Let $F$ be a field and let $f,g \in F[x]$ be two coprime polynomials. I need to prove that for every $h \in F[x]$, there exist $h_{1}, h_{2} \in F[x]$ such that $$ h_{1}f + h_{2}g+h=0$$
What I attempted is the following:
Suppose $F$ is a field and $f,g\in F[x]$ are coprime polynomials. Then $\gcd(f,g)=1$.
Since $F$ is a field, we have that $F[x]$ is a Euclidean domain. Moreover, we also have results that say that the ring of polynomials with one variable over a field is a principal ideal domain, and hence a unique factorization domain.
Therefore, we have that if $1 = \gcd(f(x),g(x))$, then $$ 1 = r_{1}(x)f(x) + r_{2}(x)g(x) $$ for some $r_{1}(x), r_{2}(x) \in F[x]$.
Since $F[x]$ is a principal ideal domain, it is also an integral domain, and thus has no zero divisors. Therefore, we are able to multiply the last equation through by any $k(x)\in F[x]$ to obtain $$k(x) = k(x)r_{1}(x)f(x) + k(x)r_{2}(x)g(x) $$ Now, subtracting $k(x)$ from both sides, we obtain $$ k(x)r_{1}(x)f(x) + k(x)r_{2}(x)g(x) - k(x) = 0 $$ Now, define $$ h(x):=-k(x) $$ $$ h_{1}(x):=k(x)r_{1}(x)$$ $$ h_{2}(x):=k(x)r_{2}(x) $$ Then, the last equation becomes $$ h_{1}(x)f(x) + h_{2}(x)g(x)+h(x) = 0. $$ Since $k(x)$ was any arbitrary element of $F[x]$, for any $h \in F[x]$, we were able to find $h_{1}$, $h_{2} \in F[x]$ (depending on $h$) such that $h_{1}f+h_{2}g+h=0$.
Is what I did correct? If not, what can I do to fix it?
It just seemed too easy to me - we have everything in place we need to get the $ 1 = r_{1}(x)f(x) + r_{2}(x)g(x)$, and from there to get to the $h_{1}f+h_{2}g+h=0$. So, I thought I would ask just to make sure.
I thank you for your help!