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How to prove :

$A/m^n$ is Artinian for all $n\geq 0$ if $A$ is a Noetherian ring and $m$ maximal ideal.

Any suggestions ?

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3 Answers 3

Hint: Show that a Noetherian ring with only $1$ prime ideal is Artinian. What are the prime ideals of $A/m^n$?

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You need to prove that $A/\mathfrak{m}^n$ is of finite length as a module over itself, or equivalently, as an $A$-module (since ideals of $A/\mathfrak{m}^n$ are the same as $A$-submodules). Consider the chain $A/\mathfrak{m}^n\supseteq \mathfrak{m}/\mathfrak{m}^n\supseteq\cdots\supseteq\mathfrak{m}^{n-1}/\mathfrak{m}^n\supseteq 0$ of $A$-submodules of $A/\mathfrak{m}^n$. As $A$ is Noetherian, these are all finitely generated $A$-modules. In particular, the quotients by consecutive terms are finitely generated $A$-modules killed by $\mathfrak{m}$, i.e., finite-dimensional $A/\mathfrak{m}$-modules. A vector space over a field is of finite length over that field if and only if it is of finite dimension, and the $A$-length of an $A$-moduled killed by $\mathfrak{m}$ is the same as the $A/\mathfrak{m}$-dimension. Thus the quotients in this filtration are all finite length over $A$. By induction, using that length is additive in short exact sequences, $A/\mathfrak{m}^n$ itself is of finite length over $A$.

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Suggestion/hint: Hopkins-Levitzki

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