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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!

user237522
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1 Answers1

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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 $t-c$.

Aphelli
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  • 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 $Q \in \mathbb{C}^{\times}$? – user237522 Dec 15 '18 at 22:51
  • I mistook the parentheses for brackets. The main point should hold though: $(f,g)$ is injective because of that relation. – Aphelli Dec 15 '18 at 22:54
  • Thanks; please, could you slightly elaborate your answer? – user237522 Dec 15 '18 at 22:55
  • I edited. Is there enough detail? – Aphelli Dec 15 '18 at 22:59
  • Thank you. I think I now understand your answer, very nice! – user237522 Dec 15 '18 at 23:07
  • More precisely: $u: t \mapsto (f(t),g(t))$ is not injective, but injective except for a finite set (the zeros of $Q(f(t),g(t))$). For example, $f(t)=t^2+1$, $g(t)=t^3-4t$. We have $\frac{P}{Q}=\frac{g(t)}{f(t)-5}=\frac{t^3-4t}{t^2+1-5}=\frac{t(t^2-4)}{t^2-4}=t$. $u$ is not injective, since $u(2)=(f(2),g(2))=(5,0)=(f(-2),g(-2))=u(-2)$. Indeed, ${2,-2}$ is the zero set of $Q=f(t)-5$. – user237522 Dec 17 '18 at 00:52
  • Further assume that $f(t)=\sum_{i \neq 1}a_it^i +\lambda t$ and $g(t)=\sum_{i \neq 1}b_it^i+\mu t$, where $\lambda,\mu \in k-{0}$. In this case, $f'(0)=\lambda$ and $g'(t)=\mu$. Question: When is it possible to take $c=0$? (when there exists no such $d$). Plausible answer: If there is no $e$ such that $f'(e)=0=g'(e)$. Am I right? Thank you very much. – user237522 Oct 14 '19 at 19:29
  • I suspect that we can always find $Q(U,V)$ (satisfying $t=P(f,g)/Q(f,g)$) such that $Q(f(d),g(d)) \neq 0$. – user237522 Oct 14 '19 at 20:06
  • Please, what do you think about https://math.stackexchange.com/questions/3404163/finding-a-b-in-k-such-that-deg-gcd-f-1-a-g-1-b-deg-gcd-f-2-a-g-2 Thank you very much! – user237522 Oct 22 '19 at 16:07
  • Please, can you take a look at https://math.stackexchange.com/questions/4723640/generalizing-a-result-in-mathbbcx-to-a-result-in-mathbbcx-y?noredirect=1#comment10008781_4723640 – user237522 Jun 22 '23 at 18:49