Exercise 7.3.5 (Fixed fields)

Question

Find another example of Theorem 7.3.3.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
We follow 	extbf{Example 7.3.4}. If we have $\kappa: \mathbb{Q}(\sqrt2) ightarrow \mathbb{Q}(\sqrt2)$ representing complex conjugation,
then $H=\{id, \kappa\}$ is a subgroup of $Aut(\mathbb{Q}(\sqrt2))$. By 	extbf{Theorem 7.3.3}, we have that $[\mathbb{Q}(\sqrt2): 	ext{Fix}(H)] \le |H| = 2$.
Since $	ext{Fix}(H) = \mathbb{Q}$, and we know that $[\mathbb{Q}(\sqrt2):\mathbb{Q}]=2$, the inequality holds.

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} 
\begin{itemize}
\item Consider the field $M = \Q(\sqrt{2})$. Then $(1,\sqrt{2})$ is a linear basis of $M$ over $\Q$, and $\sigma : M 	o M$ defined by $\sigma(a + b \sqrt{2}) = a- b \sqrt{2}$ for all $(a,b)\in \Q^2$ is an automorphism of $M$, such that $\sigma \circ \sigma = 1_M$.
$$H = \{1_M, \sigma\} \subset \mathrm{Aut}(M)$$
is a subgroup of $ \mathrm{Aut}(M)$ (in fact, $H = \Gal(M:\Q) = \mathrm{Aut}(M)$).
If $\alpha = a+ b \sqrt{2} \in \mathrm{Fix}(H)$, then $a + b \sqrt{2} = a- b \sqrt{2}$, thus $b= 0$ and $\alpha \in \Q$. Moreover $\Q \subset \mathrm{Fix}(H)$, thus 
$$\mathrm{Fix}(H) = \Q.$$
Here $[M : \mathrm{Fix}(H)] = [\Q(\sqrt{2}) : \Q] = 2 = |H|$.

\item For a more sophisticated example, see [Cox] Example 7.5.2 and 7.5.4 page 173, 174:
																														
Let $M = \C(t)$, where $t$ is an indeterminate. Then $t \mapsto 1/t$ induces an automorphism $\sigma$ of $M$, given by
$$\sigma(p) = \sigma\left( \sum_{k=0}^d a_k t^kight) =  \sum_{k=0}^d a_k \frac{1}{t^k}$$
 if $p =\sum_{k=0}^d a_k t^k\in \C[t]$ is a polynomial, and $\sigma(f) = \sigma(p)/\sigma(q)$, if $f = p/q \in \C(t)$.
 
 Then $\sigma \circ \sigma = 1_M$ and $H = \{1_M, \sigma\}$ is a subgroup of $\mathrm{Aut}(M)$.
 
 Here $t+t^{-1} \in \mathrm{Fix}(H)$, so 
 $$\C(t+t^{-1})\subset \mathrm{Fix}(H) \subset \C(t).$$
 
 Since $t$ is a root of $(x-t)(x-t^{-1}) = x^2 - (t+t^{-1}) x + 1 \in \C(t+t^{-1})[x]$, $\C(t)$ is obtained by adjoining $t$ to $\C(t+t^{-1})$, and
 $$[\C(t): \C(t+t^{-1})] \leq 2.$$
 
 By Digression 7.3.6 (or Cox, Theorem 7.5.3), $[\C(t) : \mathrm{Fix}(H)] = 2$. The tower theorem shows that $[\mathrm{Fix}(H) : \C(t+t^{-1})] = 1$, therefore
 $$\mathrm{Fix}(H) = \C(t+t^{-1}).$$
\end{itemize}

\end{proof}

Exercise 7.3.5 (Fixed fields)

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Exercise 7.3.5 (Fixed fields)

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