Suppose $F \subset K$ are both fields. Let $f \in F[x]$. Suppose $f$ is irreducible in $K[x]$. Prove $f$ is also irreducible in $F[x]$.
Want to prove by contradiction.
$f$ is reducible in $F[x]$. So $f\in F[x]\subset K[x]$ so the same operations apply in $K[x]$ making $f$ reducible in $K[x]$. But $f$ is irreducible in $K[x]$.
It cannot be that easy? Missing something?
Thanks for the comments
Proof:(By contradiction)
f is reducible in $F[x]$. $$ \exists \text{ non unit, non associate } g(x),q(x) \in F[x]:f(x)=g(x)q(x)$$
Since $f\in F[x] \subset K[x]$ so $f(x),g(x),q(x)\in K[x]$ same operation operate in $K[x]$. Making $f$ reducible in $K[x]$. Contradiction $f$ is irreducible in $K[x]$