Let \(k\) be an algebraically closed field of characteristic zero. Consider the morphism \[ \begin{aligned} f:\AA^2_k&\longrightarrow\AA^2_k,\\ (x,y)&\longmapsto(x,xy). \end{aligned} \] Is this finite? Dominant? Open? Closed? What are the fibers?
Solution
Write \((u,v)\) for the coordinates on the target \(\AA^2_k\).
1. For \((a,b)\in\AA^2(k)\), the fibers are\[ \boxed{ f^{-1}(a,b) = \begin{cases} \{(a,b/a)\},&a\neq0,\\ \{0\}\times\AA^1_k,&a=0,\ b=0,\\ \emptyset,&a=0,\ b\neq0. \end{cases}} \]Consequently,\[ f(\AA^2_k) = D(u)\cup\{(0,0)\}. \]
A point \((x,y)\) belongs to the fiber over \((a,b)\) exactly when \[ x=a, \qquad xy=b. \] If \(a\neq0\), these equations force \[ x=a, \qquad y=b/a, \] so the fiber is a singleton. If \(a=0\), the second equation becomes \[ 0=b. \] Thus for \((a,b)=(0,0)\) every point \((0,y)\) lies in the fiber, while for \(a=0\) and \(b\neq0\) the fiber is empty. The displayed description of the image follows immediately.
2. The morphism \(f\) is dominant.
By step 1, \[ D(u)\subseteq f(\AA^2_k). \] The affine plane is irreducible, so its nonempty open subset \(D(u)\) is dense (density of nonempty Zariski-open subsets). Therefore \[ \overline{f(\AA^2_k)} \supseteq \overline{D(u)} = \AA^2_k. \] Hence the image of \(f\) is dense, which is exactly dominance.
3. The morphism \(f\) is not finite.
By step 1, \[ f^{-1}(0,0)=\{0\}\times\AA^1_k, \] which is a positive-dimensional, hence infinite, fiber. A finite morphism has finite fibers. Therefore \(f\) is not finite.
4. The morphism \(f\) is not open.
The whole source \(\AA^2_k\) is open. If \(f\) were an open map, its image \[ f(\AA^2_k)=D(u)\cup\{(0,0)\} \] would be open in the target.
Let \[ L=V(u)\cong\AA^1_k \] be the vertical coordinate line. Step 1 gives \[ f(\AA^2_k)\cap L=\{(0,0)\}. \] If \(f(\AA^2_k)\) were open, this singleton would be open in \(L\). But \(L\) is an infinite curve whose Zariski topology is cofinite (cofinite topology on curves), so a singleton is not open. This contradiction proves that \(f\) is not open.
5. The morphism \(f\) is not closed.
Consider the closed hyperbola \[ H=V(xy-1)\subseteq\AA^2_k. \] For \((x,y)\in H\) one has \(xy=1\), so \[ f(x,y)=(x,1). \] Conversely, for every \(a\in k^\times\), \[ f(a,a^{-1})=(a,1). \] Hence \[ f(H) = \{(a,1):a\in k^\times\} = V(v-1)\setminus\{(0,1)\}. \] The line \(V(v-1)\cong\AA^1_k\) has the cofinite Zariski topology, so deleting one point gives a dense proper subset. Thus \(f(H)\) is not closed in the line, and therefore is not closed in \(\AA^2_k\). Since \(H\) is closed, \(f\) is not a closed map.
6. The four requested properties are\[ \boxed{ \text{$f$ is dominant, but it is neither finite, open, nor closed.} } \]
5. The fibers were computedin step
1.
7. Q.E.D.
Steps <1>1–<1>6 answer every part of the problem.