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Notes on QE for ACVF Notes on QE for ACVF Fall 2016 1 Preliminaries on Valuations We begin with some very basic definitions and examples. An excellent survey of this material with can be found in van den Dries [3]. Engler and Prestel's book [4] is a excellent text on valuation theory, though the basic results we need can be found in most good graduate algebra texts. My favorite is Lang [7]. Definition 1.1 Let R be an integral domain, (Γ; +; 0; <) an ordered abelian group, a valuation is a map v : R× ! Γ such that: i) v(ab) = v(a) + v(b); ii) v(a + b) ≥ min(v(a); v(b)). By convention, we say v(0) = 1. A valued field is a field with a valuation. Exercise 1.2 a) v(1) = 0. [Hint: Consider v(1 · 1)] b) v(−1) = 0. [Hint: Ordered groups are torsion free.] c) v(x) = v(−x); d) If K is a valued field and x 6= 0, then v(1=x) = −v(x). e) If v : R ! Γ is a valuation and v(a) < v(b), then v(a + b) = v(a). [Hint: Consider v(a + b − b)] Definition 1.3 Let K be a valued field. Let O = fx 2 K : v(x) ≥ 0g and let m = fx 2 K : v(x) > 0g. Exercise 1.4 a) x 2 O is a unit if and only if v(x) = 0. b) For any x 2 K at least one of x; 1=x is in O. c) m is the unique maximal ideal of O. Definition 1.5 In general, if K is a field a proper subring A is called a valuation ring if for all x 2 K× at least one of x; 1=x 2 A. We have argued that in a valued field O = fx : v(x) ≥ 0g is a valuation ring, but in fact every valuation ring arises this way. Suppose A is a valuation ring of K. Let U be the units of A, as a multiplicative group. Let G be the group K×=U and let τ : K× ! G be the natural homomorphism. Order G by τ(x) < τ(y) if and only if y=x 2 A. If we rewrite the group G additively with identity 0, then this is a surjective valuation and A = fx : v(x) ≥ 0g. 1 Exercise 1.6 Suppose v : K ! Γ is a surjective valuation, then the valued field constructed above is isomorphic to (K; Γ; v). From a model theoretic point of view, this says there are two ways to view valued fields, with surjective valuations. We can either view them as two-sorted structures (K; Γ; v) or as one-sorted structures (K; O) since the valuation ring O is definable in (K; Γ; v) and the valuation v : K× ! Γ is interpretable in (K; O). In some important cases we can define the valuation ring in the field language Exercise 1.7 Show that Zp is definable in Qp in the field language. 2 2 a) If p 6= 2 show that Zp = fx : 9y y = px + 1g 2 2 b) Show that in Q2, Z2 = fx : 9y = 8x + 1g. Definition 1.8 If K is a valued field with valuation ring O and maximal ideal m, then the residue field of K is k = O=m. For x 2 O we let x denote the residue x=m. Lemma 1.9 Suppose K is an algebraically closed valued field. Then the residue field is algebraically closed and the value group is divisible. Proof Suppose p(X) 2 k[X] is of degree d we can find a0; : : : ; ad 2 O such d that v(ad) = 0 and p(X) = adX + ··· + a1X + a0. There is α 2 K such that P i d aiα = 0. We need to show that α 2 O. Suppose v(α) < 0. Then v(adα ) = i P i dv(α) < v(aiα ) for i < d. Thus v( aiα ) = dv(α) < 0, a contradiction. Suppose γ 2 Γ. Let a 2 K such that v(a) = γ. There is b 2 K such that n γ b = a. But then v(b) = n . Exercise 1.10 If (K; v) is a valued field and (L; w) is a valued field extension, where L=K is an algebraic extension, then kL=kK is an algebraic extension and that the value group of L is contained in the divisible hull of v(K). The following result gives a much finer version of this. See [3] 3.19. Theorem 1.11 If (K; v) ⊆ (L; w) is a valued field extension and L=K is alge- braic, the [L : K] ≤ [kL : kK ][ΓL :ΓK ]; where [ΓL :ΓK ] is the index of ΓK in ΓL. The valuation topology If v : K× ! Γ is a valuation a 2 K and γ 2 Γ let Bγ (a) = fx 2 K : v(x − a) > γg be the open ball centered at a of radius γ and let Bγ (a) = fx 2 K : v(x − a) ≥ γg be the closed ball of radius γ centered at a. 2 Lemma 1.12 If b 2 Bγ (a), then Bγ (a) = Bγ (b) and the same is true for closed balls. In other words, every point in a ball is the center of the ball. Proof Let b 2 Bγ (a). If v(x − a) > γ, then v(x − b) ≥ min(v(x − a); v(a − b)) > γ: Examples Example 1.13 Let K be any field. The trivial valuation is v : K× ! f0g. We will only consider non-trivial valuations p-adic valued fields × n × Example 1.14 For a 2 Z let vp(a) = maxfn : p jag. Define vp : Q ! Z, by vp(a=b) = vp(a) − vp(b). In this case the residue field is Fp and the value group is Z. The p-adic valuation on Q gives rise to a metric, −vp(x−y) dp(x; y) = p for x 6= y: Here we have the ultrametric inequality dp(x; y) ≤ min dp(x; z); dp(y; z): Example 1.15 Let Qp be the completion of Q with the p-adic metric. More concretely, we can think of Qp as the set of formal sums 1 X i aip where ai = 0; : : : ; p − 1 and n 2 Z i=n and we carry when we add. There is a natural extension of the p-adic valuation P1 i to Qp namely if an 6= 0, then vp( i=n aip ) = n. The value group is Z and the residue field is Fp. We let Zp be the valuation ring of Qp. Exercise 1.16 Show that Zp is compact. We can always extend valuations to field extensions. See [7] XII S4 Theorem 1. Theorem 1.17 If (K; v) is a valued field and L=K is an extension field, then there is an extension of v to a valuation on K. More can be said for extensions of the p-adics. See, for examples, Cassel's [1] S7 and 8. 3 Theorem 1.18 a) If L=Qp is a finite extension, then there is unique extension of the valuatin to L and L still a complete valued field. alg b) There is a unique extesnsion of the valuation to Qp , but it is not com- plete. alg c) Let Cp be the completion of Qp , the Cp is algebraically closed. power series fields Example 1.19 Let k be a field and let k((t)) be the field of Laurent series 1 X i ait where ai 2 k for all i: i=n P1 i There is a natural valuation v : k((t)) ! Z such that if b = i=n ait where an 6= 0, then v(b) = n. The residue b = an so the residue field is k. By this construction we can build valued fields where the field and the residue field have the same characteristic. Note that if K is a valued field with residue field k, then either K and k have the same characteristic (the equi-characteristic case) or K has characteristic 0 and k has prime characteristic. Example 1.20 Let k be a field. The field of Puiseux series over k is 1 [ 1 k((t m )): m=1 1 n P m If x = i=n ant and an 6= 0, then v(x) = n=m. We now have a valued field with residue field k and value group Q. Theorem 1.21 If k is algebraically closed of characteristic 0, then the field of Puiseux series over k is algebraically closed. For a proof see, for example, Walker [10] IV S3. This does not work in characteristic p. This doesn't work in characteristic p > 0. The series solution to xp − x = t should be of the form 1=p 1=p2 1 a + t + t + ··· + t pn + ::: where a 2 Fp, which is not a Puiseux series. Kedlaya in [6] gives a characteri- alg zation of the algebraic closure of Fp ((t)). Exercise 1.22 Suppose k is real closed. a) Show that the field of Puiseux series over k is real closed. [Hint: If we take the algebraic extension of the Puiseux series by adjoining i, we get the Puiseux series over k(i).] 1 n P m b) Show that if b = i=n ant , then b > 0 if and only if an > 0, where < is the unique ordering of the real closed field of Puiseux series. Conclude that t is infinitesimal.
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