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

[iccanobif]

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

(2) = \mathfrak{p}^2_2
(3) = \mathfrak{p}^2_3
(5) = \mathfrak{p}^2_5

where \mathfrak{p}_n = (n, \sqrt{30}).

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

Moreover, I've found the relation

\mathfrak{p}_2 \mathfrak{p}_3 \mathfrak{p}_5 = (30).

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

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

Can anyone help me? I'm really confused!

Thanks!

 

[DonAntonio]

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

(2) = \mathfrak{p}^2_2
(3) = \mathfrak{p}^2_3
(5) = \mathfrak{p}^2_5

where \mathfrak{p}_n = (n, \sqrt{30}).

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

Moreover, I've found the relation

\mathfrak{p}_2 \mathfrak{p}_3 \mathfrak{p}_5 = (30).

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

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

Can anyone help me? I'm really confused!

Thanks!

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

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

DonAntonio

 

[iccanobif]

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

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

DonAntonio

My confusion comes from the fact that, although I know it's true what you said above (each \mathfrak{p}_n is its inverse in Cl_K, and also \mathfrak{p}_n \mathfrak{p}_m is principal), my problem is that

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

I must have made a mistake somewhere, but I don't know where!

 

[DonAntonio]

My confusion comes from the fact that, although I know it's true what you said above (each \mathfrak{p}_n is its inverse in Cl_K, and also \mathfrak{p}_n \mathfrak{p}_m is principal), my problem is that

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

I must have made a mistake somewhere, but I don't know where!

Well, can you describe why you thin \,\,\mathfrak{p}_2\mathfrak{p}_3\mathfrak{p}_5=1\,\, ? 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]

(2) = \mathfrak{p}^2_2
(3) = \mathfrak{p}^2_3
(5) = \mathfrak{p}^2_5

\mathfrak{p}_2 \mathfrak{p}_3 \mathfrak{p}_5 = (30).

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