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Krull–Akizuki theorem

From Wikipedia, the free encyclopedia

In commutative algebra, the Krull–Akizuki theorem states the following: Let A be a one-dimensional reduced noetherian ring,[1] K its total ring of fractions. Suppose L is a finite extension of K.[2] If and B is reduced, then B is a one-dimensional noetherian ring. Furthermore, for every nonzero ideal of B, is finite over A.[3][4]

Note that the theorem does not say that B is finite over A. The theorem does not extend to higher dimension. One important consequence of the theorem is that the integral closure of a Dedekind domain A in a finite extension of the field of fractions of A is again a Dedekind domain. This consequence does generalize to a higher dimension: the Mori–Nagata theorem states that the integral closure of a noetherian domain is a Krull domain.

Proof

First observe that and KB is a finite extension of K, so we may assume without loss of generality that . Then for some . Since each is integral over K, there exists such that is integral over A. Let . Then C is a one-dimensional noetherian ring, and , where denotes the total ring of fractions of C. Thus we can substitute C for A and reduce to the case .

Let be minimal prime ideals of A; there are finitely many of them. Let be the field of fractions of and the kernel of the natural map . Then we have:

and .

Now, if the theorem holds when A is a domain, then this implies that B is a one-dimensional noetherian domain since each is and since . Hence, we reduced the proof to the case A is a domain. Let be an ideal and let a be a nonzero element in the nonzero ideal . Set . Since is a zero-dim noetherian ring; thus, artinian, there is an such that for all . We claim

Since it suffices to establish the inclusion locally, we may assume A is a local ring with the maximal ideal . Let x be a nonzero element in B. Then, since A is noetherian, there is an n such that and so . Thus,

Now, assume n is a minimum integer such that and the last inclusion holds. If , then we easily see that . But then the above inclusion holds for , contradiction. Hence, we have and this establishes the claim. It now follows:

Hence, has finite length as A-module. In particular, the image of there is finitely generated and so is finitely generated. The above shows that has dimension zero and so B has dimension one. Finally, the exact sequence of A-modules shows that is finite over A.

References

  1. ^ In this article, a ring is commutative and has unity.
  2. ^ If are rings, we say that B is a finite extension of A if B is a finitely generated A module.
  3. ^ Bourbaki 1989, Ch VII, §2, no. 5, Proposition 5
  4. ^ Swanson, Irena; Huneke, Craig (2006). Integral Closure of Ideals, Rings, and Modules. Cambridge University Press. pp. 87–88.
This page was last edited on 8 October 2023, at 17:36
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