Let $\zeta=\zeta_{11}$.Show that the only subfield of $\Bbb Q(\zeta)$ with degree $2$ over $\Bbb Q$ is $\Bbb Q(a)$, with $a=\zeta+\zeta^3+\zeta^4+\zeta^5+\zeta^9$. We khow that $|G=Gal(\Bbb Q(\zeta):\Bbb Q)|=10$, and $G=\langle \sigma \rangle $, which $\sigma(\zeta)=\zeta^2$ so G has unique subgroup with order 5, $\langle \sigma^2 \rangle$.We want to show that $\Bbb Q(a)=Fix(\langle \sigma^2 \rangle)$.It is easy to show that $\Bbb Q(a)\subseteq Fix(\langle \sigma^2 \rangle)$ because $$a=\zeta+\sigma^2(\zeta)+\sigma^4(\zeta)+\sigma^6(\zeta)+\sigma^8(\zeta)\in Fix(\langle \sigma^2\rangle)$$ We khow that $[Fix((\langle \sigma^2 \rangle):\Bbb Q]=[G:\langle \sigma^2 \rangle]=2$ and if we show so that $[\Bbb Q(a):\Bbb Q]=2$ we have that $\Bbb Q(a)=Fix(\langle \sigma^2 \rangle)$, but how can i show that $[\Bbb Q(a):\Bbb Q]=2$. My first idea was to show that $a\notin \Bbb Q$ but we don't khow if $a$ is algebraic of $\Bbb Q$ and if $a$ is algebraic how can we show that $a\notin \Bbb Q$?

  • $\begingroup$ It follows from the fact that $a$ has exactly two conjugates (namely $a$ itself and $\sigma(a)=\zeta^2+\zeta^6+\zeta^7+\zeta^8+\zeta^{10})$ that $a$ is of degree 2 over $\mathbf{Q}$. $\endgroup$ – rae306 Jun 16 '20 at 18:33
  • $\begingroup$ Sure you're right i ''ll change it. $\endgroup$ – KBi7700 Jun 16 '20 at 18:53
  • $\begingroup$ Good. For extra credit, you can try and show that $\Bbb{Q}(a)=\Bbb{Q}(\sqrt{-11})$. Adapt the method from here :to find a quadratic equation satisfied by $a$ (and $\sigma(a)$). $\endgroup$ – Jyrki Lahtonen Jun 16 '20 at 19:53
  • $\begingroup$ Thank you very much it was very useful method. $\endgroup$ – KBi7700 Jun 16 '20 at 22:10

What you want to show is that $Fix(\langle \sigma^{2} \rangle) \subset \mathbb{Q}(a)$.

Let's find with some calculus (which results to be a general approach to this type of problems with small cyclotomic extensions) the fixed field of the automorphism $\phi := \sigma^{2} : \zeta \longmapsto \zeta^{4}$.

A basis of the splitting field over $\mathbb{Q}$ is $\left\lbrace 1,\zeta,\cdots, \zeta^{9} \right\rbrace$ with the known relation $1+\zeta+\cdots +\zeta^{10} = 0$

If $\alpha = a_{0} + a_{1}\zeta + \cdots a_{9}\zeta^{9}, \hspace{0.3cm} a_{i} \in \mathbb{Q}$


$$\phi(\alpha) = \sum\limits_{i=0}^{9}\phi(a_{i}\zeta^{i}) = \sum\limits_{i=0}^{9}a_{i}\sigma(\sigma(\zeta^{i}))= \sum\limits_{i=0}^{9}a_{i} \sigma(\zeta^{2i}) = \sum\limits_{i=0}^{9}a_{i} \zeta^{4i}$$

Which results to be

$$\phi(\alpha) = a_{0} + a_{1}\zeta^{4} + a_{2}\zeta^{8} + a_{3}\zeta + a_{4}\zeta^{5}+a_{5}\zeta^{9}+a_{6}\zeta^{2}+a_{7}\zeta^{6}+a_{8}\zeta^{10}+a_{9}\zeta^{3}$$

Since $$\alpha = a_{0} + \cdots + a_{9}\zeta^{9}$$

We have that $\alpha$ lies in the fixed subfield if $\alpha = \phi(\alpha)$ if and only if $a_{1} = a_{3} = a_{4} = a_{5} = a_{9},a_{2} = a_{6} = a_{7} = a_{8} = 0$

(This follows if justified from the fact a linear combination of a basis representing the null vector must $0$ to all the coefficients)

Which translates into $K^{\sigma^{2}} = \left\lbrace a + b(\zeta+\zeta^{3}+\zeta^{4}+\zeta^{5}+\zeta^{9}): a,b \in \mathbb{Q}\right\rbrace$, which is indeed what we were looking for.


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