3
$\begingroup$

If $A$ is a Noetherian local ring with maximal ideal $\mathfrak m$, how do you show that $\mathfrak m^{i}/\mathfrak m^{i+1}$ is a finitely-generated $A/\mathfrak m$-module/vector space?

I know each $\mathfrak m^i$ (and each $\mathfrak m^{i}/\mathfrak m^{i+1}$) is a f.g. $A$-module...

$\endgroup$
  • $\begingroup$ As a comment, you can generalize the statement. It hold for any ring $A$, and a maximal ideal $m$ that is finitely generated. Note that this follows because if an ideal $I$ is finitely generated over a ring, then all powers of $I$ are also finitely generated over that ring. $\endgroup$ – Rankeya Nov 1 '12 at 21:17
2
$\begingroup$

Since you edited your question, I will post a different answer. You know that $\mathfrak m^i$ is a finitely generated ideal because $A$ is Noetherian. Let $m_1, \dots, m_n$ be a finite set of generators of $\mathfrak m^i$ as an $A$-module. Try to show that the images of these generators of $m^i$ under the canonical projection map $\mathfrak m^i \rightarrow \mathfrak m^i/\mathfrak m^{i+1}$ generate $\mathfrak m^i/\mathfrak m^{i+1}$ as an $A/\mathfrak m$-module (after all what is the $A/\mathfrak m$-module action on $\mathfrak m^i/\mathfrak m^{i+1}$?). Now $A/\mathfrak m$ is a field, since $\mathfrak m$ is a maximal ideal. What is a module over a field?

$\endgroup$
  • $\begingroup$ And, in the time I was posting my answer, you edited the question again. Anyway, my answer should still stand. $\endgroup$ – Rankeya Nov 1 '12 at 21:11
0
$\begingroup$

It's easy to check the result: Let $R$ be a ring, $M$ a finite module. If $I\subset R$ is an ideal such that $IM=0$ then $M$ is finite $R/I-$module. Apply to the case above with $\mathfrak{m}(\mathfrak{m}^i/\mathfrak{m}^{i+1})=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.