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Let $X$ and $Y$ be schemes of finite type over $\mathbb{C}$. Is it true that if a morphism $f:X\rightarrow Y$ is injective, then $f$ is universally injective? (I recall having read this somewhere, but cannot find the reference.) If not, then under what conditions will injective imply universally injective?

(If you like, you may take $X$ and $Y$ to be smooth quasi-affine varieties, and $f:X\rightarrow Y$ to be a smooth morphism.)

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  • $\begingroup$ it seems to be a duplicate of mathoverflow.net/questions/114400/… $\endgroup$
    – Jérémy Blanc
    Commented May 26, 2013 at 19:04
  • $\begingroup$ Hi Jeremy, did you happen to find a definitive statement in the literature to the effect that for varieties over an algebraically closed field, an injective etale morphism is universally injective? $\endgroup$
    – Peter Crooks
    Commented May 26, 2013 at 20:04
  • $\begingroup$ Not really, and it is true that it would be good to have a reference. $\endgroup$
    – Jérémy Blanc
    Commented May 26, 2013 at 21:55
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    $\begingroup$ A map of schemes $f:X \rightarrow Y$ is universally injective if and only if the diagonal $\Delta_f:X \rightarrow X \times_Y X$ is bijective on points valued in any field. But $\Delta_f$ is a locally closed immersion, so it is equivalent to say that $\Delta_f$ is surjective. For schemes locally of finite type over an algebraically closed field $k$, a locally closed subscheme is the entire space if and only if it contains all $k$-valued points, so it follows that when $f$ is a $k$-map between such $k$-schemes then it is universally injective if and only if it is injective on $k$-valued points. $\endgroup$
    – user29283
    Commented May 27, 2013 at 0:18
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    $\begingroup$ If $f$ is a $k$-map between locally finite type $k$-schemes with $k$ an arbitrary field (not assumed algebraically closed) then $f$ is universally injective if and only if it is injective and $k(x)$ is purely inseparable over $k(f(x))$ for all closed points $x \in X$. The idea is that one looks for failure of injectivity on $\overline{k}$-points and unravels its consequences at the level of closed points over $k$, using that $f$ is injective. $\endgroup$
    – user29283
    Commented May 27, 2013 at 0:24

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A morphism $f: X \to Y$ of schemes is universally injective (radicial) iff it is injective and for all $y = f(x)$ the extension $k(y) \subseteq k(x)$ is a purely inseparable algebraic extension. (EGA I, (3.5.8))

So for schemes of finite type over $\mathbf{C}$ , an injective morphism is universally injective iff for all $y = f(x)$ with $x$ non-closed the extension $k(y) \subseteq k(x)$ is a purely inseparable algebraic extension, i.e. $k(y) = k(x)$. Perhaps this is automatically the case if it holds for all closed points and everything is of finite type, so we have Jacobson schemes.

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