# 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 $a_{i}x_{i}$ is integral over A. Let $C=A[a_{1}x_{1},\dots ,a_{n}x_{n}]$. 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 ${\mathfrak {p}}_{i}$ 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:
    $A/{{\mathfrak {p}}_{i}}\subset B/{I_{i}}\subset K_{i}$ 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 $B/{I_{i}}$ is and since $B\simeq \prod B/{I_{i}}$. Hence, we reduced the proof to the case A is a domain. Let $0\neq I\subset B$ be an ideal and let a be a nonzero element in the nonzero ideal $I\cap A$. Set $I_{n}=a^{n}B\cap A+aA$. Since $A/aA$ is a zero-dim noetherian ring; thus, artinian, there is an $l$ such that $I_{n}=I_{l}$ for all $n\geq l$. 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 $n\geq l$ 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. The above shows that $B/{aB}$ 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. $\square$