# What's the definition of “geometrically injective”?

I don't know the meaning of geometrically injective morphism f of schemes.

What's the definition of "geometrically injective"?

Thanks.

-
Not sure, but I think this is another word for "radical morphism" or "universally injective". – Steven Gubkin Oct 14 '11 at 16:24
Nitpick: that should be a radicial morphism. – Dan Petersen Oct 14 '11 at 16:27
Indeed.  – Steven Gubkin Oct 14 '11 at 16:33

A map of schemes $f \colon X \to Y$ is geometrically injective if it is injective on geometric points, i.e. points with values in an algebraic closed field. In more detail, let $K$ be an algebraically closed field. For all pairs of maps ($K$-valued points) $x, y \colon \operatorname{Spec}(K) \to X$ such that they have the same image on $Y$, i.e $f \circ x = f \circ y$ then $x = y$.
In other words the map $$\operatorname{Hom}(\operatorname{Spec}(K), X) \longrightarrow \operatorname{Hom}(\operatorname{Spec}(K), Y)$$ given by composition with $f$, is injective for every algebraically closed field $K$.
In general, when you see the word geometrically in front of another property (irreducible, reduced, etc.), it is referring to that property for the fiber product of the given scheme with the algebraic closure of the base field. For instance, if $X$ is a scheme over a field $k$, then $X$ is geometrically irreducible if $\bar{X}:=X \times_{k} \bar{k}$ is irreducible. Similarly, a $k$-morphism $X \rightarrow Y$ is geometrically injective if the associated $\bar{k}$-morphism $\bar{X} \rightarrow \bar{Y}$ is injective. – Parsa Oct 14 '11 at 20:13