Transcendence in Positive Characteristic

Transcendence in Positive Characteristic Galois Group Examples and Applications W. Dale Brownawell Matthew Papanikolas Penn State University Texas A&M University Arizona Winter School 2008 March 18, 2008 AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 1 / 28

Outline 1 Preliminaries 2 Carlitz logarithms 3 Carlitz zeta values 4 Rank 2 Drinfeld modules AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 2 / 28

Preliminaries Notation Transcendence degree theorem AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 3 / 28

Scalar quantities Let p be a fixed prime; q a fixed power of p. A := F q [θ] Z k := F q (θ) Q k Q k := F q ((1/θ)) R C := k C f = q deg f AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 4 / 28

Functions Rational functions: F q (t), k(t), C (t). Analytic functions: { } T := a i t i C [[t]] a i 0. and i 0 i 0 L := fraction field of T. Entire functions: { E := a i t i i } C [[t]] ai 0,. [k (a 0, a 1, a 2,... ) : k ] < AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 5 / 28

Galois groups and transcendence degree Theorem (P. 2008) Let M be a t-motive, and let Γ M be its associated group via Tannakian duality. Suppose that Φ GL r (k(t)) Mat r (k[t]) represents multiplication by σ on M and that det Φ = c(t θ) s, c k. Let Ψ be a rigid analytic trivialization of Φ in GL r (T) Mat r (E). That is, Ψ ( 1) = ΦΨ. Finally let Then L = k(ψ(θ)) k. tr. deg k L = dim Γ M (= dim Γ Ψ ). AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 6 / 28

Carlitz logarithms Difference equations for Carlitz logarithms Calculation of the Galois group Algebraic independence An explicit example: log C (ζ θ ) AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 7 / 28

Carlitz logarithms Recall the Carlitz exponential: exp C (z) = z + i=1 z qi (θ qi θ)(θ qi θ q ) (θ qi θ qi 1 ). AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 8 / 28

Carlitz logarithms Recall the Carlitz exponential: exp C (z) = z + i=1 Its formal inverse is the Carlitz logarithm, log C (z) = z + z qi (θ qi θ)(θ qi θ q ) (θ qi θ qi 1 ). i=1 z qi (θ θ q )(θ θ q2 ) (θ θ qi ). log C (z) converges for z < θ q/(q 1) and satisfies θ log C (z) = log C (θz) + log C (z q ). AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 8 / 28

The function L α (t) For α k, α < θ q/(q 1), we define L α (t) = α + i=1 Connection with Carlitz logarithms: Functional equation: α qi (t θ q )(t θ q2 ) (t θ qi ) T, L α (θ) = log C (α). L ( 1) α = α ( 1) + L α t θ. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 9 / 28

Difference equations for Carlitz logarithms Suppose α 1,..., α r k, α i < θ q/(q 1) for each i. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 10 / 28

Difference equations for Carlitz logarithms Suppose α 1,..., α r k, α i < θ q/(q 1) for each i. If we set t θ 0 0 α ( 1) Φ = 1 (t θ) 1 0......, α ( 1) r (t θ) 0 1 then Φ represents multiplication by σ on a t-motive M with 0 C M 1 r 0. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 10 / 28

Difference equations for Carlitz logarithms Suppose α 1,..., α r k, α i < θ q/(q 1) for each i. If we set t θ 0 0 α ( 1) Φ = 1 (t θ) 1 0......, α ( 1) r (t θ) 0 1 then Φ represents multiplication by σ on a t-motive M with We let Then 0 C M 1 r 0. Ω 0 0 ΩL α1 1 0 Ψ =....... ΩL αr 0 1 Ψ ( 1) = ΦΨ. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 10 / 28

Specialize Ψ at t = θ and find 1/π q 0 0 log C (α 1 )/π q 1 0 Ψ(θ) =....... log C (α r )/π q 0 1 AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 11 / 28

Specialize Ψ at t = θ and find 1/π q 0 0 log C (α 1 )/π q 1 0 Ψ(θ) =....... log C (α r )/π q 0 1 Thus we can determine tr. deg k k(π q, log C (α 1 ),..., log C (α r )) by calculating dim Γ Ψ. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 11 / 28

Calculating Γ Ψ Set Ψ 1, Ψ 2 GL r+1 (L k(t) L) so that (Ψ 1 ) ij = Ψ ij 1, (Ψ 2 ) ij = 1 Ψ ij, and set Ψ = Ψ 1 1 Ψ 2 GL r+1 (L k(t) L). AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 12 / 28

Calculating Γ Ψ Set Ψ 1, Ψ 2 GL r+1 (L k(t) L) so that (Ψ 1 ) ij = Ψ ij 1, (Ψ 2 ) ij = 1 Ψ ij, and set Ψ = Ψ 1 1 Ψ 2 GL r+1 (L k(t) L). Define an F q (t)-algebra map, µ = (X ij Ψ ij ) : F q (t)[x, 1/ det X] L k(t) L, which defines the F q (t)-subgroup scheme Γ Ψ GL r+1/fq(t). In our case, this implies first that {[ ]} 0 Γ Ψ GL id r+1/fq(t). r Thus we can consider the coordinate ring of Γ Ψ to be a quotient of F q (t)[x 0,..., X r, 1/X 0 ]. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 12 / 28

The vector group V The homomorphism of F q (t)-group schemes [ ] α 0 δ id r coincides with the surjection, α : Γ Ψ pr G m Γ Ψ Γ C. (Γ C = Gm ). AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 13 / 28

The vector group V The homomorphism of F q (t)-group schemes [ ] α 0 δ id r coincides with the surjection, α : Γ Ψ pr G m Γ Ψ Γ C. (Γ C = Gm ). Thus we have exact sequence of group schemes over F q (t): 0 V Γ Ψ pr G m 0, and we can consider V (G a ) r over F q (t). AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 13 / 28

Consider α G m (F q (t)) and a lift γ Γ Ψ (F q (t)). AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 14 / 28

Consider α G m (F q (t)) and a lift γ Γ Ψ (F q (t)). For any u = [ ] 1 0 v I r V (Fq (t)), we find that [ ] γ 1 1 0 uγ = V (F αv q (t)). I r AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 14 / 28

Consider α G m (F q (t)) and a lift γ Γ Ψ (F q (t)). For any u = [ ] 1 0 v I r V (Fq (t)), we find that [ ] γ 1 1 0 uγ = V (F αv q (t)). Thus V (F q (t)) is a vector subspace of F q (t) r. I r AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 14 / 28

Consider α G m (F q (t)) and a lift γ Γ Ψ (F q (t)). For any u = [ ] 1 0 v I r V (Fq (t)), we find that [ ] γ 1 1 0 uγ = V (F αv q (t)). Thus V (F q (t)) is a vector subspace of F q (t) r. Now V is smooth over F q (t) because pr : Γ Ψ G m is surjective on Lie algebras. I r AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 14 / 28

Consider α G m (F q (t)) and a lift γ Γ Ψ (F q (t)). For any u = [ ] 1 0 v I r V (Fq (t)), we find that [ ] γ 1 1 0 uγ = V (F αv q (t)). Thus V (F q (t)) is a vector subspace of F q (t) r. Now V is smooth over F q (t) because pr : Γ Ψ G m is surjective on Lie algebras. It follows that defining equations for V are linear forms in X 1,..., X r over F q (t). I r AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 14 / 28

Definining equations for Γ Ψ Pick b 0 F q (t) \ F q. Lift (use Hilbert Thm. 90) to b 0 0 0 b 1 1 0 γ =...... Γ Ψ(F q (t)). b r 0 1 We can use γ to create defining equations for Γ Ψ using defining forms for V. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 15 / 28

Theorem Suppose F = c 1 X 1 + + c r X r, c 1,..., c r F q (t), is a defining linear form for V. Then G = (b 0 1)F F(b 1,..., b r )(X 0 1) is a defining polynomial for Γ Ψ. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 16 / 28

Theorem Suppose F = c 1 X 1 + + c r X r, c 1,..., c r F q (t), is a defining linear form for V. Then G = (b 0 1)F F(b 1,..., b r )(X 0 1) is a defining polynomial for Γ Ψ. In particular, if we take t = θ, (b 0 (θ) 1) r c i (θ) log C (α i ) i=1 r c i (θ)b i (θ)π q = 0. Every k-linear relation among π q, log C (α 1 ),..., log C (α r ) is a k-linear combination of relations of this type. i=1 AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 16 / 28

Theorem Suppose F = c 1 X 1 + + c r X r, c 1,..., c r F q (t), is a defining linear form for V. Then G = (b 0 1)F F(b 1,..., b r )(X 0 1) is a defining polynomial for Γ Ψ. In particular, if we take t = θ, (b 0 (θ) 1) r c i (θ) log C (α i ) i=1 r c i (θ)b i (θ)π q = 0. Every k-linear relation among π q, log C (α 1 ),..., log C (α r ) is a k-linear combination of relations of this type. Let N be the k-linear span of π q, log C (α 1 ),..., log C (α r ). Then i=1 dim Γ Ψ = dim k N. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 16 / 28

Algebraic independence of Carlitz logarithms Starting with α 1,..., α r k (suitably small), we found Φ Mat r (k[t]) and Ψ Mat r (E) so that 1/π q 0 0 log C (α 1 )/π q 1 0 Ψ(θ) =....... log C (α r )/π q 0 1 Since tr. deg k k(π q, log C (α 1 ),... log C (α r )) = dim Γ Ψ = dim k N, we can prove the following the following theorem. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 17 / 28

Algebraic independence of Carlitz logarithms Starting with α 1,..., α r k (suitably small), we found Φ Mat r (k[t]) and Ψ Mat r (E) so that 1/π q 0 0 log C (α 1 )/π q 1 0 Ψ(θ) =....... log C (α r )/π q 0 1 Since tr. deg k k(π q, log C (α 1 ),... log C (α r )) = dim Γ Ψ = dim k N, we can prove the following the following theorem. Theorem (P. 2008) Suppose log C (α 1 ),..., log C (α r ) are linearly independent over k = F q (θ). Then they are algebraically independent over k AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 17 / 28

An Example Recall ζ θ = q 1 θ, exp C ( πq /θ ) = ζ θ, log C (ζ θ ) = π q θ. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 18 / 28

An Example Recall We take ζ θ = q 1 θ, exp C ( πq /θ ) = ζ θ, log C (ζ θ ) = π q θ. [ ] t θ 0 Φ = ζ 1/q, Ψ = θ (t θ) 1 [ ] Ω 0. ΩL ζθ 1 AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 18 / 28

An Example Recall We take ζ θ = q 1 θ, exp C ( πq /θ ) = ζ θ, log C (ζ θ ) = π q θ. [ ] t θ 0 Φ = ζ 1/q, Ψ = θ (t θ) 1 We have a relation over k on the entries of [ ] 1/πq 0 Ψ(θ) =, 1/θ 1 [ ] Ω 0. ΩL ζθ 1 namely θx 21 + 1 = 0. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 18 / 28

So dim Γ Ψ = 1. (It s at least 1 since Γ Ψ G m.) Question: What are its defining equations? AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 19 / 28

So dim Γ Ψ = 1. (It s at least 1 since Γ Ψ G m.) Question: What are its defining equations? We begin with matrices in GL r (L k(t) L): Ψ 1 = [ ] Ω 1 0, Ψ ΩL ζθ 1 1 2 = [ ] 1 Ω 0. 1 ΩL ζθ 1 AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 19 / 28

So dim Γ Ψ = 1. (It s at least 1 since Γ Ψ G m.) Question: What are its defining equations? We begin with matrices in GL r (L k(t) L): Ψ 1 = [ ] Ω 1 0, Ψ ΩL ζθ 1 1 2 = [ ] 1 Ω 0. 1 ΩL ζθ 1 Then the defining equations over F q (t) for Γ Ψ will be precisely relations among the entries of [ Ψ 1 1 Ψ 1 2 = Ω Ω 0 ]. L ζθ Ω + 1 ΩL ζθ 1 AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 19 / 28

So dim Γ Ψ = 1. (It s at least 1 since Γ Ψ G m.) Question: What are its defining equations? We begin with matrices in GL r (L k(t) L): Ψ 1 = [ ] Ω 1 0, Ψ ΩL ζθ 1 1 2 = [ ] 1 Ω 0. 1 ΩL ζθ 1 Then the defining equations over F q (t) for Γ Ψ will be precisely relations among the entries of [ Ψ 1 1 Ψ 1 2 = Ω Ω 0 ]. L ζθ Ω + 1 ΩL ζθ 1 Consider the identity of functions (check!), ζ θ (t θ)ω(t) tω(t)l ζθ (t) 1 = 0, and substitute into the lower left entry of Ψ 1 1 Ψ 2. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 19 / 28

[ Ψ 1 1 Ψ 1 2 = Ω Ω 0 ] L ζθ Ω + 1 ΩL ζθ 1 ζ θ (t θ)ω tωl ζθ 1 = 0 AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 20 / 28

[ Ψ 1 1 Ψ 1 2 = Ω Ω 0 ] L ζθ Ω + 1 ΩL ζθ 1 Lower left entry of Ψ 1 1 Ψ 2 is ζ θ (t θ)ω tωl ζθ 1 = 0 L ζθ Ω + 1 ΩL ζθ = ( 1 t ζ θ(t θ) 1 tω) Ω + 1 1 t (ζ θ(t θ)ω 1) = 1 t ζ θ(t θ) Ω + 1 tω Ω + 1 1 t ζ θ(t θ)ω 1 1 t = 1 ( 1 t Ω Ω) 1 t (1 1). AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 20 / 28

[ Ψ 1 1 Ψ 1 2 = Ω Ω 0 ] L ζθ Ω + 1 ΩL ζθ 1 Lower left entry of Ψ 1 1 Ψ 2 is ζ θ (t θ)ω tωl ζθ 1 = 0 L ζθ Ω + 1 ΩL ζθ = ( 1 t ζ θ(t θ) 1 tω) Ω Therefore, Γ Ψ is defined by + 1 1 t (ζ θ(t θ)ω 1) = 1 t ζ θ(t θ) Ω + 1 tω Ω + 1 1 t ζ θ(t θ)ω 1 1 t = 1 ( 1 t Ω Ω) 1 t (1 1). Γ Ψ : tx 12 X 11 + 1 = 0. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 20 / 28

Carlitz zeta values Brief review of Carlitz zeta values Algebraic independence theorem of Chang-Yu Theorem of Chang-P.-Yu for varying q AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 21 / 28

Applications to Carlitz zeta values ζ C (n) = a F q[θ] a monic 1 a n k, n = 1, 2,.... AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 22 / 28

Applications to Carlitz zeta values ζ C (n) = a F q[θ] a monic 1 a n k, n = 1, 2,.... As you may recall from the 2nd lecture, using the theory of Anderson and Thakur, one can construct a system of difference equations Ψ ( 1) = ΦΨ so that ζ C (n) appears in Ψ(θ). AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 22 / 28

Applications to Carlitz zeta values ζ C (n) = a F q[θ] a monic 1 a n k, n = 1, 2,.... As you may recall from the 2nd lecture, using the theory of Anderson and Thakur, one can construct a system of difference equations Ψ ( 1) = ΦΨ so that ζ C (n) appears in Ψ(θ). Known algebraic relations over k among ζ C (n): (q 1) n ζ C (n) = r n π n q, r n F q (θ), (Euler-Carlitz) ζ C (np) = ζ C (n) p, (Frobenius). AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 22 / 28

The Chang-Yu Theorem Algebraic independence of ζ C (n) Theorem (Chang-Yu 2007) For any positive integer n, the transcendence degree of the field k(π q, ζ C (1),..., ζ C (n)) over k is n n p n q 1 + n p(q 1) + 1. Question: What can we say about Carlitz zeta values if we allow q to vary? AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 23 / 28

The Chang-Yu Theorem Algebraic independence of ζ C (n) Theorem (Chang-Yu 2007) For any positive integer n, the transcendence degree of the field k(π q, ζ C (1),..., ζ C (n)) over k is n n p n q 1 + n p(q 1) + 1. Question: What can we say about Carlitz zeta values if we allow q to vary? Answer: Even then, the Euler-Carlitz relations and the Frobenius p-th power relations tell the whole stoty... AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 23 / 28

Zeta values with varying constant fields For m 1, we set ζ m (n) = a F p m [θ] a monic 1, n = 1, 2,.... an Theorem (Chang-P.-Yu) For any positive integers s and d, the transcendence degree of the field k ( d m=1 {π p m, ζ m(1),..., ζ m (s)} ) over k is d m=1 ( s s p s p m 1 + s p(p m 1) ) + 1. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 24 / 28

Rank 2 Drinfeld modules Periods and quasi-periods A Galois group example Algebraic independence in the non-cm case AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 25 / 28

Periods and quasi-periods of rank 2 Drinfeld modules Recall that for a rank 2 Drinfeld module ρ : F q [t] k[f] with ρ(t) = θ + κf + F 2, we can take [ ] [ ] [ ] 0 1 0 1 s (1) Φ = t θ κ 1/q, Ψ = 1 s (2) 1 1 1 κ s (1) 2 s (2). 2 Furthermore, [ ] Ψ(θ) 1 ω1 η = 1, ω 2 η 2 where ω 1, ω 2, η 1, η 2 are the periods and quasi-periods for ρ. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 26 / 28

An example (κ = θ + θ q ) Assume p 2. Consider the Drinfeld module ρ with ρ(t) = θ + ( θ + θ q) F + F 2. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 27 / 28

An example (κ = θ + θ q ) Assume p 2. Consider the Drinfeld module ρ with ρ(t) = θ + ( θ + θ q) F + F 2. After going through the Galois group calculation, we find in this case {[ ]} α βt Γ Ψ =. β α AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 27 / 28

An example (κ = θ + θ q ) Assume p 2. Consider the Drinfeld module ρ with ρ(t) = θ + ( θ + θ q) F + F 2. After going through the Galois group calculation, we find in this case {[ ]} α βt Γ Ψ =. β α Thus dim Γ Ψ = 2. However, here ρ has complex multiplication by F q [ t], where t acts by θ + F. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 27 / 28

An example (κ = θ + θ q ) Assume p 2. Consider the Drinfeld module ρ with ρ(t) = θ + ( θ + θ q) F + F 2. After going through the Galois group calculation, we find in this case {[ ]} α βt Γ Ψ =. β α Thus dim Γ Ψ = 2. However, here ρ has complex multiplication by F q [ t], where t acts by θ + F. Theorem (Thiery 1992) The period matrix of a Drinfeld module of rank 2 over k with CM has transcendence degree 2 over k. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 27 / 28

Rank 2 Drinfeld modules without CM In general, we say that a Drinfeld module ρ does not have complex multiplication if End(ρ) = F q [t]. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 28 / 28

Rank 2 Drinfeld modules without CM In general, we say that a Drinfeld module ρ does not have complex multiplication if End(ρ) = F q [t]. Theorem (Chang-P.) Suppose that p 2. Let ρ be a Drinfeld module of rank 2 over k without CM. Then Γ ρ = GL2. In particular, the periods and quasi-periods of ρ, ω 1, ω 2, η 1, η 2, are algebraically independent over k. AWS 2008 (Lecture 4) Galois Group Examples and Applications March 18, 2008 28 / 28