Let $f:X\rightarrow Y$ be a regular map of smooth connected algebraic varieties (say over an algebraically closed field). I know that the image $f(X)$ is only a constructible set, in general, but I am interested in conditions that ensure $f(X)$ being an algebraic variety.


**A precise question:** suppose the differential $df$ has a constant rank. Is $f(X)$ an algebraic variety (or a locally closed subvariety of $Y$)?



When is the image of a regular map an algebraic variety?
I am correcting the first sentence of the comment.


Even if the rank of $df$ is constant, the image may be only **constructible**. Let $X$ be the complement of the $s$-axis in the $(s,t)$-affine plane, $X=\{(s,t): t\neq 0\}.$ Let $Y$ be the affine $3$-space with coordinates $(u,v,w).$ Let $f$ be the function $f(s,t)=(s^2-1,s(s^2-1),s+t)$. The image of $f$ is the disjoint union of two locally closed subsets whose union is not locally closed. The first locally closed set is $\text{Zero}(v^2-u^2(u+1))\setminus \text{Zero}(u-(w^2-1),v-w(w^2-1))$. The second locally closed set is $\text{Zero}(u,v)$. The union of these two locally closed sets is not locally closed. It is only **constructible**.


