Krull–Akizuki theorem

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,[1] K its total ring of fractions. Suppose L is a finite extension of K.[2] If A B L {\displaystyle A\subset B\subset L} and B is reduced, then B is a one-dimensional noetherian ring. Furthermore, for every nonzero ideal I {\displaystyle I} of B, B / I {\displaystyle B/I} is finite over A.[3][4]

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 B K B {\displaystyle A\subset B\subset KB} and KB is a finite extension of K, so we may assume without loss of generality that L = K B {\displaystyle L=KB} . Then L = K x 1 + + K x n {\displaystyle L=Kx_{1}+\cdots +Kx_{n}} for some x 1 , , x n B {\displaystyle x_{1},\dots ,x_{n}\in B} . Since each x i {\displaystyle x_{i}} is integral over K, there exists a i A {\displaystyle a_{i}\in A} such that a i x i {\displaystyle a_{i}x_{i}} is integral over A. Let C = A [ a 1 x 1 , , a n x n ] {\displaystyle C=A[a_{1}x_{1},\dots ,a_{n}x_{n}]} . Then C is a one-dimensional noetherian ring, and C B Q ( C ) {\displaystyle C\subset B\subset Q(C)} , where Q ( C ) {\displaystyle Q(C)} denotes the total ring of fractions of C. Thus we can substitute C for A and reduce to the case L = K {\displaystyle L=K} .

Let p i {\displaystyle {\mathfrak {p}}_{i}} be minimal prime ideals of A; there are finitely many of them. Let K i {\displaystyle K_{i}} be the field of fractions of A / p i {\displaystyle A/{{\mathfrak {p}}_{i}}} and I i {\displaystyle I_{i}} the kernel of the natural map B K K i {\displaystyle B\to K\to K_{i}} . Then we have:

A / p i B / I i K i {\displaystyle A/{{\mathfrak {p}}_{i}}\subset B/{I_{i}}\subset K_{i}} and K K i {\displaystyle 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 {\displaystyle B/{I_{i}}} is and since B B / I i {\displaystyle B\simeq \prod B/{I_{i}}} . Hence, we reduced the proof to the case A is a domain. Let 0 I B {\displaystyle 0\neq I\subset B} be an ideal and let a be a nonzero element in the nonzero ideal I A {\displaystyle I\cap A} . Set I n = a n B A + a A {\displaystyle I_{n}=a^{n}B\cap A+aA} . Since A / a A {\displaystyle A/aA} is a zero-dim noetherian ring; thus, artinian, there is an l {\displaystyle l} such that I n = I l {\displaystyle I_{n}=I_{l}} for all n l {\displaystyle n\geq l} . We claim

a l B a l + 1 B + A . {\displaystyle 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 m {\displaystyle {\mathfrak {m}}} . Let x be a nonzero element in B. Then, since A is noetherian, there is an n such that m n + 1 x 1 A {\displaystyle {\mathfrak {m}}^{n+1}\subset x^{-1}A} and so a n + 1 x a n + 1 B A I n + 2 {\displaystyle a^{n+1}x\in a^{n+1}B\cap A\subset I_{n+2}} . Thus,

a n x a n + 1 B A + A . {\displaystyle a^{n}x\in a^{n+1}B\cap A+A.}

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

B / a B a l B / a l + 1 B ( a l + 1 B + A ) / a l + 1 B A / ( a l + 1 B A ) . {\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).}

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

References

  1. ^ In this article, a ring is commutative and has unity.
  2. ^ If A B {\displaystyle A\subset B} are rings, we say that B is a finite extension of A if B is a finitely generated A module.
  3. ^ Bourbaki 1989, Ch VII, §2, no. 5, Proposition 5
  4. ^ Swanson, Irena; Huneke, Craig (2006). Integral Closure of Ideals, Rings, and Modules. Cambridge University Press. pp. 87–88.
  • Bourbaki, Nicolas (1989). Commutative algebra. Berlin Heidelberg: Springer. ISBN 978-3-540-64239-8.