1

Wikipedia article about Krull-Akizuki theorem gives a very general version of theorem: 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.

Unfortunately I can't find anywhere else such a statement. Typically statement is as above but for noetherian domains and finite extensions of fields of fractions. That includes Bourbaki's Commutative algebra, that is given as a reference in Wikipedias article, Matsumura Commutative ring theory and number of others.

Proof that is given in Wikipedia is only for the case L = K and is taken from Huneke & Swanson Integral Closure of Ideals, Rings, and Modules. There theorem stated for noetherian reduced rings but again only for L = K.

So is general statement from Wikipedia correct? Are there some references or am I missing something and it can be reduced to case of domains?

Thank you!

user468985
  • 81
  • I hope you mean that $A\subset B$, otherwise what you state is false. Assuming this, can you reduce it to the case when $L=K$? – Mohan Jul 17 '19 at 18:09
  • Yes, Mohan $A \subset B$ (there is given that B contains A). Still statement in Wikipedia is not correct, is it? If A is not domain its total ring of fractions is not a field, but it is artinian and has dimension 0. In Huneke & Swanson there is condition that B is not equal to K. – user468985 Jul 17 '19 at 19:18
  • Reducing it to the case when L = K is what I'm trying to do. Idea is that if L is finite ring extension then there is finite generating set $a_1,...,a_n$ of L as K-module. Looking at their multiplication table and multiplying denominators it is possible to take $b_i = pa_i$ where p is not zero divisor of A and A-module generated by 1 and $b_i$ will be a ring. So will be B-module generated by 1 and $b_i$. They will be finite extensions so will be noetherian iff A, B are and will have the same dimension. Also L is total ring of fractions of this extension of A. – user468985 Jul 17 '19 at 19:48
  • So thats how it an be reduced to the case L = K. Is it correct? – user468985 Jul 17 '19 at 19:51
  • Good, you have the idea. – Mohan Jul 17 '19 at 20:19

0 Answers0