$K,L$ are fields, $K\subseteq L$. $f,g \in K[x]$. Suppose that $f,g$ are relatively prime as elements of $K[x]$. Prove they remain relatively prime in $L[x]$.
I've tried everything I can think of. I feel like working with the contrapositive may be helpful but that's just a feeling.
Hint: Note that Bezout's identity holds for polynomial rings in one variable over a field, since such rings are principal ideal domains (PIDs):
$$f,g\in F[x]\text{ relatively prime }\iff \exists a,b\in F[x]\text{ such that }af+bg=1.$$
Use this both with $F=K$ and $F=L$.
Zev's answer is in some sense the canonical one, but here is another point of view, which is less elegant, but perhaps more intuitive.
We can embed $L$ into its algebraic closure $\overline{L}$; the algebraic closure of $K$ in $\overline{L}$ is then an algebraic closure of $K$.
Now $f$ and $g$ coprime in $K[x]$ means that they have distinct roots in $\overline{K}$. But these are also the roots of $f$ and $g$ in $\overline{L}$, and so $f$ and $g$ have distinct roots in $\overline{L}$. Thus $f$ and $g$ are coprime in $L[x]$ as well.
A different approach, without using Bezout's Identity,
We will use the result:
If $a(x)$ and $b(x)$ have a common root c in some extension of F(a field), they have a common factor of positive degree in $F[x]$.
(To see this, observe if we take the substitution function $\sigma_c$, then $a(x),b(x)\in ker(\sigma_c)$, but as $ker(\sigma_c)$ is an ideal over $F[x]$ it is principal ideal and hence $\exists$ a common factor of positive degree).
To proceed with the proof, if $f(x),g(x)$ are relatively prime in $K$, suppose they are not relatively prime in $L$. That means they have a common root in $L$ or some extension of $L$(Guaranteed by Basic Extension Theorem). But that would imply $f(x), g(x)$ have common factor of positive degree in $K$$\Longrightarrow\Longleftarrow$
So, they must be relatively prime in the extension $K$.
