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

What's the definition of "geometrically injective"?

I can't find it. I hope your answer.


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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

2 Answers 2

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 $K$.

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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
@Parsa: Does this really coincide with Leo's definition? It seems to me that set-theoretic injectivitiy is weaker than the injectivity for morphisms defined on spectra of fields. –  Martin Brandenburg Oct 17 '11 at 6:24
@Parsa, @Martin Brandenburg. I guess that talking about injectivity in this context means injectivity for rational points, therefore both notions are not so far away. In any case "geometric point" in algebraic geometry means point form the spectrum of a field. –  Leo Alonso Oct 17 '11 at 8:59

I don't find link to add comment. You can find the various equivalent condition for radicial morphism and its proof in "Altman & Kleiman, Introduction to Grothendieck Duality Theory" on page 119.

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