Exercise 13.2.18

Question

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

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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 is concerned with the polynomials (13.28). As in the Historical Notes, we will assume that they lie in $\Q[x]$ and are irreducible.
\be
\item[(a)] Show that $\sqrt[5]{Q^2/P} + (P/Q)\sqrt[5]{Q^2/P}^2$ is a root of $x^5 - 5Px^2 - 5Qx - Q^2/P -P^3/Q$.
\item[(b)] Prove that the Galois group of $x^5 - 5Px^2 - 5Qx - Q^2/P - P^3/Q$ over $\Q$ is isomorphic to $AGL(1,\F_5)$.
\item[(c)] Prove that over $\Q(\sqrt{5})$, the first two polynomials of (13.28) have cyclic Galois group while the third has Galois group isomorphic to $\mathrm{AGL}(1,\F_5) \cap A_5$.
\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} \be \item[(a)] As in Cardan's method, we substitute $u+v$ to $x$ in $$f(x)= x^5 - 5Px^2 - 5Qx - \frac{Q^{2}}{P} -\frac{P^{3}}{Q}.$$ We obtain \begin{align*} f(u+v) &= u^{5} + 5 \, u^{4} v + 10 \, u^{3} v^{2} + 10 \, u^{2} v^{3} + 5 \, uv^{4} + v^{5} \ &\phantom{= }- 5 \, P u^{2} - 10 \, P u v - 5 \, P v^{2} - 5 \, Q u - 5 \, Q v - \frac{Q^{2}}{P} -\frac{P^{3}}{Q}\ &= \left(u^5 + v^5 - \frac{Q^{2}}{P} -\frac{P^{3}}{Q} ight)+ 5 \, {\left(u^{3} v + u^{2} v^{2} + u v^{3} - P u - P v - Q ight)} {\left(u + v ight)}.\ \end{align*} Then we verify that $u = \sqrt[5]{Q^2/P} = P^{-\frac{1}{5}} Q^{\frac{2}{5}}, v = (P/Q)\sqrt[5]{Q^2/P}^2 = P^{\frac{3}{5}} Q^{-\frac{1}{5}}$ is a solution of the system \begin{align*} 0 &= u^5 + v^5 - \frac{Q^{2}}{P} -\frac{P^{3}}{Q},\ 0 &= u^{3} v + u^{2} v^{2} + u v^{3} - P u - P v - Q. \end{align*} Indeed $$u^5 +v^5 -\frac{Q^{2}}{P} -\frac{P^{3}}{Q} = \frac{Q^2}{P} + \frac{P^3}{Q}- \frac{Q^{2}}{P} -\frac{P^{3}}{Q} = 0,$$ and, since $P = uv^2, Q = u^3v$, \begin{align*} u^{3} v + u^{2} v^{2} + u v^{3} - P u - P v - Q=u^3 v +u^2v^2+uv^3 - u^2v^2 - uv^3 - u^3v = 0. \end{align*} Therefore $$u + v = \sqrt[5]{Q^2/P} + (P/Q)\sqrt[5]{Q^2/P}^2$$ is a root of $x^5 - 5Px^2 - 5Qx - Q^2/P -P^3/Q$. If we replace $\sqrt[5]{Q}$ by another fifth root $\zeta^k \sqrt[5]{Q},\ k=1,2,3,4,$ (where $\zeta = e^{2i\pi/5}$), in the equation $$0 = u^{3} v + u^{2} v^{2} + u v^{3} - P u - P v - Q, \qquad ext{where } u= \sqrt[5]{P}^{-1}\sqrt[5]{Q}^2, v = \sqrt[5]{P}^3 \sqrt[5]{Q}^{-1},$$ then $u$ is replaced by $\zeta^{2k}u$, and $v$ is replace by $\zeta^{4k} v$, therefore $\zeta^{2k} u , \zeta^{4k} v$ is a solution of the preceding system, so $\zeta^{2k} u + \zeta^{4k} v$ is also a root of $f$. So $$u+v,\quad \zeta^2 u + \zeta^4 v, \quad\zeta^4 u + \zeta^3 v,\quad \zeta u + \zeta^2 v,\quad \zeta^3 u + \zeta v,$$ are roots of $f$, where $$u = \sqrt[5]{Q^2/P} ,\qquad v = (P/Q)\sqrt[5]{Q^2/P}^2 .$$ These roots are the five roots of $f$, as proved in the following expansion: \begin{align*} & (x - (u+v))(x- (\zeta^2 u + \zeta^4 v))(x- (\zeta^4 u + \zeta^3 v))(x-(\zeta u + \zeta^2 v))(x- (\zeta^3 u + \zeta v))\ &= x^5 -5 uv^2 x^2 -5 u^3v x -u^5-v^5. \end{align*} Since $$P = uv^2, \qquad Q = u^3 v,$$ we obtain $$u^5 = \frac{Q^2}{P},\qquad v^5 = \frac{P^3}{Q},$$ so \begin{align*} f &= x^5 - 5Px^2 - 5Qx - \frac{Q^{2}}{P} -\frac{P^{3}}{Q}\ &=(x - (u+v))(x- (\zeta^2 u + \zeta^4 v))(x- (\zeta^4 u + \zeta^3 v))(x-(\zeta u + \zeta^2 v))(x- (\zeta^3 u + \zeta v)) \end{align*} Therefore the roots of $f$ are $\zeta^{2k} u + \zeta^{4k} v,\ k=0,1,2,3,4$. This was perhaps the Euler's starting point. \item[(b)] We obtain the discriminant of $f$ with \begin{verbatim} R. = QQ[] f = x^5 - 5*P*x^2 - 5*Q*x -e; Delta = f.discriminant(x).subs(e = Q^2/P + P^3/Q).factor() Delta \end{verbatim} $$\left(3125 ight) \cdot Q^{-4} \cdot P^{-4} \cdot (- P^{8} - 11 P^{4} Q^{3} + Q^{6})^{2} $$ so $$\Delta = 5^5 \frac{(P^8 +11 P^4 Q^3 -Q^6)^2}{P^4Q^4}.$$ Thus $\Delta$ is not a square in $\Q$. Using the resolvent() function of Exercise 14, we obtain the sextic resolvent: \begin{verbatim} K. = QQ[] S. =PolynomialRing(K, order = 'degrevlex') f = x^5 - 5*P*x^2 - 5*Q*x -e theta = resolvent(f)[0]; theta.subs(e = Q^2/P + P^3/Q) \end{verbatim} \begin{align*} heta_f(y) &= { \left( 100 \, Q y^{2} + y^{3} + 2000 \, {\left( 3 \,Q^{2}-\left(P^4+Q^3 ight)/Q ight)} y ight)}^{2} - 1024 \, \Delta(f) y\ &= { \left( y^{3} +100 \, Q y^{2} - 2000 \, {\left(\frac{P^{4}}{Q} - 2 \,Q^{2} ight)} y ight)}^{2} - 1024 \, \Delta(f) y, \end{align*} which has root $0 \in \Q$ ($b_6 = 0$). By Section 13.2, the Galois group of $f$ is $\mathrm{AGL}(1,\F_5)$ up to conjugacy. \item[(c)] By Exercise 13, we know that the Galois group of $x^5 - 2$ over $\Q(\sqrt{5})$ is $\mathrm{AGL}(1,\F_5)\cap A_5$ which is not cyclic, so there is a misprint in the sentence. \be \item[$\bullet$] The polynomial $f = x^5 - D \in \Q(\sqrt{5})[x], D \in \Q$, is irreducible over $\Q$ by hypothesis. Let $\alpha$ be a root of $f$. Since $[\Q(\alpha) : \Q] = 5$ and $[\Q(\sqrt{5}) : \Q] = 2$, we obtain that $10 = 2 imes 5$ divides $[\Q(\sqrt{5},\alpha) : \Q]$, where $[\Q(\sqrt{5},\alpha) : \Q] \leq 10$. Therefore $$10 = [\Q(\sqrt{5},\alpha) : \Q] = [\Q(\sqrt{5},\alpha) : \Q(\sqrt{5})] [\Q(\sqrt{5}):\Q],$$ so $[\Q(\sqrt{5},\alpha) : \Q(\sqrt{5})] = 5$. If $p$ is the minimal polynomial of $\alpha$ over $\Q(\sqrt{5})$, then $\deg(p) = [\Q(\sqrt{5},\alpha) : \Q(\sqrt{5})] = 5$, and $p$ divides $f$, therefore $p = f$, and we have proved that $f$ is irreducible over $\Q(\sqrt{5})$. Moreover $$\Delta(f) = 5^5 D^4 = (5^2 D^2 \sqrt{5})^2$$ is a square in $\Q(\sqrt{5})$, and $0 \in \Q(\sqrt{5})$ is a root of the resolvent $$\Theta_f(y) = y^6 - \Delta(f) y.$$ If $\alpha = \sqrt[5]{D} \in \R$, and $\zeta = \zeta_5$, then $$x^5 - D = (x - \alpha)(x-\zeta \alpha)(x-\zeta^2 \alpha)(x-\zeta^3 \alpha)(x-\zeta^4 \alpha).$$ But $\zeta \alpha ot \in \R$, so $\zeta \alpha ot \in \Q(\sqrt{5})(\alpha) \subset \R$. $f$ doesn't split completely over $F(\alpha)$, where $\alpha$ is a root of $f$. By Theorem 13.2.6 and the table 13.2.C, $$\Gal(L/\Q(\sqrt{5})) \simeq \mathrm{AGL}(1,\F_5)\cap A_5,$$ where $L$ is the splitting field of $x^5 - D$. \item[$\bullet$] Now $f= x^5 - 5Px^2-5Qx-\frac{Q^2}{P} - \frac{P^3}{Q}$, and $f$ is assumed irreducible over $\Q$. With the same proof as in the first bullet, $f$ remains irreducible over $\Q(\sqrt{5})$. By part (b), $$\Delta = 5^5 \frac{(P^8 +11 P^4 Q^3 -Q^6)^2}{P^4Q^4}$$ is a square in $\Q(\sqrt{5})$, and $$ heta_f(y) = { \left( y^{3} +100 \, Q y^{2} - 2000 \, {\left(\frac{P^{4}}{Q} - 2 \,Q^{2} ight)} y ight)}^{2} - 1024 \, \Delta(f) y,$$ has root $0 \in \Q(\sqrt{5})$. By part (a), \begin{align*} f &= x^5 - 5Px^2 - 5Qx - \frac{Q^{2}}{P} -\frac{P^{3}}{Q}\ &=(x - (u+v))(x- (\zeta^2 u + \zeta^4 v))(x- (\zeta^4 u + \zeta^3 v))(x-(\zeta u + \zeta^2 v))(x- (\zeta^3 u + \zeta v)) \end{align*} We prove that $\alpha_2 = \zeta^2u + \zeta^4v$ is not real. If $\alpha_2 \in \R$, then $\alpha_2 = \overline{\alpha_2}$, so $(\zeta^2 - \zeta^3)u + (\zeta^4 - \zeta) v = 0$. Then, using $\sqrt{5} = \zeta - \zeta^2 - \zeta^3 + \zeta^4$ and $\zeta^2 + \zeta^3 = \frac{-1 + \sqrt{5}}{2}$, \begin{align*} \frac{v}{u} &= \frac{\zeta^2 - \zeta^3}{\zeta - \zeta^4} = \frac{\zeta - \zeta^2}{1 - \zeta^3}\ &=\frac{(\zeta - \zeta^2)(1 - \zeta^2)}{(1 - \zeta^3)(1-\zeta^2)}\ &= \frac{\zeta - \zeta^2 - \zeta^3 + \zeta^4}{1 - (\zeta^2+\zeta^3) + \zeta^5}\ &= \frac{\sqrt{5}}{2 +\frac{1+\sqrt{5}}{2}}\ &= \frac{\sqrt{5}-1}{2} \end{align*} Since $\frac{v}{u} = \frac{P}{Q} \sqrt[5]{\frac{Q^2}{P}} $, we would have $\sqrt[5]{\frac{Q^2}{P}} \in \Q(\sqrt{5})$, and then the root $\alpha_1 = u + v\in \Q(\sqrt{5})$, in contradiction with the irreducibility of $f$ over $\Q(\sqrt{5})$. So $\alpha_2 ot \in \R$ is not in the field $\Q(\sqrt{5})(\alpha_1)$, and $f$ doesn't split completely over $\Q(\sqrt{5})(\alpha_1)$. By the table 13.2.C, $\Gal(L/\Q(\sqrt{5}) \simeq \mathrm{AGL}(1,\F_5)\cap A_5$. \item[$\bullet$] The third Euler's polynomial is $f = x^5 - 5Px^3 + 5P^2 x - D$. If $p = -5P,q = -D$, we obtain $f = x^5+ p x^3 +\frac{1}{5} p^2x +q$, which is Example 13.2.10. We know from this example and from Exercise 14 that the Galois group of $f$ over $\Q$ is $\mathrm{AGL}(1,\F_5)$. With the same proof as in the first bullet, $f$ remains irreducible over $\Q(\sqrt{5})$. By Exercise 14, $$ \Delta(f) =\frac{1}{5^5} \cdot (4 p^{5} + 3125 q^{2})^{2} $$ is a square in the field $\Q(\sqrt{5})$, and $$ heta_f(y) = \left(y^3-7 \, p^{2}y^2+11 p^{4}y+\frac{3}{25} \, p^{6} + 4000 \, p q^{2} ight)^2 - 2^{10}\Delta(f) y $$ has the root $5p^2 \in \Q$. It remains to know if $f$ splits completely over $\Q(\sqrt{5})$. Note that $$f(u+v) = u^5 + v^5 + q + (u^2 + uv + v^2 +\frac{p}{5})(5 uv +p)(u+v).$$ Therefore if $uv = -\frac{p}{5}$, $f(u + v) = u^5 +v^5 +q$, so we find (see Exercise 14) that $$f\left(z - \frac{p}{5z} ight) = z^5 - \frac{p^5}{5^5 z^5} + q.$$ So $u+v$ is a root of $f$ if $$ \left\{ \begin{array}{rcc} u^5 + v^5 & = & -q = D \ uv & = & -\frac{p}{5} =P \end{array} ight. $$ Therefore $u^5, v^5$ are the roots of $x^2 + qx -(\frac{p}{5})^5$ and satisfy $uv = -p/5 \in \Q$. If $u+v$ is a root, so is $\zeta^k u + \zeta^{-k} v, \ k \in \Z$ (where $\zeta = \zeta_5 = e^{2i\pi/5}$). Conversely \begin{align*} &(x - u-v) (x - \zeta u - \zeta^{-1}v)(x - \zeta^2 u - \zeta^{-2}v)(x - \zeta^3 u - \zeta^{-3}v)(x - \zeta^4 u - \zeta^{-4}v) \ &= x^5 -5 uv x^3 + 5u^2v^2 x - u^5 -v^5\ &= x^5 -5 Px^3 + 5 P^2 x - D \end{align*} So the roots of $f =x^5 -5 Px^3 + 5 P^2 x - D = x^5+ p x^3 +\frac{1}{5} p^2x +q$ are $$u + v,\quad \zeta u + \zeta^{-1}v,\quad \zeta^2 u + \zeta^{-2}v, \quad \zeta^3 u + \zeta^{-3}v, \quad \zeta^4 u + \zeta^{-4}v,$$ where $(u,v)$ is a solution of the system $$ uv = P = -\frac{p}{5},\qquad u^5 + v^5 = D = -q,$$ so \begin{align*} u &= \sqrt[5] {-\frac{q}{2} + \sqrt{\left(\frac{q}{2} ight)^2 + \left(\frac{p}{5} ight)^5}},\ v &= \sqrt[5] {-\frac{ q}{2} - \sqrt{\left(\frac{q}{2} ight)^2 + \left(\frac{p}{5} ight)^5}},\ \end{align*} (where we choose the real roots, so $u,v \in \R$). We obtained the factorization of $f$: \begin{align*} f &= x^5 -5 Px^3 + 5 P^2 x - D\ &= (x - u-v) (x - \zeta u - \zeta^{-1}v)(x - \zeta^2 u - \zeta^{-2}v)(x - \zeta^3 u - \zeta^{-3}v)(x - \zeta^4 u - \zeta^{-4}v) \end{align*} If the root $\alpha_2 = \zeta u + \zeta^{-1}v$ is real, then $u=v$, so $\left(\frac{q}{2} ight)^2 + \left(\frac{p}{5} ight)^5 = 0$, and this imply $\Delta(f) = 0$, in contradiction with the assumed separability of $f$. Therefore $\alpha_2 ot \in \Q(\sqrt{5})(\alpha_1) \subset \R$, where $\alpha_1= u + v \in \R$ is a root of $f$, so $f$ doesn't split completely over $\Q(\sqrt{5})(\alpha_1)$. $\Gal(L/\Q(\sqrt{5}))$ is isomorphic to $\mathrm{AGL}(1,\F_5)\cap A_5$. \ee We obtained the same Galois group for the 3 Euler's polynomials of (13.28). \ee \end{proof}

Exercise 13.2.18

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Exercise 13.2.18

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