Show two irreducible polynomials from a polynomial field cannot share a root The problem is as follows:
Let $F$ be a field and $K \leqslant F$. Let $f(x)$ and $g(x) \in K[x]$, where $f$ and $g$ are irreducible and not associates of one another.
Show that $f$ and $g$ do not share any roots in $F$.
Here is my attempt:
AFSOC $\exists a \in F$ s.t. $f(a) = g(a) = 0$. 
Then $\exists b(x), c(x) \in F[x]$ s.t. $b(x)(x-a) = f(x)$, and $c(x)(x-a) = g(x)$.
Here is where I am a little unsure of my work. Since $b(x)$ and $c(x)$ need not be in $K[x]$, I am not sure if I can make the following claim, but anyways I will proceed.
Since $x-a$ cannot be a unit, it must be the case that both $b(x)$ and $c(x)$ are units in $F$.
Thus $f(x) = b(x)c^{-1}(x)c(x)(x-a) = b(x)c^{-1}(x)g(x)$
Contradicting that $f(x)$ and $g(x)$ are not associates.
Any hints/feedback would be much appreciated :)
 A: By Bezout, $\,f,g$ coprime $\,\Rightarrow  h f + h' g = 1\ $ so evaluating at $\,x=a\,$ yields $\,0 = 1\,$ in $F\, \Rightarrow\!\Leftarrow$
A: Well I'll be honest, your work doesn't make much sense past a certain point.
Specifically not only do $b(x)$ and $c(x)$ not need to be in $K$, they probably aren't since they're polynomials in $F[x]$, so they're not in $K$ unless they're of degree 0 and also in $K$ rather than $F$. The bigger issue is when you conclude that $b$ and $c$ (I'm dropping the $(x)$ notation, since I find it clunky) are units. That's not true (a priori), since $f$ and $g$ are irreducible in $K[x]$ not $F[x]$, and ignoring the issue of whether $f$ and $g$ are in $K[x]$, $(x-a)$ is not in $K[x]$ either, so the factorizations $f=(x-a)b$ and $g=(x-a)c$ are not factorizations in $K[x]$.
There are several approaches here, depending on your level.
The first is to use minimal polynomials. Since $a\in F$ is the root of some polynomial over $K$ (being the root of both $f$ and $g$), it has some nonzero polynomial of smallest degree in $K[x]$ of which it is a root. Using the division algorithm, you can prove that this polynomial, $m$ has the property that it divides any polynomial with coefficients in $K$ for which $a$ is a root. Thus $m\mid f$ and $m\mid g$, but since $m$ is nonzero and has a root it has positive degree, so since $m\mid f$, and $f$ is irreducible, $m$ and $f$ are associates, and similarly $m$ and $g$ are associates. Contradiction.
The other approach is essentially the same, but phrased in more abstract language.
Let $\phi_a : K[x]\to F$ be the evaluation at $a$ homomorphism, noting that it is nonzero, since $\phi_a(1)=1$. $f$ and $g$ are both in the kernel of this map and both irreducible, so they must generate the kernel, since $K[x]$ is a PID (and they're irreducible). Thus $f\mid g$ and $g\mid f$, contradiction.
