Let $K$ be a field and consider $K[[x]]$ as a $K[x]$-module. Determine if it is Artinian/Noetherian.

I used the following propositions:

If $M$ is an $R$-module and $N\subseteq M$ a submodule, then $M$ is Artinian/Noetherian if and only if both $N$ and $M/N$ are Artinian/Noetherian.

A ring $R$ is Artinian if and only if $R$ is Noetherian and every prime ideal of $R$ is maximal.

First the ring $K[x]$. All its ideals are finitely generated, so it's Noetherian. But the ideal $(0)$ is prime, but not maximal. So $K[x]$ is not Artinian.

I had then planned to use that $K[x]/(x^n)\cong K[[x]]/(x^n)$. Using this relation we can see that either $K[[x]]/(x^n)$ or $(x^n)$ isn't Artinian, since $K[x]$ is not either. This means $K[[x]]$ is not Artinian.
Using the same relation we can see that both $K[[x]]/(x^n)$ and $(x^n)$ are Noetherian since $K[x]$ is. This would mean $K[[x]]$ is Noetherian too.

Could anyone verify this proof.

  • 1
    $\begingroup$ In $K[x]/(x^n)$, $(x^n)$ means $x^nK[x]$, but in $K[[x]]/(x^n)$ it means $x^nK[[x]]$, and these are different (and non-isomorphic as $K[x]$-modules). But I think you're assuming they're the same. $\endgroup$ – Jeremy Rickard Jun 9 '15 at 12:03
  • $\begingroup$ There is a problem: you are denoting with the same symbol $(x^n)$ two distinct objects. Note that $(x^n) \subset K[x] \subset K[[x]]$ is NOT na ideal of $K[[x]]$ since for instance $\sum_{k \ge n}x^k$ is not in $K[x]$. $\endgroup$ – Crostul Jun 9 '15 at 12:03
  • $\begingroup$ @JeremyRickard That is indeed where I screwed up. $\endgroup$ – gebruiker Jun 9 '15 at 12:10
  1. $K[[X]]$ is not a noetherian $K[X]$-module since it's not finitely generated.

  2. $K[[X]]$ is not an artinian $K[X]$-module: $XK[X]\supset X^2k[X]\supset\cdots$ is a strictly descending chain of $K[X]$-submodules. (You can also argue as follows: every submodule of an artinian module is artinian, so if $K[[X]]$ is an artinian $K[X]$-module, then $K[X]$ is an artinian $K[X]$-module, that is, $K[X]$ is an artinian ring, a contradiction.)


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