Almost every place admits a prescribed integral basis #
Let F' / F be a finite separable extension and fix an F-basis b of F'. This file proves
that b is an integral basis over the valuation ring of all but finitely many places of an
algebraic function field F / k. This is Stichtenoth, Algebraic Function Fields and Codes,
2nd ed., Theorem 3.3.6.
There are two finiteness steps. First, any fixed element of F' is integral over 𝒪_P for all
but finitely many P: outside the poles of the finitely many coefficients of its minimal
polynomial over F, that polynomial is defined over 𝒪_P. Second, apply this simultaneously to
the vectors of b and its trace-dual basis. If both bases are integral at P, the trace formula
for the coordinates in b shows that every integral element has integral coordinates;
integrality of the vectors of b proves the converse.
This theorem is the finiteness input for the different divisor: a single basis can be used to compute the complementary module away from finitely many places.
Main results #
TauCeti.Place.finite_setOf_not_isIntegral: an element integral overFis integral over the valuation rings of all but finitely many places.TauCeti.Place.finite_setOf_not_isIntegralBasis: every basis is an integral basis at all but finitely many places (Stichtenoth, Theorem 3.3.6).
References #
- H. Stichtenoth, Algebraic Function Fields and Codes, 2nd ed., GTM 254, Springer, 2009, Theorem 3.3.6.
Elements integral at almost every place #
A fixed element integral over F is integral at all but finitely many places.
Indeed, outside the poles of the coefficients of its minimal polynomial over F, that monic
polynomial has coefficients in 𝒪_P and witnesses integrality over 𝒪_P.
Bases integral at almost every place #
Every basis of a finite separable extension is an integral basis at all but finitely many places (Stichtenoth, Theorem 3.3.6).