instSubsingletonAlgHomOfIsPurelyInseparable

English

If E/F is purely inseparable, then for every reduced L, there is at most one F-algebra hom from E to L.

Русский

Если расширение E/F чисто бесприводимо, то для любого снижения L существует не более одного F-алгеброморфизма E → L.

LaTeX

$$$(E \to_{F} L)$ является Subsingleton$$

Lean4

/-- If `E / F` is purely inseparable, then for any reduced `F`-algebra `L`, there exists at most one
`F`-algebra homomorphism from `E` to `L`. -/
instance instSubsingletonAlgHomOfIsPurelyInseparable [IsPurelyInseparable F E] (L : Type w) [CommRing L] [IsReduced L]
    [Algebra F L] : Subsingleton (E →ₐ[F] L) where
  allEq f
    g :=
    AlgHom.coe_ringHom_injective <|
      IsPurelyInseparable.injective_comp_algebraMap F E L (by simp_rw [AlgHom.comp_algebraMap])

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