I'm trying to show that, if a polynomial in $F[x]$ is irreducible, then it has no repeated roots.
I showed that $(f, f') = 1$ iff $f$ has no repeated roots. I know that we can then write
$$ f(x)g(x) + f'(x)h(x) = (f, f') $$ for some $g, h\in F[x]$.
I'm unsure why we can claim that $f(x)$ is then reducible because of this?
My confusion stems from here: if the GCD is independent of the field (if $f, g\in F[x]$, and $K$ is a field extension of $F$, and we consider $f, g$ as elements in $K[x]$ by denoting them $f', g'$, then $(f, g) = (f', g')$), then how do we know that $(f, f')$ is necessarily a polynomial in $F[x]$ and not $K[x]$ for some field extension?
If I can convince myself that the GCD of two polynomials always lies in the same polynomial ring, then the proof is finished because we have a contradiction wherein we have $$0 < \text{deg(gcd(}(f, f'))) \leq \text{deg}(f) - 1,$$ i.e., we have a nontrivial factor.
But part of me is unconvinced that there aren't irreducible real-valued polynomials both having $(x-i)$ as a factor. Or, in general, why $(x - \alpha)$ can't be a factor of some irreducible polynomial $f(x)\in F[x]$ where $K$ is a field extension of $F$ and $\alpha\in K$ but $\alpha\notin F$.
