Problem calculating the power of an ideal in quadratic field

Question

I want to prove that $h_K=2$ if $K=\mathbb{Q}[\sqrt{-6}]$. Using Minkowski Theorem I have that $Cl_K=\{(1),(3,\sqrt{-6}),(2,\sqrt{-6})\}$, and I thought it was a good idea to use Lagrange Theorem (order of an element divides order of the gorup).

The main problem is that I can't reduce $(2,\sqrt{-6})^2$:

$$(2,\sqrt{-6})(2,\sqrt{-6})=(4,2\sqrt{-6},2\sqrt{-6},-6)=(2,\sqrt{-6})$$

so $I^2=I$, but I think there is a mistake, since I should end up with $I^2=(1)$.

Answer 1

I will display the answer as an answer, rather than as a simple comment. The above computation line is "almost ok", correctly:

$$(2,\sqrt{-6})(2,\sqrt{-6})=(4,2\sqrt{-6},2\sqrt{-6},-6)=(2)\ .$$

For the last equality, we have to show the double inclusion. It is clear that from $4,-6$ we can exhibit the $2$ by sum. Conversely, $2$ is dividing all generators.

$\blacksquare$

Comment: Note that (for instance) sage can perform/check this and similar computations using a mathematically affine thinking and notation. or instance:

sage: K.<a> = QuadraticField(-6)
sage: K
Number Field in a with defining polynomial x^2 + 6
sage: a^2
-6
sage: K.class_group()
Class group of order 2 with structure C2 of Number Field in a with defining polynomial x^2 + 6
sage: K.class_group().gens()
(Fractional ideal class (2, a),)
sage: J = K.ideal( 2,a )
sage: J^2
Fractional ideal (2)

(There are many computer algebra systems, that can be used as an aid for calculations in number fields, sage is a free one, and collects "batteries" from all related free maths soft, so that there is a seamless way to switch between fields without contortion.)

The comment is here, since having its answer, we also have the answer to the posted question.

Answer 2

$$(2,\sqrt{-6})^2=(2,2\sqrt{-6},\,-6)=(2)\ldots$$

Answer 3

It's also worth noting that the ideals $$ (3, \sqrt{-6}) \quad\text{and}\quad (2, \sqrt{-6}) $$ belong to the same ideal class in $\mathrm{Cl}(K)$ since $$ (3, \sqrt{-6})(2, \sqrt{-6}) = (6, 2\sqrt{-6}, 3\sqrt{-6}, 6) = (\sqrt{-6}) $$ and both ideals are equal to their inverse in the class group, since they're of order $2$.

Thus your $(3, \sqrt{-6})$ generator in $\mathrm{Cl}(K)$ is redundant. This will give you $$ \mathrm{Cl}(K) = \{(1), (2, \sqrt{-6})\} $$ so that $h_{k} = 2$, as required.