Given a field $K$ and $f(X) \in K[x]$ is a polynomial with degree of a prime number. Suppose that for all extensions $L$ of $K$, if $f$ has a root in $L$ then $f$ splits in $L$. Prove that either $f$ be irreducible on $K$ or $f$ splits on $K$.
Let $p$ be the degree of $f$. Cases of $p = 2$ and $p = 3$ are quite obvious. Let $p \ge 5$.
In the attempt of solving this problem, I call $M$ to be the decomposition field of $f$ on $K$. Let $$f(x) = a(x - u_1)(x - u_2)...(x - u_p)$$ where $u_i \in M$. We clearly have that $K(u_1)$ is an extension of $K$ where $f$ has a root in $K(u_1)$. Thus, $f$ splits on $K(u_1)$. This implies $M \le K(u_1)$.
But on the other hand, we have $K(u_1) \le K(u_1,u_2,...,u_p) = M$. Thus, $M = K(u_1)$.
If $u_1 \in K$, it follows that $f$ splits in $K = K(u_1)$.
So it can be assumed that $u_i \notin K$ for all $i = \overline{1,p}$.
From this point, I have no clear idea to follow. I tried to assume that $f$ is reduccible on $K$ as $f = gh$ where $\deg g \ge 2$, $\deg h \ge 2$, but can't find a way to use the assumption that $\deg f$ be a prime.
Please give me a hint. Anything is greatly appreciated.
If $f$ has a root in $K$, then $f$ splits in $K$ by assumption. Suppose $f$ has no roots in $K$, and let $f=c\cdot f_1\dots f_r$ be the irreducible factorization of $f$ over $K$ (i.e. $f_1,\dots,f_r\in K[X]$ irreducible and monic, and $c\in K^{\times}$). Also, let $\Omega$ be an algebraic closure of $K$. Because of your assumption, the splitting field in $\Omega$ of the $f_i$ over $K$ agree. Since the $f_i$ are irreducible and of degree bigger than one, this implies that $f_1=\dots=f_r$. But then $f=f_1^r$ and thus $\deg f=r\cdot \deg f_1$. Since $\deg f_1>1$ and $\deg f$ is prime, it follows that $r=1$. Therefore $f=f_1$ is irreducible over $K$.