The Absolute Galois Group of C(t)

Let

C

be an algebraically closed field of cardinality

m

,

x

an indeterminate,

E

a finite extension of

C

(

x

) of genus

g

, and

S

a set of prime divisors of

E

/

C

. We denote the maximal extension of

E

ramified at most over

S

by

E

S

. If

X

is a smooth projective model of

E

/

C

, then we interpret

S

as a subset of

X

(

C

), call Gal(

E

S

/

E

) the

fundamental group of

X

\

S

, and denote it by

π

1

(

X

\

S

). Starting from the fundamental group of the corresponding Riemann surface and applying the Riemann existence theorem, one proves that when

r

=card(

S

)<∞, Gal(

E

S

/

E

) is the free profinite group generated by

r

+2

g

elements

σ

1

,…,

σ

r

,

τ

1

,

τ

′

1

,…,

τ

g

,

τ

′

g

with the unique defining relation

σ

1

⋅⋅⋅

σ

r

[

τ

1

,

τ

′

1

]⋅⋅⋅[

τ

g

,

τ

′

g

]=1 (Proposition 9.1.2). Using Grothendieck’s specialization theorem, we generalize that result to an arbitrary algebraically closed field

C

of characteristic 0 (Proposition 9.1.5). In particular, if

r

≥1, then

$\mathrm{Gal}(E_{S}/E)\cong \hat{F}_{r+2g-1}$

. When

m

=card(

S

) is infinite, we take the limit on all finite subsets of

S

to conclude that

$\mathrm{Gal}(E_{S}/E)\cong \hat{F}_{m}$

(Corollary 9.1.9). In particular, if

S

is all of the prime divisors of

E

/

C

, then card(

S

)=card(

C

) and we find that

$\mathrm{Gal}(E)\cong \hat{F}_{m}$

(Corollary 9.1.10). In particular, Gal(

E

) is projective (Corollary 9.1.11).

The situation is quite different when char(

C

) is a positive prime number

p

. We can not use the Riemann existence theorem to determine the structure of Gal(

E

S

/

E

). Indeed, if

S

is nonempty and of cardinality less than that of

C

, then Gal(

E

S

/

E

) is even not a free profinite group (Proposition 9.9.4) as is the case in characteristic 0. What we do know is the structure of the Galois group Gal(

E

S

,

p

′

/

E

), where

E

S

,

p

′

is the maximal Galois extension of

E

ramified at most over

S

and of degree not divisible by

p

. Using Grothendieck’s lifting to characteristic 0, one proves that the latter group is just the maximal quotient of order not divisible by

p

of the corresponding group in characteristic 0 (Proposition 9.2.1). But, this does not help us to compute Gal(

E

). Instead, we prove by algebraic means that Gal(

E

) is a free profinite group of cardinality

m

. This proof works over every algebraically closed field and does not use the Riemann existence theorem.

The first step is to prove that Gal(

E

) is projective (Proposition 9.4.6). Our proof applies some basic properties of the cohomology of profinite groups. Then we use that every finite split embedding problem for Gal(

E

) has

m

solutions (Proposition 8.6.3) to conclude that

$\mathrm{Gal}(E)\cong\hat{F}_{m}$

(Corollary 9.4.9).

Interesting enough, the same arguments work if

E

is a finite extension of

K

(

x

), where

K

is a field of cardinality

m

of positive characteristic

p

and Gal(

K

) is a pro-

p

group. Thus, even in this case

$\mathrm{Gal}(E)\cong\hat{F}_{m}$

(Theorem 9.4.8).

Next we prove for each nonempty set of prime divisors of

E

/

C

that Gal(

E

S

/

E

) is projective (Corollary 9.5.8). In addition to the projectivity of Gal(

E

), the main tool used in the proof is the Jacobian variety of a smooth projective model Γ of

E

/

C

. The same tool helps us to prove that Gal(

E

S

/

E

) is not projective if

S

is empty (Proposition 9.6.1). The latter group can be interpreted as the fundamental group of Γ.

Finally we consider the case where

E

=

C

(

x

) and apply algebraic patching to solve each split embedding problems

m

times in

E

S

, first in the case that

C

is complete under an ultrametric absolute value and then when

C

is an arbitrary algebraically closed field. This proves that

$\mathrm{Gal}(E_{S}/E)\cong\hat{F}_{m}$

if card(

S

)=

m

(Theorem 9.8.5). This is an optimal result in characteristic

p

. In that case, Gal(

E

S

/

E

) is not free if card(

S

)<

m

(Proposition 9.9.4).