Let $f,g,h \in\mathbb{F}[x]$, with $f$ and $g$ are relatively prime. If $f\mid h$ and $g\mid h$, prove that $fg\mid h$.
What I've done so far: Experimenting with natural numbers, I suspect that $h=fg\gcd(a,b)$, where $fa=gb=h$. Define $q=\gcd(a,b)$, then there exists $u,v\in\mathbb{F}[x]$ such that $q=au=bv$.
$fg(au+bv)=fgau+fgbv=hgu+hfv=h(gu+fv)$. This is where I got stuck. After experimenting with natural numbers, it seems like $gu+fv=1$ is true. We were given that $f,g$ are relatively prime, so we know that $\exists m,n$ such that $1=fm+gn$. But how can we connect $m,n$ to $u,v$? Thanks.
You're quite close to the standard proof of this fact, which works both in $\Bbb Z$ and in $F[x]$.
Since $f$ and $g$ are relatively prime, there exist $m,n$ such that $mf+ng=1$. Then $h=1h=(mf+ng)h = mfh+ngh$. Since $h=fa$ and $h=gb$, we get $h = mf(gb) + ng(fa) = (mb+na)fg$, and so $fg$ divides $h$.