I am struggeling to prove the above. What I did so far:
Let $R:=\mathbb{Z}[\sqrt{-6}]$. The Minkowskibound yields 3.11…, hence I know that every idealclass contains an ideal of norm 1,2 or 3. The case 1 yields the principal class.
Let a be an ideal of Norm $2$, hence a contains the ideal $2R$, so we have to analyze what happens to $2R$ in $R$. But since $2$ divides $-6$, it is known that $2R$ is ramified, namely $$2R=(2,\sqrt{-6})^2$$
The same is true for $3R$, namely
$$3R=(3,\sqrt{-6})^2$$
Now we want to show, that $(2,\sqrt{-6})$ is not a principal ideal. If so, since $(2,\sqrt{-6})$ has norm $2$, there would be an $\alpha\in R$ with $(2,\sqrt{-6})=(\alpha)$ and $N(\alpha)=2$. But since $a^2 +6b^2=2$ has no solutions in $\mathbb{Z}^2$, there is no such element and hence $(2,\sqrt{-6})$ is not principal. Same is true for $(3,\sqrt{-6})$.
To prove that the classnumber is $2$, all I have to show now is that $(2,\sqrt{-6})$ and $(3,\sqrt{-6})$ are in the same idealclass.
Please correct me if anything I did so far was wrong!
However, at this point I am not sure how to actually show that these ideals are in the same class. I simply tried to apply the defintion, but I did not come to a point where I could deduce anything.
This also arises the question for me. Is there a general approach to check if two ideals lie in the same class?
I really appreciate any help on this!
Best Answer
We can do this just using the prime ideal decompositions $(2)= P^2$ and $(3) = Q^2$ for some prime ideals $P$ and $Q$ without even knowing generators for $P$ and $Q$. Since $$ (6) = P^2Q^2 = (PQ)^2 = (\sqrt{-6})^2, $$ by unique factorization of ideals we have $PQ = (\sqrt{-6})$. Thus in the ideal class group, $[PQ] = [(1)]$, so $[P] = [Q]^{-1}$. Since $Q^2$ is principal, $[Q]^2 = [(1)], so $[Q]^{-1} = [Q]$. Thus $[P] = [Q]$.
If you wanted to do this with explicit generators, $$ (2,\sqrt{-6})(3,\sqrt{-6}) = (6,2\sqrt{-6},3\sqrt{-6},-6) = (\sqrt{-6})(\sqrt{-6},2,3,\sqrt{-6}) $$ and the second ideal on the right is $(1)$ since it contains $2$ and $3$. Thus $(2,\sqrt{-6})(3,\sqrt{-6}) = (\sqrt{-6})$, which is a concrete way of saying $PQ = (\sqrt{-6})$ above.