Krull–Akizuki theorem

In commutative algebra, the Krull–Akizuki theorem states the following: Let A be a one-dimensional reduced noetherian ring, K its total ring of fractions. Suppose L is a finite extension of K. If $$A\subset B\subset L$$ and B is reduced, 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
First observe that $$A\subset B\subset KB$$ and KB is a finite extension of K, so we may assume without loss of generality that $$L=KB$$. Then $$L=Kx_1+\cdots+Kx_n$$ for some $$x_1,\dots,x_n\in B$$. Since each $$x_i$$ is integral over K, there exists $$a_i\in A$$ such that $$a_ix_i$$ is integral over A. Let $$C=A[a_1x_1,\dots,a_nx_n]$$. Then C is a one-dimensional noetherian ring, and $$C\subset B\subset Q(C)$$, where $$Q(C)$$ denotes the total ring of fractions of C. Thus we can substitute C for A and reduce to the case $$L = K$$.

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:
 * $$A/{\mathfrak{p}_i} \subset B/{I_i} \subset K_i$$ and $$K\simeq\prod K_i$$.

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 $$B \simeq \prod B/{I_i}$$. 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. The above shows that $$B/{aB}$$ has dimension zero and so B has dimension one. Finally, the exact sequence $$B/aB\to B/I\to (0)$$ of A-modules shows that $$B/I$$ is finite over A. $$\square$$