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Koecher–Vinberg theoremIn 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 $$L = K. Let \mathfrak{p}_i$$ 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. $$\square$$

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