Exercise 7.3.8

Question

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Let $\alpha = \zeta_7 + \zeta_7^{-1}$, where $\zeta_7 = e^{2 \pi i/7}$.
\be
\item[(a)] Show that the minimal polynomial of $\alpha$ over $\Q$ is $x^3+x^2-2x-1$.
\item[(b)] Use Exercise 7 to show that the splitting field of $x^3+x^2-2x-1$ over $\Q$ is a Galois extension of degree 3 with Galois group isomorphic to $\Z/3\Z$.
\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} \begin{enumerate} \item[(a)] Let $\zeta = \zeta_7$ and $\alpha = \zeta +\zeta^{-1}$. We compute the minimal polynomial of $\alpha$. We have shown in Exercice 7 that \begin{align*} \zeta + \zeta^{-1} &= \alpha\ \zeta^2+\zeta^{-2} &= \alpha^2-2\ \zeta^3 + \zeta^{-3} &= \alpha^3-3\alpha. \end{align*} Thus \begin{align*} 0 &=1+\zeta + \zeta^2+\zeta^3+\zeta^4+\zeta^5+\zeta^6\ &=1 + (\zeta+\zeta^{-1}) + (\zeta^2+\zeta^{-2})+ (\zeta^3 + \zeta^{-3})\ &=1+\alpha+(\alpha^2-2)+(\alpha^3-3\alpha)\ &=\alpha^3+\alpha^2-2\alpha-1 \end{align*} $\alpha$ is so a root of $p = x^3+x^2-2x-1$. We could verify directly the irreducibility of $p$, but it is more simple to proceed so: $\bullet$ As $p(\alpha)=0$, the minimal polynomial $q$ of $\alpha$ over $\mathbb{Q}$ divides $p$: $q \mid p$, $\bullet$ $\Q(\alpha) =L_H$ is the fixed field of $H = \{e,\sigma\}$ (Exercice 6). Then $\Gal(L/L_H) = H$, and $[L:L_H] = \vert H \vert = 2$, so $$[\Q(\alpha) : \Q] = [L_H:\Q] = [L:\Q]/[L : L_H] = [L:\Q]/ \vert H \vert = 6/2 =3,$$ thus $\deg(q) = [\Q(\alpha) : \mathbb{Q}] = 3 = \deg(p)$, $\bullet$ Moreover $p,q$ are monic. Consequently $p=q$, and so $p$ is irreducible over $\Q$, and $\alpha$ is a root of $p$. Conclusion: $p=x^3+x^2-2x-1$ is the minimal polynomial of $\alpha = \zeta+\zeta^{-1}$ over $\Q$. Note: as an alternative method, to find the minimal polynomial of $\alpha$, we can use the Lagrange's construction described in the proof of Theorem 7.1.1: $3$ is a generator of the cyclic group $(\Z/7\Z)^*$ ($3^2 = 2,3^3 = -1$), so $\Gal(L/\Q) = \{e,\sigma, \sigma^2,\sigma^3 ,\sigma^4, \sigma^5\}$, where $\sigma$ is characterized by $\sigma(\zeta) = \zeta^3$ (then $\sigma^k(\zeta) = \zeta^{3^k} = \zeta,\zeta^3,\zeta^2,\zeta^{-1},\zeta^{-3}, \zeta^{-2}$ for $k = 0,1,2,3,4,5$). The {\it distinct} images of $\alpha$ by the automorphisms of $G$ are so $\zeta + \zeta^{-1}, \zeta^2+\zeta^{-2},\zeta^3+\zeta^{-3}$, so the minimal polynomial of $\alpha$ over $\Q$ is (see the proof of Th. 7.1.1) $$(x-\zeta- \zeta^{-1})(x-\zeta ^{2} - \zeta^{-2})(x-\zeta^3 - \zeta^{-3}).$$ To expand this polynomial, we use the following Sage instructions: \begin{verbatim} K. = NumberField(1+x+x^2+x^3+x^4+x^5+x^6) R. = PolynomialRing(QQ) f = (t-zeta - zeta^(-1))*(t-zeta^2-zeta^(-2))*(t-zeta^3-zeta^(-3));f \end{verbatim} $$t^{3} + t^{2} - 2 t - 1$$ which gives the minimal polynomial \begin{align*} p &= x^3+x^2 -2x-1\ &=(x-\zeta- \zeta^{-1})(x-\zeta ^{2} - \zeta^{-2})(x-\zeta^3 - \zeta^{-3})\ &=(x - 2 \cos(2\pi/7))(x- 2 \cos(4\pi/7))(x-2 \cos(6 \pi/7)) \end{align*} \item[(b)] By Exercise 6, $\Q(\alpha) = L_H $ is associate to $H$ of order 2 in the Galois correspondence. As $G =\Gal(L/F) \simeq (\Z/7\Z)^*$ is Abelian, $H$ is normal in $G$, so $\Q \subset L_H$ is a Galois extension. Consequently all the roots $\alpha,\beta, \gamma$ of $p$ are in $\Q(\alpha)$, thus $ \mathbb{Q}(\alpha) = \mathbb{Q}(\alpha,\beta,\gamma)$ is the splitting field of $p$ over $\Q$. (In fact, the other roots of $p$ are $\sigma(\zeta+ \zeta^{-1}) = \zeta^3+\zeta^{-3} = \alpha^3 - 3 \alpha$, and $\sigma^2(\zeta+ \zeta^{-1}) = \zeta^2 + \zeta^{-2} = \alpha^2 - 2$, and are all in $\Q(\zeta+\zeta^{-1})$.) Conclusion: the splitting field of $p=x^3+x^2-2x-1$ over $\Q$ is $E= \mathbb{Q}(\zeta_7) \cap \mathbb{R} = \mathbb{Q}(\zeta_7 + \zeta_7^{-1})$, and $\Q \subset E$ is a Galois extension of degree 3. Moreover (Theorem 7.2.7), $\Gal(E/\Q) \simeq \Gal(L/\Q)/\Gal(L/E) = G/H$. As $G$ is cyclic and $\vert H \vert = 2$, $G/H$ is the quotient group of a cyclic group, so is cyclic, of order 3, isomorphic to $\Z/3\Z$. $$\Gal(\Q(\zeta_7+\zeta_7^{-1})/\Q) \simeq \Z/3\Z.$$ \end{enumerate} \end{proof}

Exercise 7.3.8

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Exercise 7.3.8

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