# Can someone explain to me the last two steps of finding the ideal class group

First I have to know the degree of the ring of algebraic integers, which is easy because I seem to be limited to quadratic integer rings.

The degree is needed to calculate the Minkowski bound, which is especially easy for imaginary quadratic integer rings.

Then factorize the rational primes below the Minkowski bound. In a principal ideal domain, the primes that split or ramify as ideals also split or ramify as numbers. We conclude the ideal class group is trivial. It seems a little magical, but it also feels like it would make sense if I had a younger, more flexible mind, and everyone seems to be consistent with the PID examples.

The real trouble for me is when the class number is greater than $1$. I've looked at examples in the Alaca & Williams book, I've looked at a Lenstra paper online, and I've looked at the "similar questions" on here, but it seems like everyone takes some mysterious different steps depending on what $d$ is.

Both Alaca & Williams and Lenstra demonstrate $\mathbb{Z}[\sqrt{-14}]$. Minkowski bound tells us to look at $\langle 2 \rangle$ and $\langle 3 \rangle$. And then for some reason I don't understand, they both start talking about $2 - \sqrt{-14}$.

To preface the $\mathbb{Z}[\sqrt{-65}]$ example, Alaca & Williams then say something about calculating $k + \sqrt{d}$ for $k = 1, 2$ up to some unspecified bound. The Lenstra example for $\mathcal{O}_{\mathbb{Q}(\sqrt{-23}}$ is also confusingly different from the $\mathbb{Z}[\sqrt{-14}]$.

I could be wrong, but $$\langle 2 \rangle = \langle 2, \sqrt{-14} \rangle^2$$ and $$\langle 3 \rangle = \langle 3, 1 - \sqrt{-14} \rangle \langle 3, 1 + \sqrt{-14} \rangle,$$ and we needn't worry about $\langle 2, \sqrt{-14} \rangle \langle 3, 1 \pm \sqrt{-14} \rangle$ because those ideals have a norm of $6$ and that's above the Minkowski bound.

According to Alaca & Williams, if I'm understanding correctly, the next step here is to figure out the class generators of the ideals $\langle 2, \sqrt{-14} \rangle$ and $\langle 3, 1 \pm \sqrt{-14} \rangle$. The index has been no help to find out about "class generators." The information is buried in the book somewhere that it seemed unmotivated. Help, anyone?

The advantage of this method is the speed and accuracy with which one may find the Gauss reduced forms of a discriminant.

I wrote a blog post on this, MSE has discontinued the blog.

A binary quadratic form, with integer coefficients, is some $$f(x,y) = A x^2 + B xy + C y^2.$$ The discriminant is $$\Delta = B^2 - 4 A C.$$ We will abbreviate this by $$\langle A,B,C \rangle.$$ It is primitive if $\gcd(A,B,C)=1.$

Here is the mapping from forms to ideals: given $\langle A,B,C \rangle,$ drop the letter $C.$ That's it. $$\langle A,B,C \rangle \mapsto \left[ A, \frac{B + \sqrt \Delta}{2} \right].$$ Oh, why is this an ideal, rather than just some $\mathbf Z$-lattice? Because, given $\alpha,\beta$ rational integers, $$\left[ \alpha, \frac{\beta + \sqrt \Delta}{2} \right]$$ is an ideal if and only if $$4 \alpha | ( \Delta - \beta^2 ).$$

You mention some square roots, $-14,$ $-65,$ $-23.$

The first is discriminant $-56,$ principal genus $$\langle 1, 0, 14 \rangle, \; \; \langle 2, 0, 7 \rangle.$$ The other genus is $$\langle 3, 2, 5 \rangle, \; \; \langle 3, -2, 5 \rangle.$$

The second is discriminant $-260,$ with four genera. There is an answer by Lubin at Ideal class group of $\mathbb{Q}(\sqrt{-65})$ Principal genus $$\langle 1, 0, 65 \rangle, \; \; \langle 9,8,9 \rangle.$$ The next genus is $$\langle 5,0,13 \rangle, \; \; \langle 2,2,33 \rangle,$$ next $$\langle 3, 2, 22 \rangle, \; \; \langle 3, -2, 22 \rangle,$$ next $$\langle 6, 2, 11 \rangle, \; \; \langle 6, -2, 11 \rangle,$$

Last discriminant is $-23,$ there is just one genus as $23$ is prime $$\langle 1, 1, 6 \rangle, \; \; \langle 2,1,3 \rangle, \; \; \langle 2,-1,3 \rangle.$$

The book people like on this material is Cox, Primes of the Form $x^2 + n y^2.$