Question regarding field extensions and basis Given a = $e^\frac{2i\pi}{18}$, field extension $|\Bbb Q(a):\Bbb Q| = 6$, I was wondering what the basis for $\Bbb Q(a)$ being, or how the elements are represented. I found the minimal polynomial to be $x^6-x^3+1$, which is degree 6 in $\Bbb Q$, so I'm guessing the basis for $\Bbb Q(a)$ would $\{1,a,a^2,a^3,a^4,a^5\}$. Does that make any sense? Normally the bases I have dealt with were degree $2$ so the elements were just like $a+bi$ or something with basis $\{1,i\}$
 A: Your finding of the minimal polynomial and the basis are fine, it proves that you have well understood the setup. A trick to retrieve the Galois group is to find another root for $x^6-x^3+1$. Some trial and error gives you that $-a^2$ is such a root. So there is an element $\sigma$ of the Galois group that maps $a \mapsto -a^2$, so we can then ask what is the image $\sigma(-a^2)$? It's $--a^4$ and we verify that it is indeed another root! we can go on and find the three other roots: $\sigma(-a^4) = -a^8 = a^2-a^5$, $\sigma(a^2-a^5) = \sigma(a^2)-\sigma(a^5) = -a^4 + a^{10} = -a^4 -a$ and $\sigma(-a^4 -a) = \sigma(-a^4) - \sigma(a) = a^2 + a^5 + a^2$. If we change basis to these roots 
$$
\{a, -a^2,-a^4,a^2-a^5,a^4-a, a^5\}
$$
we have an easy expression for $\sigma$ as a matrix.
When we revert this matrix to the original basis $\{1,a,a^2,a^3,a^4,a^5\}$ this yields the matrix
$$
\sigma = \begin{pmatrix}1&0&0&1&0 &0\\
                        0&0&0&0&0 &1\\
                        0&-1&0&0&-1&0\\
                        0&0&0&-1 & 0&0\\
                        0&0&1&0&0&0\\
                        0&0&0&0&1&0
                        \end{pmatrix}
$$
This matrix generates the Galois group, a cyclic group of order $6$.
