Let $f=f(t), g=g(t) \in k[t]$, $k$ is a field of characteristic zero. Assume that:
(i) $k(f,g)=k(t)$.
(ii) $k(f',g')=k(t)$.
(iii) $\langle f'',g'' \rangle = k[t]$.
Is it true that (iv) $k[f',g']=k[t]$? I guess that there exists a counterexample..
For now I have checked the following examples:
(1) $f=t^4, g=t^3$. So $f'=4t^3, g'=3t^2$, and $f''=12t^2, g''=6t$. Conditions (i) and (ii) are satisfied. Condition (iii) is not satisfied. (iv) does not hold: $k[f',g']=k[t^2] \subsetneq k[t]$.
(2) $f=t^3-t, g=t^3-4t$. So $f'=3t^2-1, g'=3t^2-4$, and $f''=6t, g''=6t$. Condition (i) is satisfied. Conditions (ii) and (iii) are not satisfied. (iv) does not hold: $k[f',g']=k[t^2] \subsetneq k[t]$.
(3) $f=t^3-5t^2+6t=t(t^2-5t+6)=t(t-2)(t-3)$, $g=t^3-t=t(t^2-1)=t(t-1)(t+1)$. All three conditions are satisfied, and also the fourh condition. The difficult thing is to show that condition (i) is satisfied, and this is immediate by the answer to this question.
But this of course does not prove anything generally..
Thank you very much!
Edit: Perhaps the following is a counterexample: $f=t^{10}+t^2$, $g=t^5$. It is not sifficult to check that the three conditions are satisfied (condition (ii) holds by the answer to the quoted question in (3)). What about the fourth condition? Is it true or false that $k[10t^9+2t,5t^4] = k[t]$? I suspect that this is false. I should apply Theorem 2.1.
No.
Take $f=\frac{t^4}{4}-\frac{t^3}{3}+\frac{t^2}{2}$, $g=\frac{t^3}{3}-\frac{t^2}{2}$.
Then $\langle f’’,\,g’’\rangle=\langle 3t^2-2t+1,\,2t-1\rangle=k[t]$.
Moreover, $f’+g’=t^3$ so that $t^3 \in k(f’,g’)$, and hence $t^3,t^2-t \in k(f’,g’)$. So $t^4+t^2=(t^2-t)^2+2t^3 \in k(f’,g’)$, thus $t^4+t=t(t^3+1) \in k(f’,g’)$. As $t^3+1 \in k(f’,g’)$, $k(f’,g’)=t$.
Also, $\langle f’,\,g’\rangle=\langle t^3-t^2+t,\,t^2-t\rangle \subset tk[t]$.
Finally, let $K=k(f,g)$, then $t^4=4(f+g) \in K$. Therefore, $t^{-4}g^2=\left(\frac{t^2}{9}-\frac{t}{3}+\frac{1}{4}\right) \in K$, thus $t^2-3t \in K$. We already know $2t^3-3t^2 \in K$, so that $2t^3-9t \in K$. But $(2t^3-9t)^2=4t^6+81t^2-36t^4$ so that $4t^6+81t^2=t^2(4t^4+81) \in K$. As $t^4 \in K$, $t^2 \in K$. Since $g \in K$, $t^3 \in K$ and thus $K=k(t)$.