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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
    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
    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.
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