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Let $f=f(t),g=g(t),h=h(t) \in k[t]$, $k$ is a field of characteristic zero. Denote $\deg(f)=a$, $\deg(g)=b$, $\deg(h)=c$. Assume that $a \geq 2, b \geq 2, c \geq 2$.

Assume that $k[f,g] \neq k[t]$, $k[f,h]\neq k[t]$ and $k[g,h]\neq k[t]$, but $k[f,g,h]=k[t]$.

What can be said about $a$, $b$ and $c$?

For example, $f=t^2+t$, $g=t^3$, $h=t^6+t^2$; here $a=2$,$b=3$,$c=6$. (How to show that $k[f,h]\neq k[t]$?).

Of course, the inspiration for this question is Abhyankar-Moh-Suzuki theorem. See also this question.

Any comments are welcome!

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  • $\begingroup$ What would you like to say? For example, it is easy to arrange that $h$ can have any degree which is sufficiently large, so if you are looking for some divisibility, you are out of luck. $\endgroup$
    – Mohan
    Jun 14, 2018 at 2:54
  • $\begingroup$ Thanks for the comment. I am not sure what exactly I would like to obtain. (divisibility is one option). $\endgroup$
    – user237522
    Jun 14, 2018 at 9:46

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