The algebraic constant field of a finite function field #
For a finite separable extension N / K(t), define its algebraic constant
field to be the relative algebraic closure of K in N. When K is finite,
this constant field is finite-dimensional over K.
The finiteness proof is place-theoretic. Algebraic constants are integral over
K[X], hence map into the finite integral closure. Reduction modulo any
height-one prime is injective on the constant field, because it is a field.
Choosing a prime above (X) embeds the constants into a finite quotient.
The algebra structure from K to N used in the function field constant field
construction.
Equations
- BGS.HasseWeil.functionFieldConstantAlgebra K N = ((algebraMap (RatFunc K) N).comp (algebraMap K (RatFunc K))).toAlgebra
Instances For
The algebra structure from K[X] to N used in the function field constant field
construction.
Equations
- BGS.HasseWeil.functionFieldConstantPolynomialAlgebra K N = ((algebraMap (RatFunc K) N).comp (algebraMap (Polynomial K) (RatFunc K))).toAlgebra
Instances For
The algebra structure from K to (RatFuncFiniteIntegralClosure K N) used in the
function field constant field construction.
Equations
Instances For
Reduction modulo a finite place, restricted to the algebraic constant field.
Equations
Instances For
Reduction at a finite place is injective on algebraic constants.
The residue field of a rational-function finite place is finite over a finite constant field.
Every finite-place residue field of a finite separable function field is finite over a finite constant field.
Over a finite base field, the algebraic constant field of a finite separable function field is finite-dimensional.
The algebraic constant field is a finite type when the original constants are finite.
The algebraic constant field is Galois over the finite base field.