This article is about the statement of a simple but indispensable lemma in commutative algebra
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This article defines a result where the base ring (or one or more of the rings involved) is Noetherian
View more results involving Noetherianness or Read a survey article on applying Noetherianness
This fact is an application of the following pivotal fact/result/idea: Hilbert basis theorem
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is an essentially -adic filtration (in other words, there exists such that for all , ).
Then the filtration of given by:
is also essentially -adic.
The key idea is the following:
- To any filtration, find an associated module over a Noetherian ring such that the filtration being essentially -adic is equivalent to the module being finitely generated
- Prove that intersecting the filtration with a submodule is equivalent to taking a submodule of the associated module.
Setting up the modules
Denote by the blowup algebra of the ideal in ; in other words, the ring:
where the multiplication of graded components is just the usual multiplication as elements of <mth>R</math>.
We now describe a way to associate, to any filtration of a module, an associated module over the blowup algebra. Suppose the filtration is:
We define the associated module over as:
where the multiplication is defined in the usual way.
The crucial observation
The main step of the proof is to observe that is a finitely generated module over if and only if the filtration of is essentially -adic.
Induced filtration on submodule gives submodule on blowup
The module associated for the induced filtration on the submodule of , is clearly a submodule of the module .
Applying the Hilbert basis theorem
Since is a Noetherian ring, the ideal is a finitely generated ideal. Since the blowup algebra is, by construction, generated by its elements of degree zero and one, we see that the blowup algebra is a finitely generated algebra over . Thus, the blowup algebra is a quotient of a polynomial ring over . By the Hilbert basis theorem and the fact that quotients of Noetherian rings are Noetherian, we obtain that is a Noetherian ring.
Thus, if is a finitely generated module over , so is the submodule for the induced filtration on .
Putting the pieces together
Here's the summary of the proof:
- If is an essentially -adic filtration on , then the corresponding module is finitely generated over .
- Since is Noetherian, is Noetherian.
- The module corresponding to the induced filtration on is a -submodule of . Since the big module is finitely generated and the ring is Noetherian, the submodule is also finitely generated.
- Finally, since the module corresponding to the induced filtration is finitely generated, the induced filtration itself is essentially -adic.
- Book:Eisenbud, Page 151-152