Mg Problem 30 32

FINALS

30. Thm: Let $p$ be prime. For any natural number, $n$, and elements $x,y$ of the finite field $\mathbb{F}_p$, $(x+y)^{p^n} = x^{p^n} + y^{p^n}$.

Pf: Prove by induction on n. For base case $n=1$, $(x+y)^{p^n} = (x+y)^p$. The binomial theorem tells us that this expression is equal to the summation $\sum_{i=0}^{p} \binom{p}{i}x^{p-i}y^{i}$. The combination $\binom{p}{i}$ is equal to $\frac{p!}{i!(p-i)!}$. For $0<i<p$, $\frac{p!}{i!(p-i)!}$ will be a multiple of $p$. Therefore, every term except the first, $x^p$, and the last, $y^p$, in the binomial expansion for any $p$ will have a coefficient that is a multiple of $p$ and will therefore be equivalent to $0$ in the field $\mathbb{F}_p$. Thus, $(x+y)^p = x^{p}+y^{p}$, and the base case is proven. Then assume that for some $n=k$, $(x+y)^{p^k} = x^{p^k} + y^{p^k}$. Suffice it to show that $(x+y)^{p^{k+1}} = x^{p^{k+1}} + y^{p^{k+1}}$. By properties of exponents, $(x+y)^{p^{k+1}} = ((x+y)^{p^k})^p$. By inductive hypothesis, this expression is equivalent to $(x^{p^k} + y^{p^k})^p$. By the same binomial expansion logic as was used to prove the base case, this expression simplifies to $(x^{p^k})^p + (y^{p^k})^p = x^{p^{k+1}} + y^{p^{k+1}}$.

QED

31(a) Let $f: R\rightarrow{S}$ be a ring homomorphism, let $A$ and $I$ be a subring and ideal of $R$ respectively, and let $B$ and $J$ be a subring and ideal of $S$ respectively. For each question below, either prove the assertion or provide a counter example.

i. Is the image $f(A)$ of $A$ in $S$ necessarily a subring?

Yes!
Since $f(A)$ is clearly in $S$, suffice it to show that $f(A)$ is a ring, i.e. that it is closed under addition and multiplication.

Addition: From the properties of a homomorphism, we know that $f(a+b) = f(a) + f(b)$ for arbitrary $a,b \in A$. $a+b$ is in the subring $A$ because of closure under addition. Therefore, for arbitrary $f(a),f(b) \in f(A)$, $f(a)+f(b) \in f(A)$ because $f(a)+f(b) = f(a+b)$, with $f(a+b)$ the sum of two arbitrary elements of $f(A)$ and also the image of $a+b$, an element of $A$. Thus, $f(A)$ is closed under addition.

Multiplication: Similarly, $f(ab)=f(a)f(b)$. Since $a,b,ab \in A$, and the product of $f(a)$ and $f(b)$ is $f(ab) \in A$, $A$ is closed under multiplication.

Thus, $f(A)$ is a subring. QED

ii. Is the image $f(I)$ necessarily an ideal?

No!
While $f(I)$ is an ideal of $f(R)$, since the homomorphism is not necessarily surjective, we cannot say that $f(I)$ is necessarily an ideal of $S$.
Counterexample:

$f: \mathbb{Z} \rightarrow \mathbb{Z}[x]$. The ideal $2\mathbb{Z}$ maps to the even constant polynomials. This is clearly not an ideal, since $2x = 2_{\mathbb{Z}[x]}\cdot x$, for example, is not in $f(2\mathbb{Z})$, indicating that $f(2\mathbb{Z})$ is not superclosed under multiplication.

iii. Is the preimage $f^{-1}(B)$ of $B$ in $R$ necessarily a subring?

Yes!
Consider $a,b \in f^{-1}(B)$. Then $f(a),f(b) \in B$. Since $B$ is a ring, $(f(a)+f(b)), (f(a)f(b)) \in B$. By the properties of homomorphisms, $f(a) + f(b) = f(a+b) \in B$ and $f(a)f(b) = f(ab)\in B$. But then $a+b$ and $ab$ are in $f^{-1}(B)$. Therefore, $f^{-1}(B)$ is closed under both addition and multiplication, and is therefore a ring. Since it is contained in $R$, $f^{-1}(B)$ is a subring. QED

iv. Is the preimage $f^{-1}(J)$ necessarily an ideal?

Yes!
For arbitrary $x,y \in f^{-1}(J)$, $f(x),f(y) \in J$, so $f(x)-f(y) \in J = f(x-y) \in J$ by the properties of homomorphisms. But since $f(x-y)\in J$, $x-y \in f^{-1}(J)$. So $f^{-1}(J)$ is closed under addition. Suffice it to show that $f^{-1}(J)$ is superclosed under multiplication. Consider arbitrary $a \in R$ and $x \in f^{-1}(J)$. $f(a)f(x) = f(ax) \in J$ by the superclosure of $J$ under multiplication. Therefore, $ax \in f^{-1}(J)$. QED

31(b) First Isomorphism Theorem for Rings: If $\phi : R \rightarrow S$ is a ring homomorphism, then $R/ker{\phi} \cong \phi(R)$.

32. Let $p$ be a prime. It is a fact that for each $k \geq 1$ there is a unique (up to isomorphism) finite field of cardinality $p^k$, denoted $\mathbb{F}_{p^k}$. Prove that there is no surjective ring homomorphism from $\mathbb{Z}$ onto $\mathbb{F}_{p^k}$ for $k>1$.

Suppose there does exist a surjective ring homomorphism $\phi : \mathbb{Z} \rightarrow \mathbb{F}_{p^k}$. We know $\mathbb{F}_{p^k} \cong \mathbb{Z}/{p^k}\mathbb{Z}$, so we would have $\phi : \mathbb{Z} \rightarrow \mathbb{Z}/{p^k}\mathbb{Z}$. If $k>1$ then

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