# Lattices, the geometry of numbers, and adeles

A while ago, Sameer asked why Minkowski’s theorem is so powerful in number theory, and I realized I had no idea. Number theory is, certainly, about a strange sort of geometry (sorry, Sameer), but Minkowski’s lattices and plump convex bodies aren’t exactly reminiscent of Diophantine geometry. Yet Minkowski’s theorem is unreasonably useful, giving slick proof of existence results like Lagrange’s four-square theorem. More startlingly, the standard proofs of finiteness of class group and Dirichlet’s unit theorem both use it. These are deep and fundamental results, proved with a technique which at first seemed to me like a trick out of an Olympiad problem, a sleight of hand instead of a deep insight.

It turns out both Dirichlet’s unit theorem and the finiteness of class number, when looked at the right way, have all their “deepness” hidden in a notion of compactness in a certain lattice, and this can be made explicit using the adeles (roughly, a way to glue together all the places of a global field). This is the beginning of many different long, strange stories, none of which I really understand yet; hopefully, this will be the first of many posts detailing my exploration.

This post already assumes a basic familiarity with ${p}$-adics, places, and valuation theory. A secondary aim of this post is to serve as an introduction to adeles and the adelic perspective. It is somewhat atypical, in that it will not prove or even discuss many of the elementary properties of adeles other guides will, but I think that this introduction via application and intuition is in many ways preferable. Optimally, it would be read in conjunction with a more conventional introduction to adelic number theory, which will march through all the basic properties I gloss over.

I. Finiteness of class number and Dirichlet’s unit theorem, revisited

Let’s reiterate the proofs of finiteness of class number and Dirichlet’s unit theorem quickly, phrasing them in a manner so as to emphasize the role of compactness. We’ll work over the (hopefully) familiar case of a number field ${K}$, though will keep an eye towards the end goal of generalizing to any global field.

Aside: a global field, if you’re unfamiliar with the term, is either a number field, i.e. a finite extension of ${\mathbb{Q}}$, or a function field of an algebraic curve over a finite field, i.e. finite extension of ${\mathbb{F}_p(t)}$. This might seem pretty arbitrary, but there’s a reason for it beyond the (in)famous finite function field/number field analogy: the completions of such fields at their places give rise to all the local fields, which are precisely the locally compact topological fields. Working backwards, we can use this to give an abstract characterization of global fields in terms of the product formula for valuations. Indeed, I think this is probably the strongest justification for the number field/finite function field analogy in the first place.

Theorem. Let ${K}$ be a number field, with ring of numbers ${\mathcal{O}_K}$, and let ${J(K)}$ be the associated fractional ideal group. Let the class group be defined as ${\text{Cl}(K)=J(K)/K^\times}$; i.e. fractional ideals modulo principal ones. Then ${|\text{Cl}(K)|}$ is finite.

Aside: this is also a familiar construction in algebraic geometry: the divisor group of formal linear combinations of height-1 prime ideals (so, all of them, in a Dedekind domain, but in general, codimension 1 subvarieties, geometrically speaking) by the subgroup of principal divisors, formal linear combinations arising from an element of ${K^\times}$ (principal divisors), where we include in the linear combination each prime appearing in the factorization of the ideal generated by the element, multiplied by its multiplicity in the factorization. It is easy to see how this matches up with the more number theoretic definition. For “nice” schemes, including the ones we’re working with, the class group is actually naturally isomorphic to ${\text{Pic}(K)}$, the Picard group of “line bundles” (locally free rank one modules) over the structure sheaf of ${K}$ with tensor products, so the two are sometimes used interchangeably. In the global field context, “line bundles” are associated with fractional ideals, linear equivalence by multiplication by a principal ideal, and all ideals are flat modules, so tensor products are just ideal multiplication, so this correspondence is obvious.

Proof. There are two cleanly separated steps: to show that every ideal class contains an integral ideal with uniformly bounded norm, and to show that there are finitely many integral ideals with bounded norm. The latter is obvious for number fields, since a prime ideal lying over ${(p)\subset \mathbb{Q}}$ has norm divisible by ${p}$.

The former is (generally) proven using Minkowski’s theorem, applied to the lattice created by the embedding ${\sigma}$ into Minkowski space ${\mathbb{R}\otimes K}$ (considered as ${\mathbb{Q}}$-algebras).

Aside: You will also see Minkowski space described more explicitly in a non-canonical way involving choosing one of each pair of complex embeddings, which is workable but less neat – the volume calculations of the fundamental parallelpiped in terms of the Lebesgue measure on the resulting space introduces an awkward power of two, which of course cancels out in the end, but is still bothersome. A canonical alternate description from Wikipedia is the pithy “fixed subspace of the embedding into ${\mathbb{C}^n}$ under complex conjugation” (where ${n=s+2r}$ is the degree of the extension), which is pretty misleading, since it means complex conjugation acting simultaneously on the coordinates themselves, and to permute the coordinates by acting on the associated embeddings. It is an easy exercise to see how this identifies with the tensor definition, and that the natural volume forms on the two agree.

Then it is straightforward to see that the square root of the determinant ${\sqrt{|d_K|}}$ is the volume of the fundamental parallelpiped (considering ${\mathcal{O}_K}$ as the lattice), and the lattice determined by any given ideal can be calculated in terms of its index (ideal norm). The standard way to finish from here is to take the lattice associated to the inverse of an arbitrary fractional ideal, then to use a suitably scaled convex body to find a lattice point corresponding to an element of ${\mathcal{O}_K}$ that multiplies the ideal to become integral, yet has sufficiently small norm to uniformly bound the norm resulting representative of the ideal class. The technical details are boring and meaningless.

So what’s actually happening with this Minkowski space business? The perspective we need to take is that each coordinate corresponds to a distinct real/complex place of ${K}$ – that is, an equivalence class (equivalence under induced topology) of absolute values on it. In this case, we are only considering the archimedean (non-${p}$-adic, with the Archimedean property) places, so by Ostrowski’s theorem there is a unique one on ${\mathbb{Q}}$, corresponding to the usual absolute value. By extension, each real embedding and each Galois orbit of complex embeddings corresponds to a unique absolute value and hence place on general ${K}$. Then looking at Minkowski space as a product of ${\mathbb{R}}$s and ${\mathbb{C}}$s, in general, the space should be a product of completions of ${K}$ with respect to its various places, with ${K}$ canonically embedded inside.

We will need to extend our theory of “norm” here, but the already-present absolute values are suggestive enough of the general idea that we can await for this to arise naturally in the adelic perspective later. More pressing is the need to connect this to the class group. Heuristically, the idea that we can find ideal class representatives with bounded norm will correspond to compactness of the class group, and finiteness of integral ideals with bounded norm will correspond to discreteness; the proper way to think of finiteness of ${\text{Pic}(K)}$, then, is as a general topological consequence of compactness and discreteness.

To precisely give the correct link between ${\text{Pic}(K)}$ as a topological group and Minkowski space, we require adelic language and concepts, but we can give a preview here. Elements of the fractional ideal group, since we can write them as unique products of prime ideals, can be identified with a kind of “${p}$-adic lattice” of their valuations at every prime (or, their nonarchimedean places). (Equivalently, the fractional ideal group is the free abelian group generated by the primes – again, this is basically the idea of associated divisors from algebraic geometry.) So this group is already discrete! (Discrete under what topology? Well, I did say this would be an imprecise preview.) And one can clearly see the link between the discreteness of the prime lattice, and the argument we saw above for finiteness of ideals with bounded norm.

To get the class group, we just have to quotient out by the sublattice of principal ideals, induced by the embedding of ${K^\times}$. If we can prove the result is compact, we’re done – so we have isolated the compactness argument.

To do this, we find a continuous surjection from a compact space onto it: we use the subset of our lattice which corresponds to ideals with bounded norm! The role of the Minkowski’s theorem argument in our original proof is to demonstrate we can find this subset; this will be replaced by a similar adelic result later. The oddly circumspect route (finding a lattice point in Minkowski space corresponding to inverse ideal, then multiplying and seeing that the dependence on ideal norms cancels) of the classical proof will be streamlined in the adelic result, which considers the archimedean topology of Minkowski space and the non-archimedean properties of the ideals simultaneously, as we will see.

Adopting a slightly more general perspective on the ring of integers ${\mathcal{O}_K}$ and the class group ${\text{Pic}(K)}$ will help throw this into better relief, by way of explicitly involving the non-archimedean (${p}$-adic) places of ${K}$. Let ${S}$ be a finite set of places of ${K}$, containing all the archimedean places. Then analogously to ${\mathcal{O}_K}$, we define ${\mathcal{O}_{K,S}}$ to be the ring of integers of ${K}$ localized at all the places of ${S}$, so that the primes corresponding to non-archimedean ${S}$ become units. (${S}$ exactly the set of archimdean places corresponds to the usual ${\mathcal{O}_K}$.) We can analogously define the class group ${\text{Cl}_S(K)}$ or ${\text{Pic}_S(K)}$. It is not hard to demonstrate the following exact sequence, reproduced from Neukirch’s wonderful book:

${1\rightarrow \mathcal{O}_K^\times\rightarrow \mathcal{O}_{K,S}^\times \rightarrow \oplus_{\mathfrak{p}\not \in S}K^*/\mathcal{O}_{\mathfrak{p}}\rightarrow \text{Pic}(\mathcal{O}_K)\rightarrow \text{Pic}_S(\mathcal{O}_K)\rightarrow 1}$

(Note: the middle term has a more natural expression adelically, which we’ll see later. Each summand is also isomorphic to ${\mathbb{Z}}$, by the associated valuation; here’s our “${p}$-adic lattice”!) This sequence by itself shows that ${\text{Pic}_S(K)}$ is finite given the result for the usual case. But our compactness approach generalizes easily to prove the result for all ${S}$ simultaneously. This is really just two slightly different algebraic manifestations of the same point of view, because ${\text{Pic}_S(K)}$ can be identified with a quotient of the “${p}$-adic lattice” we talked about earlier, eliminating the dimensions corresponding to the places of ${S}$. Hence in fact there is a surjection from the normal class group into the ${S}$-class group, as we see in the above exact sequence, and we are given another hint of what its kernel, the middle term in the exact sequence, is adelically.

Notice this makes sense, since the ${S}$-ring of integers has fewer ideals, since we turned some elements into units.

Aside: You may be wondering, as I did: is there a way for this to make sense if ${S}$ does not contain all the archimedean places? No, obviously not, stupid.

Our treatment of Dirichlet’s unit theorem will be brief by contrast, since the compactness and adelic viewpoint there are practically begging to be discovered.

Theorem. Given a number field ${K}$ with ${r}$ real embeddings and ${s}$ pairs of complex embeddings, ${\mathcal{O}_K^\times}$ is a finitely generated abelian group; in particular, it is the product of a finite torsion group (the roots of unity) with a free abelian group of rank ${r+s-1}$.

We again have the classical embedding of ${K}$ and hence of ${\mathcal{O}_K^\times}$ into Minkowski space; take log absolute values of each coordinate (the kernel is clearly the roots of unity, which we can prove are finite in number with a separate argument). By considering the norm condition on a unit, the image lies on a codimension ${1}$ hyperplane. If we can prove that the fundamental domain associated to the lattice (the cokernel of the map) is compact, we have that it is a full lattice on this dim-${r+s-1}$ hyperplane, so we’re done.

The usual way to prove this using Minkowski’s theorem, again. Every algebraic number theory text I’ve seen does this in a somewhat different (though, modulo convolution, basically equivalent) way, but many of them obscure what’s actually going on behind seemingly arbitrary ideas and strange constructions – except for Neukirch, another reason it’s the only ANT book worth its salt. The idea which Neukirch lays bare is that we directly show that the fundamental domain is compact by showing it lies in the image of a compact set, the same trick as our earlier proof.

To see how this works, let’s go back and look at our embedding more closely. What we did was form the chain ${K\rightarrow \mathbb{R} \otimes K \rightarrow \mathbb{R}^{r+s}}$, where the second arrow is taking logarithms of absolute values (and also where the roots of unity disappear into the kernel). Inside ${V}$ we have our norm-one hyperplane ${H}$, whose preimage in ${K}$ is a codimension-one hypersurface ${S}$, where the unit group is injectively embedded – indeed, we could have done the entire thing (besides throwing out roots of unity) sans log-linearizing by thinking of the unit group as a “lattice” (discrete abelian subgroup) on the abelian variety ${S}$. So we construct our preimage as the compact “fundamental domain” of ${S}$ (modulo the unit “lattice”) inside Minkowski space, as this is the more natural setting to work in.

Aside: this is the reason why most books’ treatment of this material is very bad: the log-linearization is nice because it handles the roots of unity, and we get the lattice and hyperplanes, etc. to work with so we don’t need a theory of multiplicative abelian varieties and the “lattice” on it. So books just work everything out inside ${V}$, often using superfluous linear algebraic ideas. But Minkowski space is the more natural setting to make the geometric/number-theoretic connections apparent.

The idea is that we simply take a “box” in ${\mathbb{R}\otimes K}$ where the absolute value of each coordinate is bounded by a certain number, such that the product of the bounds is large enough so we can apply Minkowski’s theorem. Then we take the union of translates of the box over a collection of algebraic integers which can correspond to every possible norm less than that product. (Again, we see the finiteness of integral ideals of a given norm in action.) The intersection of these boxes with our surface gives us our bounded covering of the fundamental domain, after a little technical legwork, and we are done.

To generalize to ${S}$-units ${\mathcal{O}_{K,S}^\times}$, the idea is very easy: just add on the part of the “${p}$-adic lattice” corresponding to the non-archimedean places of ${S}$ onto Minkowski space, and repeat an almost identical argument. Indeed, this is basically captured by the exact sequence from earlier: there is an injection ${mathcal{O}_K^\times\rightarrow \mathcal{O}_{K,S}^\times}$, whose image is the kernel of ${\mathcal{O}_{K,S}^\times \rightarrow \oplus_{\mathfrak{p}\not \in S}K^*/\mathcal{O}_{\mathfrak{p}}}$. This latter map has a finite kernel since the class group is finite, so the kernel only affects the torsion part of ${\mathcal{O}_{K,S}^\times}$. Hence the rank of the free part is just ${r+s-1+|\{\mathfrak{p}\in S\}|=|S|-1}$ (since ${r}$ and ${s}$ account for the archimedean valuations, and the primes account for the non-archimedean).

In summary, we saw that finiteness of class number and Dirichlet’s unit theorem have proofs which can be seen in a very similar way, as the compactness of a quotient of the “${p}$-adic lattice” and the compactness of a quotient of a hypersurface in Minkowski space, respectively. Look back at our exact sequence and otice that extending these results to ${S}$-rings entails “subtracting dimensions” from the former and “adding dimensions” to the latter corresponding to the non-archimedean places of ${S}$, a clear complementary relationship. In the adelic formulation, which glues together the archimedean places of Minkowski space with the non-archimedean ${p}$-lattice, then, we should expect these to correspond to the two ends of an exact sequence in which everything is compact, which is indeed precisely what will happen.

Let ${K}$ be a global field. The adele ring ${\mathbb{A}_K}$ is simply what we get when we take a product of all the completions of ${K}$ with respect to its various places – almost. Actually, we have to take a restricted product in the following sense: let ${K_v}$ be the completion of ${K}$ with respect to place/valuation ${v}$; then the restricted product ${\prod_v' K_v}$ is the set of all elements of the full product for which all but finitely many of the coordinates are in that coordinate’s respective discrete valuation ring (with nonnegative valuation; so e.g. the ${p}$-adic integers for the place ${v_p}$ of ${K=\mathbb{Q}}$). So it’s somewhere in between a direct product and a direct sum.

Note this restriction only makes sense for non-archimedean places, because archimedean places don’t give rise to discrete valuations rings. But this is fine, because there are always only finitely many archimedean places: this is well-known for number fields, and for function fields of algebraic curves over finite fields, there are no archimedean places, and indeed the concept is not even really applicable.

Aside: Here’s a somewhat involved explanation of what’s going on with these various valuations; this is very skippable. The fact that there are no archimedean places of function fields over a finite field is an immediate consequence of the general version of Ostrowski’s theorem, which says that all such places are inherited from an embedding into ${\mathbb{C}}$, i.e. there are none when the base field has nonzero characteristic. Indeed, in general, when we have a function field ${K}$ over ${k}$, we enforce valuations of the transcendental extension ${K/k}$ to be trivial on the latter (though this is only an extra restriction in the zero-characteristic case). What happens is that we have the typical places corresponding to the prime ideals of ${k[t]}$ (generated by irreducible polynomials), and then a number of infinite places.

The easiest way to see the infinite places is when our function field is precisely ${k(t)}$. Then there is just one infinite place, which is given by the total degree (numerator minus denominator) of a rational function. This might seem out of place and similar to the exceptional archimedean places of a number field, but in fact it is no different in character from the other places! To see this, consider how the map ${t\mapsto 1/t}$ permutes the places – one can consider the infinite place generated by the “prime” ${1/t}$! This is legitimate because taking ${t\mapsto 1/t}$ amounts to just choosing a different valuation ring inside ${k(t)}$, which was an arbitrary choice to begin with. Heuristically, in the ${k}$ algebraically closed case, ${t\mapsto 1/(t-c)}$ can swap the infinite place with any one of the finite places (though this doesn’t actually work in practice unless we pass to completions); the infinite place is hence associated to any of the “primes” ${1/(t-c)}$, since they are the same asymptotically. In the ${1/t}$ case, we can actually take ${1/t}$ as the uniformizing parameter of the associated DVR.

To put this precisely, when ${k}$ is algebraically closed, the projective linear group ${\text{PGL}(k)}$ acts freely and transitively on the places of ${k(t)}$. In fact, we can see this very geometrically when ${k=\mathbb{C}}$: the places of ${\mathbb{C}(t)}$ trivial on ${\mathbb{C}}$ can be identified with ${\text{Spec } \mathbb{C}[x]}$ plus a point at infinity – nothing more than the projective line ${\mathbb{P}^1}$! This can be extended to (projective) complex algebraic curves to give an abstract nonsingular model of all such curves in terms of their discrete places, the work of Andre Weil in the pre-EGA era. We can push this to higher-dimension varieties with function fields of higher transcendence dimension, but the theory becomes significantly less nice.

It is harder to get a geometric intuition, but this applies just as well to positive characteristic. If we take ${k=\overline{\mathbb{F}}_p}$, the result still applies and we obtain projective curves over ${\overline{\mathbb{F}}_p}$. The places over a finite field, like the places over e.g. ${\mathbb{R}}$, have a more complex structure since there are places of different “degrees,” corresponding to irreducible nonlinear polynomials. Then the resulting “geometry” has not just the base projective curve, but various higher-degree points which correspond to orbits under the absolute Galois group. This is again a spectral space; in fact still the projective closure of ${\mathbb{Spec}}$ of the function field. Notice in analogue with the classical theory of algebraic curves that this interpretation in terms of projective closure gives a geometric reason why we need to take $S$ to be nonempty even in the function field case: if $S=\emptyset$, then we just get the full divisor class group of a projective curve over a field, which is $\mathbb{Z}$ by the degree homomorphism, so our result about the finiteness of class number doesn’t actually hold.

The asymmetry of archimedean places in the two types of global fields makes the arithmetic case more difficult to work with (among other things). Intuitively, the archimedean-ness of exceptional places of ${\mathbb{Q}}$ means unlike the function field case, in which we can take all the non-archimedean places in the nice form of a projective closure, it isn’t easy to make ${\text{Spec } \mathbb{Z}}$ a complete scheme. This precludes us from using certain tools which exist in the highly-structured, close-to-analytic, projective context. (This is a common theme in the global field analogy; e.g. there is no known arithmetic derivative with nice properties either, which is a major roadblock to e.g. the abc conjecture. I wonder if there’s actually a way to connect these ideas of lack of analyticity.) This line of thinking is the beginning of Arakelov theory.

This allows us actually to give a tensor-product definition of adeles, as ${K \otimes \prod_w K_w\times \prod_v \mathcal{O}_{K,v}}$, where ${\mathcal{O}_{K,v}}$ is the discrete valuation ring associated to the completion at ${v}$, and the product is split up into the archimedean and non-archimedean places. This isn’t particularly important; I just thought I’d include it to show the usefulness of tensor products and demonstrate that we can give a definition without the somewhat arbitrary-seeming restricted product.

The adeles have a natural topology as a profinite group, since before tensoring with ${K}$, the non-archimedean portion of the product is naturally the profinite completion of ${K}$. Then we take the product topology with the usual topology on the archimedean completions (if applicable); this gives the integral adeles. In the full adele ring, after tensoring with ${K}$, we consider the integral adeles to be an open subring. Note this is much finer than the subspace topology of the product topology on the unrestricted product.

Another way to look at this for the categorically is to see ${\mathbb{A}_{K}}$ as glued together from products where a fixed finite subset of the terms are allowed to be the whole field, which is a useful perspective because it allows us to reintroduce the concept of ${S}$ – finite sets of places containing the archimedean ones. Indeed, we can set the ${S}$-adeles ${\mathbb{A}_{K,S}= \prod_{v\in S} K_v \times \prod_{v\not\in S} \mathcal{O}_v}$, with the product topology. Then there are obvious inclusion morphisms from ${S}$-adeles to ${T}$-adeles, where ${S\subset T}$, so we get a directed system, and can set ${\mathbb{A}_K}$ as the colimit ${\varinjlim \mathbb{A}_{K,S}}$.

Aside: If you’re unfamiliar with the colimit (a more general categorical term for what is referred to as a direct or inductive limit in narrower settings), just think of it as “gluing together”. All the objects in the directed system (the partial ${S}$-adeles) are like pieces of the whole adele ring, and the further down the chain you go (the bigger ${S}$ is) the closer you are to the full ring, since we want any finite collection of places to be the full field, not just fixed ${S}$. There is a dual categorical concept called the limit (or the projective limit, or the inverse limit) which “glues together” in a slightly different way, by a kind of approximation. The best references in this case are actually Wikipedia: 1, 2. An important element (or, really, the defining element) of these two constructions is their universal property: for an inverse limit, every object with maps into the objects of the directed system (which commute with the maps between them) sees all those maps factor through the inverse limit, and for a direct limit, every object with maps from the inverse system sees all those maps factor through the direct limit.

If you’re familiar with the colimit, you might be wondering about the strange juxtaposition between finding the adele ring to be the colimit of the ${S}$-adeles, while simultaneously having the “main chunk” of it be a profinite group (and hence a limit). There’s not really a reason here; the two limit/colimit constructions are basically unrelated. In some sense, the ${S}$-adele direct limit is a pretty trivial gluing which just propagates the ${K_v}$-coordinates to whatever finite coordinates we want, while the profinite completion of ${\mathcal{O}_K}$ structure actually constructs the heart and soul of the adeles by bringing out the non-archimedean places.

However, there is a great example of limit-colimit adjunction related to the adeles, in the identification of ${\mathbb{A}_{\mathbb{Q}}/\mathbb{Q}}$ with the solenoid on one hand and with the character group of ${\mathbb{Q}}$ on the other. See Keith Conrad’s notes for an exposition of the latter identification.

In this way, the adeles become a locally compact topological group; this is not difficult to prove with a general topology argument. Multiplication is also continuous, so they are in fact a topological ring. Notice we have a natural embedding ${K\rightarrow \mathbb{A}_K}$ given by taken the image in each completion, which sits inside the restricted product roughly because there are only finitely many places in the denominator of an element of ${K}$.

The following result is fundamental. Not strictly necessary for our purposes, so it will only be a sketch, but it’s a good illustration of the kinds of arguments the topology of the adeles give us:

Theorem. The image of ${K}$ is discrete and cocompact in ${\mathbb{A}_K}$. (Cocompact: the quotient is compact.)

Proof. For the discreteness, it suffices to find an open neighborhood isolating ${0\in K}$, since we can translate it using the group operation. This is very simple; bound the norm of each coordinate by ${1}$; strictly in the ${\infty}$-norm case. Check that this is open, and that this isolates zero (break it into the number field and function field case).

For cocompactness, we need to find a compact set which maps surjectively to the quotient in ${\mathbb{A}_K}$; that is, a compact set with a representative of each equivalence class. (How familiar!) We can take actually the same open set as above but with the ${\infty}$-norm restricted to ${1/2}$ instead of ${1}$, non-strictly, since then we have a product of compact balls coordinate-wise which is hence compact by Tychonoff’s theorem. This amounts to saying that there is an element of ${K}$ which matches all the valuations fairly well, which is a consequence of the well-known weak approximation theorem for Dedekind domains. See here for an exposition of this theorem which mixes a sorta-pre-adelic viewpoint with a classical perspective. ${\square}$

There are many other properties of adeles (self-duality with character group, realization as a solenoid, strong approximation, etc.) which are important in the theory, but we will not cover them here. Consult Pete Clark’s notes for a sample.

Now we pass to the idele group. Element-wise, it’s just the unit group ${\mathbb{A}_K^\times}$, but the topology is not the subspace topology, since, again, inversion is not continuous. The standard way to remedy this is give it the finer induced topology as a subspace of ${\mathbb{A}_K\times \mathbb{A}_K}$, identified with ordered pairs ${(x,y)}$ with ${xy=1}$.

This seems rather arbitrary, but actually it’s just a case of poor exposition – this is actually the natural topology of the ideles when identified with ${\text{GL}(1,\mathbb{A}_K)\cong \mathbb{A}_K^\times\subset \text{Hom}_{K}(\mathbb{A}_K,\mathbb{A}_K)}$.

${K}$ embeds into ${\mathbb{A}_K^\times}$ in a very understandable way.

Theorem. (Product formula) Let the adelic (or idelic) norm ${||\cdot||_K}$ on ${\mathbb{A}_K}$ be given as ${||x||_K=||(x_v)||_K = \prod_v ||x_v||_v}$, and let ${(\mathbb{A}_K^\times)^1}$ be the subgroup of ${\mathbb{A_K}^\times}$ given by elements of norm ${1}$. Then the natural embedding of ${K}$ into ${\mathbb{A}_K}$ (coordinatewise, since there are of course natural embeddings ${K\rightarrow K_v}$) lands in ${(\mathbb{A}_K^\times)^1}$.

(Note that this makes sense for global fields, since by construction there are only countably many places, and eventually all the norms are at most ${1}$.)

(Note two: the absolute value on ${p}$-adic completions is familiar for number fields, being ${p^{-v_p}}$ on ${\mathbb{Q}_p}$, and being the normalized version of this (extension of ${\mathbb{Q}_p}$ norm with “denominators cleared”) in general. For function fields, it’s similarly defined as some constant ${c^{-v_p}}$, but the constant ${c}$ is uniform across all primes, and basically arbitrary. It is usually taken as the size ${q}$ of the finite field we are working over – I’m not sure if there’s any time this choice is actually significant.)

Proof. This is of course just a fancy way of saying that ${\prod_v ||x||_v=1}$ (notice this is actually always essentially a finite product) for all ${x\in K}$, where the product is over all places. A moment’s thought on the case of ${\mathbb{Q}}$ will give an idea of “why” this is true – the ${p}$-adic valuations will switch the powers of ${p}$ between the numerator and the denominator, so together they will cancel out with the infinite place.

There are two common ways to prove this: the elementary method is to use the theory of norms of field extensions to reduce the problem to simply the cases of ${\mathbb{F}_p(t)}$ and ${\mathbb{Q}}$, which are then obvious by explicit reasoning as above. There is subtlety involved in the calulations; recall the note above on the correct normalizations of the absolute values involved. ${\square}$

The other method is using something called the Haar measure, which you may have encountered before if you have done any study of locally compact topological groups. It is simply a measure (in the sense of measure theory) adapted to groups in the most natural way. This is the last piece of set-up we need.

Definition. A Haar measure ${\mu}$ on a topological group ${G}$ is a measure on the Borel algebra of its open subsets which is invariant under translation, i.e. ${\mu(g+A)=\mu(A)}$ for all open ${A}$. (Notice ${+}$: we’re dealing with an abelian group under addition, which makes our life a bit easier, since right vs. left translates aren’t a thing.)

The key thing that one needs to know about Haar measure is Haar’s theorem, which states that for locally compact ${G}$, there is a unique (up to multiplication) nontrivial Haar measure satisfying some nice properties – inner and outer regularity and finiteness on compact subsets. The only one really relevant to us is the last one. There are more thorough treatments of the (in my opinion, tedious) theory behind Haar measure readily available; I like this more algebraic treatment by Terry Tao more than some of the others, but you can look around if you’re interested in technical tedium.

The point is, there is a Haar measure with these nice properties on ${\mathbb{A}_K}$, and it is an extremely important part of adelic theory. We’re not going to really do too much with it, but the ability to integrate over the adeles allows one to do quite a bit of impressive (cf. “beautiful” – Jack) analysis in the unified adelic context, which is the the basis of Tate’s famous thesis.

Denote by ${\mu}$ the Haar measure on ${\mathbb{A}_K}$, scaled (since uniqueness is only up to a constant) so that the induced measure on the compact quotient (“fundamental domain”) ${\mathbb{A}_K/K}$ is ${1}$, so that the measure “agrees with the counting measure on ${K}$” in a nice way. (Think in analogy with a measure on ${\mathbb{R}^2}$ with the typical lattice points with fundamental domain the unit square, if it helps.)

The following lemma is the last piece of “niceness” we need to know about adelic constructs to proceed to the main proof:

Lemma. For any ${x\in \mathbb{A}_K}$, ${\mu(xS)=||x||_K\mu(S)}$.

Proof. This almost follows from the analogous statement, of scaling by ${||x_v||_v}$, on the natural Haar measure (think ${p}$-adic metric topology in the non-archimedean case, think Euclidean topology in the archimedean case) of each ${K_v}$ factor, which is easy to prove directly – you should work this out yourself if this doesn’t seem natural to you. This is, intuitively, breaking the problem into local cases at each place. (“Local” in the same sense as “local field”!) If ${\mathbb{A}_K}$ were a true product space so that the Haar measure could be taken as the product of the individual Haar measures, we would be done – however, recall it’s actually a restricted product, so there is subtlety involved.

Instead of naively taking the product of the measures associated to our familiar ${p}$-adic and Euclidean topologies all at once, we have to build it up through finite approximations, by using our ${\mathbb{A}_{K,S}}$ construction from earlier, each of which is a genuine product space of locally compact topological groups, where the ${\mathcal{O}_{K,S}}$ factors simply are simply given the restricted Haar measure, where they have volume ${1}$. Hence we have natural Haar measures ${\mu_S}$ on each of them which are genuine product measures. Then it is a theorem (which is not actually difficult to prove, given uniqueness of Haar measure!) that the pullback of the measure ${\mu}$ by the inclusion ${\mathbb{A}_{K,S}\rightarrow \mathbb{A}_K}$ restricts it to ${\mu_S}$. More conceptually, this is saying is that we can construct the Haar measure ${\mu}$ through the direct limit, because the category of measured locally compact topological spaces has all limits.

In particular, for any particular ${x\in \mathbb{A}_K}$, simply consider some ${S}$ “big enough” so that all the places where ${x}$ has nonzero valuation are included – possible by definition of the adeles. In particular, we can choose ${S}$ so that ${x\in \mathbb{A}^\times_{K,S}}$. Then the scaling effect of ${x}$ in ${A_{K,S}}$ is clearly ${||x||_{K,S}=||x||_K}$ (because ${x\in \mathbb{A}^\times_{K,S}}$), and since ${x\in \mathbb{A}^\times_{K,S}}$ and ${\mathbb{A}_{K,S}}$ is an open subgroup of the whole adele ring, this coincides with the general scaling effect. ${\square}$

The product formula is immediately a trivial consequence of this lemma. (This is the other, more conceptual way to prove it.)

But more importantly, with the topological and metric machinery all set up, we’re ready for the main course.

IV. Compactness, and two finiteness theorems

The ultimate result will be this:

Theorem. The following statements are equivalent, for any nonempty finite set of places ${S}$ containing all the archimedean ones:

1) ${(\mathbb{A}_{K}^\times)^1/K^\times}$ is compact.

2) ${\text{Pic}_S(K)}$ is finite and the rank of ${\mathcal{O}_{S,K}^\times}$ is ${|S|-1}$.

This begs the question: what exactly is the topology we’re considering ${(\mathbb{A}_K^\times)^1}$? Is it inherited from the ideles or the adeles? It turns out that the subspace topologies inherited from the ideles and the adeles coincide, and this is important. But more on this later.

Let’s work backwards and put the finiteness of class number and Dirichlet’s unit theorem into adelic terms.

Let ${S}$ be a nonempty finite set of places, containing all the archimedean ones. (By automorphism, we can actually assume that it contains the infinite place of a function field.) To get a handle on ${\text{Pic}_S(K)}$, we identify the mythical “${p}$-adic lattice” we discussed so much earlier: it is nothing more than ${\mathbb{A}_K^\times / \mathbb{A}_{K,S}^\times}$! Indeed, we need a group in which each element is identified with a certain valuation on every place not in ${S}$. In ${\mathbb{A}_K^\times / \mathbb{A}_{K,S}^\times}$, the places in ${S}$ are indeed quotiented out, since the ${S}$-adeles have ${K_v}$ in the product for ${v\in S}$. On the other hand, the places not in ${S}$ only have a factor ${\mathcal{O}_{K,v}}$ in the ${S}$-adeles, which becomes just the ${\mathcal{O}_{K,v}}$-units upon passing to the unit group. Hence the quotient on those coordinates can be identified simply by giving the valuation of the element. Finally, finiteness is assured by the definition of the adeles.

But then immediately, from our earlier discussion, we obtain the isomorphism ${\text{Pic}_S(K)\cong \mathbb{A}_K^\times / (\mathbb{A}_{K,S}^\times\cdot K^\times)}$. (Exercise: put our informal discussion rigorously by giving the explicit isomorphism.) We thus need this group to be compact and discrete.

It will help to write the group a different way. It is in fact isomorphic to ${(\mathbb{A}_K^\times)^1 / ((\mathbb{A}_{K,S}^\times)^1\cdot K^\times)}$! (Here, we work a priori with the inherited topology from the ideles, since we don’t know that the induced topologies are the same yet.) Indeed, we need only prove that modulo ${\mathbb{A}_{K,S}^\times}$, we can always find a idelic-norm-${1}$ representative. This is obvious for number fields, since we can alter the archimedean places however we want to achieve overall norm ${1}$. It is only slightly less obvious for function fields. Since the absolute values in the function field case are all just exponentiated valuations with the same base, all we really need is to have the valuations sum to zero, which is of course possible: since ${S}$ is nonempty, alter any given representative of a class to have whatever valuation you want at a place of ${S}$ by multiplying by an adele in ${\mathbb{A}_{K,S}}$ with the appropriate valuation there, and units everywhere else.

Aside: A not entirely obvious point is that we can find a uniformizing parameter at each non-archimedean place: that is, given non-archimedean ${v}$, some ${t\in K_v}$ with ${v(t)=1}$. This is of course necessary to get arbitrary-valuation elements in ${K_v}$. Think about the case of a number field for simplicity: if our place corresponds to a non-principal prime like ${(3,\sqrt{7}-1)}$, it’s not trivial that there is a uniformizing parameter.

To see that this is indeed the case, think about it this way: for a non-archimedean place ${v}$ corresponding to a prime ${\mathfrak{p}}$, we have a ring of integers with respect to ${v}$ in ${K}$, ${\mathcal{O}_v}$. It’s not hard to prove that this will be a Dedekind domain for any global field. Then to get a uniformizing parameter, it suffices to obtain a uniformizing parameter for the localization ${(\mathcal{O}_v)_\mathfrak{p}}$, since the field of fractions of the completion of this ring (under the adic topology) is ${K_v}$. Hence we simply need to prove that this is a DVR, or equivalently a PID since it’s a local ring. But the fact that localizations of Dedekind domains are DVRs is a standard algebraic fact, e.g. see here, so we’re done.

An intuitive way to think of this is as follows: in a Dedekind domain, we have unique ideal factorization. If we have an element ${u}$ so that the exponent of ${\mathfrak{p}}$ in ${(u)}$ is ${1}$, once we pass to the localization, ${u}$ will be a uniformizing parameter! So now it suddenly seems very plausible that we can find such a ${u}$, since we just need to make ${\mathfrak{p}}$ principal by combining it with other primes. But given finiteness of class number of global fields, every prime is torsion with respect to becoming principal, so we can just find a bunch of other ideals not divisible by ${\mathfrak{p}}$ but in its ideal class, which when multiplied together, give our desired principal ideal and uniformizing parameter.

Obviously this is not a way we should actually prove it, because it is circular logic (in the context of this blog post), and more generally, uses a more powerful result to prove a less powerful one. (A specific case of a less powerful one, no less – notice this proof doesn’t work for general Dedekind domains, which can have infinite class group!) But hopefully it is a useful way to think about it.

Aside double combo: The group ${\mathbb{A}^\times_K/K^\times}$ is, importantly, not necessarily compact. It is known as the idele class group (notice the analogy with the ideal class group, which is of course a quotient group), and is very important in global class field theory. Indeed, the (arguably) fundamental theorem of the theory is that finite-index closed/open (equivalent, since the topology is profinite) subgroups of the idele class group are in bijection with finite abelian extensions of ${K}$ (inside some separable closure of the field). The quotients by these subgroups are the Galois group of the corresponding extension. In fact, the ${S}$-class groups provide examples of these; each of them gives an extension of ${K}$! What a world.

Aside triple combo: A little more about arithmetic surfaces: taking the number field-function field dictionary even further, the fact that we have the isomorphism ${\text{Pic}_S(K)\cong \mathbb{A}_K^\times / (\mathbb{A}_{K,S}^\times\cdot K^\times)\cong (\mathbb{A}_K^\times)^1 / ((\mathbb{A}_{K,S}^\times)^1\cdot K^\times)}$ correspond to the fact that for global fields, the connected component of the identity in the Picard group is the entire group. This connected component is generally denoted ${\text{Pic}_0(V)}$ for a scheme ${V}$. This connected component on traditional algebraic curves is also known as the Jacobian ${J(V)}$, which classifies degree ${0}$ line bundles. So what we are really seeing is that ideal classes are isomorphism classes of line bundles, and that all of them are of degree ${0}$. This is, heuristically, the product formula for global fields.

Disclaimer: terminology is used loosely in this aside, and I’m not sure if everything said here is technically true. But they are all things which should be true.

So now compactness is easy to deal with: ${(\mathbb{A}_K^\times)^1 / ((\mathbb{A}_{K,S}^\times)^1\cdot K^\times)}$ is obviously a quotient group of ${(\mathbb{A}_K^\times)^1/K^\times}$. So we immediately have half of one way of our implication: the latter’s compactness implies the the finiteness of class number.

One last thing on class number: can we relate this to our ad hoc Minkowski argument for number fields? Sort of; the surjection from a bounded subset of the Minkowski lattice is a bit like our isomorphism with the adelic-norm-${1}$ version. Of course, the latter doesn’t actually bound the norms of the archimedean places, since the non-archimedean ones still can have arbitrarily small contribution, which is why there’s still work to do.

Before tackling Dirichlet’s unit theorem, we are finally forced to prove the technical lemma regarding ${(\mathbb{A}_K^\times)^1}$‘s inherited topology.

Theorem. ${(\mathbb{A}_K^\times)^1}$ inherits the same topology as a closed subgroup (resp. subspace) of the ideles and the adeles.

Proof. First we check that ${(\mathbb{A}_K^\times)^1)}$ is a closed subgroup of the adeles – the topology of the ideles is strictly finer, so this will suffice for both closedness statements. Let ${x=(x_v)\not\in (\mathbb{A}_K^\times)^1}$. Take a finite set of places ${S}$ so ${x\in \mathbb{A}_{K,S}}$. In particular, ${S}$ contains all the places of ${x}$ where ${x}$ has norm greater than ${1}$ in that place.

If the adelic norm of ${x}$ is less than ${1}$, then take open balls in every place where ${||x||_v> 1}$ with radius as close to the respective norm of ${x}$ in that place as possible. By definition of adelic norm, if we take larger and larger finite subsets of the remainder of the places where ${x}$ has norm less than ${1}$, . The product of all these open balls with the unit balls in the remaining places (just all the rings of integers so it’s a genuine open subset of the adeles) is by construction an open neighborhood of ${x}$ disjoint from the unit ideles.

If the adelic norm of ${x}$ is greater than ${1}$, it’s a little trickier. We use the convergence of the product ${\prod ||x_v||_v}$ to see that we can take a sufficiently large ${S}$ (with, minimally, the same stipulations as earlier) so that for any ${v}$ outside of this set, the exponent ${q}$ of the valuation (that is, the ${q}$ such that ${||\cdot||_v = q^v(\cdot)}$) is greater than ${\epsilon ||x||_K}$: this is clear because of the increasing exponents of number fields, and the fact that function fields only actually have finitely many places to begin with. We can similarly make ${S}$ large enough so that ${||x||_S}$ (defined in the obvious way) is in ${(1,\epsilon||x||_K)}$. Once again by taking sufficiently small neighborhoods in the places of ${S}$ of ${x_v}$ and allowing ${\mathcal{O}_{K,v}}$ elsewhere, a simple computation shows that if any place in this neighborhood outside of ${S}$ has norm ${<1}$, the whole idele has norm less than ${1}$, and otherwise of course the idele has norm greater than ${1}$. So we again have an open bounding ${x}$ away from the unit ideles. Hence between our two cases, we have the desired closedness.

(There’s some philosophizing to be done about the dichotomy of the approaches in this last paragraph between the function field/number field case once again, but I’ll leave you to think about it, since I don’t have any particularly fresh points to make.)

Since we know that the idelic subspace topology on ${(\mathbb{A}_K^\times)^1}$ is a priori at least as fine as the adelic subspace topology, it remains to show. So let ${N\cap (\mathbb{A}_K^\times)^1}$ be an open subset of the unit ideles in the idelic topology; we need to show that it contains an open subset in the adelic topology. We can assume WLOG that ${N}$ contains ${1}$ by multiplicative translation (which is continuous even in the adelic topology), and by contraction that ${N}$ is a product of neighborhoods ${N_v}$ for ${v\in S}$ for some finite set of places ${S}$, with ${\mathcal{O}_{K,v}^\times}$ in every other coordinate.

${U=\prod N_v \times \prod \mathcal{O}_{K,v}}$ is an open set in the adelic topology. If we intersect ${U}$ with ${(\mathbb{A}_K^\times)^1}$, we don’t necessarily get something in ${N\cap (\mathbb{A}_K^\times)^1}$ since the coordinates outside of ${S}$ might not be units. But we can ensure that they are if we make the ${S}$-coordinates have sufficiently small norm, so that the non-${S}$-coordinates must have exactly norm ${1}$ (rather than just ${\le 1}$, as is a priori the case) to ensure it lies in the unit adeles. But this can be done by just shrinking the ${N_v}$s suitably. The details of the computation are left for if you’re into that kind of thing. ${\square}$

Now Dirichlet’s unit theorem falls out easily. In place of the log embedding of just multiplicative Minkowski space, we have an “${S}$-log embedding” in general: ${\mathscr{L}_S: \prod_{v\in S} K_v^\times \rightarrow \mathbb{R}^{|S|}}$, given by taking ${x=(x_v)\mapsto \sum \log ||x_v||_v}$. We can then take the composite ${\mathcal{O}_{K,S}^\times \rightarrow \prod_{v\in S} K_v^\times \rightarrow \mathbb{R}^{|S|}}$, where the first map is the obvious map in to the ${S}$-places.

Obviously, the image of the first arrow lies in ${(\prod_{v\in S}K_v^\times)^1}$, which maps to the hyperplane with coordinates summing to zero. Then as before, we need to make a few claims to show this does what it’s supposed to. We only sketch the proofs, since they’re all straightforward:

1) The first arrow is a discrete embedding, and the composite image is discrete as well.

That it is an embedding is clear enough, since it’s an embedding in each coordinate. Hence it suffices to establish discreteness. By the previous theorem, it suffices to prove that the “additive version” ${\mathcal{O}_{K,S}\rightarrow \prod_{v\in S} K_v}$ is discrete (check out the subspace topologies yourself). In other words, we need a finite neighborhood of ${0\in \prod_{v\in S} K_v}$ which contains finitely many elements of ${\mathcal{O}_{K,S}}$. It is easy to explicitly construct such a neighborhood, but also note that it follows from discreteness of ${K}$ in ${\mathbb{A}_K}$ by passing to open subrings and then dropping the other factors. (Actually we also easily find that the analogous statement about cocompactness of ${\mathcal{O}_{K,S}}$ in ${\prod_{v\in S}}$ is true as well.)

To prove the composite image is discrete, we note that the map ${\mathcal{L}_S}$ is proper: take a basic compact set; the product of bounded closed intervals bounded each ${|x_v|_v}$ above and below. The result is simply an annulus in ${\prod_{v\in S} K_v}$ (a compact closed ball minus an open ball), so the result follows. Then the image of a discrete group under a proper map of locally compact topological groups is discrete (easy result; prove this! it’s not true for general spaces), so we’re done.

2) The kernel is finite and torsion.

We saw in the number field case that it was precisely the torsion elements; the roots of unity in ${\mathcal{O}_{K,S}^\times}$, which are finite in number. The kernel for the function field is the (multiplicative group of) functions in ${\mathcal{O}_{K,S}}$ (which, as you should’ve intuited, are functions generically regular, with their only possible poles at points corresponding to places of ${S}$) which are also regular and nonzero on the places of ${S}$. But then these functions can have no poles anywhere, hence no zeros by the product formula, and hence are nothing more than constants in the coefficient field ${\mathbb{F}_q}$. So again: finite, and torsion.

Aside: It feels a little weird to call the nonzero coefficients of the function field case “roots of unity,” since it’s obvious a priori that all nonzero coefficient field constants are roots of ${1}$. As it turns out, you should get over that weirdness, because this link between the units of the coefficient field and the roots of unity turns up in plenty of places. The number of these roots of unity, with degree relatively prime to the characteristic associated to the place, in an archimedean completion is always one greater than the residue field (think about what this means for, e.g., a ${p}$-adic extension, vs. the function fields – it’s almost vacuous in the latter case).

In class field theory, cyclotomic extensions and extensions of the coefficient field (for finite function fields) both give abelian extensions for their respective global fields. The celebrated Kronecker-Weber theorem tells us that every abelian extension is contained in a cyclotomic field, and Hilbert’s twelfth problem asks us to generalize this to arbitrary number field; essentially, we need to find nice descriptions of the maximal abelian extension ${K^{ab}}$. Besides special cases where nice abelian varieties (like the unit circle!) show up in the form of complex multiplication (read Kevin’s ongoing posts), the problem remains open. (Actually, there’s a nice adelic proof of Kronecker-Weber, using the idele class group machinery mentioned earlier. This even gives a description of ${K^{ab}}$ in the general case, though the abstraction of the description is not particularly useful for computation.) But for function fields, the correct formulation has been found in the Kronecker-Weber-Hayes theorem: it’s not an exact analogue; the maximal extension of the “roots of unity” coefficient field is not the whole story, but it’s an important piece.

Finally, Pete Clark claims that counterexamples to one of Malle’s asymptotics conjectures in the number field/function field correspond to these extensions arising from “roots of unity,” but I can’t really find anything on the function field case.

3) The quotient by the composite is compact.

This is really the crux of it, since with this, we obtain that the lattice is full rank in the hyperplane, so we finally get that the image of the unit group, hence the unit group, has rank ${|S|-1}$. Denote by ${H}$ the hyperplane in ${\mathbb{R}^S}$. Then it’s not hard to see we can pass to the restricted quotient map ${\mathscr{L}'_S: (\prod_{v\in S} K_v^\times)^1/\mathcal{O}_{K,S}^\times\rightarrow H/\mathscr{L}_S(\mathcal{O}_{K,S}^\times)}$.

Dirichlet’s unit theorem is now that the target is compact. We want to prove that this is equivalent to the source being compact, to get closer to our desired equivalence into adelic language. It’s not hard to check that ${\mathscr{L}_S'}$, like the map it’s descended from, is proper, so one direction is quick. In the other, the image of the target is compact by general topology, hence a closed subgroup in the Hausdorff target. It is a classic fact (and a good exercise) that if we have an exact sequence of locally compact topological groups, compactness is “additive in exact sequences”: what this means in this case is that ${H/\mathscr{L}_S(\mathcal{O}_{K,S}^\times)}$ is compact if and only if ${(\prod_{v\in S} K_v^\times)^1/\mathcal{O}_{K,S}^\times}$ and the quotient

${\displaystyle(H/\mathscr{L}_S(\mathcal{O}_{K,S}^\times))/\left(\left(\prod_{v\in S} K_v^\times\right)^1/\mathcal{O}_{K,S}^\times\right)\cong H/\mathscr{L}_S\left(\prod_{v\in S} K_v^\times\right)}$

is compact. Now we really just need to get our hands dirty and show that ${\mathscr{L}_S(\prod_{v\in S} K_v^\times)}$ contains ${|S|-1}$ linearly independent vectors. This is a bit of straightforward greasework, reminiscent of the classical proof, which I will leave to you.

Hence this last point (3) is equivalent to the statement “${(\prod_{v\in S} K_v^\times)^1/\mathcal{O}_{K,S}^\times}$ is compact.” We take it just one more step and note there is an obvious quotient map ${(\mathbb{A}_{K,S}^\times)^1/\mathcal{O}_{K,S}^\times\rightarrow (\prod_{v\in S} K_v^\times)^1/\mathcal{O}_{K,S}}$ with kernel the product of all those other unit groups not in ${S}$, which is compact by Tychonoff’s. So we are reduced from Dirichlet’s unit theorem to compactness of ${(\mathbb{A}_{K,S}^\times)^1/\mathcal{O}_{K,S}}$.

Now, finally:

Proof of equivalence.

${\displaystyle 0\rightarrow (\mathbb{A}_{K,S}^\times)^1/\mathcal{O}_{K,S}^\times\rightarrow (\mathbb{A}_K^\times)^1/K^\times \rightarrow (\mathbb{A}_K^\times)^1 / ((\mathbb{A}_{K,S}^\times)^1\cdot K^\times).}$ ${\blacksquare}$

Let’s take a big-picture view of what we’ve accomplished.

The class group is seen to be precisely a quotient of the (discrete) ${p}$-adic lattice, which has a natural adelic description as ${\mathbb{A}_K^\times / (\mathbb{A}_{K,S}^\times\cdot K^\times)}$. We are able to find norm-${1}$ representatives of every equivalence class, so the quotiented ${p}$-adic lattice is also ${(\mathbb{A}_K^\times)^1 / ((\mathbb{A}_{K,S}^\times)^1\cdot K^\times)}$. Hence finiteness of class number is equivalent to compactness of this.

Geometrically, it is obvious that the unit group (once you remove torsion) embeds nicely as a lattice in a hyperplane of dimension ${|S|-1}$, so the Dirichlet unit theorem is equivalent to the quotient of the hyperplane by the lattice – which can be pulled back to the quotient of the abelian variety ${(\prod_{v\in S} K_v^\times)^1/\mathcal{O}_{K,S}^\times}$. (Indeed, as in the classical case, everything can be done here, on the quotient of the “hypersurface,” with a little more care and theory, but the embedding provides a nice way to get rid of the torsion and put it in a linear algebraic context.) This is the “fundamental domain” of the adelic-norm-1 hypersurface under the lattice given by the unit group. Tacking on the (compact) product of the unit balls of all the places not in ${S}$ allows us to replace this with the more adelic ${(\mathbb{A}_{K,S}^\times)^1/\mathcal{O}_{K,S}^\times}$ (a “fattened fundamental domain”), and so we are reduced from Dirichlet to the compactness of this last group.

We are then able to glue these two groups together (in the sense of short exact sequence, ${\text{Ext}}$-style gluing) so that their individual compactness was equivalent to the compactness of their gluing. On a technical level, it’s clear that we have the short exact sequence where the glued object is ${(\mathbb{A}^\times_K)^1/K^\times}$. But intuitively, why do the “fattened fundamental domain” and the “quotiented ${p}$-adic lattice” fit together nicely like this? Well, the latter’s lattice structure is precisely on the places not in ${S}$; one can imagine it as a finite torsion grid-like structure along those coordinates. The former has the “fattening” – more or less extraneous product of unit balls – in the non-${S}$ coordinates, to fill in the gaps of the grid, and then a simple fundamental domain along the dimensions/coordinates corresponding to the ${S}$-places. It ${(\mathbb{A}^\times_K)^1/K^\times}$ also as hypersurface of sorts in the full idelic space, quotiented by the lattice ${K^\times}$. The first arrow in the exact sequence is then just the immersion of a fundamental polygon which is extant along some coordinates (the places of ${S}$) but not along others (the complement).

At long last, it’s time to actually prove that any of these things are actually compact. And naturally, we will need a Minkowski lemma of sorts.

Theorem. (Adelic Minkowski) For a fixed element ${c=(c_v)\in \mathbb{A}_K^\times}$, consider the set ${X_c}$ of adeles ${x=(x_v)}$ with ${||x_v||_v\le ||c_v||_v}$ at every place ${v}$. Then there exists a constant ${C}$ depending only on the global field ${K}$ so that for any ${c}$ with ${||c||_K\ge C}$, ${X_c\cap K}$ contains a nonzero element.

Proof. This is basically an exercise in manipulating Haar measure, but the idea of the proof is very geometric (as one would expect, since it’s basically the same as the proof of classical Minkowski). It’s not essential to understand the technical details, but the main thread should be easy to follow.

Let ${\mu}$ be the nicely scaled Haar measure we had from earlier. Let ${Z}$ be the product of the closed unit discs in the non-archimedean places, with the unit disc of radius ${1/2}$ (diameter ${1}$) in the archimedean places: the key is that any two elements in ${Z}$ can’t differ by more than ${1}$ in valuation at any place. ${Z}$ is compact and contains an open neighborhood, so ${\mu(Z)}$ is finite and nonzero.

Take ${C>1/\mu(Z)}$, and ${c}$ with ${||c||_K\ge C}$. It is fairly clear then that to prove our desired statement, it suffices to prove that ${cZ}$ contains two points with the same equivalence class modulo ${K}$.

Indeed, we have that ${\mu(cZ)>1}$ (by the lemma about scaling effect of multiplying by constants on the Haar measure from way back), and if we take the projection map ${\pi: cZ \rightarrow \mathbb{A}_K/K}$, we can proceed with a simple volume argument:

${\displaystyle \mu(cZ)=\int_{cZ}d\mu = \int_{\mathbb{A}_K/K}|\pi^{-1}(x)|d\mu(x)}$

where we implicitly exchange a sum with the integral, and abuse notation to let ${\mu}$ denote both the Haar measure on both the adeles and the induced one on the fundamental domain. Hence if all the ${|\pi^{-1}(x)|}$ are ${\le 1}$, ${\mu(cZ) \le \int_{\mathbb{A}_K/K}d\mu = 1}$; contradiction.

Take a moment to note how the above proof is pretty much exactly like the proof of classical Minkowski. Really, all we’re doing is changing the lattice, if we just broaden our understanding of “lattice” to “discrete cocompact subgroup of a locally compact topological abelian group with a nice norm associated to its Haar measure”. Indeed, it is possible to generalize the statement to this context, which then includes all versions of Minkowski. ${\square}$

Let’s wrap things up.

Theorem. For a global field ${K}$, ${(\mathbb{A}_K^\times)^1/K^\times}$ is compact.

Proof. Recall that the unit ideles inherit the topology from the adeles. So we just need a compact set ${W}$ of ${\mathbb{A}_K}$ which projects surjectively onto ${(\mathbb{A}_K^\times)^1/K^\times}$. But just choose ${W}$ to be the set of adeles bounded at each place by the norm at that place of ${c}$, where ${||c||_K>C}$ as in the adelic Minkowski lemma. Then for any unit idele ${x}$, there is some nonzero ${t\in K^\times}$ so that ${||t||_v\le ||x^{-1} c||_K}$ by the lemma, so ${xt\in W}$ projects onto ${t}$.

Dirichlet’s unit theorem and the finiteness of class number are immediate consequences. ${\blacksquare}$

Let’s conclude with a sort of bird’s-eye view of what this all means. It will be helpful to have read the asides to understand this.

So hopefully we have answered Sameer’s question adequately: the importance of Minkowski’s theorem and lattice geometry is not an accident; a general notion of “lattice” is extremely important because this idea manifests itself in algebraic/arithmetic groups suited to analyzing global fields, like the adeles/ideles. The classical “geometry of numbers” and Minkowski’s theorem arise because of the special phenomenon of archimedean places of number fields, whose corresponding completions give us copies of ${\mathbb{R}}$ and ${\mathbb{C}}$. When analyzing ${\mathbb{Z}}$, which is the ring of integers corresponding to ${S}$ just the archimedean places, Minkowski space then naturally becomes an important object of study. ${\mathbb{R}}$ and ${\mathbb{C}}$ have special structure that can be exploited to obtain results in this manner.

On the other hand, the presence of these places is a difficulty in some ways, since unlike the function field case, these infinite places can’t be thought of “projective completion” of the spectrum of a number field, and behave very differently, which is why a lot of results are much easier for function fields than number fields. (I mean, the Riemann hypothesis for one.) The main working approach to this is the same attitude which inspired the geometry of numbers: to work in the archimedean places, take them as they are, and use their familiar analytic structure in conjunction with the nicer non-archimedean places to make progress. This is the foundation of Arakelov theory, a modern approach to number theory. We can define Arakelov divisors, which are like elements of our “${p}$-adic lattice” with archimedean real/complex components tacked on, and even the Arakelov class group, a kind of “fattened class group,” which very much resembles in spirit our “fattened fundamental domain” from the adelic discussion of Dirichlet’s unit theorem. (Remember “obviously not, stupid?” Sorry, that was a lie; that’s precisely what this is.) We can obtain analogues of a lot of algebraic geometry this way, including Riemann-Roch, sheaves, and intersection theory.

Relatedly, but with almost the opposite attitude there is a view that there should be a field with one element, a mythical object over which ${\text{Spec }\mathbb{Z}}$ would be a curve, whose projective completion would somehow bypass, or perhaps reveal hidden depths of, the archimedean obstacle. The field with one element is a fascinating aspect of mathematical folklore which has many, many more interesting hypothetical connections and properties than the ones arising here; see the Wikipedia page and MathOverflow for more. It should be noted that quite a few pretenders exist, but none are fully satisfying.

Some of the terminology used here was nonstandard; to further get into adelic applications in class field theory and Arakelov theory, some of this should be clarified. An “${S}$-class group” is more typically described as a ray class group – though the latter concept is slightly more general. A set ${S}$ of places is a kind of modulus, which figures in Artin reciprocity, a fundamental result of global class field theory. Moduli can count places with multiplicity, however.

Huge credit to Brian Conrad’s notes, from which the bulk of this material was taken. Indeed, this post is basically a more fleshed out version of those notes, recast as a general philosophical overview/introduction to adeles.