Exercise 13.1.13

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
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ewcommand{\ssi} {si et seulement si }

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Suppose that $f \in F[x]$ satisfies the hypothesis of part (c) of Theorem 13.1.1, and let $\alpha$ be a root of $f$. Prove that $G \simeq \Z/4\Z$ if $f$ splits completely over $F(\alpha)$, and $G\simeq D_8$ otherwise. This gives a version of part (c) that doesn't use resolvents. Since we can factor over extension fields by Section 4.2, this method is useful in practice.

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} With the hypothesis of part (c), $\Delta(f) ot \in F^2$, so $\Delta(f) e 0$ and $f$ is separable. \be \item[$\bullet$] If $G \simeq \Z/4\Z$, then $G = \langle \sigma angle \subset S_4$, where $\sigma$ corresponds to $ ilde{\sigma} \in \Gal(L/F)$. Write $G_\alpha = \mathrm{Stab}_G(\alpha)$. Since $f$ is irreducible, ${\cal O}_\alpha = \{\alpha_1,\alpha_2,\alpha_3,\alpha_4\}$ is the set of the four roots of $f$, therefore $4 = |{\cal O}_\alpha| = (G: G_\alpha)$, so $G_{\alpha} = \{e\}$. Hence $ ilde{\sigma}^i(\alpha) e ilde{\sigma}^j(\alpha)$ if $1\leq i < j \leq 4$. We choose the numbering of the roots such that $\alpha_1 = \alpha$, and $ ilde{\sigma}(\alpha_1) = \alpha_3, ilde{\sigma}(\alpha_3) = \alpha_2, ilde{\sigma}(\alpha_2) = \alpha_4$ are the four distinct roots of $f$, so $\sigma = (1\,3\,2\,4)$. $$f = (x-\alpha_1)(x-\alpha_3)(x-\alpha_2)(x-\alpha_4) = (x-\alpha)(x- ilde{\sigma}(\alpha))(x- ilde{\sigma}^2(\alpha))(x- ilde{\sigma}^3(\alpha)).$$ As $\Delta(f) ot \in F^2$, $F(\sqrt{\Delta(f)})$ is a quadratic extension of $F$. Since the only subgroup of $G$ are $\{e\} \subset H =\langle \sigma^2 angle \subset G = \langle \sigma angle$, by the Galois correspondence, the only intermediate fields of $F\subset L$ are $F \subset F(\sqrt{\Delta(f)})\subset L$, and the fixed field of $H = \langle \sigma^2 angle$ is $L_H = F(\sqrt{\Delta(f)})$. If $F(\alpha) \subset F(\sqrt{\Delta(f)})$, then $\alpha \in F(\sqrt{\Delta(f)}) = L_H$, therefore $\sigma^2(\alpha) = \alpha$, and so $\alpha_2 = \alpha_1$, in contradiction with the separability of $f$. Hence $F(\alpha) ot \subset F(\sqrt{\Delta(f)})$, so $$F(\alpha) = L = F(\alpha_1,\alpha_2,\alpha_3,\alpha_4).$$ Then $f$ splits completely over $F(\alpha)$. \item[$\bullet$] If $G ot \simeq \Z/4\Z$, then by Theorem 13.1.1, $G \simeq D_8$. Therefore $[L : F] = |G| = 8$, and $[F(\alpha) : F] = \deg(f) = 4$, which implies $F(\alpha) e L = F(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$. Therefore one of the roots $\alpha_i$ is not in $F(\alpha)$, and so $f =(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)$ doesn't splits completely over $F(\alpha)$. \ee Conclusion. Let $f$ be a quadratic polynomial, and let $\alpha$ be a root of $f$. If $\Delta(f) ot \in F^2$ and $ heta_f(y)$ is reducible over $F$, then \begin{align*} f ext { splits completely over } F(\alpha) &\iff \Gal_F(f) \simeq \Z/4\Z,\ f ext { doesn't split completely over } F(\alpha) &\iff \Gal_F(f) \simeq D_8. \end{align*} \end{proof} Example 1: $f = x^4-12x^2+18$ over $\Q$. \begin{verbatim} R. = QQ[] f = x^4-12*x^2 + 18 print(f.is_irreducible()) factor(f.discriminant()), f.discriminant().is_square() \end{verbatim} $$ ext{True}$$ $$(2^{11} \cdot 3^6, ext{False}).$$ \begin{verbatim} l = f.coefficients(sparse=False); c1 = -l[3]/l[4]; c2 = l[2]/l[4];c3 = -l[1]/l[4]; c4 = l[0]/l[4]; S. = QQ[] theta_f = y^3 -c2*y^2 +(c1*c3-4*c4)*y - c3^2-c1^2*c4 + 4*c2*c4; factor(theta_f) \end{verbatim} $$(y + 12) \cdot (y^{2} - 72)$$ \begin{verbatim} K.= NumberField(f) S. = K[] f = x^4-12*x^2 + 18 factor(f) \end{verbatim} $$(x - a) \cdot (x + a) \cdot (x - \frac{1}{3} a^{3} + 3 a) \cdot (x + \frac{1}{3} a^{3} - 3 a)$$ These results prove that the Galois group of $f = x^4-12x^2+18$ over $\Q$ is isomorphic to $\Z/4\Z$. Verification with Sage: \begin{verbatim} R. = QQ[] f = x^4-12*x^2 + 18 f.galois_group().gens() \end{verbatim} $$ [(1,2,3,4)]$$ \begin{verbatim} f.galois_group().structure_description() \end{verbatim} $$C4$$ \bigskip Example 2: $f = x^4 - 2$ over $\Q$. \begin{verbatim} R. = QQ[] f = x^4-2 print(f.is_irreducible()) factor(f.discriminant()), f.discriminant().is_square() \end{verbatim} $$ ext{True}$$ $$(-1\cdot 2^{11}, ext{False})$$ \begin{verbatim} l = f.coefficients(sparse=False); c1 = -l[3]/l[4]; c2 = l[2]/l[4];c3 = -l[1]/l[4]; c4 = l[0]/l[4]; S. = QQ[] theta_f = y^3 -c2*y^2 +(c1*c3-4*c4)*y - c3^2-c1^2*c4 + 4*c2*c4; factor(theta_f) \end{verbatim} $$y \cdot (y^{2} + 8)$$ \begin{verbatim} K.= NumberField(f) S. = K[] f = x^4-2 factor(f) \end{verbatim} $$(x - a) \cdot (x + a) \cdot (x^{2} + a^{2})$$ Thus the Galois group of $x^4 - 2$ over $\Q$ is $D_8$. Verification with Sage: \begin{verbatim} R. = QQ[] f = x^4-2 f.galois_group().gens() \end{verbatim} $$ [(1,2,3,4), (1,3)]$$ \begin{verbatim} f.galois_group().structure_description() \end{verbatim} $$D4$$ \bigskip Example 3: $f = x^4-18x^2+9$ over $\Q$. \begin{verbatim} R. = QQ[] f = x^4-18*x^2 + 9 print(f.is_irreducible()) factor(f.discriminant()), f.discriminant().is_square() \end{verbatim} $$ ext{True}$$ $$(2^{14}\cdot 3^6, ext{True})$$ \begin{verbatim} l = f.coefficients(sparse=False); c1 = -l[3]/l[4]; c2 = l[2]/l[4];c3 = -l[1]/l[4]; c4 = l[0]/l[4]; S. = QQ[] theta_f = y^3 -c2*y^2 +(c1*c3-4*c4)*y - c3^2-c1^2*c4 + 4*c2*c4; factor(theta_f) \end{verbatim} $$ (y - 6) \cdot (y + 6) \cdot (y + 18) $$ The Galois group of $f = x^4-18x^2+9$ over $\Q$ is isomorphic to $\Z/2\Z imes \Z/2\Z$. Verification with Sage: \begin{verbatim} R. = QQ[] f = x^4-18*x^2 + 9 f.galois_group().gens() \end{verbatim} $$ [(1,2)(3,4), (1,4)(2,3)]$$ \begin{verbatim} f.galois_group().structure_description() \end{verbatim} $$C2 imes C2$$

Exercise 13.1.13

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Exercise 13.1.13

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