Let $F$ be a field, and let $f(x) \in F[x]$ be a polynomial of prime degree. Suppose for every field extension $K$ of $F$ that if $f$ has a root in $K$, then $f$ splits over $K$. Prove that either $f$ is irreducible over $F$ or $f$ has a root (and hence splits) in $F$.
Idea. Suppose that $f$ is not irreducible over $F$. So, $f=p_{1}^{r_{1}}...p_{n}^{r_{n}}$ where $0< \deg p_{i}^{r_{i}} < \deg f = p$ (in $F$). By absurd, suppose that $f$ has no roots in $F$, then $\deg p_{i} > 1$. Let $\alpha_{1},...,\alpha_{k}$ be the roots of $p_{1}$ and consider the splitting field of $p_{1}$, that is, $K_{1} = F(\alpha_{1},...,\alpha_{k})$.
Note that $\deg f = p = \underbrace{(r_{1}\deg p_{1})}_{a_{1}} + ... + \underbrace{(r_{n}\deg p_{n})}_{a_{n}}$, then $\gcd(a_{1},...,a_{n}) = 1$. So, there exists $p_{s}$ such that $\gcd (r_{s}\deg p_{s},r_{1}\deg p_{1}) = 1$.
By hypothesis, if $p_{i}$ has a root in $K_{1}$, then $p_{i}$ splits on $K_{1}$. But, $f$ has a root in $K_{1}$ since $p_{1}$ has too, then $f$ splits on $K_{1}$, then $p_{s}$ splits too. Thus, $K_{s} \subset K_{1}$. Therefore, $p_{s}$ divides $p_{1}$ and $\gcd(\deg p_{s},\deg p_{1})=\deg p_{s} > 1 = \gcd(r_{s}\deg p_{s},r_{1}\deg p_{1})$, an absurd.$\square$
Is this correct? Thanks for support!
The conclusion that $\gcd(a,b,c) = 1$ implies there are mutually coprime factors is wrong. For example take $\gcd(6,15,10) = 1$, but $\gcd(6,15) = 3, \gcd(10,15) = 15, \gcd(6,10) = 2$. Moreover the claim $K_s \subset K_1 \implies p_s \mid p_1$ is wrong. I mean take $p_s=x^2+1$ and $p_1 = x^4+1$ over $\mathbb{Q}$, then $K_s = \mathbb{Q}(i) \subset \mathbb{Q}(i,\sqrt{2}) = K_1$, but $p_s \not \mid p_1$
Anyway the idea is more or less OK and below is how you can prove the problem with your idea.
Let $\alpha_1$ be a root of $p_1$ in some extension field $F \subset K$, where $f=p_{1}^{r_{1}}...p_{n}^{r_{n}}$. Then we have that $f$ has a root in $F(\alpha_1)$ and by the condition it must split in this field and so we have that $F(\alpha_1)$ is a splitting field of $f$ over $F$. Now we have that $[F(\alpha_1):F] = \deg p_1$
But similarly if $\alpha_2$ is a root of $p_2$ then we have that $F(\alpha_2)$ is a splitting field of $f$ over $F$, so $F(\alpha_2) \cong F(\alpha_1)$. Hence:
$$\deg p_2 = [F(\alpha_2):F]=[F(\alpha_1):F] = \deg p_1$$
So in general we have that all irreducible factors of $f$ have the same degree. Hence:
$$p=\deg f = (r_1 + r_2 + \cdots + r_n)\deg p_1$$
As $p$ is prime we have that either $\deg p_1=1$, which means $f$ has a root in $F$ or $n=1$ and $r_1=1$, so $f$ is irreducible in $F[x]$.