Totally split algebras over a totally split algebra #
Let K be a field and A → B a map of totally split K-algebras (Algebra.IsFiniteSplit), with
B flat over A. Geometrically, Spec B → Spec A is a finite flat morphism between finite
disjoint unions of copies of Spec K. Above every K-point χ of A there are exactly as many
K-points of B as the rank of B at the prime ker χ: the fibre K ⊗[A] B is a quotient of
B, hence totally split, so its points are counted by its dimension.
When this rank is a constant d, the points of B are labelled by pairs of a point of A and an
index in Fin d, compatibly with the projection to the points of A, and B is the trivial
cover Fin d → A of A. Here flatness is automatic, since A is a semisimple ring. This splits
a finite flat cover of constant rank of a finite étale
scheme over a field once both schemes are split over the base field, for instance the branches
of a nodal curve over its nodes.
Main results #
Algebra.IsFiniteSplit.card_algHom_comp_eq_rankAtStalk: the number of points ofBabove a pointχofAis the rank ofBatker χ.Algebra.IsFiniteSplit.nonempty_algEquiv_fun_of_rankAtStalk_eq: ifBhas constant rankdoverA, thenB ≃ₐ[A] (Fin d → A).
The points of a totally split algebra above a point of its base number its rank there.
If B is a totally split K-algebra, flat over the K-algebra A, then the
K-points of B restricting to a K-point χ of A are as many as the rank of B at the prime
ker χ of A.
A totally split algebra of constant rank over a totally split algebra is a trivial cover.
Let A and B be totally split algebras over a field K, with B an A-algebra of constant
rank d. Then B is A-isomorphic to Fin d → A: the K-points of B are labelled by pairs
of a K-point of A below them and an index in Fin d. No flatness hypothesis is needed: A
is a finite product of copies of K, a semisimple ring, over which every module is flat.