TANT 20 - Class field theory in terms of ideals

Hello there! These are notes for the twentieth class of the course "Topics in algebra and number theory" held in Block 4 of the academic year 2017/18 at the University of Copenhagen.

In the previous lecture we used the idelic class group $C_K$ to give the statement of the main theorem of global class field theory. Moreover, we related this group to the "traditional" class group $\mathfrak{C}_K$ by proving that $$\mathfrak{C}_K \cong \frac{C_K}{(\widehat{\mathcal{O}_K^{\times}} \cdot \mathbb{I}_{K,\infty}) \cdot K^{\times}}.$$

In this lecture we will use this isomorphism to state the main theorem of global class field theory in terms of ideal classes. In order to do so, we need to prove that the groups $\overline{U}_K^{\mathfrak{m}} \subseteq C_K$ that we defined in the last lecture are of finite index in $C_K$, which is what we will do in the second section.

The approximation theorem

Before dealing with ray class groups we need a theorem which we could have proved in the second lecture of this course but we didn't need until now, which says that we can find elements of a field which are arbitrary close to other elements in different absolute values.

 Theorem 1   Let $K$ be any field and let $\phi_1,\dots,\phi_n$ be non-equivalent and non-trivial absolute values on $K$. Then the image of the diagonal embedding $\iota \colon K \hookrightarrow \prod_{j = 1}^n (K,\phi_j)$ is dense in the product topology.

 Proof   We will prove the theorem by induction. If $n = 1$ then we have nothing to prove, so we may assume that $n \geq 2$ and that the theorem holds for any number of absolute values less than $n$. Then we can find an element $a \in K$ such that $\phi_1(a) > 1$ and $\phi_j(a) < 1$ for all $j \in \{2,\dots,n - 1\}$ and analogously we can find an element $b \in K$ such that $\phi_1(b) > 1$ and $\phi_n(b) < 1$. Suppose now that $\phi_n(a) \leq 1$. Then there exists $m \in \mathbb{N}$ such that $\phi_j(a^m b) < 1$ for all $j \in \{ 2,\dots,n \}$. We also have that $\phi_1(a^m b) > 1$ and this implies that $$\lim_{k \to +\infty} \iota\left(\frac{(a^m b)^k}{1 + (a^m b)^k}\right) = (1,0,\dots,0)$$ i.e. that $(1,0,\dots,0) \in \overline{\iota(K)}$. If $\phi_n(a) > 1$ then there exists $m \in \mathbb{N}$ such that $\phi_j((a^m)/(a^m+1) b) < 1$ for all $j \in \{ 2,\dots,n \}$, and thus we have that $$\lim_{k \to +\infty} \iota\left(\frac{((a^m)/(a^m+1) b)^k}{1 + ((a^m)/(a^m+1) b)^k}\right) = (1,0,\dots,0)$$ which still implies that $(1,0,\dots,0) \in \overline{\iota(K)}$. We could play a similar game to see that every vector of the canonical basis of $\prod_{j = 1}^n (K,\phi_j)$ lies in $\overline{\iota(K)}$. But since the sum and product by elements of $K$ are continuous we have that the closure $\overline{\iota(K)} \subseteq \prod_{j = 1}^n (K,\phi_j)$ is a $K$-linear subspace, and thus we conclude that $\iota(K)$ is dense in $\prod_{j = 1}^n (K,\phi_j)$. Q.E.D.

The ray class groups

 Definition 2   Let $\mathfrak{m}$ be a modulus for a global field $K$. We define the ideal group associated to this modulus as the subgroup $\mathcal{I}_K^{\mathfrak{m}} \subseteq \mathcal{I}_K$ of fractional ideals such that $\mathfrak{p} \nmid \mathfrak{a}$ if $\mathfrak{m}(v_{\mathfrak{p}}) \neq 0$. We define moreover the subgroup $$K^{\mathfrak{m}} := \bigcap_{\substack{v \in \Sigma_K \\ \mathfrak{m}(v) \neq 0}} (U_K^{\mathfrak{m}}(v) \cap K^{\times}) \subseteq K^{\times}$$ and the ray associated to this modulus as $\mathcal{R}^{\mathfrak{m}}_K := \{ \alpha \mathcal{O}_K \mid \alpha \in K^{\mathfrak{m}} \}$. Finally we define the ray class group associated to $\mathfrak{m}$ as the quotient $\mathfrak{C}_K^{\mathfrak{m}} := \mathcal{I}_K^{\mathfrak{m}}/\mathcal{R}^{\mathfrak{m}}_K$.

Clearly we have that $\mathfrak{C}_K^{\mathfrak{m}} = \mathfrak{C}_K$ when $\mathfrak{m}$ is the trivial modulus. Moreover, we can generalize the relation between the idelic and the traditional class group to this context in the following way.

Let $K$ be a global field and let $\mathfrak{m}$ be a modulus for $K$. Then we can define a subgroup $$\mathbb{I}_K^{\mathfrak{m}} := \{ (a_v) \mid a_v \in U_K^{\mathfrak{m}}(v), \ \forall \ v \ \text{s.t.} \ \mathfrak{m}(v) \neq 0 \} \subseteq \mathbb{I}_K$$ and observe that the norm map $\mathbb{I}_K \to \mathcal{I}_K$ restricts to a surjective map $\mathbb{I}_K^{\mathfrak{m}} \twoheadrightarrow \mathcal{I}_K^{\mathfrak{m}}$ whose kernel is clearly equal to $U_K^{\mathfrak{m}}$. Moreover we have that the diagonal embedding $K^{\times} \hookrightarrow \mathbb{I}_K$ induces an embedding $K^{\mathfrak{m}} \hookrightarrow \mathbb{I}_K^{\mathfrak{m}}$, and the composition of this embedding with the map $\mathbb{I}_K^{\mathfrak{m}} \twoheadrightarrow \mathcal{I}_K^{\mathfrak{m}}$ sends $K^{\mathfrak{m}}$ to $\mathcal{R}^{\mathfrak{m}}_K$. Finally, observe that $\mathbb{I}_K^{\mathfrak{m}} \cdot K^{\times} = \mathbb{I}_K$, as you can see by applying Theorem 1. Putting all this together we get an isomorphism $$\mathfrak{C}_K^{\mathfrak{m}} \cong \frac{\mathbb{I}_K^{\mathfrak{m}}}{U_K^{\mathfrak{m}} \cdot K^{\mathfrak{m}}} \cong \frac{C_K}{\overline{U}_K^{\mathfrak{m}}}$$ which generalizes the isomorphism above because $U_K^{\mathfrak{m}} = \widehat{\mathcal{O}_K^{\times}} \cdot \mathbb{I}_{K,\infty}$ if $\mathfrak{m}$ is the trivial modulus.

Observe finally that for every modulus $\mathfrak{m}$ we have an exact sequence $$\{ 1 \} \to \frac{(\mathcal{O}_K/\mathfrak{f}_{\mathfrak{m}})^{\times} \times \prod_{\substack{v \in \Sigma_{K,\infty} \\ \mathfrak{m}(v) \neq 0}} \{ \pm 1\}}{\overline{\mathcal{O}_K^{\times}}} \to \mathfrak{C}_K^{\mathfrak{m}} \to \mathfrak{C}_K \to \{ 1 \}$$ where $\mathfrak{f}_{\mathfrak{m}} := \prod_{\mathfrak{p}} \mathfrak{p}^{\mathfrak{m}(v_{\mathfrak{p}})} \subseteq \mathcal{O}_K$ is the ideal associated to $\mathfrak{m}$. In the previous sequence the map $\mathfrak{C}_K^{\mathfrak{m}} \to \mathfrak{C}_K$ is induced by the inclusion $\mathcal{I}_K^{\mathfrak{m}} \to \mathcal{I}_K$, and one can see that this map is surjective by applying Theorem 1. More specifically, let $$K_{\mathfrak{m}} := \bigcap_{\substack{v \in \Sigma_K^{\infty} \\ \mathfrak{m}(v) \neq 0}} (A_{K_v}^{\times} \cap K^{\times}) \qquad \text{and} \qquad \mathcal{P}_K^{\mathfrak{m}} := \left\{ x \mathcal{O}_K \mid x \in K_{\mathfrak{m}} \right \} \cong \frac{K_{\mathfrak{m}}}{\mathcal{O}_K^{\times}}$$ and observe that we have an exact sequence $$\{ 1 \} \to \frac{\mathcal{P}_K^{\mathfrak{m}}}{\mathcal{R}_K^{\mathfrak{m}}} \to \frac{\mathcal{I}_K^{\mathfrak{m}}}{\mathcal{R}_K^{\mathfrak{m}}} \to \frac{\mathcal{I}_K^{\mathfrak{m}}}{\mathcal{P}_K^{\mathfrak{m}}} \to \{ 1 \}$$ where the term in the middle equals $\mathfrak{C}_K^{\mathfrak{m}}$ by definition and the term on the right equals $\mathfrak{C}_K$ by Theorem 1. From the definition of $K_{\mathfrak{m}}$ we see that there is an inclusion $K_{\mathfrak{m}} \hookrightarrow (\mathcal{O}_K)_{\mathfrak{p}}^{\times}$ for every prime $\mathfrak{p}$ such that $\mathfrak{m}(v_{\mathfrak{p}}) \neq 0$. This yields a surjection $K_{\mathfrak{m}} \to (\mathcal{O}_K/\mathfrak{f}_{\mathfrak{m}})^{\times}$, which in turn yields an isomorphism $(\mathcal{P}_K^{\mathfrak{m}})/(\mathcal{R}_K^{\mathfrak{m}}) \cong (\mathcal{O}_K/\mathfrak{f}_{\mathfrak{m}})^{\times}/(\overline{\mathcal{O}_K^{\times}})$.

Finiteness of the class group

The previous short exact sequence tells us that all the ray class groups $\mathfrak{C}_K^{\mathfrak{m}}$ are finite abelian groups, because the quotient $(\mathcal{O}_K/\mathfrak{f}_{\mathfrak{m}})^{\times}/(\overline{\mathcal{O}_K^{\times}})$ is clearly finite and we know that the class group $\mathfrak{C}_K$ is also finite. This last result is a classical result in algebraic number theory, and it is usually proved using Archimedean analysis (more specifically, Minkowski's convex body theorem and the geometry of numbers).

We will give in this section a different proof of this result, which uses the relations between the groups $C_K$ and $\mathfrak{C}_K$ that we just described. In order to do so we need a third "norm" on the group of idèles $\mathbb{I}_K$, which is defined by $$\begin{align} \lvert \cdot \rvert \colon \mathbb{I}_K &\to \mathbb{R}_{> 0} \\ (a_v) &\mapsto \prod_{v \in \Sigma_K} \lvert a_v \rvert_v \end{align}$$ and is called "idelic absolute value". Here the absolute values $\lvert \cdot \rvert_v$ are the normalized ones induced by Haar's measure (see Remark 2 from the previous lecture). We define the group of 1-idèle as $\mathbb{I}_K^1 := \ker(\lvert \cdot \rvert)$. It is not obvious at all that $K^{\times} \subseteq \mathbb{I}_K^1$, as it is shown in the following lemma.

 Lemma 3   Let $K$ be a global field. Then $\prod_{v \in \Sigma_K} \lvert x \rvert_v = 1$ for all $x \in K^{\times}$.

 Proof   Observe first of all that the normalization that we are taking now for the absolute values implies that for every finite extension of global fields $K \subseteq L$ and every place $w \in \Sigma_L$ which restricts to $v \in \Sigma_K$ we have that $\lvert x \rvert_w = \lvert \operatorname{N}_{L_w/K_v}(x) \rvert_v$ for every $x \in L$. Combining this with the fact that $\operatorname{N}_{L/K}(x) = \prod_{w \mid v} \operatorname{N}_{L_w/K_v}(x)$ (as we proved in the previous lecture) we get that $$\prod_{w \in \Sigma_L} \lvert x \rvert_w = \prod_{v \in \Sigma_K} \prod_{w \mid v} \lvert \operatorname{N}_{L_w/K_v}(x) \rvert_v = \prod_{v \in \Sigma_K} \lvert \operatorname{N}_{L/K}(x) \rvert_v$$ for every $x \in L$. This shows that if the product formula $\prod_{v \in \Sigma_K} \lvert x \rvert_v = 1$ holds for $K$ it holds also for $L$.

Hence to conclude it is sufficient to prove the product formula for $K = \mathbb{Q}$ and $K = \mathbb{F}_q(T)$. In both cases we have that $\Sigma_K = \{ p \}_{p \in \mathcal{O}_K} \cup \{ \infty \}$, where $p$ varies among the prime elements of $\mathcal{O}_K$ and $\infty$ is the place corresponding to the embedding $\mathbb{Q} \hookrightarrow \mathbb{R}$ if $K = \mathbb{Q}$ and to the absolute value $\lvert \cdot \rvert_{\infty}$ associated to the degree (see Example 12 of the second lecture). In both cases $\mathcal{O}_K$ is a principal ideal domain and thus we can write every $x \in K^{\times}$ as $x = p_1^{a_1} \cdots p_n^{a_n}$ for some prime elements $p_1,\dots,p_n \in \mathcal{O}_K$ and some $a_1,\dots,a_n \in \mathbb{Z}$. Hence we get that $$\prod_{v \in \Sigma_K} \lvert x \rvert_v = \left( \prod_{q \in \mathcal{O}_K} \lvert x \rvert_q \right) \cdot \lvert x \rvert_{\infty} = \left( \prod_{j = 1}^n (\lvert p_j \rvert_{p_j})^{a_j} \right) \cdot \left( \prod_{j = 1}^n (\lvert p_j \rvert_{\infty})^{a_j} \right) = 1$$ where the last equality comes from the fact that $\lvert p_j \rvert_{q} = 1$ if $q \neq p_j$ and moreover $$\lvert p_j \rvert_{p_j} = \begin{cases} p_j^{-1}, \ \text{if} \ K = \mathbb{Q} \\ q^{-\deg(p_j)}, \ \text{if} \ K = \mathbb{F}_q(T) \end{cases} \qquad \text{and} \qquad \lvert p_j \rvert_{\infty} = \begin{cases} p_j, \ \text{if} \ K = \mathbb{Q} \\ q^{\deg(p_j)}, \ \text{if} \ K = \mathbb{F}_q(T) \end{cases}$$ for every $j \in \{ 1,\dots,n  \}$. Q.E.D.

We can now relate the class group $\mathfrak{C}_K$ to this new group $\mathbb{I}_K^1$. Indeed the restriction of the surjective map $$\begin{align} \mathbb{I}_K &\twoheadrightarrow \mathcal{I}_K \\ (a_v) &\mapsto \prod_{v \in \Sigma_K^{\infty}} \mathfrak{p_v}^{v(a_v)} \end{align}$$ to $\mathbb{I}_K^1$ is also surjective. Indeed in the number field case we can modify the values of any idèle $(a_v) \in \mathbb{I}_K$ at the Archimedean places to obtain a new idèle $(a_v') \in \mathbb{I}_K^{1}$ which has the same image under the map $\mathbb{I}_K \twoheadrightarrow \mathcal{I}_K$. This gives us a map $\mathbb{I}_K^1/K^{\times} \twoheadrightarrow \mathfrak{C}_K$. In the function field case we have that $\mathfrak{C}_K = \mathfrak{D}_K^0$ is the group of divisors of degree zero, and then we have a similar surjective map $\mathbb{I}_K^1/K^{\times} \twoheadrightarrow \mathfrak{C}_K$.  

This shows that the class group $\mathfrak{C}_K$ (considered with the discrete topology) is a continuous image of the group $C^1_K := \mathbb{I}_K^1/K^{\times}$. Thus to conclude it is sufficient to prove the following theorem.

 Theorem 4   The group $C_K^1$ is compact.

 Proof   We will only give an idea of the proof of this theorem. First of all, one proves that $\mathbb{I}_K^1 \subseteq \mathbb{A}_K$ has the subspace topology, which is not trivial because a priori $\mathbb{I}_K^1$ is endowed with the subspace topology from the inclusion $\mathbb{I}_K^1 \subseteq \mathbb{I}_K$, and $\mathbb{I}_K$ does not have the subspace topology inherited from the inclusion $\mathbb{I}_K \subseteq \mathbb{A}_K$.

Having done that, it is sufficient to find a compact subset $W \subseteq \mathbb{I}_K$ such that the canonical projection map $\pi \colon W \cap \mathbb{I}_K^1 \twoheadrightarrow C_K^1$ is surjective. In order to do so one proves first of all an equivalent of Minkowski's lemma, which says that there exists a constant $\lambda_K \in \mathbb{R}_{> 0}$ (which depends only on $K$) such that for every $\alpha \in \mathbb{I}_K$ with $\lvert \alpha \rvert > \lambda_K$ one can find $\beta \in K^{\times}$ such that $\lvert \beta \rvert_v \leq \lvert \alpha_v \rvert_v$ for all $v \in \Sigma_K$.

If we prove this fact then we can take any $(a_v) \in \mathbb{I}_K$ with $\lvert (a_v) \rvert > \lambda_K$ and define $$W := \{ (b_v) \in \mathbb{I}_K \mid \lvert b_v \rvert_v \leq \lvert \alpha_v \rvert_v, \ \forall v \in \Sigma_K \} \subseteq \mathbb{I}_K$$ which is clearly compact because it is a product of closed balls in all the different completions of $K$. Then for every $(b_v) \in \mathbb{I}_K^1$ we have that $$\lvert (b_v^{-1} \cdot a_v)_v \rvert = \lvert (b_v)_v \rvert^{-1} \cdot \lvert (a_v)_v \rvert = \lvert (a_v)_v \rvert > \lambda_K$$ which implies that there exists $c \in K^{\times}$ such that $\lvert c \rvert_v \leq \lvert b_v^{-1} a_v \rvert_v$. This implies that $c \cdot (b_v)_v \in W$, and thus this shows that we can multiply every 1-idèle by an element of $K^{\times}$ to obtain an element of $W$. Hence the map $\pi$ is surjective and we are done. Q.E.D.

For a proof of the two claims in the previous result you can look at these notes by Brian Conrad, and in particular at Theorem 2.1 and Lemma 3.1.

The previous theorem finally proves that the ray class groups $\mathfrak{C}_K^{\mathfrak{m}}$ are finite and thus that the groups $\overline{U}_K^{\mathfrak{m}}$ have finite index in $C_K$. This shows that for every finite abelian extension $K \subseteq L$ there is a modulus $\mathfrak{m}$ such that $L \subseteq K(\mathfrak{m})$ and thus the Galois group $\operatorname{Gal}(L/K)$ is a quotient of $\operatorname{Gal}(K(\mathfrak{m})/K) \cong \mathfrak{C}_K^{\mathfrak{m}}$.

 Example 5   Let $K = \mathbb{Q}$, and let $\infty \in \Sigma_{\mathbb{Q}}$ be the unique Archimedean place. Then for any modulus $\mathfrak{m}$ we have that $\mathfrak{f}_{\mathfrak{m}} = n_{\mathfrak{m}} \mathbb{Z}$ for some $n_{\mathfrak{m}} \in \mathbb{Z}$. Since $\mathfrak{C}_{\mathbb{Q}} = \{ 1 \}$ we have isomorphisms $$\mathfrak{C}_{\mathbb{Q}}^{\mathfrak{m}} \cong \begin{cases} (\mathbb{Z}/n_{\mathfrak{m}} \mathbb{Z})^{\times}, \ \text{if} \ \mathfrak{m}(\infty) = 1 \\ (\mathbb{Z}/n_{\mathfrak{m}} \mathbb{Z})^{\times}/\{\pm 1\} \ \text{if} \ \mathfrak{m}(\infty) = 0 \end{cases}$$ which implies that the ray class fields of $\mathbb{Q}$ are the cyclotomic fields $\mathbb{Q}(\zeta_n)$ and their totally real subfields $\mathbb{Q}(\zeta_n + \zeta_n^{-1})$.

Conclusions and references 

In this lecture we managed to:

• prove the approximation theorem;
• define ray class groups and relate them to the group $C_K$;
• give (an idea of) another proof of the finiteness of the class group, which uses idèles.

References for this lecture include:

• Chapter 8 of these notes by Peter Stevenhagen;
• these notes, these notes and these notes by Andrew V. Sutherland;
• Chapter V of the notes "Class field theory" by James S. Milne;
• Chapter VI of the book "Algebraic number theory" by Jürgen Neukirch;
• Section III.9 of the book "Class field theory - The Bonn Lectures" by Jürgen Neukirch.