# Find all the middle fields of the extension $\mathbb{Q}/\mathbb{Q}(\exp({2\pi i\over 7}))$

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 $$x$$.

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

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 $$(\zeta+\zeta^2+\zeta^4)$$:

$$(\zeta+\zeta^2+\zeta^4)^2=\zeta^8+2\zeta^6+2\zeta^5+\zeta^4+2\zeta^3+\zeta^2=2\zeta^6+2\zeta^5+\zeta^4+2\zeta^3+\zeta^2+\zeta$$

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