 # Krull–Akizuki theorem

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In algebra, the Krull–Akizuki theorem states the following: let A be a one-dimensional reduced noetherian ring, K its total ring of fractions. If B is a subring of a finite extension L of K containing A and is not a field, then B is a one-dimensional noetherian ring. Furthermore, for every nonzero ideal I of B, B / I is finite over A.

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

Here, we give a proof when L = K . Let p i be minimal prime ideals of A; there are finitely many of them. Let K i be the field of fractions of A / p i and I i the kernel of the natural map B K K i . Then we have:

A / p i B / I i K i .

Now, if the theorem holds when A is a domain, then this implies that B is a one-dimensional noetherian domain since each B / I i is and since B = B / I i . Hence, we reduced the proof to the case A is a domain. Let 0 I B be an ideal and let a be a nonzero element in the nonzero ideal I A . Set I n = a n B A + a A . Since A / a A is a zero-dim noetherian ring; thus, artinian, there is an l such that I n = I l for all n l . We claim

a l B a l + 1 B + A .

Since it suffices to establish the inclusion locally, we may assume A is a local ring with the maximal ideal m . Let x be a nonzero element in B. Then, since A is noetherian, there is an n such that m n + 1 x 1 A and so a n + 1 x a n + 1 B A I n + 2 . Thus,

a n x a n + 1 B A + A .

Now, assume n is a minimum integer such that n l and the last inclusion holds. If n > l , then we easily see that a n x I n + 1 . But then the above inclusion holds for n 1 , contradiction. Hence, we have n = l and this establishes the claim. It now follows:

B / a B a l B / a l + 1 B ( a l + 1 B + A ) / a l + 1 B A / a l + 1 B A .

Hence, B / a B has finite length as A-module. In particular, the image of I there is finitely generated and so I is finitely generated. Finally, the above shows that B / a B has zero dimension and so B has dimension one.

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