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

(1) $\deg(f) \geq 2$ and $\deg(g) \geq 2$.

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

In this question it was mentioned that in that case, there exist $a,b,c \in \mathbb{C}$ such that $\gcd(f(t)-a,g(t)-b)=t-c$.

Unfortunately, I do not see why this is true.

Perhaps Theorem 2.1 (about resultants) or this question (about subresultants) may somehow help (perhaps no).

Edit: Just to make sure:

Is it true that there exist infinitely many $a \in \mathbb{C}$ and infinitely many $b \in \mathbb{C}$ such that $\gcd(f(t)-a,g(t)-b)=t-c$, for (infinitely many) $c \in \mathbb{C}$?

Choose $c \in \mathbb{C}$ such that $f'(c) \neq 0$ etc. (as in the answer). Clearly, there are infinitely many such $c$'s. Let $a:=f(c)$ and $b:=g(c)$.

Asumme that there exist finitely many $a \in \mathbb{C}$ or finitely many $b \in \mathbb{C}$ such that $\gcd(f(t)-a,g(t)-b)=t-c$, $c \in \mathbb{C}$.

W.l.o.g., there exist finitely many $a \in \mathbb{C}$ such that $\gcd(f(t)-a,g(t)-b)=t-c$, $c \in \mathbb{C}$.

By the pigeon hole principle, there exist $a_0$ (among those finitely many $a$'s), such that for infinitely many $c$'s, we have $a_0=f(c)$.

This is impossible from the following reason: Let $h(t):=f(t)-a_0$. Then $h(c)=f(c)-a_0=0$, so $c$ is a root of $h(t)$, and trivially every polynomial can have only finitely many different roots.

So after all, I think that I have proved that there exist infinitely many $a \in \mathbb{C}$ and infinitely many $b \in \mathbb{C}$ such that $\gcd(f(t)-a,g(t)-b)=t-c$, $c \in \mathbb{C}$.

Any hints are welcome!

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We may assume that $f$ and $g$ are monic. There exists some nonzero two-variable polynomials $P,Q$ such that $P(f(t),g(t))=tQ(f(t),g(t))$, and $Q(f,g)(t)=0$ only finitely many times (else the composition $P/Q (f,g)$ is not defined because $Q(f,g)=0$).

Let $c$ be such that $f’(c) \neq 0$, and there exists no $d$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$.

Then $f(t)-f(c)$ and $g(t)-g(c)$ have only $c$ as a common root, because any root $d$ satisfies $(f,g)(d)=(f,g)(c)$, thus $d=(P/Q)(f(d),g(d))=(P/Q)(f(c),g(c))=c$. Moreover, $c$ is a simple root of $f-f(c)$. So the gcd of the polynomials is $t-c$.

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  • $\begingroup$ Thank you very much! I understand that there exist two-variable polynomials $P$ and $Q$ such that $\frac{P(f(t),g(t))}{Q(f(t),g(t))}=t$. Please, why $Q \in \mathbb{C}^{\times}$? $\endgroup$ – user237522 Dec 15 '18 at 22:51
  • $\begingroup$ I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation. $\endgroup$ – Mindlack Dec 15 '18 at 22:54
  • $\begingroup$ Thanks; please, could you slightly elaborate your answer? $\endgroup$ – user237522 Dec 15 '18 at 22:55
  • $\begingroup$ I edited. Is there enough detail? $\endgroup$ – Mindlack Dec 15 '18 at 22:59
  • $\begingroup$ Thank you. I think I now understand your answer, very nice! $\endgroup$ – user237522 Dec 15 '18 at 23:07

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