Question

Prove that the set of nilpotent elements in a commutative ring $\mathit{R}$ is an ideal.

Short answer

The set of nilpotent elements in a commutative ring $R$is an ideal.

Step-by-step solution

Exercise 44 in section 3.2

Let a and b be nilpotent elements in a commutative ring $R$ then a + b and ab are also nilpotent.

N is an ideal

Let N be the set of all nilpotent elements in a commutative ring $R$.

Since, role="math" localid="1654239405045" $R$is commutative it is enough to show that for any role="math" localid="1654239409053" $a\in R$ and role="math" localid="1654239412666" $n\in N$ implies role="math" localid="1654239418389" $an\in N$.

Let $k\in {\mathbb{Z}}_{+}$ such that $n^K=0$then,

$$\begin{gathered} {\left(an\right)}^k=a^k{n}^k \\ =a^{k}0 \\ =0 \end{gathered}$$

Thus, $\mathrm{an}\in \mathrm{N}$ and N is an ideal of role="math" localid="1654239556504" $R$.