I have calculated some real quadratic field 's Hilbert class field with class number $$2$$,and I found they were satisfied $$Gal(H_{K}/Q)\cong Z/2Z\oplus Z/2Z$$,here $$H_{K}$$ is the Hilbert class field of a real quadratic field $$K$$ whose class number is $$2$$.Is it true for all quadratic field with class number $$2$$? How to prove it?(ps:I'm a beginner of algebraic number theory ,and I'm very interested in Hlibert class field .)
• If there exist a prime $p$,that $Q(\sqrt p)$ has class number $2$,this could be a counterexample of my problem. – fool rabbit Mar 29 at 13:02
This is a consequence of "genus theory". If $$K / \mathbf{Q}$$ is an abelian extension, then the "genus field" of $$K$$ is the maximal extension $$L / K$$ such that $$L/K$$ is unramified and $$L/\mathbf{Q}$$ is abelian. You can read more about genus theory here: https://en.wikipedia.org/wiki/Genus_field.
In your case, the fact that $$H_K$$ is Galois over $$\mathbf{Q}$$ and $$[H_K : \mathbf{Q}] = 4$$ implies that $$H_K$$ must be abelian over $$\mathbf{Q}$$ (since there are no nonabelian groups of order 4); so $$H_K$$ coincides with the genus field $$L_K$$. The genus field of of a quadratic field is always a composite of quadratic fields, so $$H_K$$ must have Galois group $$C_2 \times C_2$$.