Let $k$ be a field of characteristic zero, $k \in \{\mathbb{R},\mathbb{C}\}$. Ler $f=f(t),g=g(t) \in k[t]$ such that:
(1) The ideal in $k[t]$ generated by $f',g'$ (formal derivatives of $f,g$, respectively) is $k[t]$.
(2) $k(f,g)=k(t)$ (actually, it is possible to show that this follows from (1)).
(3) $\deg(f)=dn$, $\deg(g)=dm$, where $n < m$ and $d=\gcd(\deg(f),\deg(g))>1$. ($\gcd(n,m)$ is not known; we wish to obtain $\gcd(n,m)=n$).
(4) Both $f$ and $g$ have monomials of all possible degrees, namely, $f=\sum_{i=0}^{dn} a_i t^i$, $g=\sum_{i=0}^{dm} b_i t^j$, where $a_i,b_j \in k-\{0\}$ for all $0 \leq i \leq dn, 0 \leq j \leq dm$.
Question 1: What additional conditions are required (hopefully, none) in order to guarantee that there exist $u=u(t),v=v(t) \in k[t]$, such that:
(i) $u,v \in k[f,g]$.
(ii) $\deg(u)=dn-c, \deg(v)=dm-c$, where $1 \leq c \leq \min(dn,dm)$.
It is possible to show that if conditions (i) and (ii) are satisfied, then $\deg(f)$ divides $\deg(g)$.
Remarks:
I (2) is equivalent to: The sub-resultant of $f$ and $g$ is non-zero; see Theorem 2.1
II I think it is allowed that $c=dn$, and then (ii) becomes: $\deg(u)=0, \deg(v)=dm-dn$.
Independently of conditions (i) and (ii) of Question 1, we can ask:
Question 2: What additional conditions are required in rder to guarantee that $\deg(f)$ divides $\deg(g)$.
Perhaps the following nice answers may help here: 1 and 2.
Any hints and comments are welcome!