It is straightforward to show, for example, that the set of zero divisors of a (commutative unital) reduced Noetherian ring is precisely the union of its minimal primes. When else can we say that the set of zero divisors is equal to the union of the minimal primes? Are there other useful cases where this is true? Is there a structure theory for such rings? I'm primarily looking for conditions that do not assume that the ring is Noetherian.

Edit: I messed up this question and received a correct answer for the wrong assumptions, so I have accepted the answer. The reducedness assumption was for the example, not for the general question.

The correct question is in there, but I accidentally changed the title, so I apologize for the confusion.

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    $\begingroup$ I believe you mean that the set of zero-divisors is equal to the union of the associated primes. Example: $R=k[x,y]/(x^2,xy)$; $y$ is a zerodivisor, though the unique minimal prime is $(x)$. If your question is "When is the set of associated primes equal to the set of minimal primes?" then the answer is "When the ring is unmixed." $\endgroup$ – Graham Leuschke Apr 9 '10 at 12:20
  • $\begingroup$ Yes, I forgot the condition "reduced". $\endgroup$ – Harry Gindi Apr 9 '10 at 12:39

The answer is: always, and argument is pretty simple:

Let $R$ be a reduced commutative unital ring. If $a\in R$ is a zero divisor, then $ab=0,$ for some $b\neq 0.$ Hence $b\not \in 0 = \text{nil}(R)= \bigcap \text{Spec}(R) =\bigcap \text{Specmin}(R),$ where $\text{Specmin}(R)$ states for family of all minimal prime ideals of $R.$ So there exists a minimal prime ideal $\mathfrak{p}\triangleleft R$ s.t. $b\not \in \mathfrak{p}.$ But $\mathfrak{p}$ is prime, and $ab\in \mathfrak{p},$ thus $a\in \mathfrak{p}.$ Hence the set of zero divisors is contained in union of minimal prime ideals.

On the other hand it is well-known fact that minimal prime ideals in commutative unital rings consist of zero divisors, so the set of zero divisors in reduced commutative unital ring is exactly the union of minimal prime ideals.

  • $\begingroup$ Why does belonging to every minimal prime imply belonging to every prime? Don't you need noetherian induction for that? $\endgroup$ – Owen Biesel Apr 9 '10 at 14:46
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    $\begingroup$ Lemma: Every commutative ring has a minimal prime ideal. Proof: Zorn's lemma. Lemma: Every prime ideal contains a minimal prime. Proof: Apply the last lemma to the localization at p. The result follows by inspection. $\endgroup$ – Harry Gindi Apr 9 '10 at 15:00

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