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.

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