# How to find the minimal polynomial if I know its minimal polynomial in a subfield.

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?

• Also is cube root of unity the same thing as primitive cube root of unity? – Gabi23 Apr 26 '20 at 23:21
• No, that needs to be specified. The cube roots of $1$ are $1,\omega,\omega^2$. – quasi Apr 26 '20 at 23:21
• $1$ is also a cube root of $1$, but not a primitive cube root of $1$ – J. W. Tanner Apr 26 '20 at 23:22
• What is primitive cube root of one? – Gabi23 Apr 26 '20 at 23:23
• 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 $1$. – 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 $$\mathbb{Q}(\sqrt{2})$$.
• 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. – quasi Apr 27 '20 at 0:16