# Computation in cyclotomic field $\mathbb{Q}(\zeta_{7})$ over $\mathbb{Q}$.

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.

• A curiosity: $2(\zeta+\zeta^2+\zeta^4)+1=i\sqrt 7$. So $E=\Bbb{Q}(\sqrt{-7})$. See for example here or better yet, google up (quadratic) Gauss sum. – Jyrki Lahtonen May 11 at 5:26

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 $$k\in\{1,...,6\}$$.
• But also $\;\{1,a\}\;$ is a basis in any extension $\;\Bbb Q(a)/\Bbb Q\;$ of degree two, just as the OP wrote. – DonAntonio May 10 at 10:37
• @Surb, Set $\alpha=\zeta+\zeta^{2}+\zeta^{4}$, then is it false that $\mathbb{Q}(\alpha)$ is a vector space over $\mathbb{Q}$ having a basis as $\{1,\alpha\}$? – Primavera May 11 at 3:25
• @DonAntonio, is it possible to show that $$E=\{\beta=a_{0}+a_{1}\zeta+\cdots+a_{5}\zeta^{5}\in\mathbb{Q}(\zeta)\,:\,\sigma^{2}(\beta)=\beta)\}\stackrel{?}{=}\{c_{1}\cdot1+c_{2}\cdot(\zeta+\zeta^{2}+\zeta^{4})\,:\,c_{1},c_{2}\in\mathbb{Q}\}.$$ – Primavera May 11 at 3:31
• Nitpick: either $\zeta^2$ or $\zeta^6=\zeta^{-1}$ generates an extension of degree six. Their minimal polynomial being $\Phi_7(x)=x^6+x^5+\cdots+x+1$. – Jyrki Lahtonen May 11 at 5:20
• @Primavera We can write all the elements of $E=\Bbb{Q}(\zeta+\zeta^2+\zeta^4)$ as $\Bbb{Q}$-linear combinations of $1$ and $\zeta+\zeta^2+\zeta^4$. But neither $\zeta^3$ nor $\zeta^6$ is an element of $E$. That is the problem as Surb explained. – Jyrki Lahtonen May 11 at 5:23