HW14

Let $F=\mathbb{Q}[\sqrt[3]{3},\sqrt[4]{2}]$.

• (a) Show that for any element $\sigma \in {\mathrm{Aut}}_{\mathbb{Q}}F$, we have $\sigma (\sqrt[3]{3})=\sqrt[3]{3}$.
• (b) Find all the elements of ${\mathrm{Aut}}_{\mathbb{Q}}F$.
• (c) Show that $F/\mathbb{Q}$ is not a Galois extension. Find some $E/F$ some finite extension such that $E/\mathbb{Q}$ is a Galois extension.
• (d) Find the minimal possible $E$ (called the Galois closure of $F$ ) in Item (c) above. (give reasons why it is the minimal one).

\begin{proof} a) Since $\mathrm{Irr}(\sqrt[3]{3},F)=x^3-3$ has three roots $S:=\{\sqrt[3]{3},\sqrt[3]{3}{e}^{2\pi i/3},\sqrt[3]{3}{e}^{4\pi i/3}\}$ in $\mathbb{C}$, we have $\sigma (\sqrt[3]{3})\in S\cap F=\{\sqrt[3]{3}\}$ for any $\sigma \in {\mathrm{Aut}}_{\mathbb{Q}}(F)$. Thus, $\sigma (\sqrt[3]{3})=\sqrt[3]{3}$ for all $\sigma \in {\mathrm{Aut}}_{\mathbb{Q}}F$.

b) Similarly, since $\mathrm{Irr}(\sqrt[4]{2},F)=x^4-2$ has only two roots $\{\sqrt[4]{2},-\sqrt[4]{2}\}$ in $F$, we have $\sigma (\sqrt[4]{2})=\pm \sqrt[4]{2}$ for any $\sigma \in {\mathrm{Aut}}_{\mathbb{Q}}F$. Therefore, ${\mathrm{Aut}}_{\mathbb{Q}}F=⟨{\sigma }_0⟩\simeq {\mathbb{Z}}_2$ where ${\sigma }_0{|}_{\mathbb{Q}}=\mathrm{id}$, ${\sigma }_0(\sqrt[3]{3})=\sqrt[3]{3}$ and ${\sigma }_0(\sqrt[4]{2})=-\sqrt[4]{2}$.

c) Assume that $F/\mathbb{Q}$ is a Galois extension, then $[F:\mathbb{Q}]=|{\mathrm{Aut}}_{\mathbb{Q}}(F)|$ by $[F:\mathbb{Q}]<\infty$. However, $[F:\mathbb{Q}]=[F:\mathbb{Q}(\sqrt[4]{2})][\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]=12\ne 2$, contradiction. Therefore, $F/\mathbb{Q}$ is not Galois. Define $E=\mathbb{Q}[\sqrt[3]{3},e^{2\pi i/3},\sqrt[4]{2},e^{2\pi i/4}]$, then $E/\mathbb{Q}$ is normal and separable because it is a splitting field of $f(x)=(x^3-3)(x^4-2)$. Thus $E/\mathbb{Q}$ is a Galois extension.

d) We claim that $E$ defined in c) is the minimal one. Since $E/\mathbb{Q}$ is a Galois extension, $E/\mathbb{Q}$ is normal and separable. Since $E\supseteq F$, $E$ contains $\sqrt[3]{3}$ and $\sqrt[4]{2}$. Hence $E\supseteq \mathbb{Q}[\sqrt[3]{3},e^{2\pi i/3},\sqrt[4]{2},e^{2\pi i/4}]:=E_0$. On the other hand, in c) we have proved that $E_0$ is Galois, and so $E_0=\mathbb{Q}[\sqrt[3]{3},e^{2\pi i/3},\sqrt[4]{2},e^{2\pi i/4}]$ is what we desired. \end{proof}

Let $F$ be a finite dimensional Galois extension of $K$ and let $L$ and $M$ be two intermediate fields.

• (a) ${\mathrm{Aut}}_{LM}F={\mathrm{Aut}}_{L}F\cap {\mathrm{Aut}}_{M}F$;
• (b) ${\mathrm{Aut}}_{L\cap M}F={\mathrm{Aut}}_{L}F\vee {\mathrm{Aut}}_{M}F$;
• (c) What conclusion can be drawn if ${\mathrm{Aut}}_{L}F\cap {\mathrm{Aut}}_{M}F=1$ ?

The notation in RHS of (b) means the subgroup generated by the two groups.

\begin{proof} Since $F/K$ is Galois, $F/L$ and $F/M$ are also Galois.

a) For any $\sigma \in {\mathrm{Aut}}_{L}F\cap {\mathrm{Aut}}_{M}F$, $\sigma {|}_L=\mathrm{id}$ and $\sigma {|}_M=\mathrm{id}$ yield that $\sigma {|}_{LM}=\mathrm{id}$. Thus ${\mathrm{Aut}}_{LM}F\supseteq {\mathrm{Aut}}_{L}F\cap {\mathrm{Aut}}_{M}F$. For any $\sigma \in {\mathrm{Aut}}_{LM}F$, $\sigma {|}_{LM}=\mathrm{id}$ yields that $\sigma {|}_L=\mathrm{id}$ and $\sigma {|}_M=\mathrm{id}$. Thus $\sigma \in {\mathrm{Aut}}_{L}F\cap {\mathrm{Aut}}_{M}F$ and so ${\mathrm{Aut}}_{LM}F\subseteq {\mathrm{Aut}}_{L}F\cap {\mathrm{Aut}}_{M}F$. Therefore, we have ${\mathrm{Aut}}_{LM}F={\mathrm{Aut}}_{L}F\cap {\mathrm{Aut}}_{M}F$.

b) For any $\sigma \in {\mathrm{Aut}}_{L}F\vee {\mathrm{Aut}}_{M}F$, it is easy to check that $\sigma {|}_{L\cap M}=\mathrm{id}$ and so ${\mathrm{Aut}}_{L\cap M}F\supseteq {\mathrm{Aut}}_{L}F\vee {\mathrm{Aut}}_{M}F$. On the other hand, note that $F^{{\mathrm{Aut}}_{L}F\vee {\mathrm{Aut}}_{M}F}⩽F^{{\mathrm{Aut}}_{L}F}=L$ and $F^{{\mathrm{Aut}}_{L}F\vee {\mathrm{Aut}}_{M}F}⩽F^{{\mathrm{Aut}}_{M}F}=M$, then $F^{{\mathrm{Aut}}_{L}F\vee {\mathrm{Aut}}_{M}F}⩽M\cap L=F^{{\mathrm{Aut}}_{L\cap M}F}$. It deduces that ${\mathrm{Aut}}_{L\cap M}F\subseteq {\mathrm{Aut}}_{L}F\vee {\mathrm{Aut}}_{M}F$. Therefore, ${\mathrm{Aut}}_{L\cap M}F={\mathrm{Aut}}_{L}F\vee {\mathrm{Aut}}_{M}F$.

c) If ${\mathrm{Aut}}_{L}F\cap {\mathrm{Aut}}_{M}F=1$, then ${\mathrm{Aut}}_{LM}F=\{1\}$. Furthermore, $F=LM$ as $F/LM$ is Galois. \end{proof}

If $F$ is a finite dimensional Galois extension of $K$ and $E$ is an intermediate field, then there is a unique smallest field $L$ such that $E\subset L\subset F$ and $L$ is Galois over $K$; furthermore

$${\mathrm{Aut}}_{L}F=\bigcap _{\sigma }\sigma \left({\mathrm{Aut}}_{E}F\right){\sigma }^{-1},$$

where $\sigma$ runs over ${\mathrm{Aut}}_{K}F$.

\begin{proof} Define $L^{\prime }={\cap }_{\sigma \in {\mathrm{Aut}}_{K}F}\sigma ({\mathrm{Aut}}_{E}F){\sigma }^{-1}$, then $L^{\prime }$ is a subset of ${\mathrm{Aut}}_{K}F$. Note that $\mathrm{id}\in L^{\prime }$, and $g_1{g}_2,g_1^{-1}\in L^{\prime }$ for any $g_1,g_2\in L^{\prime }$. Hence $L^{\prime }$ is a subgroup of ${\mathrm{Aut}}_{K}F$. For any $\sigma \in {\mathrm{Aut}}_{K}F$, $\sigma L^{\prime }{\sigma }^{-1}=L^{\prime }$ yields that $L^{\prime }⊲{\mathrm{Aut}}_{K}F$. By the fundamental theorem of Galois theorem, $L:=F^{L^{\prime }}$ is a field extension of $K$ and $L/K$ is Galois. It deduces that $L^{\prime }={\mathrm{Aut}}_{L}F$. Since ${\mathrm{Aut}}_{L}F⩽{\mathrm{Aut}}_{E}F$, we have $L\supseteq E$.

It remains to show $L$ is unique. Otherwise, there exist Galois extensions $L_1\ne L_2$ such that ${\mathrm{Aut}}_{L_1}F={\mathrm{Aut}}_{L_2}F$, which is impossible by the fundamental theorem of Galois theorem. \end{proof}

Remark. I forget to show $L$ is the smallest one.