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I am interested in morphisms of algebraic varieties $X\to Y$ over a field $k$, and want to have examples of injective morphisms which are not universally injective. If $k$ is not algebraically closed, it is easy to construct plenty of examples, so I am only interested in the case where $k$ is algebraically closed.

The motivation of the question is that open immersion = étale + universally injective. This latter condition is equivalent to radicial, or to the fact that it is injective under a field extension $K/k$. I would thus be even more interested in examples with étale morphisms.

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(answer to your comment below) $x\to x^p$ is not étale: it would then be smooth and then its differential would be non vanishing but in fact $d(x^p)=p\cdot x^{p-1}dx=0$ because $p=0$. Apart from that, you are right about EGA: it talks about every alg. closed field, not just one so it is weaker than Angelo's statement. I removed my comment. –  Damian Rössler Nov 25 '12 at 16:12
Thanks! Sorry for my ignorance about characteristic $p$. I still don't get why EGA talks about all alg. closed fields if only one is enough. In Altman/Kleiman "Introduction to Grothendieck Duality Theory" Prop. 5.2, they also make many equivalence definitions of radicial and always include to check all fields. Seems weird not to say that one can only check on one algebraically closed field. –  Jérémy Blanc Nov 25 '12 at 16:19
Notice that Angelo's proof below uses the fact that the schemes are locally of finite type over a field. EGA makes no assumption on the schemes. –  Damian Rössler Nov 25 '12 at 19:17
Yes. This is probably too specialised for EGA, even if it corresponds, for me, to the "geometric" case. –  Jérémy Blanc Nov 25 '12 at 22:13

1 Answer 1

up vote 6 down vote accepted

Every morphism of schemes locally of finite type over an algebraically closed field that is injective on $k$-points is universally injective.

Let $f: X \to Y$ be finite, étale and injective on $k$-points, where $Y$ is integral. For any $y \in Y$, the degree of $f^{-1}(y)$ over $k(y)$ is constant, and equals the degree of $f$. If $y$ is a closed point, then $k(y) = k$, and since $k$ is algebraically closed the number of points of $f^{-1}(y)$ equals $d$. This means that $d=1$, and $f$ is an isomorphism.

The general case reduces to this. Suppose that $p \in X$; we need to show that $k(p)$ is a purely inseparable extension of $k(f(p))$ By restriction to the closures of $p$ and $f(p)$, we may assume that $X$ and $Y$ are integral, $f$ is dominant, and $k(X)$ is purely inseparable on $k(Y)$. If the dimension of $X$ is larger then the dimension of $Y$, the fibers are positive-dimensional, hence have infinitely many $k$-points. So $\dim X = \dim Y$, and $k(X)$ is finite over $k(Y)$. By restriction $Y$ we may assume that $X$ and $Y$ are normal, and $f$ is finite.

Let $Z$ be the normalization of $Y$ in the separable closure of $K(Y)$; we can factor $f$ as $X \to Z \to Y$. We have that $X\to Y$ is finite and dominant, hence surjective, so $Z \to Y$ is injective in $k$-points. Since $Z$ is generically étale over $Y$, by restricting $Y$ we may assume that $Z$ is étale over $Y$. Hence it has degree 1, which shows that $k(Z) = k(Y)$, which is what we want.

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@Angelo Thanks for the answer. I do not get all the points. 1) For your first case, you say that finite, étale and injective implies isomorphism. What about x-> x^p in characteristic p? It seems finite, étale and injective but is not an isomorphism. What did I miss? 2) You say "we need to show that k(p) is not a purely inseparable extension of k(f(p))". For me, we need to show the converse. –  Jérémy Blanc Nov 25 '12 at 16:05
@ Damian: I had already read EGA before asking the question (especially 3.5.5 and 3.5.11) and I do not find that it answer the question (but maybe I missed something). For me, it just says that radicial is equal to universally injective and that condition is only necessary to be checked on ALL extensions K where K is algebraically closed. It seems implicit from what is said in EGA that one algebraically closed field is not sufficient. (This is indeed why I asked the question). But again, I could have missed something. –  Jérémy Blanc Nov 25 '12 at 16:06
To Jérémy: the map $x \mapsto x^p$ is certainly not étale in characteristic $p$. A finite flat map is étale if and only if it non-ramificated, that is, if the fibers are reduced. About your other remark, you are absolutely right, I edited the post. –  Angelo Nov 25 '12 at 16:45
"Universally injective" is equivalent to "injective and all the maps on the residue fields are radicial" EGA I, 3.7.1. –  anon Nov 25 '12 at 20:23
Dear Jérémy, sorry, I don't read anything, therefore I never know what is in the literature. The result is certainly well know; it is the sort of thing I would use without mention. –  Angelo Nov 26 '12 at 12:32

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