Computable Fields and their Algebraic Closures

Transcription

1 Computable Fields and their Algebraic Closures Russell Miller Queens College & CUNY Graduate Center New York, NY. Workshop on Computability Theory Universidade dos Açores Ponta Delgada, Portugal, 6 July 2010 Slides available at qc.edu/ rmiller/slides.html Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 1 / 17

2 Classical Algebraic Closures Theorem Every field F has an algebraic closure F: a field extension of F which is algebraically closed and algebraic over F. This algebraic closure of F is unique up to F-isomorphism. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 2 / 17

3 Classical Algebraic Closures Theorem Every field F has an algebraic closure F: a field extension of F which is algebraically closed and algebraic over F. This algebraic closure of F is unique up to F-isomorphism. The theory Th(ACF m ) of algebraically closed fields of characteristic m is κ-categorical for every uncountable κ, and has countable models F m F m (X 0 ) F m (X 0, X 1 ) F m (X 0, X 1, X 2,...). So ACF s of characteristic m are indexed by their transcendence degrees. (Here F 0 = Q and F p = Z/(pZ) for prime p.) Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 2 / 17

4 Classical Algebraic Closures Theorem Every field F has an algebraic closure F: a field extension of F which is algebraically closed and algebraic over F. This algebraic closure of F is unique up to F-isomorphism. The theory Th(ACF m ) of algebraically closed fields of characteristic m is κ-categorical for every uncountable κ, and has countable models F m F m (X 0 ) F m (X 0, X 1 ) F m (X 0, X 1, X 2,...). So ACF s of characteristic m are indexed by their transcendence degrees. (Here F 0 = Q and F p = Z/(pZ) for prime p.) Fact All countable ACF s are computably presentable. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 2 / 17

5 Splitting Algorithms Theorem (Kronecker, 1882) The field Q has a splitting algorithm: it is decidable which polynomials in Q[X] have factorizations in Q[X]. Let F be a computable field of characteristic 0 with a splitting algorithm. Every primitive extension F(x) of F also has a splitting algorithm, which may be found uniformly in the minimal polynomial of x over F (or uniformly knowing that x is transcendental over F). Recall that for x E algebraic over F, the minimal polynomial of x over F is the unique monic irreducible p(x) F[X] with p(x) = 0. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 3 / 17

6 Splitting Algorithms Theorem (Kronecker, 1882) The field Q has a splitting algorithm: it is decidable which polynomials in Q[X] have factorizations in Q[X]. Let F be a computable field of characteristic 0 with a splitting algorithm. Every primitive extension F(x) of F also has a splitting algorithm, which may be found uniformly in the minimal polynomial of x over F (or uniformly knowing that x is transcendental over F). Recall that for x E algebraic over F, the minimal polynomial of x over F is the unique monic irreducible p(x) F[X] with p(x) = 0. Corollary For any algebraic computable field F, every finitely generated subfield Q(x 1,..., x n ) or F p (x 1,..., x n ) has a splitting algorithm, uniformly in the tuple x 1,..., x d. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 3 / 17

7 Computable Algebraic Closures We want a presentation of F with F as a recognizable subfield. Defn. For a computable field F, a Rabin embedding of F consists of a computable field E and a field homomorphism g : F E such that: E is algebraically closed; E is algebraic over the image g(f); and g is a computable function. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 4 / 17

8 Computable Algebraic Closures We want a presentation of F with F as a recognizable subfield. Defn. For a computable field F, a Rabin embedding of F consists of a computable field E and a field homomorphism g : F E such that: E is algebraically closed; E is algebraic over the image g(f); and g is a computable function. Rabin s Theorem (1960); see also Frohlich & Shepherdson (1956) Every computable field F has a Rabin embedding. Moreover, for every Rabin embedding g : F E, the following are Turing-equivalent: the image g(f), as a subset of E; the splitting set S F = {p F[X] : p factors nontrivially in F[X]}; the root set R F = {p F[X] : p has a root in F}. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 4 / 17

9 Proof of Rabin s Theorem R F T S F Given p(x), an S F -oracle allows us to find the irreducible factors of p in F[X]. But p R F iff p has a linear factor. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 5 / 17

10 Proof of Rabin s Theorem R F T S F Given p(x), an S F -oracle allows us to find the irreducible factors of p in F[X]. But p R F iff p has a linear factor. S F T g(f) Given a monic p(x) F[X], find all its roots r 1,..., r d E. Factorizations of its image p g in E[X] are all of the form p g (X) = h(x) j(x) = (Π i S (X r i )) (Π i / S (X r i )) for some S {1,..., d}. Check if any of these factors lies in g(f)[x]. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 5 / 17

11 Proof of Rabin s Theorem R F T S F Given p(x), an S F -oracle allows us to find the irreducible factors of p in F[X]. But p R F iff p has a linear factor. S F T g(f) Given a monic p(x) F[X], find all its roots r 1,..., r d E. Factorizations of its image p g in E[X] are all of the form p g (X) = h(x) j(x) = (Π i S (X r i )) (Π i / S (X r i )) for some S {1,..., d}. Check if any of these factors lies in g(f)[x]. g(f) T R F Given x E, find some p(x) F[X] for which p g (x) = 0. Find all roots of p in F: if p R F, find a root r 1 F, then check if p(x) X r 1 R F, etc. Then x g(f) iff x is the image of one of these roots. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 5 / 17

12 Different Presentations of F Theorem Let F = F be two computable presentations of the same field. Assume that F is algebraic (over its prime subfield Q or F p ). Then R F T R F. Proof: Given p(x) F[X], find q(x) F m [X] divisible by p(x). Use R F to find all roots of h(q)(x) in F. Then find the same number of roots of q(x) in F, and check whether any one is a root of p(x). F = F h : F m F m Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 6 / 17

13 Comparing R F, S F, and g(f ) We know that R F T S F T g(f). Is there any way to distinguish the complexity of these sets? Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 7 / 17

14 Comparing R F, S F, and g(f ) We know that R F T S F T g(f). Is there any way to distinguish the complexity of these sets? Recall: A 1 B if there is a 1-1 computable f such that: ( x)[x A f (x) B]. A wtt B if there are Φ e and a computable bound g with: ( x)φ B g(x) e (x) = χ A (x). Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 7 / 17

15 Comparing R F, S F, and g(f ) We know that R F T S F T g(f). Is there any way to distinguish the complexity of these sets? Recall: A 1 B if there is a 1-1 computable f such that: ( x)[x A f (x) B]. A wtt B if there are Φ e and a computable bound g with: ( x)φ B g(x) e (x) = χ A (x). Theorem (M, 2010) For all algebraic computable fields F, S F 1 R F. However, there exists such a field F with R F 1 S F. Problem: Given a polynomial p(x) F[X], compute another polynomial q(x) F[X] such that p(x) splits q(x) has a root. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 7 / 17

16 p(x) splits q(x) has a root. Let F t be the subfield F m [a 0,..., a t 1 ]. So every F t has a splitting algorithm. For a given p(x), find an t with p F t [X]. Check first whether p splits there. If so, pick its q(x) to be a linear polynomial. If not, find the splitting field K t of p(x) over F t, and the roots r 1,..., r d of p(x) in K t. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 8 / 17

17 p(x) splits q(x) has a root. Let F t be the subfield F m [a 0,..., a t 1 ]. So every F t has a splitting algorithm. For a given p(x), find an t with p F t [X]. Check first whether p splits there. If so, pick its q(x) to be a linear polynomial. If not, find the splitting field K t of p(x) over F t, and the roots r 1,..., r d of p(x) in K t. Proposition For F t L K t, p(x) splits in L[X] iff there exists S {r 1,..., r d } such that L contains all elementary symmetric polynomials in S. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 8 / 17

18 p(x) splits q(x) has a root. Let F t be the subfield F m [a 0,..., a t 1 ]. So every F t has a splitting algorithm. For a given p(x), find an t with p F t [X]. Check first whether p splits there. If so, pick its q(x) to be a linear polynomial. If not, find the splitting field K t of p(x) over F t, and the roots r 1,..., r d of p(x) in K t. Proposition For F t L K t, p(x) splits in L[X] iff there exists S {r 1,..., r d } such that L contains all elementary symmetric polynomials in S. Effective Theorem of the Primitive Element Each finite algebraic field extension is generated by a single element, which we can find effectively. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 8 / 17

19 Procedure to Compute q(x) For each intermediate field F t L S K t generated by the elementary symmetric polynomials in S, let x S be a primitive generator. Let q(x) be the product of the minimal polynomials q S (X) F t [X] of each x S. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 9 / 17

20 Procedure to Compute q(x) For each intermediate field F t L S K t generated by the elementary symmetric polynomials in S, let x S be a primitive generator. Let q(x) be the product of the minimal polynomials q S (X) F t [X] of each x S. : If p(x) splits in F [X], then F contains some L S. But then x S F, and q S (x S ) = 0. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 9 / 17

21 Procedure to Compute q(x) For each intermediate field F t L S K t generated by the elementary symmetric polynomials in S, let x S be a primitive generator. Let q(x) be the product of the minimal polynomials q S (X) F t [X] of each x S. : If p(x) splits in F [X], then F contains some L S. But then x S F, and q S (x S ) = 0. : If q(x) has a root x F, then some q S (x) = 0, so x is F t -conjugate to some x S. Then some σ Gal(K t /F t ) maps x S to x. But σ permutes the set {r 1,..., r d }, so x generates the subfield containing all elementary symmetric polynomials in σ(s). Then F contains the subfield L σ(s), so p(x) splits in F[X]. Thus S F 1 R F. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 9 / 17

22 No Reverse Reduction Theorem (Steiner, 2010) There exists a computable algebraic field F with R F wtt S F. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 10 / 17

23 No Reverse Reduction Theorem (Steiner, 2010) There exists a computable algebraic field F with R F wtt S F. Proof uses the following distinction between R F and S F : Facts For every Galois extension L Q and all intermediate fields F 0 and F 1 : R F0 Q[X] = R F1 Q[X] σ Gal(L/Q)[σ(F 0 ) = F 1 ]. But there exist such L F 1 F 0 Q for which S F0 Q[X] = S F1 Q[X]. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 10 / 17

24 What about the Rabin Image g(f)? Theorem (Steiner 2010) Among the reducibilities T, wtt, m, and 1, the following are the strongest which hold for all computable algebraic fields F: S F 1 R F S F wtt g(f) R F wtt g(f) R F T S F g(f) T S F g(f) wtt R F So S F is, relatively, the easiest to compute. R F and g(f) appear the same except that we have a field F with S F 1 R F and S F 1 g(f). So R F is stronger, in a subtle way. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 11 / 17

25 What about the Rabin Image g(f)? Theorem (Steiner 2010) Among the reducibilities T, wtt, m, and 1, the following are the strongest which hold for all computable algebraic fields F: S F 1 R F S F wtt g(f) R F wtt g(f) R F T S F g(f) T S F g(f) wtt R F So S F is, relatively, the easiest to compute. R F and g(f) appear the same except that we have a field F with S F 1 R F and S F 1 g(f). So R F is stronger, in a subtle way. Remaining work: for isomorphic computable algebraic fields F = F, how do these sets compare? Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 11 / 17

26 Noncomputable Algebraic Fields Now let F be any field algebraic over Q (or over F p ), but not necessarily computable. We wish to consider the spectrum of F: Spec(F) = {T-degrees d : K = F[deg(K ) = d]}. Problem: Describe Spec(F). Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 12 / 17

27 Noncomputable Algebraic Fields Now let F be any field algebraic over Q (or over F p ), but not necessarily computable. We wish to consider the spectrum of F: Spec(F) = {T-degrees d : K = F[deg(K ) = d]}. Problem: Describe Spec(F). Now if K = F and deg(k ) = d, then d can enumerate Q K K. Moreover, d can compute the (unique) isomorphism from Q F onto a fixed computable copy of Q. Moreover, every x K has a minimal polynomial over Q, and d can find it. (Kronecker!) Thus d can enumerate (Q[X] R F ). Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 12 / 17

28 Noncomputable Algebraic Fields Now let F be any field algebraic over Q (or over F p ), but not necessarily computable. We wish to consider the spectrum of F: Spec(F) = {T-degrees d : K = F[deg(K ) = d]}. Problem: Describe Spec(F). Now if K = F and deg(k ) = d, then d can enumerate Q K K. Moreover, d can compute the (unique) isomorphism from Q F onto a fixed computable copy of Q. Moreover, every x K has a minimal polynomial over Q, and d can find it. (Kronecker!) Thus d can enumerate (Q[X] R F ). Theorem (Frolov, Kalimullin, & M 2009) For any algebraic field extension F Q, Spec(F) = {d : d can enumerate Q[X] R F }. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 12 / 17

29 Useful Field Fact The proof of the inclusion uses: Fact For algebraic fields F and K, the following are equivalent: F = K. F K and K F. Every f.g. subfield of each field embeds into the other field. Let Q = K 0 K 1 K 2 = K, and f s : K s F. By algebraicity, there are only finitely many possible embeddings of each K s into F. So let g 0 = f 0 and g s be any extension of g s 1 such that t s[ f t K s = g s ]. This is noneffective, but then g = s g s embeds K into F. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 13 / 17

30 And for Algebraic Closures... Now let F be a computable copy of the algebraic closure of the algebraic field F. We have another notion of the spectrum: DgSp F (F) = {deg(g(f)) : g : F E is an isomorphism & E T }. Problem: Describe DgSp F (F). Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 14 / 17

31 And for Algebraic Closures... Now let F be a computable copy of the algebraic closure of the algebraic field F. We have another notion of the spectrum: DgSp F (F) = {deg(g(f)) : g : F E is an isomorphism & E T }. Problem: Describe DgSp F (F). Theorem (Frolov, Kalimullin, & M 2009) For any algebraic field extension F Q, either DgSp F (F) = { deg(q[x] R F ) } or DgSp F (F) = {d : d can compute Q[X] R F }. So we have a contrast. For F as a field, the spectrum was really an upper cone of e-degrees. For F as a relation on F, the spectrum is an upper cone of Turing degrees. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 14 / 17

32 Galois Groups Bad news: the automorphism group of a countable algebraic field can be uncountable! (E.g. Aut(Q) has size 2 ω.) So there is no hope that the Galois group of a computable field extension might always be computably presentable. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 15 / 17

33 Galois Groups Bad news: the automorphism group of a countable algebraic field can be uncountable! (E.g. Aut(Q) has size 2 ω.) So there is no hope that the Galois group of a computable field extension might always be computably presentable. Idea: name elements σ Aut(F) the way computable analysts name real numbers: by giving approximations σ n = σ {0, 1,..., n}. From such approximations to any σ, τ Aut(F), we can likewise approximate (τ σ). Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 15 / 17

34 Galois Actions So, to give an effective presentation of Aut(F) in this manner, we need to be able to compute (or at least enumerate) the set A F = { a 0,..., a n : b 0,..., b n : ( σ Aut(F))( i) σ(a i ) = b i }. This is the full Galois action of F. Equivalently, we need to compute or enumerate the orbit relation (or Galois action) on F : B F = { a, b : σ Aut(F) σ(a) = b}. The Galois action has recently proven useful in attempts (with Shlapentokh) to characterize computable categoricity for computable algebraic fields. Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 16 / 17

35 Standard References on Computable Fields Yu.L. Ershov; Theorie der Numerierungen III, Zeitschrift für Mathematische Logik und Grundlagen der Mathematik 23 (1977) 4, A. Frohlich & J.C. Shepherdson; Effective procedures in field theory, Phil. Trans. Royal Soc. London, Series A 248 (1956) 950, G. Metakides & A. Nerode; Effective content of field theory, Annals of Mathematical Logic 17 (1979), M. Rabin; Computable algebra, general theory, and theory of computable fields, Transactions of the American Mathematical Society 95 (1960), V. Stoltenberg-Hansen & J.V. Tucker; Computable rings and fields, in Handbook of Computability Theory, ed. E.R. Griffor (Amsterdam: Elsevier, 1999), These slides available at qc.edu/ rmiller/slides.html Russell Miller (CUNY) Algebraic Closures of Computable Fields WCT 2010 Azores 17 / 17