# Nilpotency of an element and its prime divisor.

An element $x$ in a ring $R$ is said to be nilpotent if $x^m = 0$ for some $m$ in $\mathbb{Z}^+$.

If $a \in \mathbb{Z}$ is an integer show that the element $\bar{a} \in \mathbb{Z}/ n \mathbb{Z}$ is nilpotent if and only every prime divisors of $n$ are prime divisors of $a$. I have done this proof. It is given below.

As $\bar{a}$ is nilpotent in $\mathbb{Z}/ n\mathbb{Z}$ $\exists$ an element $k \in \mathbb{Z}$ s.t. $\bar{a}^k \equiv 0$ (mod $n$). So $n|a^k$. Let $p$ be a prime divisor of $n$ then $p|a^k$ i.e. $p|a$. Conversely if every prime divisors of $a$ be a prime divisor of $n$. Then by Fundamental theorem of arithmetic we may calculate an integer $k$ s.t. $n | a^k$. So $a^k \equiv 0$ (mod $n$). Hence the proof. Please check if my attempt is accurate.

My doubt is here. Take $n = 72$. Here prime divisors are $2$ and $3$ only. So all the integer $a$ $0 < a <72$ multiple of both $2$ and $3$ will be a nilpotent element. But $\bar{2}$ is a nilpotent element as $\bar{2}^{36} = 0$ but $\bar{2}$ is not a multiple of $3$. I can not understand the fact. Is there any mistake on the statement of the result?

• Did you by any means use a $32$-bit binary computer to compute $2^{36}$? That would give $0$, which is divisible by $3$ (and by $72$). Without such computational errors, $2^{36}$ should not have acquired a factor$~3$, and hence not be divisible by$~72$. Sep 14, 2013 at 16:04
$2^{36}\equiv 64^6\equiv (-8)^6\equiv 2^{18}\equiv 64^3\equiv -8^3\equiv -2^9 \equiv 64\cdot 8\equiv64\not\equiv 0\mod72$.