# Show that $[\Bbb Q(a):\Bbb Q]=2$

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 $$a\notin \Bbb Q$$?

• 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 $\mathbf{Q}$. – rae306 Jun 16 '20 at 18:33
• Sure you're right i ''ll change it. – KBi7700 Jun 16 '20 at 18:53
• 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 $\sigma(a)$). – Jyrki Lahtonen Jun 16 '20 at 19:53
• Thank you very much it was very useful method. – 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 $$\alpha = a_{0} + a_{1}\zeta + \cdots a_{9}\zeta^{9}, \hspace{0.3cm} a_{i} \in \mathbb{Q}$$

Then

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

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.