I have some question about computation in cyclotomic field $K=\mathbb{Q}(\zeta)$, where $\zeta$ is a primitive $7$th root of unity.
I know that the subfield $E=\mathbb{Q}(\zeta+\zeta^{2}+\zeta^{4})$ of $\mathbb{Q}(\zeta)$ is of degree $2$ over $\mathbb{Q}$.
Actually, I found the primitive element of $E$ as $\zeta+\zeta^{2}+\zeta^{4}$ using the fact that $E$ is the fixed field of $\langle\sigma^{2}\rangle$, where $\sigma(\zeta)=\zeta^{3}$.
Now, considering $E$ as a vector space over $\mathbb{Q}$, then $E$ has a $\mathbb{Q}$-basis as $\{1,\zeta+\zeta^{2}+\zeta^{4}\}$. (Is it possible?)
If it possible, how to write certain elements, for example, $\zeta^{3}$ and $\zeta^{6}$, as a linear combination of elements of the previous $\mathbb{Q}$-basis?
I tried to use the fact that $\zeta^{7}=1$ and $\Phi_{7}(\zeta)=0$, but i can't find any relation.
Can anyone help me? Thank you.
Best Answer
No, it's not possible because $1\in \mathbb Q$. Notice that $\zeta +\zeta ^2+\zeta ^4$ has order at least $2$, and since $E$ has degree $2$, then $\zeta +\zeta ^2+\zeta ^4$ has degree $2$. This mean that a basis is $$\{\zeta +\zeta ^2+\zeta ^4,(\zeta +\zeta ^2+\zeta ^4)^2\}.$$ Now, neither $\zeta ^2$ nor $\zeta ^6$ are in $E$ (because those elements has degree $7$). In fact $\zeta ^k\notin E$ for all $k\in\{1,...,6\}$, because $\zeta ^k$ is a primitive root for all $k\in\{1,...,6\}$.