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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.

Thanks.

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  • $\begingroup$ Not sure, but I think this is another word for "radical morphism" or "universally injective". $\endgroup$ Oct 14, 2011 at 16:24
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    $\begingroup$ Nitpick: that should be a radicial morphism. $\endgroup$ Oct 14, 2011 at 16:27
  • $\begingroup$ Indeed. $ $ $\endgroup$ Oct 14, 2011 at 16:33

2 Answers 2

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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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    $\begingroup$ 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. $\endgroup$
    – Parsa
    Oct 14, 2011 at 20:13
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    $\begingroup$ @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. $\endgroup$ Oct 17, 2011 at 6:24
  • $\begingroup$ @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. $\endgroup$
    – Leo Alonso
    Oct 17, 2011 at 8:59
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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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