Krull–Akizuki theorem

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

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

${\displaystyle A/{{\mathfrak {p}}_{i}}\subset B/{I_{i}}\subset K_{i}}$ and ${\displaystyle K\simeq \prod K_{i}}$.

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

${\displaystyle a^{l}B\subset a^{l+1}B+A.}$

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

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

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

${\displaystyle B/{aB}\simeq a^{l}B/a^{l+1}B\subset (a^{l+1}B+A)/a^{l+1}B\simeq A/(a^{l+1}B\cap A).}$

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

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