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%-------------------------------------------------------
\begin{document}
\title{On germs of hypersurfaces with multiplicity 3}
\author{EAB, MGV, ELC}
\maketitle
We propose the following general method to study germs of multiplicity 3 singular points of hypersurfaces. Firstly, consider an equation of the germ given by a monic polynomial $F = z^3 + a_1z + a_2z + a_3 \in \CC[[{\bf x}]][z]$ of degree 3 with respect to $z$. By means of a Tschirnhaus transformation, reduce $F$ to a monic equation $z^3 + b_2z + b_3$ that does not have terms of degree 2 with respect to $z$. Secondly, consider an embedded resolution $\pi: X \rightarrow \CC^d$ of the reduced divisor given by the product $b_2b_3$ of the coefficients of degree one and zero with respect to $z$. Consider the usual stratification of the pair $(X, (\pi \circ F)^{-1} \{F=0\})$ defined in terms of decomposition $\cup_{j \in J} E_j$ of $(\pi \circ F)^{-1} \{F=0\}$ into irreducible components. Pulling back $F$ under the map $\pi \times id$, we obtain for each stratum $E_I^\circ$ a quasi-trinomial local expression for $F$.
The local quasi-trinomial expression will be non degenerate with respect to its Newton polyhedron unless the monomials $M_1$, $M_2$ and their coefficients $c_1$ and $c_2$ satisfy a precise condition.
\section{The quasi-trimonial case}
In this section we consider the case of
$$F = z^3 + M_1(c_1+A)z + M_2(c_2 + M_2) \in \CC[[{\bf x}]][z]$$
where $M_1={\bf x}^{m_1}$ and $M_2={\bf x}^{m_2}$ are monomials, $c_1,c_2 \in \CC^\ast$ and $A, B \in ({\bf x})\CC[[{\bf x}]]$
We distinguish the following two cases
\begin{itemize}
\item Assume that the monomials $M_1$ and $M2$ are not related by the divisibility relation, i.e., neither $M_1 \nmid M_2$ nor $M_1 \nmid M_2$.
In this case, the Newton polyhedra of $F$ contains a unique two dimensional compact face determined by the monomials $z^3$, $M_1z$ and $M_2$. Moreover, $F$ is non degenerate with respect to its Newton polyhedra.
\
\begin{tikzpicture}
\fill [yellow] (1,0) rectangle (3,-2);
\fill [yellow] (-2,0) rectangle (1,2);
\draw[gray, thick] (-2,-2) -- (-2,2);
\draw[gray, thick] (-2,-2) -- (3,-2);
\draw[gray, thick] (-2,0) -- (3,0);
\draw[gray, thick] (1,-2) -- (1,2);
\filldraw[black] (-2,-2) circle (2pt) node[anchor=west]{};
\filldraw[black] (1,0) circle (2pt) node[anchor=south west]{$m_2$};
\filldraw[black] (2,-1) circle (2pt) node[anchor=south west]{$m_1$};
\draw[gray, thick] (6,0,0) -- (6,0,7);
\draw[gray, thick] (6,0,0) -- (6,4,0);
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\draw[gray, thick] (6,0,4) -- (13,0,4);
\draw[gray, thick] (11,0,0) -- (11,0,7);
\draw[gray, thick] (6,3,0) -- (11,0,4);
\draw[gray, thick] (9,1,6) -- (11,0,4);
\draw[gray, thick] (6,3,0) -- (9,1,6);
\draw[gray, dashed] (9,0,0) -- (9,0,6);
\draw[gray, dashed] (6,0,6) -- (9,0,6);
\draw[gray, dashed] (9,1,0) -- (9,1,6);
\draw[gray, dashed] (6,1,6) -- (9,1,6);
\draw[gray, dashed] (9,0,6) -- (9,1,6);
\draw[gray, dashed] (6,1,0) -- (6,1,6);
\draw[gray, dashed] (6,1,0) -- (9,1,0);
\draw[gray, dashed] (6,0,6) -- (6,1,6);
\draw[gray, dashed] (9,0,0) -- (9,1,0);
\filldraw[black] (6,3,0) circle (2pt) node[anchor=south west]{$z^3$};
\filldraw[black] (11,0,4) circle (2pt) node[anchor=north west]{$M_2$};
\filldraw[black] (9,0,6) circle (2pt) node[anchor=north]{$M_1$};
\filldraw[black] (9,1,6) circle (2pt) node[anchor=south west]{$M_1z$};
\end{tikzpicture}
\
\item Assume that the monomials $M_1$ and $M_2$ are related by the divisibility relation, i.e., either $M_1 | M_2$ or $M_1 | M_2$. In this case the Newton polyhedron of $F$ is a polygonal path. We distinguish the following cases.
\
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\fill [yellow] (9,0) rectangle (11,2);
\draw[gray, thick] (6,-2) -- (6,2);
\draw[gray, thick] (6,-2) -- (11,-2);
\draw[gray, thick] (6,0) -- (11,0);
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\filldraw[black] (6,-2) circle (2pt) node[anchor=west]{};
\filldraw[black] (9,0) circle (2pt) node[anchor=north west]{$m_2$};
\filldraw[black] (9.5, 1.3) circle (2pt) node[anchor=north west]{$m_1$};
\end{tikzpicture}
\
\begin{enumerate}
\item If the point $3m_1$ does not belong to the set $2 m_2 + (\RR_{\geq 0})^2$, the Newton polyhedron of $F$ consists of two segments. The first one is determined by the monomials $z^3$ and $M_1z$ and the second is determined by the monomials $M_1z$ and $M_2$. it follows that $F$ is non degenerate with respect to its newton polyhedron.
\
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\draw[gray, thick] (6,0,4) -- (13,0,4);
\draw[gray, thick] (11,0,0) -- (11,0,7);
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\draw[gray, dashed] (8,0,0) -- (8,0,3);
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\draw[gray, dashed] (6,0,3) -- (8,0,3);
\draw[gray, dashed] (28/3,0,8/3) -- (28/3,0,4);
\draw[gray, dashed] (28/3,0,8/3) -- (11,0,8/3);
\filldraw[black] (6,3,0) circle (2pt) node[anchor=south west]{$z^3$};
\filldraw[black] (11,0,4) circle (2pt) node[anchor=north west]{$M_2$};
\filldraw[black] (8,0,3) circle (2pt) node[anchor=north]{$M_1$};
\filldraw[black] (8,1,3) circle (2pt) node[anchor=south west]{$M_1z$};
\end{tikzpicture}
\
\item If the point $3m_1$ belongs to the set $2 m_2 + (\RR_{\geq 0})^2$, the Newton polyhedron of $F$ consists of one segment with vertex corresponding to the monomials $z^3$ and $M_2$. We distinguish to further cases.
\begin{enumerate}
\item If $M_1^3 \ne M_2^2$, the point corresponding to the monomial $M_1z$ is an interior point of the Newton polyhedron of $F$. It follows that $F$ is non degenerate with respect to its Newton polyhedron.
\begin{tikzpicture}
\draw[gray, thick] (6,0,0) -- (6,0,7);
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\filldraw[black] (6,0,0) circle (2pt) node[anchor=west]{};
\draw[gray, thick] (6,0,4) -- (13,0,4);
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\draw[gray, thick] (6,3,0) -- (11,0,4);
\draw[gray, thick] (6,3,0) -- (10.5,1,3) -- (12.75, 0, 4.5);
\draw[gray, thick] (6,3,0) -- (6,3,7);
\draw[gray, thick] (6,3,0) -- (13,3,0);
\draw[gray, dashed] (6,1,3) -- (6,0,3);
\draw[gray, dashed] (10.5,0,0) -- (10.5,0,3);
\draw[gray, dashed] (6,1,3) -- (6,1,0);
\draw[gray, dashed] (10.5,1,0) -- (10.5,0,0);
\draw[gray, dashed] (6,1,0) -- (10.5,1,0);
\draw[gray, dashed] (10.5,1,0) -- (10.5,1,3);
\draw[gray, dashed] (10.5,1,3) -- (6,1,3);
\draw[gray, dashed] (10.5,1,3) -- (10.5,0,3);
\draw[gray, dashed] (10.5,0,3) -- (6,0,3);
\draw[gray, dashed] (28/3,0,8/3) -- (28/3,0,4);
\draw[gray, dashed] (28/3,0,8/3) -- (11,0,8/3);
\filldraw[black] (6,3,0) circle (2pt) node[anchor=south west]{$z^3$};
\filldraw[black] (11,0,4) circle (2pt) node[anchor=north west]{$M_2$};
\filldraw[black] (10.5,0,3) circle (2pt) node[anchor=north]{};
\filldraw[black] (10.5,1,3) circle (2pt) node[anchor=south west]{$M_1z$};
\filldraw[black] (12.75,0,4.5) circle (2pt) node[anchor=north]{};
\end{tikzpicture}
\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 $3c_1 \ne 2c_2$.
\begin{tikzpicture}
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\filldraw[black] (6,0,0) circle (2pt) node[anchor=west]{};
\draw[gray, thick] (6,0,4) -- (13,0,4);
\draw[gray, thick] (11,0,0) -- (11,0,7);
\draw[gray, thick] (6,3,0) -- (11,0,4);
\draw[gray, thick] (6,3,0) -- (6,3,7);
\draw[gray, thick] (6,3,0) -- (13,3,0);
\draw[gray, dashed] (6,1,3) -- (6,0,8/3);
\draw[gray, dashed] (28/3,0,0) -- (28/3,0,8/3);
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\draw[gray, dashed] (6,1,0) -- (28/3,1,0);
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\draw[gray, dashed] (28/3,1,8/3) -- (6,1,8/3);
\draw[gray, dashed] (28/3,1,8/3) -- (28/3,0,8/3);
\draw[gray, dashed] (28/3,0,8/3) -- (6,0,8/3);
\draw[gray, dashed] (28/3,0,8/3) -- (28/3,0,4);
\draw[gray, dashed] (28/3,0,8/3) -- (11,0,8/3);
\filldraw[black] (6,3,0) circle (2pt) node[anchor=south west]{$z^3$};
\filldraw[black] (11,0,4) circle (2pt) node[anchor=north west]{$M_2$};
\filldraw[black] (28/3,0,8/3) circle (2pt) node[anchor=north]{};
\filldraw[black] (28/3,1,8/3) circle (2pt) node[anchor=south west]{$M_1z$};
\end{tikzpicture}
\end{enumerate}
\end{enumerate}
\end{itemize}
\bibliographystyle{amsplain}
\bibliography{biblio}
\end{document}

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