Lemma 10.28.12. Let $R$ be a ring.

1. An ideal maximal among the ideals which do not contain a nonzerodivisor is prime.

2. If $R$ is nonzero and every nonzero prime ideal in $R$ contains a nonzerodivisor, then $R$ is a domain.

Proof. Consider the set $S$ of nonzerodivisors. It is a multiplicative subset of $R$. Hence any ideal maximal with respect to not intersecting $S$ is prime, see Lemma 10.28.9. Thus, if every nonzero prime ideal contains a nonzerodivisor, then $(0)$ is prime, i.e., $R$ is a domain. $\square$

## Comments (4)

Comment #67 by Rankeya on

I think the first item in the formulation of the lemma is incorrect. The term zerodivisor should be replaced by nonzerodivisor.

Comment #68 by on

Haha, yes! Of course, since every ideal contains 0 the statement of part (1) wasn't wrong, it was just completely useless. Thanks!

Comment #3541 by Laurent Moret-Bailly on

In (2), $R$ should be assumed nonzero.

There are also:

• 6 comment(s) on Section 10.28: A meta-observation about prime ideals

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