The question:

Let $F:=\mathbb{Q}, E:=\mathbb{Q}(\zeta)$ where $\zeta :=\exp({2\pi i\over 7})$. Find all the middle fields of the extension $E/F$.

My attempt:

The roots of $f(x):=\text{irr}(\zeta,F)=x^6+\dots+1$ are $\zeta^k$ with $1\leq k\leq 6$. Let $G:=\text{Aut}(E/F)$.

$f$ is separable so $|G|=[E:F]=\deg(f)=6$. Thus, $$ G=\{\sigma_k|1\leq k\leq 6,\sigma_k=(\zeta\mapsto\zeta^k)\} $$ Notice that$$ |\sigma_1|=1,|\sigma_6|=2,|\sigma_2|=|\sigma_4|=3,|\sigma_3|=|\sigma_5|=6$$ Thus $G\cong \mathbb{Z}_6$. Thus, the subgroups of $G$ are $$ H_1=\{\sigma_1,\sigma_6\},H_2=\{\sigma_1,\sigma_2,\sigma_4\}$$ From the fundamental theorem of Galois theory, the midterm fields of the extension are $K_1=E^{H_1}$ and $K_2=E^{H_2}$.

Consider $K_1$. Let $E\ni x=\sum_{i=0}^6 a_i\zeta ^i$. If $x\in K_1$ then $x=\sigma_6(x)$: $$ a_0+a_1\zeta+a_2\zeta^2+a_3\zeta^3+a_4\zeta^4+a_5\zeta^5+a_6\zeta^6= \\ a_0+a_1\zeta^6+a_2\zeta^5+a_3\zeta^4+a_4\zeta^3+a_5\zeta^2+a_6\zeta \\ \Rightarrow a_1=a_6,a_2=a_5,a_3=a_4 \\ \Rightarrow K_1=\mathbb{Q}(\cos({2\pi\over7})) $$

With similar argument we get that if $x\in K_2$ then $$a_1=a_2=a_4,\\a_3=a_5=a_6$$ and thus, $$x=\beta_0+\beta_1(\zeta+\zeta^2+\zeta^4)+\beta_2(\zeta^3+\zeta^5+\zeta^6)$$

Here I stuck, I dont know how to identify $K_2$ with that characterisation of $x$.

  • 2
    $\begingroup$ Why don't you put $\zeta+\zeta^2+\zeta^4$ into your calculator and see if you recognise it? $\endgroup$ Jun 1, 2019 at 17:11

2 Answers 2


Maybe it would be worth mentioning that $\zeta+\zeta^2+\zeta^4$ is an element which is fixed by the generator of the Group $H_2$. It follows that $\mathbb{Q}(\zeta+\zeta^2+\zeta^4)$ is a subfield of the fixed field, and by the main theorem of galois theory you know that it must be equal to the fixed field.

In case you want to find $\zeta+\zeta^2+\zeta^4$ in terms of something which looks nicer then proceed as follows:

you know that $\zeta^6+....+\zeta+1=0$. You also know that $\mathbb{Q}(\zeta+\zeta^2+\zeta^4):\mathbb{Q}$ is a degree 2 extension. Therefore, motivated by the previous results, we square $(\zeta+\zeta^2+\zeta^4)$:


Finally if you put $\alpha=\zeta+\zeta^2+\zeta^4$, then we have $\alpha^2+\alpha=2\zeta^6+2\zeta^5+2\zeta^4+2\zeta^3+2\zeta^2+2\zeta=-2$

Therefore, $\alpha$ is a root of $x^2+x+2$, so you only need to adjoin to $\mathbb{Q}$ a root of this quadratic.


You already know that the cyclotomic field $E=\mathbf Q(\zeta_7)$ is cyclic over $\mathbf Q$, with Galois group $G\cong C_2=C_2 \times C_3$, hence $E$ contains two and only two subextensions $K$ and $L$, which are cyclic of respective degrees $3$ and $2$ over $\mathbf Q$. By Galois theory, $K$ is the fixed field of the complex conjugation contained in $G$, so $K$ must be $\mathbf Q(\zeta_7 +{\zeta_7}^{-1})=\mathbf Q(cos \frac {2\pi}7)$, as you have seen. The determination of $L$ is a general problem. For $p$ odd, the unique quadratic subextension of $\mathbf Q(\zeta_p)$ is $\mathbf Q(\sqrt {p^*})/\mathbf Q$, where $p^*=(-1)^{\frac {p-1}2}p $ (this comes from a discriminant computation and classically constitutes the first step of one of the numerous proofs of the quadratic reciprocity law). Here $L=\mathbf Q(\sqrt{-7})$.

NB. One may naturally ask where $\mathbf Q(sin \frac {2\pi}7)$ lives. If you are curious, see e.g. https://math.stackexchange.com/a/2896166/300700


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