# Show $Gal(L|Q)\cong (Z/7Z)^*$ and determine $L_H$

Let $$ζ_7 = e^{2πi/7}$$ and consider the extension $$L = Q(ζ_7)$$ of $$Q$$. I want to prove $$Gal(L|Q)\cong (Z/7Z)^*$$ and also given $$H$$ is a subgroup of $$G$$ with $$|H| = 2$$, I want to determine $$L_H$$ . The answer I think is $$Q(ζ_7 + ζ_7^{−1})$$ but I am not sure how to show. To start off, I know that $$L$$ is the splitting field of $$f(x) = x^6 + x^5 + x^4 + x^3 + x^2 + x + 1$$ over $$Q$$ and that $$f(x) = min_Q(ζ_7)$$ but then how to proceed~

Note that the six roots of $$f(x)$$ are $$\zeta_{7},\zeta_{7}^{2},\zeta_{7}^{3},\zeta_{7}^{4},\zeta_{7}^{5}$$ and $$\zeta_{7}^{6}$$. Let $$\sigma\in G=\operatorname{Gal}(L/\mathbb{Q})$$ be given. Then $$\sigma(\zeta_{7})^{7}=\sigma(\zeta_{7}^{7})=\sigma(1)=1,$$ but $$\sigma(\zeta_{7})^{a}\neq 1$$ for $$1\leq a\leq 6$$, so $$\sigma(\zeta_{7})$$ is a 7th root of unity, i.e. there exists a unique $$1\leq\alpha\leq 6$$ such that $$\sigma(\zeta_{7})=\zeta_{7}^{\alpha}$$.

From this we conclude that the elements of $$G$$ are $$\sigma_{i}$$ for $$1\leq i\leq 6$$ given by $$\sigma_{i}(\zeta_{7})=\zeta_{7}^{i}$$ and we note that $$G$$ is cyclic and generated by $$\sigma_{3}$$ (check this). Define the map $$\varphi: G\rightarrow (\mathbb{Z}/7\mathbb{Z})^{\times}$$ by $$\varphi(\sigma_{i})=i\mod{7}.$$ Then $$\sigma_{i}\circ\sigma_{j}(\zeta_{7})=\sigma_{i}(\zeta_{7}^{i})=\zeta_{7}^{ij},$$ which proves that $$\varphi$$ is a group homomorphism. If $$\sigma_{j}\in\ker(\varphi)$$ then $$\sigma_{j}\equiv 1\mod{7}$$, which implies that $$\sigma_{j}=7k+1$$ for some $$k\in\mathbb{N}$$, whence $$\sigma_{j}(\zeta_{j})=\zeta_{7}^{7k+1}=\zeta_{7}$$. Thus $$\sigma_{j}$$ is the identity because it fixes $$\mathbb{Q}$$ and $$\zeta_{7}$$, whence $$\varphi$$ is injective. Because both groups have the same cardinality the homomorphism is infact an isomorphism, i.e. $$\operatorname{Gal}(L/\mathbb{Q})\simeq(\mathbb{Z}/7\mathbb{Z})^{\times}$$. Now, the subgroup $$H$$ with $$|H|=2$$ is given by $$\lbrace\sigma_{1},\sigma_{6}\rbrace$$ so noting that $$\sigma_{6}(\zeta_{7}+\zeta_{7}^{-1})=\sigma_{6}(\zeta_{7}+\zeta_{7}^{6})=\zeta_{7}^{6}+\zeta_{7}^{36}=\zeta_{7}+\zeta_{7}^{6}=\zeta_{7}+\zeta_{7}^{-1},$$ implies that $$\mathbb{Q}(\zeta_{7}+\zeta_{7}^{-1})\subseteq L^{H}$$. By the fundamental theorem of Galois theory we see that $$[L^{H}:\mathbb{Q}]=[G:H]=3,$$ which implies that either $$\mathbb{Q}(\zeta_{7}+\zeta_{7}^{-1})=\mathbb{Q}$$ or $$\mathbb{Q}(\zeta_{7}+\zeta_{7}^{-1})=L^{H}$$. If the former is true, then $$\sigma_{2}(\zeta_{7}+\zeta_{7}^{6})=\zeta_{7}+\zeta_{7}^{6},$$ but we see that $$\sigma_{2}(\zeta_{7}+\zeta_{7}^{6})=\zeta_{7}^{2}+\zeta_{7}^{5}\neq\zeta_{7}+\zeta_{7}^{6},$$ whence we conclude that $$\mathbb{Q}(\zeta_{7}+\zeta_{7}^{-1})=L^{H}$$, as desired.

• Thank you for the very clear answer. I just wanted to check the prev part: to show f(x) is the minimal polynomial, does it suffice to show it is irreducible by Eisenstein criterion? Oct 15 '18 at 13:12
• To show that it is the minimal polynomial of $\zeta_{7}$ you have to show two things: 1) $\zeta_{7}$ is a root 2) It is irreducible over $\mathbb{Q}$. Oct 15 '18 at 14:02
• Ah I understand it fully now, thanks so much! Just having another simple question, if you could help as well :) math.stackexchange.com/questions/2956624/… Oct 15 '18 at 15:01
• Sorry, may I quickly check what is the easiest way to show f(x) is irreducible over Q? Oct 15 '18 at 18:22
• You can apply the Eisenstein criterion with $p=7$ to show that $g(x)=f(x+1)$ is irreducible, which implies that $f(x)$ is irreducible. That's the standard method of proving that the cyclotomic polynomials are irreducible. Oct 15 '18 at 19:11