Nilpotent element in $\mathbb{Z}/12\mathbb{Z}$ - Ideal
By Gabriel Cooper •
Let $A$ a commutative ring with the unit element $1 \not= 0$. $a \in A$ is a nilpotent element if there exists $n \in \mathbb{N}$ such that $a^n=0$.
I have already prove that the set of nilpotent elements of $A$ is an ideal of $A$. How could determine all nilpotent element in $\mathbb{Z}/12\mathbb{Z}$ from what I got?
$\endgroup$ 11 Answer
$\begingroup$You could assume that $x \in \mathbb{Z}/\mathbb{12Z}$ is nilpotent elemet. Then there exists an integer $n$ such that $x^n=0$, i.e., $x^n \equiv 0 \pmod {12}$, in other words, $x^n$ is divisible by $12$. Therefore, $x=0$ or $\bar{6}$. Generally speaking, $x=p^{l_1}_{1}p^{l_2}_{2}...p^{l_n}_{n}$, where $p_i$, $1 \leq i \leq n$, are all the distinct prime factors of $A$, $l_i \in \mathbb{Z}_{\geq 1}$, $1 \leq i \leq n$.
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