Suppose that $m(t)$ is irreducible over $K$ and $\alpha$ has minimal polynomial $m(t)$ over $K$. Does $m(t)$ necessarily factorise over $K(\alpha)$ into linear (Degree 1) polynomials?

Thinking of concrete examples, I'm trying $K=\mathbb{Q}$ and $\alpha$ is the real cube root of 2...so I want to say no. Is this the right way to think?

  • 1
    $\begingroup$ Your counterexample is correct. $\endgroup$
    – user281392
    Sep 14, 2016 at 1:22
  • $\begingroup$ I cannot speak for the moral rectitude of your thought processes $\ddot{\smile}$. However, in your example, $\Bbb{Q}(\alpha)$ is a subfield of $\Bbb{R}$ that cannot contain the other two cube roots of $2$, so the minimal polynomial of $\alpha$ won't split in $\Bbb{Q}(\alpha)$. $\endgroup$
    – Rob Arthan
    Sep 14, 2016 at 1:27
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    $\begingroup$ @Ldog327 With no accepted answer out of 28 questions, you are one of the top thankless users on this site. Congratulations! $\endgroup$
    – user26857
    Sep 14, 2016 at 18:01

1 Answer 1


Your counterexample is correct as $\mathbb{Q}[\sqrt[3]{2}]$ is a real field but the other roots of $x^3 - 2$ are nonreal.


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