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

iccanobif

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!

DonAntonio

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

iccanobif

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!

DonAntonio

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

Hurkyl

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