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.
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 $\mathbb Q(e^{\pi i/3})$. |
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