Localization of two rings which is an integral extension, then integral extension still holds?

Question

Question seems simple, but I just can't find the solution.

Let $A/B$ be an integral ring extension and let $P$ be a prime ideal of $B$. By lying-over theorem, there is $Q$, a prime ideal of $A$, lying over $P$. Then the ring of fractions of $A$ localized on $Q$ is still integral on that of $B$ localized on $P$?

score 2 · Answer 1 · edited Dec 20 '13 at 16:06

2

Hint: consider an example.

Here is one: $A = \mathbb Z[i] \supset \mathbb Z = B$, $P = 5\mathbb Z$, $Q = (2+i)\mathbb Z[i].$

The element $1/(2-i)$ lies in $A_Q$. Is it integral over $B_P$?

edited Dec 20 '13 at 16:06

answered Dec 19 '13 at 04:38

Matt E

123,735

thanks for your exemple. I give a proof in the case for the field of fractions, when this proposition is true. link – user83008 Dec 20 '13 at 00:19

Localization of two rings which is an integral extension, then integral extension still holds?

1 Answers1