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Intersection of an ideal with a subring (B)

  1. Jun 14, 2014 #1
    In a previous thread, I asked a question different from that I actually intended to ask. Since this question is licit and was answered by micromass, I open this new thread.

    The right question is in fact:

    If R is an integral domain, and R' is INTEGRAL 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, the maximal ideals of R' to the maximal ideals of R, and (not nessarily surjectively) a non prime ideal of R' to a non-prime ideal of R. [NOTE: THIS LAST CLAIM WAS PROVED TO BE FALSE IN THIS THREAD]

    It would be nice if it could be proved that it sends SURJECTIVELY a non-prime ideal to a non-prime ideal, or equivalently, 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'; the example of micromass in the previous thread does not fit here since Z_(2) is not integral over Z. Any ideas for a proof or a counter example ?

    N.B: It is 2:00 in my country, so, I will react to possible rapid answers in several hours.
     
    Last edited: Jun 15, 2014
  2. jcsd
  3. Jun 14, 2014 #2

    micromass

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    Let ##K## be a field. Let ##R = K[x]## the polynomial ring. Let ##R^\prime = K[x,y]/(y^2 - y,xy-1)##. This is an extension of ##R## which is integral since ##y## is the root of the polynomial ##Z^2 - Z= 0##.

    Consider the ideal ##I = (x)## in ##R##. Any ideal in ##R^\prime## which contains ##I## must contain the invertible element ##x##. Thus this ideal must be entire ##R^\prime##.
     
  4. Jun 14, 2014 #3
    It is not true that non-prime ideals get sent to non-prime ideals under this map even for an integral extension. The contraction of a prime ideal will be prime again but this does not imply the contraction of a non-prime ideal is non-prime. As an example just take the rings [itex] R'=\mathbb{Z}[/itex] and [itex] R=\mathbb{Z} [/itex]. Then the ideal [itex] I=(2)_{\mathbb{Z}} [/itex] gets sent to the ideal [itex] \mathfrak{p}=(2)_{\mathbb{Z}} [/itex] when intersected with [itex] \mathbb{Z} [/itex] but [itex] \mathfrak{p} [/itex] is prime whereas [itex] I [/itex] is not a prime ideal.

    So while the map definitely is not a surjective map from non-prime ideals to non-prime ideals, this is not equivalent to saying that we don't get a surjective map from all ideals to all ideals. In the example of the Gaussian integers I gave the map is surjective on all ideals since any ideal is principle and we always have [itex] (a)_{\mathbb{Z}}\cap \mathbb{Z}=(a)_{\mathbb{Z}} [/itex] for any [itex] a\in \mathbb{Z}[/itex].

    [edited out the rest since the above post gives a counterexample.]
     
  5. Jun 15, 2014 #4
    TERANDOL : Indeed, I made a stupid mistake in my proof. I have corrected my post in the previous thread so that further readers will not be influenced by this mistake, and also added a note in the subject of this thread.

    MICROMASS : I don't understand something in your example : clearly, in R', xy=1 and [itex]y^2-y = 0 [/itex], hence multiplying this last equation by x2 leads to [itex]x^2y^2-x^2y=0= 1 - x [/itex]; so x = y = 1. How can R' be an extension of K[x] ?

    Also, in order to give the maximal extent to the counter example I am looking for, let me formulate my question in a stronger way :

    Let R be an integrally closed integral domain, K its fraction field, L a finite algebraic extension of K, and R' the integral closure of R in L. To prove or to disprove : the function that sends an ideal I' of R' to the ideal I' ∩ R of R is surjective from the set of ideals of R' to the set of ideals of R.
     
    Last edited: Jun 15, 2014
  6. Jun 15, 2014 #5

    micromass

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    Yes, you are right. It is not a valid counterexample. I'll think about it tomorrow if nobody else has posted since then.
     
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