Why is it Artinian?

Question

The following is a part of the section entitled Samuel functions in the book Commutative Ring Theory by Hideyuki Matsumura:

Let $A$ be a Noetherian semilocal ring, and $\mathfrak{m}$ the Jacobson radical of $A$. If $I$ is an ideal of $A$ such that for some $\nu >0$, we have $\mathfrak{m}^\nu \subset I \subset \mathfrak{m}$, then the ring $A/I$ is Artinian.

I can't understand why $A/I$ is Artinian. If $A$ is a local ring, which is the case in Introduction to Commutative Algebra by Atiyah and MacDonald, $\mathfrak{m}$ becomes the maximal ideal, and so $I$ becomes $\mathfrak{m}$-primary ideal. In this case I can understand why it is Artinian, since the dimension of $A/I$ is $0$. But, when $A$ is just a semilocal ring, ...

Help me!

Answer 1

Let $p$ be a prime ideal containing $I$. Since $m^{\nu}\subseteq p$ it follows that $m\subseteq p$. But $m=\cap_{i=1}^n m_i\supseteq \prod_{i=1}^n m_i$, so there is an $i$ such that $p\supseteq m_i$, hence equality. This shows that all primes in $A/I$ are maximal, so $\dim A/I=0$. (In fact, we have proved that $m_i/I$ are the only prime ideals of $A/I$.)

Answer 2

If you are familiar with the Hopkins Levitzki theorem, then an alternative way is to note that $R/I$ is also semilocal and Noetherian, and has nilpotent radical $m/I$, and so by H-L it is also Artinian.

It's a wonderful theorem about the equivalence of the ascending and descending chain conditions for modules over semiprimary rings.

(user26857 elementary answer is to be preferred if you don't know H-L or are otherwise unwilling to use it.)