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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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 $F=L$.

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I still fail to see how I connect the two fields – Tim anderson Nov 18 '11 at 13:44
@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
Rephrase @ZevChonoles's comment: If $f$ and $g$ are relatively in $K[x]$ $\Rightarrow$ $\exists a,b\in K[x]$ such that $af+bg=1\in K[x]$ $\Rightarrow$ $\exists a,b\in L[x]$ such that $af+bg=1\in L[x]$ $\Rightarrow$ $f$ and $g$ are relatively prime in $L[x]$. – bfhaha Sep 27 '15 at 13:47

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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Shouldn't this imply that being separable is same as being square-free? I am not sure why the following Wikepedia website says the otherwise: en.wikipedia.org/wiki/Separable_polynomial – Gil Jul 23 '15 at 6:11