To show that $\mathfrak{m}B \neq B$ for all maximal ideals $\mathfrak{m} \subset A$, where $B$ is finitely generated as $A$-module

Asked Dec 13 '17 at 18:47

Active Dec 13 '17 at 20:23

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Let $B \supset A$ be an extension of commutative rings such that $B$ is finitely generated as $A$-module. Prove that $\mathfrak{m}B \neq B$ for all maximal ideals $\mathfrak{m} \subset A$.

I have no idea how to proceed. I think it should be enough to show proper inclusion. Any help?

edited Dec 13 '17 at 20:23

user26857

52,094

asked Dec 13 '17 at 18:47

Naweed G. Seldon

2,635

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Did you try this: https://en.wikipedia.org/wiki/Nakayama%27s_lemma? – user26857 Dec 13 '17 at 18:52
Ah! Didn't strike me. Lemme give it a go. Thanks! – Naweed G. Seldon Dec 13 '17 at 19:56

To show that $\mathfrak{m}B \neq B$ for all maximal ideals $\mathfrak{m} \subset A$, where $B$ is finitely generated as $A$-module

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