Exercise 7.3.9

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}}

Let $F$ be a field of characteristic different from 2, and let $F\subset L$ be a finite extension. Prove that the following are equivalent:
\be
\item[(a)] $L$ is a Galois extension of $F$ with $\Gal(L/F)\simeq \Z/2\Z	imes \Z/2\Z$.
\item[(b)] $L$ is the splitting field of a polynomial of the form $(x^2-a)(x^2-b)$, where $a,b \in F$ but $\sqrt{a},\sqrt{b},\sqrt{ab}$ do not lie in $F$.
\ee

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} $\bullet$ Suppose (b): $L$ is the splitting field of $f=(x^2-a)(x^2-b)$, where $a,b \in F$, but $\sqrt{a},\sqrt{b},\sqrt{ab}$ do not lie in $F$. The splitting field of $f$ is $F(\sqrt{a}, -\sqrt{a},\sqrt{b},-\sqrt{b}) = F(\sqrt{a},\sqrt{b})$ : $$L = F(\sqrt{a},\sqrt{b}).$$ Consider the ascending chain of fields: $$F \subset F(\sqrt{a}) \subset F(\sqrt{a}, \sqrt{b}).$$ As $\sqrt{a} ot \in F, [ F(\sqrt{a}):F] eq 1$, and $[ F(\sqrt{a}):F] \leq 2$ since $\sqrt{a}$ is a root of $x^2-a \in F[x]$, thus $[ F(\sqrt{a}):F] = 2$. We prove that $\sqrt{b} ot \in F(\sqrt{a})$. Suppose, at the contrary, that $\sqrt{b} \in F(\sqrt{a})$. Then $$\sqrt{b} = u+v\sqrt{a}, \ u,v \in F.$$ By squaring this equality, $b =u^2+av^2+2uv\sqrt{a}$. If $uv eq 0$, then $\sqrt{b} = \frac{b - u^2 -av^2}{2uv} \in F$, in contradiction with the hypothesis, so $uv=0$. If $v=0$, $\sqrt{b} = u \in F$: this is excluded. If $u=0$, $\sqrt{b} = v \sqrt{a}$, so $\sqrt{ab} = va \in F$: this is also excluded. This proves that $\sqrt{b} ot \in F(\sqrt{a})$, and $\sqrt{b}$ is a root of $x^2-b \in F(\sqrt{a})[x]$, thus $$[F(\sqrt{a},\sqrt{b}):F(\sqrt{a})]=2.$$ Finally $$[L:F] = [F(\sqrt{a},\sqrt{b}):F] = [F(\sqrt{a},\sqrt{b}):F(\sqrt{a})]\ [ F(\sqrt{a}):F] = 4.$$ As the characteristic of $F$ is not 2, $\sqrt{a} eq -\sqrt{a}$, otherwise $\sqrt{a}=0 \in F$, and the same is true for $b$. Moreover $\sqrt{a} e \pm \sqrt{b}$, otherwise $\sqrt{ab} = \pm b \in F$, so $$f = (x-\sqrt{a})(x+\sqrt{a})(x-\sqrt{b})(x+\sqrt{b})$$ is a separable polynomial, and the splitting field $L$ of the separable polynomial $f \in F[x]$ is a Galois extension of $F$. Therefore, $$\vert \Gal(L/F) = [L:F] = 4.$$ If $\sigma \in G = \Gal(L/F)$, since $a$ is a root of $x^2 -a \in F[x]$, $\sigma(a)$ also, thus $\sigma(\sqrt{a}) =(-1)^k \sqrt{a},\ 0 \leq k \leq 1$. Similarly $\sigma(\sqrt{b}) =(-1)^l \sqrt{b}, \ 0 \leq l \leq 1$. As $\sigma$ is uniquely determined by the images of $\sqrt{a}, \sqrt{b}$, there are at most 4 $F$-automorphisms of $L$. As $\vert \Gal(L/F) \vert = 4$, these 4 possibilities occur, and give an element of the Galois group $\Gal(L/F)$, otherwise this group would have less than 4 elements. Then $G = \{e, \sigma, au, \zeta\}$, where \begin{align*} \sigma(\sqrt{a}) = -\sqrt{a}, \qquad&\sigma(\sqrt{b}) = \phantom{-}\sqrt{b},\ au(\sqrt{a}) = \phantom{-}\sqrt{a}, \qquad& au(\sqrt{b}) = -\sqrt{b},\ \zeta(\sqrt{a}) = -\sqrt{a}, \qquad&\zeta(\sqrt{b}) = -\sqrt{b}. \end{align*} As $\sigma, au, \zeta$ are of order 2, $$\Gal(L/F) \simeq \Z/2\Z imes \Z/2\Z.$$ $\bullet$ Conversely, suppose (a): \begin{center} $L/F$ is a Galois extension of $F$, and $\Gal(L/F) \simeq \Z/2\Z imes \Z/2\Z$. \end{center} Then $$[L:F] = \Gal(L/F) = 4.$$ Write $e,\sigma, au,\zeta$ the elements of $G =\Gal(L/F) $, where $e$ the identity of $G$. As $G = \{e,\sigma, au, \zeta\} \simeq \Z/2\Z imes \Z/2\Z$, $\zeta = \sigma \circ au$ and all the elements different from $e$ are of order 2. The only non trivial subgroups have cardinality 2: they are $\langle \sigma angle = \{e,\sigma\}, \langle au angle = \{e, au\},\langle \zeta angle = \{e,\zeta\}$. \bigskip \begin{tikzpicture} ode (S3) at (4,4) {$\{e\}$}; ode (A3) at (1,2) {$\langle \sigma angle$}; ode(n) at (4,2) {$\langle \sigma \circ au angle$}; ode (t23) at (7,2) {$\langle au angle $}; ode (id) at (4,0) {$\Gal(L/F)$}; \draw[->] (S3) edge (A3) edge (n) edge (t23); \draw[<-] (id) edge (A3) edge (n) edge (t23); \end{tikzpicture} \begin{tikzpicture} ode (S3) at (4,4) {$L$}; ode (A3) at (1,2) {$K_\sigma$}; ode(n) at (4,2) {$K_\zeta$}; ode (t23) at (7,2) {$K_ au$}; ode (id) at (4,0) {$F$}; \draw[<-] (S3) edge (A3) edge (n) edge (t23); \draw[->] (id) edge (A3) edge(n) edge (t23); \end{tikzpicture} \bigskip The intermediate field corresponding with these subgroups are the fixed fields $$K_\sigma = L_{\langle \sigma angle},K_ au=L_{\langle au angle},K_\zeta = L_{\langle\sigma \circ au angle}.$$ As the index in $G$ of these three subgroups is 2, $K_\sigma,K_ au,K_\zeta $ are quadratic extensions of $F$ (by Theorem 7.3.1$[L_H:F] = [\Gal(L/F) : H]$). Since $[L:F] = \Gal(L/F) = 4$, $L$ is a quadratic extension of each of them. As the characteristic of $F$ is different from 2, the Exercise 7.1.12 shows that $K_\sigma = F(\alpha)$, where $a=\alpha^2 \in F,\alpha ot \in F$. Write $\alpha = \sqrt{a}$, then $K_\sigma = F(\sqrt{a}),a\in F,\sqrt{a} ot \in F$. Similarly $K_ au = F(\sqrt{b}), b \in F,\beta = \sqrt{b} ot \in F$. $\alpha = \sqrt{a} \in L_{\langle \sigma angle}$, so $\sigma(\sqrt{a}) = \sqrt{a}$. $K_\sigma \cap K_ au =L_{\langle \sigma angle} \cap L_{\langle au angle} = L_{\langle \sigma, au angle} = L_G = F$ by Theorem 7.1.1(b), so $$K_\sigma \cap K_ au = F.$$ Since $\sqrt{a} \in K_\sigma \setminus F$, $\sqrt{a} ot \in K_ au$, thus $ au(\sqrt{a}) eq \sqrt{a}$. Moreover $\sqrt{a}$ is a root of $x^2 -a \in F[x]$, thus $ au(\sqrt{a}) \in \{\sqrt{a}, - \sqrt{a}\}$. Consequently $ au(\sqrt{a}) = -\sqrt{a}$. $$\sigma(\sqrt{a}) = \sqrt{a}, \qquad au(\sqrt{a}) = - \sqrt{a},$$ and similarly$$\sigma(\sqrt{b}) = -\sqrt{b}, \qquad au(\sqrt{b}) = \sqrt{b}.$$ As $(\alpha \beta)^2 = ab$, write $\alpha \beta = \sqrt{ab} = \sqrt{a} \sqrt{b}$. Then $$\sigma(\sqrt{ab}) = -\sqrt{ab}, \qquad au(\sqrt{ab}) = -\sqrt{ab}.$$ Thus $\sqrt{ab}$ lies not in the fixed field of $G$, so $\sqrt{ab} ot \in F$. The intermediate extension $E = F(\sqrt{a},\sqrt{b})$ contains $K_\sigma = F(\sqrt{a})$ and $K_ au = F(\sqrt{b})$, so $E \supset L_{\langle \sigma angle}, E \supset L_{\langle au angle}$. Therefore, by the Galois correspondence, $\Gal(L/E) \subset \Gal(L/ L_{\langle \sigma angle}) = \langle \sigma angle$ and $\Gal(L/E) \subset \langle au angle$, thus $\Gal(L/E) \subset \langle \sigma angle \cap \langle au angle = \{e\}$. Thus $\Gal(L/E) = \{e\}$, and so $E = L$. $$L = F(\sqrt{a}, \sqrt{b}).$$ As $f =(x^2-a)(x^2-b) = (x-\sqrt{a})(x+\sqrt{a})(x-\sqrt{b})(x+\sqrt{b}) \in F[x]$ splits completely in $L$, the splitting field of $f$ is $F(\sqrt{a},\sqrt{b}) =L$. The equivalence (a)$\iff$(b) is proved. \end{proof}

Exercise 7.3.9

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Exercise 7.3.9

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