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.

  • $\begingroup$ 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? $\endgroup$
    – Homaniac
    Oct 15 '18 at 13:12
  • $\begingroup$ 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 $\mathbb{Q}$. $\endgroup$ Oct 15 '18 at 14:02
  • $\begingroup$ 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/… $\endgroup$
    – Homaniac
    Oct 15 '18 at 15:01
  • $\begingroup$ Sorry, may I quickly check what is the easiest way to show f(x) is irreducible over Q? $\endgroup$
    – Homaniac
    Oct 15 '18 at 18:22
  • $\begingroup$ 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. $\endgroup$ Oct 15 '18 at 19:11

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