Krull–Akizuki theorem explained
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
and
B is reduced,then
B is a one-dimensional noetherian ring. Furthermore, for every nonzero ideal
of
B,
is finite over
A.
[3] 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
and
KB is a finite extension of
K, so we may assume without loss of generality that
.Then
for some
.Since each
is integral over
K, there exists
such that
is integral over
A.Let
.Then
C is a one-dimensional noetherian ring, and
, where
denotes the total ring of fractions of
C.Thus we can substitute
C for
A and reduce to the case
.
Let
be minimal prime ideals of
A; there are finitely many of them. Let
be the field of fractions of
and
the kernel of the natural map
. Then we have:
A/{ak{p}i}\subsetB/{Ii}\subsetKi
and
.Now, if the theorem holds when
A is a domain, then this implies that
B is a one-dimensional noetherian domain since each
is and since
. Hence, we reduced the proof to the case
A is a domain. Let
be an ideal and let
a be a nonzero element in the nonzero ideal
. Set
. Since
is a zero-dim noetherian ring; thus,
artinian, there is an
such that
for all
. We claim
Since it suffices to establish the inclusion locally, we may assume
A is a local ring with the maximal ideal
. Let
x be a nonzero element in
B. Then, since
A is noetherian, there is an
n such that
and so
an+1x\inan+1B\capA\subsetIn+2
. Thus,
Now, assume
n is a minimum integer such that
and the last inclusion holds. If
, then we easily see that
. But then the above inclusion holds for
, contradiction. Hence, we have
and this establishes the claim. It now follows:
B/{aB}\simeqalB/al+1B\subset(alB+A)/al+1B\simeqA/(alB\capA).
Hence,
has finite length as
A-module. In particular, the image of
there is finitely generated and so
is finitely generated. The above shows that
has dimension zero and so
B has dimension one. Finally, the exact sequence
of
A-modules shows that
is finite over
A.
References
- Book: Bourbaki . Nicolas . Nicolas Bourbaki . Commutative algebra . 1989 . Springer . Berlin Heidelberg . 978-3-540-64239-8.
Notes and References
- In this article, a ring is commutative and has unity.
- If
are rings, we say that B is a finite extension of A if B is a finitely generated A module.
- Book: Swanson . Irena . Huneke . Craig . Integral Closure of Ideals, Rings, and Modules . 2006 . Cambridge University Press . 87–88.