When $\mathbb{C}(f,g)=\mathbb{C}(f',g')=\mathbb{C}(t)$.

35 Views Asked by At

Let $f=f(t),g=g(t) \in \mathbb{C}[t]$ be two non-constant polynomials with $\deg(f)=n$, $\deg(g)=m$. Denote the first derivative of $f,g$ by $f',g'$, and the second derivative of $f,g$ by $f'',g''$, respectively.

Assume that the following three conditions are satisfied:

(i) $\mathbb{C}(f,g)=\mathbb{C}(t)$, namely, the field of fractions of $\mathbb{C}[f,g]$ equals the field of fractions of $\mathbb{C}[t]$.

(ii) $\mathbb{C}(f',g')=\mathbb{C}(t)$, namely, the field of fractions of $\mathbb{C}[f',g']$ equals the field of fractions of $\mathbb{C}[t]$.

(iii) $\langle f'',g'' \rangle= \langle 1 \rangle= \mathbb{C}[t]$, namely, the ideal in $\mathbb{C}[t]$ generated by $f'',g''$, is the unit ideal.

Is it possible to find the 'exact' form of such $f,g$ or find a restriction on $n,m$?

Non-example: $f=t^4,g=t^3$. Then $f'=4t^3,g'=3t^2$, $f''=12t^2, g''=6t$; conditions (i) and (ii) are satisfied: $\mathbb{C}(t)=\mathbb{C}(t^4,t^3)=\mathbb{C}(4t^3,3t^2)$, while condition (iii) is not satisfied.

This question may be relevant; its answer says the following: $\mathbb{C}(u,v)=\mathbb{C}(t)$ if and only if there exist $a,b,c \in \mathbb{C}$ such that $\gcd(u-a,v-b)=t-c$.

Any comments are welcome!