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.