User:Zelmerszoetrop/Test1

1.1) Define $H=\left\{{(x,x):x\in G}\right\}\$. We are asked whether $H<G\$. Observe that since $H\$ is not a subset of $G\$ it cannot possibly be a subgroup of $G\$. However, if what is meant is "Is $H\$ a subgroup of $G\times G\$," then the answer is yes, because it

a) contains the identity $(1,1)\in H\$,

b) is closed under multiplication: $(x,x)(y,y)=(xy,xy)\in H\$, and

C) contains inverses: $(x,x)^{-1}=(x^{-1},x^{-1})\in H\$.

1.2) We are asked if $H<G\$ is cyclic. Again, since $H\$ is not a subset of $G\$, it cannot possibly be a cyclic subgroup of $G\$. If what is meant is "Is $H\$ a cyclic subgroup of $G\times G\$," the answer is still no. Suppose $G\$ is not cyclic. Observe that the map $\phi :H\to G\$ defined $\phi ((x,x))\mapsto x\$ is an isomorphism, since it is necessarily bijective, and satisfies $\phi ((x,x)(y,y))=\phi ((xy,xy))=xy=\phi ((x,x))\phi ((y,y))\$. Then $H\$ is not cyclic.

1.5) Observe that for all $(x,x)\in H,\ (y,z)\in G$, we have $(y,z)(x,x)(y^{-1},z^{-1})=(yxy^{-1},zxz^{-1})\$. If $G\$ is commutative, then this is just $(xyy^{-1},xzz^{-1})=(x,x)\in H\$. So, $G\$ commutative $\implies H\triangleleft G\times G\$.

Now suppose $(yxy^{-1},zxz^{-1})\in H\$. Then $\forall x,y,z\in G,\ yxy^{-1}=zxz^{-1}\$. In particular, $yyy^{-1}=zyz^{-1}\$ and $yzy^{-1}=zzz^{-1}\$ for all $y,z\in G\$. But this is just

$y=zyz^{-1}\$ and $yzy^{-1}=z\$.

or

$yz=zy\ \forall y,z\in G\$.

Hence G is commutative, so $H\triangleleft G\times G\implies G\$ is commutative.

Therefore, $H\triangleleft G\times G\iff G\$ is commutative.

1.3, 1.4) By 1.5, these are necessarily false.

2.1) We are asked if every nontrivial group $G\$ contains a non-trivial proper subgroup $H\$. This is false. Consider the group $\mathbb {Z} _{3}\$. By Lagrange's theorem, a subgroup of this must have order either 1 or 3. But this means a subgroup must be either trivial, or the group itself, ie, not proper.

2.2) We are asked if every nontrivial group $G\$ contains a normal nontrivial proper subgroup $H\$. This is false. See problem 1.

2.3) We are asked if every group $H\$ can be embedded in a larger group $G\$ such that $H\$ is normal in $G\$. This is true. Let $H\$ be any group with operation $*\$ and identity $e\$, and define the set $G=H\times \left\{{0,1}\right\}\$ and the operation $\circ :G\times G\to G\$ as

$(x,a)\circ (y,b)=(x*y,a+b\ {\text{mod}}\ 1)\$

Observe that $G\$ is necessarily closed under $\circ \$, since $G\$ is closed under $*\$ and $\left\{{0,1}\right\}\$ is closed under $+\ {\text{mod}}\ 1$.

Furthermore, since both these operations are associative, $\circ \$ is as well.

We certainly have an identity, namely $(e,0)\$: observe $(x,a)\circ (e,0)=(x*e,a+0)=(x,a)\$, and inverses defined $(x,a)^{-1}=(x^{-1},a)\$:

$(x^{-1},a)\circ (x,a)=(x^{-1}*x,a+a\ {\text{mod}}\ 1)=(e,0)=(x*x^{-1},a+a\ {\text{mod}}\ 1)=(x,a)\circ (x^{-1},a)\$.

Hence, $G\$ is a group.

Define the map $\phi :H\to G\$ as $x\mapsto (x,0)\$. Observe that

$\phi (x)\circ \phi (y)=(x,0)\circ (y,0)=(x*y,0)=\phi (x*y)\$,

and so $\phi \$ is a homomorphism. Now suppose $x\neq y\in H\$. Then $\phi (x)=(x,0)\neq (y,0)=\phi (y)\$, and so $\phi \$ is an injective homomorphism, with $\phi (H)\cong H\$ and $(x,0)\in \phi (H)\ \forall (x,0)\in G\$.

Let $(x,0)\in \phi (H)\$ be any element. Observe that for any element $(y,a)\in G\$, we have

$(y,a)\circ (x,0)\circ (y,a)^{-1}=(yxy^{-1},a+a)=(yxy^{-1},0)\in \phi (H)\$.

Hence, $\phi (H)\triangleleft G\$.

3) Let $R,S\$ be commutative rings with identity, $f:R\to S\$ a ring homomorphism, and $I\subset {\text{ker}}(f)\subset R\$ an ideal in $R\$. We aim to show there is a unique ring homomorphism $g:R/I\to S\$ such that $g\circ \pi =f\$, where $\pi \$ is the canonical homomorphism $\pi :R\to R/I\$ defined $a\mapsto {\overline {a}}\$.

We construct the commutative diagram

This suggests the definition $g=f\circ \pi ^{-1}\$. Note that if this is a well-defined homomorphism, then we automatically have $g\circ \pi =f\circ \pi ^{-1}\circ \pi =f\$.

Let ${\overline {a}}\in R/I\$ be any element. We can write this as $a+I\$. Then $g({\overline {a}})=(f\circ \pi ^{-1})(a+I)=f(\pi ^{-1}(a+I))=f(a+\pi ^{-1}(I))=f(a+0)=f(a)\$, and so this is a well-defined function. Furthermore, we have

$g({\overline {ab}})=g((a+I)(b+I))=g(ab+(a+b)I+II)=g(ab+I)=f(ab)=f(a)f(b)=g({\overline {a}})g({\overline {b}})\$,

since $f\$ is a homomorphism, and

$g({\overline {a}}+{\overline {b}})=g(a+b+I)=f(a+b)=f(a)+f(b)=g({\overline {a}})+g({\overline {b}})\$,

again, since $f\$ is a homomorphism. Therefore, $g:R/I\to S\$ is a well-defined ring homomorphism.

Now suppose $\phi :R/I\to S\$ is any ring homomorphism and that $\phi \$ satisfies $\phi \circ \pi =f\$.

Then $\phi \circ \pi \circ \pi ^{-1}=f\circ \pi ^{-1}\implies \phi =f\circ \pi ^{-1}=g\$. So such $g\$ is unique.

workspace

4) We are asked to determine which rings are isomorphic:

$A=\mathbb {C} [x]/(x^{2})\$,

$B=\mathbb {C} [x]/((x-1)^{2})\$,

$C=\mathbb {C} [x]/(x^{3})\$,

$D=\mathbb {C} [x]/(x^{2}+1)\$,

$E=\mathbb {C} \times \mathbb {C} \$.

For $A\$ and $B\$, consider the map $\phi :\mathbb {C} [x]\to \mathbb {C} [x]\$ defined $\phi \left({\Sigma _{k=1}^{n}a_{k}x^{k}}\right)\mapsto \Sigma _{k=1}^{n}a_{k}(x-1)^{k}\$.

This map is clearly an automorphism of $\mathbb {C} [x]\$ sending $(x^{2})\mapsto ((x-1)^{2})$. Let $\pi \$ be the canonical homomorphism $\pi :\mathbb {C} [x]\to \mathbb {C} [x]/((x-1)^{2})\$.

Observe that ${\text{ker}}(\pi )=((x-1)^{2})\implies {\text{ker}}(\pi \circ f)=(x^{2})\$. Since $\pi \circ f:\mathbb {C} [x]\to \mathbb {C} [x]/((x-1)^{2})\$ is a homomorphism with kernel $(x^{2})\$, we must have

$\mathbb {C} [x]/(x^{2})\cong \mathbb {C} [x]/((x-1)^{2})\$.

So $A\cong B\$.

Observe $A=\left\{{ax+b:a,b\in \mathbb {C} }\right\}\$, considered as a vector space over $\mathbb {C} \$, has dimension 2, while $C=\left\{{ax^{2}+bx+c:a,b,c\in \mathbb {C} }\right\}\$ has dimension 3 as a vector space. Since $A\not \cong C\$ as vector spaces, they cannot be isomorphic as rings: $A\not \cong C\$.

For $A\$ and $D\$, observe that by the division algorithm, any polynomial $p(x)\in \mathbb {C} [x]\$ can be written $p(x)=(x^{2}+1)q(x)+r(x)\$ where ${\text{deg}}{r}<2\$, and so $D\$ is isomorphic to $\left\{{ax+b:a,b\in \mathbb {C} }\right\}\$. Hence, $D\cong A\$.

We show that $A=\mathbb {C} [x]/(x^{2})=\left\{{a+bx:a,b\in \mathbb {C} }\right\}\$ is not ring isomorphic to $E=\mathbb {C} \times \mathbb {C} \$ using a proof by contradiction. Suppose these two rings are isomorphic. Then there is a ring homomorphism $\phi :\mathbb {C} [x]\to \mathbb {C} \times \mathbb {C} \$ such that ${\text{ker}}(\phi )=(x^{2})\$.

Observe that since $\phi \$ is a homomorphism, we must also have

$\forall a,b\in \mathbb {C} ,\ \phi (ax)\phi (bx)=\phi (abx^{2})=(0,0)\$.

Therefore, either $\forall a\in \mathbb {C} ,\ ax\in {\text{ker}}(\phi )\$ or $\forall a\in \mathbb {C} ,\ \phi (ax)\in \mathbb {C} \times \mathbb {C} \$ is a zero divisor. Since $ax\not \in (x^{2})\$, the first possibility contradicts our requirement that ${\text{ker}}(\phi )=(x^{2})\$.

The only zero divisors in $\mathbb {C} \times \mathbb {C} \$ are of the forms $(0,z),\ (z,0)\$, so we must have $\forall a\in \mathbb {C} ,\ \phi (ax)\in \left\{{(0,z),(z,0)}\right\}\$ for some $z\in \mathbb {C} \$.

But we may also observe that $\forall a,b\in \mathbb {C} ,\ \phi (ax)+\phi (bx)=\phi ((a+b)x)\$ which must also be a zero divisor. Since zero divisors in $\mathbb {C} \times \mathbb {C} \$ can satisfy $(0,y)+(z,0)=(z,y)\$, which is not a zero divisor, we can say that either $\forall a\in \mathbb {C} ,\ \phi (ax)=(z,0)\$ or $\forall a\in \mathbb {C} ,\ \phi (ax)=(0,z)\$. Without loss of generality, we can assume the first case is true.

Since $\phi |_{\left\{{ax}\right\}}\subseteq (\mathbb {C} ,0)\cong \mathbb {C} \$, and since $\phi \$ is a homomorphism, observe that

$\forall a,b,\exists y,z:\ (0,0)=\phi (abx^{2})=\phi (ax)\phi (bx)=(y,0)(z,0)\implies yz=0\$.

Since $ax,bx\not \in {\text{ker}}(\phi )\$, we must have $y,z\$ as zero divisors in $\mathbb {C} \$. But $\mathbb {C} \$ has no zero divisors, and so no such homomorphism $\phi \$ exists. Hence, $A\not \cong E\$.

Observe that the ring $C=\mathbb {C} [x]/(x^{3})=\left\{{ax^{2}+bx+c:a,b,c\in \mathbb {C} }\right\}\$ has dimension 3 as a vector space. Since $E=\mathbb {C} \times \mathbb {C} \$ has dimension 2 as a vector space, they cannot be isomorphic as vector spaces, and so they cannot be isomorphic as rings: $C\not \cong E\$.

Since we have $A\cong B,\ A\cong D\$, we must have $B\cong D\$. Since $C\not \cong A\not \cong E\$, we must have $C\not \cong B\not \cong E,\ C\not \cong D\not \cong E\$. We have also shown $E\not \cong C\$. We can summarize these results in a table:

•	A	B	C	D	E
A	$\cong$	$\cong$	$\not \cong$	$\cong$	$\not \cong$
B	$\cong$	$\cong$	$\not \cong$	$\cong$	$\not \cong$
C	$\not \cong$	$\not \cong$	$\cong$	$\not \cong$	$\not \cong$
D	$\cong$	$\cong$	$\not \cong$	$\cong$	$\not \cong$
E	$\not \cong$	$\not \cong$	$\not \cong$	$\not \cong$	$\cong$

workspace 2

5) With $A,B,C,D,E\$ defined as in problem 4, we are asked to find all pairs of isomorphic vector spaces. From (4), and the fact that if two rings are isomorphic, they are isomorphic as vector spaces, we have

$A\cong B\cong D\$.

Define the map $\phi :E\to A\$ as $\phi ((a,b))=ax+b\$.

end workspace

6.1) Let $F\$ be a finite field of characteristic $p>0\$ and define the map $\phi :F\to F\$ as $a\mapsto a^{p}\$. We are given that this is a homomorphism; we aim to show that this is an automorphism.

To show that $\phi \$ is an automorphism, it suffices to show that it is a bijection from $F\to F\$.

Injection:

Suppose $x,y\in F\$ and $\phi (x)=\phi (y)\$. We aim to show that $x=y\$, meaning $F\$ is an injection. Our assumption $\phi (x)=\phi (y)\$ means $x^{p}=y^{p}\$. By multiplying both sides by $x^{-p}\$, we have $1=y^{p}x^{-p}\$, which we can rewrite as $1=(y^{-1}x)^{-p}\$. Raising both sides to the $p\$ power, we have $1^{p}=y^{-1}x\$, but since $1^{p}=1\$, this is just $1=y^{-1}x\$. Since $y^{-1}\$ is a multiplicative inverse of both $x,y\$, and since multiplicative inverses are unique, we must have $y=x\$. Hence $\phi \$ is an injection.

Surjection:

Let $y\in F\$ be any element. We aim to show $\exists x\in F:\phi (x)=x^{p}=y\$. This would demonstrate $\phi \$ is a surjection.

Since $F\$ is an injection, we can be sure that $a\neq b\implies \phi (a)\neq \phi (b)\$.

This means that $|\phi (F)|=|F|\$, because $|\phi (F)|<|F|\implies \exists a\neq b:\phi (a)=\phi (b)$, which we know to be false. But then since $F\$ is finite, $|\phi (F)|=|F|\implies \forall y\in F,\ \exists x\in F:\phi (x)=y$. Hence $\phi \$ is a surjection.

6.2) Note that the surjective proof in 6.1 fails if $|F|=\infty \$, because then $|\phi (F)|=|F|\not \implies \forall y\in F,\ \exists x\in F:\phi (x)=y$. We can find a specific counterexample in $Q\$, the field of fractions of $F[x]\$, where $F\$ is a finite field of characteristic $p\$.

Observe that since $F\$ is a field, $F[x]\$ is an integral domain, and so by Theorem 15, pg 261 of Dummit and Foote, $Q\$ is a field.

Further observe that $\underbrace {1_{Q}+\dots +1_{Q}} _{p\ {\text{times}}}=\underbrace {1_{F}+\dots +1_{F}} _{p\ {\text{times}}}=1\$, so ${\text{char}}(Q)=p\$.

Observe that if $g(x)\$ is an irreducible polynomial in $F[x]\$, then the statement "$\exists f=a(x)/b(x)\in Q:\phi (f)=g\$" is equivalent to saying $\exists a(x),b(x):a^{2}(x)=g(x),\ b^{2}(x)=1\$," since $a(x)/b(x)\$ in lowest terms implies $a^{2}(x)/b^{2}(x)\$ is in lowest terms. Since we have assumed $g(x)\$ is irreducible, this is a contradiction, and hence $g\$ cannot have a pre-image in $Q\$.