Eakin–Nagata theorem

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In abstract algebra, the Eakin–Nagata theorem states: given commutative rings AB such that B is finitely generated as a module over A, if B is a Noetherian ring, then A is a Noetherian ring.[1] (Note the converse is also true and is easier.)

The theorem is similar to the Artin–Tate lemma, which says that the same statement holds with "Noetherian" replaced by "finitely generated algebra" (assuming the base ring is a Noetherian ring).

The theorem was first proved in Paul M. Eakin's thesis Template:Harv and later independently by Template:Harvs.[2] The theorem can also be deduced from the characterization of a Noetherian ring in terms of injective modules, as done for example by David Eisenbud in Template:Harv; this approach is useful for a generalization to non-commutative rings.

Proof

The following more general result is due to Edward W. Formanek and is proved by an argument rooted to the original proofs by Eakin and Nagata. According to Template:Harv, this formulation is likely the most transparent one.

Template:Math theorem

Proof: It is enough to show that M is a Noetherian module since, in general, a ring admitting a faithful Noetherian module over it is a Noetherian ring.[3] Suppose otherwise. By assumption, the set of all IM, where I is an ideal of A such that M/IM is not Noetherian has a maximal element, I0M. Replacing M and A by M/I0M and A/Ann(M/I0M), we can assume

  • for each nonzero ideal IA, the module M/IM is Noetherian.

Next, consider the set S of submodules NM such that M/N is faithful. Choose a set of generators {x1,,xn} of M and then note that M/N is faithful if and only if for each aA, the inclusion {ax1,,axn}N implies a=0. Thus, it is clear that Zorn's lemma applies to the set S, and so the set has a maximal element, N0. Now, if M/N0 is Noetherian, then it is a faithful Noetherian module over A and, consequently, A is a Noetherian ring, a contradiction. Hence, M/N0 is not Noetherian and replacing M by M/N0, we can also assume

  • each nonzero submodule NM is such that M/N is not faithful.

Let a submodule 0NM be given. Since M/N is not faithful, there is a nonzero element aA such that aMN. By assumption, M/aM is Noetherian and so N/aM is finitely generated. Since aM is also finitely generated, it follows that N is finitely generated; i.e., M is Noetherian, a contradiction.

References

Template:Reflist

Further reading