A problem about decomposition field of a polynomial with prime degree

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.

• I guess an idea would be to consider the group $Aut(M/K)$. I would say this group must have order 1 or $p$ and you can conclude – AlexL Oct 30 '18 at 19:20
• Sir, I think that the $M/K$ you mention here is a field, but it need to be proved. I tried but I can't prove that there are no internediate field between $K$ and $M = K(u_1)$. Will your argument work when $M/K$ is only a ring? Thank you. – ElementX Oct 30 '18 at 20:35

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 $$K$$.
• Please tell me if I get your idea right: Let $M_1$ be the splitting field of $f_1$ over $K$. Then $M_1$ must contain a root of $f_1$ which is also a root of $f$. Thus $M_1$ agree with $M$ by the assumtions. Thank you, sir. – ElementX Oct 30 '18 at 20:48