Formally unramified morphisms #
A morphism of schemes f : X ⟶ Y is formally unramified if for each affine U ⊆ Y and
V ⊆ f ⁻¹' U, the induced map Γ(Y, U) ⟶ Γ(X, V) is formally unramified.
We show that these properties are local, and are stable under compositions and base change.
If S is a formally unramified R-algebra, essentially of finite type, the diagonal is an
open immersion.
A morphism of schemes f : X ⟶ Y is formally unramified if for each affine U ⊆ Y and
V ⊆ f ⁻¹' U, The induced map Γ(Y, U) ⟶ Γ(X, V) is formally unramified.
See FormallyUnramified.hom_ext and FormallyUnramified.of_hom_ext
for the infinitesimal lifting criterion.
- formallyUnramified_appLE {U : Y.Opens} : IsAffineOpen U → ∀ {V : X.Opens}, IsAffineOpen V → ∀ (e : V ≤ (TopologicalSpace.Opens.map f.base).obj U), (CommRingCat.Hom.hom (Scheme.Hom.appLE f U V e)).FormallyUnramified
Instances
Alias of AlgebraicGeometry.FormallyUnramified.formallyUnramified_appLE.
Alias of AlgebraicGeometry.FormallyUnramified.formallyUnramified_appLE.
Alias of AlgebraicGeometry.FormallyUnramified.formallyUnramified_appLE.
f : X ⟶ S is formally unramified if X ⟶ X ×ₛ X is an open immersion.
In particular, monomorphisms (e.g. immersions) are formally unramified.
The converse is true if f is locally of finite type.
The diagonal of a formally unramified morphism of finite type is an open immersion.
Given any commuting diagram
Z' --→ X
| |
↓ ↓
Z --→ Y
With X ⟶ Y formally unramified and Z' ⟶ Z an infinitesimal thickening, there exists at most
one arrow Z ⟶ X making the diagram commute.
To show that f : X ⟶ Y is formally unramified,
it suffices to check for that every following commuting diagram
Spec R --→ X
| |
↓ ↓
Spec S --→ Y
with S = R/I for some I² = 0, there exists at most one arrow Spec S ⟶ X making
the diagram commute.