Krull–Akizuki theorem

Short description: About extensions of one-dimensional Noetherian rings (commutative algebra)

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 [math]\displaystyle{ A\subset B\subset L }[/math] and B is reduced, then B is a one-dimensional noetherian ring. Furthermore, for every nonzero ideal [math]\displaystyle{ I }[/math] of B, [math]\displaystyle{ B/I }[/math] 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 [math]\displaystyle{ A\subset B\subset KB }[/math] and KB is a finite extension of K, so we may assume without loss of generality that [math]\displaystyle{ L=KB }[/math]. Then [math]\displaystyle{ L=Kx_1+\cdots+Kx_n }[/math] for some [math]\displaystyle{ x_1,\dots,x_n\in B }[/math]. Since each [math]\displaystyle{ x_i }[/math] is integral over K, there exists [math]\displaystyle{ a_i\in A }[/math] such that [math]\displaystyle{ a_ix_i }[/math] is integral over A. Let [math]\displaystyle{ C=A[a_1x_1,\dots,a_nx_n] }[/math]. Then C is a one-dimensional noetherian ring, and [math]\displaystyle{ C\subset B\subset Q(C) }[/math], where [math]\displaystyle{ Q(C) }[/math] denotes the total ring of fractions of C. Thus we can substitute C for A and reduce to the case [math]\displaystyle{ L = K }[/math].

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

[math]\displaystyle{ A/{\mathfrak{p}_i} \subset B/{I_i} \subset K_i }[/math] and [math]\displaystyle{ K\simeq\prod K_i }[/math].

Now, if the theorem holds when A is a domain, then this implies that B is a one-dimensional noetherian domain since each [math]\displaystyle{ B/{I_i} }[/math] is and since [math]\displaystyle{ B \simeq \prod B/{I_i} }[/math]. Hence, we reduced the proof to the case A is a domain. Let [math]\displaystyle{ 0 \ne I \subset B }[/math] be an ideal and let a be a nonzero element in the nonzero ideal [math]\displaystyle{ I \cap A }[/math]. Set [math]\displaystyle{ I_n = a^nB \cap A + aA }[/math]. Since [math]\displaystyle{ A/aA }[/math] is a zero-dim noetherian ring; thus, artinian, there is an [math]\displaystyle{ l }[/math] such that [math]\displaystyle{ I_n = I_l }[/math] for all [math]\displaystyle{ n \ge l }[/math]. We claim

[math]\displaystyle{ a^l B \subset a^{l+1}B + A. }[/math]

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

[math]\displaystyle{ a^n x \in a^{n+1} B \cap A + A. }[/math]

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

[math]\displaystyle{ 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). }[/math]

Hence, [math]\displaystyle{ B/{aB} }[/math] has finite length as A-module. In particular, the image of [math]\displaystyle{ I }[/math] there is finitely generated and so [math]\displaystyle{ I }[/math] is finitely generated. The above shows that [math]\displaystyle{ B/{aB} }[/math] has dimension zero and so B has dimension one. Finally, the exact sequence [math]\displaystyle{ B/aB\to B/I\to (0) }[/math] of A-modules shows that [math]\displaystyle{ B/I }[/math] is finite over A. [math]\displaystyle{ \square }[/math]

References

• Bourbaki, Nicolas (1989). Commutative algebra. Berlin Heidelberg: Springer. ISBN 978-3-540-64239-8.