Let $f=f(t),g=g(t) \in k[t]$, $k$ is a field of characteristic zero.
Assume that the following two conditions are satisfied:
(i) $\deg(f)=2$ and $\deg(g)=3$.
(ii) $k(f,g)=k(t)$.
Question: Is it possible to find a general form of such $f$ and $g$?
The above question is a special case of this question (in the current notation, in the linked question we have $\deg(f)=n$ and $\deg(g)=n+1$).
An attempt for an answer: Write $f=t^2+at+b$ and $g=t^3+ct^2+dt+e$, where $a,b,c,d,e \in k$. By Theorem 3.2, the $D$-resultant of $f$ and $g$ is nonzero. However, I am not sure if computing the $D$-resultant and showing it is nonzero yields anything.
Another way is to apply this result, and obtain that there exist $\lambda,\mu,\nu \in k$ such that $\gcd(f-\lambda,g-\mu)=t-\nu$. Again I do not see how this helps in finding a general form of $f$ and $g$.
Remark: By A-M-S theorem, $k[f,g] \subsetneq k[t]$. (Then, if I am not wrong, there exist $\lambda, \mu \in k$ such that $\deg(\gcd(f-\lambda,g-\mu)) \neq 1$).
Edit: A computation of the $D$-resultant of $f=t^2+at+b$ and $g=t^3+ct^2+dt+e$ (by using Proposition 1.3) shows that their $D$-resultant is $s^2+as+(d-ac+a^2)$. Notice that this seems plausible, since it is independent of $b,e$ as it should be.
Thank you very much!
If $\deg f=2, \deg g=3$, then $k(f,g)=k(t)$ without any further hypothesis. Notice that $[k(t):k(f)]=2$ and $[k(t):k(g)]=3$ and so $[k(t):k(f.g)]$ divides both $2,3$ and so must be one.