Today I Learned

March 20th

Today I learned the proof that a number field $K$ is discrete and cocompact in its adele ring $\mathbb A_K.$ We begin by showing this for $\QQ\subseteq\AA_\QQ$ and then will expand to the general case by viewing $K$ as a finite-dimensional $\QQ$-vector space.

For this, we exhibit a compact fundamental domain for $\AA_\QQ/\QQ$ as\[D:=[-1/2,1/2]\times\prod_{p \lt \infty}\ZZ_p.\]It is quick to see that $D$ is compact, but we can't appeal to Tychonoff because $\AA_\QQ$ isn't equipped with the product topology. There might be a way to do this because each nonarchimedean projection of $D$ is in $\ZZ_p,$ which is compact, but this trick is unnecessary.

Fix $\{U_\alpha\}_{\alpha\in\lambda}$ an open cover of $D$; there is at least one open set, say $U_\beta,$ and its projections to $\QQ_p$ are $\ZZ_p$ for all but finitely many $\nu.$ That is, there exists a finite set $S$ of (say, nonarchimedean) places such that\[\prod_{p\notin S}\ZZ_p\subseteq U_\beta.\]Note we are being somewhat sloppy with our notation because the left-hand side does not technically have all of the coordinates of $\AA_\QQ,$ but we also don't care about them yet. Indeed, this single $U_\beta$ covers most of $D$ already; it remains to cover\[[-1/2,1/2]\times\prod_{p\in S}\ZZ_p.\]However, each of these sets is compact, and our open cover $\{U_\alpha\}_{\alpha\in\lambda}$ must cover them. Extracting the corresponding open cover for each of the (finitely many) above components and directly unioning them with $U_\beta$ gives us our finite subcover of $D.$

To finish, we take the following claim to show $D$ is a fundamental domain.

Quickly, we show this is enough. Automatically this implies that each element of $\AA_\QQ/\QQ$ has exactly one representative in $D,$ so $D\cong\AA_\QQ/\QQ.$ So $D$ being compact tells us $\QQ\subseteq\AA_\QQ$ is cocompact. Additionally, we note that certainly $0\in D,$ so the above claim asserts it is the only element. In fact, $|0|_\infty \lt 1/2,$ so\[0\in\op{Int}D=\{a\in\AA_\QQ:|a_\infty| \lt 1/2\text{ and }|a_p|_p\le1\text{ for }p \lt \infty\}\subseteq D,\]which is an open subset of $\AA_\QQ$ containing $0$ as its only rational. It follows the translate $q+\op{Int}D$ is an open set containing $q$ as its only rational, so $\QQ\subseteq\AA_\QQ$ is discrete; else $q+\op{Int}D-q=\op{Int}D$ would contain another rational.

Anyways, we show the lemma now. This is a group homomorphism because\[(d_1,q_1)+(d_2,q_2)\to(d_1+q_1)+(d_2+q_2)\leftarrow(d_1+d_2,q_1+q_2).\]So for injectivity, it suffices for trivial kernel; that is, $d+q=0$ if and only if $d=0$ and $q=0.$ Well, $d+q=0$ implies $d=-q,$ so it suffices to show $D\cap\QQ=\{0\}.$ Certainly $0\in D\cap\QQ,$ and for any other $q\in D\cap\QQ,$ then $|q|_p\le1$ for $p \lt \infty$ requires $\nu_p(q)\ge0,$ so in fact $q\in\ZZ.$ But then $q\in(-1/2,1/2)$ forces $q=0,$ as desired.

Surjectivity is somewhat harder. Fix an adele $a\in\AA_\QQ$ that we want to hit. We have to cover the poles of $a$ with the rational, but there are only finitely many. Let $S$ be a finite set of nonarchimedean places containing the poles of $a,$ and we use the Chinese remainder theorem to fix\[q\equiv a_p\pmod{p^{\nu_p(a_p)}}\]for each finite prime $p\in S.$ The Chinese remainder theorem applies because $S$ contains only finitely many places. It follows that $a+q\in\ZZ_p$ for each finite prime $p.$

To push this into $D,$ it remains to deal with $a_\infty.$ Choose $n:=\lfloor a_\infty\rceil\in\ZZ$ to be the closest integer to $a_\infty.$ Then\[a+q_0+n\in[-1/2,1/2]\times\prod_{p \lt \infty}\ZZ_p=D.\]Thus, $a=(a+q+n)-q-n$ is in the image of $D\oplus\QQ.$ This completes the lemma and the proposition. $\blacksquare$

It remains to extend this to arbitrary number fields. As promised, fix $\{\alpha_1,\ldots,\alpha_n\}$ be a basis of $K/\QQ.$ The main lemma to proceed is that this same basis can be lifted to a basis of $\AA_K$ over $\AA_\QQ,$ but we have to pay attention to topology.

We're not going to be terribly formal with this lemma. The map sends $(a_k)_{k=1}^n$ to $\sum_ka_k\alpha_k.$ This map is linear and has trivial kernel, which we won't show explicitly; track the basis backwards. Similarly, surjectivity follows because we can represent some $a_\nu\in K_\nu$ with that basis and then Cauchy sequences move backwards.

Anyways, the difficult portion of the statement is the topological part. It is a fact that the diagonal embedding of the basis $\{\alpha_\bullet\}$ induces a topological isomorphism\[\alpha_p:\QQ_p^n\to\prod_{\mf p\mid p}K_\mf p\]for each finite place $p$ of $\QQ,$ and in fact, for all but finitely many $p$ (including $p=\infty$), $\alpha_p$ restricts to an isomorphism $\ZZ_p^n\cong\prod_{\mf p\mid p}\mathcal O_{K_\mf p}.$ Letting $\mf m_0$ be the (formal) product of this finite set of places, we note that any $\mf m_0\mid\mf m$ has\[\AA_{\QQ,\mf m}^n=\prod_{p\mid\mf m}\QQ_p^n\times\prod_{p\nmid\mf m}\ZZ_p^n.\]Using our $\alpha_p,$ we have that\[\AA_{\QQ,\mf m}^n\cong\left(\prod_{p\mid\mf m}\prod_{\nu\mid p}K_\nu\right)\times\left(\prod_{p\nmid\mf m}\prod_{\nu\mid p}K_\nu\right)\cong\AA_{K,\mf m}.\]So we have that our basis induces an isomorphism $\AA_{\QQ,\mf m}^n\cong\AA_{K,\mf m}$ for any modulus $\mf m$ divisible by $\mf m_0.$ This isomorphisms agree on restriction, so gluing them together as $\mf m$ varies lets us piece together a full isomorphism $\AA_\QQ^n\cong\AA_K$ (the $\AA_{K,\mf m}$ cover $\AA_K$). Note we are not being terribly rigorous here. $\blacksquare$

Now we can prove the main theorem with relative ease. Pulling back the topological isomorphism $\AA_\QQ^n\to\AA_K$ means that showing $K$ is discrete/cocompact in $\AA_K$ is equivalent to showing that $\QQ^n$ is discrete/cocompact in $\AA_\QQ.$ Explicitly, the conditions for discreteness and cocompactness can be written using only open sets, so we can pull them back along the homeomorphism.

But now, showing $\QQ^n$ is discrete/cocompact in $\AA_\QQ^n$ can be shown by merely showing $\QQ$ is discrete/cocompact in $\AA_\QQ,$ by coordinate projection. In particular, the product is finite, so we're dealing with the box topology on $\AA_\QQ^n,$ so it suffices to check only one coordinate and then multiply the open sets checking discrete/cocompact.

At this point, appealing to the proposition we showed at the beginning finishes.

We remark that the proof holds for arbitrary global fields without too much extra work. Changing $\QQ$ out for $\FF_q[t]$ and $p=\infty$ to $p=(t^{-1})$ for our infinite prime does most of the work. We do not show this here.