# How to calculate the ideal class group of a quadratic number field?

The books I use to study Algebraic Number Theory are rather thin on the ground with concrete examples, so I make my own and check the results with Sage.

To get some more hands on experience I want to calculate the order, generators and structure of class groups of quadratic number fields. But how do I start? Is there a step-by-step approach? If so, please list the steps ( or point to a reference ) and illustrate with an example.

Sage results:

• Minkowski's bound is usually a good place to start. See en.wikipedia.org/wiki/Minkowski%27s_bound – Alex Wertheim Dec 1 '14 at 16:13
• Yes, I also read the article on the Ideal class group where the minkowski bound is mentioned. Explaining the example in that article (Ideal class group) might be an answer I dont get that either. – nilo de roock Dec 1 '14 at 16:21
• You might want to check out Chapter 5 in Daniel A. Marcus's Number Fields. – clear Jul 21 at 10:24

This is covered in the lecture notes on Number Rings by Peter Stevenhagen.

For quadratic number fields $K = \mathbf{Q}(\sqrt{d})$ we know the ring of integers $\mathcal{O}_K = \mathbf{Z}[\alpha]$ with $\alpha = (1+\sqrt{d})/2$ if $d \equiv 1 \pmod 4$ and $\alpha = \sqrt{d}$ otherwise, with discriminant $\Delta_K = d$ and $\Delta_K = 4d$ respectively. The ideal class group is generated by the primes of norm less than the Minkowski bound, which is $M_K = \frac{2}{\pi}\sqrt{|\Delta_K|}$ if $d < 0$ and $M_K = \frac{1}{2}\sqrt{|\Delta_K|}$ if $d > 0$.

These primes in $\mathbf{Z}[\alpha]$ lie above (are factors of) rational primes less than $M_K$. Rational primes can be factored using the Kummer-Dedekind theorem (3.1 in the notes). For quadratic fields we in fact have a description of the factorization of rational primes in $\mathbf{Z}[\alpha]$ in terms of the Legendre symbol: see corollary 3.11 in the notes and exercise 5 following it.

It remains to find relations between these primes, which can be done by factoring ideals of the form $(k-\alpha)$ with $k \in \mathbf{Z}$, which have norm $N_{K/\mathbf{Q}}(k-\alpha) = f^\alpha_\mathbf{Q}(k)$. Since we need relations between prime ideals of small norm, we take values of $k$ for which $f^\alpha_{\mathbf{Q}}(k)$ is small. No rational primes divide $k - \alpha$, so in the quadratic case only primes $\mathfrak{p}$ of degree $f(\mathfrak{p}/p) = 1$ can occur, and if such a prime occurs then the other prime $\mathfrak{q}|p$ does not occur.

To show that a prime has a particular order in the class group, we have to show that all lower powers are not principal. In an imaginary quadratic field ($d < 0$), the norm equation can often be used to show that an ideal is not principal: assume that it is principal and consider the norm of a generator.

Let me illustrate this for $K = \mathbf{Q}(\sqrt{-30})$, which is the second example in your screenshot. Since $-30 \equiv 2 \pmod 4$ we have $\mathcal{O}_K = \mathbf{Z}[\sqrt{-30}]$ and $\sqrt{-30}$ has minimal polynomial $f = x^2 + 30$. The Minkowski constant in this case is $M_K = \frac{2}{\pi}\sqrt{30} < 9$, so we only have to factor the primes $2, 3, 5, 7$. Corollary 3.11 says that $p$ is split in $\mathbf{Z}[\sqrt{d}]$ for $(\frac{d}{p}) = 1$, inert for $(\frac{d}{p}) = -1$ and ramified for $(\frac{d}{p}) = 0$. Hence $2, 3$ and $5$ are ramified and $7$ is inert. We have $(2) = (2,\sqrt{-30})^2 = \mathfrak{p}_2^2$, $(3) = (3,\sqrt{-30})^2 = \mathfrak{p}_3^2$ and $(5) = (5,\sqrt{-30})^2 = \mathfrak{p}_5^2$ by Kummer-Dedekind. The class group is generated by $\mathfrak{p}_2, \mathfrak{p}_3$ and $\mathfrak{p}_5$.

From $f(0) = 30 = 2\cdot 3 \cdot 5$ we deduce $(\sqrt{-30}) = \mathfrak{p}_2\mathfrak{p}_3\mathfrak{p}_5$, or $[\mathfrak{p}_2] + [\mathfrak{p}_3] + [\mathfrak{p}_5] = 0$ in additive notation (where the brackets denote the class), so $\mathfrak{p}_2$ and $\mathfrak{p}_3$ alone generate the class group. We show that their classes are not equal and both are non-principal, so that the class group has order $4$ with structure $C_2 \times C_2$.

If $[\mathfrak{p}_2] = [\mathfrak{p}_3]$ then $\mathfrak{p}_3 = (\beta)\mathfrak{p}_2$ for some $\beta \in \mathbf{Q}(\sqrt{-30})$, and upon squaring we obtain $(3) = (2\beta^2)$. Since the equation $x^2 + 30y^2 = 1$ has only the solutions $(\pm 1, 0)$, the unit group of $\mathbf{Z}[\sqrt{-30}]$ is $\{\pm 1\}$, so we would have $3 = \pm 2\beta^2$, which is not possible for any $\beta \in \mathbf{Q}(\sqrt{-30})$.

If we would have $\mathfrak{p}_2 = (2,\sqrt{-30}) = (\beta)$ for some $\beta = x + y\sqrt{-30}$, then we would have $\beta|2$ and $\beta \neq \pm 2$, so $2 = N(\beta) = x^2 + 30y^2$, but this has no integral solutions $(x,y)$ so $\mathfrak{p}_2$ is non-principal.

Similarly, $\mathfrak{p}_3$ is non-principal because $x^2 + 30y^2 = 3$ has no integral solutions.

More examples can be found in the notes; particularly in chapter 7.

• An (insignificant?) errata: actually $\Delta_K=-120$, and $M_K\approx6.97$, so we only have to consider primes 2, 3, and 5 in $\mathbb Z$. – The Great Seo Dec 11 '16 at 13:07
• Hi, the class group is a multiplicative group but you are saying that $[\mathfrak{p}_2] + [\mathfrak{p}_3] + [\mathfrak{p}_5] = 0$ "in the additive notation". Can you please explain what do you mean by that and what is the reason for mentioning it? – Ninja Dec 20 '17 at 18:01