Reduced Noetherian implies zero divisor in minimal prime

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

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

Breakdown without both assumptions

 * Noetherian not implies zero divisor in minimal prime
 * Reduced not implies zero divisor in minimal prime

Converse
The converse is true for any Noetherian ring.

Applications

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

Facts used

 * 1) uses::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)
 * 2) uses::Equivalence of definitions of nilradical: In a commutative unital ring, the nilradical, defined as the set of nilpotent elements, also equals the intersection of all prime ideals.

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 in some minimal prime.

Proof: There exists $$0 \ne b \in A$$ such that $$ab = 0$$.


 * 1) By fact (1), every prime ideal contains a minimal prime ideal. Hence, the intersection of all prime ideals equals the intersection of all minimal prime ideals. By fact (2), the intersection of all prime ideals is the set of nilpotent elements. Finally, since the ring is reduced, it has no nonzero nilpotents, so the intersection of all primes is the zero ideal. Thus, the intersection of all minimal primes is the zero ideal.
 * 2) Clearly, we have that $$ab$$ is in every minimal prime. For every minimal prime, either $$a$$ is in the minimal prime or $$b$$ is in the 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.
 * 3) Combining steps (1) and (2), we obtain that $$b = 0$$.