Composition of rational functions

Given a rational function $R\in\Bbb R(x_1,\dots,x_n)$ with multilinear numerator and denominator, is there always a rational function $G\in\Bbb R(x)$ such that $G\circ R\in\Bbb R[x_1,\dots,x_n]$, $G(0)=0$, $G(1)=1$? In situations where such $G$ exists, what is the minimal degree of rational function $G\in\Bbb R(x)$ such that $G\circ R\in\Bbb R[x_1,\dots,x_n]$, $G(0)=0$, $G(1)=1$ where degree of rational function is maximum of degrees of numerator and denominator in lowest forms?

Since $G(0)=0,G(1)=1$, $G(x)=x\frac{C(x)}{D(x)}=x\frac{(x-1)A(x)+r}{(x-1)B(x)+r}$ holds with $r\neq0$.

By multilinear, I mean only monomials of form $ax_{i_1}x_{i_2}x_{i_3}\dots x_{i_d}$ with $i_j\in\{1,\dots,n\}$, $a\in\Bbb R$ are present.

• Is there some editing issue? Which question are you asking? – Dylan Thurston May 11 '15 at 21:50
• Are you asking a total of four different questions? – Dylan Thurston May 11 '15 at 21:52
• no two questions. question is essentially about degree. – T.... May 11 '15 at 21:53
• Re your edit: can two $i_j$ be the same? – Todd Trimble May 12 '15 at 2:57
• no they cannot be. – T.... May 12 '15 at 3:00

I am not sure I know what multilinear means. If the following $R$ is multilinear, it is a counterexample. Take $R(x_1, x_2, x_3, x_4) = (x_1 x_2+x_3 x_4)/(x_2 x_4) = x_1/x_4 + x_3/x_2$. I claim that there is no nonconstant rational function $G$ for which $G \circ R$ is polynomial. Proof: If there were, then $G(R(t,t,1,1))$ would be polynomial in $t$, which is to say, $G(t+t^{-1})$ would be polynomial in $t$.
Suppose that there is a rational function $G$ for which $G(t+t^{-1})$ is polynomial. For any complex number $t$, the function $G$ cannot have a pole at $t+t^{-1}$. But, as $t$ ranges over $\mathbb{C}$, the function $t+t^{-1}$ ranges over the entire Riemann sphere $\mathbb{C} \cup \{ \infty \}$. So $G$ has no poles at all, and is a constant.
• Is there a way to classify for which such $R$, a $G$ exists? – T.... May 12 '15 at 2:52
• Multilinear seems to be a distraction, so let's just ask, for which rational functions $R$ on $\mathbb{C}^n$, there is a $G \in \mathbb{C}(u)$ such that $G \circ R$ is polynomial. As noted above, $R(\mathbb{C}^n)$ must omit some point of the Riemann sphere. Conversely, if $R(\mathbb{C}^n)$ omits $a$, then $1/(R-a)$ omits $\infty$, and hence is polynomial. – David E Speyer May 12 '15 at 3:31