**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.13**

An element $x \in R$, $R$ a ring, is called *nilpotent* if $x^m = 0$ for some positive integer m.

(1) Show that if $n = a^kb$ for some integers $a,b$, then $\overline{ab}$ is nilpotent in $\mathbb{Z}/(n)$.

(2) If $a$ is an integer, show that the element $\overline{a} \in \mathbb{Z}/(n)$ is nilpotent if and only if every prime divisor of $n$ also divides $a$. Find the nilpotent elements of $\mathbb{Z}/(72)$ explicitly.

(3) Let $X$ be a nonempty set and $F$ a field, and let $R = {}^XF$ be the ring of functions $X \rightarrow F$. Prove that $R$ contains no nonzero nilpotent elements.

Solution:

(1) Suppose $n = a^kb$, where $k \geq 1$. Now $$(ab)^k = a^kb^k = (a^kb)b^{k-1} = nb^{k-1} \equiv 0 \pmod n.$$ (2) ($\Rightarrow$) Suppose $\overline{a} \in \mathbb{Z}/(n)$ is nilpotent. Then $a^m = nk$ for some $m$ and $k$. Now if $p$ is a prime dividing $n$, then $p$ divides $a^m$, so that it divides $a$. Thus every prime dividing $n$ divides $a$.

($\Leftarrow$) Let $n = p_1^{e_1} \cdots p_k^{e_k}$ and $a = p_1^{d_1} \cdots p_k^{d_k}m$, where $1 \leq e_i, d_i$ for all $i$ and $m$ is some integer. Let $t = \max \{ e_i \}$. Then $$a^t = (p_1^{d_1} \cdots p_k^{d_k}m)^t = p_1^{d_it} \cdots p_k^{d_it}m^t,$$ where $d_it \geq e_i$ for each $i$. Thus $a^t = nb$ for some integer $b$, and we have $a^t \equiv 0 \pmod n$.

The nilpotent elements of $\mathbb{Z}/(72)$, where $72 = 2^3 \cdot 3^2$, are as follows: 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66.

(3) Suppose $\alpha \in R$ is nilpotent. If $\alpha \neq 0$, then there exists $x \in X$ such that $\alpha(x) \neq 0$. Let $m$ be minimal such that $\alpha(x)^m = 0$; note that $m \geq 1$. Then $\alpha(x) \alpha(x)^{m-1} = 0$, where $\alpha(x) \alpha(x)^{m-1}$ are not zero. Thus $F$ contains zero divisors, a contradiction. Thus no nonzero element of $R$ is nilpotent.