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Let $f_1,g_1,f_2,g_2 \in k[t]$, $k$ a field of characteristic zero.

(Edit: Assume that $k(f_1,g_1)=k(t)$ and $k(f_2,g_2)=k(t)$).

By this question, there exist $a_1,b_1 \in k$ such that $\deg(\gcd (f_1-a_1,g_1-b_1)) =1$, and there exist $a_2,b_2 \in k$ such that $\deg (\gcd (f_2-a_2,g_2-b_2)) =1$.

Remark: I do not care about $c_1,c_2 \in k$ such that $\gcd (f_1-a_1,g_1-b_1) = t-c_1$ and $\gcd (f_2-a_2,g_2-b_2) = t-c_2$. (Namely, it is fine with me if $c_1 \neq c_2$).

Is it possible to find $a,b \in k$ such that $\deg (\gcd (f_1-a,g_1-b)) =1$, and
$\deg (\gcd (f_2-a,g_2-b)) =1$?

From what I understand, we can take $a:=a_1=a_2$, but I am not sure if we can find the same $b$.

A relevant comment can be found here = the last comment of Gerry Myerson: "... 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 $\gcd(f−a,g−b)=t−c$". The problem is that, after fixing a 'good' $a$ for both $(f_1,g_1)$ and $(f_2,g_2)$, I am not sure if we can find 'good' $b_1=b_2$.

Even if there exist infinitely many 'good' pairs for $(f_1,g_1)$, $U:=\{(a_1,b_1)| \deg(\gcd(f_1-a_1,g_1-b_1))=1\}$, and infinitely many 'good' pairs for $(f_2,g_2)$, $V:=\{(a_2,b_2)| \deg(\gcd(f_2-a_2,g_2-b_2))=1\}$, I guess that there is no guarantee that $U \cap V$ is nonempty.

Any hints are welcome!

user237522
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What does $\deg(\gcd(f-a,g-b))=1$ mean? It means $\gcd(f-a,g-b)=t-c$ for some $c$ in $k$, which implies that $f(c)=a$ and $g(c)=b$.

So, for your question, you want $c_1,c_2$ in $k$ such that $f_1(c_1)=f_2(c_2)=a$ and $g_1(c_1)=g_2(c_2)=b$.

But given polynomials $f_1,f_2$ there's no reason why there should exist $c_1,c_2$ such that $f_1(c_1)=f_2(c_2)$. E.g., over the reals, if $f_1(t)=t^2+1$ and $f_2(t)=-t^2-1$ then the ranges of $f_1$ and $f_2$ are disjoint.

Gerry Myerson
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  • Very nice and clear answer, thank you very much! – user237522 Oct 23 '19 at 10:14
  • Please, do you think that there exists a mild condition that guarantees a positive answer to my question? – user237522 Oct 23 '19 at 10:27
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    I don't know, but I'm not optimistic. It's not much to ask that the range of $f_1$ and the range of $f_2$ not be disjoint, so there will be $c_1,c_2$ such that $f_1(c_1)=f_2(c_2)$, but then it's a bit much to expect that also $g_1(c_1)=g_2(c_2)$. – Gerry Myerson Oct 23 '19 at 10:59
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    It’s probably not going to be mild, even in the algebraically closed case. Indeed, a necessary condition is that the curves of parametric equations $(f_1,g_1)$ and $(f_2,g_2)$ meet. These (using resultants) correspond to two unidimensional subvarieties of an affine plane. So either the curves are equal (which is pretty strong), or their intersection is finite (which seemingly enables all sorts of weird algebraic behavior at the intersection). – Aphelli Oct 23 '19 at 11:28
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    Thank you very much both of you. Interesting comments. – user237522 Oct 23 '19 at 12:43