I have a question in that I have to prove that being $f,g\in k[x,y]$ irreducible polynomials distincts, there is $a,b\in k[x,y]$ and $c(x)\neq 0$ s.t.
$$af+bg=c(x)$$
Well, as $k(x)$ is a principal ideal domain, so $k(x)[y]$ is also a PID.
So, in $k(x)[y]$, $f$ and $g$ have a mcd, say $c_1\in k(x)[y]$, and there's $a_1,b_1\in k(x)[y]$ s.t. $a_1f+b_1g=c_1$. So, multiplying for some polynomial in $x$, there's $a,b,c\in k[x,y]$ s.t.
$$af+bg=c.$$
Note that by Gauss' Lema there's $c_2$ s.t. $c_2c_1$ is a mcd to $f$ and $g$ in $k[x,y]$.
However, I do not know as I can conclude that $c=c(x)$, ie, independs of $y$.
Many thanks in advance.
Perhaps I'm having trouble reading your post, but I can't figure out how you haven't already answered your question yourself.
I'll walk through what I think you're saying, and you can comment if at some point we differ.
Since $k(x)$ is a field, $k(x)[y]$ is a PID. By Gauss' lemma, $f,g$ being irreducible in $k[x,y]$ tells us that they are irreducible in $k(x)[y]$. Thus, since they are distinct (presumably including up to units), there exist $A,B\in k(x)[y]$ such that $$ Af + Bg = 1.$$ Then the coefficients of $A$ and $B$ have a least common denominator $c(x)\in k[x]$, so that $a:=Ac, b:=Bc\in k[x,y]$. Then multiplying by $c$, we get $$ af + bg = c,$$ with $c\in k[x]$, as desired.
EDIT: I found it, you've got a $c_1$ instead of $1$ where I have $Af+Bg=1$, you have $Af+Bg=c_1$. Since $f$ and $g$ are irreducible and distinct, they generate the unit ideal, thus you can take $c_1$ to be $1$, since that is their greatest common divisor.