Exercise 13.3.6

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 $D=\sum_{\sigma \in A_4} \sigma \cdot x_1^3x_2^2x_3 \in F[x_1,x_2,x_3,x_4]$.
\be
\item[(a)] Prove that $D=\frac{1}{2}(\sigma_1\sigma_2\sigma_3 -3 \sigma_1^2 \sigma_4 -3\sigma_3^2 + 4\sigma_2 \sigma_4)+\frac{1}{2}\sqrt{\Delta}$ in characteristic $
e 2$.
\item[(b)] Prove that $\sqrt{\Delta}=D-(12)\cdot D$ in all characteristics.
\ee
\begin{proof}
\item[(a)] Since $$\sum_{\sigma \in S_4} \sigma =\sum_{\sigma \in A_4} \sigma + \sum_{\sigma \in S_4 \setminus A_4} \sigma =\sum_{\sigma \in A_4} \sigma + (12)\cdot \sum_{\sigma \in  A_4} \sigma $$ and based on results of Ex.2.2.3, we have:
$$\sum_{\sigma \in S_4} \sigma \cdot x_1^3x_2^2x_3 = D+(12)\cdot D =\sigma_1\sigma_2\sigma_3 -3 \sigma_1^2 \sigma_4 -3\sigma_3^2 + 4\sigma_2 \sigma_4 $$
Sage verification:
\begin{verbatim}
    R.<x1,x2,x3,x4,y1,y2,y3,y4> = PolynomialRing(QQ, order = 'degrevlex')
    elt = SymmetricFunctions(QQ).e()
    e = [elt([i]).expand(4).subs(x0=x1, x1=x2, x2=x3, x3 = x4) for i in range(5)]
    J = R.ideal(e[1]-y1, e[2]-y2, e[3]-y3,e[4]-y4)
    G = J.groebner_basis()
    D = x1^3*x2^2*x3;
    D=D+D.subs(x1=x3,x2=x4,x3=x1)+D.subs(x1=x2,x2=x1,x3=x4)+D.subs(x1=x4,x2=x3,x3=x2)
    D=D+D.subs(x1=x2,x2=x3,x3=x1)+D.subs(x1=x2,x2=x3,x3=x1).subs(x1=x2,x2=x3,x3=x1)
    u=D+D.subs(x1=x2,x2=x1)
    var('sigma_1,sigma_2,sigma_3,sigma_4')
    u.reduce(G).subs(y1=sigma_1, y2 = sigma_2,y3=sigma_3,y4=sigma_4)
\end{verbatim}
$$\sigma_1\sigma_2\sigma_3 -3 \sigma_1^2 \sigma_4 -3\sigma_3^2 + 4\sigma_2 \sigma_4 $$

Assuming that $D-(12)\cdot D=\sqrt{\Delta}$ is valid, then $2D 
e 0$ in characteristic $
e 2$ and
$$2D=2(\frac{1}{2}(\sigma_1\sigma_2\sigma_3 -3 \sigma_1^2 \sigma_4 -3\sigma_3^2 + 4\sigma_2 \sigma_4)+\frac{1}{2}\sqrt{\Delta}).$$
\item[(b)] We use Sage to prove that $D-(12)\cdot D-\sqrt{\Delta}=0$:

\begin{verbatim}
    Delta = (x1-x2)*(x1-x3)*(x1-x4)*(x2-x3)*(x2-x4)*(x3-x4)
    D-D.subs(x1=x2,x2=x1)-Delta==0
\end{verbatim}
\begin{center}
 {\it True}
\end{center}
Hence in all characteristics  $$\sqrt{\Delta}=D-(12)\cdot D.$$
\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 13.3.6

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Exercise 13.3.6

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