The well-known theorem of [Abhyankar–Moh–Suzuki](https://en.wikipedia.org/wiki/Abhyankar%E2%80%93Moh_theorem) 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](https://math.stackexchange.com/questions/2733573/why-mathbbcf-1t-f-2t-mathbbct-iff-f-1t-f-2t-neq0-and?noredirect=1&lq=1).


**(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?](https://mathoverflow.net/questions/304189/is-there-a-converse-of-abhyankar-moh-suzuki-theorem) and [Generalizations of Abhyankar-Moh theorem (embeddings of the line in the plane)](https://mathoverflow.net/questions/295158/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 variation on Abhyankar–Moh–Suzuki theorem
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
$\mathbb{C}[f,g]=\mathbb{C}[f-g,g]=\mathbb{C}[t^2,t^3+1]=\mathbb{C}[t^2,t^3]\subsetneq \mathbb{C}[t]$.


