A Blueprint for Fermat’s Last Theorem

8 Miniproject: Adeles

8.1 Status

This is an active miniproject.

8.2 The goal

There are several goals to this miniproject.

  1. Define the adeles \(\mathbb {A}_K\) of a number field \(K\) and give them the structure of a \(K\)-algebra (status: now in mathlib thanks to Salvatore Mercuri);

  2. Prove that \(\mathbb {A}_K\) is a locally compact topological ring (status: also proved by Mercuri but not yet in mathlib);

  3. Base change: show that if \(L/K\) is a finite extension of number fields then the natural map \(L\otimes _K\mathbb {A}_K\to \mathbb {A}_L\) is an isomorphism; (status: not formalized yet, but there is a plan – see the project dashboard);

  4. Prove that \(K \subseteq \mathbb {A}_K\) is a discrete subgoup and the quotient is compact (status: not formalized yet, but there is a plan – see the project dashboard);

  5. Get this stuff into mathlib (status: (1) done, (2)–(4) not done).

We briefly go through the basic definitions. Let \(K\) be a number field. Let \(\widehat{\mathbb {Z}}=\projlim _{N\geq 1}(\mathbb {Z}/N\mathbb {Z})\) be the profinite completion of \(\mathbb {Z}\), equipped with the projective limit topology.

A cheap definition of the finite adeles \(\mathbb {A}_K^\infty \) of \(K\) is \(K\otimes _{\mathbb {Z}}\widehat{\mathbb {Z}}\), equipped with the \(\widehat{\mathbb {Z}}\)-module topology. A cheap definition of the infinite adeles \(K_\infty \) of \(K\) is \(K\otimes _{\mathbb {Q}}\mathbb {R}\) with the \(\mathbb {R}\)-module topology (this is a finite-dimensional \(\mathbb {R}\)-vector space so this is just the usual topology on \(\mathbb {R}^n\)). A cheap definition of the adeles of \(K\) is \(\mathbb {A}_K^\infty \times K_\infty \) with the product topology. This is a commutative topological ring.

However in the literature different definitions are often given. The finite adeles of \(K\) are usually defined in the books as the so-called restricted product \(\prod '_{\mathfrak {p}}K_{\mathfrak {p}}\) over the completions \(K_{\mathfrak {p}}\) of \(K\) at all maximal ideals \(\mathfrak {p}\subseteq \mathcal{O}_K\) of the integers of \(K\). Here the restricted product is the subset of \(\prod _{\mathfrak {p}}K_{\mathfrak {p}}\) consisting of elements which are in the integers \(\mathcal{O}_{K,\mathfrak {p}}\) of \(K_{\mathfrak {p}}\) for all but finitely many \(\mathfrak {p}\). This is the definition given in mathlib. Mathlib also has the proof that they’re a topological ring; furthermore the construction of the finite adeles in mathlib works for any Dedekind domain (this was pointed out to me by María Inés de Frutos Fernández; the adeles are an arithmetic object, but the finite adeles are an algebraic object).

Similarly the infinite adeles of a number field \(K\) are usually defined as \(\prod _v K_v\), the product running over the archimedean completions of \(K\), and this is the mathlib definition.

The adeles of a number field \(K\) are the product of the finite and infinite adeles, and mathlib knows that they’re a \(K\)-algebra and a topological ring.

8.3 Local compactness

As mentioned above, Salvatore Mercuri has a complete formalisation of the proof that the adele ring is locally compact as a topological space. His work is in his own repo which I don’t want to have as a dependency of FLT, because this work should all be in mathlib.

Theorem 8.1

The adeles of a number field are locally compact.

Proof

See this line in Mercuri’s repo. The proof is: a restricted product of topological spaces \(X_i\) over compact open subspaces \(C_i\) is locally compact, because \(\prod _i C_i\) is open and compact (we should have this result in mathlib in general rather than the ad hoc approach which we currently have for adeles). Hence \(\mathbb {A}_K^\infty \) is locally compact. The vector space \(\mathbb {R}^n\) is locally compact. Hence \(K_{\infty }\) is locally compact. Hence their product \(\mathbb {A}_K\) is locally compact.

8.4 Base change

The “theorem” we want is that if \(L/K\) is a finite extension of number fields, then \(\mathbb {A}_L=L\otimes _K\mathbb {A}_K\). This isn’t a theorem though, this is actually a definition (the map between the two objects) and a theorem about the definition (that it’s an isomorphism). In fact the full claim is that it is both a homeomorphism and an \(L\)-algebra isomorphism. Before we can prove the theorem, we need to make the definition.

Recall that the adeles \(\mathbb {A}_K\) of a number field is a product \(\mathbb {A}_K^\infty \times K_\infty \) of the finite adeles and the infinite adeles. So our “theorem” follows immediately from the “theorem”s that \(\mathbb {A}_L^\infty =L\otimes _K\mathbb {A}_K^\infty \) and \(L_\infty =L\otimes _KK_\infty \). We may thus treat the finite and infinite results separately.

8.4.1 Base change for finite adeles

As pointed out above, the theory of finite adeles works fine for Dedekind domains. So in this subsection let \(A\) be a Dedekind domain. Recall that the height one spectrum of \(A\) is the nonzero prime ideals of \(A\). Note that because we stick to the literature, rather than to common sense, fields are Dedekind domains in mathlib, and the height one spectrum of a field is empty. The reason I don’t like allowing fields to be Dedekind domains is that geometrically the definition of Dedekind domain in the literature is “smooth affine curve, or a point”. But many theorems in algebraic geometry begin “let \(C\) be a smooth curve”, rather than “let \(C\) be a smooth curve or a point”.

Let \(K\) be the field of fractions of \(A\). If \(v\) is in the height one spectrum of \(A\), then we can put the \(v\)-adic topology on \(A\) and on \(K\), and consider the completions \(A_v\) and \(K_v\). The finite adele ring \(\mathbb {A}_{A,K}^\infty \) is defined to be the restricted product of the \(K_v\) with respect to the \(A_v\), as \(v\) runs over the height one spectrum of \(A\).

Now let \(L/K\) be a finite separable extension, and let \(B\) be the integral closure of \(A\) in \(L\). We want to relate the finite adeles of \(K\) and of \(L\). We work place by place, starting by fixing one place upstairs and analysing its relation to the place downstairs.

So let \(w\) be a nonzero prime ideal of \(B\). Say \(w\) lies over \(v\), a prime ideal of \(A\). Then we can put the \(w\)-adic topology on \(L\) and the \(v\)-adic topology on \(K\). Furthermore we can equip \(K\) with an additive \(v\)-adic valuation, that is, a function also called \(v\) fron \(K\) to \(\mathbb {Z}\cup \{ \infty \} \) normalised so that if \(\pi \) is a uniformiser for \(v\) then \(v(\pi )=1\). Similarly we consider \(w\) as a function from \(L\) to \(\mathbb {Z}\cup \{ \infty \} \). The next lemma explains how these valuations are related.

Lemma 8.2

If \(i:K\to L\) denotes the inclusion then for \(k\in K\) we have \(e\times w(i(k))=v(k)\), where \(e\) is the ramification index of \(w/v\).

Proof

Standard (and formalized).

Definition 8.3

There’s a natural ring map \(K_v\to L_w\) extending the map \(K\to L\). It is defined by completing the inclusion \(K\to L\) at the finite places \(v\) and \(w\) (which can be done because the previous lemma shows that the map is uniformly continuous for the \(v\)-adic and \(w\)-adic topologies).

Giving \(L_w\) the \(K_v\)-module structure coming from the natural map \(K_v\to L_w\), the \(w\)-adic topology on \(L_w\) is the \(K_v\)-module topology.

Proof

Any basis for \(L\) as a \(K\)-vector space spans \(L_w\) as a \(K_v\)-module, so \(L_w\) is finite-dimensional over \(K_v\) and the module topology is the same as the product topology. So we need to establish that the product topology on \(L_w=K_v^n\) is the \(w\)-adic topology. But the \(w\)-adic topology is induced by the \(w\)-adic norm, which makes \(L_w\) into a normed \(K_v\)-vector space, and (after picking a basis) the product norm on \(L_w=K_v^n\) also makes \(L_w\) into a normed \(K_v\)-vector space. So the result follows from the standard fact (see for example the lemma on p52 of Cassels-Froelich) that any two norms on a finite-dimensional vector space over a complete field are equivalent (and thus induce the same topology).

Now instead of fixing \(w\) upstairs, we fix \(v\) downstairs. So say \(v\) is in the height one spectrum of \(A\).

Definition 8.5

The product of the maps \(K_v\to L_w\) for \(w|v\) is a natural ring map \(K_v\to \prod _{w|v}L_w\) lying over \(K\to L\). Here the product runs over the height one primes of \(B\) which pull back to \(v\).

Because \(K_v\to \prod _{w|v}L_w\) lies over \(K\to L\), there’s an induced \(L\)-algebra map \(L\otimes _KK_v\to \prod _{w|v}L_w\).

The induced \(L\)-algebra homomorphism \(L\otimes _KK_v\to \prod _{w|v}L_w\) is an isomorphism of rings.

Proof

We follow Theorem 5.12 on p21 of these notes. We may write \(L=K(\alpha )\) as a finite separable extension is simple. Let \(f(x)\) be the minimum polynomial of \(\alpha \). Factor \(f(x)=f_1(x)f_2(x)\cdots f_r(x)\) into monic irreducibles \(K_v[x]\); these are distinct by separability. We have \(L=K[x]/(f)\) so \(L\otimes _KK_v=K_v[x]/(f)=K_v[x]/(\prod _i f_i)=\oplus _i K_v[x]/(f_i)\). Write \(L_i=K_v[x]/(f_i)\). We need to show that the \(L_i\) correspond naturally to the completions \(L_w\) for \(w|v\).

First note that \([L_i:K_v]\leq [L:K]{\lt}\infty \) and so there’s a unique extension of the \(v\)-adic norm on \(K_v\) to \(L_i\). The restriction of this norm to \(L\) must be equivalent to the \(w\)-adic norm for some \(w|v\) by Ostrowski; then we must have \(L_i=L_w\) because \(L\) is dense in both and both are complete.

Next note that if \(i\not=j\) then \(L_i\) and \(L_j\) cannot be isomorphic as \(L\otimes _KK_v\)-algebras, because such an isomorphism would send \(x\) to \(x\) and thus show \(f_i=f_j\). Hence the map from \(L_i\) to the \(w\) dividing \(v\) is injective.

For surjectivity, note that if \(w|v\) then \(L_w\) is an \(L\otimes _KK_v\)-algebra and hence admits a map from \(L\otimes _K K_v\) which must factor through one of the \(L_i\). This gives an injection \(L_i\to L_w\). But \(L_i\) is complete, as it’s a finite extension of \(K_v\), and the image is dense because \(L\) is dense in \(L_w\), hence \(L_i=L_w\).

For \(v\) fixed, the product topology on \(\prod _{w|v}L_w\) is the \(K_v\)-module topology.

Proof

This is a finite product of \(K_v\)-modules each of which has the \(K_v\)-module topology by 8.4, and the product topology is the module topology for a finite product of modules each of which has the module topology (this is in mathlib).

If we give \(L\otimes _KK_v\) the \(K_v\)-module topology then the \(L\)-algebra isomorphism \(L\otimes _K K_v\cong \prod _{w|v}L_w\) is also a homeomorphism.

Proof

Indeed, is a \(K_v\)-algebra isomorphism between two modules each of which have the module topology, and any module map is automorphically continuous for the module topologies.

We now start thinking about what’s going on at the integral level.

The isomorphism \(L\otimes _KK_v\to \prod _{w|v}L_w\) induces an isomorphism \(B\otimes _AA_v\to \prod _{w|v}B_w\) for all but finitely many \(v\) in the height one spectrum of \(A\).

Proof

Certainly the image of the integral elements are integral. The argument in the other direction is more delicate. One approach (following Cassels–Froehlich, Cassels’ article “Global fields”, section 12 lemma, p61) is the following. Choose a \(K\)-basis \(\omega _1,\omega _2,\ldots ,\omega _n\) for \(L/K\) with all \(\omega _i\in A\). Then \(\omega _1A\oplus \cdots \oplus \omega _nA\) and \(B\) are two \(A\)-lattices in \(L\), so they agree at almost all primes. In particular \(\omega _1A_v\oplus \omega _2A_v\oplus \cdots \oplus \omega _nA_v=B\otimes _AA_v\) for all but finitely many primes \(v\) of \(A\). If we define the discriminant map \(D\) on \(L/K\) by \(D(\gamma _1,\gamma _2,\ldots ,\gamma _n)=det_{i,j}(trace_{L/K}(\gamma _i\gamma _j))\) then it’s well-known that \(d:=D(\omega _1,\omega _2,\ldots ,\omega _n)\) is nonzero (here we use separability), and hence a \(v\)-adic unit for all but finitely many \(v\). Furthermore if \(\gamma _i\in \prod _{w|v}B_w\) for all \(i\) then \(D(\gamma _1,\gamma _2,\ldots ,\gamma _n)\in A_v\) as all of the traces are in \(A_v\). Now say we have an element of \(\prod _{w|v}B_w\), and write it as \(\sum _i b_i\omega _i\) with \(b_i\in K_v\). Then for each \(i\) we have \(D(\omega _1,\omega _2,\ldots ,\omega _{i-1},b,\omega _{i+1},\ldots ,\omega _n)\in A_v\) but it is also \(b_i^2d\). Because \(d\) is a \(v\)-adic unit for almost all \(v\), we see that the \(b_i\) must also be in \(A_v\) for almost all \(v\).

We can take the product of the maps \(K_v\to \prod _{w|v}L_w\) over \(v\).

Definition 8.10

There’s a natural \(K\)-algebra homomorphism \(\prod _v K_v\to \prod _w L_w\), where the products run over the height one spectra of \(A\) and \(B\) respectively.

Note that we make no claim about continuity; such a claim wouldn’t help us because the adeles do not get the subspace topology.

Theorem 8.11
#

This map induces a natural \(K\)-algebra homomorphism \(\mathbb {A}_{A,K}^\infty \to \mathbb {A}_{B,L}^\infty \).

Proof

This does not need the hard direction of theorem 8.9. the definition of the map is straightforward using only that integers map to integers.

Theorem 8.12

If we give \(L\otimes _K\mathbb {A}_{A,K}^\infty \) the “module topology”, coming from the fact that \(L\otimes _K\mathbb {A}_{A,K}^\infty \) is an \(\mathbb {A}_{A,K}^\infty \)-module, then the induced \(L\)-algebra morphism \(L\otimes _K\mathbb {A}_{A,K}^\infty \to \mathbb {A}_{B,L}^\infty \) is a topological isomorphism.

Proof

Existence of the map follows from theorem 8.9. Surjectivity follows from theorem 8.9. The fact that it is a topological isomorphism surely follows from the fact that \(L\otimes _K K_v=\oplus _{w|v}L_w\) and that this identification identifies the subgroup \(\mathcal{O}_L\otimes _K K_v\) with \(\oplus _{w|v}\mathcal{O}_{L_w}\) for all but finitely many \(v\). Although I’m wondering whether it’s easier to prove that the map \(\mathbb {A}_{A,K}^\infty \to \mathbb {A}_{B,L}^\infty \) makes \(\mathbb {A}_{B,L}^\infty \) into an \(\mathbb {A}_{A,K}^\infty \)-module and the claim (also undoubtedly true, although I am not sure of the best level of abstraction here) that the topology on \(\mathbb {A}_{B,L}^\infty \) is the \(\mathbb {A}_{A,K}^\infty \)-module topology.

8.4.2 Base change for infinite adeles

Recall that if \(K\) is a number field then the infinite adeles of \(K\) are defined to be the product \(\prod _{v\mid \infty } K_v\) of all the completions of \(K\) at the infinite places.

The result we need here is that if \(L/K\) is a finite extension of number fields, then the map \(K\to L\) extends to a continuous \(K\)-algebra map \(K_\infty \to L_\infty \), and thus to a continuous \(L\)-algebra isomorphism \(L\otimes _KK_\infty \to L_\infty \). Perhaps a cheap proof would be to deduce it from the fact that \(K_\infty =K\otimes _{\mathbb {Q}}\mathbb {R}\).

Theorem 8.13

If \(K\to L\) is a ring homomorphism between two number fields then there is a natural isomorphism (both topological and algebraic) \(L\otimes _KK_\infty \cong L_\infty \).

Proof

Standard.

8.4.3 Base change for adeles

From the previous results we deduce immediately that if \(L/K\) is a finite extension of number fields then there’s a natural (topological and algebraic) isomorphism \(L\otimes _K\mathbb {A}_K\to \mathbb {A}_L\).

Theorem 8.14
#

If \(K\to L\) is a ring homomorphism between two number fields then there is a natural isomorphism (both topological and algebraic) \(L\otimes _K\mathbb {A}_K\cong \mathbb {A}_L\).

Proof

Follows from the previous results.

Something else we shall need:

Theorem 8.15

If \(K\to L\) is a ring homomorphism between two number fields then the topology on \(\mathbb {A}_L\) is the \(\mathbb {A}_K\)-module topology, where the module structure comes from the natural map \(\mathbb {A}_K\to \mathbb {A}_L\).

Proof

Because \(\mathbb {A}_L=\mathbb {A}_K\otimes _{K}L\) we know that \(\mathbb {A}_L\) is a finite free \(\mathbb {A}_K\)-module, and a standard fact (being PRed to mathlib) about the module topology is that the module topology on a finite free module is just the product topology. It thus suffices to show that the topology on \(\mathbb {A}_L\) is the product topology after picking an \(\mathbb {A}_K\)-basis for \(\mathbb {A}_L\) but this is standard.

8.5 Discreteness and compactness

We need that if \(K\) is a number field then \(K\subseteq \mathbb {A}_K\) is discrete, and the quotient (with the quotient topology) is compact. Here is a proposed proof.

Theorem 8.16
#

There’s an open subset of \(\mathbb {A}_{\mathbb {Q}}\) whose intersection with \(\mathbb {Q}\) is \(\{ 0\} \).

Proof

Use \(\prod _p{\mathbb {Z}_p}\times (-1,1)\). Any rational \(q\) in this set is a \(p\)-adic integer for all primes \(p\) and hence (writing it in lowest terms as \(q=n/d\)) satisfies \(p\nmid d\), meaning that \(d=\pm 1\) and thus \(q\in \mathbb {Z}\). The fact that \(q\in (-1,1)\) implies \(q=0\).

Theorem 8.17
#

There’s an open subset of \(\mathbb {A}_{K}\) whose intersection with \(K\) is \(\{ 0\} \).

Proof

By a previous result, we have \(\mathbb {A}_K=K\otimes _{\mathbb {Q}}\mathbb {A}_{\mathbb {Q}}\). Choose a basis of \(K/\mathbb {Q}\); then \(K\) can be identified with \(\mathbb {Q}^n\subseteq (\mathbb {A}_{\mathbb {Q}})^n\) and the result follows from the previous theorem.

Theorem 8.18
#

The additive subgroup \(K\) of \(\mathbb {A}_K\) is discrete.

Proof

If \(x\in K\) and \(U\) is the open subset in the previous lemma, then it’s easily checked that \(K\cap U=\{ 0\} \) implies \(K\cap (U+x)=\{ x\} \), and \(U+x\) is open.

For compactness we follow the same approach.

Theorem 8.19
#

The quotient \(\mathbb {A}_{\mathbb {Q}}/\mathbb {Q}\) is compact.

Proof

The space \(\prod _p\mathbb {Z}_p\times [0,1]\subseteq \mathbb {A}_{\mathbb {Q}}\) is a product of compact spaces and is hence compact. I claim that it surjects onto \(\mathbb {A}_{\mathbb {Q}}/\mathbb {Q}\). Indeed, if \(a\in \mathbb {A}_{\mathbb {Q}}\) then for the finitely many prime numbers \(p\in S\) such that \(a_p\not\in \mathbb {Z}_p\) we have \(a_p\in \frac{r_p}{p^{n_p}}+\mathbb {Z}_p\) with \(r_p/p^{n_p}\in \mathbb {Q}\), and if \(q=\sum _{p\in S}\frac{r_p}{p^{n_p}}\in \mathbb {Q}\) then \(a-q\in \prod _p\mathbb {Z}_p\times \mathbb {R}\). Now just subtract \(\lfloor a_{\infty }-q\rfloor \) to move into \(\prod _p\mathbb {Z}_p\times [0,1)\) and we are done.

Theorem 8.20
#

The quotient \(\mathbb {A}_K/K\) is compact.

Proof

We proceed as in the discreteness proof above, by reducing to \(\mathbb {Q}\). As before, choosing a \(\mathbb {Q}\)-basis of \(K\) gives us \(\mathbb {A}_K/K\cong (\mathbb {A}_{\mathbb {Q}}/\mathbb {Q})^n\) so the result follows from the previous theorem.