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Let $f=f(t),g=g(t) \in k[t]$, whrer $k$ is a field of characteristic zero.

Assume that:

(i) $k(f,g)=k(t)$. Therefore, there exist $P(X,Y),Q(X,Y) \in k[X,Y]$, such that $t=\frac{P(f,g)}{Q(f,g))}$.

(ii) $f=\sum_{i=1}^{n}a_it^i$, $g=\sum_{j=1}^{m}b_jt^j$, where $a_i,b_j \in k$ and $a_1b_1 \neq 0$. Therefore, $f'(0)=a_1$, $g'(0)=b_1$, $f(0)=0$, $g(0)=0$.

By questions 1 and 2, there exist $a,b,c \in k$ such that $\gcd(f-a,g-b)=t-c$.

When is it possible to find $a,b \in k$ such that $\gcd(f-a,g-b)=t$? namely, when can we can take $c=0$?

According to the answer in 2, $c=0$ is possible, if there exist no $d \in k$ such that $Q(f(d),g(d))=0$ and $f(d)=f(c)$. Here, $f(c)=0$, so if $d$ is another root of $f$, we must have $Q(0,g(d))=Q(f(d),g(d)) \neq 0$. However, I do not know what is the form of such $Q$; if we knew that $Q(0,g(d) \neq 0$, then we were done).

Remark: Here $t$ divides both $f$ and $g$, but perhaps there exist additional common divisors of $f$ and $g$. Notice that $t^2$ does not divide $f$ nor $g$, since otherwise, $f'(0)=g'(0)=0$, but here $f'(0)=a_1 \neq 0$ and $g'(0)=b_1 \neq 0$.

Thank you very much!

user237522
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  • If such $a,b$ exist, then it must be $a=f(0)$ and $b=g(0)$. – Gerry Myerson Oct 15 '19 at 21:48
  • Thank you! Please, can you explain? (So here $a=f(0)=0, b=g(0)=0$, the original $f$ and $g$). – user237522 Oct 15 '19 at 21:50
  • If $t$ is a divisor of $f(t)-a$, then $a$ must be the constant term of $f(t)$; that is, $a$ must be $f(0)$. – Gerry Myerson Oct 15 '19 at 21:52
  • @GerryMyerson, what if we replace the requirement $\gcd(f-a,g-b)=t$ by $\gcd(f-a,g-b)=1$? – user237522 Oct 15 '19 at 21:52
  • That should be true for all but finitely many pairs $a,b$. – Gerry Myerson Oct 15 '19 at 21:53
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    @GerryMyerson, nice and easy explanation. Thank you very much. – user237522 Oct 15 '19 at 21:54
  • @GerryMyerson, thank you for your second answer. I guess that the explanation is somewhat similar (probably easier) to the explanation for $\gcd(f-a,g-b)=t-c$ presented in [2]? – user237522 Oct 15 '19 at 21:57
  • Please, I do not understand: Shouldn't $\deg(\gcd (f-a,g-b))=0$ true for all but finitely many pairs $(a,b) \in k^2$ contradict https://math.stackexchange.com/questions/3042016/why-mathbbcft-gt-mathbbct-implies-that-gcdft-a-gt-b-t-c?noredirect=1&lq=1, namely, $\deg(\gcd (f-a,g-b))=1$ true for all but finitely many pairs $(a,b) \in k^2$? – user237522 Oct 18 '19 at 13:13
  • I think you're misunderstanding that other question. I think it says for (almost) every $c$ there exist $a,b$ such that $\gcd(f-a,g-b)=t-c$. So for (almost) every $a$ there exist $b,c$ such that $\gcd(f-a,g-b)=t-c$. But that doesn't imply that for (almost) every pair $a,b$ there exists $c$ such that $\gcd(f-a,g-b)=t-c$. – Gerry Myerson Oct 19 '19 at 11:34
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    Thank you very much for the clarification. (I guess you are right). – user237522 Oct 20 '19 at 07:34
  • @GerryMyerson, 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 16:06

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