Exercise 7.3.6

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

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}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

This exercise will work out the Galois correspondance for the splitting field of $x^4 - 4x^2 + 2$ over $\Q$. In Exercise 6 of section 5.1 you showed that $L = \Q(\sqrt{2+\sqrt{2}})$ and that $\Gal(L/\Q) \simeq \Z/4\Z$. Now, similar to Example 7.3.4, determine all subgroups of $\Gal(L/\Q)$ and the corresponding intermediate fields of $\Q \subset L$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

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}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
Let $L$ be the splitting field of  $x^4 - 4x^2 + 2$ over $\Q$.
We proved in Ex. 5.1.6 and Ex. 6.3.4 that $L = \Q(\alpha, \beta) = \Q(\alpha)$, where
$$\alpha = \sqrt{2 + \sqrt{2}}, \beta = \sqrt{2 - \sqrt{2}},$$
and
$$\Gal(L/\Q) = \langle \sigma angle \simeq \Z/4\Z,$$
where $\sigma$ is the only $\Q$-automorphisme of $L$ such that $\sigma(\alpha) = \beta$ (and then $\sigma(\beta) = -\alpha$).

Since $\Gal(L/\Q) = \langle \sigma angle$ is cyclic of order 4, as in Ex. 7.3.5, the only non trivial subgroup $H$ is of order 2, and $H = \langle \sigma^2angle$.

By the Fundamental Theorem of Galois Theory, there exists thus a unique intermediate field distinct of $L$ and $\Q$, which is thus $\Q(\sqrt{2})$ :
$$L_H = L_{\langle \sigma^2 angle} = \Q(\sqrt{2}).$$
The correspondance is between the two chains:
$$
\begin{array}{lll}
\Q &\subset \Q(\sqrt{2}) &\subset \Q(\sqrt{2+ \sqrt{2}}) = L,\
G = \langle\sigmaangle &\supset \langle \sigma^2 angle &\supset \{e\}.
\end{array}
$$

\end{proof}

Exercise 7.3.6

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Exercise 7.3.6

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