Hopefully today we can finish this topic off. We’ll jump right in. Let be Noetherian and finite, then is primary if and only if consists of a single element.
We’ll use the second formulation of primary. Suppose . Then by last time , so we have . Suppose is a zero divisor for . Then from AP I we get that . Thus by the second formulation, is a primary submodule.
For the reverse, suppose is a primary submodule and . Then every element of is a zero-divisor for . So . Thus . By definition of associated prime we get , and primes are radical so taking radicals of both sides get the other inclusion and so . i.e. the only element of is .
We now note that is actually a primary ideal. Let and suppose but that . Then but . So is a zero-divisor for which gives . Thus the ideal is -primary.
Thus we make the definition for modules that if then the submodule is –primary (sometimes called a primary submodule belonging to ).
Note that the intersection of any two -primary submodules is again -primary. This is seen by embedding .
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 as an irredundant primary decomposition with . Since the intersection of any finite number of -primary submodules is again -primary we can group that intersection together and consider it as just a single submodule. In this way we get a decomposition in which when . 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 is a primary submodule. Suppose is not primary. We can assume without loss of generality by replacing with . Then by the first theorem in this post we get that has at least two elements . i.e. contains submodules isomorphic to , so which means 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 , then where .
We’ll again assume WLOG that . Thus is isomorphic to a submodule of . i.e. .
Now by being irredundant . So pick a non-zero element . Then . But we have is primary belonging to , so for some . Thus which gives that for some we have and . Choose a non-zero element . Then .
But note that , so we have . Since the submodule is primary so giving . If we do this for all the other we get that 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 is a minimal associated prime of , then the -primary component of is where . Thus it is uniquely determined given the data and . Non-minimal are not unique: Take the ring , then .