Let $k$ be a field of characteristic zero, for example $k=\mathbb{R}$ or $k=\mathbb{C}$.


Of course, $k(x^2,x^3)=k(x)$, since $x=\frac{x^3}{x^2}$.


Let $f\_1,\ldots,f\_n,g\_1,\ldots,g\_m \in k[x]$, $n,m \geq 1$.


Denote: $F(T)=f\_nT^n+\cdots+f\_1T+x^2$
and $G(T)=g\_mT^m+\cdots+g\_1T+x^3$.


Let $h \in k[x]$.



> 
> Is it possible to characterize
> $f\_1,\ldots,f\_n,g\_1,\ldots,g\_m \in k[x]$, $n,m \geq 1$,
> such that for *every* $h \in k[x]$ we have: $k(F(h),G(h))=k(x)$?
> 
> 
> 


I could not find such $f\_i,g\_j,n,m$.


**For example:**
$F(T)=T+x^2$, $G(T)=T+x^3$.
For $h=x^3$ we get $F(h)=x^3+x^2$, $G(h)=x^3+x^3=2x^3$,
so clearly, $k(F(h),G(h))=k(x)$.
For $h=x^4-x^3$ we get $F(h)=x^4-x^3+x^2$, $G(h)=x^4-x^3+x^3=x^4$,
abd then, $k(F(h),G(h)) \subsetneq k(x)$.


I think that for arbitrary fixed $f\_i,g\_j,n,m$, there exists $h \in k[x]$ such that
$k(F(h),G(h)) \subsetneq k(x)$, but not sure how to prove this.


I am familiar with D-resultant, but it seems difficult to apply it.
The theorem relevant here is that
$D-\operatorname{Res\_s}(\frac{(F(h))(s)-F(h)}{s-x},\frac{(G(h))(s)-G(h)}{s-x})$
is a non-zero element of $k[x]$ if and only if $k(F(h),G(h))=k(x)$.
See [this paper](https://www.academia.edu/29672226/Jacobian_pairs_D_resultants_and_automorphisms_of_the_plane).


If we apply this to $x^2,x^3$ we obtain that their D-resultant is $x^2$, and indeed
$k(x^2,x^3)=k(x)$.


Also, I wonder if algebraic geometry could help solve my question.


**In addition:**
**(1)** I do not mind to further assume that for every $h \in k$,
$k(F(h),G(h)) = k(x)$.
In the above example, this is satisfied.
**(2)** $f\_1,\ldots,f\_n,g\_1,\ldots,g\_m \in k[x]$,
not necessarily in $k$.


Any comments are welcome; thank you!
Also asked in [MSE](https://math.stackexchange.com/questions/4245950/a-variation-on-kx2-x3-kx) with no comments.



A variation on $k(x^2,x^3)=k(x)$
Such $F, G$ do exist, unless I missed a criterion.


One possibility is to take $F(T)$ a polynomial such that $F(h)$ is never zero, such as $T^3 + x^2$, and then $G(T) = x F(T)$. Then you can use the same $x = G(T)/F(T)$ trick.


