# Maximal ideals of commutative Artinian rings

I would like some help on an exercise I thought I had done correctly at first glance, but obviously have doubts about. The question is;

Let $R$ be a commutative Artinian ring. Then R has finitely many maximal ideals.

The exercise has two previous subsections, showing that $Ann_R(R/m)=m$ for maximal ideals $m$, and $R/m_1 \cong R/m_2$ iff $m_1 = m_2$.

Also, a hint for the question says to consider the semi-simple case, and then generalise.

My proof so far;

Letting $J = J(R)$ be the Jacobson radical, because of commutativity it is the intersection of all maximal two sided ideals. I know that $R/J$ is semi-simple as $R$ is artinian and $R/J$ has Jacobson radical $0$. So $R/J$ has finitely many ideals, it has finitely many maximal ideals and is isomorphic to a direct sum of quotients of maximal ideals in $R/J$. I also know these quotients to be simple $R$-modules . Then, the quotient map $R/J\to R/m$ for maximal ideal $m$ is well defined and by some simple arguments, we have $R/m$ is module isomorphic to a quotient in the decomposition of $R/J$. By the exercise above, we then have only a finite number of choices of $m$.

But I feel I am brushing over the decomposition part of the proof. I do not believe it is as complicated as what I have written, but I do not see what I am missing.

• I believe I am wanted to use the correspondence of simple $R$-modules and simple $R/J$-modules, and apply the fact that all simple $R/J$-modules are isomorphic to one in $R/J$'s decomposition. Then use the exercise. – Rhys Evans Apr 6 '15 at 9:53

If you already know that $R/J$ has only finite many maximal ideals, then the next step is to recognize those correspond exactly to the maximal ideals of $R$.
Do you see why? The maximal ideals of $R/I$ are the maximal ideals of $R$ containing $I$. But in the case of $J$...?