# Looking for example of a commutative non-unital ring in which every maximal ideal is a prime ideal

Give example of a commutative non-unital ring in which every maximal ideal is a prime ideal.

The motivation for this question is : It is known that if $R$ is a commutative ring with identity $1 \ne 0$ and $M$ is a maximal ideal then $R/M$ is a field ; I have seen that if $R$ is a ring and $M$ is a maximal and prime ideal then also $R/M$ becomes a field ; also it is known that if $R$ is comm, ring with a multiplicative identity then every maximal ideal is prime ; hence arises the question of existence of a commutative ring without identity in which every maximal ideal is prime ...

• @Souvik Thanks for adding more context. That's enough (for me anyhow) to take the question seriously. I hope you'll apply similar effort to future questions :) – rschwieb Dec 12 '14 at 14:43
• Souvik Dey: I see that you create the tags (maximal-ideals) and (prime-ideals). I thought that it would be polite to let you know that I have just made a post on meta asking whether these two tags are actually needed. – Martin Sleziak Dec 16 '14 at 8:24
• @MartinSleziak: Ok , if you guys think that it is not important enough you might remove it and keep using "ideals" tag , I have no issue with that – Souvik Dey Dec 16 '14 at 14:31

## 1 Answer

In a ring without maximal ideals (which necessarily has no identity) the condition is vacuously satisfied. This page by Patrick Morandi is the most efficient way to get examples of these to you.

Here's another non-vacuous example. Let $F_2$ be the field of two elements, and consider the ideal $\oplus_{i=1}^\infty F_2\subseteq \prod_{i=1}^\infty F_2$. The right hand side is a ring with identity, of course, but the left hand side is a ring without identity. It's easy to see both rings are von Neumann regular rings. Let's denote the left hand ring by $R$.

If $M$ is a maximal ideal of $R$, then $M+(x)=R$ for any $x\notin M$. Since $R$ is von Neumann regular, $(x)=(e)$ for an idempotent element $e$. It easily follows that $R/M$ is a field with identity $e+M$, so $M$ is prime.