PROPOSITION
Let $f$ and $g$ be a non-zero polynomial in $k(t)$ show that :
there exists a unique polynomial such that
(1) $d$ divides $f$ and $g$
(2) if $d$ divides $f$ and $g$ then $d'$ divides $d$ 
proof:
(1) Define $d$ as a non-zero polynomial
of smallest possible degree such that it is of the form $a(x)f(X)+b(x)g(x)$ for some
polynomials $a(X)$ and $b(x)$. Now assume that $d$ monic by multiplying with
a constant.
using division with remainder, we find $q(x)$ and $r(x)$ such that $f(x)=q(x)d+r(x)$ with  $deg(r(x))<deg(d(x))$
now $r(x)=f(x)-q(x)(a(x)f(x)+b(x)g(x)=(1-q(x)a(x))f(x)+(1-q(x)b(x))g(X)$ .
Now because of the minimality of the degree of d, we must have $r(x)=0$. this shows that
d divides
$f(x)$.we can proof that $d$ divides $g(x)$ in a similar way.
(2)if $d'$ is any polynomial dividing both $f(x)$ and $g(x)$ then $d'$ also divides
$d=a(x)f(x)+b(x)g(x)$. This also
shows that $d$ is a common divisor of $f(x)$ and $g(x)$ of largest possible degree.
if $d$ is another monic common divisor of $f(x)$ and $g(x)$ then $d'$ divides $d$ 
and since both are monic of the same degree,we get $d'=d$. This shows the uniqueness.