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Confusion: computation of class number of K=Q(sqrt(30))

  1. Apr 27, 2012 #1
    As an exercise, I'm trying to compute the class number of [itex]K = \mathbb{Q}(\sqrt{30})[/itex]. By the Minkowski bound, I just need to consider the prime ideals which divide 2,3,5.
    I've found that

    [itex] (2) = \mathfrak{p}^2_2 [/itex]
    [itex] (3) = \mathfrak{p}^2_3 [/itex]
    [itex] (5) = \mathfrak{p}^2_5 [/itex]

    where [itex] \mathfrak{p}_n = (n, \sqrt{30}) [/itex].

    Also, using Legendre symbols it's easy to see that none of the [itex] \mathfrak{p}_n, n = 2,3,5 [/itex], is principal.

    Moreover, I've found the relation

    [itex] \mathfrak{p}_2 \mathfrak{p}_3 \mathfrak{p}_5 = (30) [/itex].

    Now, these relations make me think that the class group is of order 4 (generated by [itex] \mathfrak{p}_2, \mathfrak{p}_3 [/itex], with the two generators in different ideal classes because of the last relation and the fact that [itex] \mathfrak{p}_5 [/itex] is not principal).

    BUT: I know that the class number should be 2.

    Can anyone help me? I'm really confused!

    Thanks!
     
  2. jcsd
  3. Apr 27, 2012 #2


    An idea: it's not hard to show that [itex]\mathfrak{p}_k^{-1}=\mathfrak{p}_k\,,\,\,k=2,3,5[/itex], so you'd only need to show

    that all these ideal cosets are the in fact the same, and indeed [tex](2,\sqrt{30})(3,\sqrt{30})=(\sqrt{30}) \Longrightarrow \mathfrak{p}_2=\mathfrak{p}_3\,\,\,and\,\,\,etc.[/tex]

    DonAntonio
     
    Last edited: Apr 28, 2012
  4. Apr 28, 2012 #3
    My confusion comes from the fact that, although I know it's true what you said above (each [itex]\mathfrak{p}_n[/itex] is its inverse in [itex]Cl_K[/itex], and also [itex]\mathfrak{p}_n \mathfrak{p}_m[/itex] is principal), my problem is that

    [itex]\mathfrak{p}_2 \mathfrak{p}_3 = 1[/itex]
    and
    [itex]\mathfrak{p}_2 \mathfrak{p}_3 \mathfrak{p}_5 = 1[/itex]
    imply that
    [itex]\mathfrak{p}_5 = 1[/itex]
    which it's not true, as [itex]\mathfrak{p}_5[/itex] is not principal.

    I must have made a mistake somewhere, but I don't know where!
     
  5. Apr 28, 2012 #4


    Well, can you describe why you thin [itex]\,\,\mathfrak{p}_2\mathfrak{p}_3\mathfrak{p}_5=1\,\,[/itex] ? As far as I can see, the product of the first

    two already is 1, so how come when the third one comes you still get 1?

    DonAntonio
     
  6. Apr 28, 2012 #5

    Hurkyl

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    That can't be right. Your equations above say that (30) is the square of the left hand side.
     
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