# A direct proof that minimal primes are associated

It is a well-known theorem that, for a Noetherian ring $A$, the minimal primes of $A$ are among the associated primes of $A$; i.e., for every minimal prime $\mathfrak{p}$ of $A$, there is an element $f \in A$ such that $Ann(f) = \mathfrak{p}$. Morally, one might think of $f$ as a "characteristic function" of the irreducible component Spec $A/\mathfrak{p}$ of Spec $A$.

The usual proof of this fact is to look at the set of associated primes of $A$ (or more generally of an $A$-module), together with one or more other sets of primes of $A$, and then go on proving various things that finally conclude with "these collections of primes have the same minimal elements." I find this proof not particularly satisfactory. What I would like is a proof that starts off "Let $\mathfrak{p}$ be a minimal prime of $A$" and then proceeds to "construct" an element $f \in A$ such that $Ann(f) = \mathfrak{p}$. Unfortunately, I have not had much success with proceeding along these lines. Is there anyone who knows (or is able to invent) such a proof?

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If you are willing to assume that $0$ has a primary decomposition, then take one which is minimal say $0 = \bigcap_{i = 1}^n Q_i$ with radicals $P_i = \sqrt{Q_i}$. Now the minimal primes are the $P_i$.

Let me show that $P = P_1$ is associated. Let $R = \bigcap_{i = 2}^n Q_i$, $Q = Q_1$. So $0 = Q \cap R$, and

$$R = \frac{R}{Q \cap R} \cong \frac{R + Q}{Q} \subset \frac{A}{Q}.$$

So $\mathop{Ass}(R) \subset \mathop{Ass}(A/Q) = \{ P \}$, where the last equality uses the assumption that $A$ is noetherian. Again, since $A$ is Noetherian, $\mathop{Ass}(R) \neq \emptyset$, so $P \in \mathop{Ass}(R) \subset \mathop{Ass}(A)$.

Until now, the proof is non-constructive. Now, what it the element $f$ you are looking for? It all lies in the claim that since $A$ is Noetherian, $\mathop{Ass}(R) \neq \emptyset$.

To show this, consider the set of ideals $X = \{ \mathop{Ann}(f) \mid f \in R \}$. Since $A$ is Noetherian, $X$ has a maximal element, and it is not difficult to show that such maximal element has to be prime. Indeed let $I \in X$ be maximal, say $I = \mathop{Ann}(f)$, and let $ab \in I$. Then $abf = 0$, and either $bf = 0$, in which case $b \in I$, or $a \in \mathop{Ann}(bf) \supset I$. By maximality, $\mathop{Ann}(bf) = I$, so $a \in I$.

In conclusion your desired element $f$ is defined by the condition that $\mathop{Ann}(f)$ is maximal in $X$. I leave to you the choice whether this is or not more constructive than the proofs you have seen. In any case the very Noetherian condition guarantees a non-constructive existence of a maximal ideal, so it is entirely possible thatone cannot get anything more explicit than this.

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I appreciate the effort you've put into this answer, but I'm still hoping for something a bit more explicit. – Charles Staats Jul 6 '10 at 17:57
Don't worry, not a big effort :-) But as I said, the fact that this is true only for Noetherian rings makes me think that the non-constructivity is inherent. – Andrea Ferretti Jul 6 '10 at 18:52
Maybe the proof above would look more explicit if you substitute the ascending chain condition for the maximal ideal condition. So you take $f_1 \in R$; if $\mathop{Ann}(f_1)$ is not maximal you take $f_2 \in R$ such that $\mathop{Ann}(f_2) \supset \mathop{Ann}(f_1)$ and so on, until you have to stop by the ACC. The last $f_n \in R$ that you take is your desired element. – Andrea Ferretti Jul 6 '10 at 18:53
The proof that "every module over a Noetherian ring has an associated prime" is fine (although you need to restrict $f$ to be nonzero). It's all the stuff leading up to that where I would like to see a more explicit discussion of $f$. (Part of the issue is that the proof of primary decompositions is typically wrapped up in the same "non-explicit" reasoning I discuss in the question.) – Charles Staats Jul 6 '10 at 19:15
But you only need existence, which is easy and is as explicit (or not explicit) as the other step. If 0 is irreducible, you are done. Otherwise write $0 = I \cap J$. Now if both $I$ and $J$ are irreducible you stop, otherwise write $I = K \cap L$, and so on. This process will stop in a Noetherian ring, and when it stops, you have a primary decomposition on $0$. You can make it minimal by removing some components. – Andrea Ferretti Jul 6 '10 at 19:31