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.