$f(x) \in F[x]$ reducible,prime degree,for any extension $K/F$,if $f(x)$ has a root in $K$ then $f(x)$ splits in $K$,then $f(x)$ has a root in $F$? Let $f(x) \in F[x]$ be a polynomial , which is not irreducible in $F[x]$ , of prime degree such that for any extension $K/F$ , if $f(x)$ has a root in $K$ then $f(x)$ splits in $K$ ; then is it true that $f(x)$ has a root in $F$ ? 
 A: The condition "if $ f(x) $ has a root in $ K $, then $ f $ splits into linear factors in $ K $" implies that the splitting field $ L/F $ of $ f $  is generated by any one of its roots, and since the degree of this extension doesn't depend on the choice of primitive element, it follows that every root of $ f $ has equal degree over $ F $, i.e $ f $ splits into irreducible factors of equal degree in $ F[X] $. However, $ f $ has prime degree and is reducible; so the only way this can happen is if $ f $ splits into linear factors in $ F[X] $, so trivially it has a root in $ F $.
A: Let's try by absurd.
Assume that $f$ is not irreducible in $F[X]$ and satisfies the property (i.e. that any extension of $F$ contains either no roots of $f$, either all the roots of $f$).
Write $f=P_1^{k_1}...P_r^{k_r}$ with $P_i$ irreducible polynomials and $k_i\ge 1$.
Assume that $d_i:=\mathrm{deg}(P_i)>1$ for all $i$ (i.e. $f$ has no roots in $F$). 
Let $K/F$ be the splitting field of $P_1$: it has degree $d_1$ over $F$ by assumption on $f$, because adding one root of $P_1$ is the same as adding all of them. Take $i_0\in\{1, ..., r\}$ such that $d_{i_0}$ is prime to $d_1$ (such a $i_0$ exists, otherwise $\mathrm{deg}(f)=\sum_{i=1}^r k_id_i$ wouldn't be prime).
The hypothesis on $f$ tells us that $P_{i_0}$ also splits in $K$, which implies that the splitting field of $P_{i_0}$ is a subextension of $K$, hence $d_{i_0}$ divides $d_{1}$ and contradicts $(d_{i_0}, d_{1})=1$ ...
So there exists a $i_1\in\{1, ..., r\}$ such that $d_{i_1}=1$, and $f$ has a root in $F$.
