Reduced Noetherian implies zero divisor in minimal prime

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

Statement

Suppose $A$ is a reduced Noetherian ring: in other words, it has no nilpotents, and every ideal is finitely generated. Then, any zero divisor in $A$ must be inside some minimal prime ideal of $A$.

(For non-Noetherianness, we can only say that every zero divisor must be inside some prime ideal, since minimal primes are not guaranteed to exist).

Applications

• Reduced Noetherian local one-dimensional implies Cohen-Macaulay
• Reduced Noetherian one-dimensional implies Cohen-Macaulay

Facts used

• Noetherian ring has finitely many minimal primes and every prime contains a minimal prime (and thus, in particular, the nilradical is an intersection of those finitely many minimal primes)

Converse

The converse is true for any Noetherian ring. Further information: Noetherian implies every element in minimal prime is zero divisor

Generalizations

The result is true if we replace the assumption of Noetherianness by the assumption that every prime contains a minimal prime.

Proof

Given: Suppose $a\in A$ is a zero divisor

To prove: $a$ is not in any minimal prime

Proof: There exists $0\neq b\in A$ such that $ab=0$. Clearly, we have that $ab$ is in every minimal prime. If $a$ is not in any minimal prime, then $b$ must be in every minimal prime, hence $b$ must be in the intersection of all minimal primes. But the intersection of all minimal primes is the intersection of all primes, which is the nilradical. Since we're in a reduced ring, this is zero, so $b=0$, a contradiction.