Krull–Akizuki theorem

In algebra, the Krull–Akizuki theorem states the following: let A be a one-dimensional reduced noetherian ring, K its total ring of fractions. If B is a subring of a finite extension L of K containing A and is not a field, then B is a one-dimensional noetherian ring. 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

Here, we give a proof when $L=K$. 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:

$A/{{\mathfrak {p}}_{i}}\subset B/{I_{i}}\subset K_{i}$.

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 $B=\prod B/{I_{i}}$. 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. Finally, the above shows that $B/{aB}$ has zero dimension and so B has dimension one. $\square$

References

• Nicolas Bourbaki, Commutative algebra