Exercise 13.4.4

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}} Consider the polynomial $s_u(y)$ when $f=x^3+x^2-2x-1$ from Example 13.4.3. \be \item[(a)] Compute $s_u(y) \in \mathbb{Q}[u_1,u_2,u_3,y]$, and derive factorization given in Example 13.4.3. \item[(b)] Let h be the first factor of $s_u(y)$ given in Example 13.4.3, multiplied by -1 so that it is monic in y. Using Sage, write h as a polynomial in y so that its coefficients are of the form \begin{align*} m a~symmetric~polynomial~in~ \it u_1,u_2,u_3 m ~+~ a~remainder~in~ \it u_1,u_2,u_3. \end{align*} This should give the formula for h given in Example 13.4.3. \ee \begin{proof} \be \item[(a)] The following Sage instructions: \begin{verbatim} R. = PolynomialRing(QQ, order = 'lex') elt = SymmetricFunctions(QQ).e() e = [elt([i]).expand(3).subs(x0=x1, x1=x2,x2=x3) for i in range(4)] J = R.ideal(e[1]-sigma1, e[2]-sigma2, e[3]-sigma3) G = J.groebner_basis() S = y - (u1*x1 + u2*x2 + u3*x3) S1 = S * S.subs(x1=x1,x2=x3,x3=x2) *S .subs(x1=x2,x2=x1,x3=x3) S1 = S1 * S.subs(x1=x2,x2=x3,x3=x1) * S.subs(x1=x3,x2=x1,x3=x2) S = S1 * S.subs(x1=x3,x2=x2,x3=x1) s = S.reduce(G).polynomial(y) sf = s.subs(sigma1=-1,sigma2=-2,sigma3=1) dec = sf.factor(); dec \end{verbatim} give the same output as in Example 13.4.3: \begin{align*} s_u(y)&= (-y^{3} + \left(- u_{1} - u_{2} - u_{3} ight) y^{2} + \left(2 u_{1}^{2} - 3 u_{1} u_{2} + 2 u_{2}^{2} - 3 u_{1} u_{3} - 3 u_{2} u_{3} + 2 u_{3}^{2} ight) y \ &+ u_{1}^{3} - 4 u_{1}^{2} u_{2} + 3 u_{1} u_{2}^{2} + u_{2}^{3} + 3 u_{1}^{2} u_{3} - u_{1} u_{2} u_{3} - 4 u_{2}^{2} u_{3} - 4 u_{1} u_{3}^{2} + 3 u_{2} u_{3}^{2} + u_{3}^{3})\ &\cdot (-y^{3} + \left(- u_{1} - u_{2} - u_{3} ight) y^{2} + \left(2 u_{1}^{2} - 3 u_{1} u_{2} + 2 u_{2}^{2} - 3 u_{1} u_{3} - 3 u_{2} u_{3} + 2 u_{3}^{2} ight) y\ &+ u_{1}^{3} + 3 u_{1}^{2} u_{2} - 4 u_{1} u_{2}^{2} + u_{2}^{3} - 4 u_{1}^{2} u_{3} - u_{1} u_{2} u_{3} + 3 u_{2}^{2} u_{3} + 3 u_{1} u_{3}^{2} - 4 u_{2} u_{3}^{2} + u_{3}^{3}) \end{align*} \item[(b)] The following Sage instruction: \begin{verbatim} h = (-1)* dec[1][0]; h \end{verbatim} gives the second (it's the good one) irreducible factor $h$ of $s_u(y)$, given in Example 13.4.3, multiplied by -1: \begin{align*} h=& y^{3} + \left(u_{1} + u_{2} + u_{3} ight) y^{2} + \left(-2 u_{1}^{2} + 3 u_{1} u_{2} - 2 u_{2}^{2} + 3 u_{1} u_{3} + 3 u_{2} u_{3} - 2 u_{3}^{2} ight) y \ &- u_{1}^{3} - 3 u_{1}^{2} u_{2} + 4 u_{1} u_{2}^{2} - u_{2}^{3} + 4 u_{1}^{2} u_{3} + u_{1} u_{2} u_{3} - 3 u_{2}^{2} u_{3} - 3 u_{1} u_{3}^{2} + 4 u_{2} u_{3}^{2} - u_{3}^{3} \end{align*} A second reduction doesn't give quite the expected result \begin{verbatim} R. = PolynomialRing(QQ, order = 'lex') elt = SymmetricFunctions(QQ).e() e = [elt([i]).expand(3).subs(x0=u1, x1=u2,x2=u3) for i in range(4)] J = R.ideal(e[1]-sigma1, e[2]-sigma2, e[3]-sigma3) G = J.groebner_basis() g = R(h).reduce(G).polynomial(y); g \end{verbatim} $$y^{3} + \sigma_{1} y^{2} + \left(-2 \sigma_{1}^{2} + 7 \sigma_{2} ight) y + 21 u_{2} u_{3}^{2} - 14 u_{2} u_{3} \sigma_{1} - 7 u_{3}^{2} \sigma_{1} + 7 u_{3} \sigma_{1}^{2} - \sigma_{1}^{3} + 7 u_{2} \sigma_{2} - 7 u_{3} \sigma_{2} + 7 \sigma_{3} $$ But this last instructions verify the formula of the text: \begin{verbatim} S.=ZZ[] A=S(h(0)) B = 7*sigma3 -sigma1^3 + 7*(u1*u2^2 + u1^2*u3 + u2*u3^2) B = B.subs(sigma1 = e[1],sigma2=e[2], sigma3=e[3]) B = S(B) A-B \end{verbatim} $$0$$ This is a verification of \begin{align*} h&=y^{3}+ (u_{1} + u_{2} + u_{3}) y^{2} + (7(u_{1} u_{2} + u_{1} u_{3} + u_{2} u_{3})-2(u_{1} + u_{2} + u_{3})^{2}) y\ & +7u_{1} u_{2} u_{3}-(u_{1} + u_{2} + u_{3})^{3}+ 7(u_{1} u_{2}^{2} + u_{1}^{2} u_{3} + u_{2} u_{3}^{2}). \end{align*} \ee\end{proof}

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}}

Exercise 13.4.4

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Exercise 13.4.4

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