HW13

Consider the extension $\mathbb{Q}\subset F=\mathbb{Q}[w]$ where $w^3=1$ is a primitive cubic root of 1. Compute the group ${\mathrm{Aut}}_{\mathrm{Q}}F$. Prove that $F/\mathbb{Q}$ is a Galois extension.

\begin{proof} Define $\tau =e^{2\pi i/3}$, then $F=\mathbb{Q}[\tau ]$ and $F$ is a $2$-dimensional vector space over $\mathbb{Q}$ with a basis $\{1,\tau \}$. For any $\phi \in {\mathrm{Aut}}_{\mathbb{Q}}F$, $\phi (\tau {)}^3=\phi ({\tau }^3)=1$ and so $\phi (\tau )\in \{1,\tau ,{\tau }^2\}$. Since $\phi$ is an automorphism, $\phi (\tau )\ne 1$. Define $\varphi :F\to F,\tau \mapsto {\tau }^2$ such that $\varphi {|}_{\mathbb{Q}}=\mathrm{id}$, then ${\mathrm{Aut}}_{\mathbb{Q}}F=⟨\varphi ⟩\simeq \mathbb{Z}/2\mathbb{Z}$. Note that $\tau \notin F^{{\mathrm{Aut}}_{\mathbb{Q}}F}$ as $\varphi (\tau )\ne \tau$ and $\varphi {|}_{\mathbb{Q}}=\mathrm{id}$, then $F^{{\mathrm{Aut}}_{\mathbb{Q}}F}=\mathbb{Q}$. Therefore, $F/\mathbb{Q}$ is a Galois extension. \end{proof}

Consider the extension $\mathbb{Q}\subset F=\mathbb{Q}[\zeta ]$ where ${\zeta }^5=1$ is a primitive $5$-th root of 1 . Show that the group ${\mathrm{Aut}}_{Q}F\simeq \mathbb{Z}/4\mathbb{Z}$. Prove that $F/\mathbb{Q}$ is a Galois extension.

\begin{proof} Take $\phi \in {\mathrm{Aut}}_{\mathbb{Q}}F$, then $\phi (\zeta )$ is also a primitive $5$-th root of $1$. Define $\tau =e^{2\pi i/5}$, then $\phi (\tau )\in \{\tau ,{\tau }^2,{\tau }^3,{\tau }^4\}$ and so ${\mathrm{Aut}}_{\mathbb{Q}}F=⟨\varphi ⟩$ where $\varphi :F\to F,\tau \to {\tau }^2$ with $\tau {|}_{\mathbb{Q}}=\mathrm{id}$. Note that $|\varphi |=4$, then ${\mathrm{Aut}}_{\mathbb{Q}}F=⟨\varphi ⟩\simeq \mathbb{Z}/4\mathbb{Z}$. Also note that $[F:\mathbb{Q}]=4$ and $F$ is a vector space over $\mathbb{Q}$ with basis $\{1,\tau ,{\tau }^2,{\tau }^3\}$. For any $v:=a+b\tau +c{\tau }^2+d{\tau }^3\in F^{{\mathrm{Aut}}_{\mathbb{Q}}F}$ with $a,b,c,d\in \mathbb{Q}$, we have $b=c=d=0$ because $v^{\varphi }=v$. Therefore, $F^{{\mathrm{Aut}}_{\mathbb{Q}}F}=\mathbb{Q}$ and $F/\mathbb{Q}$ is a Galois extension. \end{proof}

(20pts) Let $F/K$ be an extension of degree $2$.

• (1) If char $K=0$, then $F/K$ is a Galois extension.
• (2) Suppose char $K=2$. Give an example where $F/K$ is NOT Galois.
• (3) (10pts)(might be challenging) Suppose char $K=p\ne 2$. Then $F/K$ is a Galois extension.

\begin{proof} (1) Take $e\in F\setminus K$, then $f:=\mathrm{Irr}(e,K)$ is degree $2$. Thus, there exists $g\in F[x]$ such that $f(x)=(x-e)g(x)$. Since $\mathrm{deg}g=1$, another root $e^{\prime }$ of $f$ is also contained in $F$. Since $\mathrm{char}K=0$, $e\ne e^{\prime }$. Define $\phi :F\to F,e\mapsto e^{\prime }$ with $\phi {|}_{\mathbb{Q}}=\mathrm{id}$. We can verify that $\phi$ is a field automorphism, then ${\mathrm{Aut}}_{K}F\ne \{\mathrm{id}\}$. Since $|{\mathrm{Aut}}_{K}F|⩽[F:K]=2$, there is ${\mathrm{Aut}}_{K}F\simeq {\mathbb{Z}}_2$. Since $F$ is a vector space over $K$ with a basis $\{1,e\}$ and $\phi (e)\ne e$, $F^{{\mathrm{Aut}}_{K}F}=K$. That is, $F/K$ is a Galois extension.

(2) Let $K={\mathbb{F}}_2(t)$ and $F={\mathbb{F}}_2(\sqrt{t})$. Then $\mathrm{Irr}(\sqrt{t},K)=x^2-t$ and so $[F:K]=2$. Since $\sqrt{t}$ is the only root of $x^2-t$, ${\mathrm{Aut}}_{K}F=\{\mathrm{id}\}$ and then $F^{{\mathrm{Aut}}_{K}F}=F$. Thus, $F/K$ is not Galois.

(3) Similarly as (1), take $e\in F\setminus K$ and then $f=\mathrm{Irr}(e,K)$ has two roots $e,e^{\prime }\in K$. Since $\mathrm{deg}f=2$ and $\mathrm{char}K\ne 2$, $f^{\prime }=2x+(e+e^{\prime })\ne 0$ and then $f^{\prime }\ne 0$. Thus $f$ is separable and so $e\ne e^{\prime }$. Define $\phi :F\to F,e\mapsto e^{\prime }$ with $\phi {|}_{\mathbb{Q}}=\mathrm{id}$ and we still can verify $\phi \in {\mathrm{Aut}}_{K}F\ne \{\mathrm{id}\}$. It deduces that ${\mathrm{Aut}}_{K}F\simeq {\mathbb{Z}}_2$ and $F/K$ is Galois. \end{proof}

${\mathrm{Aut}}_{\mathbb{Q}}(\mathbb{R})$ is the identity group. (Hint: Since every positive element of $\mathbb{R}$ is a square, it follows that an automorphism of $\mathbb{R}$ sends positives to positives and hence that it preserves the order in $\mathbb{R}$. Trap a given real number between suitable rational numbers.)

\begin{proof} For any $\phi \in {\mathrm{Aut}}_{\mathbb{Q}}(\mathbb{R})$, there is $\phi {|}_{\mathbb{Q}}=\mathrm{id}$. For any $r>0$, $r=m^2$ for some $m\in \mathbb{R}$ and $\phi (r)=\phi (m^2)=(\phi (m){)}^2>0$. For any $r^{\prime }\in \mathbb{R}\setminus \mathbb{Q}$, there exists $\{p_n/q_n\}$ and $\{s_n/t_n\}$ with $p_n,q_n,s_n,t_n\in \mathbb{Q}$ and $p_n/q_n\uparrow r^{\prime }$ and $s_n/t_n\downarrow r^{\prime }$. Thus $p_n/q_n=\phi (p_n/q_n)⩽\phi (r^{\prime })⩽\phi (s_n/t_n)=s_n/t_n$ by $p_n/q_n⩽r^{\prime }⩽s_n/t_n$. By Sandwich Theorem, $\phi (r^{\prime })=r^{\prime }$. Therefore, $\phi =\mathrm{id}$ for all $\phi \in {\mathrm{Aut}}_{\mathbb{Q}}(\mathbb{R})$ and so ${\mathrm{Aut}}_{\mathbb{Q}}(\mathbb{R})=\{\mathrm{id}\}$. \end{proof}