Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer

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

share|improve this answer
How did you come up with $e^{ \pi i/3}$? – Breezy May 1 at 21:25
@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 1 at 21:37
In this example, you can also explicitly find all the roots with the help of the trusty quadratic formula. – Michael Joyce May 1 at 22:08
ohh I always thought the $n$th roots of unity were $\zeta$. – Breezy May 3 at 21:15
@Breezy $\zeta_n=e^{2i\pi / n}$ by definition. – Alex Becker May 4 at 1:00
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.