Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Intersection of ideals with subring

  1. Jun 14, 2014 #1

    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. jcsd
  3. Jun 14, 2014 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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
  4. Jun 14, 2014 #3
    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.
  5. Jun 14, 2014 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Sorry, there was a mistake in my post. Check it again please.
  6. Jun 14, 2014 #5
    Simple and nice !
  7. Jun 14, 2014 #6
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Intersection of ideals with subring
  1. Subring Test (Replies: 4)

  2. Stuck on a Subring (Replies: 3)

  3. Verifying a subring? (Replies: 1)