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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 $k$ is a perfect field (there are no Frobenius issues and all finite extensions involved are separable), which covers the characteristic 0 case at leastautomatically 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 perfectness)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.

2 added 141 characters in body

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 case when $k$ is a perfect field (all finite extensions are separable), which covers the characteristic 0 case at least. (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 perfectness), 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.

1

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 case when $k$ is a perfect field (all finite extensions are separable), which covers the characteristic 0 case at least. (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 perfectness), this means (I think) that $P(x,y)-f(z)$ is reducible in $k(x,y,z)$ for some non-trivial polynomial $f$. 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.