If $\omega$ is the cube root of unity then I know it’s minimal polynomial over $\mathbb{Q}$ is $x^2 + x +1$. How do I know this is still the minimal polynomial of $\omega$ over $\mathbb{Q}(7^{1/3})$?

I know the only roots of $x^2+x+1$ are $\omega$ and $\omega ^2$. And so I thought this still wouldn’t split in $\mathbb{Q}(7^{1/3})$ and therefore must still be irreducible in the larger field.

But does this logic work in general? If a polynomial is irreducible over a field and none of its roots are in the field extension does that mean it is still irreducible in the field extension?

  • $\begingroup$ Also is cube root of unity the same thing as primitive cube root of unity? $\endgroup$ – Gabi23 Apr 26 '20 at 23:21
  • $\begingroup$ No, that needs to be specified. The cube roots of $1$ are $1,\omega,\omega^2$. $\endgroup$ – quasi Apr 26 '20 at 23:21
  • $\begingroup$ $1$ is also a cube root of $1$, but not a primitive cube root of $1$ $\endgroup$ – J. W. Tanner Apr 26 '20 at 23:22
  • $\begingroup$ What is primitive cube root of one? $\endgroup$ – Gabi23 Apr 26 '20 at 23:23
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    $\begingroup$ If $n$ is a positive integer, we say$\;x\in\mathbb{C}\;$is a primitive $n$-th root of unity if $x^n=1$, and $x^m\ne 1$ for any positive integer $m < n$.$\;$So if $w=exp(2i\pi/3)$, then $w$ and $w^2$ are both primitive cube roots of $1$. $\endgroup$ – quasi Apr 26 '20 at 23:27

For a simple counterexample, let $f(x)=x^4-2$.

Then $f$ is irreducible over $\mathbb{Q}$, but over the field $\mathbb{Q}(\sqrt{2})$, $f$ factors as $(x^2+\sqrt{2})(x^2-\sqrt{2})$.

However none of the roots of $x^4-2$ are elements of $\mathbb{Q}(\sqrt{2})$.

  • $\begingroup$ So why did my argument work the first time? $\endgroup$ – Gabi23 Apr 27 '20 at 0:13
  • $\begingroup$ Because if a a polynomial of degree $2$ or $3$ factors over a field, then it must have a linear factor, hence must have a root in the field. Moreover if a quadratic polynomial factors over a field, then the field must contain both roots. $\endgroup$ – quasi Apr 27 '20 at 0:16
  • $\begingroup$ Thank you very much. $\endgroup$ – Gabi23 Apr 27 '20 at 0:36

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