Quotient of Noetherian local ring with $\mathfrak{m}$-primary ideal

Question

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!

Answer 1

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.

Answer 2

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 $A/\mathfrak q.$

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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.

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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, $\mathfrak m/\mathfrak q.$