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About integral extensions

  1. Mar 4, 2014 #1
    I think I have proved the following theorem:

    "If R is an integral domain, x is integral of degree n over R, and K is the fraction field of R, then the intersection of R[x] with K is included in 1/n R[X] (the set of elements r/n where r belongs to R). Furthermore, if R is Dedekind, then this intersection is R."

    My simple proof can be found here:
    https://upload.wikimedia.org/wikipedia/commons/d/de/Bensimhoun-1.lemma_in_Galois_Theory-2.RxInterQuotR-3.conjugates_of_polynomial.pdf [Broken] (pp. 3--5).

    My question is: Where can I found reliable references to this theorem, or at least reliable references to another theorem implying it ?

    Thx.
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 4, 2014 #2

    micromass

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    The statement about Dedekind domains seems to be the same as saying that Dedekind domains are integrally closed. Chapter 9 of Atiyah-MacDonald should be relevant here.
     
  4. Mar 4, 2014 #3

    micromass

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    So let's take an integral domain ##R## and take any ##\theta\in K## which is integral over ##R##. Since ##\theta\in K##, it follows that ##n=1##. So your theorem proves that

    [tex]R[\theta]\cap K\subseteq R[/tex]

    So your theorem implies that any integral domain is integrally closed. There are counterexamples to that, so where did I go wrong?
     
  5. Mar 4, 2014 #4
    If x belongs to K and is integral over R, then obviously x belongs to R. so, the intersection of R[X] with R is R. In your statement, you infer from a particular case to the general case, and this is what is wrong.
     
  6. Mar 4, 2014 #5

    micromass

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    Consider ##R= \mathbb{C}[t^2,t^3]##, then ##K = \mathbb{C}(t)##. But then ##t\in K## is integral over ##R## since it's the root of

    [tex]X^2 - t^2 = 0[/tex]

    but ##t## is not in ##R##.
     
  7. Mar 4, 2014 #6
    Sorry, you are right about the fact that an element of K integral over R needs not belong to R. In fact, the problem in your counterexample above is that you assert that n=1, which is, of course wrong unless x belong to R.
     
  8. Mar 4, 2014 #7

    micromass

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    I just take an arbitrary element ##\theta## in ##K\setminus R## that is integral over ##R##. My previous post shows that such situation exist.
    Then since ##\theta\in K##, it follows that ##n=1## (since your definition of ##n## was the degree of ##\theta## over ##K##). Why is this wrong?
     
  9. Mar 4, 2014 #8
    You are, once more right. I have understood my mistake, and that my theorem is false. Thank you so many for your help.
     
  10. Mar 4, 2014 #9

    micromass

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    You did prove something. So I think that you should carefully check the proof and see where your error lies. Maybe with some additional assumptions, it might still work?
     
  11. Mar 4, 2014 #10
    Of course, if it is supposed that the minimal polynomial of theta is also monic with coefficients in R, this works. But this makes the theorem much more uninteresting.
     
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