Class group of a number field \(K\) is a group denotes as \[\begin{equation} \mathop{\mathrm{Cl}}(K) = \{ \text{ Fractional ideals in $\mathop{\mathrm{\mathcal{O}_K}}$}\}/\sim, \end{equation}\] where ideal \(\mathcal{I}_1 \sim \mathcal{I}_2\) if and only if for some \(k \in K^{\times}\), \(\mathcal{I}_1 = k \mathcal{I}_2\).
It is a group under the product of fractional ideals.
Here is an argument for the finiteness of the class group. First a lemma.
There are finitely many ideals of bounded norm in \(\mathop{\mathrm{\mathcal{O}_K}}\).
For each prime ideal \(\mathcal{P}\subseteq \mathop{\mathrm{\mathcal{O}_K}}\), there is a unique prime \(\langle p \rangle = \mathbb{Z} \cap \mathcal{P}\). Also, there are at most \([K:\mathbb{Q}]\) prime ideals above a given prime \(p \in \mathbb{Z}_{\geq 1}\), each of which has norm at least \(p\). Every ideal can be written as a product of powers of finitely many prime ideals. This concluded the proof.
We can then use this argument to prove the following.
The class group \(\mathop{\mathrm{Cl}}(K)\) is finite.
For each fractional ideal \(\mathcal{I}\), we can multiply it with some element of \(K^{\times}\) and find a proper ideal of \(\mathop{\mathrm{\mathcal{O}_K}}\) in the same ideal class. For the proof, we will show that every proper ideal is equivalent to some other ideal of norm \(\leq C\) for some constant \(C>0\).
Let \(e_1,\dots,e_n \in \mathop{\mathrm{\mathcal{O}_K}}\) be a basis of \(\mathop{\mathrm{\mathcal{O}_K}}\) as a \(\mathbb{Z}\)-module. Then, for any \(u = \sum b_i e_i\), we know that \[\begin{equation} \mathop{\mathrm{N}}(u) = \mathop{\mathrm{N}}( \sum b_i e_i) \end{equation}\] must be some polynomial in the \(b_i \in \mathbb{Z}\) of total degree \(n = [K:\mathbb{Q}]\). The coefficients of the polynomial in \(b_i\) is independent of \(u\) and let \(C\) be the sum of the absolute values of those coefficients.
Fix an ideal class \([\mathcal{I}]\) in \(\mathop{\mathrm{Cl}}(K)\) and let \([\mathcal{J}]=[\mathcal{I}^{-1}]\) be such that \(\mathcal{J} \subseteq \mathop{\mathrm{\mathcal{O}_K}}\) is a proper ideal. Consider the finite set \[\begin{equation} S = \left\{ \sum r_i e_i \mid 1 \leq r_i \leq \lfloor \mathop{\mathrm{N}}(\mathcal{J})^{\frac{1}{n}} + 1\rfloor\right\}. \end{equation}\] It is clear that the cardinality of \(S\) is strictly greater than \(\mathop{\mathrm{N}}(\mathcal{J})= \mathop{\mathrm{\#}}\mathop{\mathrm{\mathcal{O}_K}}/\mathcal{J}\). Therefore, it must contain at least two distinct elements of \(\mathop{\mathrm{\mathcal{O}_K}}\) that are equivalent modulo \(\mathcal{J}\). Hence, by taking their difference, we are able to produce an element \(x\) in \(\mathcal{J}\ \{0\}\) and also in \(S-S\). Note that if \(x = \sum b_i e_i\), we can write that \(|b_i| \leq N(\mathcal{J})^{\frac{1}{n}}\).
Now that \(\mathcal{J} \subseteq x \mathop{\mathrm{\mathcal{O}_K}}\) would have an ideal decomposition in \(\mathop{\mathrm{\mathcal{O}_K}}\), we know that there exists a unique integral ideal \(\mathcal{I}'\) such that \[\begin{equation} \mathcal{I}' \mathcal{J} = x \mathop{\mathrm{\mathcal{O}_K}}. \end{equation}\] It is clear that \(\mathcal{I}'\) is a representative of the ideal class \([\mathcal{I}]\). This means that \[\begin{equation} \mathop{\mathrm{N}}(\mathcal{I}) \mathop{\mathrm{N}}(\mathcal{J}) = \mathop{\mathrm{N}}(x) \leq C \max\{ |b_1|,\dots,|b_n| \}^{n} \leq C \mathop{\mathrm{N}}(\mathcal{J}). \end{equation}\] So we get \(\mathop{\mathrm{N}}(\mathcal{I}) \leq C\).
It is the adelic quotient
\[\begin{equation} \tilde{\mathop{\mathrm{Cl}}}(K) = \widehat{ \mathcal{O}_{K} } \backslash \mathbb{G}_m(\mathbb{A}_K)^{(1)} / K^{\times } \end{equation}\] where \(\mathbb{G}_{m}(\mathbb{A}_K)^{(1)}\) is the group of unit norm ideles and \[\begin{equation} \widehat{\mathcal{O}_K} = \prod_{\mathcal{P} \nmid \infty} \mathcal{O}_{\mathcal{P}}. \end{equation}\]
Another equivalent description is that it is the set of \[\begin{equation} (g,\mathcal{I}) \end{equation}\] where \(g \in ({K}\otimes \mathbb{R} )^{\times}\) and \(\mathcal{I}\) is a fractional ideal in \(K\). We then identify all such pairs by \[\begin{equation} (g_1 , \mathcal{I}_1 ) \sim (g_2,\mathcal{I}_2) \Leftrightarrow g_1 \mathcal{I}_1 = c {g_2} \mathcal{I}_2 \text{ for some } c>0, \end{equation}\] where the equivality on the right side is as lattices in \(K \otimes \mathbb{R}\).
It is actually the extension of a torus and a finite group: \[\begin{equation} 0 \rightarrow ( K \otimes {\mathbb{R}} )^{(1)}/\mathop{\mathrm{\mathcal{O}_K}}^{\times} \rightarrow \tilde{\mathop{\mathrm{Cl}}}(K)\rightarrow \mathop{\mathrm{Cl}}(K) \rightarrow 0 \end{equation}\] Topologically, this means that it is finitely many disjoint copies of a torus. The paper [1] presents a very interesting connection of random walks on these groups with worst-case to average-case reductions in Ideal-SVP problem.
This page was updated on November 13, 2023.
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