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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.

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  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen May 11 at 5:26
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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 $k\in\{1,...,6\}$.

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  • $\begingroup$ But also $\;\{1,a\}\;$ is a basis in any extension $\;\Bbb Q(a)/\Bbb Q\;$ of degree two, just as the OP wrote. $\endgroup$ – DonAntonio May 10 at 10:37
  • $\begingroup$ @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 $\{1,\alpha\}$? $\endgroup$ – Primavera May 11 at 3:25
  • $\begingroup$ @DonAntonio, is it possible to show that $$E=\{\beta=a_{0}+a_{1}\zeta+\cdots+a_{5}\zeta^{5}\in\mathbb{Q}(\zeta)\,:\,\sigma^{2}(\beta)=\beta)\}\stackrel{?}{=}\{c_{1}\cdot1+c_{2}\cdot(\zeta+\zeta^{2}+\zeta^{4})\,:\,c_{1},c_{2}\in\mathbb{Q}\}.$$ $\endgroup$ – Primavera May 11 at 3:31
  • $\begingroup$ Nitpick: either $\zeta^2$ or $\zeta^6=\zeta^{-1}$ generates an extension of degree six. Their minimal polynomial being $\Phi_7(x)=x^6+x^5+\cdots+x+1$. $\endgroup$ – Jyrki Lahtonen May 11 at 5:20
  • $\begingroup$ @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. $\endgroup$ – Jyrki Lahtonen May 11 at 5:23

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