# 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

$A/{{\mathfrak {p}}_{i}}\subset B/{I_{i}}\subset K_{i}$ .
$a^{l}B\subset a^{l+1}B+A.$ 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. $\square$ 