## An ideal which is finitely generated by nilpotent elements is nilpotent

Prove that if $R$ is a commutative ring and $N = (A)$ is a finitely generated ideal, where $A = \{a_1,\ldots,a_n\}$ and each $a_i$ is nilpotent, then $N$ is a nilpotent ideal. Deduce that if the nilradical $\mathfrak{N}(R)$ of $R$ is finitely generated then it is a nilpotent ideal.

Say $a_i^{m_i} = 0$ for each $a_i \in A$. Now let $M = \sum m_i$.

By this previous exercise, we know that $N^M = (B)$, where $B = \{ \prod a_i^{k_i} \ |\ \sum k_i = M \}$. Then it must be the case that for each element $\prod a_i^{k_i}$ of $B$, some $k_i$ is at least $m_i$. Thus $B = 0$, and we have $N^M = 0$. Thus $N$ is a nilpotent ideal.

If $\mathfrak{N}(R)$ is finitely generated, then each generator is (by definition) nilpotent. Thus $\mathfrak{N}(R)$ is a nilpotent ideal.