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In the Galois group of $\mathbb C(x)/\mathbb C$ we can identify the dihedral group $D_n=\langle \tau,\sigma \rangle $ where $\tau (x)=\frac{1}{x}$ and $\sigma(x)=\zeta_n x$.

Now, one can easily see that $\tau(x^n+x^{-n})=x^n+x^{-n}$ as well as $\sigma(x^n+x^{-n})=x^n+x^{-n}$, so the fixed field $\mathcal F(D_n)$ surely contains $\mathbb C(x^n+x^{-n})$. I want to show that it is also contained in the given set, so $\mathcal F(D_n)=\mathbb C(x^n+x^{-n})$.

Let's take an element $\frac{p(x)}{q(x)}\in\mathcal F(D_n)$. Since $\mathbb C$ is algebraically closed we can write $p(x)=\prod_{i=1}^n (x-\alpha_i)$ and $q(x)=\prod_{j=1}^m (x-\beta_j)$. If it is in the fixed field, we further obtain the relation $$\frac{(x-\alpha_1)\cdots(x-\alpha_n)}{(x-\beta_1)\cdots (x-\beta_m)}=\tau\left(\frac{(x-\alpha_1)\cdots(x-\alpha_n)}{(x-\beta_1)\cdots (x-\beta_m)}\right)=\frac{(x^{-1}-\alpha_1)\cdots(x^{-1}-\alpha_n)}{(x^{-1}-\beta_1)\cdots (x^{-1}-\beta_m)}$$

and we can do the same for $\sigma$: $$\frac{(x-\alpha_1)\cdots(x-\alpha_n)}{(x-\beta_1)\cdots (x-\beta_m)}=\sigma\left(\frac{(x-\alpha_1)\cdots(x-\alpha_n)}{(x-\beta_1)\cdots (x-\beta_m)}\right)=\frac{(\zeta_n x-\alpha_1)\cdots(\zeta_nx-\alpha_n)}{(\zeta_nx-\beta_1)\cdots (\zeta_nx-\beta_m)}$$

Does this help in any way? How can I show that $\frac{p(x)}{q(x)}$ looks like $x^n+x^{-n}$ for some $n\in\mathbb Z$?

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Since $\mathcal{F}(D_n)$ is defined as the fixed subfield of $\mathbb{C}(x)$ by a finite group of automorphism, it is a Galois extension, with Galois group isomorphic to $D_n$ and degree $[\mathbb{C}(x):\mathcal{F}(D_n)]=|D_n|=2n$.

Let $y=x^n+x^{-n}$. Then $x^ny=x^{2n}+1$, so $\mathbb{C}(x)$ is a finite extension of $\mathbb{C}(y)$, with $[\mathbb{C}(x):\mathbb{C}(y)]\leq 2n$. But then considering the tower of extension $\mathbb{C}(y)\leq \mathcal{F}(D_n)\leq \mathbb{C}(x)$, one gets $$2n\geq [\mathbb{C}(x):\mathbb{C}(y)]=[\mathbb{C}(x):\mathcal{F}(D_n)]\cdot [\mathcal{F}(D_n):\mathbb{C}(y)]= 2n [\mathcal{F}(D_n):\mathbb{C}(y)],$$ so that $[\mathcal{F}(D_n):\mathbb{C}(y)]=1$ and thus $\mathcal{F}(D_n)=\mathbb{C}(y)$.

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  • $\begingroup$ Thank you for your answer. Can you explain your last $\geq$? Is this really necessary or can we just write $=$? $\endgroup$ – Buh Jun 12 '18 at 12:25
  • $\begingroup$ @Buh Yes, sorry, it's an equality. I don't know why I only put a $\geq$, I've edited it now. $\endgroup$ – Arnaud D. Jun 12 '18 at 12:37
  • $\begingroup$ Actually, do you mind adding a little more explanation as to why $\text{Gal}((\mathbb C(x)/\mathcal F(D_n))\cong D_n$? Maybe it's trivial, but I don't think I understand it entirely. $\endgroup$ – Buh Jun 12 '18 at 17:47
  • $\begingroup$ @Buh It's a general result : for any field $K$ and finite subgroup $G$ of $Aut(K)$, the subfield $K^G$ of elements fixed by $G$ is such that $K/K^G$ is a Galois extension and $Gal(K/K^G)\simeq G$. $\endgroup$ – Arnaud D. Jun 12 '18 at 20:16
  • $\begingroup$ It's a classic theorem of characterization of Galois extensions, so you should be able to find it in most books or lecture notes about Galois theory. $\endgroup$ – Arnaud D. Jun 12 '18 at 21:06

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