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) =. 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 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 = {x K : v(x) 0} and let m = {x K : v(x) > 0}. Exercise 1.4 a) x O is a unit if and only if v(x) = 0. b) For any x 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 K at least one of x, 1/x A. We have argued that in a valued field O = {x : v(x) 0} 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 A. If we rewrite the group G additively with identity 0, then this is a surjective valuation and A = {x : v(x) 0}. 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 Z p is definable in Q p in the field language. a) If p 2 show that Z p = {x : y y 2 = px 2 + 1} b) Show that in Q 2, Z 2 = {x : y 2 = 8x 2 + 1}. 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 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) k[x] is of degree d we can find a 0,..., a d O such that v(a d ) = 0 and p(x) = a d X d + + a 1 X + a 0. There is α K such that ai α i = 0. We need to show that α O. Suppose v(α) < 0. Then v(a d α d ) = dv(α) < v(a i α i ) for i < d. Thus v( a i α i ) = dv(α) < 0, a contradiction. Suppose γ Γ. Let a K such that v(a) = γ. There is b K such that b n = a. But then v(b) = γ 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 k L /k K 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 algebraic, the [L : K] [k L : k K ][Γ L : Γ K ], where [Γ L : Γ K ] is the index of Γ K in Γ L. The valuation topology If v : K Γ is a valuation a K and γ Γ let B γ (a) = {x K : v(x a) > γ} be the open ball centered at a of radius γ and let B γ (a) = {x K : v(x a) γ} be the closed ball of radius γ centered at a. 2

Lemma 1.12 If b B γ (a), then B γ (a) = B γ (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 B γ (a). If v(x a) > γ, then Examples v(x b) min(v(x a), v(a b)) > γ. Example 1.13 Let K be any field. The trivial valuation is v : K {0}. We will only consider non-trivial valuations p-adic valued fields Example 1.14 For a Z let v p (a) = max{n : p n a}. Define v p : Q Z, by v p (a/b) = v p (a) v p (b). In this case the residue field is F p and the value group is Z. The p-adic valuation on Q gives rise to a metric, Here we have the ultrametric inequality d p (x, y) = p vp(x y) for x y. d p (x, y) min d p (x, z), d p (y, z). Example 1.15 Let Q p be the completion of Q with the p-adic metric. More concretely, we can think of Q p as the set of formal sums a i p i where a i = 0,..., p 1 and n Z i=n and we carry when we add. There is a natural extension of the p-adic valuation to Q p namely if a n 0, then v p ( i=n a ip i ) = n. The value group is Z and the residue field is F p. We let Z p be the valuation ring of Q p. Exercise 1.16 Show that Z p is compact. 1. We can always extend valuations to field extensions. See [7] XII S4 Theorem 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/Q p is a finite extension, then there is unique extension of the valuatin to L and L still a complete valued field. b) There is a unique extesnsion of the valuation to Q alg p, but it is not complete. c) Let C p be the completion of Q alg p, the C p is algebraically closed. power series fields Example 1.19 Let k be a field and let k((t)) be the field of Laurent series a i t i where a i k for all i. i=n There is a natural valuation v : k((t)) Z such that if b = i=n a it i where a n 0, then v(b) = n. The residue b = a n 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 m=1 k((t 1 m )). If x = i=n a nt n m and a n 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 x p x = t should be of the form a + t 1/p + t 1/p2 + + t 1 p n +... where a F p, which is not a Puiseux series. Kedlaya in [6] gives a characterization of the algebraic closure of F alg p ((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).] b) Show that if b = i=n a nt n m, then b > 0 if and only if a n > 0, where < is the unique ordering of the real closed field of Puiseux series. Conclude that t is infinitesimal. 4

Example 1.23 Let k be a field and let (Γ, +, <, 0) be an ordered abelian group. Consider all formal sums f = γ Γ a γ t γ. The support of f is {γ : a γ 0}. The field of Hahn series k(((γ))) is the collection of all formal sums with well ordered support. We define ( a γ t γ )( b γ t γ ) = ( γ 1+γ 2=γ a γ1 b γ2 ) t γ. The definition of Hahn series is set up so that {(γ 1, γ 2 ) : γ 1 + γ 2 = γ} is a finite set and multiplication is well defined (note: you still need to show the support of the product is well ordered see [3] 3.5). We can define a valuation letting v(f) be the minimal element of the support of f. The residue field is k and the value group is Γ. Exercise 1.24 k(((γ))) has no proper extension to a valued field with value group Γ and residue field k. If K is real closed or algebraically closed with equi-characteristic residue field, the residue field is k and the value group is Γ, then there is a valuation preserving embedding of K into k(((γ))). (See [5] for more details). Other examples Example 1.25 Let a C and let M a be the field of germs of meromorphic functions at a. Consider the map ord(f) where if f is defined at a, then ord(f) is the order of zero at a, and if a is a pole, ord(f) is minus the order of pole at a. Then ord: M a Z is a valuation with residue field C. Exercise 1.26 Let F be an ordered field with infinite elements and let O be a proper convex subring. Show that O is a valuation ring. Exercise 1.27 Suppose F is a real closed field with infinite elements and O is the ring of elements bounded in absolute value by a standard natural number. a) Show that the maximal idea m is the set of infinitesimals. b) Show that the residue field k is a real closed subfield of R, there is an embedding of k into K and the residue map is the standard part map. c) Show that the value group is divisible. 2 Quantifier Elimination References for this section include [2] and [3] Let L d be the language {+,,, 0, 1, } where is a binary relation symbol and if (K, v) is a valued field then K = x y if and only if v(x) v(y), i.e., if and only if y x O. Note that the valuation ring O is quantifier free definable in L d and is definable in (K, O). 5

Theorem 2.1 (Robinson) The theory of algebraically closed fields with a nontrivial valuation admits quantifier elimination in the language L d. 1 Quantifier elimination will follow from the following Proposition. Proposition 2.2 Suppose (K, v) and (L, w) are algebraically closed fields with non-trivial valuation and L is K + -saturated. Suppose R K is a subring and f : R L is and L d -embedding. Then f extends to a valued field embedding g : K L. Exercise 2.3 Show that the proposition implies quantifier elimination. [Hint: See [8] 14.3.28. We will prove the Proposition via a series of lemmas. First, we show that without loss of generality we can assume R is a field. Lemma 2.4 Suppose (K, v) and (L, w) are valued fields, R K is a subring and f : R L is and L d -embedding. Then f extends to a valuation preserving embedding of K 0, the fraction field of R into L. Proof Extend f to K 0, by f(a/b) = f(a)/f(b). Suppose x, y K 0. There are a, b, c R such that x = a c and y = b c. Then v(x) v(y) v(a) v(b) R = a b L = f(a) f(b) v(f(x)) v(f(y)). We next show that we can extend embedding from fields to algebraically closed fields. Lemma 2.5 Suppose (K, v) and (L, w) are valued fields, K 0 K is a field and f : K 0 L is a valuation preserving embedding. Then f extends to a valuation preserving embedding of K alg 0, the algebraic closure of K 0 into L. The proof of Lemma 2.5 will require one important algebraic result from valuation theory. Recall that an algebraic extension F/K is normal if every polynomial p K[X] with one root in F has all roots in F. (i.e., F is a splitting field for any polynomial over K having a zero in F. In characteristic 0, this is equivalent to F/K is Galois, but in general a normal extension is a Galois extension followed by a purely inseparable extension. The following theorem is proved in [3] 3.15 or [7] XII S4. Theorem 2.6 Let (K, v) be a valued field, let F/K be normal and suppose O 1 and )O 2 are valuation rings of F such that O i K = O K. Then there is σ Gal(F/K) such that σ(o 1 ) = O 2., i.e., any two extensions of the valuation to F over K are conjugate. 1 Actually, Robinson only proved model completeness, but his methods extend to prove quantifier elimination. 6

Proof of Lemma 2.5 It suffices to show that if x K \ K 0 is algebraic over K, then we can extend f to K(x). Let K 0 (x) F K with F/K 0 normal. There is a field embedding g : F L with g f and g(v) gives rise to a valuation on g(f ) extending f(v K 0 ). Then g(v F ) and w g(f ) are valuations on g(f ) extending f(v K 0 ) on f(k 0 ). By the Theorem above there is σ Gal(g(F )/f(k 0 )) mapping g(v F ) to w g(f ). Thus σ g is the desired valued field embedding of F in to L extending f. Thus in proving Proposition 2.2 it suffices to show that if we have (K, v) and (L, w) non-trivially valued algebraically closed fields, L K + -saturated, K 0 K algebraically closed and f : K 0 L a valued preserving embedding, then we can extend f to K. There are three cases to consider. Case 1: Suppose x K, v(x) = 0 and x is transcendental over k K0. We will show that we can extend f to K 0 [x], then use Lemmas 2.4 and 2.5 to extend to K 0 (x) acl. Since L is K + -saturated, there is y L such that y is transcendental over k f(k0). We will send x to y. Suppose a = m 0 + a 1 x + + m n x n, where m i K 0. Suppose m l has minimal valuation. Then a = m l ( b i x i ) where v(b i ) 0 and b l = 1. Then v( b i x i ) 0. If v( b i x i ) > 0, then taking residues we see that bi x i = 0, but b l = 1, so this is a nontrivial polynomial and x is algebraic over k K0. Thuse v( b i x i ) = 0 and v(a) = m l. Thus v(a) = min{v(m i ) : i = 0,..., n}. Similarly, in L, w( f(m i )y i ) = min{w(f(m i )) : i = 0,..., n}. Thus the extension of f to K 0 [x] is and L d - embedding. Case 2: Suppose x K and v(x) v(k 0 ). Let γ = v(x). Suppose a, b K 0, i < j are in N, and v(a) + iγ = v(b) + jγ. Since K 0 is algebraically closed there is c K 0 such that c j i = a b, but then γ = v(c) v(k 0 ). Suppose a K 0 [x] and a = m 0 + m 1 x +... m n x n. Since the v(m i ) + iγ are distinct, v(a) = minv(m i ) + iγ. Since L is K + -saturated, there is y L realizing the type {w(f(a)) < w(y) : a K 0, v(a) < v(x)} {w(y) < w(f(b)) : v(x) < v(a)} Then v(a)+iv(x) < v(b)+jv(x) if and only if w(f(a))+iw(y) < w(f(b))+jw(y) for all a, b K 0 and the extension of f to K 0 [x] sending x to y is and L d - embedding. Case 3: Suppose x K \ K 0, v(k 0 (x)) = v(k 0 ) and k K0(x) = k K0, i.e., K 0 (x) is an immediate extension of K 0. Let C = {v(x a) : a K 0 }. Since v(k 0 (x)) = v(k 0 ), C v(k 0 ). We claim that C has no maximal element. Suppose v(b) C is maximal. Then 7

v( x a b ) = 0 and, since k K 0 = k K0(x), there is c K 0 such that x a b c = ɛ where v(ɛ) > 0. But then, a contradiction. Consider the tuple v(x a bc) = v(bɛ) > v(b), Σ(y) = {w(y f(a)) = w(b) : a, b K 0, v(x a) = v(b). We claim that Σ is finitely satisfiable. Suppose a 1,..., a n, b 1,..., b n K 0 and v(x a i ) = v(b i ). Because f is valuation preserving it suffices to find c K 0 with v(c a i ) = v(b i ) for i = 1,..., n. Since C has no maximal element, there is c K 0 such that v(x c) > v(b i ) for i = 1,..., n. Then v(c a i ) = v(x a i ) = v(b i ). By sending x to y we can extend f to an ring isomorphism between K 0 [x] and f(k 0 )[y]. For a K 0, there is p(x) K 0 [X] such that d = p(x). Factoring p into linear factors over the algebraically closed field K 0, there is a 0,..., a n such that d = p(x) = a 0 n i=1 (x a i ). For each i we can find b i K 0 such that v(x a i ) = v(b i ). Thus By choice of y, we also have thus f preserves. v(d) = v(a 0 ) + w(f(d)) = w(f(a 0 ) + n v(b i ) i=1 n w(f(b i )), i=1 This concludes the proof of quantifier elimination. Quantifier Elimination in multi-sorted languages It is useful to have some multi-sorted versions of quantifier elimination. Let L d,γ be a language with two sorts, a sort for a field K and a sort for the value group Γ. On the field sort we have L d. On the value group sort we have {+,, <, 0} and we have the valuation map v : K Γ. Let L d,γ,k be a language with three sorts where we have an additional k for the residue field. On the residue field we have the usual language of fields {+,,, 0, 1} and we have an additional map r : K 2 k where r(a, b) is the residue of a b if v(a) v(b) and b 0. Proving quantifier elimination in L d,γ,k, say, requires two extra steps to make sure the residue map and the valuation are surjective. Note that the steps below 8

that passed from a subring to a substructure that is an algebraically closed field, go through without significant changes. Exercise 2.7 Suppose (K 0, Γ 0, k 0 ) is an L d,γ,k substructure of a non-trivially valued field K, K 0 is an algebraically closed field, f : (K 0, Γ 0, k 0 ) L is an L d,γ,k -embedding and γ Γ 0 \ v(k 0 ). Then we can extend f so that γ is in the value group of the domain. [Hint: Similar to case 2]. Exercise 2.8 Suppose (K 0, Γ 0, k 0 ) is an L d,γ,k substructure of a non-trivially valued field K, K 0 is an algebraically closed field, f : (K 0, Γ 0, k 0 ) L is an L d,γ,k -embedding and a k 0 \ res(k 2 ). Then we can extend f so that a is in the residue field of the domain. References [1] J. W. S. Cassels, Local Fields, Cambridge University Press, 1986. [2] Z. Chatzidakis, Théorie des Modèls des corps valués, http://www.math.ens.fr/ zchatzid/papiers/cours08.pdf [3] L. P. D. van den Dries, Lectures on the Model Theory of Valued Fields, Model Theory in Algebra, Analysis and Arithmetic, H. D. Macpherson and C. Toffalori ed., Springer, 2010. [4] A. J. Engler and A. Prestel, Valued Fields, Springer, 2005. [5] I. Kaplansky, Maximal fields with valuations. Duke Math. J. 9, (1942). 303-321. [6] K. Kedlaya, The algebraic closure of the power series field in positive characteristic. Proc. Amer. Math. Soc. 129 (2001), no. 12, 3461-3470. [7] S. Lang, Algebra, Addison-Wesley, 1971. [8] H. D. Macpherson, Model Theory of Valued Fields, http://www1.maths.leeds.ac.uk/pure/staff/macpherson/modnetval4.pdf [9] D. Marker, Model Theory: An Introduction, Springer, 2002. [10] R. Walker, Algebraic Curves, Springer-Verlag, 1978. 9