Let I be an ideal in a Noetherian ring R; letM be a finitely generatedR-module and let N a submodule of M. Then there exists an integer k ≥ 1 so that, for n ≥ k,
Proof
The lemma immediately follows from the fact that R is Noetherian once necessary notions and notations are set up. For any ring R and an ideal I in R, we set We say a decreasing sequence of submodules is an I-filtration if ; moreover, it is stable if for sufficiently largen. If M is given an I-filtration, we set ; it is a graded module over. Now, let M be a R-module with the I-filtration by finitely generated R-modules. We make an observation Indeed, if the filtration is I-stable, then is generated by the first terms and those terms are finitely generated; thus, is finitely generated. Conversely, if it is finitely generated, say, by some homogeneous elements in, then, for, each f in can be written as with the generators in. That is,. We can now prove the lemma, assuming R is Noetherian. Let. Then are an I-stable filtration. Thus, by the observation, is finitely generated over. But is a Noetherian ring since R is. Thus, is a Noetherian module and any submodule is finitely generated over ; in particular, is finitely generated when N is given the induced filtration; i.e.,. Then the induced filtration is I-stable again by the observation.
Besides the use in completion of a ring, a typical application of the lemma is the proof of the Krull's intersection theorem, which says: for a proper idealI in a commutative Noetherian ring that is either a local ring or an integral domain. By the lemma applied to the intersection, we find ksuch that for, But then. Thus, if A is local, by Nakayama's lemma. If A is an integral domain, then one uses the determinant trick : In the setup here, take u to be the identity operator on N; that will yield a nonzero element x in A such that, which implies. For both a local ring and an integral domain, the "Noetherian" cannot be dropped from the assumption: for the local ring case, see local ring#Commutative case. For the integral domain case, take to be the ring of algebraic integers. If is a prime ideal of A, then we have: for every integer. Indeed, if, then for some complex number. Now, is integral over ; thus in and then in, proving the claim.