I am trying to understand some concepts via random exercises I found from past papers but this particular one, I am not sure even where to start.

There aren't any solutions for the paper so would someone be kind enough to provide me with a standard solution, so I can study the flow and try figuring out the ideas myself? I might comment and ask if simply I cannot digest a part of it

Let $\omega = e^{\frac{2 \pi i}{5}}$. So then the $\mathbb{Q}(\omega)$ consists of all elements of the form $p+q \omega+r\omega^2+s\omega^3+t\omega^4 $. Let the $\mathbb{Q}$-automorphisms be

$$\alpha_1: p+q \omega+r\omega^2+s\omega^3+t\omega^4 \rightarrow p+q \omega+r\omega^2+s\omega^3+t\omega^4 $$

$$\alpha_2: p+q \omega+r\omega^2+s\omega^3+t\omega^4 \rightarrow p+s \omega+q\omega^2+t\omega^3+r\omega^4 $$

$$\alpha_3: p+q \omega+r\omega^2+s\omega^3+t\omega^4 \rightarrow p+r \omega+t\omega^2+q\omega^3+s\omega^4 $$

$$\alpha_4: p+q \omega+r\omega^2+s\omega^3+t\omega^4 \rightarrow p+t \omega+s\omega^2+r\omega^3+q\omega^4 $$

Take the subgroup $\{\alpha_1, \alpha_4\}$ and find its fixed field.

So I am starting with the Galois group $\{\alpha_1, \alpha_2, \alpha_3, \alpha_4\}$, and considering a particular subgroup. I don't have a method to proceed, so it would be very much appreciated if I can find a way to answer it, thank you in advance!

  • $\begingroup$ Actually, $\Bbb Q(\omega)$ consists of all the elements $p+q\omega+r\omega^2+s\omega^3$. Including the $t\omega^4$ term is not exactly wrong, but it is superfluous. $\endgroup$
    – Lubin
    Feb 13, 2016 at 2:00
  • $\begingroup$ For a different point of view you may observe that $\alpha_4$ is the usual complex conjugation. $\endgroup$ Feb 13, 2016 at 14:34

2 Answers 2


Here’s the method I use for this very special case, and other similar ones. You have $\omega^4+\omega^3+\omega^2+\omega+1=0$, and you can rewrite this as $$ 0=\omega^2+\omega+1+\omega^{-1}+\omega^{-2}\,. $$ Now, your nontrivial automorphism interchanges $\omega$ and $\omega^{-1}$, so has to have $\Bbb Q(\omega+\omega^{-1},\omega\omega^{-1})$ as its fixed field. Of course the second quantity is automatically fixed, so let’s see what kind of equation $\xi=\omega+\omega^{-1}$ might satisfy. \begin{align} \xi^2&=\omega^2+2+\omega^{-2}\\ &=-\omega+1-\omega^{-1}\qquad\text{(subtracting zero)}\\ &=-\xi+1\,, \end{align} which tells you your equation.


Let $E/F$ be a finite Galois extension fields of characteristic zero. Let $n$ be the degree of the extension, $G$ the Galois group. Suppose that $\{x_1,\ldots,x_n\}$ is basis for $E$ over $F$. Let $H$ be a subgroup of $G$, and consider the elements $y_i = \sum_{h\in H} h\cdot x_i$. Show that $\{y_1,\ldots,y_n\}$ span the fixed field of $H$.


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