I have to compute the splitting field of $x^6-1 \in Q[x]$. I know that $x^6-1=(x+1)(x-1)(x^2-x+1)(x^2+x+1)$ but I don't know what to do after that.
Please help me.
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You want to find the smallest field containing $\mathbb Q$ which contains all the roots of $x^6-1$. For concreteness, we can work in $\mathbb C$. The roots of $x^6-1$ are $e^{k\pi i/3}$ for $k=0,\ldots,5$. It is easy to see that these are all powers of $e^{\pi i/3}$, so $\mathbb Q(e^{\pi i/3})$ contains both $\mathbb Q$ and all roots of $x^6-1$. Thus it contains the splitting field. Since it is by definition the smallest field containing both $\mathbb Q$ and $e^{\pi i/3}$, the splitting field is contained in it as well. Thus the splitting field is $\mathbb Q(e^{\pi i/3})$.
Alex Becker
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How did you come up with $e^{ \pi i/3}$? – May 01 '13 at 21:25
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@Breezy It's a well-known fact that the $n$-th roots of unity are $e^{2k\pi i/n}$ for $k=0,\ldots,n-1$. – Alex Becker May 01 '13 at 21:37
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In this example, you can also explicitly find all the roots with the help of the trusty quadratic formula. – Michael Joyce May 01 '13 at 22:08
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ohh I always thought the $n$th roots of unity were $\zeta$. – Breezy May 03 '13 at 21:15
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@Breezy $\zeta_n=e^{2i\pi / n}$ by definition. – Alex Becker May 04 '13 at 01:00
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oh. well that makes more sense now. thank you! – Breezy May 04 '13 at 23:22