Fermat's Last Theorem Blueprint

6. Stating the modularity lifting theorems🔗

I think that a nice and accessible goal (which will maybe take a month or two) would be to state the modularity lifting theorems which we'll be formalising. There are in fact two; one (the "minimal case") is proved using an extension of the original Taylor--Wiles techniques, and the other is deduced from it using various more modern tricks which were developed later. This chapter (currently work in progress) will contain a detailed discussion of all the things involved in the statement of the theorem.

6.1. Automorphic forms and analysis🔗

Modular forms were historically the first nontrivial examples of automorphic forms, but by the 1950s or so it was realised that they were special cases of a very general notion of an automorphic form, as were Dirichlet characters! Modular forms are holomorphic automorphic forms for the group \GL_2/\Q, and Dirichlet characters are automorphic forms for the group \GL_1/\Q. It's possible to make sense of the notion of an automorphic form for the group G/k. Here k is a "global field" -- that is, a field which is either a finite extension of \Q (a number field) or a finite extension of (\Z/p\Z)(T) (a function field), and G is a connected reductive group variety over k.

The reason that the definition of a modular form involves some analysis (they are holomorphic functions) is that if you quotient out the group \GL_2(\R) by its centre and the maximal compact subgroup O_2(\R), you get something which can be naturally identified with the upper half plane, a symmetric space with lots of interesting differential operators associated to it (for example a Casimir operator). However if you do the same thing with \GL_1(\R) then you get a one point set, which is why a Dirichlet character is just a combinatorial object; it's a group homomorphism (\Z/N\Z)^\times\to\bbC^\times where N is some positive integer. It turns out that there are many other connected reductive groups where the associated symmetric space is 0-dimensional, and in these cases the definition of an automorphic form is again combinatorial. An example would be the group variety associated to the units of a totally definite quaternion algebra over a totally real field. In this case, the analogue of \GL_2(\R) would be the units \bbH^\times in the Hamilton quaternions, a maximal compact subgroup would be the quaternions of norm 1 (homeomorphic to the 3-sphere S^3) and quotienting out \bbH^\times by its centre \R^\times and S^3 again just gives you 1 point.

Before we talk about quaternion algebras, let's talk about central simple algebras.

6.2. Central simple algebras🔗

Convention: in this section, fields are commutative, but algebras over a field may not be.

Recall that a central simple algebra over a field K is a nonzero K-algebra D such that K is the centre of D and that D has no nontrivial two-sided ideals.

Another way of saying that D has no nontrivial two-sided ideals: every surjective ring homomorphism D\twoheadrightarrow A to any ring A is either an isomorphism, or the zero map to the zero ring. Note that this latter condition has nothing to do with K.

Lemma6.1
groupXL∃∀Nused by 0

If n\geq1 then the n\times n matrices M_n(K) are a central simple algebra over K.

Proof

We prove more generally that matrices with coefficients in K and indexed by an arbitrary nonempty finite type are a central simple algebra over K.

They are clearly an algebra over K, with K embedded via scalar matrices as usual (the injectivity of the map from K comes from nonemptiness of the finite index type). The centre clearly contains K; to show that it equals K, we argue as follows. Let e(i,j) be the matrix with a 1 in the ith row and jth column, and zeros everywhere else. An element Z=(Z_{s,t})_{s,t} of the centre commutes with all matrices e(i,j) for i\not=j and these equations immediately imply that Z_{i,j}=0 if i\not=j and that Z_{i,i}=Z_{j,j}.

It suffices to prove that any nonzero two-sided ideal I is all of M_n(K). So say 0\not=M\in I and let us fix (i,j) such that M_{i,j}\not=0. One easily checks that M_{i,j} \mathrm{id} = \sum_k e(k,i)\times M\times e(j,k)\in I, where \mathrm{id}\in M_n(K) is the identity matrix. Therefore, \mathrm{id}\in I, so I=M_n(K).

The definition also requires that the ring be nonzero, but this follows from the index type being nonempty.

Lemma6.2
groupXL∃∀Nused by 0

If D is a central simple algebra over K and L/K is a field extension, then L\otimes_KD is a central simple algebra over L.

Proof

This is not too hard: it's lemma b of section 12.4 in Peirce's "Associative algebras".

Next: define trace and norm.