I am calculating the class group of $\mathbb Q(\sqrt 6)$. My working is as follows:

The Minkowski bound is $\lambda(6)=\sqrt 6<3$ so we only need to look at prime ideals of norm $2$. $2$ divides the discriminant, $24$, of $\mathbb Q(\sqrt 6)$, so $2$ ramifies, and so for any prime ideal $\mathfrak p$ of norm $2$, $\mathfrak p^2$ is principal. But $(2, \sqrt 6)^2=(2)$, so $\mathfrak p=(2, \sqrt 6)$ has norm $2$. If I can show $\mathfrak p$ is not principal, then I can conclude that $Cl(\mathbb Q(\sqrt 6))\cong C_2.$ But if there was an element of $\mathcal O_{\mathbb Q(\sqrt 6)}$ of norm 2, this would correspond to a solution of the diophantine equation $x^2-6y^2=2$. Clearly $x$ is even, so writing $x=2n$ we get $2n^2-3y^2=1$. Now reducing modulo $3$, we get $2n^2\equiv 1 \pmod 3$ which has no soultions. Hence $\mathfrak p$ is not principal and $Cl (\mathbb Q(\sqrt 6)) \cong C_2.$

However, this link says that the class group is trivial. Where have I gone wrong? What am I misunderstanding? Is $(2, \sqrt 6)$ principal?


There are indeed no elements of norm $2$, but there are elements of norm $-2$.

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    $\begingroup$ $(2+\sqrt 6)(\sqrt 6-2)=2 \in (2+\sqrt 6)$ and $2+\sqrt 6 - 2=\sqrt 6 \in (2+\sqrt 6)$. So $(2+\sqrt 6)=(2,\sqrt 6)$ $\endgroup$ – James Apr 28 '15 at 10:46

Where you went wrong is where you overlooked the Diophantine equation $x^2 - 6y^2 = -2$. The fundamental unit of this ring is $5 + 2 \sqrt 6$, which has a norm of 1, not $-1$. This means that if $x^2 - 6y^2 = n$ has no solutions, it does not rule out that $x^2 - 6y^2 = -n$ might.

And indeed $x^2 - 6y^2 = -2$ does have infinitely many solutions, with the most obvious one being $x = 2$, $y = 1$, which you yourself later discovered. Then $(2 + \sqrt 6)(5 + 2 \sqrt 6)$ gives another, and from there you can figure out how to get as many as you want.

More importantly to your query, however, $(2 - \sqrt 6)(2 + \sqrt 6) = -2$, and since $-2 \in \langle 2 \rangle$, combined with the fact that $(2 + \sqrt 6)(3 - \sqrt 6) = \sqrt 6$, we find that both $\langle 2 \rangle$ and $\langle \sqrt 6 \rangle$ are properly contained in $\langle 2 + \sqrt 6 \rangle$, which is not only a principal ideal but a prime ideal after all.

You were right that 2 ramifies, as we have $\langle 2 \rangle = \langle 2 + \sqrt 6 \rangle^2$. In general, if $\mathbb Q(\sqrt{pq})$ (with $p$ and $q$ distinct primes) is a principal ideal domain, you can find factorizations for $\pm p$, $\pm q$ in that domain, and consequently that $\sqrt{pq}$ is reducible.


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