Intersection of ideals with subring

  • Thread starter Thread starter coquelicot
  • Start date Start date
  • Tags Tags
    Intersection
Click For Summary
The discussion revolves around a theorem concerning integral domains and their ideals, specifically regarding the surjective function that maps ideals from an integrally closed ring R' to those in a base ring R. The author acknowledges a previous contribution from a user named Micromass that helped clarify the theorem's implications. They express a desire to prove that this function is a surjection from the ideals of R' to those of R, while noting a specific example involving the ideals of R and R'. The author also corrects a misunderstanding about the conditions of the theorem, clarifying that R' should be integral over R, not merely integrally closed. The conversation highlights the importance of precise definitions in mathematical discussions.
coquelicot
Messages
301
Reaction score
68
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:
Physics news on Phys.org
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:
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.
 
coquelicot said:
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.

Sorry, there was a mistake in my post. Check it again please.
 
  • Like
Likes 1 person
Simple and nice !
 
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.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 5 ·
Replies
5
Views
925
  • · Replies 3 ·
Replies
3
Views
858
  • · Replies 31 ·
2
Replies
31
Views
2K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
5K
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
5K