Constants and restriction for the function-field Galois closure #
Equip the chosen normal closure of L / K(t) with its induced K-algebra
structure, take the algebraic constant field, and restrict every
K(t)-automorphism to those constants. Over a finite base field the constant
field is finite Galois and its automorphism group is cyclic, generated by
Frobenius. The restriction map is surjective: a constant fixed by its image
is fixed by the full Galois group over K(t), hence lies in K(t); because it
is algebraic over K, the exact constant-field theorem for K(t) puts it in
K.
Equations
The algebraic constant field of the chosen Galois closure.
Equations
Instances For
Restriction of a K(t)-automorphism of the normal closure to its algebraic
constant field.
Equations
- One or more equations did not get rendered due to their size.
Instances For
The kernel consists exactly of automorphisms fixing every algebraic constant.
Every automorphism of the algebraic constant field extends to an
automorphism of the normal closure over K(t).
The proof uses fixed fields rather than constructing a constant compositum.
If an algebraic constant is fixed by the image of the restriction map, then
it is fixed by every automorphism of the normal closure over K(t), so the
Galois fixed-field theorem puts it in K(t). The rational function field has
no elements algebraic over K except K itself. Thus the image has trivial
fixed field and is the full constant-field Galois group.
After enlarging the base constants to the algebraic constants of the normal closure, that field is the exact constant field: no further element of the normal closure is algebraic over it.
The module structure of (FunctionFieldNormalClosureConstantField K L) over K used in
the function field normal closure constants construction.
Equations
Instances For
Frobenius on the finite algebraic constant field has order equal to its
degree over K.