# 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 :) Commented Dec 12, 2014 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. Commented Dec 16, 2014 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 Commented Dec 16, 2014 at 14:31

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.