# Associated Primes I

I’d like to go over associated primes in general rather than just the Noetherian ring form of primary decomposition of ideals. The natural generalization is to modules, since ideals are sub-modules over the ring treated as a module over itself. We’ll need to define a few things first.

Let M be an R-module. Let P be a prime ideal of R. Then we call this an associated prime of M if $P=ann(x)$ for some $x\in M$. The set of associated primes is denoted $Ass_R(M)$ (since R will be understood, we’ll just drop that from here on).

Now suppose $I\subset R$ an ideal. Then the elements of $Ass(A/I)$ are called the “prime divisors” of I.

Now we’ll get some basics out of the way. Suppose that R is Noetherian. First off, $Ass(M)\neq \emptyset$ when $M\neq 0$. We show this by showing that any maximal element of the family $\mathcal{F}=\{ann(x) : 0\neq x\in M\}$ is an associated prime. This is an important fact on its own.

Note that all we really are trying to show is that a maximal element of $\mathcal{F}$ is prime as an ideal, since it will already be of the form $ann(x)$. Let $A\in\mathcal{F}$ be a maximal element. Suppose $A=ann(x)$ and that $ab\in A$ and that $b\notin A$. Then $abx=0$ but $bx\neq 0$. Thus $ann(x)\subset ann(bx)$. But by maximality, $ann(x)\subset ann(bx)$. Thus $ann(x)=ann(bx)$ which means $a\in A$. Thus $A$ is prime and hence an associated prime of M.

The other fact we need is that the set of zero-divisors for M is precisely the set $\displaystyle \bigcup_{P\in Ass(M)}P$.

If $a\in \displaystyle \bigcup_{P\in Ass(M)}P$, then this just says there is some $x\in M$ such that $a\in ann(x)\Rightarrow ax=0$ and hence $a$ is a zero-divisor. The reverse inclusion just uses the previous fact. Let $a$ be such that $ax=0$ for some $x$. So we have $a\in ann(x)$. Then take a maximal element $P\in\mathcal{F}$ containing $ann(x)$. By the last fact $P\in Ass(M)$ and hence $a\in \displaystyle \bigcup_{P\in Ass(M)}P$.

For the theorem of the day, you may need a refresher on the spectrum of a ring, and on localization.

We no longer assume R is Noetherian. Let $S\subset R$ be multiplicative, and $N$ an $R_S$-module. Viewing $Spec(R_S)\subset Spec(R)$, then we have $Ass_R(N)=Ass_{R_S}(N)$. In general, if $R$ is Noetherian, then for any R-module M, we have $Ass(M_S)=Ass(M)\cap Spec(R_S)$.

Let $x\in N$. Then $ann_R(x)=ann_{R_S}(x)\cap R$. This is just because the elements of R that kill x, are just the fractions that kill x that are “actually” in R. This immediately gives us one inclusion, since if $P\in Ass_{R_S}(N)$ then $P\cap R\in Ass_R(N)$.

Now suppose $Q\in Ass_R(N)$. Then there is some $x\in N$ such that $Q=ann_R(x)$. Thus $x\neq 0$ giving $Q\cap S=\emptyset$. Thus $QR_S\in Spec(R_S)$ with $QR_S=ann_{R_S}(x)$. This proves the first statement.

We now show the second statement about M. Suppose $P\in Ass(M)\cap Spec(R_S)$. Thus we again get that $P\cap S=\emptyset$ and $P=ann_R(x)$ for some non-zero $x\in M$. Suppose that that $(r/s)x=0$ in $M_S$. Thus there is some $t\in S$ such that $trx=0$ in $M$. But we’ve already noted that $t\notin P$ and $tr\in P$, thus by primality $r\in P$. So $PR_S=ann_{R_S}(x)$. Thus $PR_S\in Ass(M_S)$ giving one inclusion.

For the reverse, suppose $Q\in Ass(M_S)$. By clearing the denominator we can assume that for a non-zero $x\in M$ we have $Q=ann_{R_S}(x)$. Let $Q^c=Q\cap R$. Then $Q=Q^cR_S$. We have that $Q^c$ is finitely generated since $R$ is Noetherian, so there is some $t\in S$ such that $Q^c=ann_R(tx)$. Thus $Q^c\in Ass_R(M)$ which gives the reverse inclusion.

A nice little corollary is that for Noetherian rings a prime ideal $P\in Ass_R(M)$ if and only if $PR_P\in Ass_{R_P}(M_P)$.

1. I’m wondering about your setting. Is R supposed to be Noetherian from the very beginning? If not, I don’t know why Ass(M) must be nonempty, because in your proof, it seems unclear to me that $\frak{F}$ has a maximal element.
2. oops, i meant $\mathcal{F}$.