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It is well known that in commutative rings, maximal ideals are prime.

Can we give an example of a ring $R$ such that every prime ideal of $R$ is maximal with $|\operatorname{Max}(R)|=\infty?$

user26857
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Angel
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2 Answers2

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This condition is equivalent to having Krull dimension zero. Every Boolean ring has this property (quotients of Boolean rings are Boolean, and the only Boolean ring which is also an integral domain is $\mathbb{F}_2$, which is a field), and a Boolean ring has finitely many maximal ideals iff it's finite. So an example is given by any infinite Boolean ring such as $\prod_I \mathbb{F}_2$ for any infinite set $I$.

The spectrum of this ring can be identified with the space $\beta I$ of ultrafilters on $I$. For more details see, for example, this blog post.

Qiaochu Yuan
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If $R=\mathbb F_2^{\mathbb N}$ then $\dim R=0$ (see here) and since it is not noetherian must have infinitely many maximal ideals (why?).

Remark.

Let $R$ be a commutative ring with $\dim R=0$. Then $R$ is semilocal iff $R/N(R)$ is noetherian.

If $R$ is semilocal, then $N(R)$ is a finite intersection of maximal ideals, so $R/N(R)$ is isomorphic to a finite direct product of fields.
Conversely, if $R/N(R)$ is noetherian, then $R/N(R)$ is artinian, so it is semilocal. But $R$ and $R/N(R)$ have the same prime ideals, so $R$ is semilocal.

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user26857
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    In fact, a commutative ring $R$ is zero-dimensional iff $R/N(R)$ is von Neumann regular. If you suppose $R$ reduced, then this is equivalent to $R$ von Neumann regular. In particular, every boolean ring is such an example, and this explains why I've chosen that example. – user26857 Dec 06 '15 at 18:57