In the previous post, I talked about constructing the p-adic completions of Q and a bit about the analogous concept for number fields, the P-adic completions. It’s worth also talking about function fields (which, remember, are finite extensions of F(p)(x)), which exhibit striking similarities to number fields, so I’ll be more abstract than the usual here.
First, one of the features of function fields that number fields don’t have is lack of rigidity. Q is a very rigid object, in the sense that there are exactly two ring homomorphisms from Q to itself – the zero homomorphism, and the identity. A number field of degree n has, besides the zero homomorphism, at most n homomorphisms to itself, all of which are in fact isomorphisms. Function fields have infinitely many ring homomorphisms to themselves: for instance, for every nonzero integer k, there’s the homomorphism on F(p)(x) that sends x to x^k.
That creates a lot of weird features, like function field extensions that are isomorphic to the base field. For instance, F(p)(x) has an extension to F(p)(SQRT(x)), which clearly has the same ring structure as F(p)(x).
To simplify matters, I’m going to write F(p) as F, because usually the value of p isn’t important. The base function field is then F(x). Its completion with respect to the ideal (x) is F((x)), the field of Laurent polynomials. For an explanation, first consider F[x], the ring of polynomials over F; it can be extended to F[[x]], the ring of possibly infinite polynomials over F, called power series. F[[x]] is a local ring with maximal ideal (x), so the fraction field of F[[x]], F((x)), is the field of all power series, possibly with finitely many negative powers of x. The natural valuation on F((x)) is of course p^(-deg(f)), where deg(f) is the degree of the lower power of x occurring in f. If it’s hard to understand, think of F((x)) as an analog of Q(p).
Although there are infinitely many distinct valuations on F(x), they all generate extensions of F((x)), unlike in the case of Q. The valuation defined by the degree yields F((1/x)), the field of Laurent polynomials with finitely many positive powers of x but possibly infinitely many negative ones; this is clearly isomorphic to F((x)). The valuations defined by linear terms other than x yield the same structure as F((x)), since sending f(x) to f(x–a) is an isomorphism. The valuations defined by higher order polynomials in F[x] yield field extensions of F((x)), since sending f(x) to f(g(x)) is a homomorphism, which allows us to identify a subfield of the completion with F((x)).
So it makes sense to define a local field as a finite extension of either F((x)) for some finite field F, or of Q(p) for some prime p.