Non-dominant polynomial maps in the plane 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.
 A: (Too long to fit in comments.) An alternative proof for the second question ($k$ is any field and $P(x,y)-c$ is supposed to be reducible for infinitely many $c\in k$ and $P(x,y)$ is not constant). The generic fiber $X$ of $f: \mathbb A^2_k \to \mathbb A^1_k$ is not geometrically irreducible because otherwise there would only be finitely many $c$ for which $f^{-1}(c)$ is not geometrically irreducible (EGA IV.9.7.7). Note that $X$ is an integral variety over $K:=k(P)$ and its field of rational functions is $k(x,y)$. Now $X$ being non geometrically irreducible over $K$ is easily seen to be equivalent to $K$ not being separably closed in $k(x,y)$ (this is a property of field extensions).Let $t\in k(x,y)$ generate a non-trivial finite extension of $K$. One can arrange to have $t$ integral over $k[P]$. Hence $t=T(x,y)\in k[x,y]$. As $P \in k[t]$, there exists a polynomial $R\in k[Y]$ such that $P=R(t)=R(T(x,y))$. We have $\deg R >1$ because $t\notin k(P)$, then the total degree of $T(x,y)$ is strictly smaller than that of $P(x,y)$. 
This works also for polynomials in more variables. 
[Edit] The proof above is incomplete when $k$ has positive characteristic. The problem is that $P(x,y)-c$ reducible does not mean the fiber $f^{-1}(c)$ is reducible, but is not integral. 
The proof can be repaired when $k$ is perfect as follows. We saw that if $X$ is not geometrically irreducible over $K=k(P)$ then $P=R(T)$ as desired. When $k$ has characteristic $0$, if $X$ is geometrically irreducible over $K=k(P)$, then it is geometrically integral, hence all but finitely many fibers of $f: \mathbb A^2_k\to \mathbb A^1_k$ are integral by EGA, loc. cit., contradiction. Suppose now that $k$ has characteristic $p>0$ and that $X$ is geometrically irreducible over $K$. Then for all but finitely many $c$, $f^{-1}(c)$ is irreducible. Combining with the initial hypothesis, there are infinitely many $c$ such that $f^{-1}(c)$ is not integral but irreducible, so $P(x,y)-c=Q_c(x,y)^{r_c}$ for some $r_c >1$. As $r_c \le \deg P$, there at least two $c_1, c_2$ with the same $r=r_{c_1}$. This implies immediately that $p \mid r$ (in fact that $r$ is a power of $p$) and, when $k$ is perfect, that $P$ is a $p$-th power and we are done. 
When $k$ is imperfect, the statement is not true if we only suppose that $P-c$ is reducible for infinitely many $c$. For example, take $c_0\in k \setminus k^p$ and $P(x,y)=x^p+c_0$. Then $P-c$ is reducible for all $c\in c_0 + k^p$, but $P$ is not of the form $R(T)$. However if we suppose that $P-c$ is reducible for all but finitely many $c\in k$, then the above proof works (almost all $r_c$ are divisible by $p$). 
Conclusion: the answer to the second question is yes for any field $k$ if $P(x,y)-c$ is reducible for all but finitely many $c\in k$, and it is enough to have this condition for infinitely many $c$ if $k$ is perfect.  
A: My questions are now answered, but just for sake of completeness I wanted to record my own "low tech" approach to the questions which seem to solve the problem in the characteristic zero case, when there are no Frobenius issues and all finite extensions involved are automatically separable.  (But my algebraic geometry/Galois theory is so lousy that there may be some mistakes here.)  It may be that the other answers are basically a modernised version of these arguments that don't rely on separability.
My idea was to use the two questions to solve each other by induction on the degree of the polynomials involved.  If one has a positive answer to the second question, one can deduce the first as follows.  We can take $P$ to have smaller or equal degree to $Q$.  There are two cases, depending on whether $P(x,y)-c$ is generically reducible or generically irreducible.  (We can of course assume $P$ non-constant.) In the former case, one can use the second question to replace $P$ by a polynomial $T$ of lower degree and use the induction hypothesis.  In the latter case, if $(x,y) \to (P(x,y),Q(x,y))$ is not dominant, then its image is in a curve, and so for generic $c$, the locus of $P(x,y)-c$ must be contained in a finite union of loci of $Q(x,y)-c'$ for some $c'$ depending on $c$.  But $P(x,y)-c$ is irreducible, so by the nullstellensatz this shows that $Q(x,y)-c'$ is some multiple of $P(x,y)-c$ for some $c'$, and we can again use the induction hypothesis to finish up.
Now we use the first question to solve the second.  If $P(x,y)-c$ is reducible, then for generic $c$ I think one can use abstract nonsense to factor $P(x,y)-c$ in $K(x,y)$ where $K$ is some finite extension of $k(c)$.  Using the primitive element theorem (here is where I need separability and thus characteristic zero), this means (I think) that $P(x,y)-f(z)$ is reducible in $k(x,y,z)$ for some non-trivial polynomial $f$.  (It may be that $f$ is merely a rational function rather than a polynomial, but it does not seem to really affect the argument either way.)  But, by viewing $P(x,y)-f(z)$ as a polynomial in $z$ with coefficients in $k(x,y)$, we see that all factors of $P(x,y)-f(z)$ (again viewed as polynomials of $z$ with coefficients in $k(x,y)$) must have coefficients that lie in the algebraic closure of $k(P)$.  Pick a non-constant coefficient $Q(x,y)$ of this type, then $(x,y) \mapsto (P(x,y),Q(x,y))$ is not dominant, and $Q$ has degree strictly less than $P$, and then the first question gives the desired representation $P=R(T)$.
Combining the two implications with an induction on degree seems to give the claim.
A: 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.
