# Intersection of ideals with subring

1. Jun 14, 2014

### coquelicot

Hello,

Thanks to the help of micromass in a previous thread, I am now able to prove the following theorem (which can be seen as a (somewhat improved) version of the "going up" and "going down" theorems):

If R is an integral domain, and R' is integrally closed over R, then the function f which assigns to an ideal I' of R' the ideal I = I' ∩ R, sends surjectively the prime ideals of R' to the prime ideals of R and the maximal ideals of R' to the maximal ideals of R.

It would be nice if it could be proved that f is a surjection from the set of ideals of R' to the set of ideals of R. But for the moment, I can only see that every ideal I of R is included in a maximal ideal of R'; any ideas for a proof or a counter example ?

Last edited: Jun 15, 2014
2. Jun 14, 2014

### micromass

Let $R=\mathbb{Z}$, $R^\prime = \mathbb{Z}_{(2)} = \{a/b\in \mathbb{Q}~\vert~\textrm{gcd}(a,b)=1,~2~\text{does not divide}~b\}$ and $I = 6\mathbb{Z}\subseteq R$.

Then any ideal $J$ of $R^\prime$ which contains $I$ also contains $2$. So the intersection $J\cap R$ must contain $2$ and can thus not equal $6\mathbb{Z}$.

Also, the going up/down theorems are a bit more general than what you are stating here. They deal with sequences of prime ideals.

Last edited: Jun 14, 2014
3. Jun 14, 2014

### coquelicot

Thanks again Micromass, but in fact, I supposed implicitly that R' contains R, a condition that I have forgotten to write explicitly in my question (sorry). So, your example does not fit with this additional condition.

Following your remark, I will also examine what is exactly the going up and going down theorem.

4. Jun 14, 2014

### micromass

Sorry, there was a mistake in my post. Check it again please.

5. Jun 14, 2014

### coquelicot

Simple and nice !

6. Jun 14, 2014

### coquelicot

After reconsidering the example, I realized that I have, once more, asked a question different from what I meant. I wrote "R' integrally closed over R", while I actually meant "R' integral over R". I'm really sorry for this mistake, but after all, this is a licit question and micromass found an answer; so, since may be usefull to other persons, I will open a new thread and ask the right question.