proof}\n\n    First off, let us suppose that $(u, v)=0$.\n\nNow, let $a \\in \\mathbb{F}$. Next, $u, a v$ are orthogonal.\n\nThe Pythagorean theorem thus implies that\n\n$$\n\n\\begin{aligned}\n\n\\|u+a v\\|^2 & =\\|u\\|^2+\\|a v\\|^2 \\\\\n\n& \\geq\\|u\\|^2\n\n\\end{aligned}\n\n$$\n\nSo, by taking the square roots, this will now give us $\\|u\\| \\leq\\|u+a v\\|$.\n\nNow, to prove the implication in the other direction, we must now let $\\|u\\| \\leq$ $\\|u+a v\\|$ for all $a \\in \\mathbb{F}$. Squaring this inequality, we get both:\n\n$$\n\n\\begin{gathered}\n\n\\|u\\|^2 a n d \\leq\\|u+a v\\|^2 \\\\\n\n=(u+a v, u+a v) \\\\\n\n=(u, u)+(u, a v)+(a v, u)+(a v, a v) \\\\\n\n=\\|u\\|^2+\\bar{a}(u, v)+a \\overline{(u, v)}+|a|^2\\|v\\|^2 \\\\\n\n\\|u\\|^2+2 \\Re \\bar{a}(u, v)+|a|^2\\|v\\|^2\n\n\\end{gathered}\n\n$$\n\nfor all $a \\in \\mathbb{F}$.\n\nTherefore,\n\n$$\n\n-2 \\Re \\bar{a}(u, v) \\leq|a|^2\\|v\\|^2\n\n$$\n\nfor all $a \\in \\mathbb{F}$. In particular, we can let $a$ equal $-t(u, v)$ for $t>0$. Substituting this value for $a$ into the inequality above gives\n\n$$\n\n2 t|(u, v)|^2 \\leq t^2|(u, v)|^2\\|v\\|^2\n\n$$\n\nfor all $t>0$.\n\nStep 4\n\n4 of 4\n\nDivide both sides of the inequality above by $t$, getting\n\n$$\n\n2|(u, v)|^2 \\leq t \\mid(u, v)^2\\|v\\|^2\n\n$$\n\nfor all $t>0$. If $v=0$, then $(u, v)=0$, as desired. If $v \\neq 0$, set $t$ equal to $1 /\\|v\\|^2$ in the inequality above, getting\n\n$$\n\n2|(u, v)|^2 \\leq|(u, v)|^2,\n\n$$\n\nwhich implies that $(u, v)=0$.\n\n\\end{proof}"}
-{"id": "Axler|exercise_6_7", "formal_statement": "theorem exercise_6_7 {V : Type*} [inner_product_space \u2102 V] (u v : V) :\n  \u27eau, v\u27eb_\u2102 = (\u2225u + v\u2225^2 - \u2225u - v\u2225^2 + I*\u2225u + I\u2022v\u2225^2 - I*\u2225u-I\u2022v\u2225^2) / 4 :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that if $V$ is a complex inner-product space, then $\\langle u, v\\rangle=\\frac{\\|u+v\\|^{2}-\\|u-v\\|^{2}+\\|u+i v\\|^{2} i-\\|u-i v\\|^{2} i}{4}$ for all $u, v \\in V$.", "nl_proof": "\\begin{proof}\n\nLet $V$ be an inner-product space and $u, v\\in V$. Then \n\n$$\n\n\\begin{aligned}\n\n\\|u+v\\|^2 & =\\langle u+v, v+v\\rangle \\\\\n\n& =\\|u\\|^2+\\langle u, v\\rangle+\\langle v, u\\rangle+\\|v\\|^2 \\\\\n\n-\\|u-v\\|^2 & =-\\langle u-v, u-v\\rangle \\\\\n\n& =-\\|u\\|^2+\\langle u, v\\rangle+\\langle v, u\\rangle-\\|v\\|^2 \\\\\n\ni\\|u+i v\\|^2 & =i\\langle u+i v, u+i v\\rangle \\\\\n\n& =i\\|u\\|^2+\\langle u, v\\rangle-\\langle v, u\\rangle+i\\|v\\|^2 \\\\\n\n-i\\|u-i v\\|^2 & =-i\\langle u-i v, u-i v\\rangle \\\\\n\n& =-i\\|u\\|^2+\\langle u, v\\rangle-\\langle v, u\\rangle-i\\|v\\|^2 .\n\n\\end{aligned}\n\n$$\n\nThus $\\left(\\|u+v\\|^2\\right)-\\|u-v\\|^2+\\left(i\\|u+i v\\|^2\\right)-i\\|u-i v\\|^2=4\\langle u, v\\rangle.$\n\n\\end{proof}"}
+{"id": "Axler|exercise_6_2", "formal_statement": "theorem exercise_6_2 {V : Type*} [add_comm_group V] [module \u2102 V]\n  [inner_product_space \u2102 V] (u v : V) :\n  \u27eau, v\u27eb_\u2102 = 0 \u2194 \u2200 (a : \u2102), \u2016u\u2016  \u2264 \u2016u + a \u2022 v\u2016 :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $u, v \\in V$. Prove that $\\langle u, v\\rangle=0$ if and only if $\\|u\\| \\leq\\|u+a v\\|$ for all $a \\in \\mathbf{F}$.", "nl_proof": "\\begin{proof}\n\n    First off, let us suppose that $(u, v)=0$.\n\nNow, let $a \\in \\mathbb{F}$. Next, $u, a v$ are orthogonal.\n\nThe Pythagorean theorem thus implies that\n\n$$\n\n\\begin{aligned}\n\n\\|u+a v\\|^2 & =\\|u\\|^2+\\|a v\\|^2 \\\\\n\n& \\geq\\|u\\|^2\n\n\\end{aligned}\n\n$$\n\nSo, by taking the square roots, this will now give us $\\|u\\| \\leq\\|u+a v\\|$.\n\nNow, to prove the implication in the other direction, we must now let $\\|u\\| \\leq$ $\\|u+a v\\|$ for all $a \\in \\mathbb{F}$. Squaring this inequality, we get both:\n\n$$\n\n\\begin{gathered}\n\n\\|u\\|^2 a n d \\leq\\|u+a v\\|^2 \\\\\n\n=(u+a v, u+a v) \\\\\n\n=(u, u)+(u, a v)+(a v, u)+(a v, a v) \\\\\n\n=\\|u\\|^2+\\bar{a}(u, v)+a \\overline{(u, v)}+|a|^2\\|v\\|^2 \\\\\n\n\\|u\\|^2+2 \\Re \\bar{a}(u, v)+|a|^2\\|v\\|^2\n\n\\end{gathered}\n\n$$\n\nfor all $a \\in \\mathbb{F}$.\n\nTherefore,\n\n$$\n\n-2 \\Re \\bar{a}(u, v) \\leq|a|^2\\|v\\|^2\n\n$$\n\nfor all $a \\in \\mathbb{F}$. In particular, we can let $a$ equal $-t(u, v)$ for $t>0$. Substituting this value for $a$ into the inequality above gives\n\n$$\n\n2 t|(u, v)|^2 \\leq t^2|(u, v)|^2\\|v\\|^2\n\n$$\n\nfor all $t>0$.\n\nStep 4\n\n4 of 4\n\nDivide both sides of the inequality above by $t$, getting\n\n$$\n\n2|(u, v)|^2 \\leq t \\mid(u, v)^2\\|v\\|^2\n\n$$\n\nfor all $t>0$. If $v=0$, then $(u, v)=0$, as desired. If $v \\neq 0$, set $t$ equal to $1 /\\|v\\|^2$ in the inequality above, getting\n\n$$\n\n2|(u, v)|^2 \\leq|(u, v)|^2,\n\n$$\n\nwhich implies that $(u, v)=0$.\n\n\\end{proof}"}
+{"id": "Axler|exercise_6_7", "formal_statement": "theorem exercise_6_7 {V : Type*} [inner_product_space \u2102 V] (u v : V) :\n  \u27eau, v\u27eb_\u2102 = (\u2016u + v\u2016^2 - \u2016u - v\u2016^2 + I*\u2016u + I\u2022v\u2016^2 - I*\u2016u-I\u2022v\u2016^2) / 4 :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that if $V$ is a complex inner-product space, then $\\langle u, v\\rangle=\\frac{\\|u+v\\|^{2}-\\|u-v\\|^{2}+\\|u+i v\\|^{2} i-\\|u-i v\\|^{2} i}{4}$ for all $u, v \\in V$.", "nl_proof": "\\begin{proof}\n\nLet $V$ be an inner-product space and $u, v\\in V$. Then \n\n$$\n\n\\begin{aligned}\n\n\\|u+v\\|^2 & =\\langle u+v, v+v\\rangle \\\\\n\n& =\\|u\\|^2+\\langle u, v\\rangle+\\langle v, u\\rangle+\\|v\\|^2 \\\\\n\n-\\|u-v\\|^2 & =-\\langle u-v, u-v\\rangle \\\\\n\n& =-\\|u\\|^2+\\langle u, v\\rangle+\\langle v, u\\rangle-\\|v\\|^2 \\\\\n\ni\\|u+i v\\|^2 & =i\\langle u+i v, u+i v\\rangle \\\\\n\n& =i\\|u\\|^2+\\langle u, v\\rangle-\\langle v, u\\rangle+i\\|v\\|^2 \\\\\n\n-i\\|u-i v\\|^2 & =-i\\langle u-i v, u-i v\\rangle \\\\\n\n& =-i\\|u\\|^2+\\langle u, v\\rangle-\\langle v, u\\rangle-i\\|v\\|^2 .\n\n\\end{aligned}\n\n$$\n\nThus $\\left(\\|u+v\\|^2\\right)-\\|u-v\\|^2+\\left(i\\|u+i v\\|^2\\right)-i\\|u-i v\\|^2=4\\langle u, v\\rangle.$\n\n\\end{proof}"}
 {"id": "Axler|exercise_6_16", "formal_statement": "theorem exercise_6_16 {K V : Type*} [is_R_or_C K] [inner_product_space K V]\n  {U : submodule K V} : \n  U.orthogonal = \u22a5  \u2194 U = \u22a4 :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $U$ is a subspace of $V$. Prove that $U^{\\perp}=\\{0\\}$ if and only if $U=V$", "nl_proof": "\\begin{proof}\n\n    $V=U \\bigoplus U^{\\perp}$, therefore $U^\\perp = \\{0\\}$ iff $U=V$. \n\n\\end{proof}"}
 {"id": "Axler|exercise_7_6", "formal_statement": "theorem exercise_7_6 {V : Type*} [inner_product_space \u2102 V]\n  [finite_dimensional \u2102 V] (T : End \u2102 V)\n  (hT : T * T.adjoint = T.adjoint * T) :\n  T.range = T.adjoint.range :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Prove that if $T \\in \\mathcal{L}(V)$ is normal, then $\\operatorname{range} T=\\operatorname{range} T^{*}.$", "nl_proof": "\\begin{proof}\n\n    Let $T \\in \\mathcal{L}(V)$ to be a normal operator.\n\nSuppose $u \\in \\operatorname{null} T$. Then, by $7.20$,\n\n$$\n\n0=\\|T u\\|=\\left\\|T^* u\\right\\|,\n\n$$\n\nwhich implies that $u \\in \\operatorname{null} T^*$.\n\nHence\n\n$$\n\n\\operatorname{null} T=\\operatorname{null} T^*\n\n$$\n\nbecause $\\left(T^*\\right)^*=T$ and the same argument can be repeated.\n\nNow we have\n\n$$\n\n\\begin{aligned}\n\n\\text { range } T & =\\left(\\text { null } T^*\\right)^{\\perp} \\\\\n\n& =(\\text { null } T)^{\\perp} \\\\\n\n& =\\operatorname{range} T^*,\n\n\\end{aligned}\n\n$$\n\nwhere the first and last equality follow from items (d) and (b) of 7.7.\n\nHence, range $T=$ range $T^*$.\n\n\\end{proof}"}
 {"id": "Axler|exercise_7_10", "formal_statement": "theorem exercise_7_10 {V : Type*} [inner_product_space \u2102 V]\n  [finite_dimensional \u2102 V] (T : End \u2102 V)\n  (hT : T * T.adjoint = T.adjoint * T) (hT1 : T^9 = T^8) :\n  is_self_adjoint T \u2227 T^2 = T :=", "src_header": "import .common \n\nopen set fintype complex polynomial submodule linear_map finite_dimensional\nopen module module.End inner_product_space\n\nopen_locale big_operators\n\n", "nl_statement": "Suppose $V$ is a complex inner-product space and $T \\in \\mathcal{L}(V)$ is a normal operator such that $T^{9}=T^{8}$. Prove that $T$ is self-adjoint and $T^{2}=T$.", "nl_proof": "\\begin{proof}\n\n    Based on the complex spectral theorem, there is an orthonormal basis of $\\left(e_1, \\ldots, e_n\\right)$ of $V$ consisting of eigenvectors of $T$. Now, let $\\lambda_1, \\ldots, \\lambda_n$ be the corresponding eigenvalues. Therefore,\n\n$$\n\nT e_1=\\lambda_j e_j\n\n$$\n\nfor $j=1 \\ldots n$.\n\n\n\nNext, by applying $T$ repeatedly to both sides of the equation above, we get $T^9 e_j=\\left(\\lambda_j\\right)^9 e_j$ and rei =8ej. Thus $T^8 e_j=\\left(\\lambda_j\\right)^8 e_j$, which implies that $\\lambda_j$ equals 0 or 1 . In particular, all the eigenvalues of $T$ are real. This would then imply that $T$ is self-adjoint.\n\n\n\nNow, by applying $T$ to both sides of the equation above, we get\n\n$$\n\n\\begin{aligned}\n\nT^2 e_j & =\\left(\\lambda_j\\right)^2 e_j \\\\\n\n& =\\lambda_j e_j \\\\\n\n& =T e_j\n\n\\end{aligned}\n\n$$\n\nwhich is where the second equality holds because $\\lambda_j$ equals 0 or 1 . Because $T^2$ and $T$ agree on a basis, they must be equal.\n\n\\end{proof}"}
@@ -118,7 +118,7 @@
 {"id": "Herstein|exercise_2_11_22", "formal_statement": "theorem exercise_2_11_22 {p : \u2115} {n : \u2115} {G : Type*} [fintype G] \n  [group G] (hp : nat.prime p) (hG : card G = p ^ n) {K : subgroup G}\n  [fintype K] (hK : card K = p ^ (n-1)) : \n  K.normal :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that any subgroup of order $p^{n-1}$ in a group $G$ of order $p^n$ is normal in $G$.", "nl_proof": "\\begin{proof}\n\nProof: First we prove the following lemma.\n\n\n\n\\textbf{Lemma:} If $G$ is a finite $p$-group with $|G|>1$, then $Z(G)$, the center of $G$, has more than one element; that is, if $|G|=p^k$ with $k\\geq 1$, then $|Z(G)|>1$.\n\n\n\n\\textit{Proof of the lemma:} Consider the class equation\n\n$$\n\n|G|=|Z(G)|+\\sum_{a \\notin Z(G)}[G: C(a)],\n\n$$\n\nwhere $C(a)$ denotes the centralizer of $a$ in $G$. If $G=Z(G)$, then the lemma is immediate. Suppose $Z(G)$ is a proper subset of $G$ and consider an element $a\\in G$ such that $a\\notin Z(G)$. Then $C(a)$ is a proper subgroup of $G$. Since $C(a)$ is a subgroup of a $p$-group, $[G:C(a)]$ is divisible by $p$ for all $a\\notin Z(G)$. This implies that $p$ divides $|G|=|Z(G)|+\\sum_{a\\notin Z(G)} [G:C(a)]$.\n\n\n\nSince $p$ also divides $|G|$, it follows that $p$ divides $|Z(G)|$. Hence, $|Z(G)|>1$. $\\Box$\n\n\n\nThis proves our \\textbf{lemma}.\n\n\n\nWe will prove the result by induction on $n$.\n\nIf $n=1$, the $G$ is a cyclic group of prime order and hence every subgroup of $G$ is normal in $G$. Thus, the result is true for $n=1$.\n\nSuppose the result is true for all groups of order $p^m$, where $1 \\leq m<n$.\n\nLet $H$ be a subgroup of order $p^{n-1}$.\n\nConsider $N(H)=\\{g \\in H: g H=H g\\}$.\n\nIf $H \\neq N(H)$, then $|N(H)|>p^{n-1}$. Thus, $|N(H)|=p^n$ and $N(H)=G$.\n\nIn this case $H$ is normal in $G$.\n\nLet $H=N(H)$. Then $Z(G)$, the center of $G$, is a subset of $H$ and $Z(G) \\neq$ $\\{e\\}$.\n\nBy Cauchy's theorem and the above Claim, there exists $a \\in Z(G)$ such that $o(a)=p$.\n\nLet $K=\\langle a\\rangle$, a cyclic group generated by $a$.\n\nThen $K$ is a normal subgroup of $G$ of order $p$. Now, $|H / K|=p^{n-2}$ and $|G / K|=p^{n-1}$.\n\nThus, by induction hypothesis, $H / K$ is a normal subgroup of $G / K$.\n\n\\end{proof}"}
 {"id": "Herstein|exercise_4_1_19", "formal_statement": "theorem exercise_4_1_19 : infinite {x : quaternion \u211d | x^2 = -1} :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Show that there is an infinite number of solutions to $x^2 = -1$ in the quaternions.", "nl_proof": "\\begin{proof}\n\nLet $x=a i+b j+c k$ then\n\n$$\n\nx^2=(a i+b j+c k)(a i+b j+c k)=-a^2-b^2-c^2=-1\n\n$$\n\nThis gives $a^2+b^2+c^2=1$ which has infinitely many solutions for $-1<a, b, c<1$.\n\n\\end{proof}"}
 {"id": "Herstein|exercise_4_2_5", "formal_statement": "theorem exercise_4_2_5 {R : Type*} [ring R] \n  (h : \u2200 x : R, x ^ 3 = x) : comm_ring R :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $R$ be a ring in which $x^3 = x$ for every $x \\in R$. Prove that $R$ is commutative.", "nl_proof": "\\begin{proof}\n\nTo begin with\n\n$$\n\n2 x=(2 x)^3=8 x^3=8 x .\n\n$$\n\nTherefore $6 x=0 \\quad \\forall x$.\n\nAlso\n\n$$\n\n(x+y)=(x+y)^3=x^3+x^2 y+x y x+y x^2+x y^2+y x y+y^2 x+y^3\n\n$$\n\nand\n\n$$\n\n(x-y)=(x-y)^3=x^3-x^2 y-x y x-y x^2+x y^2+y x y+y^2 x-y^3\n\n$$\n\nSubtracting we get\n\n$$\n\n2\\left(x^2 y+x y x+y x^2\\right)=0\n\n$$\n\nMultiply the last relation by $x$ on the left and right to get\n\n$$\n\n2\\left(x y+x^2 y x+x y x^2\\right)=0 \\quad 2\\left(x^2 y x+x y x^2+y x\\right)=0 .\n\n$$\n\nSubtracting the last two relations we have\n\n$$\n\n2(x y-y x)=0 .\n\n$$\n\nWe then show that $3\\left(x+x^2\\right)=0 \\forall x$. You get this from\n\n$$\n\nx+x^2=\\left(x+x^2\\right)^3=x^3+3 x^4+3 x^5+x^6=4\\left(x+x^2\\right) .\n\n$$\n\nIn particular\n\n$$\n\n3\\left(x+y+(x+y)^2\\right)=3\\left(x+x^2+y+y^2+x y+y x\\right)=0\n\n$$\n\nwe end-up with $3(x y+y x)=0$. But since $6 x y=0$, we have $3(x y-y x)=0$. Then subtract $2(x y-y x)=0$ to get $x y-y x=0$.\n\n\\end{proof}"}
-{"id": "Herstein|exercise_4_2_9", "formal_statement": "theorem exercise_4_2_9 {p : \u2115} (hp : nat.prime p) (hp1 : odd p) :\n  \u2203 (a b : \u2124), a / b = \u2211 i in finset.range p, 1 / (i + 1) \u2192 \u2191p \u2223 a :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $p$ be an odd prime and let $1 + \\frac{1}{2} + ... + \\frac{1}{p - 1} = \\frac{a}{b}$, where $a, b$ are integers. Show that $p \\mid a$.", "nl_proof": "\\begin{proof}\n\n    First we prove for prime $p=3$ and then for all prime $p>3$.\n\nLet us take $p=3$. Then the sum\n\n$$\n\n\\frac{1}{1}+\\frac{1}{2}+\\ldots+\\frac{1}{(p-1)}\n\n$$\n\nbecomes\n\n$$\n\n1+\\frac{1}{3-1}=1+\\frac{1}{2}=\\frac{3}{2} .\n\n$$\n\nTherefore in this case $\\quad \\frac{a}{b}=\\frac{3}{2} \\quad$ implies $3 \\mid a$, i.e. $p \\mid a$.\n\nNow for odd prime $p>3$.\n\nLet us consider $f(x)=(x-1)(x-2) \\ldots(x-(p-1))$.\n\nNow, by Fermat, we know that the coefficients of $f(x)$ other than the $x^{p-1}$ and $x^0$ are divisible by $p$.\n\nSo if,\n\n$$\n\n\\begin{array}{r}\n\nf(x)=x^{p-1}+\\sum_{i=0}^{p-2} a_i x^i \\\\\n\n\\text { and } p>3 .\n\n\\end{array}\n\n$$\n\nThen $p \\mid a_2$, and\n\n$$\n\nf(p) \\equiv a_1 p+a_0 \\quad\\left(\\bmod p^3\\right)\n\n$$\n\nBut we see that\n\n$$\n\nf(x)=(-1)^{p-1} f(p-x) \\text { for any } x,\n\n$$\n\nso if $p$ is odd,\n\n$$\n\nf(p)=f(0)=a_0,\n\n$$\n\nSo it follows that:\n\n$$\n\n0=f(p)-a_0 \\equiv a_1 p \\quad\\left(\\bmod p^3\\right)\n\n$$\n\nTherefore,\n\n$$\n\n0 \\equiv a_1 \\quad\\left(\\bmod p^2\\right) .\n\n$$\n\nHence,\n\n$$\n\n0 \\equiv a_1 \\quad(\\bmod p) .\n\n$$\n\nNow our sum is just $\\frac{a_1}{(p-1) !}=\\frac{a}{b}$.\n\nIt follows that $p$ divides $a$. This completes the proof.\n\n\\end{proof}"}
+{"id": "Herstein|exercise_4_2_9", "formal_statement": "theorem exercise_4_2_9 {p : \u2115} (hp : nat.prime p) (hp1 : odd p) :\n  \u2203 (a b : \u2124), (a / b : \u211a) = \u2211 i in finset.range p, 1 / (i + 1) \u2192 \u2191p \u2223 a :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $p$ be an odd prime and let $1 + \\frac{1}{2} + ... + \\frac{1}{p - 1} = \\frac{a}{b}$, where $a, b$ are integers. Show that $p \\mid a$.", "nl_proof": "\\begin{proof}\n\n    First we prove for prime $p=3$ and then for all prime $p>3$.\n\nLet us take $p=3$. Then the sum\n\n$$\n\n\\frac{1}{1}+\\frac{1}{2}+\\ldots+\\frac{1}{(p-1)}\n\n$$\n\nbecomes\n\n$$\n\n1+\\frac{1}{3-1}=1+\\frac{1}{2}=\\frac{3}{2} .\n\n$$\n\nTherefore in this case $\\quad \\frac{a}{b}=\\frac{3}{2} \\quad$ implies $3 \\mid a$, i.e. $p \\mid a$.\n\nNow for odd prime $p>3$.\n\nLet us consider $f(x)=(x-1)(x-2) \\ldots(x-(p-1))$.\n\nNow, by Fermat, we know that the coefficients of $f(x)$ other than the $x^{p-1}$ and $x^0$ are divisible by $p$.\n\nSo if,\n\n$$\n\n\\begin{array}{r}\n\nf(x)=x^{p-1}+\\sum_{i=0}^{p-2} a_i x^i \\\\\n\n\\text { and } p>3 .\n\n\\end{array}\n\n$$\n\nThen $p \\mid a_2$, and\n\n$$\n\nf(p) \\equiv a_1 p+a_0 \\quad\\left(\\bmod p^3\\right)\n\n$$\n\nBut we see that\n\n$$\n\nf(x)=(-1)^{p-1} f(p-x) \\text { for any } x,\n\n$$\n\nso if $p$ is odd,\n\n$$\n\nf(p)=f(0)=a_0,\n\n$$\n\nSo it follows that:\n\n$$\n\n0=f(p)-a_0 \\equiv a_1 p \\quad\\left(\\bmod p^3\\right)\n\n$$\n\nTherefore,\n\n$$\n\n0 \\equiv a_1 \\quad\\left(\\bmod p^2\\right) .\n\n$$\n\nHence,\n\n$$\n\n0 \\equiv a_1 \\quad(\\bmod p) .\n\n$$\n\nNow our sum is just $\\frac{a_1}{(p-1) !}=\\frac{a}{b}$.\n\nIt follows that $p$ divides $a$. This completes the proof.\n\n\\end{proof}"}
 {"id": "Herstein|exercise_4_3_25", "formal_statement": "theorem exercise_4_3_25 (I : ideal (matrix (fin 2) (fin 2) \u211d)) : \n  I = \u22a5 \u2228 I = \u22a4 :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $R$ be the ring of $2 \\times 2$ matrices over the real numbers; suppose that $I$ is an ideal of $R$. Show that $I = (0)$ or $I = R$.", "nl_proof": "\\begin{proof}\n\n    Suppose that $I$ is a nontrivial ideal of $R$, and let\n\n$$\n\nA=\\left(\\begin{array}{ll}\n\na & b \\\\\n\nc & d\n\n\\end{array}\\right)\n\n$$\n\nwhere not all of $a, b, c d$ are zero. Suppose, without loss of generality -- our steps would be completely analogous, modulo some different placement of 1 s in our matrices, if we assumed some other element to be nonzero -- that $a \\neq 0$. Then we have that\n\n$$\n\n\\left(\\begin{array}{ll}\n\n1 & 0 \\\\\n\n0 & 0\n\n\\end{array}\\right)\\left(\\begin{array}{ll}\n\na & b \\\\\n\nc & d\n\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n\na & b \\\\\n\n0 & 0\n\n\\end{array}\\right) \\in I\n\n$$\n\nand so\n\n$$\n\n\\left(\\begin{array}{ll}\n\na & b \\\\\n\n0 & 0\n\n\\end{array}\\right)\\left(\\begin{array}{ll}\n\n1 & 0 \\\\\n\n0 & 0\n\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n\na & 0 \\\\\n\n0 & 0\n\n\\end{array}\\right) \\in I\n\n$$\n\nso that\n\n$$\n\n\\left(\\begin{array}{ll}\n\nx & 0 \\\\\n\n0 & 0\n\n\\end{array}\\right) \\in I\n\n$$\n\nfor any real $x$. Now, also for any real $x$,\n\n$$\n\n\\left(\\begin{array}{ll}\n\nx & 0 \\\\\n\n0 & 0\n\n\\end{array}\\right)\\left(\\begin{array}{ll}\n\n0 & 1 \\\\\n\n0 & 0\n\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n\n0 & x \\\\\n\n0 & 0\n\n\\end{array}\\right) \\in I .\n\n$$\n\nLikewise\n\n$$\n\n\\left(\\begin{array}{ll}\n\n0 & 0 \\\\\n\n1 & 0\n\n\\end{array}\\right)\\left(\\begin{array}{ll}\n\n0 & x \\\\\n\n0 & 0\n\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n\n0 & 0 \\\\\n\n0 & x\n\n\\end{array}\\right) \\in I\n\n$$\n\nand\n\n$$\n\n\\left(\\begin{array}{ll}\n\n0 & 0 \\\\\n\n0 & x\n\n\\end{array}\\right)\\left(\\begin{array}{ll}\n\n0 & 0 \\\\\n\n1 & 0\n\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n\n0 & 0 \\\\\n\nx & 0\n\n\\end{array}\\right)\n\n$$\n\nThus, as\n\n$$\n\n\\left(\\begin{array}{ll}\n\na & b \\\\\n\nc & d\n\n\\end{array}\\right)=\\left(\\begin{array}{ll}\n\na & 0 \\\\\n\n0 & 0\n\n\\end{array}\\right)+\\left(\\begin{array}{ll}\n\n0 & b \\\\\n\n0 & 0\n\n\\end{array}\\right)+\\left(\\begin{array}{ll}\n\n0 & 0 \\\\\n\nc & 0\n\n\\end{array}\\right)+\\left(\\begin{array}{ll}\n\n0 & 0 \\\\\n\n0 & d\n\n\\end{array}\\right)\n\n$$\n\nand since all the terms on the right side are in $I$ and $I$ is an additive group, it follows that\n\n$$\n\n\\left(\\begin{array}{ll}\n\na & b \\\\\n\nc & d\n\n\\end{array}\\right)\n\n$$\n\nfor arbitrary $a, b, c, d$ is in $I$, i.e. $I=R$\n\nNote that the intuition for picking these matrices is that, if we denote by $E_{i j}$ the matrix with 1 at position $(i, j)$ and 0 elsewhere, then\n\n$$\n\nE_{i j}\\left(\\begin{array}{ll}\n\na_{1,1} & a_{1,2} \\\\\n\na_{2,1} & a_{2,2}\n\n\\end{array}\\right) E_{n m}=a_{j, n} E_{i m}\n\n$$\n\n\\end{proof}"}
 {"id": "Herstein|exercise_4_5_16", "formal_statement": "theorem exercise_4_5_16 {p n: \u2115} (hp : nat.prime p) \n  {q : polynomial (zmod p)} (hq : irreducible q) (hn : q.degree = n) :\n  \u2203 is_fin : fintype $ polynomial (zmod p) \u29f8 ideal.span ({q} : set (polynomial $ zmod p)), \n  @card (polynomial (zmod p) \u29f8 ideal.span {q}) is_fin = p ^ n \u2227 \n  is_field (polynomial $ zmod p):=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Let $F = \\mathbb{Z}_p$ be the field of integers $\\mod p$, where $p$ is a prime, and let $q(x) \\in F[x]$ be irreducible of degree $n$. Show that $F[x]/(q(x))$ is a field having at exactly $p^n$ elements.", "nl_proof": "\\begin{proof}\n\n    In the previous problem we have shown that any for any $p(x) \\in F[x]$, we have that\n\n$$\n\np(x)+(q(x))=a_{n-1} x^{n-1}+\\cdots+a_1 x+a_0+(q(x))\n\n$$\n\nfor some $a_{n-1}, \\ldots, a_0 \\in F$, and that there are $p^n$ choices for these numbers, so that $F[x] /(q(x)) \\leq p^n$. In order to show that equality holds, we have to show that each of these choices induces a different element of $F[x] /(q(x))$; in other words, that each different polynomial of degree $n-1$ or lower belongs to a different coset of $(q(x))$ in $F[x]$.\n\n\n\nSuppose now, then, that\n\n$$\n\na_{n-1} x^{n-1}+\\cdots+a_1 x+a_0+(q(x))=b_{n-1} x^{n-1}+\\cdots+b_1 x+b_0+(q(x))\n\n$$\n\nwhich is equivalent with $\\left(a_{n-1}-b_{n-1}\\right)^{n-1}+\\cdots\\left(a_1-b_1\\right) x+\\left(a_0-b_0\\right) \\in(q(x))$, which is in turn equivalent with there being a $w(x) \\in F[x]$ such that\n\n$$\n\nq(x) w(x)=\\left(a_{n-1}-b_{n-1}\\right)^{n-1}+\\cdots\\left(a_1-b_1\\right) x+\\left(a_0-b_0\\right) .\n\n$$\n\nDegree of the right hand side is strictly smaller than $n$, while the degree of the left hand side is greater or equal to $n$ except if $w(x)=0$, so that if equality is hold we must have that $w(x)=0$, but then since polynomials are equal iff all of their coefficient are equal we get that $a_{n-1}-b_{n-1}=$ $0, \\ldots, a_1-b_1=0, a_0-b_0=0$, i.e.\n\n$$\n\na_{n-1}=b_{n-1}, \\ldots, a_1=b_1, a_0=b_0\n\n$$\n\nwhich is what we needed to prove.\n\n\\end{proof}"}
 {"id": "Herstein|exercise_4_5_25", "formal_statement": "theorem exercise_4_5_25 {p : \u2115} (hp : nat.prime p) :\n  irreducible (\u2211 i : finset.range p, X ^ p : polynomial \u211a) :=", "src_header": "import .common \n\nopen set function nat fintype real subgroup ideal polynomial submodule zsqrtd \nopen char_p mul_aut matrix\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "If $p$ is a prime, show that $q(x) = 1 + x + x^2 + \\cdots x^{p - 1}$ is irreducible in $Q[x]$.", "nl_proof": "\\begin{proof}\n\n    Lemma: Let $F$ be a field and $f(x) \\in F[x]$. If $c \\in F$ and $f(x+c)$ is irreducible in $F[x]$, then $f(x)$ is irreducible in $F[x]$.\n\nProof of the Lemma: Suppose that $f(x)$ is reducible, i.e., there exist non-constant $g(x), h(x) \\in F[x]$ so that\n\n$$\n\nf(x)=g(x) h(x) .\n\n$$\n\nIn particular, then we have\n\n$$\n\nf(x+c)=g(x+c) h(x+c) .\n\n$$\n\nNote that $g(x+c)$ and $h(x+c)$ have the same degree at $g(x)$ and $h(x)$ respectively; in particular, they are non-constant polynomials. So our assumption is wrong.\n\nHence, $f(x)$ is irreducible in $F[x]$. This proves our Lemma.\n\n\n\nNow recall the identity\n\n$$\n\n\\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+\\ldots \\ldots+x^2+x+1 .\n\n$$\n\nWe prove that $f(x+1)$ is $\\$$ |textbffirreducible in $\\mathbb{Q}[x]$ and then apply the Lemma to conclude that $f(x)$ is irreducible in $\\mathbb{Q}[x] .3 \\$$ Note that\n\n$$\n\n\\begin{aligned}\n\n& f(x+1)=\\frac{(x+1)^p-1}{x} \\\\\n\n& =\\frac{x^p+p x^{p-1}+\\ldots+p x}{x} \\\\\n\n& =x^{p-1}+p x^{p-2}+\\ldots .+p .\n\n\\end{aligned}\n\n$$\n\nUsing that the binomial coefficients occurring above are all divisible by $p$, we have that $f(x+1)$ is irreducible $\\mathbb{Q}[x]$ by Eisenstein's criterion applied with prime $p$. \n\n\n\nThen by the lemma $f(x)$ is irreducible $\\mathbb{Q}[x]$. This completes the proof.\n\n\\end{proof}"}
@@ -181,5 +181,5 @@
 {"id": "Dummit-Foote|exercise_9_1_6", "formal_statement": "theorem exercise_9_1_6 : \u00ac is_principal \n  (ideal.span ({X 0, X 1} : set (mv_polynomial (fin 2) \u211a))) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $(x, y)$ is not a principal ideal in $\\mathbb{Q}[x, y]$.", "nl_proof": "\\begin{proof}\n\n    Suppose, to the contrary, that $(x, y)=p$ for some polynomial $p \\in \\mathbb{Q}[x, y]$. From $x, y \\in$ $(x, y)=(p)$ there are $s, t \\in \\mathbb{Q}[x, y]$ such that $x=s p$ and $y=t p$.\n\nThen:\n\n$$\n\n\\begin{aligned}\n\n& 0=\\operatorname{deg}_y(x)=\\operatorname{deg}_y(s)+\\operatorname{deg}_y(p) \\text { so } \\\\\n\n& 0=\\operatorname{deg}_y(p) \\\\\n\n& 0=\\operatorname{deg}_x(y)=\\operatorname{deg}_x(s)+\\operatorname{deg}_x(p) \\text { so } \\\\\n\n& 0=\\operatorname{deg}_x(p) \\text { so }\n\n\\end{aligned}\n\n$$\n\nFrom : $\\quad 0=\\operatorname{deg}_y(p)=\\operatorname{deg}_x(p)$ we get $\\operatorname{deg}(p)=0$ and $p \\in \\mathbb{Q}$.\n\nBut $p \\in(p)=(x, y)$ so $p=a x+b y$ for some $a, b \\in \\mathbb{Q}[x, y]$\n\n$$\n\n\\begin{aligned}\n\n\\operatorname{deg}(p) & =\\operatorname{deg}(a x+b y) \\\\\n\n& =\\min (\\operatorname{deg}(a)+\\operatorname{deg}(x), \\operatorname{deg}(b)+\\operatorname{deg}(y)) \\\\\n\n& =\\min (\\operatorname{deg}(a)+1, \\operatorname{deg}(b)+1) \\geqslant 1\n\n\\end{aligned}\n\n$$\n\nwhich contradicts $\\operatorname{deg}(p)=0$.\n\nSo we conclude that $(x, y)$ is not principal ideal in $\\mathbb{Q}[x, y]$\n\n\\end{proof}"}
 {"id": "Dummit-Foote|exercise_9_3_2", "formal_statement": "theorem exercise_9_3_2 {f g : polynomial \u211a} (i j : \u2115)\n  (hfg : \u2200 n : \u2115, \u2203 a : \u2124, (f*g).coeff = a) :\n  \u2203 a : \u2124, f.coeff i * g.coeff j = a :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that if $f(x)$ and $g(x)$ are polynomials with rational coefficients whose product $f(x) g(x)$ has integer coefficients, then the product of any coefficient of $g(x)$ with any coefficient of $f(x)$ is an integer.", "nl_proof": "\\begin{proof}\n\n    Let $f(x), g(x) \\in \\mathbb{Q}[x]$ be such that $f(x) g(x) \\in \\mathbb{Z}[x]$.\n\nBy Gauss' Lemma there exists $r, s \\in \\mathbb{Q}$ such that $r f(x), s g(x) \\in \\mathbb{Z}[x]$, and $(r f(x))(s g(x))=r s f(x) g(x)=f(x) g(x)$. From this last relation we can conclude that $s=r^{-1}$.\n\n\n\nTherefore for any coefficient $f_i$ of $f(x)$ and $g_j$ of $g(x)$ we have that $r f_i, r^{-1} g_j \\in$ $\\mathbb{Z}$ and by multiplicative closure and commutativity of $\\mathbb{Z}$ we have that $r f_i r^{-1} g_j=$ $f_i g_j \\in \\mathbb{Z}$\n\n\\end{proof}"}
 {"id": "Dummit-Foote|exercise_9_4_2b", "formal_statement": "theorem exercise_9_4_2b : irreducible \n  (X^6 + 30*X^5 - 15*X^3 + 6*X - 120 : polynomial \u2124) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $x^6+30x^5-15x^3 + 6x-120$ is irreducible in $\\mathbb{Z}[x]$.", "nl_proof": "\\begin{proof}\n\n    $$\n\nx^6+30 x^5-15 x^3+6 x-120\n\n$$\n\nThe coefficients of the low order.: $30,-15,0,6,-120$\n\nThey are divisible by the prime 3 , but $3^2=9$ doesn 't divide $-120$. So this polynomial is irreducible over $\\mathbb{Z}$. \n\n\\end{proof}"}
-{"id": "Dummit-Foote|exercise_9_4_2d", "formal_statement": "theorem exercise_9_4_2d {p : \u2115} (hp : p.prime \u2227 p > 2) \n  {f : polynomial \u2124} (hf : f = (X + 2)^p): \n  irreducible (\u2211 n in (f.support - {0}), (f.coeff n) * X ^ (n-1) : \n  polynomial \u2124) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $\\frac{(x+2)^p-2^p}{x}$, where $p$ is an odd prime, is irreducible in $\\mathbb{Z}[x]$.", "nl_proof": "\\begin{proof}\n\n$\\frac{(x+2)^p-2^p}{x} \\quad \\quad p$ is on add pprime $Z[x]$\n\n$$\n\n\\frac{(x+2)^p-2^p}{x} \\quad \\text { as a polynomial we expand }(x+2)^p\n\n$$\n\n$2^p$ cancels with $-2^p$, every remaining term has $x$ as $a$ factor\n\n$$\n\n\\begin{aligned}\n\n& x^{p-1}+2\\left(\\begin{array}{l}\n\np \\\\\n\n1\n\n\\end{array}\\right) x^{p-2}+2^2\\left(\\begin{array}{l}\n\np \\\\\n\n2\n\n\\end{array}\\right) x^{p-3}+\\ldots+2^{p-1}\\left(\\begin{array}{c}\n\np \\\\\n\np-1\n\n\\end{array}\\right) \\\\\n\n& 2^k\\left(\\begin{array}{l}\n\np \\\\\n\nk\n\n\\end{array}\\right) x^{p-k-1}=2^k \\cdot p \\cdot(p-1) \\ldots(p-k-1), \\quad 0<k<p\n\n\\end{aligned}\n\n$$\n\nEvery lower order coef. has $p$ as a factor but doesnt have $\\$ \\mathrm{p}^{\\wedge} 2 \\$$ as a fuction so the polynomial is irreducible by Eisensteins Criterion.\n\n\\end{proof}"}
+{"id": "Dummit-Foote|exercise_9_4_2d", "formal_statement": "theorem exercise_9_4_2d {p : \u2115} (hp : p.prime \u2227 p > 2) \n  {f : polynomial \u2124} (hf : f = (X + 2)^p): \n  irreducible (\u2211 n in (f.support \\ {0}), (f.coeff n) * X ^ (n-1) : \n  polynomial \u2124) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $\\frac{(x+2)^p-2^p}{x}$, where $p$ is an odd prime, is irreducible in $\\mathbb{Z}[x]$.", "nl_proof": "\\begin{proof}\n\n$\\frac{(x+2)^p-2^p}{x} \\quad \\quad p$ is on add pprime $Z[x]$\n\n$$\n\n\\frac{(x+2)^p-2^p}{x} \\quad \\text { as a polynomial we expand }(x+2)^p\n\n$$\n\n$2^p$ cancels with $-2^p$, every remaining term has $x$ as $a$ factor\n\n$$\n\n\\begin{aligned}\n\n& x^{p-1}+2\\left(\\begin{array}{l}\n\np \\\\\n\n1\n\n\\end{array}\\right) x^{p-2}+2^2\\left(\\begin{array}{l}\n\np \\\\\n\n2\n\n\\end{array}\\right) x^{p-3}+\\ldots+2^{p-1}\\left(\\begin{array}{c}\n\np \\\\\n\np-1\n\n\\end{array}\\right) \\\\\n\n& 2^k\\left(\\begin{array}{l}\n\np \\\\\n\nk\n\n\\end{array}\\right) x^{p-k-1}=2^k \\cdot p \\cdot(p-1) \\ldots(p-k-1), \\quad 0<k<p\n\n\\end{aligned}\n\n$$\n\nEvery lower order coef. has $p$ as a factor but doesnt have $\\$ \\mathrm{p}^{\\wedge} 2 \\$$ as a fuction so the polynomial is irreducible by Eisensteins Criterion.\n\n\\end{proof}"}
 {"id": "Dummit-Foote|exercise_9_4_11", "formal_statement": "theorem exercise_9_4_11 : \n  irreducible ((X 0)^2 + (X 1)^2 - 1 : mv_polynomial (fin 2) \u211a) :=", "src_header": "import .common \n\nopen set function nat int fintype real polynomial mv_polynomial\nopen subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix\n\nopen_locale pointwise\nopen_locale big_operators\nnoncomputable theory\n\n", "nl_statement": "Prove that $x^2+y^2-1$ is irreducible in $\\mathbb{Q}[x,y]$.", "nl_proof": "\\begin{proof}\n\n$$\n\np(x)=x^2+y^2-1 \\in Q[y][x] \\cong Q[y, x]\n\n$$\n\nWe have that $y+1 \\in Q[y]$ is prime and $Q[y]$ is an UFD, since $p(x)=x^2+y^2-1=x^2+$ $(y+1)(y-1)$ by the Eisenstein criterion $x^2+y^2-1$ is irreducibile in $Q[x, y]$.\n\n\\end{proof}"}