Exercise 8.3.4 (Fixed field)

Question

I took a small liberty in the sentence beginning
‘The same argument’, because it included an inequality but the
previous argument didn’t. Prove the statement made in that sentence.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
The statement is considering $\phi$ of order 2 only.
If $[\mathbb{Q}(\xi,i):\mathbb{Q}(\alpha)] = 2$, then we are in agreement with the previous argument.
The only other case is then that $[\mathbb{Q}(\xi,i):\mathbb{Q}(\alpha)] = 1$, in which case $\mathbb{Q}(\alpha) = \mathbb{Q}(\xi, i)$,
and $\phi = id$ which is of order 1, and hence, $Fix\langle \phi angle = \mathbb{Q}(\xi, i)$, by the fundamental theorem.

Richard Ganaye · Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

\begin{proof} 
Let $\varphi \in G$ be an element of order $2$, and $\alpha \in \Q(\xi,i)$ such that $[\Q(\xi,i) : \Q(\alpha)] \leq 2$ and $\varphi(\alpha) = \alpha$.

Then $\alpha \in \mathrm{Fix}(\varphi)$, thus $$\Q(\alpha) \subset \mathrm{Fix}(\varphi) \subset \Q(\xi,i).$$
By the fundamental theorem, $[\Q(\xi,i) : \mathrm{Fix}(\varphi)] = |\langle \varphi angle| = 2$.

If $d = [ \mathrm{Fix}(\varphi)  : \Q(\alpha)]$, the tower theorem shows that
$$2 d = [\Q(\xi,i) : \mathrm{Fix}(\varphi)] \,  [ \mathrm{Fix}(\varphi)  : \Q(\alpha)] \leq 2,$$
therefore $1 \leq d \leq 1$, so $d = [ \mathrm{Fix}(\varphi)  : \Q(\alpha)] = 1$. This shows that 
$$ \mathrm{Fix}(\varphi)  = \Q(\alpha).$$
\end{proof}

Exercise 8.3.4 (Fixed field)

Answers

Comments

Comments

Exercise 8.3.4 (Fixed field)

Answers

Comments

Comments

Add answer