A variation on Abhyankar–Moh–Suzuki theorem The well-known theorem of Abhyankar–Moh–Suzuki says the following:
Let $f=f(t), g=g(t) \in k[t]$, $k$ is a field of characteristic zero.
If $k[f,g]=k[t]$, then $\deg(f) \mid \deg(g)$ or $\deg(g) \mid \deg (f)$.
Let us concentrate on the case $k=\mathbb{C}$ (I wonder if there is a difference in my following question between $k=\mathbb{C}$ and $k=\mathbb{R}$).
Let $f=f(t), g=g(t) \in \mathbb{C}[t]$ satisfy the following conditions:
(1) $f=hf_1$, $g=hg_1$, where $\gcd(f_1,g_1)=1$
(= $h$ is the gcd of $f$ and $g$) and each of
$\deg(h)$, $\deg(f_1)$, $\deg(g_1)$ is $\geq 1$.
(2) $\deg(f_1)=\deg(g_1)$, hence $\deg(f)=\deg(g)$.
(3) $\mathbb{C}(f,g)=\mathbb{C}(t)$.

Should such $f$ and $g$ generate $\mathbb{C}[t]$, namely, $\mathbb{C}[f,g]=\mathbb{C}[t]$?

I have not found an example of such $f$ and $g$ not generating $\mathbb{C}[t]$, but maybe I am missing something easy.
Maybe one (or more) of the following ideas could help:
(i) The following known criterion:
$\mathbb{C}[f(t),g(t)]=\mathbb{C}[t]$ iff $(f'(t),g'(t))\neq 0$ and $t\mapsto (f(t),g(t))$ is injective, see this.
(ii) The theory of SAGBI bases (I will try to find relevant papers, I remember there are ones. I think there is a result saying that if $f', g' \in \mathbb{C}[f,g]$, then $\mathbb{C}[f,g]=\mathbb{C}[t]$).
(iii) The theory of sub-resultants.
((iv) Maybe if $\deg(h)=1$, then I can show that $\mathbb{C}[f,g]=\mathbb{C}[t]$. However, I prefer not restricting $\deg(h)$).
Somewhat relevant questions: Is there a converse of Abhyankar-Moh-Suzuki theorem? and Generalizations of Abhyankar-Moh theorem (embeddings of the line in the plane).
Any comments are welcome!
Edit: I have now found a counterexample for the case $\deg(g) \geq 2$:
$f=t^{15}+t^2$, $g=t^{15}$. Conditions (1),(2),(3) are being satisfied, but $\mathbb{C}[f,g]=\mathbb{C}[t^2] \subsetneq \mathbb{C}[t]$.
So only when $\deg(h)=1$ perhaps there is a positive answer.
Therefore, and in view of idea (i), we will add a fourth condition:
(4) $f',g'$ are not simultaneously zero.
Then, the above counterexample is not a counterexample anymore, since $f'=15t^{14}+2t$, $g'=15t^{14}$ have a common zero at $0$.
 A: The answer is no. You can simply choose $$f=t(t^2-t+1)$$
$$g=t(t^2+1)$$
These satisfy all your conditions, but $\mathbb{C}[f,g]\subsetneq \mathbb{C}[t]$.
Let us check this:
$(1)-(2)$: $f=tf_1$ and $g=tg_1$ where $f_1=t^2-t+1$, $g_1= t^2+1$ are both of degree $2$ and without common factor.
$(3)$: $\mathbb{C}(f,g)=\mathbb{C}(t)$ because $g-f=t^2$, so $\frac{g}{g-f+1}=t$.
Finally, we observe that
$\mathbb{C}[f,g]=\mathbb{C}[f-g,g]=\mathbb{C}[t^2,t^3+1]=\mathbb{C}[t^2,t^3]\subsetneq \mathbb{C}[t]$.
A: Since you (OP) seem to be interested in Lüroth’s theorem, you might want to look at a "constructive" proof that I just wrote up from Schinzel's Selected topics on polynomials. It seems using it one can show (see the bottom part of the linked page) that $\mathbb{C}(f,g) = \mathbb{C}(t)$ if $\deg(f)$ and $\deg(g)$ are relatively prime.
I am still a bit skeptical about the result - it seems too good to be true. In any event, if I have not made any mistake and it is indeed true, then this gives a lot of counterexamples to your question, since for any polynomial $h$ with no multiple roots, there are (infinitely many) $f_1, g^*_1$ such that

*

*$d := \deg(h) + \deg(f_1)$ and $e^* := \deg(h) + \deg(g^*_1)$ are relatively prime,

*$d > e^*$,

*$e^*$ does not divide $d$, and

*there is no common zero of $(f_1h)'$ and $(g^*_1h)'$.

Then setting $f := f_1h$ and $g := (f_1 + g^*_1)h$ gives a counterexample.
