Given fields $K\subseteq L$, why does $f,g$ relatively prime in $K[x]$ imply relatively prime in $L[x]$?

$K,L$ 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.

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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.

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$$f,g\in F[x]\text{ relatively prime }\iff \exists a,b\in F[x]\text{ such that }af+bg=1.$$
Use this both with $F=K$ and $F=L$.
@Tim: If $f$ and $g$ are relatively prime as elements of $K[x]$, there are $a,b\in K[x]$ such that $af+bg=1$. But $K[x]\subseteq L[x]$, so $a,b,f,g\in L[x]$, so the fact that $af+bg=1$ tells us that... –  Zev Chonoles Nov 18 '11 at 14:38