let be the ring and the minimal prime. so is exactly the set of units of hence any proper ideal of is contained in since is a minimal prime, it cannot contain any prime ideal

properly. thus is the unique prime ideal of now suppose is a non-unit. so suppose is not nilpotent. let then because we assumed that is

not nilpotent. let which is a non-empty set because apply Zorn's lemma to find a maximal element of we claim that is prime: suppose but

then therefore by maximality of so there exists such that but then:

because thus which is contradiction! so is prime and hence thus which is false because

Remark 1: in general, in a (not even necessarily commutative) ring given a multiplicatively closed set with we can always find a prime ideal which is contained in

Remark 2: if you already know about the nilradical of a commutative ring, then your problem is trivial: since is the only prime ideal of the nilradical of is but we know that the nilradical

is exactly the set of nilpotent elements of

Remark 3: the information given in the problem about zero-divisors is only used to conclude that is exactly the set of units of