# Why is it Artinian?

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!

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 $A/I$.)
• Thank you for your help. I understand that for each product $x_1 x_2 \cdots x_n$ where $x_i \in m_i$ there exists $i$ such that $x_i \in p$. But how do we know that $p \supset m_i$ for some $i$. – Aki Jun 5 '14 at 12:53
• @Aki If $p\nsupseteq m_i$ for all $i$ there is $x_i\in m_i-p$ for all $i$. Then what about the product $x_1\cdots x_n$? (It seems you are not aware that an ideal $P$ is prime iff $P\ne A$ and if $P\supseteq IJ$ then $P\supseteq I$ or $P\supseteq J$, where $I,J$ are ideals of $A$, which is an equivalent definition for a prime ideal.) – user26857 Jun 5 '14 at 13:04
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.