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?
Thank your for your help. If it is already discussed please mention the link only.
Source : Abstract Algebra by Dummit and Foot , 2nd edition Page 232 Exercise 13 b.
