index int64 1 367 | name stringlengths 17 29 | textbook stringclasses 10
values | header stringclasses 60
values | helper stringclasses 14
values | informal_stmt stringlengths 42 474 | formal_stmt stringlengths 66 545 | informal_proof stringlengths 40 3.85k | formal_proof stringclasses 1
value |
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201 | Axler_exercise_3_8 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators | Suppose that $V$ is finite dimensional and that $T \in \mathcal{L}(V, W)$. Prove that there exists a subspace $U$ of $V$ such that $U \cap \operatorname{null} T=\{0\}$ and range $T=\{T u: u \in U\}$. | theorem Axler_exercise_3_8 {F V W : Type*} [AddCommGroup V]
[AddCommGroup W] [Field F] [Module F V] [FiniteDimensional F V] [Module F W]
(L : V →ₗ[F] W) :
∃ U : Submodule F V, U ⊓ (ker L) = ⊥ ∧
(range L = range (domRestrict L U)):= by
sorry | The point here is to note that every subspace of a vector space has a complementary subspace.
In this example, $U$ will precisely turn out to be the complementary subspace of null $T$. That is, $V=U \oplus \operatorname{null} T$
How should we characterize $U$ ? This can be achieved by extending a basis $B_1=\left\{v_1,... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
202 | Axler_exercise_5_1 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators | Suppose $T \in \mathcal{L}(V)$. Prove that if $U_{1}, \ldots, U_{m}$ are subspaces of $V$ invariant under $T$, then $U_{1}+\cdots+U_{m}$ is invariant under $T$. | theorem Axler_exercise_5_1 {F V : Type*} [AddCommGroup V] [Field F]
[Module F V] {T : V →ₗ[F] V} {n : ℕ} (U : Fin n → Submodule F V)
(hU : ∀ i : Fin n, Submodule.map T (U i) ≤ U i) :
Submodule.map T (∑ i : Fin n, U i : Submodule F V) ≤
(∑ i : Fin n, U i : Submodule F V) := by
sorry | First off, assume that $U_1, \ldots, U_m$ are subspaces of $V$ invariant under $T$. Now, consider a vector $u \in$ $U_1+\ldots+U_m$. There does exist $u_1 \in U_1, \ldots, u_m \in U_m$ such that $u=u_1+\ldots+u_m$.
Once you apply $T$ towards both sides of the previous equation, we would then get $T u=T u_1+\ldots+$ $T... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
203 | Axler_exercise_5_11 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators | Suppose $S, T \in \mathcal{L}(V)$. Prove that $S T$ and $T S$ have the same eigenvalues. | theorem Axler_exercise_5_11 {F V : Type*} [Field F] [AddCommGroup V]
[Module F V] [FiniteDimensional F V] (S T : End F V) :
(S * T).Eigenvalues = (T * S).Eigenvalues := by
sorry | To start, let $\lambda \in F$ be an eigenvalue of $S T$. Now, we would want $\lambda$ to be an eigenvalue of $T S$. Since $\lambda$, by itself, is an eigenvalue of $S T$, then there has to be a nonzero vector $v \in V$ such that $(S T) v=\lambda v$.
Now, With a given reference that $(S T) v=\lambda v$, you will then ha... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
204 | Axler_exercise_5_13 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators | 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. | theorem Axler_exercise_5_13 {F V : Type*} [AddCommGroup V] [Field F]
[Module F V] [FiniteDimensional F V] {T : End F V}
(hS : ∀ U : Submodule F V, finrank F U = finrank F V - 1 →
Submodule.map T U ≤ U) : ∃ c : F, T = c • LinearMap.id := by
sorry | 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.
Next, 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{s... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
205 | Axler_exercise_5_24 | Axler | import Mathlib
open Filter Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators | 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. | theorem Axler_exercise_5_24 {V : Type*} [AddCommGroup V]
[Module ℝ V] [FiniteDimensional ℝ V] {T : End ℝ V}
(hT : ∀ c : ℝ, eigenspace T c = ⊥) {U : Submodule ℝ V}
(hU : Submodule.map T U ≤ U) : Even (finrank ℝ U) := by
sorry | 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
$$
\left.T\righ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
206 | Axler_exercise_6_3 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators InnerProductSpace | Prove that $\left(\sum_{j=1}^{n} a_{j} b_{j}\right)^{2} \leq\left(\sum_{j=1}^{n} j a_{j}{ }^{2}\right)\left(\sum_{j=1}^{n} \frac{b_{j}{ }^{2}}{j}\right)$ for all real numbers $a_{1}, \ldots, a_{n}$ and $b_{1}, \ldots, b_{n}$. | theorem Axler_exercise_6_3 {n : ℕ} (a b : Fin n → ℝ) :
(∑ i, a i * b i) ^ 2 ≤ (∑ i : Fin n, (i + 1) * a i ^ 2) * (∑ i, b i ^ 2 / (i + 1)) := by
sorry | Let $a_1, a_2, \ldots, a_n, b_1, b_2, \ldots, b_n \in R$.
We have that
$$
\left(\sum_{j=1}^n a_j b_j\right)^2
$$
is equal to the
$$
\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
$$
This can be observed as an inner pr... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
207 | Axler_exercise_6_13 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End InnerProductSpace
open scoped BigOperators ComplexConjugate | Suppose $\left(e_{1}, \ldots, e_{m}\right)$ is an orthonormal list of vectors in $V$. Let $v \in V$. Prove that $\|v\|^{2}=\left|\left\langle v, e_{1}\right\rangle\right|^{2}+\cdots+\left|\left\langle v, e_{m}\right\rangle\right|^{2}$ if and only if $v \in \operatorname{span}\left(e_{1}, \ldots, e_{m}\right)$. | theorem Axler_exercise_6_13 {V : Type*} [NormedAddCommGroup V] [RCLike F] [InnerProductSpace F V] {n : ℕ}
{e : Fin n → V} (he : Orthonormal F e) (v : V) :
‖v‖^2 = ∑ i : Fin n, ‖⟪v, e i⟫_F‖^2 ↔ v ∈ Submodule.span F (e '' Set.univ) := by
sorry | If $v \in \operatorname{span}\left(e_1, \ldots, e_m\right)$, it means that
$$
v=\alpha_1 e_1+\ldots+\alpha_m e_m .
$$
for some scalars $\alpha_i$. We know that $\alpha_k=\left\langle v, e_k\right\rangle, \forall k \in\{1, \ldots, m\}$. Therefore,
$$
\begin{aligned}
\|v\|^2 & =\langle v, v\rangle \\
& =\left\langle\alph... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
208 | Axler_exercise_7_5 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators InnerProductSpace | Show that if $\operatorname{dim} V \geq 2$, then the set of normal operators on $V$ is not a subspace of $\mathcal{L}(V)$. | theorem Axler_exercise_7_5 {V : Type*} [NormedAddCommGroup V] [RCLike F] [InnerProductSpace F V]
[FiniteDimensional F V] (hV : finrank F V ≥ 2) :
∀ U : Submodule F (End F V), U.carrier ≠
{T | T * adjoint T = adjoint T * T} := by
sorry | 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 s... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
209 | Axler_exercise_7_9 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators ComplexConjugate | Prove that a normal operator on a complex inner-product space is self-adjoint if and only if all its eigenvalues are real. | theorem Axler_exercise_7_9 {V : Type*} [NormedAddCommGroup V] [InnerProductSpace ℂ V]
[FiniteDimensional ℂ V] (T : End ℂ V)
(hT : T * adjoint T = adjoint T * T) :
IsSelfAdjoint T ↔ ∀ e : T.Eigenvalues, (e : ℂ).im = 0 := by
sorry | 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 eig... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
210 | Axler_exercise_7_11 | Axler | import Mathlib
open Fintype Complex Polynomial LinearMap FiniteDimensional Module Module.End
open scoped BigOperators | 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 $S^{2}=T$.) | theorem Axler_exercise_7_11 {V : Type*} [NormedAddCommGroup V] [InnerProductSpace ℂ V]
[FiniteDimensional ℂ V] {T : End ℂ V} (hT : T*adjoint T = adjoint T*T) :
∃ (S : End ℂ V), S ^ 2 = T := by
sorry | Let $V$ be a complex inner product space.
It is known that an operator $S \in \mathcal{L}(V)$ is called a square root of $T \in \mathcal{L}(V)$ if
$$
S^2=T
$$
Now, suppose that $T$ is a normal operator on $V$.
By the Complex Spectral Theorem, there is $e_1, \ldots, e_n$ an orthonormal basis of $V$ consisting of eigenva... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
211 | Dummit_Foote_exercise_1_1_2a | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove the the operation $\star$ on $\mathbb{Z}$ defined by $a\star b=a-b$ is not commutative. | theorem Dummit_Foote_exercise_1_1_2a : ∃ a b : ℤ, a - b ≠ b - a := by
sorry | Not commutative since
$$
1 \star(-1)=1-(-1)=2
$$
$$
(-1) \star 1=-1-1=-2 .
$$ | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
212 | Dummit_Foote_exercise_1_1_4 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that the multiplication of residue class $\mathbb{Z}/n\mathbb{Z}$ is associative. | theorem Dummit_Foote_exercise_1_1_4 (n : ℕ) (a b c : ZMod n) :
(a * b) * c = a * (b * c) := by
sorry | We have
$$
\begin{aligned}
(\bar{a} \cdot \bar{b}) \cdot \bar{c} &=\overline{a \cdot b} \cdot \bar{c} \\
&=\overline{(a \cdot b) \cdot c} \\
&=\overline{a \cdot(b \cdot c)} \\
&=\bar{a} \cdot \overline{b \cdot c} \\
&=\bar{a} \cdot(\bar{b} \cdot \bar{c})
\end{aligned}
$$
since integer multiplication is associative. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
213 | Dummit_Foote_exercise_1_1_15 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that $(a_1a_2\dots a_n)^{-1} = a_n^{-1}a_{n-1}^{-1}\dots a_1^{-1}$ for all $a_1, a_2, \dots, a_n\in G$. | theorem Dummit_Foote_exercise_1_1_15 {G : Type*} [Group G] (as : List G) :
as.prod⁻¹ = (as.reverse.map (λ x => x⁻¹)).prod := by
sorry | For $n=1$, note that for all $a_1 \in G$ we have $a_1^{-1}=a_1^{-1}$.
Now 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
$$
\left(a_1 \cdot a_2\right)^{-1}=a_2^{-1} \cdot a_1^{-1}
$$
since
$$
a_1 \cdot a_2 \cdot a_2^{-1} a_1^{-1}=1 .
$$
For the inductive ste... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
214 | Dummit_Foote_exercise_1_1_17 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $x$ be an element of $G$. Prove that if $|x|=n$ for some positive integer $n$ then $x^{-1}=x^{n-1}$. | theorem Dummit_Foote_exercise_1_1_17 {G : Type*} [Group G] {x : G} {n : ℕ+}
(hxn: orderOf x = n) :
x⁻¹ = x ^ (n - 1 : ℕ) := by
sorry | We have $x \cdot x^{n-1}=x^n=1$, so by the uniqueness of inverses $x^{-1}=x^{n-1}$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
215 | Dummit_Foote_exercise_1_1_20 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | For $x$ an element in $G$ show that $x$ and $x^{-1}$ have the same order. | theorem Dummit_Foote_exercise_1_1_20 {G : Type*} [Group G] {x : G} :
orderOf x = orderOf x⁻¹ := by
sorry | Recall that the order of a group element is either a positive integer or infinity.
Suppose $|x|$ is infinite and that $\left|x^{-1}\right|=n$ for some $n$. Then
$$
x^n=x^{(-1) \cdot n \cdot(-1)}=\left(\left(x^{-1}\right)^n\right)^{-1}=1^{-1}=1,
$$
a contradiction. So if $|x|$ is infinite, $\left|x^{-1}\right|$ must als... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
216 | Dummit_Foote_exercise_1_1_22b | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Deduce that $|a b|=|b a|$ for all $a, b \in G$. | theorem Dummit_Foote_exercise_1_1_22b {G: Type*} [Group G] (a b : G) :
orderOf (a * b) = orderOf (b * a) := by
sorry | Let $a$ and $b$ be arbitrary group elements. Letting $x=a b$ and $g=a$, we see that
$$
|a b|=\left|a^{-1} a b a\right|=|b a| .
$$ | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
217 | Dummit_Foote_exercise_1_1_29 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that $A \times B$ is an abelian group if and only if both $A$ and $B$ are abelian. | theorem Dummit_Foote_exercise_1_1_29 {A B : Type*} [Group A] [Group B] :
(∀ x y : A × B, x * y = y * x) ↔ (∀ x y : A, x * y = y * x) ∧
(∀ x y : B, x * y = y * x) := by
sorry | $(\Rightarrow)$ Suppose $a_1, a_2 \in A$ and $b_1, b_2 \in B$. Then
$$
\left(a_1 a_2, b_1 b_2\right)=\left(a_1, b_1\right) \cdot\left(a_2, b_2\right)=\left(a_2, b_2\right) \cdot\left(a_1, b_1\right)=\left(a_2 a_1, b_2 b_1\right) .
$$
Since two pairs are equal precisely when their corresponding entries are equal, we hav... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
218 | Dummit_Foote_exercise_1_3_8 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $\Omega=\{1,2,3, \ldots\}$ then $S_{\Omega}$ is an infinite group | theorem Dummit_Foote_exercise_1_3_8 : Infinite (Equiv.Perm ℕ) := by
sorry | Recall that the codomain of an injective function must be at least as large (in cardinality) as the domain of the function. With that in mind, define the function
$$
\begin{gathered}
f: \mathbb{N} \rightarrow S_{\mathbb{N}} \\
f(n)=(1 n)
\end{gathered}
$$
where $(1 n)$ is the cycle decomposition of an element of $S_{\m... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
219 | Dummit_Foote_exercise_1_6_11 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $A$ and $B$ be groups. Prove that $A \times B \cong B \times A$. | theorem Dummit_Foote_exercise_1_6_11 {A B : Type*} [Group A] [Group B] :
Nonempty (A × B ≃* B × A) := by
sorry | We know from set theory that the mapping $\varphi: A \times B \rightarrow B \times A$ given by $\varphi((a, b))=(b, a)$ is a bijection with inverse $\psi: B \times A \rightarrow A \times B$ given by $\psi((b, a))=(a, b)$. Also $\varphi$ is a homomorphism, as we show below.
Let $a_1, a_2 \in A$ and $b_1, b_2 \in B$. The... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
220 | Dummit_Foote_exercise_1_6_23 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $G$ be a finite group which possesses an automorphism $\sigma$ such that $\sigma(g)=g$ if and only if $g=1$. If $\sigma^{2}$ is the identity map from $G$ to $G$, prove that $G$ is abelian. | theorem Dummit_Foote_exercise_1_6_23 {G : Type*} [Group G] [Finite G]
(σ : MulAut G) (hs : ∀ g : G, σ g = g ↔ g = 1)
(hs2 : ∀ g : G, σ (σ g) = g) :
∀ x y : G, x * y = y * x := by
sorry | Solution: We define a mapping $f: G \rightarrow G$ by $f(x)=x^{-1} \sigma(x)$.
Claim: $f$ is injective.
Proof of claim: Suppose $f(x)=f(y)$. Then $y^{-1} \sigma(y)=x^{-1} \sigma(x)$, so that $x y^{-1}=\sigma(x) \sigma\left(y^{-1}\right)$, and $x y^{-1}=\sigma\left(x y^{-1}\right)$. Then we have $x y^{-1}=1$, hence $x=y... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
221 | Dummit_Foote_exercise_2_1_13 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $H$ be a subgroup of the additive group of rational numbers with the property that $1 / x \in H$ for every nonzero element $x$ of $H$. Prove that $H=0$ or $\mathbb{Q}$. | theorem Dummit_Foote_exercise_2_1_13 (H : AddSubgroup ℚ)
(hH : ∀ ⦃x : ℚ⦄, x ∈ H → x ≠ 0 → (1 / x) ∈ H) :
H = ⊥ ∨ H = ⊤ := by
sorry | Solution: First, suppose there does not exist a nonzero element in $H$. Then $H=0$.
Now suppose there does exist a nonzero element $a \in H$; without loss of generality, say $a=p / q$ in lowest terms for some integers $p$ and $q$ - that is, $\operatorname{gcd}(p, q)=1$. Now $q \cdot \frac{p}{q}=p \in H$, and since $q /... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
222 | Dummit_Foote_exercise_2_4_16a | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | A subgroup $M$ of a group $G$ is called a maximal subgroup if $M \neq G$ and the only subgroups of $G$ which contain $M$ are $M$ and $G$. Prove that if $H$ is a proper subgroup of the finite group $G$ then there is a maximal subgroup of $G$ containing $H$. | theorem Dummit_Foote_exercise_2_4_16a {G : Type*} [Group G] [Fintype G] {H : Subgroup G}
(hH : H ≠ ⊤) :
∃ M : Subgroup G, M ≠ ⊤ ∧
(∀ K : Subgroup G, M ≤ K → K = M ∨ K = ⊤) ∧
H ≤ M := by
sorry | If $H$ is maximal, then we are done. If $H$ is not maximal, then there is a subgroup $K_1$ of $G$ such that $H<K_1<G$. If $K_1$ is maximal, we are done. But if $K_1$ is not maximal, there is a subgroup $K_2$ with $H<K_1<K_2<G$. If $K_2$ is maximal, we are done, and if not, keep repeating the procedure. Since $G$ is fin... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
223 | Dummit_Foote_exercise_2_4_16c | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Show that if $G=\langle x\rangle$ is a cyclic group of order $n \geq 1$ then a subgroup $H$ is maximal if and only $H=\left\langle x^{p}\right\rangle$ for some prime $p$ dividing $n$. | theorem Dummit_Foote_exercise_2_4_16c {n : ℕ+} {G : Type*} [Group G] {x : G}
(hx : orderOf x = n) (hG : closure ({x} : Set G) = ⊤) (H : Subgroup G) :
(∃ p : ℕ, Prime p ∧ p ∣ n ∧ H = Subgroup.closure {x ^ p}) ↔
(H ≠ ⊤ ∧ ∀ K : Subgroup G, H ≤ K → K = H ∨ K = ⊤) := by
sorry | Suppose $H$ is a maximal subgroup of $G$. Then $H$ is cyclic, and we may write $H=\left\langle x^k\right\rangle$ for some integer $k$, with $k>1$. Let $d=(n, k)$. Since $H$ is a proper subgroup, we know by Proposition 6 that $d>1$. Choose a prime factor $p$ of $d$. If $k=p=d$ then $k \mid n$ as required.
If, however, ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
224 | Dummit_Foote_exercise_3_1_22a | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $H$ and $K$ are normal subgroups of a group $G$ then their intersection $H \cap K$ is also a normal subgroup of $G$. | theorem Dummit_Foote_exercise_3_1_22a (G : Type*) [Group G] (H K : Subgroup G)
[Normal H] [Normal K] :
Normal (H ⊓ K) := by
sorry | Suppose $H$ and $K$ are normal subgroups of $G$. We already know that $H \cap K$ is a subgroup of $G$, so we need to show that it is normal. Choose any $g \in G$ and any $x \in H \cap K$. Since $x \in H$ and $H \unlhd G$, we know $g x g^{-1} \in H$. Likewise, since $x \in K$ and $K \unlhd G$, we have $g x g^{-1} \in K$... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
225 | Dummit_Foote_exercise_3_2_8 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $H$ and $K$ are finite subgroups of $G$ whose orders are relatively prime then $H \cap K=1$. | theorem Dummit_Foote_exercise_3_2_8 {G : Type*} [Group G] (H K : Subgroup G)
[Fintype H] [Fintype K]
(hHK : Nat.Coprime (card H) (card K)) :
H ⊓ K = ⊥ := by
sorry | Solution: Let $|H|=p$ and $|K|=q$. We saw in a previous exercise that $H \cap K$ is a subgroup of both $H$ and $K$; by Lagrange's Theorem, then, $|H \cap K|$ divides $p$ and $q$. Since $\operatorname{gcd}(p, q)=1$, then, $|H \cap K|=1$. Thus $H \cap K=1$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
226 | Dummit_Foote_exercise_3_2_16 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Use Lagrange's Theorem in the multiplicative group $(\mathbb{Z} / p \mathbb{Z})^{\times}$to prove Fermat's Little Theorem: if $p$ is a prime then $a^{p} \equiv a(\bmod p)$ for all $a \in \mathbb{Z}$. | theorem Dummit_Foote_exercise_3_2_16 (p : ℕ) (hp : Nat.Prime p) (a : ℤ) :
a ^ p ≡ a [ZMOD p] := by
sorry | Solution: If $p$ is prime, then $\varphi(p)=p-1$ (where $\varphi$ denotes the Euler totient). Thus
$$
\mid\left((\mathbb{Z} /(p))^{\times} \mid=p-1 .\right.
$$
So for all $a \in(\mathbb{Z} /(p))^{\times}$, we have $|a|$ divides $p-1$. Hence
$$
a=1 \cdot a=a^{p-1} a=a^p \quad(\bmod p) .
$$ | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
227 | Dummit_Foote_exercise_3_3_3 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $H$ is a normal subgroup of $G$ of prime index $p$ then for all $K \leq G$ either $K \leq H$, or $G=H K$ and $|K: K \cap H|=p$. | theorem Dummit_Foote_exercise_3_3_3 {p : Nat.Primes} {G : Type*} [Group G]
{H : Subgroup G} [hH : H.Normal] (hH1 : H.index = p) :
∀ K : Subgroup G, K ≤ H ∨ (H ⊔ K = ⊤ ∧ (K ⊓ H).relIndex K = p) := by
sorry | Solution: Suppose $K \backslash N \neq \emptyset$; say $k \in K \backslash N$. Now $G / N \cong \mathbb{Z} /(p)$ is cyclic, and moreover is generated by any nonidentity- in particular by $\bar{k}$
Now $K N \leq G$ since $N$ is normal. Let $g \in G$. We have $g N=k^a N$ for some integer a. In particular, $g=k^a n$ for ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
228 | Dummit_Foote_exercise_3_4_4 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Use Cauchy's Theorem and induction to show that a finite abelian group has a subgroup of order $n$ for each positive divisor $n$ of its order. | theorem Dummit_Foote_exercise_3_4_4 {G : Type*} [CommGroup G] [Fintype G] {n : ℕ}
(hn : n ∣ (card G)) :
∃ (H : Subgroup G) (H_fin : Fintype H), @card H H_fin = n := by
sorry | Let $G$ be a finite abelian group. We use induction on $|G|$. Certainly the result holds for the trivial group. And if $|G|=p$ for some prime $p$, then the positive divisors of $|G|$ are 1 and $p$ and the result is again trivial.
Now assume that the statement is true for all groups of order strictly smaller than $|G|$... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
229 | Dummit_Foote_exercise_3_4_5b | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that quotient groups of a solvable group are solvable. | theorem Dummit_Foote_exercise_3_4_5b {G : Type*} [Group G] [IsSolvable G]
(H : Subgroup G) [Normal H] :
IsSolvable (G ⧸ H) := by
sorry | Next, note that
$$
H_i=G_i \cap H=\left(G_i \cap G_{i+1}\right) \cap H=G_i \cap H_{i+1} .
$$
By the Second Isomorphism Theorem, we then have
$$
H_{i+1} / H_i=H_{i+1} /\left(H_{i+1} \cap G_i\right) \cong H_{i+1} G_i / G_i \leq G_{i+1} / G_i .
$$
Since $H_{i+1} / H_i$ is isomorphic to a subgroup of the abelian group $G_{... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
230 | Dummit_Foote_exercise_4_2_8 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $H$ has finite index $n$ then there is a normal subgroup $K$ of $G$ with $K \leq H$ and $|G: K| \leq n!$. | theorem Dummit_Foote_exercise_4_2_8 {G : Type*} [Group G] {H : Subgroup G}
{n : ℕ} (hn : n > 0) (hH : H.index = n) :
∃ K ≤ H, K.Normal ∧ 0 < K.index ∧ K.index ≤ n.factorial := by
sorry | Solution: $G$ acts on the cosets $G / H$ by left multiplication. Let $\lambda: G \rightarrow S_{G / H}$ be the permutation representation induced by this action, and let $K$ be the kernel of the representation.
Now $K$ is normal in $G$, and $K \leq \operatorname{stab}_G(H)=H$. By the First Isomorphism Theorem, we have ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
231 | Dummit_Foote_exercise_4_2_9a | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $p$ is a prime and $G$ is a group of order $p^{\alpha}$ for some $\alpha \in \mathbb{Z}^{+}$, then every subgroup of index $p$ is normal in $G$. | theorem Dummit_Foote_exercise_4_2_9a {G : Type*} [Fintype G] [Group G] {p α : ℕ}
(hp : p.Prime) (ha : α > 0) (hG : card G = p ^ α) :
∀ H : Subgroup G, H.index = p → H.Normal := by
sorry | Solution: Let $G$ be a group of order $p^k$ and $H \leq G$ a subgroup with $[G: H]=p$. Now $G$ acts on the conjugates $g H g^{-1}$ by conjugation, since
$$
g_1 g_2 \cdot H=\left(g_1 g_2\right) H\left(g_1 g_2\right)^{-1}=g_1\left(g_2 H g_2^{-1}\right) g_1^{-1}=g_1 \cdot\left(g_2 \cdot H\right)
$$
and $1 \cdot H=1 H 1=H$... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
232 | Dummit_Foote_exercise_4_4_2 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $G$ is an abelian group of order $p q$, where $p$ and $q$ are distinct primes, then $G$ is cyclic. | theorem Dummit_Foote_exercise_4_4_2 {G : Type*} [Fintype G] [CommGroup G]
{p q : Nat.Primes} (hpq : p ≠ q) (hG : card G = p*q) :
IsCyclic G := by
sorry | Let $G$ be an abelian group of order $p q$. We need to prove that if $p$ and $q$ are distinct primes than $G$ is cyclic. By Cauchy's theorem there are $a, b \in G$ with $a$ of order $p$ and $b$ of order $q$. Since $(|a|,|b|)=1$ and $a b=b a$ then $|a b|=|a| \cdot|b|=p q$. Therefore $a b$ is an element of order $p q$, t... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
233 | Dummit_Foote_exercise_4_4_6b | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that there exists a normal subgroup that is not characteristic. | theorem Dummit_Foote_exercise_4_4_6b :
∃ (G : Type*) (hG : Group G) (H : @Subgroup G hG), @Normal G hG H ∧ ¬ @Characteristic G hG H := by
sorry | We have to produce a group $G$ and a subgroup $H$ such that $H$ is normal in $G$, but not characteristic. Consider the Klein's four group $G=\{ e, a, b, a b\}$. This is an abelian group with each element having order 2. Consider $H=\{ e, a\}$. $H$ is normal in $G$. Define $\sigma: G \rightarrow G$ as $\sigma(a)=b, \sig... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
234 | Dummit_Foote_exercise_4_4_8a | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $G$ be a group with subgroups $H$ and $K$ with $H \leq K$. Prove that if $H$ is characteristic in $K$ and $K$ is normal in $G$ then $H$ is normal in $G$. | theorem Dummit_Foote_exercise_4_4_8a {G : Type*} [Group G] (H K : Subgroup G)
(hHK : H ≤ K) [hHK1 : (H.subgroupOf K).Characteristic] [hK : K.Normal] :
H.Normal := by
sorry | We prove that $H$ is invariant under every inner automorphism of $G$. Consider a inner automorphism $\phi_g$ of $G$. Now, $\left.\phi_g\right|_K$ is a automorphism of $K$ because $K$ is normal in $G$. But $H$ is a characteristic subgroup of $K$, so $\left.\phi_g\right|_K(H) \subset H$, so in general $\phi_g(H) \subset ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
235 | Dummit_Foote_exercise_4_5_13 | Dummit Foote | import Mathlib
import Mathlib.Tactic.IntervalCases
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that a group of order 56 has a normal Sylow $p$-subgroup for some prime $p$ dividing its order. | theorem Dummit_Foote_exercise_4_5_13 {G : Type*} [Group G] [Fintype G]
(hG : card G = 56) :
∃ (p : ℕ) (P : Sylow p G), p.Prime ∧ (p ∣ card G) ∧ P.Normal := by
sorry | Since $|G|=56=2^{3}.7$, $G$ has $2-$Sylow subgroup of order $8$, as well as $7-$Sylow subgroup of order $7$. Now, we count the number of such subgroups. Let $n_{7}$ be the number of $7-$Sylow subgroup and $n_{2}$ be the number of $2-$Sylow subgroup. Now $n_{7}=1+7k$ where $1+7k|8$. The choices for $k$ are $0$ or $1$.... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
236 | Dummit_Foote_exercise_4_5_15 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that a group of order 351 has a normal Sylow $p$-subgroup for some prime $p$ dividing its order. | theorem Dummit_Foote_exercise_4_5_15 {G : Type*} [Group G] [Fintype G]
(hG : card G = 351) :
∃ (p : ℕ) (P : Sylow p G), p.Prime ∧ (p ∣ card G) ∧ P.Normal := by
sorry | Since $|G|=351=3^{2}.13$, $G$ has $3-$Sylow subgroup of order $9$, as well as $13-$Sylow subgroup of order $13$. Now, we count the number of such subgroups. Let $n_{13}$ be the number of $13-$Sylow subgroup and $n_{3}$ be the number of $3-$Sylow subgroup. Now $n_{13}=1+13k$ where $1+13k|9$. The choices for $k$ is $0$.... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
237 | Dummit_Foote_exercise_4_5_17 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $|G|=105$ then $G$ has a normal Sylow 5 -subgroup and a normal Sylow 7-subgroup. | theorem Dummit_Foote_exercise_4_5_17 {G : Type*} [Fintype G] [Group G]
(hG : card G = 105) :
(∃ (P : Sylow 5 G), P.Normal) ∧ (∃ (P : Sylow 7 G), P.Normal) := by
sorry | Since $|G|=105=3.5.7$, $G$ has $3-$Sylow subgroup of order $3$, as well as $5-$Sylow subgroup of order $5$ and, $7-$Sylow subgroup of order 7. Now, we count the number of such subgroups. Let $n_{3}$ be the number of $3-$Sylow subgroup, $n_{5}$ be the number of $5-$Sylow subgroup, and $n_{7}$ be the number of $7-$Sylow... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
238 | Dummit_Foote_exercise_4_5_19 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $|G|=6545$ then $G$ is not simple. | theorem Dummit_Foote_exercise_4_5_19 {G : Type*} [Fintype G] [Group G]
(hG : card G = 6545) : ¬ IsSimpleGroup G := by
sorry | Since $|G|=132=2^{2}.3.11$, $G$ has $2-$Sylow subgroup of order $4$, as well as $11-$Sylow subgroup of order $11$, and $3-$Sylow subgroup of order $3$. Now, we count the number of such subgroups. Let $n_{11}$ be the number of $11-$Sylow subgroup and $n_{3}$ be the number of $3-$Sylow subgroup. Now $n_{11}=1+11k$ wher... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
239 | Dummit_Foote_exercise_4_5_21 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $|G|=2907$ then $G$ is not simple. | theorem Dummit_Foote_exercise_4_5_21 {G : Type*} [Fintype G] [Group G]
(hG : card G = 2907) : ¬ IsSimpleGroup G := by
sorry | Since $|G|=2907=3^{2}.17.19$, $G$ has $19-$Sylow subgroup of order $19$. Now, we count the number of such subgroups. Let $n_{19}$ be the number of $19-$Sylow subgroup. Now $n_{19}=1+19k$ where $1+19k|3^{2}.17$. The choices for $k$ is $0$. Hence, there is a unique $19-$Sylow subgroup and hence is normal. so $G$ is not s... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
240 | Dummit_Foote_exercise_4_5_23 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $|G|=462$ then $G$ is not simple. | theorem Dummit_Foote_exercise_4_5_23 {G : Type*} [Fintype G] [Group G]
(hG : card G = 462) : ¬ IsSimpleGroup G := by
sorry | Let $G$ be a group of order $462=11 \cdot 42$. Note that 11 is a prime not dividing 42 . Let $P \in$ $S y l_{11}(G)$. [We know $P$ exists since $S y l_{11}(G) \neq \emptyset$]. Note that $|P|=11^1=11$ by definition.
The number of Sylow 11-subgroups of $G$ is of the form $1+k \cdot 11$, i.e., $n_{11} \equiv 1$ (mod 11... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
241 | Dummit_Foote_exercise_4_5_33 | Dummit Foote | import Mathlib
import Mathlib.Algebra.Group.Subgroup.Lattice
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $P$ be a normal Sylow $p$-subgroup of $G$ and let $H$ be any subgroup of $G$. Prove that $P \cap H$ is the unique Sylow $p$-subgroup of $H$. | theorem Dummit_Foote_exercise_4_5_33 {G : Type*} [Group G] [Fintype G] {p : ℕ}
(P : Sylow p G) [hP : P.Normal] (H : Subgroup G) [Fintype H] :
(∀ R : Sylow p H, R.toSubgroup = (H ⊓ P.toSubgroup).subgroupOf H) ∧
Nonempty (Sylow p H) := by
sorry | Let $G$ be a group and $P$ is a normal $p$-Sylow subgroup of $G .|G|=p^a . m$ where $p \nmid m$. Then $|P|=p^a$. Let $H$ be a subgroup of $G$. Now if $|H|=k$ such that $p \nmid k$. Then $P \cap H=\{e\}$. There is nothing to prove in this case. Let $|H|=p^b . n$, where $b \leq a$, and $p \nmid n$. Now consider $P H$ whi... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
242 | Dummit_Foote_exercise_7_1_2 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that if $u$ is a unit in $R$ then so is $-u$. | theorem Dummit_Foote_exercise_7_1_2 {R : Type*} [Ring R] {u : R}
(hu : IsUnit u) : IsUnit (-u) := by
sorry | Solution: Since $u$ is a unit, we have $u v=v u=1$ for some $v \in R$. Thus, we have
$$
(-v)(-u)=v u=1
$$
and
$$
(-u)(-v)=u v=1 .
$$
Thus $-u$ is a unit. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
243 | Dummit_Foote_exercise_7_1_12 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that any subring of a field which contains the identity is an integral domain. | theorem Dummit_Foote_exercise_7_1_12 {F : Type*} [Field F] {K : Subring F}
(hK : (1 : F) ∈ K) : IsDomain K := by
sorry | Solution: Let $R \subseteq F$ be a subring of a field. (We need not yet assume that $1 \in R$ ). Suppose $x, y \in R$ with $x y=0$. Since $x, y \in F$ and the zero element in $R$ is the same as that in $F$, either $x=0$ or $y=0$. Thus $R$ has no zero divisors. If $R$ also contains 1 , then $R$ is an integral domain. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
244 | Dummit_Foote_exercise_7_2_2 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators nonZeroDivisors | Let $p(x)=a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}$ be an element of the polynomial ring $R[x]$. Prove that $p(x)$ is a zero divisor in $R[x]$ if and only if there is a nonzero $b \in R$ such that $b p(x)=0$. | theorem Dummit_Foote_exercise_7_2_2 {R : Type*} [CommRing R] (p : Polynomial R) :
(∃ q : Polynomial R, q ≠ 0 ∧ p * q = 0) ↔
(∃ b : R, b ≠ 0 ∧ b • p = 0) := by
sorry | Solution: If $b p(x)=0$ for some nonzero $b \in R$, then it is clear that $p(x)$ is a zero divisor.
Now suppose $p(x)$ is a zero divisor; that is, for some $q(x)=\sum_{i=0}^m b_i x^i$, we have $p(x) q(x)=0$. We may choose $q(x)$ to have minimal degree among the nonzero polynomials with this property.
We will now show b... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
245 | Dummit_Foote_exercise_7_3_16 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $\varphi: R \rightarrow S$ be a surjective homomorphism of rings. Prove that the image of the center of $R$ is contained in the center of $S$. | theorem Dummit_Foote_exercise_7_3_16 {R S : Type*} [Ring R] [Ring S]
{φ : R →+* S} (hf : Function.Surjective φ) :
φ '' (center R) ⊆ center S := by
sorry | Suppose $r \in \varphi[Z(R)]$. Then $r=\varphi(z)$ for some $z \in Z(R)$. Now let $x \in S$. Since $\varphi$ is surjective, we have $x=\varphi y$ for some $y \in R$. Now
$$
x r=\varphi(y) \varphi(z)=\varphi(y z)=\varphi(z y)=\varphi(z) \varphi(y)=r x .
$$
Thus $r \in Z(S)$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
246 | Dummit_Foote_exercise_7_4_27 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $R$ be a commutative ring with $1 \neq 0$. Prove that if $a$ is a nilpotent element of $R$ then $1-a b$ is a unit for all $b \in R$. | theorem Dummit_Foote_exercise_7_4_27 {R : Type*} [CommRing R] (hR : (0 : R) ≠ 1)
{a : R} (ha : IsNilpotent a) (b : R) :
IsUnit (1-a*b) := by
sorry | $\mathfrak{N}(R)$ is an ideal of $R$. Thus for all $b \in R,-a b$ is nilpotent. Hence $1-a b$ is a unit in $R$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
247 | Dummit_Foote_exercise_8_2_4 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $R$ be an integral domain. Prove that if the following two conditions hold then $R$ is a Principal Ideal Domain: (i) any two nonzero elements $a$ and $b$ in $R$ have a greatest common divisor which can be written in the form $r a+s b$ for some $r, s \in R$, and (ii) if $a_{1}, a_{2}, a_{3}, \ldots$ are nonzero elem... | theorem Dummit_Foote_exercise_8_2_4 {R : Type*} [CommRing R] [IsDomain R] [GCDMonoid R]
(h1 : ∀ a b : R, a ≠ 0 → b ≠ 0 → ∃ r s : R, gcd a b = r*a + s*b)
(h2 : ∀ a : ℕ → R, (∀ i : ℕ, a i ≠ 0 ∧ a (i + 1) ∣ a i) →
∃ N : ℕ, ∀ n ≥ N, ∃ u : R, IsUnit u ∧ a n = u * a N) :
IsPrincipalIdealRing R := by
sorry | Let $I \leq R$ be a nonzero ideal and let $I / \sim$ be the set of equivalence classes of elements of $I$ with regards to the relation of being associates. We can equip $I / \sim$ with a partial order with $[x] \leq[y]$ if $y \mid x$. Condition (ii) implies all chains in $I / \sim$ have an upper bound, so By Zorn's lem... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
248 | Dummit_Foote_exercise_8_3_5a | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $R=\mathbb{Z}[\sqrt{-n}]$ where $n$ is a squarefree integer greater than 3. Prove that $2, \sqrt{-n}$ and $1+\sqrt{-n}$ are irreducibles in $R$. | theorem Dummit_Foote_exercise_8_3_5a {n : ℤ} (hn0 : n > 3) (hn1 : Squarefree n) :
Irreducible (2 : Zsqrtd $ -n) ∧
Irreducible (⟨0, 1⟩ : Zsqrtd $ -n) ∧
Irreducible (1 + ⟨0, 1⟩ : Zsqrtd $ -n) := by
sorry | Suppose $a=a_1+a_2 \sqrt{-n}, b=b_1+b_2 \sqrt{-n} \in R$ are such that $2=a b$, then $N(a) N(b)=4$. Without loss of generality we can assume $N(a) \leq N(b)$, so $N(a)=1$ or $N(a)=2$. Suppose $N(a)=2$, then $a_1^2+n a_2^2=2$ and since $n>3$ we have $a_2=0$, which implies $a_1^2=2$, a contradiction. So $N(a)=1$ and $a$ ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
249 | Dummit_Foote_exercise_8_3_6b | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Let $q \in \mathbb{Z}$ be a prime with $q \equiv 3 \bmod 4$. Prove that the quotient ring $\mathbb{Z}[i] /(q)$ is a field with $q^{2}$ elements. | theorem Dummit_Foote_exercise_8_3_6b {q : ℕ} (hq0 : q.Prime) (hq1 : q ≡ 3 [ZMOD 4]) :
IsField (GaussianInt ⧸ span ({↑q} : Set GaussianInt)) ∧
∃ finR : Fintype (GaussianInt ⧸ span ({↑q} : Set GaussianInt)),
@Fintype.card _ finR = q ^ 2 := by
sorry | The division algorithm gives us that every element of $\mathbb{Z}[i] /\langle q\rangle$ is represented by an element $a+b i$ such that $0 \leq a, b<q$. Each such choice is distinct since if $a_1+b_1 i+\langle q\rangle=a_2+b_2 i+\langle q\rangle$, then $\left(a_1-a_2\right)+\left(b_1-b_2\right) i$ is divisible by $q$, s... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
250 | Dummit_Foote_exercise_9_1_10 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal MvPolynomial
open scoped BigOperators | Prove that the ring $\mathbb{Z}\left[x_{1}, x_{2}, x_{3}, \ldots\right] /\left(x_{1} x_{2}, x_{3} x_{4}, x_{5} x_{6}, \ldots\right)$ contains infinitely many minimal prime ideals. | theorem Dummit_Foote_exercise_9_1_10 {f : ℕ → MvPolynomial ℕ ℤ}
(hf : f = fun i => MvPolynomial.X (2*i) * MvPolynomial.X (2*i+1)) :
Infinite (minimalPrimes (MvPolynomial ℕ ℤ ⧸ span (range f))) := by
sorry | Let $R=\mathbb{Z}\left[x_1, x_2, \ldots, x_n\right]$ and consider the ideal $K=\left(x_{2 k+1} x_{2 k+2} \mid k \in \mathbb{Z}_{+}\right)$in $R$.
Consider the family of subsets $X=\left\{\left\{x_{2 k+1}, x_{2 k+2}\right\} \mid k \in \mathbb{Z}_{+}\right\}$, and $Y$ the set of choice function on $X$, ie the set of func... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
251 | Dummit_Foote_exercise_9_4_2a | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that $x^4-4x^3+6$ is irreducible in $\mathbb{Z}[x]$. | theorem Dummit_Foote_exercise_9_4_2a : Irreducible (X^4 - 4*X^3 + 6 : Polynomial ℤ) := by
sorry | $$
x^4-4 x^3+6
$$
The polynomial is irreducible by Eisenstein's Criterion since the prime $2$ doesn't divide the leading coefficient 2 divide coefficients of the low order term $-4,0,0$ but 6 is not divided by the square of 2. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
252 | Dummit_Foote_exercise_9_4_2c | Dummit Foote | import Mathlib
import Mathlib.Tactic.IntervalCases
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that $x^4+4x^3+6x^2+2x+1$ is irreducible in $\mathbb{Z}[x]$. | theorem Dummit_Foote_exercise_9_4_2c : Irreducible
(X^4 + 4*X^3 + 6*X^2 + 2*X + 1 : Polynomial ℤ) := by
sorry | $$
p(x)=x^4+6 x^3+4 x^2+2 x+1
$$
We calculate $p(x-1)$
$$
\begin{aligned}
(x-1)^4 & =x^4-4 x^3+6 x^2-4 x+1 \\
6(x-1)^3 & =6 x^3-18 x^2+18 x-6 \\
4(x-1)^2 & =4 x^2-8 x+4 \\
2(x-1) & =2 x-2 \\
1 & =1
\end{aligned}
$$
$$
\begin{aligned}
& p(x-1)=(x-1)^4+6(x-1)^3+4(x-1)^2+2(x-1)+1=x^4+2 x^3-8 x^2+ \\
& 8 x-2 \\
& q(x)=x^4+... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
253 | Dummit_Foote_exercise_9_4_9 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that the polynomial $x^{2}-\sqrt{2}$ is irreducible over $\mathbb{Z}[\sqrt{2}]$. You may assume that $\mathbb{Z}[\sqrt{2}]$ is a U.F.D. | theorem Dummit_Foote_exercise_9_4_9 :
Irreducible (X^2 - C Zsqrtd.sqrtd : Polynomial (Zsqrtd 2)) := by
sorry | $Z[\sqrt{2}]$ is an Euclidean domain, and so a unique factorization domain.
We have to prove $p(x)=x^2-\sqrt{2}$ irreducible.
Suppose to the contrary.
if $p(x)$ is reducible then it must have root.
Let $a+b \sqrt{2}$ be a root of $x^2-\sqrt{2}$.
Now we have
$$
a^2+2 b^2+2 a b \sqrt{2}=\sqrt{2}
$$
By comparing the coeff... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
254 | Dummit_Foote_exercise_11_1_13 | Dummit Foote | import Mathlib
open Fintype Subgroup Set Polynomial Ideal
open scoped BigOperators | Prove that as vector spaces over $\mathbb{Q}, \mathbb{R}^n \cong \mathbb{R}$, for all $n \in \mathbb{Z}^{+}$. | theorem Dummit_Foote_exercise_11_1_13 {n : ℕ+} :
Nonempty ((Fin n → ℝ) ≃ₗ[ℚ] ℝ) := by
sorry | Since $B$ is a basis of $V$, every element of $V$ can be written uniquely as a finite linear combination of elements of $B$. Let $X$ be the set of all such finite linear combinations. Then $X$ has the same cardinality as $V$, since the map from $X$ to $V$ that takes each linear combination to the corresponding element ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
255 | Herstein_exercise_2_1_18 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $G$ is a finite group of even order, show that there must be an element $a \neq e$ such that $a=a^{-1}$. | theorem Herstein_exercise_2_1_18 {G : Type*} [Group G]
[Fintype G] (hG2 : Even (card G)) :
∃ (a : G), a ≠ 1 ∧ a = a⁻¹ := by
sorry | First note that $a=a^{-1}$ is the same as saying $a^2=e$, where $e$ is the identity. I.e. the statement is that there exists at least one element of order 2 in $G$.
Every element $a$ of $G$ of order at least 3 has an inverse $a^{-1}$ that is not itself -- that is, $a \neq a^{-1}$. So the subset of all such elements has... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
256 | Herstein_exercise_2_1_26 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $G$ is a finite group, prove that, given $a \in G$, there is a positive integer $n$, depending on $a$, such that $a^n = e$. | theorem Herstein_exercise_2_1_26 {G : Type*} [Group G]
[Fintype G] (a : G) : ∃ (n : ℕ), n > 0 ∧ a ^ n = 1 := by
sorry | Because there are only a finite number of elements of $G$, it's clear that the set $\left\{a, a^2, a^3, \ldots\right\}$ must be a finite set and in particular, there should exist some $i$ and $j$ such that $i \neq j$ and $a^i=a^j$. WLOG suppose further that $i>j$ (just reverse the roles of $i$ and $j$ otherwise). Then ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
257 | Herstein_exercise_2_2_3 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $G$ is a group in which $(a b)^{i}=a^{i} b^{i}$ for three consecutive integers $i$, prove that $G$ is abelian. | theorem Herstein_exercise_2_2_3 {G : Type*} [Group G]
(h : ∃ n : ℤ, (∀ a b : G, (a * b) ^ n = a ^ n * b ^ n) ∧
(∀ a b : G, (a * b) ^ (n + 1) = a ^ (n + 1) * b ^ (n + 1)) ∧
(∀ a b : G, (a * b) ^ (n + 2) = a ^ (n + 2) * b ^ (n + 2))) :
∀ a b : G, a * b = b * a := by
sorry | Let $G$ be a group, $a, b \in G$ and $i$ be any integer. Then from given condition,
$$
\begin{aligned}
(a b)^i & =a^i b^i \\
(a b)^{i+1} & =a^{i+1} b^{i+1} \\
(a b)^{i+2} & =a^{i+2} b^{i+2}
\end{aligned}
$$
From first and second, we get
$$
a^{i+1} b^{i+1}=(a b)^i(a b)=a^i b^i a b \Longrightarrow b^i a=a b^i
$$
From fir... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
258 | Herstein_exercise_2_2_6c | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Let $G$ be a group in which $(a b)^{n}=a^{n} b^{n}$ for some fixed integer $n>1$ for all $a, b \in G$. For all $a, b \in G$, prove that $\left(a b a^{-1} b^{-1}\right)^{n(n-1)}=e$. | theorem Herstein_exercise_2_2_6c {G : Type*} [Group G] {n : ℕ} (hn : n > 1)
(h : ∀ (a b : G), (a * b) ^ n = a ^ n * b ^ n) :
∀ (a b : G), (a * b * a⁻¹ * b⁻¹) ^ (n * (n - 1)) = 1 := by
sorry | We start with the following two intermediate results.
(1) $(a b)^{n-1}=b^{n-1} a^{n-1}$.
(2) $a^n b^{n-1}=b^{n-1} a^n$.
To prove (1), notice by the given condition for all $a, b \in G$
$(b a)^n=b^n a^n$, for some fixed integers $n>1$.
Then,
$(b a)^n=b^n a^n \Longrightarrow b .(a b)(a b) \ldots .(a b) . a=b\left(b^{n-1}... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
259 | Herstein_exercise_2_3_16 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial Subgroup
open scoped BigOperators | If a group $G$ has no proper subgroups, prove that $G$ is cyclic of order $p$, where $p$ is a prime number. | theorem Herstein_exercise_2_3_16
{G : Type*} [Group G]
(hG : ∀ H : Subgroup G, H = ⊤ ∨ H = ⊥)
(h_nontrivial : ∃ g : G, g ≠ (1 : G)) :
IsCyclic G ∧ ∃ p : ℕ, Nat.Prime p ∧ Nat.card G = p := by
sorry | Case-1: $G=(e), e$ being the identity element in $G$. Then trivially $G$ is cyclic.
Case-2: $G \neq(e)$. Then there exists an non-identity element in $G.$ Let us consider an non-identity element in $G$, say $a\neq (e)$. Now look at the cyclic subgroup generated by $a$, that is, $\langle a\rangle$. Since
$a\neq ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
260 | Herstein_exercise_2_5_23 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | 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$. | theorem Herstein_exercise_2_5_23 {G : Type*} [Group G]
(hG : ∀ (H : Subgroup G), H.Normal) (a b : G) :
∃ (j : ℤ) , b*a = a^j * b:= by
sorry | Let $G$ be a group where each subgroup is normal in $G$. let $a, b \in G$.
$$
\begin{aligned}
\langle a\rangle\triangleright G &\Rightarrow b \cdot\langle a\rangle=\langle a\rangle \cdot b . \\
& \Rightarrow \quad b \cdot a=a^j \cdot b \text { for some } j \in \mathbb{Z}.
\end{aligned}
$$
(hence for $a_1 b \in G ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
261 | Herstein_exercise_2_5_31 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | 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$. | theorem Herstein_exercise_2_5_31 {G : Type*} [CommGroup G] [Fintype G]
{p m n : ℕ} (hp : Nat.Prime p) (hp1 : ¬ p ∣ m) (hG : card G = p^n*m)
{H : Subgroup G} [Fintype H] (hH : card H = p^n) :
Subgroup.Characteristic H := by
sorry | 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 characteristic subgroup, that is $\phi(H)=H$ for any automorphism $\phi$ of $G$. Now consider $\phi(H)$. Clearly $|\phi(H)|=p^n$. ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
262 | Herstein_exercise_2_5_43 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Prove that a group of order 9 must be abelian. | theorem Herstein_exercise_2_5_43 (G : Type*) [Group G] [Fintype G]
(hG : card G = 9) :
∀ a b : G, a * b = b * a := by
sorry | 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$.
So, we have now a group $G$ of order 9. Suppose that $G$ is cyclic, then $G$ is a... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
263 | Herstein_exercise_2_5_52 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | 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. | theorem Herstein_exercise_2_5_52 {G : Type*} [Group G] [Fintype G]
(φ : G ≃* G) {I : Finset G} (hI : ∀ x ∈ I, φ x = x⁻¹)
(hI1 : (0.75 : ℚ) * card G < card I) :
(∀ x : G, φ x = x⁻¹) ∧ ∀ x y : G, x * y = y * x := by
sorry | Let us start with considering $b$ to be an arbitrary element in $A$.
1. Show that $\left|A \cap\left(b^{-1} A\right)\right|>\frac{|G|}{2}$, where
$$
b^{-1} A=\left\{b^{-1} a \mid a \in A\right\}
$$
First 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 $... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
264 | Herstein_exercise_2_7_7 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $\varphi$ is a homomorphism of $G$ onto $G'$ and $N \triangleleft G$, show that $\varphi(N) \triangleleft G'$. | theorem Herstein_exercise_2_7_7 {G : Type*} [Group G] {G' : Type*} [Group G']
(φ : G →* G') (hφ : Function.Surjective φ) (N : Subgroup G) [N.Normal] :
(Subgroup.map φ N).Normal := by
sorry | We 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'$.
Now, let $a', b' \in \varphi(N)$. Then there exist elemen... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
265 | Herstein_exercise_2_8_15 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Prove that if $p > q$ are two primes such that $q \mid p - 1$, then any two nonabelian groups of order $pq$ are isomorphic. | theorem Herstein_exercise_2_8_15 {G H: Type*} [Fintype G] [Group G] [Fintype H] [Group H]
(hG_nonab : ∃ x y : G, x * y ≠ y * x) (hH_nonab : ∃ x y : H, x * y ≠ y * x)
{p q : ℕ} (hp : Nat.Prime p) (hq : Nat.Prime q)
(h : p > q) (h1 : q ∣ p - 1) (hG : card G = p*q) (hH : card H = p*q) :
Nonempty (G ≃* H) := by
s... | 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 orde... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
266 | Herstein_exercise_2_10_1 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | 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)$. | theorem Herstein_exercise_2_10_1 {G : Type*} [Group G] (A : Subgroup G)
[A.Normal] {b : G} (hb : b ∉ A) (hp : Nat.Prime (orderOf b)) :
A ⊓ (Subgroup.closure {b}) = ⊥ := by
sorry | If $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.
... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
267 | Herstein_exercise_2_11_7 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $P \triangleleft G$, $P$ a $p$-Sylow subgroup of $G$, prove that $\varphi(P) = P$ for every automorphism $\varphi$ of $G$. | theorem Herstein_exercise_2_11_7 {G : Type*} [Group G] {p : ℕ} (hp : Nat.Prime p)
{P : Sylow p G} (hP : P.Normal) :
Subgroup.Characteristic (P : Subgroup G) := by
sorry | Let $\phi$ be an automorphism of $G$. Let $P$ be a normal sylow p-subgroup. $\phi(P)$ is also a sylow-p subgroup. But since $P$ is normal, it is unique. Hence $\phi(P)=P$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
268 | Herstein_exercise_3_2_21 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $\sigma, \tau$ are two permutations that disturb no common element and $\sigma \tau = e$, prove that $\sigma = \tau = e$. | theorem Herstein_exercise_3_2_21 {α : Type*} {σ τ : Equiv.Perm α}
(h1 : ∀ a : α, ¬(σ a ≠ a ∧ τ a ≠ a)) (h2 : τ * σ = 1) :
σ = 1 ∧ τ = 1 := by
sorry | Note that $\sigma \tau=e$ can equivalentnly be phrased as $\tau$ being the inverse of $\sigma$. Our statement is then equivalent to the statement that an inverse of a nonidentity permutation disturbs at least one same element as that permutation. To prove this, let $\sigma$ be a nonidentity permutation, then let $\left... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
269 | Herstein_exercise_4_1_34 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Let $T$ be the group of $2\times 2$ matrices $A$ with entries in the field $\mathbb{Z}_2$ such that $\det A$ is not equal to 0. Prove that $T$ is isomorphic to $S_3$, the symmetric group of degree 3. | theorem Herstein_exercise_4_1_34 :
Nonempty (Equiv.Perm (Fin 3) ≃* Matrix.GeneralLinearGroup (Fin 2) (ZMod 2)) := by
sorry | The order of $T$ is $2^4-2^3-2^2+2=6$; we now find those six matrices:
$$
\begin{array}{ll}
A_1=\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right), & A_2=\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right) \\
A_3=\left(\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right), & A_4=\left(\begin{array}{ll}
1 & 1... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
270 | Herstein_exercise_4_2_6 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $a^2 = 0$ in $R$, show that $ax + xa$ commutes with $a$. | theorem Herstein_exercise_4_2_6 {R : Type*} [Ring R] (a x : R)
(h : a ^ 2 = 0) : a * (a * x + x * a) = (a * x + x * a) * a := by
sorry | We need to show that
$$
a(a x+x a)=(a x+x a) a \text { for } a, x \in R .
$$
Now,
$$
\begin{gathered}
a(a x+x a)=a(a x)+a(x a) \\
=a^2 x+a x a \\
=0+a x a=a x a .
\end{gathered}
$$
Again,
$$
\begin{gathered}
(a x+x a) a=(a x) a+(x a) a \\
=a x a+x a^2 \\
=a x a+0=a x a .
\end{gathered}
$$
It follows that,
$$
a(a x+x a)... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
271 | Herstein_exercise_4_3_1 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $R$ is a commutative ring and $a \in R$, let $L(a) = \{x \in R \mid xa = 0\}$. Prove that $L(a)$ is an ideal of $R$. | theorem Herstein_exercise_4_3_1 {R : Type*} [CommRing R] (a : R) :
∃ I : Ideal R, {x : R | x*a=0} = I := by
sorry | First, note that if $x \in L(a)$ and $y \in L(a)$ then $x a=0$ and $y a=0$, so that
$$
\begin{aligned}
x a-y a & =0 \\
(x-y) a & =0,
\end{aligned}
$$
i.e. $L(a)$ is an additive subgroup of $R$. (We have used the criterion that $H$ is a subgroup of $G$ if for any $h_1, h_2 \in H$ we have that $h_1 h_2^{-1} \in H$.
Now... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
272 | Herstein_exercise_4_4_9 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Show that $(p - 1)/2$ of the numbers $1, 2, \ldots, p - 1$ are quadratic residues and $(p - 1)/2$ are quadratic nonresidues $\mod p$. | theorem Herstein_exercise_4_4_9 (p : ℕ) (hp : Nat.Prime p) (hodd : p % 2 = 1) :
(∃ S : Finset (ZMod p), 0 ∉ S ∧ S.card = (p-1)/2 ∧ ∀ q ∈ S, ∃ x : ZMod p, x^2 = q) ∧
(∃ S : Finset (ZMod p), 0 ∉ S ∧ S.card = (p-1)/2 ∧ ∀ q ∈ S, ¬ ∃ x : ZMod p, x^2 = q) := by
sorry | To find all the quadratic residues $\bmod p$ among the integers $1,2, \ldots, p-1$, we compute the least positive residues modulo $p$ of the squares of the integers $1,2, \ldots, p-1\}$.
Since there are $p-1$ squares to consider, and since each congruence $x^2 \equiv a (\bmod p)$ has either zero or two solutions, ther... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
273 | Herstein_exercise_4_5_23 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Let $F = \mathbb{Z}_7$ and let $p(x) = x^3 - 2$ and $q(x) = x^3 + 2$ be in $F[x]$. Show that $p(x)$ and $q(x)$ are irreducible in $F[x]$ and that the fields $F[x]/(p(x))$ and $F[x]/(q(x))$ are isomorphic. | theorem Herstein_exercise_4_5_23 {p q: Polynomial (ZMod 7)}
(hp : p = X^3 - 2) (hq : q = X^3 + 2) :
Irreducible p ∧ Irreducible q ∧
(Nonempty $ Polynomial (ZMod 7) ⧸ span ({p} : Set $ Polynomial $ ZMod 7) ≃+*
Polynomial (ZMod 7) ⧸ span ({q} : Set $ Polynomial $ ZMod 7)) := by
sorry | We have that $p(x)$ and $q(x)$ are irreducible if they have no roots in $\mathbb{Z}_7$, which can easily be checked. E.g. for $p(x)$ we have that $p(0)=5, p(1)=6, p(2)=6, p(3)=4, p(4)=6$, $p(5)=4, p(6)=4$, and similarly for $q(x)$.
We have that every element of $F[x] /(p(x))$ is equal to $a x^2+b x+c+(p(x))$, and like... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
274 | Herstein_exercise_4_6_2 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | Prove that $f(x) = x^3 + 3x + 2$ is irreducible in $Q[x]$. | theorem Herstein_exercise_4_6_2 : Irreducible (X^3 + 3*X + 2 : Polynomial ℚ) := by
sorry | Let us assume that $f(x)$ is reducible over $\mathbb{Q}[x]$.
Then there exists a rational root of $f(x)$.
Let $p / q$ be a rational root of $f(x)$, where $\operatorname{gcd}(p, q)=1$.
Then $f(p / q)=0$.
Now,
$$
\begin{aligned}
& f(p / q)=(p / q)^3+3(p / q)+2 \\
\Longrightarrow & (p / q)^3+3(p / q)+2=0 \\
\Longrightarro... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
275 | Herstein_exercise_5_1_8 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $F$ is a field of characteristic $p \neq 0$, show that $(a + b)^m = a^m + b^m$, where $m = p^n$, for all $a, b \in F$ and any positive integer $n$. | theorem Herstein_exercise_5_1_8 {p m n: ℕ} {F : Type*} [Field F]
(hp : p ≠ 0) (hn : n > 0)
(hF : CharP F p) (a b : F) (hm : m = p ^ n) :
(a + b) ^ m = a^m + b^m := by
sorry | Since $F$ is of characteristic $p$ and we have considered arbitrary two elements $a, b$ in $F$ we have
$$
\begin{aligned}
& p a=p b=0 \\
& \Longrightarrow p^n a=p^n b=0 \\
& \Longrightarrow m a=m b=0 \text {. } \\
&
\end{aligned}
$$
Now we know from Binomial Theorem that
$$
(a+b)^m=\sum_{i=0}^m\left(\begin{array}{c}
m ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
276 | Herstein_exercise_5_3_7 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $a \in K$ is such that $a^2$ is algebraic over the subfield $F$ of $K$, show that a is algebraic over $F$. | theorem Herstein_exercise_5_3_7 {K : Type*} [Field K] {F : Subfield K}
{a : K} (ha : IsAlgebraic F (a ^ 2)) : IsAlgebraic F a := by
sorry | Since $a^2$ is algebraic over $F$, there exist a non-zero polynomial $f(x)$ in $F[x]$ such that $f\left(a^2\right)=0$. Consider a new polynomial $g(x)$ defined as $g(x)=f\left(x^2\right)$. Clearly $g(x) \in F[x]$ and $g(a)=f\left(a^2\right)= 0$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
277 | Herstein_exercise_5_4_3 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If a ∈ ℂ satisfies p(a) = 0 where p(x) = x^5 + √2·x³ + √5·x² + √7·x + √11,
then a is algebraic over ℚ of degree at most 80. | theorem Herstein_exercise_5_4_3 {a : ℂ}
(ha : a ^ 5 + (Real.sqrt 2 : ℂ) * a ^ 3 + (Real.sqrt 5 : ℂ) * a ^ 2 +
(Real.sqrt 7 : ℂ) * a + (Real.sqrt 11 : ℂ) = 0) :
∃ q : Polynomial ℚ, q ≠ 0 ∧ q.natDegree ≤ 80 ∧
Polynomial.aeval a q = 0 := by
sorry | Given $a \in \mathbb{C}$ such that $p(a)=0$, where
$$
p(x)=x^5+\sqrt{2} x^3+\sqrt{5} x^2+\sqrt{7} x+\sqrt{11}
$$
Here, we note that $p(x) \in \mathbb{Q}(\sqrt{2}, \sqrt{5}, \sqrt{7}, \sqrt{11})$ and
$$
\begin{aligned}
{[Q(\sqrt{2}, \sqrt{5}, \sqrt{7}, \sqrt{11}): \mathbb{Q}] } & =[Q(\sqrt{2}, \sqrt{5}, \sqrt{7}, \s... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
278 | Herstein_exercise_5_6_14 | Herstein | import Mathlib
open Fintype Set Real Ideal Polynomial
open scoped BigOperators | If $F$ is of characteristic $p \neq 0$, show that all the roots of $x^m - x$, where $m = p^n$, are distinct. | theorem Herstein_exercise_5_6_14 {p m n: ℕ} (hp : p ≠ 0) (hn : n > 0)
{F : Type*} [Field F] [CharP F p] (hm : m = p ^ n) :
(X ^ m - X : Polynomial F).Separable := by
sorry | Let us consider $f(x)=x^m-x$. Then $f \in F[x]$.
Claim: $f(x)$ has a multiple root in some extension of $F$ if and only if $f(x)$ is not relatively prime to its formal derivative, $f^{\prime}(x)$.
Proof of the Claim: Let us assume that $f(x)$ has a multiple root in some extension of $F$. Let $y$ be a multiple root of... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
279 | Ireland_Rosen_exercise_1_30 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | Prove that $\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ is not an integer. | theorem Ireland_Rosen_exercise_1_30 (n : ℕ) (hn : 2 ≤ n) :
¬ ∃ a : ℤ, (Finset.range (n - 1)).sum (fun i => (1 : ℚ) / ((i : ℚ) + 2)) = (a : ℚ) := by
sorry | Let $2^s$ be the largest power of 2 occurring as a denominator in $H_n$, say $2^s=k \leqslant n$. Write $H_n=$ $\frac{1}{2^s}+\left(1+1 / 2+\ldots+1 /(k-1)+1 /(k+1)+\ldots+1 / n\right.$. The sum in parentheses can be written as $1 / 2^{s-1}$ times sum of fractions with odd denominators, so the denominator of the sum in... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
280 | Ireland_Rosen_exercise_2_4 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | If $a$ is a nonzero integer, then for $n>m$ show that $\left(a^{2^{n}}+1, a^{2^{m}}+1\right)=1$ or 2 depending on whether $a$ is odd or even. | theorem Ireland_Rosen_exercise_2_4 {a : ℤ} (ha : a ≠ 0)
(hf : f_a = λ n m : ℕ => Int.gcd (a^(2^n) + 1) (a^(2^m)+1)) {n m : ℕ}
(hnm : n > m) :
(Odd a → f_a n m = 2) ∧ (Even a → f_a n m = 1) := by
sorry | \begin{align*}
\operatorname{ord}_p\, n! &= \sum_{k\geq 1} \left \lfloor \frac{n}{p^{k}}\right \rfloor \leq \sum_{k\geq 1} \frac{n}{p^{k}} = \frac{n}{p} \frac{1}{1 - \frac{1}{p}} = \frac{n}{p-1}
\end{align*}
The decomposition of $n!$ in prime factors is
$n! = p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$
whe... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
281 | Ireland_Rosen_exercise_2_27a | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | Show that $\sum^{\prime} 1 / n$, the sum being over square free integers, diverges. | theorem Ireland_Rosen_exercise_2_27a :
¬ Summable (λ i : {p : ℤ // Squarefree p} => (1 : ℚ) / i) := by
sorry | Let $S \subset \mathbb{N}^*$ the set of square free integers.
Let $N \in \mathbb{N}^*$. Every integer $n, \, 1\leq n \leq N$ can be written as $n = a b^2$, where $a,b$ are integers and $a$ is square free. Then $1\leq a \leq N$, and $1\leq b \leq \sqrt{N}$, so
$$\sum_{n\leq N} \frac{1}{n} \leq \sum_{a \in S, a\leq N}\ ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
282 | Ireland_Rosen_exercise_3_4 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | Show that the equation $3 x^{2}+2=y^{2}$ has no solution in integers. | theorem Ireland_Rosen_exercise_3_4 : ¬ ∃ x y : ℤ, 3*x^2 + 2 = y^2 := by
sorry | If $3x^2+2 = y^2$, then $\overline{y}^2 = \overline{2}$ in $\mathbb{Z}/3\mathbb{Z}$.
As $\{-1,0,1\}$ is a complete set of residues modulo $3$, the squares in $\mathbb{Z}/3\mathbb{Z}$ are $\overline{0} = \overline{0}^2$ and $\overline{1} = \overline{1}^2 = (\overline{-1})^2$, so $\overline{2}$ is not a square in $\m... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
283 | Ireland_Rosen_exercise_3_10 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | /--
This technical lemma isolates two distinct factors in the factorial product so that we can
force a desired divisor `N`. It will be used repeatedly in the main argument.
-/
private lemma dvd_factorial_from_two_factors {N a b : ℕ}
(hN : 1 ≤ N) (ha_pos : 1 ≤ a) (ha_lt : a < N)
(hb_pos : 1 ≤ b) (hb_lt : b < N)... | If $n$ is not a prime, show that $(n-1) ! \equiv 0(n)$, except when $n=4$. | theorem Ireland_Rosen_exercise_3_10 {n : ℕ+} (hn0 : ¬ n.Prime) :
Nat.factorial (n-1) ≡ 0 [MOD n] ↔ n ≠ 4 := by
sorry | Suppose that $n >1$ is not a prime. Then $n = uv$, where $2 \leq u \leq v \leq n-1$.
$\bullet$ If $u \neq v$, then $n = uv \mid (n-1)! = 1\times 2 \times\cdots \times u \times\cdots \times v \times \cdots \times (n-1)$ (even if $u\wedge v \neq 1$ !).
$\bullet$ If $u=v$, $n = u^2$ is a square.
If $u$ is not prime, $u... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
284 | Ireland_Rosen_exercise_4_4 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | Consider a prime $p$ of the form $4 t+1$. Show that $a$ is a primitive root modulo $p$ iff $-a$ is a primitive root modulo $p$. | theorem Ireland_Rosen_exercise_4_4 {p t : ℕ} (hp0 : p.Prime) (hp1 : p = 4 * t + 1)
(a : ZMod p) :
IsPrimitiveRoot a (p - 1) ↔ IsPrimitiveRoot (-a) (p - 1) := by
sorry | Suppose that $a$ is a primitive root modulo $p$. As $p-1$ is even, $(-a)^{p-1}=a^{p-1} \equiv 1$ $(\bmod p)$
If $(-a)^n \equiv 1(\bmod p)$, with $n \in \mathbb{N}$, then $a^n \equiv(-1)^n(\bmod p)$.
Therefore $a^{2 n} \equiv 1(\bmod p)$. As $a$ is a primitive root modulo $p, p-1|2 n, 2 t| n$, so $n$ is even.
Hence $a^... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
285 | Ireland_Rosen_exercise_4_6 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators | If $p=2^{n}+1$ is a Fermat prime, show that 3 is a primitive root modulo $p$. | theorem Ireland_Rosen_exercise_4_6 {p n : ℕ} (hp : p.Prime) (hpn : p = 2 ^ n + 1)
(hn : 2 ≤ n) :
IsPrimitiveRoot (3 : ZMod p) (p - 1) := by
sorry | \newcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}
Write $p = 2^k + 1$, with $k = 2^n$.
We suppose that $n>0$, so $k\geq 2, p \geq 5$. As $p$ is prime, $3^{p-1} \equiv 1 \pmod p$.
In other words, $3^{2^k} \equiv 1 \pmod p$ : the order of $3$ is a divisor of $2^k$, a power of $2$.
$3$ has order $2^k$ modulo $p$ i... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
286 | Ireland_Rosen_exercise_4_11 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators Finset | Prove that $1^{k}+2^{k}+\cdots+(p-1)^{k} \equiv 0(p)$ if $p-1 \nmid k$ and $-1(p)$ if $p-1 \mid k$. | theorem Ireland_Rosen_exercise_4_11 {p : ℕ} (hp : p.Prime) (k s: ℕ)
(hs : s = ∑ n : Fin (p-1), (n + 1 : ℕ) ^ k) :
((¬ p - 1 ∣ k) → s ≡ 0 [MOD p]) ∧ (p - 1 ∣ k → s ≡ -1 [ZMOD p]) := by
sorry | Let $S_k = 1^k+2^k+\cdots+(p-1)^k$.
Let $g$ a primitive root modulo $p$ : $\overline{g}$ a generator of $\mathbb{F}_p^*$.
As $(\overline{1},\overline{g},\overline{g}^{2}, \ldots, \overline{g}^{p-2}) $ is a permutation of $ (\overline{1},\overline{2}, \ldots,\overline{p-1})$,
\begin{align*}
\overline{S_k} &= \overline... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
287 | Ireland_Rosen_exercise_5_28 | Ireland Rosen | import Mathlib
open Real
open scoped BigOperators
open NumberTheorySymbols | Show that $x^{4} \equiv 2(p)$ has a solution for $p \equiv 1(4)$ iff $p$ is of the form $A^{2}+64 B^{2}$. | theorem Ireland_Rosen_exercise_5_28 {p : ℕ} (hp : p.Prime) (hp1 : p ≡ 1 [MOD 4]) :
(∃ x, x ^ 4 ≡ 2 [MOD p]) ↔ ∃ A B, p = A ^ 2 + 64 * B ^ 2 := by
sorry | If $p\equiv 1\ [4]$ and if there exists $x \in \mathbb{Z}$ such that $x^4 \equiv 2\ [p]$, then
$$2^{\frac{p-1}{4} }\equiv x^{p-1} \equiv 1 \ [p].$$
From Ex. 5.27, where $p = a^2 +b^2, a$ odd, we know that $$f^{\frac{ab}{2}} \equiv 2^{\frac{p-1}{4} } \equiv 1 \ [p].$$
Since $f^2 \equiv -1\ [p]$, the order of $f$ m... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
288 | Ireland_Rosen_exercise_12_12 | Ireland Rosen | import Mathlib
open Real
open Polynomial
open scoped BigOperators | Show that $\sin (\pi / 12)$ is an algebraic number. | theorem Ireland_Rosen_exercise_12_12 : IsAlgebraic ℚ (sin (Real.pi/12)) := by
sorry | $$
\begin{aligned}
\sin \pi/12=\sin \left(\pi/4-\pi/6\right) & =\sin \pi/4 \cos \pi/6-\cos \pi/4 \sin \pi/6 \\
& =\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}} \\
& =\frac{\sqrt{3}-1}{2 \sqrt{2}}
\end{aligned}
$$ | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
289 | Munkres_exercise_13_3b | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that the collection $$\mathcal{T}_\infty = \{U | X - U \text{ is infinite or empty or all of X}\}$$ does not need to be a topology on the set $X$. | theorem Munkres_exercise_13_3b : ∃ X : Type,
let T_inf : Set (Set X) := {t | Set.Infinite tᶜ ∨ t = ∅ ∨ t = Set.univ}
¬ (Set.univ ∈ T_inf ∧
(∀ s ⊆ T_inf, ⋃₀ s ∈ T_inf) ∧
(∀ t₁ ∈ T_inf, ∀ t₂ ∈ T_inf, t₁ ∩ t₂ ∈ T_inf)) := by
sorry | Let $X=\mathbb{R}, U_1=(-\infty, 0)$ and $U_2=(0, \infty)$. Then $U_1$ and $U_2$ are in $\mathcal{T}_{\infty}$ but $U_1 \cup U_2=\mathbb{R} \backslash\{0\}$ is not. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
290 | Munkres_exercise_13_4a2 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | def is_topology (X : Type*) (T : Set (Set X)) :=
univ ∈ T ∧
(∀ s t, s ∈ T → t ∈ T → s ∩ t ∈ T) ∧
(∀ s, (∀ t ∈ s, t ∈ T) → sUnion s ∈ T) | If $\mathcal{T}_\alpha$ is a family of topologies on $X$, show that $\bigcup \mathcal{T}_\alpha$ does not need to be a topology on $X$. | theorem Munkres_exercise_13_4a2 :
∃ (X I : Type*) (T : I → Set (Set X)),
(∀ i, is_topology X (T i)) ∧ ¬ is_topology X (⋃ i : I, T i) :=
sorry | On the other hand, the union $\bigcup_\alpha \mathcal{T}_\alpha$ is in general not a topology on $X$. For instance, let $X=\{a, b, c\}$. Then $\mathcal{T}_1=\{\emptyset, X,\{a\}\}$ and $\mathcal{T}_2=\{\emptyset, X,\{b\}\}$ are topologies on $X$ but $\mathcal{T}_1 \cup \mathcal{T}_2=$ $\{\emptyset, X,\{a\},\{b\}\}$ is ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
291 | Munkres_exercise_13_4b2 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | def is_topology (X : Type*) (T : Set (Set X)) :=
univ ∈ T ∧
(∀ s t, s ∈ T → t ∈ T → s ∩ t ∈ T) ∧
(∀s, (∀t ∈ s, t ∈ T) → sUnion s ∈ T) | Let $\mathcal{T}_\alpha$ be a family of topologies on $X$. Show that there is a unique largest topology on $X$ contained in all the collections $\mathcal{T}_\alpha$. | theorem Munkres_exercise_13_4b2 (X I : Type*) (T : I → Set (Set X)) (h : ∀ i, is_topology X (T i)) :
∃! T', is_topology X T' ∧ (∀ i, T' ⊆ T i) ∧
∀ T'', is_topology X T'' → (∀ i, T'' ⊆ T i) → T'' ⊆ T' := by
sorry | Now we prove that there exists a unique largest topology contained in all $\mathcal{T}_\alpha$. Uniqueness of such topology is clear. Consider $\mathcal{T}=\bigcap_\alpha \mathcal{T}_\alpha$. We already know that $\mathcal{T}$ is a topology by, and clearly $\mathcal{T} \subset \mathcal{T}_\alpha$ for all $\alpha$. If $... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
292 | Munkres_exercise_13_5b | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | def is_topology (X : Type*) (T : Set (Set X)) :=
univ ∈ T ∧
(∀ s t, s ∈ T → t ∈ T → s ∩ t ∈ T) ∧
(∀s, (∀t ∈ s, t ∈ T) → sUnion s ∈ T) | Show that if $\mathcal{A}$ is a subbasis for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\mathcal{A}$. | theorem Munkres_exercise_13_5b {X : Type*} (A : Set (Set X)) :
generateFrom A = generateFrom (sInter {T | is_topology X T ∧ A ⊆ T}) := by
sorry | If we now considered $\mathcal{A}$ as a subbasis, then the elements of $\mathcal{T}$ are union of finite intersections of elements of $\mathcal{A}$. The inclusion $\mathcal{O} \subset \mathcal{T}$ is again clear and $\mathcal{T} \subset \mathcal{O}$ holds since every union of finite intersections of elements of $\mathc... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
293 | Munkres_exercise_13_8a | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Show that the collection $\{(a,b) \mid a < b, a \text{ and } b \text{ rational}\}$ is a basis that generates the standard topology on $\mathbb{R}$. | theorem Munkres_exercise_13_8a :
IsTopologicalBasis {S : Set ℝ | ∃ a b : ℚ, a < b ∧ S = Ioo ↑a ↑b} := by
sorry | Exercise 13.8. (a) First note that $\mathcal{B}$ is a basis for a topology on $\mathbb{R}$. This follows from the fact that the union of its elements is all of $\mathbb{R}$ and the intersection of two elements of $\mathcal{B}$ is either empty or another element of $\mathcal{B}$. Let $\mathcal{T}$ be the standard topolo... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
294 | Munkres_exercise_16_1 | Munkres | import Mathlib
open Filter Set TopologicalSpace Topology
open scoped Topology | Show that if $Y$ is a subspace of $X$, and $A$ is a subset of $Y$, then the topology $A$ inherits as a subspace of $Y$ is the same as the topology it inherits as a subspace of $X$. | theorem Munkres_exercise_16_1 {X : Type*} [TopologicalSpace X]
(Y : Set X) (A : Set Y) :
IsInducing (fun a : A => (a : X)) := by
sorry | Exercise 16.1. Let $\mathcal{T}$ be the topology $A$ inherits as a subspace of $Y$, and $\mathcal{O}$ be the topology it inherits as a subspace of $X$. A (standard) basis element for $\mathcal{T}$ has the form $U \cap A$ where $U$ is open in $Y$, so is of the form $(Y \cap V) \cap A=V \cap A$ where $V$ is open in $X$. ... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
295 | Munkres_exercise_16_6 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | def rational (x : ℝ) := x ∈ range ((↑) : ℚ → ℝ) | Show that the countable collection \[\{(a, b) \times (c, d) \mid a < b \text{ and } c < d, \text{ and } a, b, c, d \text{ are rational}\}\] is a basis for $\mathbb{R}^2$. | theorem Munkres_exercise_16_6 :
IsTopologicalBasis
(Set.image2 (fun s t => s ×ˢ t)
(⋃ (a : ℚ) (b : ℚ) (_ : a < b), {Set.Ioo (a : ℝ) b})
(⋃ (a : ℚ) (b : ℚ) (_ : a < b), {Set.Ioo (a : ℝ) b})) := by
sorry | We know that $\mathcal{B}=\{(a,b)|a<b, a \text{ and } b \text rational\}$ is a basis for $\mathcal{R}$, therefore the set we are concerned with in the above question is a basis for $\mathcal{R}^2$. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
296 | Munkres_exercise_18_8a | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $Y$ be an ordered set in the order topology. Let $f, g: X \rightarrow Y$ be continuous. Show that the set $\{x \mid f(x) \leq g(x)\}$ is closed in $X$. | theorem Munkres_exercise_18_8a {X Y : Type*} [TopologicalSpace X] [TopologicalSpace Y]
[LinearOrder Y] [OrderTopology Y] {f g : X → Y}
(hf : Continuous f) (hg : Continuous g) :
IsClosed {x | f x ≤ g x} := by
sorry | We prove that $U=\{x \mid g(x)<f(x)\}$ is open in $X$. Let $a \in U$, so that $g(a)<f(a)$. If there is an element $c$ between $g(a)$ and $f(a)$, then $a \in g^{-1}((-\infty, c)) \cap f^{-1}((c,+\infty))$. If there are no elements between $g(a)$ and $f(a)$, then $a \in g^{-1}\left((-\infty, f(a)) \cap f^{-1}((g(a),+\inf... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
297 | Munkres_exercise_18_13 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | Let $A \subset X$; let $f: A \rightarrow Y$ be continuous; let $Y$ be Hausdorff. Show that if $f$ may be extended to a continuous function $g: \bar{A} \rightarrow Y$, then $g$ is uniquely determined by $f$. | theorem Munkres_exercise_18_13
{X : Type*} [TopologicalSpace X] {Y : Type*} [TopologicalSpace Y] [T2Space Y]
{A : Set X} {f : X → Y} (hf : ContinuousOn f A)
(g : X → Y) (g_con : ContinuousOn g (closure A))
(gExt : ∀ x : A, g x = f x) :
∀ (g' : X → Y), (ContinuousOn g' (closure A) ∧ (∀ x : A, g' x = f x)) → (∀... | Let $h, g: \bar{A} \rightarrow Y$ be continuous extensions of $f$. Suppose that there is a point $x \in \bar{A}$ such that $h(x) \neq g(x)$. Since $h=g$ on $A$, we must have $x \in A^{\prime}$. Since $Y$ is Hausdorff, there is a neighbourhood $U$ of $h(x)$ and a neighbourhood $V$ of $g(x)$ such that $U \cap V=\emptyset... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
298 | Munkres_exercise_20_2 | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology
noncomputable section | Show that $\mathbb{R} \times \mathbb{R}$ in the dictionary order topology is metrizable. | theorem Munkres_exercise_20_2
[TopologicalSpace (ℝ ×ₗ ℝ)] [OrderTopology (ℝ ×ₗ ℝ)]
: MetrizableSpace (ℝ ×ₗ ℝ) := by
sorry | The dictionary order topology on $\mathbb{R} \times \mathbb{R}$ is the same as the product topology $\mathbb{R}_d \times \mathbb{R}$, where $\mathbb{R}_d$ denotes $\mathbb{R}$ with the discrete topology. We know that $\mathbb{R}_d$ and $\mathbb{R}$ are metrisable. Thus, it suffices to show that the product of two metri... | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip | |
299 | Munkres_exercise_21_6b | Munkres | import Mathlib
open Filter Set TopologicalSpace
open scoped Topology | abbrev I : Set ℝ := Icc 0 1 | Define $f_{n}:[0,1] \rightarrow \mathbb{R}$ by the equation $f_{n}(x)=x^{n}$. Show that the sequence $\left(f_{n}\right)$ does not converge uniformly. | theorem Munkres_exercise_21_6b
(f : ℕ → I → ℝ)
(h : ∀ x n, f n x = ↑x ^ n) :
¬ ∃ f₀, TendstoUniformly f f₀ atTop := by
sorry | The sequence $\left(f_n\right)_n$ does not converge uniformly, since given $0<\varepsilon<1$ and $N \in \mathbb{Z}_{+}$, for $x=\varepsilon^{1 / N}$ we have $d\left(f_N(x), f(x)\right)=\varepsilon$. We can also apply Theorem 21.6: the convergence is not uniform since $f$ is not continuous. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
300 | Munkres_exercise_22_2a | Munkres | import Mathlib
open Filter Set TopologicalSpace Topology | Let $p: X \rightarrow Y$ be a continuous map. Show that if there is a continuous map $f: Y \rightarrow X$ such that $p \circ f$ equals the identity map of $Y$, then $p$ is a quotient map. | theorem Munkres_exercise_22_2a {X Y : Type*} [TopologicalSpace X]
[TopologicalSpace Y] (p : X → Y) (h : Continuous p) :
(∃ (f : Y → X), Continuous f ∧ p ∘ f = id) → IsQuotientMap p := by
sorry | Let $1_Y: Y \rightarrow Y$ be the identity map in $Y$. If $U$ is a subset of $Y$ and $p^{-1}(U)$ is open in $X$, then $f^{-1}\left(p^{-1}(U)\right)=1_Y^{-1}(U)=U$ is open in $Y$ by continuity of $f$. Thus $p$ is a quotient map. | to avoid data contamination, the formal proof is in proofnet_verified_gt.zip |
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