## Examples of nilpotent elements

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.

1. Suppose $n = a^kb$, where $k \geq 1$. Now $(ab)^k = a^kb^k$ $= (a^kb)b^{k-1}$ $= nb^{k-1}$ $\equiv 0$ mod $n$.
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 $a^t \equiv 0$ mod $n$.

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.

• curious  On February 7, 2011 at 10:19 pm

Hello, don’t you mean $1 \leq e_{i} \leq d_{i}$ so that every prime divisor of $n$ divides $a$ ?

• nbloomf  On February 7, 2011 at 10:41 pm

Yes. Thanks!

• Jason  On December 7, 2011 at 11:10 pm

I think it’s not necessarily true that $1 \leq e_i \leq d_i$. For example, 6 is nilpotent in $Z/12Z$. However, I think the rest of the argument works without this assumption.

Also a minor typo, I think you meant \$a^t = (p_1^{d_1} \cdots p_k^{d_k}m)^t

• nbloomf  On December 9, 2011 at 1:24 am

Thanks!