# An example of a commutative ring with infinitely many maximal ideals

I am looking for a commutative ring $R$ with identity that has the following properties:

1) $\mathrm{Max}(R)$, the set of all maximal ideals of $R$, is infinite;

2) whenever $\{I_\alpha\}_{\alpha\in\Gamma}$ is an arbitrary family of finitely generated ideals of $R$ such that for each maximal ideal $\mathfrak m$  there exists $\beta\in\Gamma$ with $I_\beta\subseteq\mathfrak m$, then there exists a finite subfamily $\{I_1, I_2,..., I_n\}$ of $\{I_\alpha\}_{\alpha\in\Gamma}$ with $I_1 I_2\cdots I_n\subseteq J(R)$.

Here $J(R)$ is the Jacobson radical, the intersection of all maximal ideals of $R$.

• You can quotient by $J(R)$ and reduce to the case that $R$ is semiprimitive. – Kevin Casto Oct 18 '16 at 20:56

Let $R$ be the subring of $\prod_{i=1}^{\infty}\mathbb{Q}$ of sequences which are eventually constant. This ring has the "obvious" maximal ideals $M_i$ of sequences which are zero in the $i$th coordinate, along with the less obvious maximal ideal $M_{0}$ consisting of sequences which are eventually zero.
After reindexing, suppose we have a family of finitely generated ideals $\{I_{i}\}_{i=0}^{\infty}$ with $I_{i}\subseteq M_{i}$. The generators from $I_0$ have only finitely many non-zero entries, say up to position $n$. Then $I_0 I_1 \cdots I_n=0$.