# Finitely generated quotients in Noetherian local rings

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...

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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. – Rankeya Nov 1 '12 at 21:17

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?
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 $\mathfrak{m}(\mathfrak{m}^i/\mathfrak{m}^{i+1})=0$.