Noetherian implies every element in minimal prime is zero divisor
This article defines a result where the base ring (or one or more of the rings involved) is Noetherian
View more results involving Noetherianness or Read a survey article on applying Noetherianness
The converse statement is true when the ring is reduced. Further information: reduced Noetherian implies zero divisor in minimal prime
The assumption that the ring be Noetherian can be weakened to the assumption that every prime contains a minimal prime, and such that there are only finitely many minimal primes.
Given: A Noetherian ring
To prove: If lies in a minimal prime ideal of , then is a zero divisor.
Proof: Since is a Noetherian ring, it has only finitely many minimal primes, say (and without loss of generality, . Moreover, any prime ideal contains a minimal prime ideal, so the nilradical of equals the intersection of the s. In particular, the product of the s lies inside the nilradical of .
Let us next consider the product . If this product is contained in , then, one of the s is contained in , contradicting the assumption of minimality. Thus, is not contained in , so we can find elements such that .
Now consider the product . This product lies in the nilradical, hence some power of it is zero. Thus, there exists a such that:
However, the expression cannot be zero because , and therefore, is not nilpotent.
Thus, is a zero divisor, and hence, is a zero divisor (that's because the set of nonzerodivisors is a multiplicatively closed subset, so if were not a zero divisor, would also not be a zero divisor).