Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.26
Solution:
(1) We begin by showing that $\varphi(a+b) = \varphi(a) + \varphi(b)$ for nonnegative $b$ by induction. For the base case b = 0, we have $$\varphi(a+0) = \varphi(a) = \varphi(a) + 0 = \varphi(a) + \varphi(0). $$For the inductive step, suppose that for some $b \geq 0$, for all $a$, $\varphi(a+b) = \varphi(a)+\varphi(b)$. If $a+b > 0$, then $$\varphi(a+(b+1)) = \varphi((a+b)+1) = \varphi(a+b)+1 = \varphi(a) + \varphi(b) + 1 = \varphi(a) + \varphi(b+1).$$ Thus the conclusion holds for all positive $b$. Suppose now that $b < 0$. If $a \geq 0$, then $$\varphi(a+b) = \varphi(b+a) = \varphi(b)+\varphi(a) = \varphi(a) + \varphi(b).$$ If $a < 0$, then $a+b < 0$, and we have $$\varphi(a+b) = -\varphi(-a-b) = -(\varphi(-a) + \varphi(-b)) = -\varphi(-a) – \varphi(-b) = \varphi(a)+\varphi(b).$$ Thus for all integers $a$ and $b$, $\varphi(a+b) = \varphi(a) + \varphi(b)$.
To show that $\varphi(ab) = \varphi(a)\varphi(b)$, we again proceed for nonnegative $b$ by induction. For the base case, note that $$\varphi(a0) = \varphi(0) = 0 = \varphi(a) \cdot 0 = \varphi(a)\varphi(0).$$ For the inductive step, suppose the result holds for all a for some $b \geq 0$. Then $$\varphi(a(b+1)) = \varphi(ab+a) = \varphi(ab)+\varphi(a) = \varphi(a)\varphi(b) + \varphi(a) = \varphi(a)(\varphi(b)+1) = \varphi(a)\varphi(b+1).$$ Thus by induction the result holds for all nonnegative $b$. Now suppose $b < 0$. Then $$\varphi(ab) = \varphi((-a)(-b)) = \varphi(-a)\varphi(-b) = (-\varphi(a))(-\varphi(b)) = \varphi(a)\varphi(b).$$ Thus $\varphi$ is a ring homomorphism.
Now we show that $\mathsf{ker}\ \varphi = n\mathbb{Z}$, where $n = \mathsf{char}\ R$. ($\subseteq$) Suppose $a \in \mathsf{ker}\ \varphi$; then $\varphi(a) = 0$. Suppose now that $n$ does not divide $a$. Then we have $a = nq+r$ for some $r$ with $0 < |r| < n$ by the division algorithm. Note moreover that $\varphi(n) = 0$ by definition, and that $n$ is minimal with this property. Now $$0 = \varphi(a) = \varphi(qn+r) = \varphi(r),$$ a contradiction. Thus $n$ divides $a$, and $a \in n\mathbb{Z}$. ($\supseteq$) Suppose $a \in n\mathbb{Z}$. Then $a = nb$ for some $b$. Note that $\varphi(n) = 0$ by definition, so that $\varphi(a) = \varphi(nb) = 0$. Thus $a \in \mathsf{ker}\ \varphi$.
(2) Consider $\mathbb{Q}$. We claim that $\varphi = \iota$, the inclusion map. To see this, note that $\varphi(0) = 0 = \iota(0)$, that for $a \geq 0$, $$\varphi(a+1) = \varphi(a)+1 = \iota(a)+1 = a+1 = \iota(a+1),$$ and that for $a < 0$, $$\varphi(a) = -\varphi(-a) -\iota(-a) = -(-a) = a = \iota(a).$$Since $\iota = \varphi$ is injective, $\mathsf{ker}\ \varphi = 0$. Thus $\mathsf{char}\ \mathbb{Q} = 0$.
Consider $\mathbb{Z}[x]$. We claim that $\varphi = \iota$, the inclusion map. The proof of this proceeds exactly as in the previous example.
Consider $(\mathbb{Z}/(n))[x]$. Note that $\varphi(n) = \overline{n} = 0$. Thus $n\mathbb{Z} \subseteq \mathsf{ker}\ \varphi$. Now let $a \in \mathsf{ker}\ \varphi$ and suppose $n$ does not divide $a$. By the division algorithm, we have $a = qn+r$ where $0 < |r| < n$. Then $$0 = \varphi(a) = \varphi(qn+r) = \varphi(r),$$ a contradiction since $n$ is minimal. Thus $n$ divides $a$, and $\mathsf{ker}\ \varphi \subseteq n\mathbb{Z}$. Thus $\mathsf{char}\ \mathbb{Z}/n\mathbb{Z} = n$.
(3) Let $R$ be a commutative ring with characteristic $p$, and let $a,b \in R$. Note that $p$ divides $p!$ but not $k!$ or $(p-k)!$ when $0 < k < p$. Thus $p$ divides $\binom{p}{k}$ for $0 < k < p$. Using the Binomial Theorem, we have $$(a+b)^p = \sum_{k=0}^p \binom{p}{k} a^kb^{p-k} = a^p + b^p.$$
