Krull–Akizuki theorem

In 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

1. In this article, a ring is commutative and has unity.
2. Bourbaki 1989, Ch VII, §2, no. 5, Proposition 5

• Nicolas Bourbaki, Commutative algebra