$\mathbb{C}(f,g)=\mathbb{C}(t)$ and $(f'(t),g'(t)) \neq 0$, but $\mathbb{C}[f,g]\subsetneq \mathbb{C}[t]$

Actually, the following question is a consequence of this MO question.

Recall the following result: $\mathbb{C}[f(t),g(t)]=\mathbb{C}[t]$ iff $(f'(t),g'(t))\neq 0$ and $t\mapsto (f(t),g(t))$ is injective.

Let $f=f(t), g=g(t) \in \mathbb{C}[t]$ satisfy the following conditions:

(1) $\deg(f)=\deg(g) \geq 4$.

(2) $f=hf_1$, $g=hg_1$, $h$ is the gcd of $f$ and $g$, and $\deg(h) \geq 2$ and $\deg(f_1)=\deg(g_1) \geq 2$.

(3) $\mathbb{C}(f,g)=\mathbb{C}(t)$.

(4) $f',g'$ are not simultaneously zero.

Question: Is it true that $\mathbb{C}[f,g]=\mathbb{C}[t]$?

I am not able to find a counterexample or a proof.

Remark: $f=t^{15}+t^2$, $g=t^{15}$ is not a counterexample, since condition (4) is not satisfied.

Thank you very much!

$$f= (t+3)(t+2)t^2,\qquad g= (t+3)(t+2)(t+1)^2$$

• $\Bbb{C}(f/g) = \Bbb{C}( t^2/(t+1)^2)$ is a quadratic subfield and $f(t)\ne f(\frac1{-2-1/t})$ so $$\Bbb{C}(f/g)\subsetneq \Bbb{C}(f/g,f)\subseteq \Bbb{C}(t)$$ from which $\Bbb{C}(f,g)=\Bbb{C}(t)$.

• $\gcd(f',g')=1$

• If $t+3\in \Bbb{C}[f,g]$ then it must be in $f \Bbb{C}[f,g]+g \Bbb{C}[f,g]$ so that $t+3$ vanishes at $-2$ which is absurd.