$\gcd$ of polynomials over a field I have the polynomials $f,g\neq 0 $ over a field $F$. We know also that $\gcd(f,g)=1$ and 
$$
       \det \begin{pmatrix}
        a & b \\
        c & d \\
        \end{pmatrix}\neq 0.
$$
I need to prove that $\gcd(af+bg,cf+dg) = 1 $ for every $a,b,c,d \in F$. 
I really do not know how to start answer the question. Thanks for helpers!
 A: Just work through the equations of gcd (bearing in mind that you're working in $f(x)$, hence constants do not matter):
$ \gcd(af+bg, cf+dg) = \gcd(adf+bdg, cbf + bdg) = \gcd( (ad-bc)f, cbf + bdg) = \gcd( f, cbf+bdg) = \gcd(f, bdg) = \gcd(f, g) = 1$
A: HINT: What if $\gcd(af+bg,cf+dg) \neq 1$? 
Well, By using the Unique Factorization Property of the polynomial ring over a field you can find a prime polynomial $p$ such that $p \mid \gcd(af+bg,cf+dg)$. 
Does it give you any idea about how you should go ahead with this argument to get a contradiction?
2nd hint:
Well, since $p \mid af+bg$ and $p \mid cf+dg$ we can conclude that $p \mid d(af+bg)-b(cf+dg) \implies p \mid (ad-bc)f$ .
But $p$ is prime, and $ad-bc \neq 0$, therefore $\gcd(p, ad-bc) =1$ since $p$ is assumed to be a monic polynomial. so, you get $p \mid f$. the same reasoning gives you $p \mid g$. That means $p \mid \gcd(f,g)$ which gives you the contradiction.
A: I see linear combinations. Linear combinations mean linear algebra. It is very often useful to write linear combinations in terms of matrix arithmetic when you can. In this case, we get the very interesting
$$ \left[ \begin{matrix} af + bg \\ cf + dg \end{matrix} \right]
= \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \left[ \begin{matrix}f \\ g \end{matrix} \right] $$
There are a number of ways you can go from here. Here are two things to get you started.
Exercise: compare elementary matrices (i.e. elementary row operations) to steps of the Euclidean algorithm.
Exercise: What can you easily say about the relationship between $\gcd(f,g)$ and $\gcd(af+bg, cf+dg)$? (optional: rephrase it in terms of vectors and matrices, rather than pairs and linear combinations) How can you use the fact that matrix is invertible to go the other direction?
