Can we prove, without axiom of choice, that the set of all zero divisors (including $0$) of a commutative ring with unity contains a prime ideal?

Let $R$ be a commutative ring with unity , I know that assuming axiom of choice , if $A$ is the set of all zero divisors (including $0$ ) then it is a union of prime ideals so it contains a prime ideal . Can we prove , without choice , that the set of all zero divisors(including $0$ ) of a commutative ring with unity contains a prime ideal ? Is it equivalent with axiom of choice ?

• You should at least assume that the ring has prime ideals, or any axiom that shows existence of prime ideals, c.f. this, e.g.. – Ben May 14 '15 at 16:19
• @Ben : But does the fact that the set of all zero divisors is a union of prime ideals follows from cohen's result from multiplicative sets ? – user228168 May 14 '15 at 16:29
• Yes, it does. The subset $S\subset R$ consisting of all the non-zero divisors is closed under multiplication and contains one. Since any zero divisor $r\in R$ generates an ideal $Rr$ disjoint from $S$, Cohens theorem gives a prime ideal $\mathfrak{p}\subset R-S$ containing $r$. Thus, the set of all zero-divisors is the union of all these primes. – Ben May 14 '15 at 17:06
• @saundev you are disregarding the use of cohen's lemma since it uses the axiom of choice, right? Actually, it may be the case that cohens lemma is equivalent to something weaker.... – rschwieb May 15 '15 at 2:32

The comment of rschwieb made me think it's time put everything together.

The statement

A) The set of zero divisors of a commutative ring is the union of prime ideals.

requires the existence of prime ideals in any commutative ring. According to this MO answer, prime existence is equivalent to the boolean prime ideal theorem (BPI for short). Since there are models of ZF where there are rings without prime ideals, whilst BPI is weaker than the axiom of choice, the best we can do is prove A in ZF+BPI. The same MO answer also gives a reference according to which in ZF+BPI also the following holds:

B) Let $A$ be any commutative ring, $S\subset A$ a multiplicative closed subset (containing $1$), and $\mathfrak{a}\subset A$ an ideal disjoint from $S$. Then $\mathfrak{a}$ is contained in a prime ideal $\mathfrak{p}\subset A$ which is disjoint from $S$ too.

This, however, is enough to conclude A, as shown in the comments. Let me recall the argument shortly, for sake of completeness.

The set of non(!)-zero divisors $S\subset A$ contains $1$ and is multiplicative closed. Furthermore, any zero divisor $a\in A$ generates an ideal $Aa\subset A$ disjoint from $S$. By B, there is a prime ideal $\mathfrak{p}$ disjoint from $S$ and containing $Aa$. Therefore, the union of all prime ideals $\mathfrak{p}\subset A$ disjoint from $S$ is all of $A-S$, the set of zero divisors in $A$, as desired.