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.*



Examples of injective morphisms which are not universally injective
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.


