Sunday, January 30, 2011

The Global Artin Map for $\mathbb{Q}$

I'm discussing the global Artin homomorphism $\theta : \mathcal{C}_\mathbb{Q} \rightarrow \text{Gal}(\mathbb{Q}^\text{ab}/\mathbb{Q})$ used in our key statement about the Bost-Connes system, as well as the Adeles, Ideles, and other components needed to consider it.

In my last significant post I stated the global Artin map for the rationals intertwined with the Galois action on the values of Bost-Connes KMS states. I wanted to talk about the Artin map in a bit more detail. The general case for any number field $\mathbb{K}$ is a bit too complicated for me to discuss right now, but the case for $\mathbb{Q}$ isn't bad at all. Let's start with $\mathcal{C}_\mathbb{Q} = \mathbb{A}^{\ast}_\mathbb{Q}/\mathbb{Q}^{\ast}$, the Idele class group of $\mathbb{Q}$.
First recall the definition of the Adeles, $\mathbb{A}_\mathbb{Q}$.

$\displaystyle \mathbb{A}_\mathbb{Q} = \mathbb{R} \times \prod_{p \; \text{prime}}^{\prime} \mathbb{Q}_p$

Where $\mathbb{Q}_p$ is the completed field of $p$-adic numbers. The idea here is that we're looking at all possible completions of the number field $\mathbb{Q}$, including the standard absolute value and all $p$-adic valuations. Each of these completed fields is a local'' field, containing information about $\mathbb{Q}$ at a particular prime, and the ring of Adeles combines all that information into one giant ring. The $\prime$ on the product $\prod^{\prime}$ indicates that this product is restricted over $\mathbb{Z}_p$, id est, it has the condition that

$\displaystyle \mathbb{R} \times \prod_{p \; \text{prime}}^{\prime} \mathbb{Q}_p = \{ a=(a_\infty,a_2,a_3,\ldots) \in \mathbb{A}_\mathbb{Q} \; \text{iff} \; a_p \in \mathbb{Z}_p \; \text{for all but finitely many p}\$

$\mathbb{A}^{\ast}_\mathbb{Q}$ is the set of all invertible elements in the ring, we can write it $\mathbb{R}^\ast \times \prod_p^\prime \mathbb{Q}_p^\ast$. As $\mathbb{Q}_p$ is the field of fractions for $\mathbb{Z}_p$, we have an isomorphism $\mathbb{Q}_p \cong \mathbb{Z}_p[\frac{1}{p}] \cong p^\mathbb{Z} \times \mathbb{Z}_p$. If we identify $p^\mathbb{Z}$ with $\mathbb{Z}$ we can write $\mathbb{Q}_p^\ast \cong \mathbb{Z} \times \mathbb{Z}^\ast_p$ and

$\displaystyle \mathbb{A}_\mathbb{Q}^\ast = \mathbb{R}^\ast \times \prod_{p \; \text{prime}}^{\prime} \mathbb{Q}_p^\ast \cong \{\pm1\} \times \mathbb{R}_{>0} \times \prod \mathbb{Z}_p^\ast \times \bigoplus_p \mathbb{Z}$

Now for an $r \in \mathbb{Q}^\ast$, we can write $r = \text{sgn}(r) \prod_p p^n(p)$, so we can take $\mathbb{Q}^\ast$ to $\{\pm1\} \times \bigoplus_p \mathbb{Z}$ by $r \mapsto (\text{sgn}(r),n(p)_p)$. By noted that $\mathbb{R}_{>0} \cong \mathbb{R}$ by logarithm, we can write:

$\displaystyle \mathbb{A}_\mathbb{Q}^\ast \cong \mathbb{Q}^\ast \times \mathbb{R} \times \prod_p \mathbb{Z}_p^\ast$

Now recall our favourite profinite group $\hat{\mathbb{Z}} = \varprojlim_k \mathbb{Z}/k\mathbb{Z} \cong \text{End}(\mathbb{Q}/\mathbb{Z})$. Since we can write each $k$ as $p_1^{n(p_1)} \ldots p_l^{n(p_l)}$, we also have

$\displaystyle \mathbb{Z}/k\mathbb{Z} = \prod_{i=1}^{l} \mathbb{Z}/p_i^{n(p_i)} \mathbb{Z}$

After passing to the inverse limit we have:

$\displaystyle \hat{\mathbb{Z}} = \varprojlim_k \mathbb{Z}/k\mathbb{Z} = \prod_p \varprojlim_n \mathbb{Z}/p^n\mathbb{Z} = \prod_p \mathbb{Z}_p$

Thus $\mathbb{A}_\mathbb{Q}^\ast \cong \mathbb{Q}^\ast \times \mathbb{R} \times \hat{\mathbb{Z}}^\ast$ and finally:

$\displaystyle \mathcal{C}_\mathbb{Q} \cong \mathbb{R} \times \hat{\mathbb{Z}}^\ast$

Now I cannot prove it, but class field theory tells us that the map $\theta:\mathcal{C}_\mathbb{Q} \rightarrow \text{Gal}(\mathbb{Q}^\text{ab}/\mathbb{Q})$ is surjective, and the kernel is the connected component of the identity on $\mathcal{C}_\mathbb{Q}$, which we denote $\mathcal{D}_\mathbb{Q} = \mathbb{R}$. Hence $\mathcal{C}_\mathbb{Q} / \mathcal{D}_\mathbb{Q} \cong \hat{\mathbb{Z}}^\ast$ and we have an isomorphism:

$\displaystyle \theta : \hat{\mathbb{Z}}^\ast \rightarrow \text{Gal}(\mathbb{Q}^\text{ab}/\mathbb{Q})$

We can see that these two groups are isomorphic assuming only the Kronecker-Weber theorem (which says that $\mathbb{Q}^\text{ab} = \mathbb{Q}^\text{cycl}$) and some facts about inverse limits and Galois theory. $\mathbb{Q}^\text{cycl} = \bigcup_n \mathbb{Q}(\zeta_n)$, where $\zeta_n$ is the n-th root of unity. Recall that $\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong (\mathbb{Z}/n\mathbb{Z})^\ast$. Hence we have

$\displaystyle \text{Gal}(\mathbb{Q}^\text{cycl}/\mathbb{Q}) = \varprojlim_n \text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \cong \varprojlim_n (\mathbb{Z}/n\mathbb{Z})^\ast = \hat{\mathbb{Z}}^\ast$

Thanks to the following sources for help on this post: