Local Fields

Dear Readers,

Let’s examine the role of topology in the study of fields and arithmetic. A topology on a field compatible with the field operations is given by an absolute value, which in turn defines a metric. Outside of number theory, people usually mean the standard real absolute  value |x| = \pm x when they talk about an absolute value. Note that on \mathbf{Q} this absolute value carries the Archimedian property that \mathbf{Z} is an unbounded set. What about the other case?

We say an absolute value is non-Archimedian if instead of the standard triangle inequality |a + b| \le |a| + |b| we have the inequality |a+ b| \le \max\{ |a|, |b|\}(note that under this equality |n| \le 1 if n is an integer, so the integers are a bounded set under their canonical image in our field). If you are not familiar with this, I don’t know of a better gentle introduction than Fernando Gouvea’s Introduction to thep-adic numbers. Very briefly, we can define an absolute value on \mathbf{Q} by defining it on \mathbf{Z}. If p is a prime and m is a nonzero integer then we can write m = n p^r where n, r are nonzero integers and p  does not divide n. Then we can define |m|_p = p^{-r} to be the p-adic absolute value on \mathbf{Q}. So we have the real topology, the discrete topology and the p-adic topologies on \mathbf{Q}. The following important theorem says that this is it(at least for those which “respect the field operations” as we put it).

Theorem(Ostrowski’s Theorem): If |\cdot | is an absolute value on \mathbf{Q} then the topology it defines is homeomorphic to that generated by the real absolute value, the discrete absolute value or one of the p-adic absolute values.

This is a magnificent theorem which can be proved using very classical analytic methods . Certainly of interest is the topic of completions, which are well-covered in any undergraduate analysis course. With respect to any of these absolute values we can complete the rational numbers and get a new field which strictly contains the rationals and is often easier to work with. If we complete the rationals with respect to the discrete valuation… we just get the rationals again because any Cauchy sequence is already eventually constant. If we complete with respect to the real absolute value, we get \mathbf{R} and if we complete with respect to a p-adic absolute value, we get \mathbf{Q}_p.

There is another theorem of Ostrowski’s, this time on completions, which we note here:

Theorem( Ostrowski’s [Big] Theorem): A field K complete with respect to an Archimedian absolute value is either \mathbf{R} or \mathbf{C}.

I use the terminology “Big” in parallel to those two Picard Theorems in Complex Analysis, and also because we use “Little” Ostrowski to prove this theorem. The key observation is that if K has nonzero characteristic as a ring then the integers have finite image in K, and thus form a bounded set. Hence K has characteristic zero, so \mathbf{Z} and thus \mathbf{Q} inject into K. Then the absolute value restricted to \mathbf{Q} must still be archimedian and thus equivalent to the real absolute value, so \mathbf{R} injects into K. Then some analysis can show that any element of K is degree 2 algebraic over \mathbf{R}.

So what to do with non-Archimedian absolute values on a field L? Note that each defines a valuation ring R made up of the elements of size less than or equal to one and a valuation ideal of elements of size strictly less than one.  If the supremum of the absolute values of elements in the valuation ideal is not 1, we call R a discrete valuation ring. The valuation ideal is maximal since everything in L of size exactly 1 has an inverse in the valuation ring. The quotient is a field called the residue field and with this in mind we use the following (non-standard, but often used) terminology.

Definition: We call L a local field if it is complete with respect to a non-Archimedian absolute value, its valuation ring is a discrete valuation ring and its residue field is finite.

Using this definition of a local field, we can give the following complete characterization:

Theorem: If L is a local field, it is a finite extension of either \mathbf{Q}_p or \mathbb{F}_p((t))(the Laurent series field over \mathbb{F}_p).

The reason local fields are so popular is the following:

Theorem(Hensel’s Lemma): If L is a local field, R its valuation ring, P its valuation ideal and f a polynomial over R such that not all of its coefficients are in P then if f factors into relatively prime polynomials \bar{g}, \bar{h} in the residue field then there exist lifts g, h \in R[x] such that \deg(g) = \deg(\bar g), f = gh.

Just like “Local Fields” there are many definitions of “Hensel’s Lemma” of varying strengths. For instance, I do not need L to be a local field here, only complete with respect to a non-Archimedian absolute value. Possibly the strongest possible version is the case that if R is a valuation ring(not necessarily discrete) and its fraction field K is complete with respect to the given non-Archimedian absolute value and V is a variety over R, then a smooth rational point in the special fiber(over the residue field) gives way to a smooth point on the generic fiber (over K). For a proof, see Silverman’s Advanced Topics in the Arithmetic of Elliptic curves, Theorem IV.6.4.

Note also that he speaks in terms of Henselian rings and fields. That’s where we’re going next.

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3 Responses to Local Fields

  1. My favorite version of Hensel’s Lemma is the following:

    Let A be a commutative ring, complete with respect to the ideal \mathfrak{m}_A (that is, set up as the completion of a local ring), and let f(x)\in A[x]. Then if a\in A such that f(x)\equiv 0 (mod f'(a)^2\mathfrak{m}_A), then there exists b\in A such that f(b)=0 and b\equiv a (mod f'(a)\mathfrak{m}_A).

    This version, if you do the details out with A=\mathbb{R} and \mathfrak{m}_A=(1), you pretty much get Newton’s Method of approximating the roots of functions, so a Henselian ring is one where Newton’s method works.

  2. Alpha says:

    You write “where n, r are nonzero integers” in the sentence before the definition of the p-adic absolute value on Q. Are you sure that r can not be zero?

    • Jim Stankewicz says:

      Answer one: Yes, that’s what I meant. We can have r as any non-negative integer.

      Answer two: Well, since we’re just defining it on the integers to extend it to the rationals, considering integers of this form is technically good enough… but what you said is still what I meant.

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