The following question is the same as this question; I ask it here, since I have not got any comments there (I really apologize if I should have waited some more time before asking it here; it is just that recently I have asked there also several related questions without getting any comments).
Let $f=f(t),g=g(t) \in k[t]$, $k$ is a field of characteristic zero.
The well-known theorem of Abhyankar-Moh-Suzuki says the following: If $k[f,g]=k[t]$, then $\deg(f) | \deg(g)$ or $\deg(g) | \deg (f)$. (Actually, there is a version in any characteristic, but characteristic zero is good enough for me. Moreover, in view of the second remark below, perhaps it is better to assume that $k$ is also algebraically closed).
Is there some kind of converse, namely, something like: If $\deg(f) | \deg(g)$ or $\deg(g) | \deg (f)$ and $f$ and $g$ satisfy property $P$, then $k[f,g]=k[t]$?
Of course, property $P$ must include: $k(f,g)=k(t)$, but this alone is not enough, as $(f,g)=(t^6+t^2,t^3)$ shows; indeed, $k(t^6+t^2,t^3)=k(t^2,t^3)=k(t)$ but $k[t^6+t^2,t^3] \subsetneq k[t]$. Notice that the condition that $\deg(g) | \deg(f)$ is quite 'artificial' since we just added $t^6=(t^3)^2$ to $t^2$, and $\{t^2,t^3\}$ has relatively prime degrees (so by AMS theorem there is no chance that $\{t^2,t^3\}$ will generate $k[t]$) and $k[t^6+t^2,t^3]=k[t^2,t^3]$.
I had several ideas for property $P$, but unfortunately they did not guarantee that $k[f,g]=k[t]$, for example: If both $f$ and $g$ are odd polynomials = having terms of odd degrees only. Notice that if $f$ and $g$ are just odd polynomials then they can be far from generating $k[t]$, for example, $(f,g)=(t^5,t^{15})$, for which $k(f,g)=k(t^5,t^{15})=k(t^5) \subsetneq k(t)$. This question only shows that if $f$ and $g$ are odd polynomials, then $k(f,g)=k(h)$, where $h$ is a polynomial of odd degree but not necessarily of degree $1$, like here that $h=t^5$ (notice that the existence of such $h$ follows from Luroth-Noether theorem).
But even if we assume that two odd polynomials $f$ and $g$ satisfy $k(f,g)=k(t)$, there is no guarantee that $k[f,g]=k[t]$, as $(f,g)=(t^5,t^{15}+t^3)$ shows; indeed, $k(t^5,t^{15}+t^3)=k(t^5,t^3)=k(t^2,t^3)=k(t)$ but $k[t^5,t^{15}+t^3] \subsetneq k[t]$.
Remarks:
(1) There are very nice criteria for $k[f,g]=k[t]$ and $k(f,g)=k(t)$, see Theorem 2.1 of this paper. However, in practice, D-resultant seems not useful; perhaps it is useful only for very low degrees such as $(\deg(f),\deg(g)) \in \{(2,2),(3,3),(2,4)\}$.
(2) If $k$ is also algebraically closed (this is necessary!), then: $k[f,g]=k[t]$ if and only if for all $t \in k$, $(f'(t),g'(t))\neq (0,0)$ and $H: t \mapsto (f(t),g(t))$ is injective, see this paper. However, showing that $H$ is injective is difficult (except low degrees cases). Based on this result and on the answer to this question we can get: $k[f,g]=k[t]$ if and only if for all $\lambda,\mu \in k$, $\deg(\gcd(f(t)-\lambda,g(t)-\mu)) \leq 1$. (I can add a proof for this if someone will require. See also this question).
Any comments are welcome!
