Krull–Akizuki theorem

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 $A\subset B\subset L$ and ''B'' is reduced,
then ''B'' is a noetherian ring of dimension at most one. Furthermore, for every nonzero ideal $I$ of ''B'', $B/I$ 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 $A\subset B\subset KB$ and ''KB'' is a finite extension of ''K'', so we may assume without loss of generality that $L=KB$.
Then $L=Kx_1+\cdots+Kx_n$ for some $x_1,\dots,x_n\in B$.
Since each $x_i$ is integral over ''K'', there exists $a_i\in A$ such that $a_ix_i$ is integral over ''A''.
Let C=A _1x_1,\dots,a_nx_n/math>.
Then ''C'' is a one-dimensional noetherian ring, and $C\subset B\subset Q(C)$, where $Q(C)$ denotes the total ring of fractions of ''C''.
Thus we can substitute ''C'' for ''A'' and reduce to the case $L = K$.

Let $\mathfrak_i$ be minimal prime ideals of ''A''; there are finitely many of them. Let $K_i$ be the field of fractions of $A/$ and $I_i$ the kernel of the natural map $B \to K \to K_i$. Then we have:
:$A/ \subset B/ \subset K_i$ and $K\simeq\prod K_i$.
Now, if the theorem holds when ''A'' is a domain, then this implies that ''B'' is a one-dimensional noetherian domain since each $B/$ is and since $B \simeq \prod B/$. 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
:$a^l B \subset a^B + A.$
Since it suffices to establish the inclusion locally, we may assume ''A'' is a local ring with the maximal ideal $\mathfrak$. Let ''x'' be a nonzero element in ''B''. Then, since ''A'' is noetherian, there is an ''n'' such that $\mathfrak^ \subset x^ A$ and so $a^x \in a^B \cap A \subset I_$. Thus,
:$a^n x \in a^ B \cap A + A.$
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 $a^n x \in I_$. But then the above inclusion holds for $n-1$, contradiction. Hence, we have $n = l$ and this establishes the claim. It now follows:
:$B/ \simeq a^l B/a^ B \subset (a^B + A)/a^ B \simeq A/(a^B \cap A).$
Hence, $B/$ has finite length as ''A''-module. In particular, the image of $I$ there is finitely generated and so $I$ is finitely generated. The above shows that $B/$ has dimension at most zero and so ''B'' has dimension at most one. Finally, the exact sequence $B/aB\to B/I\to (0)$ of ''A''-modules shows that $B/I$ is finite over ''A''. $\square$

References

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{{DEFAULTSORT:Krull-Akizuki theorem
Theorems in algebra
Commutative algebra