Post on 22-Sep-2020
On the Galois Group ofLinear Difference-Differential Equations
Ruyong Feng
KLMM, Chinese Academy of Sciences, China
Ruyong Feng (KLMM, CAS) Galois Group 1 / 19
Contents
1 Basic Notations and Concepts
2 Problem Statement
3 Ring of Sequences
4 Main Results
5 Future Work
Ruyong Feng (KLMM, CAS) Galois Group 2 / 19
1. Basic Notations and Concepts
In this talk, all fields are of characteristic zero.
σ: shift operator; δ: differential operator (σδ = δσ)
k : σδ-field with alg. closed constant field.
Example: C(x , t) with σ(x) = x + 1 and δ = ddt
Difference-differential equations:σ(Y ) = AY ,δ(Y ) = BY
where Y = (y1, y2, · · · , yn)T , A ∈ GLn(k),B ∈ gln(k).
Integrable condition:σ(B)A = δ(A) + AB.
Ruyong Feng (KLMM, CAS) Galois Group 3 / 19
1. Basic Notations and Concepts
Example: Tchebychev polynomial
Tn(t) =n2
[ n2 ]∑
m=0
(−1)m(n −m − 1)!
m!(n − 2m)!(2t)n−2m.
Let Y = (Tn(t),Tn+1(t))T , thenY (n + 1, t) =
(0 1−1 2t
)Y (n, t),
dY (n,t)dt =
((n−1)t1−t2 − n−1
1−t2
n1−t2 − nt
1−t2
)Y (n, t).
(Hermite polynomial, Legendre polynomial, Bessel polynomial, · · · )
Ruyong Feng (KLMM, CAS) Galois Group 4 / 19
1. Basic Notations and Concepts
R: σδ-Picard Vessiot extension of k w.r.t. σ(Y ) = AY , δ(Y ) = BY if
R is a simple σδ-ring (no nontrivial σδ-ideals);
∃ Z ∈ GLn(R) s.t. σ(Z ) = AZ and δ(Z ) = BZ ;
R = k [Zi,j ,1
det(Z ) ] where Z = (Zi,j ).
Galois group of R over k :
Gal(R/k) , σδ-k -automorphism of R
Gal(R/k) is a linear algebraic group over the constant field of k .
Reference: Hardouin, C. and Singer, M. F., Differential Galois Theory ofLinear Difference Equations, Math. Ann., 342(2), 333-377, 2008.
Ruyong Feng (KLMM, CAS) Galois Group 5 / 19
2. Problem Statement
Let k = C(x , t).σ(Y ) = A(x , t)Y
δ(Y ) = B(x , t)Y Gσδ : Galois group over C(x , t)
c ∈ C, s.t. A(x , c),B(x , c) are well-defined and det(A(x , c)) 6= 0
σ(Y ) = A(x , c)Y Gσc : Galois group over C(x)
` ∈ Z, s.t. A(`, t),B(`, t) are well-defined and det(A(`, t)) 6= 0
δ(Y ) = B(`, t)Y Gδ` : Galois group over C(t)
Question: What are the relations among Gσδ,Gδ` and Gσ
c ?
Note:σ(B)A = δ(A) + AB ⇒ δ(Y ) = B(`, t)Y ∼ δ(Y ) = B(m, t)Y ⇒ Gδ
` = Gδm.
Ruyong Feng (KLMM, CAS) Galois Group 6 / 19
2. Problem Statement
Let k = C(x , t).σ(Y ) = A(x , t)Y
δ(Y ) = B(x , t)Y Gσδ : Galois group over C(x , t)
c ∈ C, s.t. A(x , c),B(x , c) are well-defined and det(A(x , c)) 6= 0
σ(Y ) = A(x , c)Y Gσc : Galois group over C(x)
` ∈ Z, s.t. A(`, t),B(`, t) are well-defined and det(A(`, t)) 6= 0
δ(Y ) = B(`, t)Y Gδ` : Galois group over C(t)
Question: What are the relations among Gσδ,Gδ` and Gσ
c ?
Note:σ(B)A = δ(A) + AB ⇒ δ(Y ) = B(`, t)Y ∼ δ(Y ) = B(m, t)Y ⇒ Gδ
` = Gδm.
Ruyong Feng (KLMM, CAS) Galois Group 6 / 19
2. Problem Statement
Let k = C(x , t).σ(Y ) = A(x , t)Y
δ(Y ) = B(x , t)Y Gσδ : Galois group over C(x , t)
c ∈ C, s.t. A(x , c),B(x , c) are well-defined and det(A(x , c)) 6= 0
σ(Y ) = A(x , c)Y Gσc : Galois group over C(x)
` ∈ Z, s.t. A(`, t),B(`, t) are well-defined and det(A(`, t)) 6= 0
δ(Y ) = B(`, t)Y Gδ` : Galois group over C(t)
Question: What are the relations among Gσδ,Gδ` and Gσ
c ?
Note:σ(B)A = δ(A) + AB ⇒ δ(Y ) = B(`, t)Y ∼ δ(Y ) = B(m, t)Y ⇒ Gδ
` = Gδm.
Ruyong Feng (KLMM, CAS) Galois Group 6 / 19
2. Problem Statement
Let k = C(x , t).σ(Y ) = A(x , t)Y
δ(Y ) = B(x , t)Y Gσδ : Galois group over C(x , t)
c ∈ C, s.t. A(x , c),B(x , c) are well-defined and det(A(x , c)) 6= 0
σ(Y ) = A(x , c)Y Gσc : Galois group over C(x)
` ∈ Z, s.t. A(`, t),B(`, t) are well-defined and det(A(`, t)) 6= 0
δ(Y ) = B(`, t)Y Gδ` : Galois group over C(t)
Question: What are the relations among Gσδ,Gδ` and Gσ
c ?
Note:σ(B)A = δ(A) + AB ⇒ δ(Y ) = B(`, t)Y ∼ δ(Y ) = B(m, t)Y ⇒ Gδ
` = Gδm.
Ruyong Feng (KLMM, CAS) Galois Group 6 / 19
2. Problem Statement
Let k = C(x , t).σ(Y ) = A(x , t)Y
δ(Y ) = B(x , t)Y Gσδ : Galois group over C(x , t)
c ∈ C, s.t. A(x , c),B(x , c) are well-defined and det(A(x , c)) 6= 0
σ(Y ) = A(x , c)Y Gσc : Galois group over C(x)
` ∈ Z, s.t. A(`, t),B(`, t) are well-defined and det(A(`, t)) 6= 0
δ(Y ) = B(`, t)Y Gδ` : Galois group over C(t)
Question: What are the relations among Gσδ,Gδ` and Gσ
c ?
Note:σ(B)A = δ(A) + AB ⇒ δ(Y ) = B(`, t)Y ∼ δ(Y ) = B(m, t)Y ⇒ Gδ
` = Gδm.
Ruyong Feng (KLMM, CAS) Galois Group 6 / 19
2. Problem Statement
Example: Consider σ(y) = ty
δ(y) = xt y
Gσδ = C∗
σ(y) = cy , c ∈ C∗ Gσc =
ξ ∈ C|ξm = 1 cm = 1
C∗ otherwise
δ(y) =`
ty , ` ∈ Z Gδ
` = 1
Gσδ = Gσc Gδ
` , if c is not a root of unity.
Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
2. Problem Statement
Example: Consider σ(y) = ty
δ(y) = xt y
Gσδ = C∗
σ(y) = cy , c ∈ C∗ Gσc =
ξ ∈ C|ξm = 1 cm = 1
C∗ otherwise
δ(y) =`
ty , ` ∈ Z Gδ
` = 1
Gσδ = Gσc Gδ
` , if c is not a root of unity.
Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
2. Problem Statement
Example: Consider σ(y) = ty
δ(y) = xt y
Gσδ = C∗
σ(y) = cy , c ∈ C∗
Gσc =
ξ ∈ C|ξm = 1 cm = 1
C∗ otherwise
δ(y) =`
ty , ` ∈ Z Gδ
` = 1
Gσδ = Gσc Gδ
` , if c is not a root of unity.
Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
2. Problem Statement
Example: Consider σ(y) = ty
δ(y) = xt y
Gσδ = C∗
σ(y) = cy , c ∈ C∗ Gσc =
ξ ∈ C|ξm = 1 cm = 1
C∗ otherwise
δ(y) =`
ty , ` ∈ Z Gδ
` = 1
Gσδ = Gσc Gδ
` , if c is not a root of unity.
Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
2. Problem Statement
Example: Consider σ(y) = ty
δ(y) = xt y
Gσδ = C∗
σ(y) = cy , c ∈ C∗ Gσc =
ξ ∈ C|ξm = 1 cm = 1
C∗ otherwise
δ(y) =`
ty , ` ∈ Z
Gδ` = 1
Gσδ = Gσc Gδ
` , if c is not a root of unity.
Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
2. Problem Statement
Example: Consider σ(y) = ty
δ(y) = xt y
Gσδ = C∗
σ(y) = cy , c ∈ C∗ Gσc =
ξ ∈ C|ξm = 1 cm = 1
C∗ otherwise
δ(y) =`
ty , ` ∈ Z Gδ
` = 1
Gσδ = Gσc Gδ
` , if c is not a root of unity.
Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
2. Problem Statement
Example: Consider σ(y) = ty
δ(y) = xt y
Gσδ = C∗
σ(y) = cy , c ∈ C∗ Gσc =
ξ ∈ C|ξm = 1 cm = 1
C∗ otherwise
δ(y) =`
ty , ` ∈ Z Gδ
` = 1
Gσδ = Gσc Gδ
` , if c is not a root of unity.
Ruyong Feng (KLMM, CAS) Galois Group 7 / 19
.
I will present partial results on the relations among Gσδ,Gδ` and Gσ
c .
To describe the relations among these groups, we would need to embedPicard Vessiot extensions of the above systems into the ring of sequences.
Ruyong Feng (KLMM, CAS) Galois Group 8 / 19
3. Ring of Sequences
F : differential field with derivation δ.
The ring of sequences over F :
SF := a = (a0,a1, · · · )|ai ∈ F/ ∼
where a ∼ b⇔ ∃ d ∈ Z≥0, s.t . ai = bi for all i ≥ d .
Define
a + b = (a0 + b0,a1 + b1, · · · ),ab = (a0b0,a1b1, · · · ),
σ((a0,a1, · · · , )) = (a1,a2, · · · , ),δ((a0,a1, · · · , )) = (δ(a0), δ(a1), · · · , ).
SF is a σδ-ring.
Ruyong Feng (KLMM, CAS) Galois Group 9 / 19
3. Ring of Sequences
Define σ|F = 1F and σ(x) = x + 1. Then F (x) becomes a σδ-field.
F (x) can be σδ-embedded into SF :
F (x) −→ SF
f (x) −→ (0, · · · ,0, f (N), f (N + 1), · · · , )
where f (i) is well-defined for all i ≥ N.
In particular,
F −→ SF
b −→ (b,b,b, · · · , ).
SF is a σδ-extension ring of F (x).
Ruyong Feng (KLMM, CAS) Galois Group 10 / 19
3. Ring of Sequences
F (x): σδ-field with alg. closed constant field.
A(x , t) ∈ GLn(F (x)),B(x , t) ∈ gln(F (x)).
Let ` ∈ Z>0 satisfy for all i ≥ `,
A(i , t),B(i , t) are well-defined;
det(A(i , t)) 6= 0.
K : quotient field of δ-PV extension of δ(Y ) = B(`, t)Y over F .
U: fundamental matrix of δ(Y ) = B(`, t)Y in GLn(K ).
Ruyong Feng (KLMM, CAS) Galois Group 11 / 19
3. Ring of Sequences
Define V = (V0,V1, · · · , ) ∈ GLn(SK ) as
V0 = · · · = V`−1 = 0,V` = U,V`+1 = A(`+ 1, t)V`,V`+2 = A(`+ 2, t)V`+1, · · · .
Theorem: F (x)[V ,1/det(V )] is a σδ-Picard Vessiot extension ofσ(Y ) = A(x , t)Y ,δ(Y ) = B(x , t)Y
over F (x).
Note: F (x)[V ,1/det(V )] is a σδ-subring of SF .
Ruyong Feng (KLMM, CAS) Galois Group 12 / 19
3. Ring of Sequences
Define V = (V0,V1, · · · , ) ∈ GLn(SK ) as
V0 = · · · = V`−1 = 0,V` = U,V`+1 = A(`+ 1, t)V`,V`+2 = A(`+ 2, t)V`+1, · · · .
Theorem: F (x)[V ,1/det(V )] is a σδ-Picard Vessiot extension ofσ(Y ) = A(x , t)Y ,δ(Y ) = B(x , t)Y
over F (x).
Note: F (x)[V ,1/det(V )] is a σδ-subring of SF .
Ruyong Feng (KLMM, CAS) Galois Group 12 / 19
4. Main Results
Let F = C(t).
Lemma: Gδ` is an algebraic subgroup of Gσδ (under isomorphism).
Proof:ψ : Gδ
` = Gal(K/C(t)) −→ σδ-Aut(SK/C(x , t))
ρ −→ ψ(ρ)
ψ(ρ)(a) = (ρ(a))
ψ(Gδ` ) −→ Gσδ
ψ(ρ) −→ ψ(ρ)|C(x,t)[V ,1/ det(V )]
Ruyong Feng (KLMM, CAS) Galois Group 13 / 19
4. Main Results
Let F = C(t).
Lemma: Gδ` is an algebraic subgroup of Gσδ (under isomorphism).
Proof:ψ : Gδ
` = Gal(K/C(t)) −→ σδ-Aut(SK/C(x , t))
ρ −→ ψ(ρ)
ψ(ρ)(a) = (ρ(a))
ψ(Gδ` ) −→ Gσδ
ψ(ρ) −→ ψ(ρ)|C(x,t)[V ,1/ det(V )]
Ruyong Feng (KLMM, CAS) Galois Group 13 / 19
4. Main Results
Let F = C(t).
Lemma: Gδ` is an algebraic subgroup of Gσδ (under isomorphism).
Proof:ψ : Gδ
` = Gal(K/C(t)) −→ σδ-Aut(SK/C(x , t))
ρ −→ ψ(ρ)
ψ(ρ)(a) = (ρ(a))
ψ(Gδ` ) −→ Gσδ
ψ(ρ) −→ ψ(ρ)|C(x,t)[V ,1/ det(V )]
Ruyong Feng (KLMM, CAS) Galois Group 13 / 19
4. Main Results
Ω: differentially closed field containing C(t).
GσδΩ : Galois group of
σ(Y ) = A(x , t)Yδ(Y ) = B(x , t)Y
over Ω(x).
Lemma: GσδΩ is a normal algebraic subgroup of Gσδ (under isomorphism).
Theorem: Gσδ = GσδΩ Gδ
` .
Gσt : Galois group of σ(Y ) = A(x , t)Y over C(t)(x).
Theorem: Gσt (Ω) is conjugate over Ω to Gσδ
Ω (Ω).
Remark: Under the conjugation, Gσt is an algebraic group defined over C and
Gσt (C) = Gσδ
Ω . In this sense, Gσδ = Gσt (C)Gδ
` .
Ruyong Feng (KLMM, CAS) Galois Group 14 / 19
4. Main Results
Ω: differentially closed field containing C(t).
GσδΩ : Galois group of
σ(Y ) = A(x , t)Yδ(Y ) = B(x , t)Y
over Ω(x).
Lemma: GσδΩ is a normal algebraic subgroup of Gσδ (under isomorphism).
Theorem: Gσδ = GσδΩ Gδ
` .
Gσt : Galois group of σ(Y ) = A(x , t)Y over C(t)(x).
Theorem: Gσt (Ω) is conjugate over Ω to Gσδ
Ω (Ω).
Remark: Under the conjugation, Gσt is an algebraic group defined over C and
Gσt (C) = Gσδ
Ω . In this sense, Gσδ = Gσt (C)Gδ
` .
Ruyong Feng (KLMM, CAS) Galois Group 14 / 19
4. Main Results
Ω: differentially closed field containing C(t).
GσδΩ : Galois group of
σ(Y ) = A(x , t)Yδ(Y ) = B(x , t)Y
over Ω(x).
Lemma: GσδΩ is a normal algebraic subgroup of Gσδ (under isomorphism).
Theorem: Gσδ = GσδΩ Gδ
` .
Gσt : Galois group of σ(Y ) = A(x , t)Y over C(t)(x).
Theorem: Gσt (Ω) is conjugate over Ω to Gσδ
Ω (Ω).
Remark: Under the conjugation, Gσt is an algebraic group defined over C and
Gσt (C) = Gσδ
Ω . In this sense, Gσδ = Gσt (C)Gδ
` .
Ruyong Feng (KLMM, CAS) Galois Group 14 / 19
4. Main Results
Ω: differentially closed field containing C(t).
GσδΩ : Galois group of
σ(Y ) = A(x , t)Yδ(Y ) = B(x , t)Y
over Ω(x).
Lemma: GσδΩ is a normal algebraic subgroup of Gσδ (under isomorphism).
Theorem: Gσδ = GσδΩ Gδ
` .
Gσt : Galois group of σ(Y ) = A(x , t)Y over C(t)(x).
Theorem: Gσt (Ω) is conjugate over Ω to Gσδ
Ω (Ω).
Remark: Under the conjugation, Gσt is an algebraic group defined over C and
Gσt (C) = Gσδ
Ω . In this sense, Gσδ = Gσt (C)Gδ
` .
Ruyong Feng (KLMM, CAS) Galois Group 14 / 19
4. Main Results
Ω: differentially closed field containing C(t).
GσδΩ : Galois group of
σ(Y ) = A(x , t)Yδ(Y ) = B(x , t)Y
over Ω(x).
Lemma: GσδΩ is a normal algebraic subgroup of Gσδ (under isomorphism).
Theorem: Gσδ = GσδΩ Gδ
` .
Gσt : Galois group of σ(Y ) = A(x , t)Y over C(t)(x).
Theorem: Gσt (Ω) is conjugate over Ω to Gσδ
Ω (Ω).
Remark: Under the conjugation, Gσt is an algebraic group defined over C and
Gσt (C) = Gσδ
Ω . In this sense, Gσδ = Gσt (C)Gδ
` .
Ruyong Feng (KLMM, CAS) Galois Group 14 / 19
4. Main Results
Example: Y (n + 1, t) =
(0 1−1 2t
)Y (n, t),
dY (n,t)dt =
(n−1)t1−t2 − n−1
1−t2
n1−t2 − nt
1−t2
Y (n, t).
Gσδ =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
⋃(0 ξη 0
)∣∣∣ ξη = 1ξ, η ∈ C
Gσδ
Ω =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Ruyong Feng (KLMM, CAS) Galois Group 15 / 19
4. Main Results
Example: Y (n + 1, t) =
(0 1−1 2t
)Y (n, t),
dY (n,t)dt =
(n−1)t1−t2 − n−1
1−t2
n1−t2 − nt
1−t2
Y (n, t).
Gσδ =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
⋃(0 ξη 0
)∣∣∣ ξη = 1ξ, η ∈ C
GσδΩ =
(ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Ruyong Feng (KLMM, CAS) Galois Group 15 / 19
4. Main Results
Example: Y (n + 1, t) =
(0 1−1 2t
)Y (n, t),
dY (n,t)dt =
(n−1)t1−t2 − n−1
1−t2
n1−t2 − nt
1−t2
Y (n, t).
Gσδ =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
⋃(0 ξη 0
)∣∣∣ ξη = 1ξ, η ∈ C
Gσδ
Ω =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Ruyong Feng (KLMM, CAS) Galois Group 15 / 19
4. Main Results
Example: Y (n + 1, t) =
(0 1−1 2t
)Y (n, t),
dY (n,t)dt =
(n−1)t1−t2 − n−1
1−t2
n1−t2 − nt
1−t2
Y (n, t).
Gσδ =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
⋃(0 ξη 0
)∣∣∣ ξη = 1ξ, η ∈ C
Gσδ
Ω =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Gσt =
(ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C(t)
Gσ
t (C) =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Ruyong Feng (KLMM, CAS) Galois Group 16 / 19
4. Main Results
Example: Y (n + 1, t) =
(0 1−1 2t
)Y (n, t),
dY (n,t)dt =
(n−1)t1−t2 − n−1
1−t2
n1−t2 − nt
1−t2
Y (n, t).
Gσδ =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
⋃(0 ξη 0
)∣∣∣ ξη = 1ξ, η ∈ C
Gσδ
Ω =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Gσ
t =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C(t)
Gσ
t (C) =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Ruyong Feng (KLMM, CAS) Galois Group 16 / 19
4. Main Results
Example: Y (n + 1, t) =
(0 1−1 2t
)Y (n, t),
dY (1,t)dt =
(1−1)t1−t2 − 1−1
1−t2
11−t2 − 1t
1−t2
Y (1, t).
Gσδ =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
⋃(0 ξη 0
)∣∣∣ ξη = 1ξ, η ∈ C
Gσδ
Ω =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Gσ
t =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C(t)
Gσ
t (C) =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Gδ1 =
(1 00 1
),(
0 11 0
)
Ruyong Feng (KLMM, CAS) Galois Group 17 / 19
4. Main Results
Example: Y (n + 1, t) =
(0 1−1 2t
)Y (n, t),
dY (1,t)dt =
(1−1)t1−t2 − 1−1
1−t2
11−t2 − 1t
1−t2
Y (1, t).
Gσδ =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
⋃(0 ξη 0
)∣∣∣ ξη = 1ξ, η ∈ C
Gσδ
Ω =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Gσ
t =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C(t)
Gσ
t (C) =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Gδ
1 =(
1 00 1
),(
0 11 0
)
Ruyong Feng (KLMM, CAS) Galois Group 17 / 19
4. Main Results
Example: Y (n + 1, t) =
(0 1−1 2t
)Y (n, t),
dY (n,t)dt =
(n−1)t1−t2 − n−1
1−t2
n1−t2 − nt
1−t2
Y (n, t).
Gσδ =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
⋃(0 ξη 0
)∣∣∣ ξη = 1ξ, η ∈ C
Gσδ
Ω =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Gσ
t =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C(t)
Gσ
t (C) =(
ξ 00 η
)∣∣∣ ξη = 1ξ, η ∈ C
Gδ
1 =(
1 00 1
),(
0 11 0
)Gσδ = Gσδ
Ω Gδ1 = Gσ
t (C)Gδ1
Ruyong Feng (KLMM, CAS) Galois Group 18 / 19
5. Future Work
Gσt : Galois group of σ(Y ) = A(x , t)Y over C(t)(x)
Gσc : Galois group of σ(Y ) = A(x , c)Y over C(x)
To give the complete results, one need to solve
Problem: What are the relations between Gσt and Gσ
c ?
Ruyong Feng (KLMM, CAS) Galois Group 19 / 19