Prime Composition Series
Throughout this article, A is a noetherian ring and all A-modules are finitely generated.
Recall (proposition 1 here) that if M is a noetherian and artinian module, we can find a sequence of submodules whose consecutive factors are simple modules. Correspondingly we have:
For finitely generated M, there exists a sequence of submodules
such that each as A-modules for prime ideals .
Assume . Since by proposition 3 here, there is an embedding of A-modules . If equality holds, we are done. Otherwise, let be the image of the map and repeat with to obtain a submodule . Repeating this process, this must eventually terminate since we cannot have an infinite ascending chain of submodules . ♦
Let be as in proposition 1 with for some prime . Then
In particular if M is finitely generated then is finite.
Repeatedly applying proposition 4 here, we have
Since each for each i, we are done. ♦
Associated Primes and Support
If , then there is a map ; upon localizing at we get an A-linear and so . Hence we have shown:
For any module M, .
For a partial reverse inclusion, we have:
If is a minimal element of then .
By proposition 5 here, we have
which is non-empty since . By lemma 1 this set lies in ; but since is minimal in , by exercise B.2 here has exactly one element: so
so by proposition 5 here again. ♦
If is not a minimal prime of , we call it an embedded prime of M.
Let k be a field, and , considered as an A-module. Note that with radical so we get (by proposition 1 here)
On the other hand, we claim that . Indeed we have:
so and is an embedded prime of M.
Conversely, we pick the chain of submodules with generated by . Then ; as shown above, . Also so by corollary 1
Find an A-module M and prime ideals with but .
Existence of Primary Decomposition
In this section we fix an ambient non-zero A-module M (finitely generated of course) and consider its submodules. For each prime , take the set of all submodules such that . Note that
Now for each prime , fix a maximal element .
We have .
If then so the above intersection only needs to be taken over , i.e. a finite number of terms.
Let . If it has an associated prime . Since we have , a contradiction. ♦
Let . Then .
By proposition 4 here, we have . Since we have . On the other hand if , then we have an injection whose image is of the form . But now
which does not contain , contradicting maximality of . Hence no such exists. ♦
For an associated prime of M, a –primary submodule of M is an such that .
A primary decomposition of M is an expression
where each is a -primary submodule of M.
The decomposition is irredundant if for any , . It is minimal if it is irredundant and all are distinct.
By lemmas 2 and 3, a minimal primary decomposition exists for every non-zero module.
It is not true that in every primary decomposition , must be a maximal element of . We will see an example in the next article.
Prove that if are -primary submodules, so is .
Hence given any primary decomposition, we can get a minimal one by first removing the redundant terms then taking the intersection of all with the same corresponding .
Properties of Primary Decomposition
Here, we will discuss properties of a general primary decomposition. Throughout this section we fix:
where is -primary in M.
Every must occur among the .
Since the canonical map is injective. Hence
If the primary decomposition is irredundant, then every is an associated prime of M.
If then among the injections and , we have since and are disjoint. Thus
and , contradicting the condition of irredundancy. ♦
Thus proposition 4 gives us a way to compute : find a primary decomposition and remove terms until it becomes irredundant. However, note that proposition 4 does not require to be distinct.
If the primary decomposition is minimal, then
has exactly n elements.
Apply propositions 3 and 4. ♦
Finally, we define the primary decomposition of submodules.
Let be a submodule. A primary decomposition of N in M is an expression
such that is a -primary submodule of M.
A primary decomposition of N in M corresponds bijectively to a primary decomposition of M/N. We say the primary decomposition of N in M is irredundant or minimal if the corresponding primary decomposition of M/N is so.