I want to prove the following:

Let $(A, \mathfrak{m})$ be an Artinian local ring. Prove that the maximal ideal $\mathfrak{m}$ is nilpotent.

I think I managed to prove it in the following way:

Consider the descending chain of ideals $$\mathfrak{m} \supset \mathfrak{m}^2 \supset \dots \supset \mathfrak{m^n} \supset \dots$$ Since $(A, \mathfrak{m})$ is Artinian, the descending chain stabilizes, so $\mathfrak{m}^k = \mathfrak{m}^{k+1} = \dots$ for som $k$. Now I want to show that $\mathfrak{m}^k = 0$ for which I use Nakayama's Lemma. Since $\mathfrak{m}^k = \mathfrak{m}^{k+1} = \mathfrak{m}\mathfrak{m}^{k}$, it follows that $\mathfrak{m}^k = 0$.

If this is fine, I'm not completely satisfied with this proof, since I think I implicitly used the fact that an Artinian ring is Noetherian in order to use Nakayama's Lemma (I need my $A$-module to be finitely generated, don't I?). So I thought I'd use the hint provided by Reid (Exercise 3.9). It says that I should suppose that $\mathfrak{m}^k \neq 0$ and consider the ideal $I$ where $I$ is the minimal ideal such that $I\mathfrak{m}^k \neq 0$. Then, show that $I = (x)$ is principal and apply Nakayama's Lemma to it.

I haven't showed yet that $I$ is principal, but I can't see why this gets us in the finitely generated $A$-module hypothesis to apply Nakayama's Lemma, so I'd like some help about that, because it seems I'm missing something but I don't know what. Thank you in advance for your help!


You are right. One cannot use Nakayama's lemma until you know that $A$ is Noetherian, since otherwise you don't know that $\mathfrak m^k$ is finitely generated. Of course this result is a step in the proof that Artinian rings are Noetherian.

The usual dodge is to consider a minimal ideal with $I\mathfrak m^k\ne0$. Obviously then there is some $x\in I$ with $x\mathfrak m^k\ne0$. So $(x)$ is an ideal and $(x)\mathfrak m^k\ne0$ and $(x)\subseteq I$. Therefore, by minimality $I=(x)$. Now also $(x)\mathfrak m^k$ is an ideal, $((x)\mathfrak m^k)\mathfrak m^k= (x)\mathfrak m^{2k}=(x)\mathfrak m^k\ne0$. Again by minimality $I=(x)\mathfrak m^k$. Now $(x)=(x)\mathfrak m^k$. Therefore $(x)=(x)\mathfrak m$ and we can apply Nakayama, as $(x)$ is finitely generated, so see that $(x)$ is killed by an element outside $\mathfrak m$ and so a unit.


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