Notes on QE for ACVF

Transcription

1 Notes on QE for ACVF Fall 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

2 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] 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

3 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

4 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

5 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

6 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] 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

7 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

8 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

9 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, [2] Z. Chatzidakis, Théorie des Modèls des corps valués, 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, [4] A. J. Engler and A. Prestel, Valued Fields, Springer, [5] I. Kaplansky, Maximal fields with valuations. Duke Math. J. 9, (1942) [6] K. Kedlaya, The algebraic closure of the power series field in positive characteristic. Proc. Amer. Math. Soc. 129 (2001), no. 12, [7] S. Lang, Algebra, Addison-Wesley, [8] H. D. Macpherson, Model Theory of Valued Fields, [9] D. Marker, Model Theory: An Introduction, Springer, [10] R. Walker, Algebraic Curves, Springer-Verlag,