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Let $\alpha := e^{2\pi i/6}$ be the sixth root of unity. Is $\mathbb Q(\alpha)$ both the splitting field of $x^6 - 1$ and $x^2 - x + 1$?

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    $\begingroup$ The two second degree factors of $x^6-1$, one of which equals your $x^2-x-1$, have the same discriminant, $-3$. Consequently $x^6-1$ has splitting field $\mathbb{Q}(\sqrt{-3})$, which is also the splitting field of $x^2-x-1$. $\endgroup$ – Erik Vesterlund Jul 28 '13 at 20:19
  • $\begingroup$ Note that of the two $x^2-x+1$ is irreducible over $\mathbb Q$. Sometimes it is important to test for irreducibility before applying apparently general theorems about extensions. $\endgroup$ – Mark Bennet Jul 28 '13 at 20:52
  • $\begingroup$ Possibly also interesting is that $\mathbb{Q}(\alpha)$ is also the splitting field of $(3x-2)^6-1$. $\endgroup$ – user14972 Jul 28 '13 at 21:28
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Hint: Alll sixth roots of unity are powers of $\alpha$. And $x^6-1=(x^3+1)(x^3-1)$.

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  • $\begingroup$ and it factors as $x^6 - 1 = (x-1)(x+1)(x^2+x+1)(x^2-x+1)$ into irreducible factors, but what does this tell me regarding my question. $\endgroup$ – StefanH Jul 28 '13 at 21:20
  • $\begingroup$ I had left it at $x^3+1$. The two roots of this are primitive $6$-th roots of unity. So one of them is $alpha$. So we have shown that the splitting fields of the two polynomials you were given are both $\mathbb{Q}(\alpha)$. $\endgroup$ – André Nicolas Jul 28 '13 at 23:51
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Clearly, $x^6-1$ splits over ${\bf Q}(\alpha)$, as does $x^2-x+1$. Furthermore, $\alpha$ is a root of $x^6-1$, so it must belong to its splitting field.

What are roots of $x^2-x+1$? Can you produce a sixth root of unity from one of them?

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  • $\begingroup$ Or both, in fact .... $\endgroup$ – Ted Shifrin Jul 28 '13 at 20:40
  • $\begingroup$ the roots of $x^2 - x + 1$ are the primitve sixth roots of unity, and so they generate all roots of $x^6 - 1$. $\endgroup$ – StefanH Jul 28 '13 at 21:19
  • $\begingroup$ @Stefan: That's correct. $\endgroup$ – tomasz Jul 28 '13 at 23:17

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