6
$\begingroup$

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!

$\endgroup$

2 Answers 2

5
$\begingroup$

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 $A/I$.)

$\endgroup$
3
  • $\begingroup$ 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 $i$. $\endgroup$
    – Aki
    Jun 5, 2014 at 12:53
  • 1
    $\begingroup$ @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.) $\endgroup$
    – user26857
    Jun 5, 2014 at 13:04
  • $\begingroup$ I didn't know the alternative definition. Thank you for that. $\endgroup$
    – Aki
    Jun 5, 2014 at 14:02
1
$\begingroup$

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.)

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .