# When is the image of a regular map an algebraic variety?

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 $Y$)?

• Even if the rank of $df$ is constant, the image may be only locally closed. Let $X$ be the complement of the $s$-axis in the $(s,t)$-affine plane, $X=\{(s,t): t\neq 0\}$. Let $Y$ be 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 contains a dense open subset of the variety $\text{Zero}(v^2-u^2(u+1))$, namely the open complement of $C=\text{Zero}(u-(w^2-1),v-w(w^2-1))$. However, the image also contains the points $f(-1,2)$ and $f(1,-2)$, and these are contained in $C$. Commented Mar 29, 2018 at 10:27
• Locally closed is good. It is an algebraic variety! Commented Mar 29, 2018 at 11:10
• Yes, I am. You whammedd me with your first sentence! Commented Mar 29, 2018 at 12:20
• The sentence should have been, "Even if the rank is constant, the image may be only constructible." Commented Mar 29, 2018 at 12:26
• I presume this is one of the reasons for the importance of proper maps. Commented Mar 30, 2018 at 1:06

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.