4 The ideal class group
Recall that in a number field \(K\), a principal fractional ideal is a fractional ideal of the form \((x)=\{ xy| y \in \mathcal{O}_K\} \) for \(x \in K\). For example, principal ideals are principal fractional ideals. Note that furthermore, if \((x)\) and \((y)\) are two principal fractional ideals, then \((x)(y)=(xy)\) is also a fractional ideal and moreover, \((x)(x^{-1})=\mathcal{O}_K\). Thus the set of principal fractional ideals forms a subgroup of \(J_K\) (the group of fractional ideal as defined in Proposition 3.3.11). Let \(P_K\) denote the subgroup of principal fractional ideals.
The class group of \(K\) is defined to be quotient group
Its elements are called ideal classes. If \(\mathfrak {a}\) is a fractional ideal, we let \([\mathfrak {a}]\) denote its class in \(\operatorname{Cl}_K\). We let \(h_K\) denote the size of \(\operatorname{Cl}_K\) (which we will see below is finite), this is called the class number of \(K\).
Note that in the class group, the identity element is given by the class of principal fractional ideals, which we denote by \([1]\) or \([(1)]\).
Furthermore, note that if \(\mathfrak {a}\) and \(\mathfrak {b}\) are fractional ideals then
means that there is some principal fractional ideal \((\alpha )\) such that \(\mathfrak {a}=(\alpha )\mathfrak {b}\). So as a SET \([\mathfrak {a}]=\{ (x)\mathfrak {a}\mid x \in K\} \).
Note that, \(\mathcal{O}_K\) is a PID if and only if \(\operatorname{Cl}_K\) is trivial. Moreover, since a Dedekind domain is a PID if and only if its a UFD, this means that \(\mathcal{O}_K\) is a UFD if and only if \(\operatorname{Cl}_K\) is trivial. So \(\operatorname{Cl}_K\) can be thought of measuring how far \(\mathcal{O}_K\) is from being a UFD.
Let \(K\) be a number field. Then \(\operatorname{Cl}_K\) is finite.
To prove this theorem we will make use of the following result, which we will prove later.
Let \(K\) be a number field with \(r_1\) real embeddings and \(r_2\) conjugate pairs of complex embeddings. Let \([K:\mathbb {Q}]=n\) and let \(\mathfrak {a}\) be an ideal of \(\mathcal{O}_K\). Then there is an element \(a \in \mathfrak {a}\) such that
The quantity \( \frac{n!}{n^n} \left( \frac{4}{\pi } \right)^{r_2} |\Delta (\mathcal{O}_K)|^{1/2}\) is known as the Minkowski bound and we will denote it by \(M_K\).
Now, we have:
Let \(K\) be a number field and let \(C\) be an ideal class in \(\operatorname{Cl}_K\). Then \(C\) contains an ideal \(\mathfrak {a}\) in \(\mathcal{O}_K\) such that
Consider the class \(C^{-1}\). This has a representative which is an ideal \(\mathfrak {b}\) of \(\mathcal{O}_K\). By Theorem 4.0.3 there is an \(x \in \mathfrak {b}\) such that \(|N_{K/\mathbb {Q}}(x)| \leq M_K N(\mathfrak {b})\).
Set \(\mathfrak {a}=\mathfrak {b}^{-1} (x)\). Then this is in the same class as \(C\) since its a multiple of \(\mathfrak {b}^{-1}\) by a principal ideal \((x)\). Moreover, since \(x \in \mathfrak {b}\) it follows that \(\mathfrak {b}^{-1}(x) \subset \mathcal{O}_K\) (recall that \(\mathfrak {b}^{-1}=\{ y \in K | y\mathfrak {b}\subset \mathcal{O}_K\} \) ), so \(\mathfrak {a}\) is a proper ideal. Now by construction
which gives the result.