Krull–Akizuki theorem

From Wikipedia, the free encyclopedia
Jump to: navigation, search

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[edit]

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[edit]

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