Questions 30-32 Pre Write

30. Let $p$ be prime. Show that for all natural numbers $n$ and elements $x,y$ of the finite field $\mathbb{F}_p$ , we have $(x+y)^{p^n}=x^{p^n}+y^{p^n}$.
Proof: We can prove the above by using induction. The base case $n=1$ is equal to $(x+y)^p = \sum \binom{p}{i}x^{p_i}y^i$ where $0 < i < p$. $\binom{p}{i} = p! / (i!(p-i)!) = (p(p-1)...(p-i)!) / (i!(p-i)!)$ This means that $i!$ is a factor of $p$.
We then assume that the above statement holds for some $n$ and $(x+y)^{p^n}$. The $n+1$ case is equal to $(x+y)^{p^{n+1}}= (x+y)^{p^{n}*p}=((x+y)^{p^n})^p$. We know that, by the inductive hypothesis $(x+y)^{p^n}$ is true and then by the base case $(x+y)^{p}$ is true.

31. Let $f:R\rightarrow S$ be a ring homomorphism, let $A$ and $I$ be a subring and ideal of $R$, and let $B$ and $J$ be a subring and ideal of $S$.
Statement: Is the image $f(A)$ of $A$ in $S$ necessarily a subring?
Proof: We must show that $f(A) \in S$ is closed under addition and multiplication. To show addition, suppose we have some $y_1, y_2 \in f(A)$ such that $\exists x_1 \in A s.t. y_1 = f(x_1)$ and $\exists x_2 \in A s.t. y_2 = f(x_2)$. It follows that $y_1 + y_2 = f(x_1) +_S f(x_2)$. As $f$ is a homomorphism, $f(x_1 +_R x_2) \in A$. Similarly, $y_1 ⋅ y_2 = f(x_1) ⋅_S f(x_2)$. As $f$ is a homomorphism, $f(x_1 ⋅_R x_2) \in A$.
Statement: Is the image $f(I)$ necessarily an ideal?
Proof: This can be proven false through contradiction by looking at the polynomial ring in the integers. $\mathbb{Z} \mapsto \mathbb{Z}[x]$. ${...-2p, -p, ...,p, 2p...} \rightarrow ^(\iota) {...-2p, -p, ...,p, 2p...}$ but $px \notin \iota(p \mathbb{Z})$ as x is a variable and does not have to be in $\mathbb{Z}$.

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 $k>1$.
Proof: $f: \mathbb{Z} \rightarrow \mathbb{F}_{p^k}$ is surjective if and only if $Im(f) \in \mathbb{Z} = \mathbb{F}_{p^k}$. Not everything in $\mathbb{Z}$ maps to a prime number. We have already show that any number can be written as a unique sum of primes, but this is not the same.

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