Krull–Akizuki theorem

Koecher–Vinberg theoremIn algebra, the Krull–Akizuki theorem states the following: let A be a one-dimensional reduced noetherian ring,[1] 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.[2]

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 \( L = K. Let \mathfrak{p}_i \) be minimal prime ideals of A; there are finitely many of them. Let \(K_i \) be the field of fractions of \( A/{\mathfrak{p}_i} \) and \( I_i \) the kernel of the natural map \(B \to K \to K_i \) . Then we have:

\(A/{\mathfrak{p}_i} \subset B/{I_i} \subset K_i. \)

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 \(B = \prod B/{I_i} \) . Hence, we reduced the proof to the case A is a domain. Let 0 \(\ne I \subset B \) be an ideal and let a be a nonzero element in the nonzero ideal \(I \cap A \) . Set \(I_n = a^nB \cap A + aA \) . Since A/aA is a zero-dim noetherian ring; thus, artinian, there is an l such that \( I_n = I_l \) for all \( n \ge l \) . We claim

\( a^l B \subset a^{l+1}B + A. \)

Since it suffices to establish the inclusion locally, we may assume A is a local ring with the maximal ideal \(\mathfrak{m} \) . Let x be a nonzero element in B. Then, since A is noetherian, there is an n such that \(\mathfrak{m}^{n+1} \subset x^{-1} A \) and so \( a^{n+1}x \in a^{n+1}B \cap A \subset I_{n+2} \) . Thus,

\( a^n x \in a^{n+1} B \cap A + A. \)

Now, assume n is a minimum integer such that \( n \ge l \) and the last inclusion holds. If n > l, then we easily see that \(a^n x \in I_{n+1} \) . But then the above inclusion holds for n-1, contradiction. Hence, we have n = l and this establishes the claim. It now follows:

\( B/{aB} \simeq a^l B/a^{l+1} B \subset (a^{l +1}B + A)/a^{l+1} B \simeq A/{a^{l +1}B \cap A} \) .

Hence, B/{aB} 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/{aB} has zero dimension and so B has dimension one. \( \square \)

References

In this article, a ring is commutative and has unity.

Bourbaki 1989, Ch VII, §2, no. 5, Proposition 5

http://books.google.com/books?id=APPtnn84FMIC&lpg=PA83&ots=2L9MiWbIYZ&dq=krull%20akizuki&pg=PA85#v=onepage&q=krull%20akizuki&f=false
Nicolas Bourbaki, Commutative algebra

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