Let $A$ be a local noetherian ring and $\mathcal{M}$ is maximal ideal of $A$. Then $A/\mathcal{M}^n$ is artinian.
How can I prove this?
Let $A$ be a local noetherian ring and $\mathcal{M}$ is the maximal ideal of $A$. Then $A/\mathcal{M}^n$ is artinian.
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commutative-algebra
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1Hint: For noetherian rings, artinian is the same as zero-dimensional. In particular for local noetherian rings, artinian is the same as saying that the maximal ideal is the only prime ideal. You should show the latter. – 2017-01-30
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0$A/M^n$ is Noetherian and has dimension 0 (read the accepted answer in the linked question), so it is Artinian. – 2017-01-30
1 Answers
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If you want to prove it by hand, consider the short exact sequence: $$0\rightarrow\mathfrak m^{n-1}/\mathfrak m^n\rightarrow A/\mathfrak m^n\rightarrow A/\mathfrak m^{n-1}\rightarrow 0,$$ and use induction on $n$. Note the left-hand term is an $A/\mathfrak m$ vector space, and remember that for a vector space, artinian is the same as noetherian is the same as finite-dimensional.
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1To really make this precise, one should note the following: If $R$ is a ring, $I$ an ideal and $M$ a $R/I$-module, then we can also view $M$ as an $R$-module and we have that $R$-submodules of $M$ are the same as $R/I$-submodules of $M$. In particular the artinian property does not depend on the choice of the two module structures. One has to use this fact multiple times when going through your suggested proof, as one switches between $A/\mathfrak m$-, $A/\mathfrak m^{n-1}$- and $A/\mathfrak m^n$-module structures during the induction step. – 2017-01-30
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1@Moos: That's right, but I didn't try to solve the question – only to give a sketch. You also might have mentioned the 3rd isomorphism theorem. – 2017-01-30