id stringlengths 17 29 | nl_statement stringlengths 43 377 | nl_proof stringlengths 0 3.81k | formal_statement stringlengths 52 344 | src_header stringclasses 10
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Pugh|exercise_2_41 | Let $\|\cdot\|$ be any norm on $\mathbb{R}^{m}$ and let $B=\left\{x \in \mathbb{R}^{m}:\|x\| \leq 1\right\}$. Prove that $B$ is compact. | \begin{proof}
Let us call $\|\cdot\|_E$ the Euclidean norm in $\mathbb{R}^m$. We start by claiming that there exist constants $C_1, C_2>0$ such that
$$
C_1\|x\|_E \leq\|x\| \leq C_2\|x\|_E, \forall x \in \mathbb{R}^m .
$$
Assuming (1) to be true, let us finish the problem. First let us show that $B$ is bounded... | theorem exercise_2_41 (m : β) {X : Type*} [normed_space β ((fin m) β β)] :
is_compact (metric.closed_ball 0 1) := | import .common
open set real filter function ring_hom topological_space
open_locale big_operators
open_locale filter
open_locale topology
noncomputable theory
|
Pugh|exercise_2_57 | Show that if $S$ is connected, it is not true in general that its interior is connected. | \begin{proof}
Consider $X=\mathbb{R}^2$ and
$$
A=([-2,0] \times[-2,0]) \cup([0,2] \times[0,2])
$$
which is connected, while $\operatorname{int}(A)$ is not connected.
To see this consider the continuous function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is defined by $f(x, y)=x+y$. Let $U=f^{-1}(0,+\infty)$ whi... | theorem exercise_2_57 {X : Type*} [topological_space X]
: β (S : set X), is_connected S β§ Β¬ is_connected (interior S) := | import .common
open set real filter function ring_hom topological_space
open_locale big_operators
open_locale filter
open_locale topology
noncomputable theory
|
Pugh|exercise_2_126 | Suppose that $E$ is an uncountable subset of $\mathbb{R}$. Prove that there exists a point $p \in \mathbb{R}$ at which $E$ condenses. | \begin{proof}
I think this is the proof by contrapositive that you were getting at.
Suppose that $E$ has no limit points at all. Pick an arbitrary point $x \in E$. Then $x$ cannot be a limit point, so there must be some $\delta>0$ such that the ball of radius $\delta$ around $x$ contains no other points of $E$ :
... | theorem exercise_2_126 {E : set β}
(hE : Β¬ set.countable E) : β (p : β), cluster_pt p (π E) := | import .common
open set real filter function ring_hom topological_space
open_locale big_operators
open_locale filter
open_locale topology
noncomputable theory
|
Pugh|exercise_3_4 | Prove that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$. | \begin{proof}
$$
\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2 \sqrt{n}}
$$
\end{proof} | theorem exercise_3_4 (n : β) :
tendsto (Ξ» n, (sqrt (n + 1) - sqrt n)) at_top (π 0) := | import .common
open set real filter function ring_hom topological_space
open_locale big_operators
open_locale filter
open_locale topology
noncomputable theory
|
Pugh|exercise_3_63b | Prove that $\sum 1/k(\log(k))^p$ diverges when $p \leq 1$. | \begin{proof}
Using the integral test, for a set $a$, we see
$$
\lim _{b \rightarrow \infty} \int_a^b \frac{1}{x \log (x)^c} d x=\lim _{b \rightarrow \infty}\left(\frac{\log (b)^{1-c}}{1-c}-\frac{\log (a)^{1-c}}{1-c}\right)
$$
which goes to infinity if $c \leq 1$ and converges if $c>1$. Thus,
$$
\sum_{n=2}^... | theorem exercise_3_63b (p : β) (f : β β β) (hp : p β€ 1)
(h : f = Ξ» k, (1 : β) / (k * (log k) ^ p)) :
Β¬ β l, tendsto f at_top (π l) := | import .common
open set real filter function ring_hom topological_space
open_locale big_operators
open_locale filter
open_locale topology
noncomputable theory
|
Herstein|exercise_2_1_21 | Show that a group of order 5 must be abelian. | \begin{proof}
Suppose $G$ is a group of order 5 which is not abelian. Then there exist two non-identity elements $a, b \in G$ such that $a * b \neq$ $b * a$. Further we see that $G$ must equal $\{e, a, b, a * b, b * a\}$. To see why $a * b$ must be distinct from all the others, not that if $a *$ $b=e$, then $a$ an... | theorem exercise_2_1_21 (G : Type*) [group G] [fintype G]
(hG : card G = 5) :
comm_group G := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_2_1_27 | If $G$ is a finite group, prove that there is an integer $m > 0$ such that $a^m = e$ for all $a \in G$. | \begin{proof}
Let $n_1, n_2, \ldots, n_k$ be the orders of all $k$ elements of $G=$ $\left\{a_1, a_2, \ldots, a_k\right\}$. Let $m=\operatorname{lcm}\left(n_1, n_2, \ldots, n_k\right)$. Then, for any $i=$ $1, \ldots, k$, there exists an integer $c$ such that $m=n_i c$. Thus
$$
a_i^m=a_i^{n_i c}=\left(a_i^{n_i}\r... | theorem exercise_2_1_27 {G : Type*} [group G]
[fintype G] : β (m : β), β (a : G), a ^ m = 1 := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_2_2_5 | Let $G$ be a group in which $(a b)^{3}=a^{3} b^{3}$ and $(a b)^{5}=a^{5} b^{5}$ for all $a, b \in G$. Show that $G$ is abelian. | \begin{proof}
We have
$$
\begin{aligned}
& (a b)^3=a^3 b^3, \text { for all } a, b \in G \\
\Longrightarrow & (a b)(a b)(a b)=a\left(a^2 b^2\right) b \\
\Longrightarrow & a(b a)(b a) b=a\left(a^2 b^2\right) b \\
\Longrightarrow & (b a)^2=a^2 b^2, \text { by cancellation law. }
\end{aligned}
$$
Again,
$$
\b... | theorem exercise_2_2_5 {G : Type*} [group G]
(h : β (a b : G), (a * b) ^ 3 = a ^ 3 * b ^ 3 β§ (a * b) ^ 5 = a ^ 5 * b ^ 5) :
comm_group G := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_2_3_17 | If $G$ is a group and $a, x \in G$, prove that $C\left(x^{-1} a x\right)=x^{-1} C(a) x$ | \begin{proof}
Note that
$$
C(a):=\{x \in G \mid x a=a x\} .
$$
Let us assume $p \in C\left(x^{-1} a x\right)$. Then,
$$
\begin{aligned}
& p\left(x^{-1} a x\right)=\left(x^{-1} a x\right) p \\
\Longrightarrow & \left(p x^{-1} a\right) x=x^{-1}(a x p) \\
\Longrightarrow & x\left(p x^{-1} a\right)=(a x p) x^... | theorem exercise_2_3_17 {G : Type*} [has_mul G] [group G] (a x : G) :
set.centralizer {xβ»ΒΉ*a*x} =
(Ξ» g : G, xβ»ΒΉ*g*x) '' (set.centralizer {a}) := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_2_4_36 | If $a > 1$ is an integer, show that $n \mid \varphi(a^n - 1)$, where $\phi$ is the Euler $\varphi$-function. | \begin{proof}
Proof: We have $a>1$. First we propose to prove that
$$
\operatorname{Gcd}\left(a, a^n-1\right)=1 .
$$
If possible, let us assume that
$\operatorname{Gcd}\left(a, a^n-1\right)=d$, where $d>1$.
Then
$d$ divides $a$ as well as $a^n-1$.
Now,
$d$ divides $a \Longrightarrow d$ divides $a^n$.
Thi... | theorem exercise_2_4_36 {a n : β} (h : a > 1) :
n β£ (a ^ n - 1).totient := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_2_5_30 | Suppose that $|G| = pm$, where $p \nmid m$ and $p$ is a prime. If $H$ is a normal subgroup of order $p$ in $G$, prove that $H$ is characteristic. | \begin{proof}
Let $G$ be a group of order $p m$, such that $p \nmid m$. Now, Given that $H$ is a normal subgroup of order $p$. 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 $|\phi(H)|=p$. Suppose $\phi(H) \neq H$,... | theorem exercise_2_5_30 {G : Type*} [group G] [fintype G]
{p m : β} (hp : nat.prime p) (hp1 : Β¬ p β£ m) (hG : card G = p*m)
{H : subgroup G} [fintype H] [H.normal] (hH : card H = p):
characteristic H := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_2_5_37 | If $G$ is a nonabelian group of order 6, prove that $G \simeq S_3$. | \begin{proof}
Suppose $G$ is a non-abelian group of order 6 . We need to prove that $G \cong S_3$. Since $G$ is non-abelian, we conclude that there is no element of order 6. Now all the nonidentity element has order either 2 or 3 . All elements cannot be order 3 .This is because except the identity elements there ... | theorem exercise_2_5_37 (G : Type*) [group G] [fintype G]
(hG : card G = 6) (hG' : is_empty (comm_group G)) :
G β* equiv.perm (fin 3) := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_2_5_44 | Prove that a group of order $p^2$, $p$ a prime, has a normal subgroup of order $p$. | \begin{proof}
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 $K \neq \{ e \}$ and $K \subseteq H$.
So, we have now a group $G$ of order $p^2$. Suppose that the group is cyclic, t... | theorem exercise_2_5_44 {G : Type*} [group G] [fintype G] {p : β}
(hp : nat.prime p) (hG : card G = p^2) :
β (N : subgroup G) (fin : fintype N), @card N fin = p β§ N.normal := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_2_6_15 | If $G$ is an abelian group and if $G$ has an element of order $m$ and one of order $n$, where $m$ and $n$ are relatively prime, prove that $G$ has an element of order $mn$. | \begin{proof}
Let $G$ be an abelian group, and let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively, where $m$ and $n$ are relatively prime. We will show that the product $ab$ has order $mn$ in $G$, which will prove that $G$ has an element of order $mn$.
To show that $ab$ has order $mn$, let $k$ be... | theorem exercise_2_6_15 {G : Type*} [comm_group G] {m n : β}
(hm : β (g : G), order_of g = m)
(hn : β (g : G), order_of g = n)
(hmn : m.coprime n) :
β (g : G), order_of g = m * n := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_2_8_12 | Prove that any two nonabelian groups of order 21 are isomorphic. | \begin{proof}
By Cauchy's theorem we have that if $G$ is a group of order 21 then it has an element $a$ of order 3 and an element $b$ of order 7. By exercise 2.5.41 we have that the subgroup generated by $b$ is normal, so there is some $i=0,1,2,3,4,5,6$ such that $a b a^{-1}=b^i$. We know $i \neq$ 0 since that imp... | theorem exercise_2_8_12 {G H : Type*} [fintype G] [fintype H]
[group G] [group H] (hG : card G = 21) (hH : card H = 21)
(hG1 : is_empty(comm_group G)) (hH1 : is_empty (comm_group H)) :
G β* H := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_2_9_2 | If $G_1$ and $G_2$ are cyclic groups of orders $m$ and $n$, respectively, prove that $G_1 \times G_2$ is cyclic if and only if $m$ and $n$ are relatively prime. | \begin{proof}
The order of $G \times H$ is $n$. $m$. Thus, $G \times H$ is cyclic iff it has an element with order n. $m$. Suppose $\operatorname{gcd}(n . m)=1$. This implies that $g^m$ has order $n$, and analogously $h^n$ has order $m$. That is, $g \times h$ has order $n$. $m$, and therefore $G \times H$ is cycli... | theorem exercise_2_9_2 {G H : Type*} [fintype G] [fintype H] [group G]
[group H] (hG : is_cyclic G) (hH : is_cyclic H) :
is_cyclic (G Γ H) β (card G).coprime (card H) := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_2_11_6 | If $P$ is a $p$-Sylow subgroup of $G$ and $P \triangleleft G$, prove that $P$ is the only $p$-Sylow subgroup of $G$. | \begin{proof}
let $G$ be a group and $P$ a sylow-p subgroup. Given $P$ is normal. By sylow second theorem the sylow-p subgroups are conjugate. Let $K$ be any other sylow-p subgroup. Then there exists $g \in G$ such that $K=g P g^{-1}$. But since $P$ is normal $K=g P g^{-1}=P$. Hence the sylow-p subgroup is unique.... | theorem exercise_2_11_6 {G : Type*} [group G] {p : β} (hp : nat.prime p)
{P : sylow p G} (hP : P.normal) :
β (Q : sylow p G), P = Q := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_2_11_22 | Show that any subgroup of order $p^{n-1}$ in a group $G$ of order $p^n$ is normal in $G$. | \begin{proof}
Proof: First we prove the following lemma.
\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 $|Z(G)|>1$.
\textit{Proof of the lemma:} Consider the class equation
$$
|G|=|Z(G)|+\sum_{a \n... | theorem exercise_2_11_22 {p : β} {n : β} {G : Type*} [fintype G]
[group G] (hp : nat.prime p) (hG : card G = p ^ n) {K : subgroup G}
[fintype K] (hK : card K = p ^ (n-1)) :
K.normal := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_4_1_19 | Show that there is an infinite number of solutions to $x^2 = -1$ in the quaternions. | \begin{proof}
Let $x=a i+b j+c k$ then
$$
x^2=(a i+b j+c k)(a i+b j+c k)=-a^2-b^2-c^2=-1
$$
This gives $a^2+b^2+c^2=1$ which has infinitely many solutions for $-1<a, b, c<1$.
\end{proof} | theorem exercise_4_1_19 : infinite {x : quaternion β | x^2 = -1} := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_4_2_5 | Let $R$ be a ring in which $x^3 = x$ for every $x \in R$. Prove that $R$ is commutative. | \begin{proof}
To begin with
$$
2 x=(2 x)^3=8 x^3=8 x .
$$
Therefore $6 x=0 \quad \forall x$.
Also
$$
(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
$$
and
$$
(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
$$
Subtracting we get
$$
2\left(x^2 y+x y x+y x^2\right)=0
$$
Multiply the ... | theorem exercise_4_2_5 {R : Type*} [ring R]
(h : β x : R, x ^ 3 = x) : comm_ring R := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_4_2_9 | Let $p$ be an odd prime and let $1 + \frac{1}{2} + ... + \frac{1}{p - 1} = \frac{a}{b}$, where $a, b$ are integers. Show that $p \mid a$. | \begin{proof}
First we prove for prime $p=3$ and then for all prime $p>3$.
Let us take $p=3$. Then the sum
$$
\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{(p-1)}
$$
becomes
$$
1+\frac{1}{3-1}=1+\frac{1}{2}=\frac{3}{2} .
$$
Therefore in this case $\quad \frac{a}{b}=\frac{3}{2} \quad$ implies $3 \mid a$, i.e. $p... | theorem exercise_4_2_9 {p : β} (hp : nat.prime p) (hp1 : odd p) :
β (a b : β€), (a / b : β) = β i in finset.range p, 1 / (i + 1) β βp β£ a := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_4_3_25 | Let $R$ be the ring of $2 \times 2$ matrices over the real numbers; suppose that $I$ is an ideal of $R$. Show that $I = (0)$ or $I = R$. | \begin{proof}
Suppose that $I$ is a nontrivial ideal of $R$, and let
$$
A=\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)
$$
where not all of $a, b, c d$ are zero. Suppose, without loss of generality -- our steps would be completely analogous, modulo some different placement of 1 s in our matrices... | theorem exercise_4_3_25 (I : ideal (matrix (fin 2) (fin 2) β)) :
I = β₯ β¨ I = β€ := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_4_5_16 | Let $F = \mathbb{Z}_p$ be the field of integers $\mod p$, where $p$ is a prime, and let $q(x) \in F[x]$ be irreducible of degree $n$. Show that $F[x]/(q(x))$ is a field having at exactly $p^n$ elements. | \begin{proof}
In the previous problem we have shown that any for any $p(x) \in F[x]$, we have that
$$
p(x)+(q(x))=a_{n-1} x^{n-1}+\cdots+a_1 x+a_0+(q(x))
$$
for some $a_{n-1}, \ldots, a_0 \in F$, and that there are $p^n$ choices for these numbers, so that $F[x] /(q(x)) \leq p^n$. In order to show that equality... | theorem exercise_4_5_16 {p n: β} (hp : nat.prime p)
{q : polynomial (zmod p)} (hq : irreducible q) (hn : q.degree = n) :
β is_fin : fintype $ polynomial (zmod p) β§Έ ideal.span ({q} : set (polynomial $ zmod p)),
@card (polynomial (zmod p) β§Έ ideal.span {q}) is_fin = p ^ n β§
is_field (polynomial $ zmod p):= | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_4_5_25 | If $p$ is a prime, show that $q(x) = 1 + x + x^2 + \cdots x^{p - 1}$ is irreducible in $Q[x]$. | \begin{proof}
Lemma: Let $F$ be a field and $f(x) \in F[x]$. If $c \in F$ and $f(x+c)$ is irreducible in $F[x]$, then $f(x)$ is irreducible in $F[x]$.
Proof of the Lemma: Suppose that $f(x)$ is reducible, i.e., there exist non-constant $g(x), h(x) \in F[x]$ so that
$$
f(x)=g(x) h(x) .
$$
In particular, then w... | theorem exercise_4_5_25 {p : β} (hp : nat.prime p) :
irreducible (β i : finset.range p, X ^ p : polynomial β) := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_4_6_3 | Show that there is an infinite number of integers a such that $f(x) = x^7 + 15x^2 - 30x + a$ is irreducible in $Q[x]$. | \begin{proof}
Via Eisenstein's criterion and observation that 5 divides 15 and $-30$, it is sufficient to find infinitely many $a$ such that 5 divides $a$, but $5^2=25$ doesn't divide $a$. For example $5 \cdot 2^k$ for $k=0,1, \ldots$ is one such infinite sequence.
\end{proof} | theorem exercise_4_6_3 :
infinite {a : β€ | irreducible (X^7 + 15*X^2 - 30*X + a : polynomial β)} := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_5_2_20 | Let $V$ be a vector space over an infinite field $F$. Show that $V$ cannot be the set-theoretic union of a finite number of proper subspaces of $V$. | \begin{proof}
Assume that $V$ can be written as the set-theoretic union of $n$ proper subspaces $U_1, U_2, \ldots, U_n$. Without loss of generality, we may assume that no $U_i$ is contained in the union of other subspaces.
Let $u \in U_i$ but $u \notin \bigcup_{j \neq i} U_j$ and $v \notin U_i$. Then, we have $... | theorem exercise_5_2_20 {F V ΞΉ: Type*} [infinite F] [field F]
[add_comm_group V] [module F V] {u : ΞΉ β submodule F V}
(hu : β i : ΞΉ, u i β β€) :
(β i : ΞΉ, (u i : set V)) β β€ := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_5_3_10 | Prove that $\cos 1^{\circ}$ is algebraic over $\mathbb{Q}$. | \begin{proof}
Since $\left(\cos \left(1^{\circ}\right)+i \sin \left(1^{\circ}\right)\right)^{360}=1$, the number $\cos \left(1^{\circ}\right)+i \sin \left(1^{\circ}\right)$ is algebraic. And the real part and the imaginary part of an algebraic number are always algebraic numbers.
\end{proof} | theorem exercise_5_3_10 : is_algebraic β (cos (real.pi / 180)) := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Herstein|exercise_5_5_2 | Prove that $x^3 - 3x - 1$ is irreducible over $\mathbb{Q}$. | \begin{proof}
Let $p(x)=x^3-3 x-1$. Then
$$
p(x+1)=(x+1)^3-3(x+1)-1=x^3+3 x^2-3
$$
We have $3|3,3| 0$ but $3 \nmid 1$ and $3^2 \nmid 3$. Thus the polynomial is irreducible over $\mathbb{Q}$ by 3 -Eisenstein criterion.
\end{proof} | theorem exercise_5_5_2 : irreducible (X^3 - 3*X - 1 : polynomial β) := | import .common
open set function nat fintype real subgroup ideal polynomial submodule zsqrtd
open char_p mul_aut matrix
open_locale big_operators
noncomputable theory
|
Artin|exercise_2_2_9 | Let $H$ be the subgroup generated by two elements $a, b$ of a group $G$. Prove that if $a b=b a$, then $H$ is an abelian group. | \begin{proof}
Since $a$ and $b$ commute, for any $g, h\in H$ we can write $g=a^ib^j$ and $h = a^kb^l$. Then $gh = a^ib^ja^kb^l = a^kb^la^ib^j = hg$. Thus $H$ is abelian.
\end{proof} | theorem exercise_2_2_9 {G : Type*} [group G] {a b : G}
(h : a * b = b * a) :
β x y : closure {x | x = a β¨ x = b}, x*y = y*x := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_2_4_19 | Prove that if a group contains exactly one element of order 2 , then that element is in the center of the group. | \begin{proof}
Let $x$ be the element of order two. Consider the element $z=y^{-1} x y$, we have: $z^2=\left(y^{-1} x y\right)^2=\left(y^{-1} x y\right)\left(y^{-1} x y\right)=e$. So: $z=x$, and $y^{-1} x y=x$. So: $x y=y x$. So: $x$ is in the center of $G$.
\end{proof} | theorem exercise_2_4_19 {G : Type*} [group G] {x : G}
(hx : order_of x = 2) (hx1 : β y, order_of y = 2 β y = x) :
x β center G := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_2_11_3 | Prove that a group of even order contains an element of order $2 .$ | \begin{proof}
Pair up if possible each element of $G$ with its inverse, and observe that
$$
g^2 \neq e \Longleftrightarrow g \neq g^{-1} \Longleftrightarrow \text { there exists the pair }\left(g, g^{-1}\right)
$$
Now, there is one element that has no pairing: the unit $e$ (since indeed $e=e^{-1} \Longleftrigh... | theorem exercise_2_11_3 {G : Type*} [group G] [fintype G]
(hG : even (card G)) : β x : G, order_of x = 2 := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_3_5_6 | Let $V$ be a vector space which is spanned by a countably infinite set. Prove that every linearly independent subset of $V$ is finite or countably infinite. | \begin{proof}
Let $A$ be the countable generating set, and let $U$ be an uncountable linearly independent set. It can be extended to a basis $B$ of the whole space. Now consider the subset $C$ of elements of $B$ that appear in the $B$-decompositions of elements of $A$.
Since only finitely many elements are involv... | theorem exercise_3_5_6 {K V : Type*} [field K] [add_comm_group V]
[module K V] {S : set V} (hS : set.countable S)
(hS1 : span K S = β€) {ΞΉ : Type*} (R : ΞΉ β V)
(hR : linear_independent K R) : countable ΞΉ := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_6_1_14 | Let $Z$ be the center of a group $G$. Prove that if $G / Z$ is a cyclic group, then $G$ is abelian and hence $G=Z$. | \begin{proof}
We have that $G / Z(G)$ is cyclic, and so there is an element $x \in G$ such that $G / Z(G)=\langle x Z(G)\rangle$, where $x Z(G)$ is the coset with representative $x$. Now let $g \in G$
We know that $g Z(G)=(x Z(G))^m$ for some $m$, and by definition $(x Z(G))^m=x^m Z(G)$.
Now, in general, if $H \... | theorem exercise_6_1_14 (G : Type*) [group G]
(hG : is_cyclic $ G β§Έ (center G)) :
center G = β€ := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_6_4_3 | Prove that no group of order $p^2 q$, where $p$ and $q$ are prime, is simple. | \begin{proof}
We may as well assume $p<q$. The number of Sylow $q$-subgroups is $1 \bmod q$ and divides $p^2$. So it is $1, p$, or $p^2$. We win if it's 1 and it can't be $p$, so suppose it's $p^2$. But now $q \mid p^2-1$, so $q \mid p+1$ or $q \mid p-1$.
Thus $p=2$ and $q=3$. But we know no group of order 36 is ... | theorem exercise_6_4_3 {G : Type*} [group G] [fintype G] {p q : β}
(hp : prime p) (hq : prime q) (hG : card G = p^2 *q) :
is_simple_group G β false := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_6_8_1 | Prove that two elements $a, b$ of a group generate the same subgroup as $b a b^2, b a b^3$. | \begin{proof}
Let $H = \langle bab^2, bab^3\rangle$. It is clear that $H\subset \langle a, b\rangle$. Note that $(bab^2)^{-1}(bab^3)=b$, therefore $b\in H$. This then implies that $b^{-1}(bab^2)b^{-2}=a\in H$. Thus $\langle a, b\rangle\subset H$.
\end{proof} | theorem exercise_6_8_1 {G : Type*} [group G]
(a b : G) : closure ({a, b} : set G) = closure {b*a*b^2, b*a*b^3} := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_10_2_4 | Prove that in the ring $\mathbb{Z}[x],(2) \cap(x)=(2 x)$. | \begin{proof}
Let $f(x) \in(2 x)$. Then there exists some polynomial $g(x) \in \mathbb{Z}$ such that
$$
f(x)=2 x g(x)
$$
But this means that $f(x) \in(2)$ (because $x g(x)$ is a polynomial), and $f(x) \in$ $(x)$ (because $2 g(x)$ is a polynomial). Thus, $f(x) \in(2) \cap(x)$, and
$$
(2 x) \subseteq(2) \cap(x... | theorem exercise_10_2_4 :
span ({2} : set $ polynomial β€) β (span {X}) =
span ({2 * X} : set $ polynomial β€) := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_10_4_6 | Let $I, J$ be ideals in a ring $R$. Prove that the residue of any element of $I \cap J$ in $R / I J$ is nilpotent. | \begin{proof}
If $x$ is in $I \cap J, x \in I$ and $x \in J . R / I J=\{r+a b: a \in I, b \in J, r \in R\}$. Then $x \in I \cap J \Rightarrow x \in I$ and $x \in J$, and so $x^2 \in I J$. Thus
$$
[x]^2=\left[x^2\right]=[0] \text { in } R / I J
$$
\end{proof} | theorem exercise_10_4_6 {R : Type*} [comm_ring R]
[no_zero_divisors R] {I J : ideal R} (x : I β J) :
is_nilpotent ((ideal.quotient.mk (I*J)) x) := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_10_7_10 | Let $R$ be a ring, with $M$ an ideal of $R$. Suppose that every element of $R$ which is not in $M$ is a unit of $R$. Prove that $M$ is a maximal ideal and that moreover it is the only maximal ideal of $R$. | \begin{proof}
Suppose there is an ideal $M\subset I\subset R$. If $I\neq M$, then $I$ contains a unit, thus $I=R$. Therefore $M$ is a maximal ideal.
Suppose we have an arbitrary maximal ideal $M^\prime$ of $R$. The ideal $M^\prime$ cannot contain a unit, otherwise $M^\prime =R$. Therefore $M^\prime \subset M$. But... | theorem exercise_10_7_10 {R : Type*} [ring R]
(M : ideal R) (hM : β (x : R), x β M β is_unit x) :
is_maximal M β§ β (N : ideal R), is_maximal N β N = M := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_11_4_1b | Prove that $x^3 + 6x + 12$ is irreducible in $\mathbb{Q}$. | \begin{proof}
Apply Eisenstein's criterion with $p=3$.
\end{proof} | theorem exercise_11_4_1b {F : Type*} [field F] [fintype F] (hF : card F = 2) :
irreducible (12 + 6 * X + X ^ 3 : polynomial F) := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_11_4_6b | Prove that $x^2+1$ is irreducible in $\mathbb{F}_7$ | \begin{proof}
If $p(x)=x^2+1$ were reducible, its factors must be linear. But no $p(a)$ for $a\in\mathbb{F}_7$ evaluates to 0, therefore $x^2+1$ is irreducible.
\end{proof} | theorem exercise_11_4_6b {F : Type*} [field F] [fintype F] (hF : card F = 31) :
irreducible (X ^ 3 - 9 : polynomial F) := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_11_4_8 | Let $p$ be a prime integer. Prove that the polynomial $x^n-p$ is irreducible in $\mathbb{Q}[x]$. | \begin{proof}
Straightforward application of Eisenstein's criterion with $p$.
\end{proof} | theorem exercise_11_4_8 {p : β} (hp : prime p) (n : β) :
irreducible (X ^ n - p : polynomial β) := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Artin|exercise_13_4_10 | Prove that if a prime integer $p$ has the form $2^r+1$, then it actually has the form $2^{2^k}+1$. | \begin{proof}
In particular, we have
$$
\frac{x^a+1}{x+1}=\frac{(-x)^a-1}{(-x)-1}=1-x+x^2-\cdots+(-x)^{a-1}
$$
by the geometric sum formula. In this case, specialize to $x=2^{2^m}$ and we have a nontrivial divisor.
\end{proof} | theorem exercise_13_4_10
{p : β} {hp : nat.prime p} (h : β r : β, p = 2 ^ r + 1) :
β (k : β), p = 2 ^ (2 ^ k) + 1 := | import .common
open function fintype subgroup ideal polynomial submodule zsqrtd char_p
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_1_1_3 | Prove that the addition of residue classes $\mathbb{Z}/n\mathbb{Z}$ is associative. | \begin{proof}
We have
$$
\begin{aligned}
(\bar{a}+\bar{b})+\bar{c} &=\overline{a+b}+\bar{c} \\
&=\overline{(a+b)+c} \\
&=\overline{a+(b+c)} \\
&=\bar{a}+\overline{b+c} \\
&=\bar{a}+(\bar{b}+\bar{c})
\end{aligned}
$$
since integer addition is associative.
\end{proof} | theorem exercise_1_1_3 (n : β€) :
β (a b c : β€), (a+b)+c β‘ a+(b+c) [ZMOD n] := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_1_1_5 | Prove that for all $n>1$ that $\mathbb{Z}/n\mathbb{Z}$ is not a group under multiplication of residue classes. | \begin{proof}
Note that since $n>1, \overline{1} \neq \overline{0}$. Now suppose $\mathbb{Z} /(n)$ contains a multiplicative identity element $\bar{e}$. Then in particular,
$$
\bar{e} \cdot \overline{1}=\overline{1}
$$
so that $\bar{e}=\overline{1}$. Note, however, that
$$
\overline{0} \cdot \bar{k}=\overlin... | theorem exercise_1_1_5 (n : β) (hn : 1 < n) :
is_empty (group (zmod n)) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_1_1_16 | Let $x$ be an element of $G$. Prove that $x^2=1$ if and only if $|x|$ is either $1$ or $2$. | \begin{proof}
$(\Rightarrow)$ Suppose $x^2=1$. Then we have $0<|x| \leq 2$, i.e., $|x|$ is either 1 or 2 .
( $\Leftarrow$ ) If $|x|=1$, then we have $x=1$ so that $x^2=1$. If $|x|=2$ then $x^2=1$ by definition. So if $|x|$ is 1 or 2 , we have $x^2=1$.
\end{proof} | theorem exercise_1_1_16 {G : Type*} [group G]
(x : G) (hx : x ^ 2 = 1) :
order_of x = 1 β¨ order_of x = 2 := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_1_1_18 | Let $x$ and $y$ be elements of $G$. Prove that $xy=yx$ if and only if $y^{-1}xy=x$ if and only if $x^{-1}y^{-1}xy=1$. | \begin{proof}
If $x y=y x$, then $y^{-1} x y=y^{-1} y x=1 x=x$. Multiplying by $x^{-1}$ then gives $x^{-1} y^{-1} x y=1$.
On the other hand, if $x^{-1} y^{-1} x y=1$, then we may multiply on the left by $x$ to get $y^{-1} x y=x$. Then multiplying on the left by $y$ gives $x y=y x$ as desired.
\end{proof} | theorem exercise_1_1_18 {G : Type*} [group G]
(x y : G) : x * y = y * x β yβ»ΒΉ * x * y = x β xβ»ΒΉ * yβ»ΒΉ * x * y = 1 := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_1_1_22a | If $x$ and $g$ are elements of the group $G$, prove that $|x|=\left|g^{-1} x g\right|$. | \begin{proof}
First we prove a technical lemma:
{\bf Lemma.} For all $a, b \in G$ and $n \in \mathbb{Z},\left(b^{-1} a b\right)^n=b^{-1} a^n b$.
The statement is clear for $n=0$. We prove the case $n>0$ by induction; the base case $n=1$ is clear. Now suppose $\left(b^{-1} a b\right)^n=b^{-1} a^n b$ for som... | theorem exercise_1_1_22a {G : Type*} [group G] (x g : G) :
order_of x = order_of (gβ»ΒΉ * x * g) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_1_1_25 | Prove that if $x^{2}=1$ for all $x \in G$ then $G$ is abelian. | \begin{proof}
Solution: Note that since $x^2=1$ for all $x \in G$, we have $x^{-1}=x$. Now let $a, b \in G$. We have
$$
a b=(a b)^{-1}=b^{-1} a^{-1}=b a .
$$
Thus $G$ is abelian.
\end{proof} | theorem exercise_1_1_25 {G : Type*} [group G]
(h : β x : G, x ^ 2 = 1) : β a b : G, a*b = b*a := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_1_1_34 | If $x$ is an element of infinite order in $G$, prove that the elements $x^{n}, n \in \mathbb{Z}$ are all distinct. | \begin{proof}
Solution: Suppose to the contrary that $x^a=x^b$ for some $0 \leq a<b \leq n-1$. Then we have $x^{b-a}=1$, with $1 \leq b-a<n$. However, recall that $n$ is by definition the least integer $k$ such that $x^k=1$, so we have a contradiction. Thus all the $x^i$, $0 \leq i \leq n-1$, are distinct. In part... | theorem exercise_1_1_34 {G : Type*} [group G] {x : G}
(hx_inf : order_of x = 0) (n m : β€) :
x ^ n β x ^ m := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_1_6_4 | Prove that the multiplicative groups $\mathbb{R}-\{0\}$ and $\mathbb{C}-\{0\}$ are not isomorphic. | \begin{proof}
Isomorphic groups necessarily have the same number of elements of order $n$ for all finite $n$.
Now let $x \in \mathbb{R}^{\times}$. If $x=1$ then $|x|=1$, and if $x=-1$ then $|x|=2$. If (with bars denoting absolute value) $|x|<1$, then we have
$$
1>|x|>\left|x^2\right|>\cdots,
$$
and in parti... | theorem exercise_1_6_4 :
is_empty (multiplicative β β* multiplicative β) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_1_6_17 | Let $G$ be any group. Prove that the map from $G$ to itself defined by $g \mapsto g^{-1}$ is a homomorphism if and only if $G$ is abelian. | \begin{proof}
$(\Rightarrow)$ Suppose $G$ is abelian. Then
$$
\varphi(a b)=(a b)^{-1}=b^{-1} a^{-1}=a^{-1} b^{-1}=\varphi(a) \varphi(b),
$$
so that $\varphi$ is a homomorphism.
$(\Leftarrow)$ Suppose $\varphi$ is a homomorphism, and let $a, b \in G$. Then
$$
a b=\left(b^{-1} a^{-1}\right)^{-1}=\varphi\left(... | theorem exercise_1_6_17 {G : Type*} [group G] (f : G β G)
(hf : f = Ξ» g, gβ»ΒΉ) :
β x y : G, f x * f y = f (x*y) β β x y : G, x*y = y*x := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_2_1_5 | Prove that $G$ cannot have a subgroup $H$ with $|H|=n-1$, where $n=|G|>2$. | \begin{proof}
Solution: Under these conditions, there exists a nonidentity element $x \in H$ and an element $y \notin H$. Consider the product $x y$. If $x y \in H$, then since $x^{-1} \in H$ and $H$ is a subgroup, $y \in H$, a contradiction. If $x y \notin H$, then we have $x y=y$. Thus $x=1$, a contradiction. Th... | theorem exercise_2_1_5 {G : Type*} [group G] [fintype G]
(hG : card G > 2) (H : subgroup G) [fintype H] :
card H β card G - 1 := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_2_4_4 | Prove that if $H$ is a subgroup of $G$ then $H$ is generated by the set $H-\{1\}$. | \begin{proof}
If $H=\{1\}$ then $H-\{1\}$ is the empty set which indeed generates the trivial subgroup $H$. So suppose $|H|>1$ and pick a nonidentity element $h \in H$. Since $1=h h^{-1} \in\langle H-\{1\}\rangle$ (Proposition 9), we see that $H \leq\langle H-\{1\}\rangle$. By minimality of $\langle H-\{1\}\rangle... | theorem exercise_2_4_4 {G : Type*} [group G] (H : subgroup G) :
subgroup.closure ((H : set G) \ {1}) = β€ := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_2_4_16b | Show that the subgroup of all rotations in a dihedral group is a maximal subgroup. | \begin{proof}
Fix a positive integer $n>1$ and let $H \leq D_{2 n}$ consist of the rotations of $D_{2 n}$. That is, $H=\langle r\rangle$. Now, this subgroup is proper since it does not contain $s$. If $H$ is not maximal, then by the previous proof we know there is a maximal subset $K$ containing $H$. Then $K$ must... | theorem exercise_2_4_16b {n : β} {hn : n β 0}
{R : subgroup (dihedral_group n)}
(hR : R = subgroup.closure {dihedral_group.r 1}) :
R β β€ β§
β K : subgroup (dihedral_group n), R β€ K β K = R β¨ K = β€ := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_3_1_3a | Let $A$ be an abelian group and let $B$ be a subgroup of $A$. Prove that $A / B$ is abelian. | \begin{proof}
Lemma: Let $G$ be a group. If $|G|=2$, then $G \cong Z_2$.
Proof: Since $G=\{e a\}$ has an identity element, say $e$, we know that $e e=e, e a=a$, and $a e=a$. If $a^2=a$, we have $a=e$, a contradiction. Thus $a^2=e$. We can easily see that $G \cong Z_2$.
If $A$ is abelian, every subgroup of $A$ ... | theorem exercise_3_1_3a {A : Type*} [comm_group A] (B : subgroup A) :
β a b : A β§Έ B, a*b = b*a := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_3_1_22b | Prove that the intersection of an arbitrary nonempty collection of normal subgroups of a group is a normal subgroup (do not assume the collection is countable). | \begin{proof}
Let $\left\{H_i \mid i \in I\right\}$ be an arbitrary collection of normal subgroups of $G$ and consider the intersection
$$
\bigcap_{i \in I} H_i
$$
Take an element $a$ in the intersection and an arbitrary element $g \in G$. Then $g a g^{-1} \in H_i$ because $H_i$ is normal for any $i \in H$
By the... | theorem exercise_3_1_22b {G : Type*} [group G] (I : Type*)
(H : I β subgroup G) (hH : β i : I, subgroup.normal (H i)) :
subgroup.normal (β¨
(i : I), H i):= | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_3_2_11 | Let $H \leq K \leq G$. Prove that $|G: H|=|G: K| \cdot|K: H|$ (do not assume $G$ is finite). | \begin{proof}
Proof. Let $G$ be a group and let $I$ be a nonempty set of indices, not necessarily countable. Consider the collection of subgroups $\left\{N_\alpha \mid \alpha \in I\right\}$, where $N_\alpha \unlhd G$ for each $\alpha \in I$. Let
$$
N=\bigcap_{\alpha \in I} N_\alpha .
$$
We know $N$ is a subgro... | theorem exercise_3_2_11 {G : Type*} [group G] {H K : subgroup G}
(hHK : H β€ K) :
H.index = K.index * H.relindex K := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_3_2_21a | Prove that $\mathbb{Q}$ has no proper subgroups of finite index. | \begin{proof}
Solution: We begin with a lemma.
Lemma: If $D$ is a divisible abelian group, then no proper subgroup of $D$ has finite index.
Proof: We saw previously that no finite group is divisible and that every proper quotient $D / A$ of a divisible group is divisible; thus no proper quotient of a divisible g... | theorem exercise_3_2_21a (H : add_subgroup β) (hH : H β β€) : H.index = 0 := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_3_4_1 | Prove that if $G$ is an abelian simple group then $G \cong Z_{p}$ for some prime $p$ (do not assume $G$ is a finite group). | \begin{proof}
Solution: Let $G$ be an abelian simple group.
Suppose $G$ is infinite. If $x \in G$ is a nonidentity element of finite order, then $\langle x\rangle<G$ is a nontrivial normal subgroup, hence $G$ is not simple. If $x \in G$ is an element of infinite order, then $\left\langle x^2\right\rangle$ is a no... | theorem exercise_3_4_1 (G : Type*) [comm_group G] [is_simple_group G] :
is_cyclic G β§ β G_fin : fintype G, nat.prime (@card G G_fin) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_3_4_5a | Prove that subgroups of a solvable group are solvable. | \begin{proof}
Let $G$ be a solvable group and let $H \leq G$. Since $G$ is solvable, we may find a chain of subgroups
$$
1=G_0 \unlhd G_1 \unlhd G_2 \unlhd \cdots \unlhd G_n=G
$$
so that each quotient $G_{i+1} / G_i$ is abelian. For each $i$, define
$$
H_i=G_i \cap H, \quad 0 \leq i \leq n .
$$
Then $H_i \... | theorem exercise_3_4_5a {G : Type*} [group G]
(H : subgroup G) [is_solvable G] : is_solvable H := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_3_4_11 | Prove that if $H$ is a nontrivial normal subgroup of the solvable group $G$ then there is a nontrivial subgroup $A$ of $H$ with $A \unlhd G$ and $A$ abelian. | \begin{proof}
Suppose $H$ is a nontrivial normal subgroup of the solvable group $G$.
First, notice that $H$, being a subgroup of a solvable group, is itself solvable. By exercise $8, H$ has a chain of subgroups
$$
1 \leq H_1 \leq \ldots \leq H
$$
such that each $H_i$ is a normal subgroup of $H$ itself and $H_... | theorem exercise_3_4_11 {G : Type*} [group G] [is_solvable G]
{H : subgroup G} (hH : H β β₯) [H.normal] :
β A β€ H, A.normal β§ β a b : A, a*b = b*a := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_4_3_26 | Let $G$ be a transitive permutation group on the finite set $A$ with $|A|>1$. Show that there is some $\sigma \in G$ such that $\sigma(a) \neq a$ for all $a \in A$. | \begin{proof}
Let $G$ be a transitive permutation group on the finite set $A,|A|>1$. We want to find an element $\sigma$ which doesn't stabilize anything, that is, we want a $\sigma$ such that
$$
\sigma \notin G_a
$$
for all $a \in A$.
Since the group is transitive, there is always a $g \in G$ such that $b=g ... | theorem exercise_4_3_26 {Ξ± : Type*} [fintype Ξ±] (ha : fintype.card Ξ± > 1)
(h_tran : β a b: Ξ±, β Ο : equiv.perm Ξ±, Ο a = b) :
β Ο : equiv.perm Ξ±, β a : Ξ±, Ο a β a := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_4_2_14 | Let $G$ be a finite group of composite order $n$ with the property that $G$ has a subgroup of order $k$ for each positive integer $k$ dividing $n$. Prove that $G$ is not simple. | \begin{proof}
Solution: Let $p$ be the smallest prime dividing $n$, and write $n=p m$. Now $G$ has a subgroup $H$ of order $m$, and $H$ has index $p$. Then $H$ is normal in $G$.
\end{proof} | theorem exercise_4_2_14 {G : Type*} [fintype G] [group G]
(hG : Β¬ (card G).prime) (hG1 : β k β£ card G,
β (H : subgroup G) (fH : fintype H), @card H fH = k) :
Β¬ is_simple_group G := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_4_4_6a | Prove that characteristic subgroups are normal. | \begin{proof}
Let $H$ be a characterestic subgroup of $G$. By definition $\alpha(H) \subset H$ for every $\alpha \in \operatorname{Aut}(G)$. So, $H$ is in particular invariant under the inner automorphism. Let $\phi_g$ denote the conjugation automorphism by $g$. Then $\phi_g(H) \subset H \Longrightarrow$ $g H g^{-... | theorem exercise_4_4_6a {G : Type*} [group G] (H : subgroup G)
[subgroup.characteristic H] : subgroup.normal H := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_4_4_7 | If $H$ is the unique subgroup of a given order in a group $G$ prove $H$ is characteristic in $G$. | \begin{proof}
Let $G$ be group and $H$ be the unique subgroup of order $n$. Now, let $\sigma \in \operatorname{Aut}(G)$. Now Clearly $|\sigma(G)|=n$, because $\sigma$ is a one-one onto map. But then as $H$ is the only subgroup of order $n$, and because of the fact that a automorphism maps subgroups to subgroups, w... | theorem exercise_4_4_7 {G : Type*} [group G] {H : subgroup G} [fintype H]
(hH : β (K : subgroup G) (fK : fintype K), card H = @card K fK β H = K) :
H.characteristic := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_4_5_1a | Prove that if $P \in \operatorname{Syl}_{p}(G)$ and $H$ is a subgroup of $G$ containing $P$ then $P \in \operatorname{Syl}_{p}(H)$. | \begin{proof}
If $P \leq H \leq G$ is a Sylow $p$-subgroup of $G$, then $p$ does not divide $[G: P]$. Now $[G: P]=[G: H][H: P]$, so that $p$ does not divide $[H: P]$; hence $P$ is a Sylow $p$-subgroup of $H$.
\end{proof} | theorem exercise_4_5_1a {p : β} {G : Type*} [group G]
{P : subgroup G} (hP : is_p_group p P) (H : subgroup G)
(hH : P β€ H) : is_p_group p H := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_4_5_14 | Prove that a group of order 312 has a normal Sylow $p$-subgroup for some prime $p$ dividing its order. | \begin{proof}
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 choi... | theorem exercise_4_5_14 {G : Type*} [group G] [fintype G]
(hG : card G = 312) :
β (p : β) (P : sylow p G), P.normal := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_4_5_16 | Let $|G|=p q r$, where $p, q$ and $r$ are primes with $p<q<r$. Prove that $G$ has a normal Sylow subgroup for either $p, q$ or $r$. | \begin{proof}
Let $|G|=p q r$. We also assume $p<q<r$. We prove that $G$ has a normal Sylow subgroup of $p$, $q$ or $r$. Now, Let $n_p, n_q, n_r$ be the number of Sylow-p subgroup, Sylow-q subgroup, Sylow-r subgroup resp. So, we have $n_r=1+r k$ such that $1+r k \mid p q$. So, in this case as $r$ is greatest $n_r$... | theorem exercise_4_5_16 {p q r : β} {G : Type*} [group G]
[fintype G] (hpqr : p < q β§ q < r)
(hpqr1 : p.prime β§ q.prime β§ r.prime)(hG : card G = p*q*r) :
nonempty (sylow p G) β¨ nonempty(sylow q G) β¨ nonempty(sylow r G) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_4_5_18 | Prove that a group of order 200 has a normal Sylow 5-subgroup. | \begin{proof}
Let $G$ be a group of order $200=5^2 \cdot 8$. Note that 5 is a prime not dividing 8 . Let $P \in$ $S y l_5(G)$. [We know $P$ exists since $S y l_5(G) \neq \emptyset$ by Sylow's Theorem]
The number of Sylow 5-subgroups of $G$ is of the form $1+k \cdot 5$, i.e., $n_5 \equiv 1(\bmod 5)$ and $n_5$ di... | theorem exercise_4_5_18 {G : Type*} [fintype G] [group G]
(hG : card G = 200) :
β N : sylow 5 G, N.normal := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_4_5_20 | Prove that if $|G|=1365$ then $G$ is not simple. | \begin{proof}
Since $|G|=1365=3.5.7.13$, $G$ has $13-$Sylow subgroup of order $13$. Now, we count the number of such subgroups. Let $n_{13}$ be the number of $13-$Sylow subgroup. Now $n_{13}=1+13k$ where $1+13k|3.5.7$. The choices for $k$ is $0$. Hence, there is a unique $13-$Sylow subgroup and hence is normal. so... | theorem exercise_4_5_20 {G : Type*} [fintype G] [group G]
(hG : card G = 1365) : Β¬ is_simple_group G := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_4_5_22 | Prove that if $|G|=132$ then $G$ is not simple. | \begin{proof}
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 ... | theorem exercise_4_5_22 {G : Type*} [fintype G] [group G]
(hG : card G = 132) : Β¬ is_simple_group G := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_4_5_28 | Let $G$ be a group of order 105. Prove that if a Sylow 3-subgroup of $G$ is normal then $G$ is abelian. | \begin{proof}
Given that $G$ is a group of order $1575=3^2 .5^2 .7$. Now, Let $n_p$ be the number of Sylow-p subgroups. It is given that Sylow-3 subgroup is normal and hence is unique, so $n_3=1$. First we prove that both Sylow-5 subgroup and Sylow 7-subgroup are normal. Let $P$ be the Sylow3 subgroup. Now, Consid... | theorem exercise_4_5_28 {G : Type*} [group G] [fintype G]
(hG : card G = 105) (P : sylow 3 G) [hP : P.normal] :
comm_group G := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_5_4_2 | Prove that a subgroup $H$ of $G$ is normal if and only if $[G, H] \leq H$. | \begin{proof}
$H \unlhd G$ is equivalent to $g^{-1} h g \in H, \forall g \in G, \forall h \in H$. We claim that holds if and only if $h^{-1} g^{-1} h g \in H, \forall g \in G, \forall h \in H$, i.e., $\left\{h^{-1} g^{-1} h g: h \in H, g \in G\right\} \subseteq H$. That holds by the following argument:
If $g^{-1}... | theorem exercise_5_4_2 {G : Type*} [group G] (H : subgroup G) :
H.normal β β
(β€ : subgroup G), Hβ β€ H := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_7_1_11 | Prove that if $R$ is an integral domain and $x^{2}=1$ for some $x \in R$ then $x=\pm 1$. | \begin{proof}
Solution: If $x^2=1$, then $x^2-1=0$. Evidently, then,
$$
(x-1)(x+1)=0 .
$$
Since $R$ is an integral domain, we must have $x-1=0$ or $x+1=0$; thus $x=1$ or $x=-1$.
\end{proof} | theorem exercise_7_1_11 {R : Type*} [ring R] [is_domain R]
{x : R} (hx : x^2 = 1) : x = 1 β¨ x = -1 := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_7_1_15 | A ring $R$ is called a Boolean ring if $a^{2}=a$ for all $a \in R$. Prove that every Boolean ring is commutative. | \begin{proof}
Solution: Note first that for all $a \in R$,
$$
-a=(-a)^2=(-1)^2 a^2=a^2=a .
$$
Now if $a, b \in R$, we have
$$
a+b=(a+b)^2=a^2+a b+b a+b^2=a+a b+b a+b .
$$
Thus $a b+b a=0$, and we have $a b=-b a$. But then $a b=b a$. Thus $R$ is commutative.
\end{proof} | theorem exercise_7_1_15 {R : Type*} [ring R] (hR : β a : R, a^2 = a) :
comm_ring R := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_7_2_12 | Let $G=\left\{g_{1}, \ldots, g_{n}\right\}$ be a finite group. Prove that the element $N=g_{1}+g_{2}+\ldots+g_{n}$ is in the center of the group ring $R G$. | \begin{proof}
Let $M=\sum_{i=1}^n r_i g_i$ be an element of $R[G]$. Note that for each $g_i \in G$, the action of $g_i$ on $G$ by conjugation permutes the subscripts. Then we have the following.
$$
\begin{aligned}
N M &=\left(\sum_{i=1}^n g_i\right)\left(\sum_{j=1}^n r_j g_j\right) \\
&=\sum_{j=1}^n \sum_{i=1}... | theorem exercise_7_2_12 {R G : Type*} [ring R] [group G] [fintype G] :
β g : G, monoid_algebra.of R G g β center (monoid_algebra R G) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_7_3_37 | An ideal $N$ is called nilpotent if $N^{n}$ is the zero ideal for some $n \geq 1$. Prove that the ideal $p \mathbb{Z} / p^{m} \mathbb{Z}$ is a nilpotent ideal in the ring $\mathbb{Z} / p^{m} \mathbb{Z}$. | \begin{proof}
First we prove a lemma.
Lemma: Let $R$ be a ring, and let $I_1, I_2, J \subseteq R$ be ideals such that $J \subseteq I_1, I_2$. Then $\left(I_1 / J\right)\left(I_2 / J\right)=I_1 I_2 / J$.
Proof: ( $\subseteq$ ) Let
$$
\alpha=\sum\left(x_i+J\right)\left(y_i+J\right) \in\left(I_1 / J\right)\left(I... | theorem exercise_7_3_37 {R : Type*} {p m : β} (hp : p.prime)
(N : ideal $ zmod $ p^m) :
is_nilpotent N β is_nilpotent (ideal.span ({p} : set $ zmod $ p^m)) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_8_1_12 | Let $N$ be a positive integer. Let $M$ be an integer relatively prime to $N$ and let $d$ be an integer relatively prime to $\varphi(N)$, where $\varphi$ denotes Euler's $\varphi$-function. Prove that if $M_{1} \equiv M^{d} \pmod N$ then $M \equiv M_{1}^{d^{\prime}} \pmod N$ where $d^{\prime}$ is the inverse of $d \bmod... | \begin{proof}
Note that there is some $k \in \mathbb{Z}$ such that $M^{d d^{\prime}} \equiv M^{k \varphi(N)+1} \equiv\left(M^{\varphi(N)}\right)^k \cdot M \bmod N$. By Euler's Theorem we have $M^{\varphi(N)} \equiv 1 \bmod N$, so that $M_1^{d^{\prime}} \equiv M \bmod N$.
\end{proof} | theorem exercise_8_1_12 {N : β} (hN : N > 0) {M M': β€} {d : β}
(hMN : M.gcd N = 1) (hMd : d.gcd N.totient = 1)
(hM' : M' β‘ M^d [ZMOD N]) :
β d' : β, d' * d β‘ 1 [ZMOD N.totient] β§
M β‘ M'^d' [ZMOD N] := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_8_3_4 | Prove that if an integer is the sum of two rational squares, then it is the sum of two integer squares. | \begin{proof}
Let $n=\frac{a^2}{b^2}+\frac{c^2}{d^2}$, or, equivalently, $n(b d)^2=a^2 d^2+c^2 b^2$. From this, we see that $n(b d)^2$ can be written as a sum of two squared integers. Therefore, if $q \equiv 3(\bmod 4)$ and $q^i$ appears in the prime power factorization of $n, i$ must be even. Let $j \in \mathbb{N... | theorem exercise_8_3_4 {R : Type*} {n : β€} {r s : β}
(h : r^2 + s^2 = n) :
β a b : β€, a^2 + b^2 = n := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_8_3_6a | Prove that the quotient ring $\mathbb{Z}[i] /(1+i)$ is a field of order 2. | \begin{proof}
Let $a+b i \in \mathbb{Z}[i]$. If $a \equiv b \bmod 2$, then $a+b$ and $b-a$ are even and $(1+i)\left(\frac{a+b}{2}+\frac{b-a}{2} i\right)=a+b i \in\langle 1+i\rangle$. If $a \not \equiv b \bmod 2$ then $a-1+b i \in\langle 1+i\rangle$. Therefore every element of $\mathbb{Z}[i]$ is in either $\langle ... | theorem exercise_8_3_6a {R : Type*} [ring R]
(hR : R = (gaussian_int β§Έ ideal.span ({β¨0, 1β©} : set gaussian_int))) :
is_field R β§ β finR : fintype R, @card R finR = 2 := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_9_1_6 | Prove that $(x, y)$ is not a principal ideal in $\mathbb{Q}[x, y]$. | \begin{proof}
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 $y=t p$.
Then:
$$
\begin{aligned}
& 0=\operatorname{deg}_y(x)=\operatorname{deg}_y(s)+\operatorname{deg}_y(p) \text { so... | theorem exercise_9_1_6 : Β¬ is_principal
(ideal.span ({X 0, X 1} : set (mv_polynomial (fin 2) β))) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_9_3_2 | Prove that if $f(x)$ and $g(x)$ are polynomials with rational coefficients whose product $f(x) g(x)$ has integer coefficients, then the product of any coefficient of $g(x)$ with any coefficient of $f(x)$ is an integer. | \begin{proof}
Let $f(x), g(x) \in \mathbb{Q}[x]$ be such that $f(x) g(x) \in \mathbb{Z}[x]$.
By Gauss' Lemma there exists $r, s \in \mathbb{Q}$ such that $r f(x), s g(x) \in \mathbb{Z}[x]$, and $(r f(x))(s g(x))=r s f(x) g(x)=f(x) g(x)$. From this last relation we can conclude that $s=r^{-1}$.
Therefore for an... | theorem exercise_9_3_2 {f g : polynomial β} (i j : β)
(hfg : β n : β, β a : β€, (f*g).coeff = a) :
β a : β€, f.coeff i * g.coeff j = a := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_9_4_2b | Prove that $x^6+30x^5-15x^3 + 6x-120$ is irreducible in $\mathbb{Z}[x]$. | \begin{proof}
$$
x^6+30 x^5-15 x^3+6 x-120
$$
The coefficients of the low order.: $30,-15,0,6,-120$
They are divisible by the prime 3 , but $3^2=9$ doesn 't divide $-120$. So this polynomial is irreducible over $\mathbb{Z}$.
\end{proof} | theorem exercise_9_4_2b : irreducible
(X^6 + 30*X^5 - 15*X^3 + 6*X - 120 : polynomial β€) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_9_4_2d | Prove that $\frac{(x+2)^p-2^p}{x}$, where $p$ is an odd prime, is irreducible in $\mathbb{Z}[x]$. | \begin{proof}
$\frac{(x+2)^p-2^p}{x} \quad \quad p$ is on add pprime $Z[x]$
$$
\frac{(x+2)^p-2^p}{x} \quad \text { as a polynomial we expand }(x+2)^p
$$
$2^p$ cancels with $-2^p$, every remaining term has $x$ as $a$ factor
$$
\begin{aligned}
& x^{p-1}+2\left(\begin{array}{l}
p \\
1
\end{array}\right) x^{p-2}... | theorem exercise_9_4_2d {p : β} (hp : p.prime β§ p > 2)
{f : polynomial β€} (hf : f = (X + 2)^p):
irreducible (β n in (f.support \ {0}), (f.coeff n) * X ^ (n-1) :
polynomial β€) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
Dummit-Foote|exercise_9_4_11 | Prove that $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$. | \begin{proof}
$$
p(x)=x^2+y^2-1 \in Q[y][x] \cong Q[y, x]
$$
We 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 $Q[x, y]$.
\end{proof} | theorem exercise_9_4_11 :
irreducible ((X 0)^2 + (X 1)^2 - 1 : mv_polynomial (fin 2) β) := | import .common
open set function nat int fintype real polynomial mv_polynomial
open subgroup ideal submodule zsqrtd gaussian_int char_p mul_aut matrix
open_locale pointwise
open_locale big_operators
noncomputable theory
|
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