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actualizar Mult3.tex

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Mult3.tex
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@ -391,14 +391,17 @@ $$ \prod_{i=1}^n h_i^{2} \prod_{j=1}^m k_j^3 \left ( \prod_{i=1}^n h_i^{3n_i-2}
\begin{proposition}
The singular point is isolated if and only if the disciminant $D_zF$ is reduced.
The singular point is isolated if and only if the discriminant $D_zF$ is reduced.
\end{proposition}
\noindent \textcolor{red}{Proof. Ask Enrique.}
\begin{proposition}\label{prop.non-deg}
The monic Weierstrass polynomial $z^3 -3b_2z + 2b_3$ is non degenerate unless $M_1(c_1+A)$ and $b_3 = M_2(c_2 + B)$ with $M_1^3 = M_2^2$ and \textcolor{red}{$2c_1=3c_2$}
\begin{proposition}\label{prop.non-deg} Let $b_2, b_3 \in \CC[[x, y]]$ with $b_2(0), b_3(0) = 0$. Let us write $b_2= M_2(c_2+A)$ and $b_3 = M_3(c_3 + B)$ with $M_2, M_3 \in \CC[x,y]$ (monic) monomials, $c_2, c_3 \in \CC^\ast$ and $A, B \in \CC[[x,y]]$ such that $A(0), B(0) = 0$.
The monic Weierstrass polynomial $z^3 -3b_2z + 2b_3$ is non degenerate unless $(c_2M_2)^3 = (c_3M_3)^2$.
\end{proposition}
\noindent \textcolor{blue}{Proof. The Newton polyhedron of $z^3 -3b_2z + 2b_3$ can be 2-dimensional, polygonal with two segments or polygonal with a unique segment. In the two first cases, it is non degenerate. The compact faces of the Newton polyhedron of $z^3 -3b_2z + 2b_3$ lie on a segment if and only if $(M_2)^3 = (M_3)^2$. The condition of $z^3 -3b_2z + 2b_3$ being non degenerate is equivalent to $z^3 -3b_2z + 2b_3 \in \CC[z]$ having a double root. Notice a triple root is not possible because the degree 2 monomial is zero. Assume that $z^3 -3b_2z + 2b_3 =(z-a)^2(z-b)$ for some $a, b \in \CC^\ast$. Equivalently, we have $2a+b=0$, and therefor $c_2=(\frac{b}{2})^2}$ and $c_3=(\frac{-b}{2})^3$.}
\section{Jung´s method for power series of multiplicity three }
@ -585,8 +588,8 @@ N^\Delta_i& 25& {\bf 26}& {\bf 52}& {\bf 78}& {\bf 104}& {\bf 130}& {\bf 156}& {
\hline
\hline
\nu_i&2&3&5&7&9&11&13&6&11\\
\chi()&-11&1&-1&0&0&0&-2&1&-1\\
E_i^2& -13& -2& -4& -2& -2& -2& -1&-2&-1\\
\chi()&-13&1&-1&0&0&0&-2&1&-1\\
E_i^2& -11& -2& -4& -2& -2& -2& -1&-2&-1\\
\hline
\end{array}$$
@ -967,8 +970,8 @@ In this case, the Newton polyhedra of $F$ contains a unique two dimensional comp
\item If $M_1^3 = M_2^2$, the point corresponding to the monomial $M_1z$ lies in the unique dimensional compact face of the Newton polyhedron of $F$. The face equation is equal to
$$z^3 + c_1z + c_2.$$
Therefore, $F$ is non degenerate with respect to its Newton polyhedron if and only if $3(c_1)^2 \ne 2(c_2)^3$.
$$z^3 -3 c_1z + 2 c_2.$$
Therefore, $F$ is non degenerate with respect to its Newton polyhedron if and only if \textcolor{blue}{$(c_1)^3 \ne (c_2)^2$}.
\begin{tikzpicture}
@ -1137,7 +1140,7 @@ $$J_{\check \tau_0}=\frac{2}{\nu_1(3m_2s + 2\nu_1+m_2)}$$
We remark that this case it is equivalent to $f=z(z^2 - x^{m_2})$.
\subsubsection{Third case $f= z^3 - x^{m_1}z^2 - x^{m_3}$ and $\omega = x^{\nu_1-1}dxdz$ such that $m_3 > m_1$}
\subsubsection{Third case $f= z^3 - x^{m_1}z^2 - x^{m_3}$ and $\omega = x^{\nu_1-1}dxdz$ such that $m_3 > 3 m_1$}
Newton polyhedra has five compact faces: the 0-dimensional faces $\tau_0$ corresponding to the monomial $z^3$, $\tau_1$ corresponding to the monomial $x^{m_1}z^2$, and $\tau_2$ corresponding to the monomial $ x^{m_3}$, and the 1-dimensional faces $\sigma_1$ contained in the hyperplanes
$$p + m_1r = 3m_1$$
and $\sigma_2$ contained in the hyperplanes
@ -1179,9 +1182,9 @@ $$J_{\check \tau_0}=\frac{m_1}{\nu_1(3m_1s + \nu_1+m_1)}$$
Similarly we have
$$\zeta_f (t) = $$
We remark that this case is equivalent to $f=(z-x^{m_1})(z^2 - x^{m_3-m_2})$
$$\zeta_f (t) = \frac{(t^{3m_1}-1)(t^{m_3}-1) }{(t-1)(t^3-1)} $$
$$\zeta_f (t) = \frac{(t^{3m_1}-1)(t^{2m_3}-1) }{(t-1)(t^3-1)(t^{m_3}-1)}$$
We remark that this case is equivalent to $f=(z-x^{m_1})(z^2 - x^{m_3-m_1})$
\subsection{Trinomial forms for strata $E^\circ_{i,j}$}
@ -1286,7 +1289,7 @@ $$Z(F_9, \omega_9)_{o_9}=Z(F_{9,31}, \omega_{9,31})_{o_{9,31}}= Z(F_{17}, \omega
$$f_7=f_{15}=z^3 - x^{78}, \quad f_{7,33}=f_{15,34}=z^3 - x^{52}z + x^{78}y,$$
and $$\omega_7=\omega_{15}=\omega_{7,33}=\omega_{15,34}= x^{13}yzdxdydz,$$
From the formulas given in the previous section; we have
$$Z(F_7, \omega_7)_{o_7}=Z(F_{7,33}, \omega_{7,33})_{o_{7,33}}= Z(F_{15}, \omega_{15})_{o_{15}}=Z(F_{15,34}, \omega_{15,34})_{o_{15,34}} = \frac{1}{26(s^2 +3/2 s + 1/2}$$
$$Z(F_7, \omega_7)_{o_7}=Z(F_{7,33}, \omega_{7,33})_{o_{7,33}}= Z(F_{15}, \omega_{15})_{o_{15}}=Z(F_{15,34}, \omega_{15,34})_{o_{15,34}} = \frac{1}{26(s^2 +3/2 s + 1/2}.$$
\end{example}
\subsection{Second reduction: Recursive elimination of exceptional $(-1)$-curve not intersecting the strict transform of the discrimiant}
@ -1296,6 +1299,9 @@ $$Z(F_7, \omega_7)_{o_7}=Z(F_{7,33}, \omega_{7,33})_{o_{7,33}}= Z(F_{15}, \omega
\begin{proposition}\label{red2}
\end{proposition}
\begin{example} The Proposition \ref{red2} can be applied twice to eliminate the curves $E_9$ and $E_8$. First, $E_9$ is an exceptional $(-1)$-curve that does not intersect the strict transform of the discriminant, and $E_8$ is an exceptional with $E_8^2=-2$ After blowing down $E_9$, the exceptional curve $E_8$ becomes a $(-1)$-curve. It does not intersect the strict transform of the discriminant.
\end{example}
\subsection{Third reduction: $\QQ$-resolution of the discriminant}