Associated Primes II

Today we’ll continue towards a primary decomposition for modules. First, I’ll list two facts without proof that may come up (they are quite straightforward to prove if you want to try). If R is any ring and $0\to M'\to M\to M''\to 0$ is an exact sequence of $R$-modules, then $Ass(M)\subset Ass(M')\cup Ass(M'')$. Secondly, if $R$ is a Noetherian ring and $M$ a non-zero finite $R$-module, then there is a chain $0=M_0\subset M_1\subset \cdots \subset M_n=M$ of submodules of $M$ such that for each $i$ we have $M_i/M_{i-1}\simeq R/P_i$ with $P_i\in Spec R$.

I don’t remember, but I may have even proved that second one when talking about Artin-Rees. Now let $R$ be Noetherian and $M$ a finite $R$-module.

I) $Ass(M)$ is finite. We induct on the length of the chain in the second fact. Suppose this is true for all $M$ having a chain of the above form of length $n-1$. If $N$ is a finite module with a chain of length $n$, then since $R$ is Noetherian and $N_{n-1}$ is a submodule it is also finite. So by the inductive hypothesis, $Ass(N_{n-1})$ is finite. Now consider the exact sequence $0\to N_{n-1} \to N \to N/N_{n-1}\to 0$. By the first fact $Ass(N)\subset Ass(N_{n-1})\cup Ass(N/N_{n-1})$. But the chain has the condition that $N/N_{n-1}\simeq R/P$ for some $P\in Spec(R)$. Since $Ass(R/P)=\{P\}$ we have that the cardinality of $Ass(N)$ can increase by at most one from the cardinality of $Ass(N_{n-1})$ which was finite.

II) $Ass(M)\subset Supp(M)$. Suppose $P\in Ass(M)$. Then $M$ contains a submodule isomorphic to $R/P$ (it is just the image of the hom $r\mapsto r\cdot x$ and apply first iso theorem). So $0\to R/P\to M$ is exact, so when we localize we still have an exact sequence $0\to R_P/PR_P\to M_P$. Since $R_P/PR_P\neq 0$, $M_P\neq 0$ which means $P\in Supp(M)$.

III) The set of minimal elements of $Ass(M)$ coincides with the minimal elements of $Supp(M)$. Well, (II) gave one inclusion so suppose $P\in Supp(M)$ is a minimal element. Then since $M_P\neq 0$ by the last post we get that $Ass(M_P)\neq \emptyset$. But we also figured out a formula for this set $Ass(M_P)=Ass(M)\cap Spec(R_P)\subset Supp(M)\cap Spec(R_P)=\{P\}$. Thus by non-emptyness we must have $P\in Ass(M)$.

Recall when we working in the Zariski topology on $Spec(R)$. We have an operator on ideals $V(I)=\{p\in Spec(R) : p\subset I\}$, and the Zariski closed sets of $Spec(R)$ are precisely those sets that are of the form $V(I)$ for some $I$.

So by definition of this operator, if $P_1, \ldots , P_r$ are the minimal elements of $Supp(M)$, then $Supp(M)=V(P_1)\cup \cdots \cup V(P_r)$. Another property of the topological space $Spec(R)$ is that a subspace is irreducible if and only if it is $V(\frak{p})$ for some minimal prime $\frak{p}$. So if we think of $Supp(M)$ as a closed subspace of $Spec(R)$, then the irreducible components are precisely $V(P_i)$. We call the $P_i$ the isolated associated primes of M. The other associated primes are called embedded primes.

Due to the above geomtric interpretation of what isolated and embedded primes are, the terms make sense. An isolated prime gives you full irreducible component of $Supp(M)$ whereas an embedded prime gives some embedded subspace of the component generated by the prime it lies over.

I’ll finish with the new definitions. Suppose $N\subset M$ is a submodule. Then we call $N$ a primary submodule if for all $r\in R$ and $x\in M$ we have the condition $x\notin N$ and $rx\in N\Rightarrow r^nM\subset N$ for some $n$.

The above condition is equivalent to the condition: if $r\in R$ is a zero-divisor for $M/N$, then $r\in \sqrt{(ann(M/N)}$. Showing these are equivalent is immediate when you write out what the definitions of all these things are. This shows that the property of being primary is dependent only on the quotient module $M/N$.

Sorry to end on some definitions, but I think if I do another theorem this post will become too long.

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