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.

1.3, 1.4, 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.

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.

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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} \$.

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.

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\$.

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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\$.