Let $P(x,y), Q(x,y)$ be polynomials of two variables over an algebraically closed field $k$. Suppose that the map $(x,y) \mapsto (P(x,y),Q(x,y))$ is not a dominant map from $k^2$ to $k^2$. Does this mean that one has $P(x,y) = R(T(x,y))$ and $Q(x,y) = S(T(x,y))$ for some polynomials $R,S,T$?


For fixed $x$, it seems to me that the curve $\{ (P(x,y), Q(x,y)): y \in k \}$ either degenerates to a point or is a curve of genus zero, which ought to kill off the problem, but I don't have enough of an understanding of the genus zero plane curves to do this.


A closely related question: if $P(x,y)-c$ is reducible for all $c \in k$ (or maybe all but finitely many $c$), does this mean that $P(x,y) = R(T(x,y))$ for some polynomials $R, T$, with $T$ having strictly smaller degree than $P$? I played around a bit with some Galois theory but was only able to convince myself that this question was more or less equivalent to the original question.



Non-dominant polynomial maps in the plane
Yes, both results are true. For the first, as you say, the image of the map is either a point, or dominates an affine curve $C$ of geometric genus 0. By standard results, it factors through the normalization $X$ of $C$; the curve $X$ is a smooth affine curve of genus 0, so it is the complement of a finite subset of $\mathrm P^1$. This finite subset cannot contain more than one point, because otherwise $X$ would have non-constant invertible regular function, which would pull back to non-constant invertible regular functions on $\mathbb A^2$, which don't exist. Hence $X = \mathbb A^1$, and you can factor your map through $\mathbb A^1$, which gives what you want.


For the second question, Quim's argument works.


