A direct proof that minimal primes are associated

Question

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 $\operatorname{Ann}(f) = \mathfrak{p}$. 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 $\operatorname{Ann}(f) = \mathfrak{p}$. 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?

Answer 1

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.

Answer 2

Let $p\subset A$ be a minimal prime. Consider the localisation map $\phi:A\to A_p$. The kernel $I=\mathop{Ker} \phi$ is the set of all $a\in A$ such that $af=0$ for some $f\in S=A\setminus p$. That is, $I=\cup_{f\in S}Ann(f)$. Since $A$ is Noetherian and $Ann(f)+Ann(g)\subset Ann(fg)$, we conclude that $I=Ann(s)$ for some $s\in S$. We also know that $I\subset p$ since $p$ is prime. If $I=p$, then we are done. If $I\subsetneq p$, then $pA_p$ is nilpotent since its the maximal ideal of the Artinial local ring $pA_p$ (here we use that $p$ is minimal). Assume that $(pA_p)^n=0$ and $(pA_p)^{n-1}\neq 0$. Pick a nonzero element $a/u$, $a\in p$, $u\in S$ in $(pA_p)^{n-1}$. Then $Ann(a/1) = pA_p$. I claim that $Ann(as)=p$. Indeed, if $b\in p$, then $\phi(ba)=0$, so $ba\in I=Ann(s)$ and $bas=0$. Thus, $p\subset Ann(as)$. On the other hand, if $t\in S$ is such that $ast=0$, then $a/1=0$ in $A_p$, which is not the case. Thus, $Ann(as)\subset p$.