Associated Primes III

Hopefully today we can finish this topic off. We’ll jump right in. Let $R$ be Noetherian and $M$ finite, then $N\subset M$ is primary if and only if $Ass(M/N)$ consists of a single element.

We’ll use the second formulation of primary. Suppose $Ass(M/N)=\{P\}$. Then by last time $Supp(M/N)=V(P)$, so we have $P=\sqrt{(ann(M/N)}$. Suppose $r\in R$ is a zero divisor for $M/N$. Then from AP I we get that $r\in P\Rightarrow r\in\sqrt{(ann(M/N))}$. Thus by the second formulation, $N$ is a primary submodule.

For the reverse, suppose $N$ is a primary submodule and $P\in Ass(M/N)$. Then every element of $P$ is a zero-divisor for $M/N$. So $r\in P\Rightarrow r\in\sqrt{ann(M/N)}$. Thus $P\subset \sqrt{ann(M/N)}$. By definition of associated prime we get $ann(M/N)\subset P$, and primes are radical so taking radicals of both sides get the other inclusion and so $\sqrt{ann(M/N)}=P$. i.e. the only element of $Ass(M/N)$ is $P$.

We now note that $I=ann(M/N)$ is actually a primary ideal. Let $r,s\in R$ and suppose $rs\in I$ but that $s\notin I$. Then $(rs)(M/N)=0$ but $s(M/N)\neq 0$. So $r$ is a zero-divisor for $M/N$ which gives $r\in P=\sqrt{I}$. Thus the ideal $I$ is $P$-primary.

Thus we make the definition for modules that if $Ass(M/N)=\{P\}$ then the submodule $N$ is $P$primary (sometimes called a primary submodule belonging to $P$).

Note that the intersection of any two $P$-primary submodules is again $P$-primary. This is seen by embedding $M/(N\cap N')\hookrightarrow (M/N)\oplus (M/N')$.

We call a submodule reducible if it can be written as such an intersection and irreducible otherwise (this is a property on submodules, not to be confused with the notion of irreducible for modules).

Any submodule of a Noetherian module can be written as a finite intersection of irreducible submodules. This is seen by applying Zorn’s lemma to the set of submodules having no such representation.

We are about to the point where we can define a primary decomposition. Of course there is not going to be a unique way of doing it, but we’ll make some contrived definitions to get it as unique as possible.

We’ll call an intersection irredundant if none of the components of the intersection can be omitted (in particular this will prevent unnecessary repetition sort of like multiplying by a bunch of 1’s in a prime factoring).

A decomposition of a submodule is an expression of the submodule as an intersection of a finite number of submodules, and if each component is irreducible, then we say it is an irreducible decomposition. Likewise, we define primary decomposition if each component is primary.

Now suppose we write $N=\cap N_i$ as an irredundant primary decomposition with $Ass(M/N_i)=\{P_i\}$. Since the intersection of any finite number of $P_i$-primary submodules is again $P_i$-primary we can group that intersection together and consider it as just a single submodule. In this way we get a decomposition in which $P_i\neq P_j$ when $i\neq j$. This makes the decomposition as short as possible.

To wrap up we need to prove that everything behaves the way we want (note that we’ve been assuming Noetherian ring and finite module).

I) An irreducible submodule of $M$ is a primary submodule. Suppose $N\subset M$ is not primary. We can assume $N=0$ without loss of generality by replacing $M$ with $M/N$. Then by the first theorem in this post we get that $Ass(M)$ has at least two elements $P_1, P_2$. i.e. $M$ contains submodules $K_i$ isomorphic to $R/P_i$, so $K_1\cap K_2=0$ which means $N=0$ is reducible.

II) Now for an important one. We want to be able to read off the associated primes from the decomposition, so If we have an irredundant primary decomposition of a proper submodule $N=\cap N_i$, then $Ass(M/N)=\{P_1, \ldots, P_r\}$ where $Ass(M/N_i)=\{P_i\}$.

We’ll again assume WLOG that $N=0=\cap N_i$. Thus $M$ is isomorphic to a submodule of $M/N_1\oplus \cdots \oplus M/N_r$. i.e. $Ass(M)\subset Ass\left(\bigoplus M/N_i\right)=\bigcup Ass(M/N_i)=\{P_1, \ldots, P_r\}$.

Now by being irredundant $N_2\cap \cdots \cap N_r\neq 0$. So pick a non-zero element $x\in N_2\cap \cdots \cap N_r$. Then $ann(x)=(0:x)=(N_1:x)$. But we have $(N_1 :M)$ is primary belonging to $P_1$, so $P_1^nM\subset N_1$ for some $n$. Thus $P_1^nx=0$ which gives that for some $i we have $P_1^ix\neq 0$ and $P_1^{i+1}x=0$. Choose a non-zero element $y\in P_1^ix$. Then $P_1y=0$.

But note that $y\in N_2\cap \cdots \cap N_r$, so we have $y\notin N_1$. Since the submodule is primary $ann(y)\subset P_1$ so $P_1=ann(y)$ giving $P_1\in Ass(M)$. If we do this for all the other $P_i$ we get that $\{P_1, \cdots , P_r\}\subset Ass(M)$ giving equality of sets.

III) Lastly we want the existence and uniqueness. Every proper submodule has a primary decomposition. This is just because we know that there is an irreducible decomposition, so apply (I) to each irreducible component.

Uniqueness is a little trickier. We must restrict our attention to minimal primes. Suppose $P$ is a minimal associated prime of $M/N$, then the $P$-primary component of $N$ is $\phi_P^{-1}(N_P)$ where $\phi_P : M\to M_P$. Thus it is uniquely determined given the data $M, N$ and $P$. Non-minimal are not unique: Take the ring $\mathbb{C}[x,y]$, then $(x^2, xy)=(x)\cap (x^2, y)=(x)\cap (x^2, xy, y^2)$.