- #1
iccanobif
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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!
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!