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.

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    $\begingroup$ 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 $\endgroup$ – AlexL Oct 30 '18 at 19:20
  • $\begingroup$ 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. $\endgroup$ – 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 $K$.

  • $\begingroup$ 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. $\endgroup$ – ElementX Oct 30 '18 at 20:48
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    $\begingroup$ Yes, that is true of course. However, the main point in the reasoning is that the irreducible monic polynomial (I have just fixed this) defining an extension of degree bigger than one is unique. $\endgroup$ – Maurizio Moreschi Oct 31 '18 at 10:57

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