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Let $A$ be noetherian local with maximal ideal $\mathfrak{m}$ and let $\mathfrak{q}$ be an $\mathfrak{m}$-primary ideal. Then why is $A/\mathfrak{q}$ an artin ring?

$A/\mathfrak{q}$ is noetherian so all we need to do is prove the image of $\mathfrak{m}$ in $A/\mathfrak{q}$ is the only prime ideal (prime and maximal ideals are the same in Artin rings) so that krull dimension of $A/\mathfrak{q}$ is 0. But how do I show this?

This claim is used in the proof of 11.4 of atiyah and macdonald. But I can't see why it is true!

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  • $\begingroup$ Remind us what $\mathfrak m-$primary means? $\endgroup$ Sep 27, 2021 at 17:48
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    $\begingroup$ @ThomasAndrews A primary ideal whose radical is $\mathfrak{m}$. $\endgroup$
    – Mark
    Sep 27, 2021 at 17:48
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    $\begingroup$ This means that there cannot be a prime ideal $\mathfrak p$ of $A$ such that $\mathfrak q\subseteq \mathfrak p\subsetneq \mathfrak m.$ $\endgroup$ Sep 27, 2021 at 17:58

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In a Noetherian ring, every ideal contains some power of its radical. Thus there is some $n$ such that $\mathfrak{m}^n\subseteq \mathfrak{q}$. It follows that $(\mathfrak{m}/\mathfrak{q})^n$ is the zero ideal in $A/\mathfrak{q}$. Now, if in a ring $R$ there are some (finitely many) maximal ideals such that their product is $\{0\}$ then $R$ is Noetherian if and only if $R$ is Artin. This is corollary $6.11$ in Atiyah-Macdonald. Since $A/\mathfrak{q}$ is obviously Noetherian, it follows that it is an Artin ring.

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Assume there is a prime ideal $\mathfrak p$ of $A$ such that $\mathfrak q\subseteq\mathfrak p\subsetneq \mathfrak m.$

Since $\mathfrak q$ is $\mathfrak m-$primary, for all $a\in\mathfrak m,$ $a^k\in \mathfrak q$ for some $k.$ Thus $a^k\in \mathfrak p.$ Since $\mathfrak p$ is prime, this means $a\in\mathfrak p.$

Thus $\mathfrak m\subseteq \mathfrak p.$ Contraditiction.

So there is no prime ideal other than $\mathfrak m/\mathfrak q$ in $A/\mathfrak q.$


This shows you only need that the radical of $\mathfrak q$ is $\mathfrak m,$ not that $\mathfrak q$ is primary.

But in a local ring, the radical condition implies the primary part, because, if $ab\in\mathfrak q$ and $b\not\in\mathfrak m,$ then $b$ is a unit in $A$ and thus $a\in \mathfrak q.$

So either $a\in\mathfrak q$ or $b^k\in \mathfrak q$ for some $k.$ Thus, by definition, $\mathfrak q$ is primary.


However, this theorem has a generalization. In any commutative ring, $A,$ if $\mathfrak m$ is a maximal ideal and $\mathfrak q\subseteq \mathfrak m$ is an ideal whose radical is $\mathfrak m,$ then $A/\mathfrak q$ has only one prime ideal, $\mathfrak m/\mathfrak q.$

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  • $\begingroup$ Hello! I would like to ask if in the second part of your answer (the one that starts with ""his shows you only need...") you also require the ring to be Noetherain, or if you are just assuming that it is local. $\endgroup$
    – kubo
    Jan 25 at 12:00
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A different approach, for future references.

It is based on Corollary $6.11$ from Atiyah-MacDonald, which states the following:

Let $A$ be a ring in which the zero ideal is a product of finitely many (not necessarily distinct) maximal ideals. Then $A$ is Noetherian iff $A$ is Artinian.

We know that $A$ is Noetherian local with maximal ideal $\mathfrak{m}$, thus $A/\mathfrak{q}$ is also Noetherian local with maximal ideal $\overline{\mathfrak{m}}$ (check this post to see why $A/\mathfrak{q}$ is also local). Corollary $7.16$ from the same book states that $\mathfrak{q}$ is $\mathfrak{m}$-primary iff $\mathfrak{m}^n\subseteq \mathfrak{q}\subseteq\mathfrak{m}$ for some $n>0$. In particular, this means that $\overline{\mathfrak{m}}^n=(0)$. Thus, by Corollary $6.11$ we have that $A/\mathfrak{q}$ is Aritinian.

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