Classification of simple 0-dimensional isolated complete intersection
singularities  Thuy Huong Pham, Gerhard Pfister, and Gert-Martin Greuel

## Abstract

The aim of this article is the classification of simple 0-dimensional isolated
complete intersection singularities in positive characteristic. As usual, a
singularity is called simple or 0-modal if there are only finitely many
isomorphism classes of singularities into which the given singularity can
deform. The notion of simpleness was introduced by V. I. Arnold and the
classification of low modality singularities has become a fundamental task in
singularity theory. Simple complex analytic isolated complete intersection
singularities (ICIS) were classified by M. Giusti. However, the classification
in positive characteristic requires different methods and is much more
involved. The final result is nevertheless similar to the classification in
characteristic 0 with some additional normal forms in low characteristic.

The theoretical results in this paper mainly concern families of ICIS that are
formal in the fiber and algebraic in the base (formal deformation theory is
not sufficient). In particular, we give a definition of modality in this
situation and prove its semicontinuity.

##  1 A criterion for non-simpleness

In this section we establish the notations, prove some preliminary results and
show a general criterion for non-simplicity. This criterion (Theorem 1.15)
allows us to significantly reduce the number of cases to be analyzed. The
classification then consists of two main tasks: a) to further reduce the list
of potentially simple singularities and b) to prove that the singularities of
the list are indeed simple (Theorem 2.6 and 3.4). In step a), the
semicontinuity of the modality is an important tool. In both steps we have to
consider small characteristics separately (where characteristic 2 is treated
in a separate section) and for the computations we make substantial use of the
system singular [DGPS15].

Let $K$ be an algebraically closed field of arbitrary characteristic $p\geq 0$
and denote by

$R=K[[x_{1},\ldots,x_{n}]]$

the formal power series ring in $n$ variables ${\bf x}=(x_{1},\ldots,x_{n})$
with maximal ideal $\mathfrak{m}=\langle x_{1},\ldots,x_{n}\rangle$. For
$f=\sum a_{\alpha}{\bf x}^{\alpha}\in R$, we denote by $f({\bf 0})$ the
coefficient $a_{0}$ of $f$ and by $supp(f)=\\{\alpha\mid a_{\alpha}\neq 0\\}$
the support of $f$. For $I\subset R$ a proper ideal, denote by
$\operatorname{mng}(I)=\dim_{K}I/\mathfrak{m}I$ the minimal number of
generators of $I$. Let $GL(m,R)$ the group of $m\times m$ invertible matrices
over $R$ and $Aut(R)$ the group of $K$-algebra automorphisms of $R$.

######  Definition 1.1.

  1. 1.

A polynomial $f\in K[x_{1},\ldots,x_{n}]$ is called quasi-homogeneous of type
$(d;{\bf a})=(d;a_{1},\ldots,a_{n})\in\mathbb{Z}_{+}\times\mathbb{Z}_{+}^{n}$
if $f$ is a $K$-linear combination of monomials ${\bf
x}^{\alpha}=x_{1}^{\alpha_{1}}\ldots x_{n}^{\alpha_{n}}$ satisfying
$\langle{\bf a},{\bf\alpha}\rangle:=\sum\limits_{i=1}^{n}a_{i}\alpha_{i}=d.$
Denote by $G_{d}^{\bf a}$ the vector space of all quasi-homogeneous
polynomials of type $(d;{\bf a})$. Then

$K[x_{1},\ldots,x_{n}]=\mathop{\oplus}\limits_{d\geq 0}G_{d}^{\bf a}.$

We set $G_{d}^{\bf a}=\\{0\\}$ if $d<0.$

  2. 2.

An element $f=(f_{1},\ldots,f_{m})\in K[x_{1},\ldots,x_{n}]^{m}$ is called
quasi-homogeneous of type $({\bf d};{\bf
a})=(d_{1},\ldots,d_{m};a_{1},\ldots,a_{n})\in\mathbb{Z}_{+}^{m}\times\mathbb{Z}_{+}^{n}$
if

$f_{i}\in G_{d_{i}}^{\bf a},\hskip 10.0pt\forall i=1,\ldots,m.$

$a_{i}$ is called the weight of $x_{i}$ and $d_{i}$ the (weighted) degree of
$f_{i}$. Set

$G_{\nu}^{{\bf d},{\bf a}}:=\\{(f_{1},\ldots,f_{m})\in
K[x_{1},\ldots,x_{n}]^{m}\mid f_{i}\in G^{\bf a}_{d_{i}+\nu},i=1,\ldots,m\\},$

which is a finite dimensional $K$-vector space and
$K[x_{1},\ldots,x_{n}]^{m}=\oplus_{\nu\in\mathbb{Z}}G^{\bf a}_{d_{i}+\nu}$.

######  Definition 1.2.

Let $({\bf d};{\bf
a})=(d_{1},\ldots,d_{m};a_{1},\ldots,a_{n})\in\mathbb{Z}_{+}^{m}\times\mathbb{Z}_{+}^{n}$.

  1. 1.

For $f\in R\smallsetminus\\{0\\}$ we define

$v_{\bf a}(f):=\inf\\{\langle{\bf a},{\bf\alpha}\rangle\mid\alpha\in
supp(f)\\}$

and call it the ${\bf a}$-order of $f$, or just the order if a=$(1,...,1)$,
denoted by $\operatorname{ord}_{\bf a}(f)$ or $\operatorname{ord}(f)$. If
$f=0$ we set $v_{\bf a}(f)=\infty$.

  2. 2.

For $f=(f_{1},\ldots,f_{m})\in R^{m}$ we set

$v_{{\bf d},{\bf a}}(f):=\inf\limits_{i=1,\ldots,m}\\{v_{\bf
a}(f_{i})-d_{i}\\}.$

If $f\neq 0$ then $v_{{\bf d},{\bf a}}(f)=\inf\\{\nu\mid
f=\sum\limits_{\nu}f^{(\nu)},\hskip 5.0pt0\neq f^{(\nu)}\in G^{{\bf d},{\bf
a}}_{\nu}\\}.$

######  Definition 1.3.

(1) Let $f=(f_{1},\ldots,f_{m})\in R^{m}$. Denote by

$T^{1}(f):=R^{m}{\Big{/}}\left(\left\langle
f_{1},\ldots,f_{m}\right\rangle\cdot R^{m}+\left\langle\frac{\partial
f}{\partial x_{1}},\ldots,\frac{\partial f}{\partial x_{n}}\right\rangle\cdot
R\right),$

$\frac{\partial f}{\partial x_{i}}=(\frac{\partial f_{1}}{\partial
x_{i}},\ldots,\frac{\partial f_{m}}{\partial x_{i}})\in R^{m}$, the Tjurina
module of $f$ and

$\tau(f):=\dim_{K}T^{1}(f)$

the Tjurina number of $f$.

(2) Two elements $f,g\in R^{m}$ are called left equivalent, if they belong to
the same orbit of the left group

$G:=GL(m,R)\rtimes Aut(R),$

acting on on $R^{m}$ by

$(U,\phi,f)\mapsto U\cdot\phi(f),$

with $U\in GL(m,R),\phi\in Aut(R)$ and

$\phi(f)=[f_{1}\left(\phi({\bf x}))\ldots f_{m}(\phi({\bf x})\right)]^{t},$

where $\phi({\bf x})=\left(\phi(x_{1}),\ldots,\phi(x_{n})\right)$.

######  Definition 1.4.

An ideal $I\subset R$ defines a complete intersection if $I$ can be gnerated
by $f_{1},\ldots,f_{m}$ with $f_{i}\in\mathfrak{m}$ for all $i$ such that
$f_{i}$ is a non-zero divisor of $R/\langle f_{1},...,f_{i-1}\rangle$ for
$i=1,...,m$. Then $\operatorname{mng}(I)=m$ and $\dim R/I=n-m\geq 0$.

Let $f=(f_{1},\ldots,f_{m})$ and $g=(g_{1},\ldots,g_{m})\in R^{m}$ be such
that $I=\langle f_{1},\ldots,f_{m}\rangle$ and $J=\langle
g_{1},\ldots,g_{m}\rangle$ define complete intersections with
$\operatorname{mng}(I)=\operatorname{mng}(J)=m$. Then $f$ is called contact
equivalent to $g$, denoted $f\sim g$, if $g\in Gf$, i.e. there is a matrix
$U\in GL(m,R)$ and an automorphism $\phi\in Aut(R)$ such that

$\left[\begin{matrix}g_{1}\\\ \vdots\\\
g_{m}\end{matrix}\right]=U\phi\left(\left[\begin{matrix}f_{1}\\\ \vdots\\\
f_{m}\end{matrix}\right]\right).$

For simplicity of notations, we also write $g=U\phi(f).$

Note that by [P16, Proposition 3.3.4], $f\sim g$ if and only if
$I\mathop{\sim}\limits^{c}J$, i.e. there is an automorphism $\phi\in Aut(R)$
such that $\phi(I)=J$. It follows easily that if $f\sim g$ then
$\tau(f)=\tau(g)$.

######  Definition 1.5.

Let $f=(f_{1},\ldots,f_{m})\in R^{m}$ be such that $I=\langle
f_{1},\ldots,f_{m}\rangle$ defines a complete intersection with
$\operatorname{mng}(I)=m$. We call $f$ (or $I$) an isolated complete
intersection singularity (ICIS) if there is a positive integer $k$ such that

$\mathfrak{m}^{k}\subset I+I_{m}(Jac(f)),$

where $Jac(f)=\left[\frac{\partial f_{i}}{\partial x_{j}}\right]_{ij}$ is the
$m\times n$ Jacobian matrix of $f$ and $I_{m}(Jac(f))$ denotes the ideal
generated by all $m\times m$ minors of $Jac(f)$. We set

$I_{m,n}:=\\{f=(f_{1},\ldots,f_{m})\in R^{m}\mid f\text{ is an ICIS with
}\operatorname{mng}(I)=m\\}.$

Note that any ICIS $f$ is finitely determined (see Proposition 1.11), in
particular, the $f_{i}$ can be chosen as polynomials (up to an automorphism of
$R$).

######  Lemma 1.6.

Let $k\in\mathbb{N}$ and $D=\langle h_{1},\ldots,h_{k}\rangle\subset R^{m},$ a
submodule, where $h_{i}\in G^{{\bf d},{\bf a}}_{\nu_{i}}$, $i=1,\ldots,k$ are
quasi homogeneous. Then we have an isomorphism of $K$-vector spaces

$R^{m}/D\cong\mathop{\prod}\limits_{\nu\in\mathbb{Z}}\left(G^{{\bf d},{\bf
a}}_{\nu}/D\cap G^{{\bf d},{\bf a}}_{\nu}\right).$

We omit the elementary proof.

######  Lemma 1.7.

Let $f\in K[x_{1},\ldots,x_{n}]^{m}\cap I_{m,n}$ be quasi homogeneous of type
$({\bf d};{\bf a})=(d_{1},\ldots,d_{m};a_{1},\ldots,a_{n})$. Then, as
$K$-vector spaces,

$T^{1}(f)\cong\mathop{\oplus}\limits_{\nu\in\mathbb{Z}}T^{1}_{\nu}(f),$

where $T^{1}_{\nu}(f)=G^{{\bf d},{\bf a}}_{\nu}\Big{/}\left(\left\langle
f_{1},\ldots,f_{m}\right\rangle\cdot R^{m}+\left\langle\frac{\partial
f}{\partial x_{1}},\ldots,\frac{\partial f}{\partial
x_{n}}\right\rangle\right)\cap G^{{\bf d},{\bf a}}_{\nu}$.

###### Proof.

Note that if $f\in K[x_{1},\ldots,x_{n}]^{m}\cap I_{m,n}$ is quasi homogeneous
of type $({\bf d};{\bf a})$ then for all $i=1,\ldots,n$, $j=1,\ldots,m$,
$\frac{\partial f_{j}}{\partial x_{i}}$ is quasi homogeneous of type
$(d_{j}-a_{i};a)$ (or zero). Therefore, $f_{j}e_{k}\in G^{{\bf d},{\bf
a}}_{d_{j}-d_{k}}$, where $\\{e_{k}\\}_{k=1,\ldots,m}$ is the canonical basis
of $R^{m}$, and $\frac{\partial f_{j}}{\partial x_{i}}\in G^{{\bf d},{\bf
a}}_{-a_{i}}$, i.e. $D:=\left\langle f_{1},\ldots,f_{m}\right\rangle\cdot
R^{m}+\left\langle\frac{\partial f}{\partial x_{1}},\ldots,\frac{\partial
f}{\partial x_{n}}\right\rangle$ is a quasi homogeneous submodule of $R^{m}.$
By Lemma 1.6,

$R^{m}/D\cong\mathop{\prod}\limits_{\nu\in\mathbb{Z}}\left(G^{{\bf d},{\bf
a}}_{\nu}/D\cap G^{{\bf d},{\bf a}}_{\nu}\right).$

Since $\tau(f)<\infty,$ we have
$R^{m}/D\cong\mathop{\oplus}\limits_{\nu\in\mathbb{Z}}\left(G^{{\bf d},{\bf
a}}_{\nu}/D\cap G^{{\bf d},{\bf a}}_{\nu}\right).$ ∎

The following proposition is quite useful for the classification of
singularities.

######  Proposition 1.8.

Let $f=(f_{1},\ldots,f_{m})\in K[x_{1},\ldots,x_{n}]^{m}\cap I_{m,n}$ be quasi
homogeneous of type $({\bf d};{\bf
a})\in\mathbb{Z}_{+}^{m}\times\mathbb{Z}_{+}^{n}$. Let
$g=(g_{1},\ldots,g_{m})\in R^{m}$ be such that

$v_{{\bf d},{\bf a}}(g)>\beta=\sup(0,\alpha),$

where $\alpha=\sup\\{i\mid T^{1}_{i}(f)\neq 0\\}$. Then $f+g\sim f$.

###### Proof.

In the complex analytic setting a proof is given in [Giu83, Proposition 1].
Using the finite determinacy of an ICIS (see Proposition 1.11), the proof
works as well for formal power series in any characteristic. We omit the
details. ∎

######  Definition 1.9.

Let $f=(f_{1},\ldots,f_{m})\in I_{m,n}$. The tangent image of the action of
$G$ on $R^{m}$ is defined as the module

$\tilde{T}_{f}(Gf):=\langle f_{1},\ldots,f_{m}\rangle
R^{m}+\mathfrak{m}\left\langle\frac{\partial f}{\partial
x_{1}},\ldots,\frac{\partial f}{\partial
x_{n}}\right\rangle\subset\mathfrak{m}R^{m},$

which has finite $K$-codimension in $R^{m}$ for an ICIS. It is easy to see
that

$T^{1,sec}(f):=\mathfrak{m}R^{m}/\tilde{T}_{f}(Gf)$

has finite $K$-dimension iff this holds for $T^{1}(f)=R^{m}/\langle
f_{1},\ldots,f_{m}\rangle R^{m}+\langle\frac{\partial f}{\partial
x_{1}},\ldots,\frac{\partial f}{\partial x_{n}}\rangle$ ([GP18, Proposition
4.2]). We denote by

$\tau^{sec}(f):=\dim_{K}T^{1,sec}(f).$

######  Remark 1.10.

(1) The tangent image was introduced in [GP18] (in a more general context) to
replace the tangent space to the orbit $Gf$ in positive characteristic. In
characteristic zero $\tilde{T}_{f}(Gf)$ coincides with the tangent space
$T_{f}(Gf)$, see [GP18, Lemma 2.8]. In general we have
$\tilde{T}_{f}(Gf)\subset T_{f}(Gf)$ and the inclusion can be strict in
positive characteristic (see Example 2.9 in [GP18]).

It was shown in [GP19, Theorem 1.4 ] that an ICIS $f$ is finitely determined
iff $\tau^{sec}(f)<\infty$ (equivalently $\tau(f)<\infty$).

(2) Let $g_{1},\ldots,g_{d}\in K[[{\bf{x}}]]^{m}$ represent a $K$-basis of
$T^{1}(f)$, resp. of $T^{1,sec}(f)$, then

$F_{\bf t}({\bf x}):=F({\bf x},t_{1},\ldots,t_{d})=f({\bf
x})+\sum\limits_{i=1}^{d}t_{i}g_{i}({\bf x})\in K[[{\bf t},{\bf x}]]$

represents a formally semiuniversal deformation of $f$, resp. a formally
semiuniversal deformation with section of $f$. See [KS72] for the first case.
The latter statement follows because every section can be trivialized (see
[GLS07, Proposition II.2.2]) and then the proof given in [KS72] or [GLS07,
Theorem II.1.16] can be adapted. If the $g_{i}$ represent a system of
generators, then $F_{\bf t}$ represents a formally versal deformation, resp.
with section.

(3) Note that the $g_{i}$ can be chosen as monomials and the $f_{i}$ as
polynomials (since the ICIS $f$ is finitely determined). Thus the
semiuniversal deformation, resp. with section, of $f$ has an algebraic
representative $F_{\bf t}({\bf x})\in K[{\bf t},{\bf x}]$ with $F_{\bf 0}({\bf
x})=f({\bf x})$. $F_{\bf t}({\bf x})$ is called an unfolding at $0\in T$ of
$f$ over $T=K^{d}$.

The algebraic unfolding $F_{\bf t}({\bf x})$ is G-complete or contact
complete, meaning that any unfolding $H_{\bf s}({\bf x})=H({\bf
x},s_{1},...,s_{e})\in K[{\bf s},{\bf x}]$ of $f$ at $s_{0}\in S=K^{e}$ with
$H_{s_{0}}({\bf x})=f({\bf x})$ is an étale pullback of $F$, i.e. there exist
an étale neighbourhood $\varphi:W,w_{0}\to S,s_{0}$ and a morphism
$\psi:W,w_{0}\to T,0$ such that $H({\bf x},\varphi(w))$ is contact-equivalent
to $F({\bf x},\psi(w))$ for all $w\in W$. This is important for the
classification of singularities in the non-analytic case (the proof of [GNg16,
Proposition 2.14] for hypersurfaces can be generalized to an ICIS).

For the classification we need to consider $k$-jets of power series. Denote by

$R^{m}_{k}=R^{m}/\mathfrak{m}^{k+1}R^{m},k\geq 0,$

the $k$-jet space of $R^{m}$ and for $f\in R^{m}$, let $j_{k}(f)$ denote the
$k$-jet of $f$, i.e., the image of $f$ in $R^{m}_{k}$. Consider the group of
$k$-jets

$G_{k}=\\{(j_{k}(U),j_{k}(\phi))\mid U\in GL(m,R),\phi\in Aut(R)\\},$

where $(j_{k}(\phi))(x_{i})=j_{k}(\phi(x_{i}))$, $i=1,\ldots,n.$ Then the
action of the left group $G$ on $R^{m}$ induces the action on the $k$-jet
spaces

$\displaystyle G_{k}\times R^{m}_{k}$ $\displaystyle\to R^{m}_{k}$
$\displaystyle(j_{k}(U),j_{k}(\phi),j_{k}(f))$ $\displaystyle\mapsto
j_{k}(U\phi(f)),$

which is an algebraic action of the algebraic group $G_{k}$ on the affine
space $R^{m}_{k}$.

We recall the finite determinacy and the semicontinuity of the Tjurina number
of a complete intersection in arbitrary characteristic. We consider families
$F_{\bf t}({\bf x})=F({\bf x},t_{1},\ldots,t_{k})\in K[\bf t][[\bf x]]$, which
are polynomial in the parameter ${\bf t}$, and denote for fixed ${\bf t}\in
K^{k}$ by $F_{\bf t}({\bf x})$ also the power series in $K[[{\bf x}]]$.

######  Proposition 1.11.

(1) Let $f\in R^{m}$ be an ICIS. Then $f$ is finitely determined, i.e. there
exists a $k$ such that each $g\in R^{m}$ with $j_{k}(g)=j_{k}(f)$ is contact
equivalent to $f$ ($f$ is then called $k$-determined). Moreover, $f$ is
$(2\tau(f)-\operatorname{ord}(f)+2)$-determined and every deformation of $f$
is $(2\tau(f)+1)$-determined.

(2) Let $F_{\bf t}({\bf x}):=F({\bf x},t_{1},\ldots,t_{k})\in K[\bf t][[\bf
x]]$ s.t. $F_{{\bf t}_{0}}$ is an ICIS in $I_{m,n}$ for some ${\bf t}_{0}\in
K^{k}$. Then there exists a Zariski open set $U\subset K^{k}$ s.t. $F_{\bf
t}\in I_{m,n}$ for all ${\bf t}\in U$. Moreover, for each $\tau\geq 0$ the
sets

$U_{\tau}=\\{{\bf t}\in U\mid\tau(F_{\bf t})\leq\tau\\}\text{ and
}U_{\tau}^{sec}=\\{{\bf t}\in U\mid\tau^{sec}(F_{\bf t})\leq\tau\\}\ $

are open in $K^{k}$. In particular, if $\tau_{min}=\text{min}\\{\tau(F_{\bf
t})\mid{\bf t}\in K^{d}\\}$, then $U_{\tau_{min}}$ is open and dense in
$K^{k}$, and similar for $\tau_{min}^{sec}$.

###### Proof.

(1) follows from [GP18, Theorem 4.6]. (2) follows for $d=1$ from [GP18,
Proposition 3.4] and for an arbitrary Noetherian ring $A$ instead of $K[\bf
t]$ (also for non-closed points) from [GPf21, Proposition 3.4 and Corollary
2.7]. ∎

In our classification we will consider families of $k$-jets of an ICIS. These
are deformations with trivial section which are versal for sufficiently large
$k$. Hence, in the following definition of simple we consider deformations
with section.

######  Definition 1.12.

Let $g_{1},\ldots,g_{d}\in K[{\bf{x}}]^{m}$ be a set of $K$-generators of
$T^{1,sec}(f).$ An element $f\in I_{m,n}$ is called simple if there is a
finite set $\\{h_{1},\ldots,h_{l}\\}\subset I_{m,n}$ such that for

$F_{\bf t}({\bf x}):=F({\bf x},t_{1},\ldots,t_{d})=f({\bf
x})+\sum\limits_{i=1}^{d}t_{i}g_{i}({\bf x}),$

there is a Zariski open neighborhood $U$ of $0$ in $K^{d}$ such that for each
${\bf t}=(t_{1},\ldots,t_{d})\in U$, there is a $j\in\\{1,\ldots,l\\}$ such
that $F_{\bf t}\sim h_{j}.$

######  Remark 1.13.

(1) By Remark 1.10 (2) and (3) the unfolding $F_{\bf t}({\bf x})\in K[{\bf
t},{\bf x}]$ defines a formally versal deformation with section of $f$ which
is contact complete. It follows that $f$ is simple iff for an arbitrary
unfolding $H_{\bf s}({\bf x})=H({\bf x},s_{1},...,s_{e})\in K[{\bf s},{\bf
x}]$ of $f$ at $s_{0}\in S=K^{e}$ there is a Zariski open neighborhood $U$ of
$s_{0}$ in $S$ such that for each ${\bf s}\in U$, there is a
$j\in\\{1,\ldots,l\\}$ such that $H_{\bf s}\sim h_{j}.$

(2) Following [GK90], we call an element $f\in\mathfrak{m}R^{m}$ 0-modular if
for each $k$ (equivalently, for sufficiently large $k$), there is a Zariski
neighborhood $U_{k}$ of $j_{k}(f)$ in $R^{m}_{k}$ such that $U_{k}$ meets only
finitely many $G_{k}$-orbits of $k$-jets. Here, an integer $k$ is sufficiently
large for $f$ with respect to $G$ if there exists a neighbourhood $U$ of
$j_{k}(f)$ in $\mathfrak{m}R^{m}_{k}$ such that every $g\in\mathfrak{m}R^{m}$
with $j_{k}(g)\in U$ is $k$-determined with respect to $G$. If $f$ is an ICIS,
then $k=(2\tau(f)+1)$ is sufficiently large by Proposition 1.11 (1).

(3) More generally, we define the modality of an ICIS $f$ as follows. Let
$U\subset R^{m}_{k}$ an open neighbourhood of $j_{k}(f)$. We set

$\displaystyle U(i)$ $\displaystyle:=$ $\displaystyle\\{g\in U\ |\
\dim_{g}(U\cap G_{k}\cdot g)=i\\},i\geq 0,$ $\displaystyle
G_{k}\text{-}\mathrm{par}(U)$ $\displaystyle:=$ $\displaystyle\max_{i\geq
0}\\{\dim U(i)-i\\},$

and call
$G_{k}\text{-}\mathrm{par}(j_{k}(f)):=\min\\{G\text{-}\mathrm{par}(U)\ |\
U\text{ a neighbourhood of }j_{k}(f)\\}$ the number of $G_{k}$-parameters of
$j_{k}(f)$ (in $R^{m}_{k}$). It can be shown that if $k$ is sufficiently large
then $G_{k}\text{-}\mathrm{par}(j_{k}(f))$ is independent of $k$ and denoted
by $G\text{-}\mathrm{par}(f)$ or $G\text{-}\mathrm{mod}(f)$ and called the
$G$-modality of $f$ (this shown in [GNg16, Definition 2.3] for hypersurfaces;
the proof can be adapted for an ICIS). $f$ is 0-modular iff
$G\text{-}\mathrm{mod}(f)=0$.

######  Proposition 1.14.

(1) An ICIS is simple iff it is 0-modular.
(2) The $G$-modality of an ICIS is semicontinuous. In particular, any
deformation of a simple ICIS is again simple.

###### Proof.

We only sketch the proof. (1) We can modify the proof in [GNg16, Proposition
2.14. and Corollary 2.17] (for hypersurfaces) and use the semi-continuity of
$\tau(f)$ and $\tau^{sec}(f)$ (see Proposition 1.11), to show that an ICIS is
simple iff it is 0-modular.
(2) It follows from Definition 1.12 and the openness of versality that any
deformation of a simple ICIS is again simple. More general, the $G$-modality
of an ICIS is semicontinuous. This can be proved as for hypersurfaces, see
[GNg16, Proposition 2.7]. ∎

The following result is the main general condition for a 0-dimensional ICIS
being not simple.

######  Theorem 1.15.

$f=(f_{1},...,f_{n})\in I_{n,n}$ is not simple if one of the following cases
occurs:
a) $n\geq 2$ and $\operatorname{ord}(f)\geq 3$.
b) $n\geq 3$ and $\operatorname{ord}(f)=2$.

###### Proof.

Set $l=\operatorname{ord}(f)=\min\\{\operatorname{ord}(f_{i})\\}$. Then the
action of $G_{l}$ on $R^{n}_{l}$ induces an algebraic action of the affine
algebraic group $G^{\prime}:=GL(n,K)\times GL(n,K)$ on the affine variety
$X:=\mathfrak{m}^{l}R^{n}/\mathfrak{m}^{l+1}R^{n}$.

Assume by contradiction that $f$ is simple. Since $f$ is 0-modular, there is a
Zariski neighborhood $U_{l}$ of $j_{l}(f)$ in $R^{n}_{l}$ which intersects
only finitely many $G_{l}$-orbits. This implies that there is Zariski
neighborhood $U$ of $j_{l}(f)$ in $X$ which intersects only finitely many
$G^{\prime}$-orbits, say $U=\mathop{\cup}\limits_{i=1}^{s}(U\cap O_{i})$.

We get that there exists an orbit $G^{\prime}h,h\in X,$ such that

$n{n-1+l\choose l}=\dim X=\dim(U\cap X)=\max\limits_{i}\\{\dim(O_{i}\cap
U)\\}\leq\dim G^{\prime}h$

(otherwise $X$ cannot be covered by finitely many orbits).
It is easy to see that the elements $\\{(a^{l}E_{n},(1/a)E_{n})\mid a\in
K^{*}\\}$, with $E_{n}$ the $n\times n$ identity matrix, are in the stabilizer
of $h$ under $G^{\prime}$, hence

$\dim G^{\prime}h\leq\dim G^{\prime}-1=2n^{2}-1.$

a) If $n\geq 2$ and $l=3$ then $\dim X=\frac{n^{2}(n+1)(n+2)}{6}\geq 2n^{2}>$
$2n^{2}-1$. If $l>3$ the l.h.s. gets bigger while the r.h.s. stays constant.
Hence $\dim X>\dim G^{\prime}h$ and $f$ cannot be simple for $n\geq 2$ and
$l\geq 3$.
b) If $n\geq 3$ and $l=2$ then $\dim X=\frac{n^{2}(n+1)}{2}>2n^{2}-1,$ showing
that $f$ is not simple.

∎

For further analysis we need the following splitting lemma in any
characteristic.

######  Definition 1.16.

Let $f\in R$. We denote by

$H(f)=\left[\frac{\partial^{2}f}{\partial x_{i}\partial
x_{j}}(0)\right]_{i,j=1,\ldots,n}\in Mat(n,K)$

the Hessian matrix of $f$ and by

$\operatorname{corank}(f)=n-\operatorname{rank}(H(f))$

the corank of $f$. $f$ is called non-degenerate if corank($f$) = 0, otherwise
it is called degenerate.

######  Lemma 1.17 (Right splitting lemma in characteristic $p\neq 2$).

Let $p\neq 2$ and $f\in\mathfrak{m}^{2}$ be such that
$\operatorname{corank}(f)=k\geq 0.$ Then

$f\mathop{\sim}\limits^{r}g(x_{1},\ldots,x_{k})+x_{k+1}^{2}+\ldots+x_{n}^{2},$

where $g\in\mathfrak{m}^{3}$ is uniquely determined up to right equivalence.
Of course, $x_{i}^{2}+x_{j}^{2}\mathop{\sim}\limits^{r}x_{i}x_{j}$.

###### Proof.

[GNg16, Lemma 3.9]. ∎

######  Lemma 1.18 (Right splitting lemma in characteristic 2).

Let $p=2$ and $f\in\mathfrak{m}^{2}\subset K[[x_{1},\ldots,x_{n}]]$, $n\geq
2$. Then there is an $l$, $0\leq 2l\leq n$, such that

$f\mathop{\sim}\limits^{r}x_{1}x_{2}+x_{3}x_{4}+\ldots+x_{2l-1}x_{2l}+g(x_{2l+1},\ldots,x_{n})$

where $g\in\langle x_{2l+1},\ldots,x_{n}\rangle^{3}\subset
K[[x_{2l+1},\ldots,x_{n}]],$ or $g=x_{2l+1}^{2}+h$ with $h\in\langle
x_{2l+1},\ldots,x_{n}\rangle^{3}\subset K[[x_{2l+1},\ldots,x_{n}]]$ if $2l<n$.
$g$ is uniquely determined up to right equivalence.

###### Proof.

[GNg16, Lemma 3.12]. ∎

##  2 Simple isolated complete intersection singularities in characteristic
$p\neq 2$

Let $f=(f_{1},...,f_{n})\in I_{n,n}$ be an ICIS. If $n=1$, then $f$ defines a
hypersurface singularity $\mathop{\sim}\limits^{c}x^{k+1}$, which is simple of
type $A_{k}$. Hence we assume $n\geq 2$. If $n\geq 3$ then
$\operatorname{ord}(f)=1$ for $f$ simple (by Theorem 1.15) and then $f\sim
g\in I_{n-1,n-1}$ by the implicit function theorem. Thus we may assume $n=2$.
By Theorem 1.15 an ICIS $(f_{1},...,f_{n})\in I_{n,n}$ can be simple only if
$n=2$ and $\operatorname{ord}(f)=2$, which we assume from now on. Let
$R=K[[x,y]]$, $\mathfrak{m}=\langle x,y\rangle$ its maximal ideal, and
$p=\operatorname{char}(K)$.

######  Proposition 2.1.

Let $p\geq 0$. Let $f=(f_{1},f_{2})\in I_{2,2}$, with
$\operatorname{ord}(f)=2$ and $j_{2}(f_{i})$ a non-degenerate quadratic form
for some $i\in\\{1,2\\}$. Then there are $m,n\geq 2$ such that

$f\sim(xy,x^{n}+y^{m})=:F^{n,m}_{n+m-1}.$

###### Proof.

By Lemma 1.17 and 1.18 we may assume that $f_{1}=xy$. We can use $xy$ to kill
in $f_{2}$ all terms divisible by $xy$. We may therefore assume that
$f_{2}=u_{1}x^{n}+u_{2}y^{m}$ with $u_{1}\in K[[x]]$ and $u_{2}\in K[[y]]$
units ($u_{1}=0$ or $u_{2}=0$ is not possible, since $(f_{1},f_{2})$ is a
complete intersection). We multiply $f_{2}$ with $u_{1}^{-1}u_{2}^{-1}$ and
kill with $xy$ the terms of $u_{1}^{-1}y^{m}$ and $u_{2}^{-1}x^{n}$ and obtain
$\alpha x^{n}+\beta y^{m}$ with $\alpha=u_{2}^{-1}(0)$ and
$\beta=u_{1}^{-1}(0)\in K$ both different from $0$. Since $K$ is algebraically
closed there exist the $n$-th root of $\alpha$ and the $m$-th root of $\beta$
and we find an automorphism mapping the ideal $\langle xy,f_{2}\rangle$ to
$\langle xy,x^{n}+y^{m}\rangle$. ∎

From now on in this section, we assume that $\operatorname{char}(K)=p\neq 2$
unless otherwise stated.

######  Lemma 2.2.

Let $f,g\in\mathfrak{m}^{2}$. Assume that $j_{2}(f)$ is degenerate. Then there
exist $s\geq 3$ and $\alpha\in\\{0,1\\}$ such that

$(f,g)\sim\left(x^{2}+\alpha y^{s},\sum\limits_{i\geq
t}a_{i}y^{i}+x\sum\limits_{j\geq q}b_{j}y^{j}\right),$

where $t\geq 2$, $q\geq 1$ and $a_{i},b_{j}\in K$.

###### Proof.

Using Lemma 1.17 we may assume that $f=x^{2}+h(y)$ with $h=0$ or $h=y^{s}\cdot
e$, $s\geq 3$, $e$ a unit. If $h\neq 0$, we can find (since $p\neq 2$) $e_{1}$
with $e_{1}^{2}=e^{-1}$. Using the automorphism $\varphi$ of $K[[x,y]]$
defined by $\varphi(x)=e_{1}x$, $\varphi(y)=y$ we may assume that
$f=x^{2}+y^{s}$. All together we have $f=x^{2}+\alpha y^{s}$ with
$\alpha\in\\{0,1\\}$. The Weierstraß Division Theorem implies that
$g=k\cdot(x^{2}+\alpha y^{s})+\sum\limits_{i\geq
t}a_{i}y^{i}+x\sum\limits_{j\geq q}b_{j}y^{j}$ for suitable $k\in K[[x,y]]$
and $a_{i}$, $b_{j}\in K$. This implies that

$(x^{2}+\alpha y^{s},g)\sim\left(x^{2}+\alpha y^{s},\sum\limits_{i\geq
t}a_{i}y^{i}+x\sum\limits_{j\geq q}b_{j}y^{j}\right).$

∎

######  Lemma 2.3.

Let $f=(f_{1},f_{2})=(x^{2}+\alpha y^{s},\sum\limits_{i\geq
t}a_{i}y^{i}+x\sum\limits_{j\geq q}b_{j}y^{j})$ be an ICIS such that $s\geq
3$, $t\geq 2$, $q\geq 1$ and $\alpha\in\\{0,1\\}$.

  1. 1.

If $a_{i}=0$ for all $i$ and $b_{q}\neq 0$ then $\alpha=1$ and
$f\sim(x^{2}+y^{s},xy^{q})$.

  2. 2.

If $b_{j}=0$ for all $j$ and $a_{t}\neq 0$ then $f\sim(x^{2}+\alpha
y^{s},y^{t})$. If $\alpha=0$ or $t\leq s$ then $f\sim(x^{2},y^{t})$.

Assume now that $a_{t}b_{q}\neq 0$.

  3. 3.

If $t\leq q$ then $f\sim(x^{2}+\alpha y^{s},y^{t})$. If additionally
$\alpha=0$ or $t\leq s$ then $f\sim(x^{2},y^{t})$.

  4. 4.

If $t>q$ and $\alpha=0$ then $f\sim(x^{2},y^{t}+xy^{q})$.

  5. 5.

Let $t>q$ and $\alpha=1$.
If $2t-2q-s\neq 0$ 111Note that for $f=(x^{2}+y^{4},y^{5}+axy^{3})$ we have
$2t-2q-s=0$ and the parameter $a$ cannot be avoided by the proof of
Proposition 2.4. and ($p=0$ or $p\nmid 2t-2q-s$) then
$f\sim(x^{2}+y^{s},y^{t}+exy^{q})$, $e\in K[[y]]$ with $e(0)=1$.
Otherwise $f\sim(x^{2}+y^{s},y^{t}+exy^{q})$ for a suitable unit $e\in
K[[y]]$.

  6. 6.

If $t=q+1$ and $p\nmid t$ then $f\sim(x^{2}+\alpha y^{s},y^{t})$.

###### Proof.

(1): $f_{2}=x\sum\limits_{j\geq q}b_{j}y^{j}=uxy^{q}$, $u$ a unit. We get
$\langle f_{1},f_{2}\rangle=\langle x^{2}+\alpha y^{s},xy^{q}\rangle$. Since
$f$ is an ICIS $\alpha$ must be $1$.

(2): $f_{2}=\sum\limits_{i\geq t}a_{i}y^{i}=uy^{t},u$ a unit. This implies
$\langle f_{1},f_{2}\rangle=\langle x^{2}+\alpha y^{s},y^{t}\rangle$. If
$\alpha=0$ or $t\leq s$ then obviously $\langle x^{2}+\alpha
y^{s},y^{t}\rangle=\langle x^{2},y^{t}\rangle$.

(3):If $t\leq q$ then $\sum\limits_{i\geq t}a_{i}y^{i}+x\sum\limits_{j\geq
q}b_{j}y^{j}=y^{t}\cdot unit$, implying $\langle f_{1},f_{2}\rangle=\langle
x^{2}+\alpha y^{s},y^{t}\rangle$. If $\alpha=0$ or $t\leq s$ then $\langle
x^{2}+\alpha y^{s},y^{t}\rangle=\langle x^{2},y^{t}\rangle$.

(4): Multiplying $\sum\limits_{i\geq t}a_{i}y^{i}+x\sum\limits_{j\geq
q}b_{j}y^{j}$ by the inverse of $\sum\limits_{i\geq t}a_{i}y^{i-t}$ we may
assume that $f=(x^{2},y^{t}+e\cdot xy^{q})$ with $e=\sum\limits_{j\geq
q}b_{j}y^{j-q}$ a unit. Consider the automorphism $\varphi$ of $K[[x,y]]$
defined by $\varphi(x)=e^{-1}x$ and $\varphi(y)=y$ then
$\varphi(f)=(e^{-2}x^{2},y^{t}+xy^{q})\sim(x^{2},y^{t}+xy^{q}).$

(5): Multiplying $\sum\limits_{i\geq t}a_{i}y^{i}+x\sum\limits_{j\geq
q}b_{j}y^{j}$ by the inverse of $\sum\limits_{i\geq t}a_{i}y^{i-t}$ we may
assume that $f=(x^{2}+y^{s},y^{t}+e\cdot xy^{q})$ with $e=\sum\limits_{j\geq
q}b_{j}y^{j-q}$ a unit. Consider the automorphism $\varphi$ of $K[[x,y]]$
defined by $\varphi(x)=ax$ and $\varphi(y)=by$ with units $a,b\in K[[x,y]]$.
Then $\varphi(f)=(a^{2}x^{2}+b^{s}y^{s},b^{t}y^{t}+\varphi(e)ab^{q}xy^{q})$.
We have to choose $a$ and $b$ such that

  * •

$a^{2}=b^{s}$,

  * •

$b^{t}=e(0)ab^{q}$.

We get the equation

$b^{t-q-\frac{s}{2}}=e(0),$

which can be solved if $p\nmid 2t-2q-s\neq 0$.

(6): If $\alpha=0$, using (4) we may assume that $f=(x^{2},y^{t}+xy^{t-1})$.
Now $y^{t}+xy^{t-1}=(y+\frac{1}{2}x)^{t}+x^{2}h$ for a suitable $h\in
K[[x,y]]$. This implies that $\langle x^{2},y^{t}+xy^{t-1}\rangle=\langle
x^{2},(y+\frac{1}{2}x)^{t}\rangle$. Using the automorphism $\varphi$ of
$K[[x,y]]$ definde by $\varphi(x)=x$ and $\varphi(y)=y-\frac{1}{2}x$ we obtain
the result.
If $\alpha=1$, using (5) we may assume that $f=(x^{2}+y^{s},y^{t}+exy^{t-1})$,
$n\geq 3$, for a suitable unit $e\in K[[x,y]]$. Now
$y^{t}+exy^{t-1}=(y+\frac{1}{t}ex)^{t}+x^{2}h$ for a suitable
$h\in\mathfrak{m}^{t-2}$. Using the automorphism $\varphi$ of $K[[x,y]]$
definde by $\varphi(x)=x$ and $\varphi(y)=y-\frac{1}{t}x$ we obtain
$f\sim(x^{2}+(y-\frac{1}{t}ex)^{s},y^{t}+x^{2}\bar{h})$ with
$\bar{h}\in\mathfrak{m}^{t-2}$. Now $\langle
x^{2}+(y-\frac{1}{t}ex)^{s},y^{t}+x^{2}\bar{h}\rangle=\langle
x^{2}+(y-\frac{1}{t}ex)^{s},y^{t}-(y-\frac{1}{t}ex)^{n}\bar{h})\rangle=\langle
x^{2}+y^{s}+\beta
xy^{s-1}+\tilde{k}_{1},y^{t}+\tilde{h}_{1}\rangle=\langle(x+\frac{1}{2}\beta
y^{n-1})^{2}+y^{s}+\tilde{k}_{1},y^{t}+\tilde{h}_{1}\rangle$ with $\beta\in
K$, $\bar{k}_{1},\tilde{k}_{1}\in\mathfrak{m}^{n+1}$ and
$\tilde{h}_{1}\in\mathfrak{m}^{t+1}$. Using the automorphism $\tilde{\varphi}$
of $K[[x,y]]$ definde by $\tilde{\varphi}(x)=x-\frac{1}{2}\beta y^{s-1}$ and
$\tilde{\varphi}(y)=y$ we obtain $f\sim(x^{2}+y^{s}+k_{1},y^{t}+h_{1})$ with
$k_{1}\in\mathfrak{m}^{s+1}$ and $h_{1}\in\mathfrak{m}^{t+1}$. Iterating this
process we obtain $f\sim(x^{2}+y^{s},y^{t})$. ∎

######  Proposition 2.4.

Let $\operatorname{char}(K)=p$.

  1. 1.

Assume that $p\neq 2$. Let $f=(x^{2}+\alpha y^{s},\sum\limits_{i\geq
t}a_{i}y^{i}+x\sum\limits_{j\geq q}b_{j}y^{j})$, $\alpha\in\\{0,1\\}$.
If $s\geq 4$, $t\geq 5$ and $q\geq 3$ then $f$ is not simple.

  2. 2.

Assume that $p=2$. Then $(x^{2}+axy,h),a\in K$, is not simple for $h\in
K[[x,y]]$ with $\operatorname{ord}(h)\geq 3$.

.

###### Proof.

The following Singular computation shows that $(x^{2}+y^{4},y^{5}+axy^{3})$ is
not equivalent to $(x^{2}+y^{4},y^{5}+bxy^{3})$, except $a^{2}=b^{2}$. In
particular $(x^{2}+y^{4},y^{5}+axy^{3})$ is not simple for each $a\in K$. We
show below that this implies that $f$ is not simple if $\alpha=1$. $f$ with
$\alpha=0$ is not simple (by Proposition 1.14 (2)), since it deforms to $f$
with $\alpha=1$ and $s\geq 4$.

Singular computation to show that $(x^{2}+y^{4},y^{5}+axy^{3})$ is not simple:

    ring R=(0,a,b,l,m0,m1,m2,m3,m4,m5,m6,m7,n0,n1,n2,n3,n4,n5,n6,n7,o,s0,s1,
                  t,u0,u1,v),(x,y),dp;
    int i;
    poly f=x2+y4;
    poly g=y5+a*xy3;
    poly h=x2+y4;
    poly k=y5+b*xy3;
    map phi=R,m0*y+m1*y2+m2*x+m3*xy+m4*y3+m5*y4+m6*x2+m7*xy2,
                       n0*y+n1*y2+n2*x+n3*xy+n4*y3+n5*y4+n6*x2+n7*xy2;
    poly F=jet(phi(f)-(s0+s1*y)*h-t*k,5);
    poly G=jet(phi(g)-(u0+u1*y)*h-v*k,5);
    // R ist graded with deg(x)=2, deg(y)=1, we compute up to degree 5

    matrix M1=coef(F,xy);
    matrix M2=coef(G,xy);

    ideal I;
    for(i=1;i<=ncols(M1);i++){I[size(I)+1]=M1[2,i];}
    for(i=1;i<=ncols(M2);i++){I[size(I)+1]=M2[2,i];}

    ring S=0,(a,b,l,m0,m1,m2,m3,m4,m5,m6,m7,n0,n1,n2,n3,n4,n5,n6,n7,o,s0,s1,
                   t,u0,u1,v,u,z),dp;
    ideal I=imap(R,I);
    I=I,(m0*n2-n0*m2)*o-1,(s0*v-u0*t)*z-1; //to make matrix and map invertible

    ideal F=eliminate(I,
    l*m0*m1*m2*m3*m4*m5*m6*m7*n0*n1*n2*n3*n4*n5*n6*n7*o*s0*s1*t*u0*u1*v*o*z);
    F;
    > F[1]=a^2-b^2

Let $s\geq 4$, $t\geq 5$ and $q\geq 3$. Set
$f_{a}:=(x^{2}+y^{4},y^{5}+axy^{3}),a\in K,$ and $F_{\lambda}:=f+\lambda
f_{a}$. $F_{\lambda}$ is a deformation of $f$ and we show that $F_{\lambda}$
is not simple for almost all $\lambda$ (i.e. all except finitely many). Then,
by the semicontinuity of the modality (Proposition 1.14), $f$ is not simple.
We have

$F_{\lambda}=(x^{2}(1+\lambda)+y^{4}(\lambda+\alpha
y^{s-4}),y^{5}(\lambda+\sum_{i\geq t-5}a_{i}y^{i})+x(\lambda
ay^{3}+\sum_{j\geq q}b_{j}y^{j}).$

For general $\lambda$ this is equivalent to

$(x^{2}+y^{4},y^{5}+xy^{3}(ba+\sum_{j\geq q}-3b^{\prime}_{j}y^{j})),$

for some $b\neq 0$. In fact, replace $x$ by $\sqrt{1+\lambda}^{-1}x$, multiply
$F_{\lambda,1}$ with the inverse of $v=(\lambda+\alpha y^{s-4})$ and replace
$x$ by $\sqrt{v}x$ and, finally, multipy $F_{\lambda,2}$ with the inverse of
$\lambda+\sum_{i\geq t-5}a_{i}y^{i}$. With $a\neq 0$ the factor of $xy^{3}$ is
a unit and we get $(x^{2}+y^{4},y^{5}+xy^{3}b_{0}a+g)$, with $b_{0}\neq 0$ and
$g=\sum_{j\geq 4}b^{\prime}_{j}y^{j}$. By the Singular computation above
$h_{0}:=(x^{2}+y^{4},y^{5}+xy^{3}b_{0}a)$ is not simple.

To see that $h_{g}:=(x^{2}+y^{4},y^{5}+xy^{3}b_{0}a+g)$ is not simple, we
cannot use the semicontinuity since $h_{g}$ is a deformation of $h_{0}$ with a
priori smaller modality. But we can prove the non-simpleness in two ways:
a) It suffices that the 5-jet of $h_{g}$ is not simple. This follows from a
similar (but longer222On a MacBook Air with M1 chip about 4,5 h and about 5
sec without the term $xy^{4}$.) Singular computation as above, which shows
that $(x^{2}+y^{4},y^{5}+axy^{3}+cxy^{4})$ is not equivalent to
$(x^{2}+y^{4},y^{5}+bxy^{3}+dxy^{4})$, except $a^{2}=b^{2}$ for arbitrary
$c,d$.
b) We notice that $h_{0}$ is quasi-homogeneous of type $(4,5;1,2)$ and it
suffices to show that the weighted 5-jet of $h_{g}$ is not simple. We make an
”Ansatz” with a general (not quasi-homogeneous) coordinate transformation up
to weighted degree 5, apply it to
$(f,g)_{a,c}=(x^{2}+y^{4},y^{5}+axy^{3}+cxy^{4})$ and compare it with
$(h,k)=:(f,g)_{b,d}$:

    map phi=R,m0*y+m1*y2+m2*x+m3*xy+m4*y3+m5*y4+m6*x2+m7*xy2,
                       n0*y+n1*y2+n2*x+n3*xy+n4*y3+n5*y4+n6*x2+n7*xy2;
    poly F=jet(phi(f)-(s0+s1*y)*h-t*k,5);
    poly G=jet(phi(g)-(u0+u1*y)*h-v*k,5);

Then $(f,g)_{a,c}\sim(f,g)_{b,d}$ implies $F=G=0$. Looking at $F$, we see that
$\tt m0$ must be 0 and hence $\tt phi(cxy^{4})$ has weighted degree bigger
than 5. Therefore $cxy^{4}$ can be deleted and the first Singular computation
above suffices to show that $h_{g}$ is not simple.

In characteristic $2$ the following Singular computation shows that
$(x^{2}+axy,cxy^{2}+dy^{3})\sim(x^{2}+bxy,cxy^{2}+dy^{3})$ implies
$a^{4}b^{2}c^{4}d+a^{2}b^{4}c^{4}d+a^{4}c^{2}d^{3}+b^{4}c^{2}d^{3}=0$, which
happens, for fixed $a,c,d$, only for finitely many $b$.
It follows that $\langle x^{2}+axy,j_{3}(h)\rangle=\langle
x^{2}+axy,cxy^{2}+dy^{3}\rangle$, $c,d$ suitable, is not simple. Hence
$(x^{2}+axy,h)$ is not simple.

    ring R=(2,a,b,c,d,m0,m1,n0,n1,o,s,t,u,v),(x,y),dp;
    int i;
    poly f=x2+a*xy;        //a the parameter of the family
    poly g=c*xy2+d*y3;     //generic element of degree 3
    poly h=x2+b*xy;
    poly k=c*xy2+d*y3;
    map phi=R,m0*y+m1*x,n0*y+n1*x;
    poly F=phi(f)-s*h-t*k;
    poly G=phi(g)-u*h-v*k;

    matrix M1=coef(F,xy);
    matrix M2=coef(G,xy);

    ideal I;
    for(i=1;i<=ncols(M1);i++){I[size(I)+1]=M1[2,i];}
    for(i=1;i<=ncols(M2);i++){I[size(I)+1]=M2[2,i];}

    ring S=2,(a,b,c,d,m0,m1,n0,n1,o,s,t,u,v,u,z),dp;
    ideal I=imap(R,I);
    I=I,(m0*n1-n0*m1)*o-1,(s*v-u*t)*z-1;

    ideal F=eliminate(I,m0*m1*n0*n1*o*s*t*u*v*o*z);
    F;
    > F[1]=a^4*b^2*c^4*d+a^2*b^4*c^4*d+a^4*c^2*d^3+b^4*c^2*d^3

∎

######  Proposition 2.5.

If $p\neq 2$, then the following ICIS are the only candidates for being
simple:

  1. 1.

$F_{s+m-1}^{s,m}:\text{ }(xy,x^{s}+y^{m})$, $s,m\geq 2$,

  2. 2.

$G_{5}^{0}:\text{ }(x^{2},y^{3})$,

  3. 3.

$G_{5}^{1}:\text{ }(x^{2},xy^{2}+y^{3})$ in $\operatorname{char}p=3$,

  4. 4.

$G_{7}:\text{ }(x^{2},y^{4})$,

  5. 5.

$H_{s+3}:\text{ }(x^{2}+y^{s},xy^{2})$, $s\geq 3$,

  6. 6.

$I_{2t-1}^{0}:\text{ }(x^{2}+y^{3},y^{t})$, $t\geq 4$,

  7. 7.

$I_{2t-1}^{1}:\text{ }(x^{2}+y^{3},y^{t}+xy^{t-1})$, $t\geq 4$ if $p\mid t$,

  8. 8.

$I_{2q+2}^{0}:\text{ }(x^{2}+y^{3},xy^{q})$, $q\geq 3$,

  9. 9.

$I_{2q+2}^{1}:\text{ }(x^{2}+y^{3},xy^{q}+y^{q+2})$, $q\geq 3$ if $p\mid
2q+3$.

###### Proof.

Let $f=(f_{1},f_{2})$ be an ICIS.

(I) If $j_{2}(f_{i})$ is non-degenerate for some $i$ we obtain $f\sim
F_{s+m-1}^{s,m}$ for suitable $s,m\geq 2$, using Proposition 2.1.

(II) If $j_{2}(f_{i})$ is degenerate for some $i$ we obtain using Lemma 2.2
that

$f\sim(x^{2}+\alpha y^{s},\sum\limits_{i\geq t}a_{i}y^{i}+x\sum\limits_{j\geq
q}b_{j}y^{j}),\ a_{i},b_{j}\in K,$

with $s\geq 3$, $t\geq 2$, $q\geq 1$ and $\alpha\in\\{0,1\\}$.

Using Proposition 2.4 and that ord$(f)=2$, we obtain that if $f$ is simple,
then

$s=3$ or $2\leq t\leq 4$ or $1\leq q\leq 2$.

Now we use these inequalities and Lemma 2.3.

  1. 1.

Let $a_{i}=0$ for all $i$ and $b_{q}\neq 0$ (this we may assume since $f$ is a
complete intersection). Then $f\sim(x^{2}+y^{s},xy^{q})$ ($\alpha=1$ since $f$
is a complete intersection). For $f$ to be simple we have $s=3$ and then we
get $I^{0}_{2q+2}$ or we have $q=2$ and then we get $H_{s+3}$.

  2. 2.

Let $b_{j}=0$ for all $j$ and $a_{t}\neq 0$. Then $f\sim(x^{2}+\alpha
y^{s},y^{t})$.
$-$ If $\alpha=0$ or $t\leq s$ then $f\sim(x^{2},y^{t})$. Then $t$ must be
$2$, $3$ or $4$. For $t=2$ we get
$(x^{2},y^{2})\sim(xy,x^{2}+y^{2})=F_{3}^{2,2}$. For $t=3,4$ we obtain
$G^{0}_{5},G_{7}$.
$-$ If $\alpha=1$ and $t>s$ then $s=3$, and we obtain $I^{0}_{2t-1}$.

  3. 3.

Assume now that $a_{t}b_{q}\neq 0$.
Lemma 2.3 implies:

    1. (a)

If $t\leq q$ then $f\sim(x^{2}+\alpha y^{s},y^{t})$.
$-$ If $\alpha=0$ or $t\leq s$ then $f\sim(x^{2},y^{t})$. We have $2\leq t\leq
4$ and we obtain $F_{3}^{2,2}$ or $G^{0}_{5}$ or $G_{7}$.
$-$ If $\alpha=1$ and $t>s$ then $s=3$, and we obtain $I^{0}_{2t-1}$.

    2. (b)

If $t>q$ then, by Lemma 2.3 (4) and (5) and using the automorphism $\varphi$
of $K[[x,y]]$ defined by $\varphi(x)=e^{-1}x$ and $\varphi(y)=y,$ we obtain

$\displaystyle f$ $\displaystyle\sim(x^{2},y^{t}+xy^{q}),\text{ if }\alpha=0,$
$\displaystyle f$
$\displaystyle\sim(x^{2}+y^{s},y^{t}+exy^{q})\sim(x^{2}+e^{2}y^{s},y^{t}+xy^{q}),\text{
if }\alpha=1,$

where $e\in K[[y]]$ a unit with $e(0)=1$ if $2t-2q-s\neq 0$ and ($p=0$ or
$p\nmid 2t-2q-s$).

      1. i.

Let $t=q+1$.
(i.1) If $p\nmid t$, then $f\sim(x^{2}+\alpha y^{s},y^{t})$, by Lemma 2.3 (6).
$-$ If $\alpha=0$ or $t\leq s$ then $f\sim(x^{2},y^{t})$, $2\leq t\leq 4$, and
we obtain $F_{3}^{2,2}$ or $G^{0}_{5}$ or $G_{7}$.
$-$ If $\alpha=1$ and $t>s$ then $s=3$, and we obtain $I^{0}_{2t-1}$.
(i.2) If $p\mid t$, then $f\sim(x^{2}+\alpha y^{s},y^{t}+exy^{t-1})$ with
$e\in K[[y]]$ a unit.
$-$ If $\alpha=0$ then $p=3$ and $t=3$ and by Lemma 2.3 (4) we get
$G^{1}_{5}$.
$-$ If $\alpha=1$ then either $s=3$ or ($t=3$ and $p=3$).
(i.2.1) For the first case ($s=3$) we obtain from Lemma 2.3 (5)

$f\sim(x^{2}+y^{3},y^{t}+exy^{t-1})\sim(x^{2}+e^{2}y^{3},y^{t}+xy^{t-1}),$

$e\in K[[y]]$ with $e(0)=1$ (since $p\mid t$ we have $p\nmid 2t-2q-s$ since
$s=3$). Let $e^{2}=\sum_{j}a_{j}y^{j}$ with $a_{0}=1$. Define
$F(z):=z\sum_{j}a_{j}y^{j}z^{j}-1$.
$F$ has the following properties:

         * •

$F(1)\in\langle y\rangle,$

         * •

$F^{\prime}(1)$ is a unit.

The implicit function theorem implies that there is a unit $z(y)$ such that

$z(y)e^{2}(z(y)x,z(y)y)=z(y)\sum_{j}a_{j}y^{j}z(y)^{j}=1.$

The automorphism $\varphi$ defined by $\varphi(x)=z(y)x$ and
$\varphi(y)=z(y)y$ gives the equivalence
$(x^{2}+e^{2}y^{3},y^{t}+xy^{t-1})\sim(x^{2}+y^{3},y^{t}+xy^{t-1}).$
If $p=t=3$ we get $G^{1}_{5}$. If $t\geq 4$ we get $I^{1}_{2t-1}$.

(i.2.2) To see the second case ($s>3$, $t=3$ and $p=3$) we start with
$f=(f_{1},f_{2})\sim(x^{2}+e^{2}y^{s},y^{3}+xy^{2}).$ Substracting
$e^{2}y^{s-3}f_{2}$ from $f_{1}$ we get

$\langle x^{2}+e^{2}y^{s},y^{3}+xy^{2})\rangle=\langle
x^{2}-e^{2}xy^{s-1},y^{3}+xy^{2}\rangle$

$=\langle(x-\frac{1}{2}e^{2}y^{s-1})^{2}-\frac{1}{4}e^{4}y^{2s-2},y^{3}+xy^{2}\rangle$.

Using the automorphism $\varphi(x)=x+\frac{1}{2}e^{2}y^{s-1}$ and
$\varphi(y)=y$ we obtain

$(x^{2}-\frac{1}{4}e^{4}y^{2s-2},y^{3}+xy^{2}+\frac{1}{2}e^{2}y^{s+1})=(x^{2}-\frac{1}{4}e^{4}y^{2s-2},uy^{3}+xy^{2})$

with $u=1+\frac{1}{2}e^{2}y^{s-2}$ a unit.
Setting $\varphi(x)=ux$ and $\varphi(y)=y$ we obtain

$(u^{2}x^{2}-\frac{1}{4}e^{4}y^{2s-2},uy^{3}+uxy^{2})$
$\sim(x^{2}-\frac{1}{4}u^{-2}e^{4}y^{2s-2},y^{3}+xy^{2}),$

which raises the power $s$ of $y$ to $2s-2$. Iterating this process we obtain
$f\sim(x^{2},y^{3}+xy^{2})$ which is $G^{1}_{5}$.

      2. ii.

Now assume that $t>q+1$.
(ii.1) If Let $\alpha=0$. Then Proposition 2.4 implies $q\leq 2$.
$-$ If $q=1$, we obtain $(x^{2},y^{t}+xy)$. Similar as below with $q=2$ this
is equivalent to $(x^{2}+y^{2t-2},xy)$, which is $F_{2t-1}^{2,2t-2},t\geq 2$.
$-$ If $q=2$ we obtain $(x^{2},y^{t}+xy^{2})$. But

$(x^{2},y^{t}+xy^{2})=(x^{2},y^{2}(y^{t-2}+x))$ $\sim((x-y^{t-2})^{2},xy^{2})$

$=(x^{2}-2xy^{t-2}+y^{2t-4},xy^{2})\sim(x^{2}+y^{2t-4},xy^{2})$, since $t\geq
4$.

We get $H_{2t-1}$.

(ii.2) Let $\alpha=1$. Using 3.(b) above we can start with
$(x^{2}+y^{s},y^{t}+exy^{q})$, with $e=e(y)=e_{0}+e_{k}y^{k}+$ higher terms,
where $e_{0}$ and $e_{k}$ are non-zero.
Then Proposition 2.4 implies $s=3$ or $2\leq t\leq 4$ or $1\leq q\leq 2$.
(ii.2.1) Let $s\geq 3$ and $q\leq 2$
$-$ Let $q=1.$ Set $m=:\min\\{s,2t-2\\}$ if $s\neq 2t-2$ or if $s=2t-2$ and
$e^{-2}(0)\neq-1$. If $s=2t-2$ and $e^{-2}(0)=-1$ set $m=:s+k$.

We get
$(x^{2}+y^{s},y^{t}+exy)\sim(x^{2}+y^{s},y(e^{-1}y^{t-1}+x)$
$\sim(x-e^{-1}y^{t-1})^{2}+y^{s},xy)\sim(x^{2}+e^{-2}y^{2t-2}+y^{s},xy)$
$\sim(x^{2}+y^{m}u,xy)\sim(x^{2}u^{-1}+y^{m},xy)$, $u(y)$ a unit.
We get $F^{2,m}_{m+1}$ since $p>2$. If $s=3$ we get $F^{2,3}_{4}$.
$-$ Let $q=2$. Consider $(x^{2}+y^{s},y^{t}+exy^{2})$,
$e(y)=e_{0}+e_{k}y^{k}+$ higher terms, a unit. Set $m=:\min\\{s,2t-4\\}$ if
$s\neq 2t-4$ or if $s=2t-4$ and $e^{-2}(0)\neq-1$. If $s=2t-4$ and
$e^{-2}(0)=-1$ set $m=:s+k$.
We get $(x^{2}+y^{s},y^{t}+exy^{2})\sim(x^{2}+y^{m},xy^{2})$, i.e. $H_{m+3}$.
This follows from

$(x^{2}+y^{s},y^{t}+exy^{2})=(x^{2}+y^{s},y^{2}(e^{-1}y^{t-2}+x))\sim((x-e^{-1}y^{t-2})^{2}+y^{s},xy^{2})$
$=(x^{2}-2e^{-1}xy^{t-2}+e^{-2}y^{2t-4}+y^{s},xy^{2})\sim(x^{2}+e^{-2}y^{2t-4}+y^{s},xy^{2})\sim(x^{2}+uy^{m},xy^{2})\sim(u^{-1}x^{2}+y^{m},xy^{2})$,
$u$ a unit, $\sim((vx)^{2}+y^{m},xy^{2})$
$(v^{2}=u^{-1})\sim(x^{2}+y^{m},v^{-1}xy^{2})$ $\sim(x^{2}+y^{m},xy^{2})$.

(ii.2.2) Let $s=3$ and $t\geq q+3$. We have

$y^{t}+exy^{q}=y^{t-3}(x^{2}+y^{3})+(e-xy^{t-q-3})xy^{q},$

with $(e-xy^{t-q-3})$ a unit. Hence
$(x^{2}+y^{3},exy^{q}+y^{t})\sim(x^{2}+y^{3},xy^{q})$.
(ii.2.3) Let $s=3$ and $t=q+2$.
We consider $(f,g)=(x^{2}+y^{3},xy^{q}+ey^{q+2})$ and, using Lemma 2.3 (5), we
may assume that $e(0)=1$. We make the Ansatz

$\varphi(x)=x+\alpha y^{2}\text{ and }\varphi(y)=y+\beta x.$
$\varphi(f)=x^{2}+y^{3}+(2\alpha+3\beta)xy^{2}+(3\beta^{2}y+\beta^{3}x)x^{2}+\alpha^{2}y^{4},$
$\varphi(g)=xy^{q}+q\beta x^{2}y^{q-1}+(\alpha+1)y^{q+2}+\text{ terms of w-deg
}>2q+4,$

with w$(x)=3$, w$(y)=2$. Then

$\varphi(g)-q\beta y^{q-1}\varphi(f)=xy^{q}+(\alpha+1-q\beta)y^{q+2}+\text{
terms of w-deg }>2q+4.$

We want to remove the term $xy^{2}$ from $\varphi(f)$ and $y^{q+2}$ from the
last equation.

If $p$ does not divide $2q+3$ then this is possible choosing

$\alpha=-\frac{3}{2q+3}\text{ and }\beta=\frac{2}{2q+3}.$

We obtain

$(f,g)\sim(x^{2}+y^{3}+(3\beta^{2}y+\beta^{3}x)x^{2}+\alpha^{2}y^{4},xy^{q}+\text{
terms of w-deg }>2q+4)$ $\sim(x^{2}+\bar{e}y^{3},xy^{q}+\text{ terms of w-deg
}>2q+4)$

for a suitable unit $\bar{e}(x,y)$ with $\bar{e}(0)=1$.
Now consider the polynomial $F(z)=\bar{e}(zx,zy)z-1$. We have

$F(1)\in\langle x,y\rangle$ and
$\frac{dF}{dz}(1)=\bar{e}(zx,zy)(1)+z\frac{d\bar{e}(zx,zy)}{dz}(1)\in
1+\langle x,y\rangle$.

The Implicit Function Theorem implies that there exists an unit $z(x,y)\in
K[[x,y]]$ such that

$F(z(x,y))=\bar{e}(z(x,y)x,z(x,y)y)z(x,y)-1=0.$

This implies

$(x^{2}+\bar{e}y^{3},xy^{q}+\text{ terms of w-deg }>2q+4)$
$\sim(x^{2}+y^{3},xy^{q}+\text{ terms of w-deg }>2q+4)$
$\sim(x^{2}+y^{3},xy^{q}+\gamma y^{l})$

for a suitable $\gamma$ and $l\geq q+3$. This case is already settled (see
ii.2.2).

If $p\mid 2q+3$ we consider for $(f,g)=(x^{2}+y^{3},xy^{q}+ey^{q+2})$ with
$e(0)=1$ the polynomial $F(z)=ze(z^{3}x,z^{2}y)-1$. We have

$F(1)\in\langle x,y\rangle$ and
$\frac{dF}{dz}(1)=e(z^{3}x,z^{2}y)(1)+z\frac{de(z^{3}x,z^{2}y)}{dz}(1)\in
1+\langle x,y\rangle$.

The Implicit Function Theorem implies that there exists an unit $z(x,y)\in
K[[x,y]]$ such that

$F(z(x,y))=e(z(x,y)^{3}x,z(x,y)^{2}y)z(x,y)-1=0.$

The map $\varphi:K[[x,y]]\longrightarrow K[[x,y]]$ define by
$\varphi(x)=z(x,y)^{3}x$ and $\varphi(y)=z(x,y)^{2}y$ applied to $(f,g)$ gives

$(z(x,y)^{6}x^{2}+z(x,y)^{6}y^{3},z(x,y)^{2q+3}xy^{q}+e(z(x,y)^{3}x,z(x,y)^{2}y)z(x,y)^{2q+4}y^{q+2})$
$\sim(x^{2}+y^{3},xy^{q}+y^{q+2}).$

∎

######  Theorem 2.6.

Let $p=$ char$(K)\neq 2$. An ICIS $f\in K[[x,y]]^{2}$ is simple iff it is
contact equivalent to one of the normal forms in Table 2.

Type |  Normal form of $f$ |
---|---|---
$F^{m,n}_{m+n-1}$, $m,n\geq 2$ | $\hskip 28.45274pt(xy,x^{m}+y^{n})$ |
$G_{5}$ |  $G_{5}^{0}\hskip 14.22636pt(x^{2},y^{3})$ and additionally |
|  $G_{5}^{1}\hskip 14.22636pt(x^{2},xy^{2}+y^{3})$ if $p=3$ |
$G_{7}$ | $\hskip 28.45274pt(x^{2},y^{4})$ |
$H_{n+3}$, $n\geq 3$ | $\hskip 28.45274pt(x^{2}+y^{n},xy^{2})$ |
$I_{2t-1}$,  $t\geq 4$ |  $I_{2t-1}^{0}\ (x^{2}+y^{3},y^{t})$ and additionally |
|  $I_{2t-1}^{1}\ (x^{2}+y^{3},y^{t}+xy^{t-1})$ if $p\mid t$ |
$I_{2q+2}$, $q\geq 3$ |  $I_{2q+2}^{0}\ (x^{2}+y^{3},xy^{q})$ and additionally |
|  $I_{2q+2}^{1}\ (x^{2}+y^{3},xy^{q}+y^{q+2})$ if $p\mid 2q+3$ |

Table 2

###### Proof.

By Proposition 2.5 we have only to show that the normal forms $f$ in Table 2
are simple. By Definition 1.12 we have to consider the semiuniversal
deformation (with section) of $f=(f_{1},f_{2})$,

$F_{\bf t}({\bf x}):=F({\bf x},t_{1},\ldots,t_{d})=f({\bf
x})+\sum\limits_{i=1}^{d}t_{i}g_{i}({\bf x}),$

where $g_{1},...,g_{d}$ is a $K$-basis of

$T^{1,sec}(f)=\mathfrak{m}K[[x,y]]^{2}/\langle f_{1},f_{2}\rangle
K[[x,y]]^{2}+\mathfrak{m}\langle{\partial f}/{\partial x},{\partial
f}/{\partial y}\rangle.$

$f$ is simple if the set $\\{F_{\bf t}({\bf x}),{\bf t}\in K^{d}\\}$
decomposes into only finitely many contact classes. Moreover, by the argument
at the beginning of Section 2 we need to consider only those $g_{i}$ with
$\operatorname{ord}(g_{i})\geq 2$ (if $\operatorname{ord}(g_{i})=1$ for some
$i$ we get an $A_{k}$ hypersurface singularity with bounded $k$).

We have the following bases of $T^{1,sec}(f)$:

  1. 1.

$F^{m,n}_{m+n-1}$, $m,n\geq 2$
$\bullet~{}{p\nmid m}~{}\textit{ or }{p\nmid n}:(x,0)$, $(y,0),$ $(0,x),$ …,
$(0,x^{m-1}),$ $(0,y),$ …, $(0,y^{n-1}).$
$\bullet~{}p\mid m,p\mid n\textit{ and }{m>n}:$ $(0,x),\ldots,(0,x^{m}),$
$(0,y),\ldots,(0,y^{n-1}).$
$\bullet~{}p\mid m~{}\textit{and}~{}m=n:$ $(x,0),(y,0),$
$(0,x),\ldots,(0,x^{m-1}),$ $(0,y),\ldots,(0,y^{n}).$

$F^{m,n}_{m+n-1}$ is of quasi-homogeneous type $(m+n,mn;m,n)$.

  2. 2.

$G_{5}^{0}$
$\bullet~{}p=3:(x,0),(y,0),(y^{2},0),$ $(0,x),(0,y),(0,xy),$
$(0,y^{2}),(0,xy^{2}).$
$\bullet~{}p\geq 5:(x,0),(y,0),(y^{2},0),$ $(0,x),(0,y),(0,xy),(0,y^{2})$.

$G_{5}^{1}$
$(x,0),(y,0),(y^{2},0),$ $(0,x),(0,y),(0,xy)$, $(0,y^{2})$.

$G_{5}$ is of quasi-homogeneous type $(2,3;1,1)$.

  3. 3.

$G_{7}$
$(x,0),(y,0),(y^{2},0),$ $(y^{3},0),(0,x),(0,y),$
$(0,xy),(0,y^{2}),(0,xy^{2}),(0,y^{3})$.

$G_{7}$ is of quasi-homogeneous type $(2,4;1,1)$.

  4. 4.

$H_{n+3}$, $n\geq 3$
$(x,0),(y,0),\ldots,(y^{n-1},0),$ $(0,x),(0,y),(0,xy),$ $(0,y^{2}),(0,y^{3})$.

$H_{n+3}$ is of quasi-homogeneous type $(2n,n+4;n,2)$.

  5. 5.

a) $I_{2t-1}^{0}$, $t\geq 4$.

$\bullet~{}p\nmid t:(x,0),(y,0),(y^{2},0),$ $(0,y),\ldots,(0,y^{t-1}),(0,x)$,
$(0,xy),\ldots,(0,xy^{t-2})$
$\bullet~{}p\mid t:$
$(x,0),(y,0),(y^{2},0),(0,y),\ldots,(0,y^{t-1}),(0,x),(0,xy),\ldots,(0,xy^{t-1}).$

b) $I_{2t-1}^{1}$, $t\geq 4$.
$\bullet~{}p\mid t$:
$(x,0),(y,0),(y^{2},0),(0,y),\ldots,(0,y^{t-1}),(0,x),(0,xy),\ldots,(0,xy^{t-2}).$

$I^{0}_{2t-1}$ is of quasi-homogeneous type $(6,2t;3,2)$.

  6. 6.

a) $I^{0}_{2q+2}$, $q\geq 3$
$\bullet~{}p\nmid 2q+3:(x,0),(y,0),(y^{2},0)$,
$(0,x),(0,y),\ldots,(0,y^{q+1})$,
$(0,xy),(0,xy^{2}),\ldots,(0,xy^{q-1})$
$\bullet~{}p\mid 2q+3:$ $(x,0),(y,0),(y^{2},0)$,
$(0,x),(0,y),\ldots,(0,y^{q+2}),$
$(0,xy),(0,xy^{2}),\ldots,(0,xy^{q-1})$

b) $I^{1}_{2q+2}$, $q\geq 3$, $p\mid 2q+3.$
$(x,0),(y,0),(y^{2},0)$,
$(0,x),(0,y),\ldots,(0,y^{q+1}),(0,xy),(0,xy^{2}),\ldots,(0,xy^{q-1})$

$I^{0}_{2q+2}$ is of quasi-homogeneous type $(6,2q+3;3,2)$.

1\. We prove that $F^{m,n}_{m+n-1}=(xy,x^{m}+y^{n})$ is simple:
Since $j_{2}(f_{1})$ is non-degenerate, this holds also for any deformation,
which is then of type $F^{p,q}_{p+q-1}$ for some $p,q$ by Proposition 2.5.
Since $\tau^{sec}(f)$ depends only on $n$ and $m$ (by (1) above) and since
$\tau^{sec}$ is semicontinuous by Proposition 1.11 there are only finitely
many $F^{p,q}_{p+q-1}$ into which $F^{m,n}_{m+n-1}$ deforms. Hence $f$ is
simple.

2\. a) We prove that $G_{5}^{0}=(x^{2},y^{3})$ is simple:
$\bullet~{}p=3$: We have to consider

$g=(x^{2}+ay^{2},by^{2}+y^{3}+cxy+dxy^{2}),$

where $a,b,c,d\in K.$

i) $c\neq 0$: By Proposition 2.1, $g$ is of type $F$.

ii) $c=0$:

1) $a\neq 0$: By Proposition 2.1, $g$ is of type $F$.

2) $a=0$:

2.1) $b\neq 0$: $g=(x^{2},by^{2}+y^{3}+dxy^{2}).$ By Lemma 2.3, ,
$g\sim(xy,x^{2}+y^{2})=F^{2,2}_{3}.$

2.2) $b=0$: $g=(x^{2},y^{3}+dxy^{2}).$

2.2.1) $d=0$: $g=G_{5}^{0}.$

2.2.2) $d\neq 0$: By the proof of Proposition 2.5,
$g\sim(x^{2},y^{3}+xy^{2})=G_{5}^{1}$.

$\bullet~{}p\geq 5$: Consider

$g=(x^{2}+ay^{2},by^{2}+y^{3}+cxy),$

where $a,b,c\in K.$

i) $c\neq 0$: By Proposition 2.1, there are $m,n\geq 2$ such that $g\sim
F^{m,n}_{m+n-1}$.

ii) $c=0$:

1) $a\neq 0$: By Proposition 2.1, there are $s,t\geq 2$ such that $g\sim
F^{s,t}_{s+t-1}$.

2) $a=0$:

2.1) $b\neq 0$: Since $p\neq 2$, let $u\in K[[y]]$ be such that $u^{2}=y+b$.
Consider the automorphism $\phi:R\to R$, $x\mapsto x$, $y\mapsto uy$. Then

$\phi^{-1}(g)=\phi^{-1}(x^{2},y^{2}u^{2})=(x^{2},y^{2})\sim(xy,x^{2}+y^{2})=F^{2,2}_{3}.$

2.2) $b=0$: $g\sim G_{5}^{0}.$

2\. b) We prove $G_{5}^{1}=(x^{2},xy^{2}+y^{3})$ is simple. Consider

$g=(x^{2}+cy^{2},ay^{2}+y^{3}+bxy+xy^{2}),$

where $a,b,c\in K$.

i) $c\neq 0$: By Proposition 2.1, there are $m,n\geq 2$ such that $g\sim
F^{m,n}_{m,n}$.

ii) $c=0$: $g=(x^{2},ay^{2}+y^{3}+bxy+xy^{2})$

1) $a\neq 0$:

1.1) $b\neq 0$: By Proposition 2.1, there are $s,t\geq 2$ such that $g\sim
F^{s,t}_{s+t-1}$.

1.2) $b=0$: $g=(x^{2},ay^{2}+y^{3}+xy^{2})$

1.2.1) $a=0$: $g=G_{5}^{1}.$

1.2.2) $a\neq 0$: By Lemma 2.3 , $g\sim F^{2,2}_{3}.$

2) $a=0$: $g=(x^{2},y^{3}+bxy+xy^{2})$

2.1) $b=0$: $g=G_{5}^{1}.$

2.2) $b\neq 0$: By Proposition 2.1, there are $k,l\geq 2$ such that $g\sim
F^{k,l}_{k+l-1}$.

3\. We prove $G_{7}=(x^{2},y^{4})$ is simple. Consider

$g=(x^{2}+cy^{2}+dy^{3},a_{2}y^{2}+a_{3}y^{3}+y^{4}+b_{1}xy+b_{2}xy^{2}),$

where $a_{i},b_{i},c,d\in K.$

i) $b_{1}\neq 0$: By Proposition 2.1, there are $m,n\geq 2$ such that $g\sim
F^{m,n}_{m+n-1}$.

ii) $b_{1}=0$:

1) $c\neq 0$: By Proposition 2.1, there are $s,t\geq 2$ such that $g\sim
F^{s,t}_{s+t-1}$.

2) $c=0$: $g=(x^{2}+dy^{3},a_{2}y^{2}+a_{3}y^{3}+y^{4}+b_{2}xy^{2}).$

2.1) $d\neq 0$: We may assume $d=1$.

2.1.1) $a_{2}\neq 0$: By Lemma 2.3, we get $g\sim F^{2,2}_{3}.$

2.1.2) $a_{2}=0$: $g=(x^{2}+y^{3},a_{3}y^{3}+y^{4}+b_{2}xy^{2}).$

2.1.2.1) $a_{3}\neq 0$:

2.1.2.1.1) $b_{2}\neq 0$: By the proof of Proposition 2.5, we get
$g\sim(x^{2},y^{3})$ if $p\geq 5$,

and $g\sim(x^{2},y^{3}+xy^{2})$ if $p=3.$

2.1.2.1.2) $b_{2}=0$: By Lemma 2.3 , we have $g\sim(x^{2},y^{3})$.

2.1.2.2) $a_{3}=0$: $g=(x^{2}+y^{3},y^{4}+b_{2}xy^{2}).$

2.1.2.2.1) $b_{2}\neq 0$: By Lemma 2.3, we get

$g\sim H_{6}=(x^{2}+y^{3},xy^{2}).$

2.1.2.2.2) $b_{2}=0$: $g=(x^{2}+y^{3},y^{4})=I^{0}_{7}.$

2.2) $d=0$: $g=(x^{2},a_{2}y^{2}+a_{3}y^{3}+y^{4}+b_{2}xy^{2}).$

2.2.1) $b_{2}\neq 0$:

2.2.1.1) $a_{2}\neq 0$: By Lemma 2.3, $g\sim F^{2,2}_{3}$.

2.2.1.2) $a_{2}=0$: $g=(x^{2},a_{3}y^{3}+y^{4}+b_{2}xy^{2}).$

2.2.1.2.1) $a_{3}\neq 0$: By the proof of Proposition 2.5,
$g\sim(x^{2},y^{3})$ if $p\geq 5$ and

$g\sim(x^{2},y^{3}+xy^{2})$ if $p=3$.

2.2.1.2.2) $a_{3}=0$: By the proof of Proposition 2.5, there is $r\geq 4$ such
that

$g=(x^{2},y^{4}+b_{2}xy^{2})\sim(x^{2}+y^{r},xy^{2})=H_{r+3}.$

2.2.2) $b_{2}=0$: $g=(x^{2},a_{2}y^{2}+a_{3}y^{3}+y^{4}).$

2.2.2.1) $a_{2}\neq 0$: By Lemma 2.3, $g\sim F^{2,2}_{3}$.

2.2.2.2) $a_{2}=0$:

2.2.2.2.1) $a_{3}\neq 0$: By Lemma 2.3, $g\sim(x^{2},y^{3}).$

2.2.2.2.2) $a_{3}=0$: $g=G_{7}=(x^{2},y^{4}).$

4\. We prove $H_{n+3}=(x^{2}+y^{n},xy^{2})$, $n\geq 3$, is simple. Consider

$g=(x^{2}+c_{2}y^{2}+\ldots+c_{n-1}y^{n-1}+c_{n}y^{n},a_{2}y^{2}+a_{3}y^{3}+bxy+xy^{2}),$

where $c_{n}=1.$

i) $b\neq 0$: By Proposition 2.1, $g$ is of type $F$.

ii) $b=0$:

1) $c_{2}\neq 0$: By Proposition 2.1, $g$ is of type $F$.

2) $c_{2}=0$:
$g=(x^{2}+c_{3}y^{3}+\ldots+c_{n-1}y^{n-1}+y^{n},a_{2}y^{2}+a_{3}y^{3}+xy^{2}).$

2.1) If $n=3$ then $g=(x^{2}+y^{3},a_{2}y^{2}+a_{3}y^{3}+xy^{2})$. If
$a_{2}\neq 0$, by Lemma 2.3, $g\sim F^{2,2}_{3}.$ If $a_{2}=0$ and $a_{3}\neq
0$, by the proof of Proposition 2.5, $g\sim(x^{2},y^{3})$ if $p\geq 5$ and
$g\sim(x^{2},y^{3}+xy^{2})$ if $p=3$. If $a_{2}=a_{3}=0$ then $g=H_{6}.$

2.2) If $n\geq 4$, set

$i=\min\\{j\mid c_{j}\neq 0\\}\geq 3.$

Set $u=c_{i}+c_{i+1}y+\ldots+y^{n-i}$. Then
$g\sim(u^{-1}x^{2}+y^{i},a_{2}y^{2}+a_{3}y^{3}+xy^{2})$. Let $v\in K[[y]]$ be
such that $v^{2}=u^{-1}$. Then $v$ is a unit. Consider the automorphism
$\phi:R\to R$, $x\mapsto vx$, $y\mapsto y$. Then

$g\sim(x^{2}+y^{i},\phi^{-1}(a_{2}y^{2}+a_{3}y^{3}+xy^{2}))=\left(x^{2}+y^{i},a_{2}y^{2}+a_{3}y^{3}+\mathop{\sum}\limits_{i\geq
2}d_{j}xy^{j}\right),$

where $d_{j}\in K$ and $d_{2}\neq 0$.

2.2.1) $a_{2}\neq 0$: By Lemma 2.3, $g\sim F^{2,2}_{3}$.

2.2.2) $a_{2}=0$:

2.2.2.1) $a_{3}\neq 0$: By the proof of Proposition 2.5, $g\sim(x^{2},y^{3})$
if $p\geq 5$ and $g\sim(x^{2},y^{3}+xy^{2})$ if $p=3$.

2.2.2.2) $a_{3}=0$: By Lemma 2.3, $g\sim H_{i+3}$.

5\. a) We prove that $I^{0}_{2t-1}=(x^{2}+y^{3},y^{t})$, $t\geq 4$, is simple.
The proofs for the cases $p\nmid t$ and $p\mid t$ are put together as below.

Consider the unfolding

$g=(x^{2}+y^{3}+cy^{2},a_{2}y^{2}+\cdots+a_{t-1}y^{t-1}+a_{t}y^{t}+b_{1}xy+\cdots+b_{t-1}xy^{t-1}),$

where $a_{t}=1$, and $b_{t-1}=0$ if $p\nmid t$.

i) $b_{1}\neq 0$: By Proposition 2.1, $g$ is of type $F$.

ii) $b_{1}=0$:

1) $c\neq 0$: By Proposition 2.1, $g$ is of type $F$.

2) $c=0$:
$g=(x^{2}+y^{3},a_{2}y^{2}+\ldots+a_{t-1}y^{t-1}+a_{t}y^{t}+b_{2}xy^{2}+\ldots+b_{t-1}xy^{t-1}).$

2.1) $a_{2}\neq 0$: By Lemma 2.3, $g\sim(x^{2},y^{2})\sim F^{2,2}_{3}$.

2.2) $a_{2}=0$:
$g=(x^{2}+y^{3},a_{3}y^{3}+\ldots+a_{t-1}y^{t-1}+a_{t}y^{t}+b_{2}xy^{2}+\ldots+b_{t-1}xy^{t-1}).$

2.2.1) $a_{3}\neq 0$:

2.2.1.1) $b_{2}\neq 0$: By the proof of Proposition 2.5, $g\sim(x^{2},y^{3})$
if $p\geq 5$, and $g\sim(x^{2},y^{3}+xy^{2})$ if $p=3$.

2.2.1.2) $b_{2}=0$: By Lemma 2.3, $g\sim(x^{2},y^{3})$.

2.2.2) $a_{3}=0$: Set $T=\min\\{i\mid a_{i}\neq 0\\}$. Then $4\leq T\leq t.$

2.2.2.1) $b_{2}\neq 0$: By the proof of Proposition 2.5, $g\sim H_{r+3}$ for
some $r\geq 3$.

2.2.2.2) $b_{2}=0$:

If $b_{j}$ are not all 0 then we set

$Q=\min\\{j\mid b_{j}\neq 0\\}.$

Then $Q\geq 3$. We consider the following cases:

2.2.2.2.1) $T\leq Q$: by Lemma 2.3, $g\sim(x^{2}+y^{3},y^{T})=I^{0}_{2T-1}$.

2.2.2.2.2) $T=Q+1$: if $p\nmid T$, by Lemma 2.3, $g\sim I^{0}_{2T-1}$. If
$p\mid T$, by the proof of Proposition 2.5, $g\sim I^{1}_{2T-1}$.

2.2.2.2.3) $T\geq Q+3$: by the proof of Proposition 2.5, $g\sim I^{0}_{2Q+2}$.

2.2.2.2.4) $T=Q+2$: by the proof of Proposition 2.5, $g\sim I^{0}_{2Q+2}$ if
$p\nmid 2Q+3$, and $g\sim I^{1}_{2Q+2}$ if $p\mid 2Q+3$.

If $b_{j}=0$ for all $j$ then by Lemma 2.3, $g\sim I^{0}_{2T-1}$.

5\. b) We prove that $I^{1}_{2t-1}=(x^{2}+y^{3},y^{t}+xy^{t-1})$, $t\geq 4$,
$p\mid t$, is simple.
Consider the unfolding

$g=(x^{2}+y^{3}+cy^{2},a_{2}y^{2}+\cdots+a_{t-1}y^{t-1}+a_{t}y^{t}+b_{1}xy+\cdots+b_{t-1}xy^{t-1}),$

where $a_{t}=b_{t-1}=1$.

i) $b_{1}\neq 0$: By Proposition 2.1, $g$ is of type $F$.

ii) $b_{1}=0$:

1) $c\neq 0$: By Proposition 2.1, $g$ is of type $F$.

2) $c=0$:
$g=(x^{2}+y^{3},a_{2}y^{2}+\ldots+a_{t-1}y^{t-1}+a_{t}y^{t}+b_{2}xy^{2}+\ldots+b_{t-1}xy^{t-1}).$

2.1) $a_{2}\neq 0$: By Lemma 2.3, $g\sim(x^{2},y^{2})\sim F^{2,2}_{3}$.

2.2) $a_{2}=0$:
$g=(x^{2}+y^{3},a_{3}y^{3}+\ldots+a_{t-1}y^{t-1}+a_{t}y^{t}+b_{2}xy^{2}+\ldots+b_{t-1}xy^{t-1}).$

2.2.1) $a_{3}\neq 0$:

2.2.1.1) $b_{2}\neq 0$: By the proof of Proposition 2.5, $g\sim(x^{2},y^{3})$
if $p\geq 5$, and $g\sim(x^{2},y^{3}+xy^{2})$ if $p=3$.

2.2.1.2) $b_{2}=0$: By Lemma 2.3, $g\sim(x^{2},y^{3})$.

2.2.2) $a_{3}=0$: Set

$T=\min\\{i\mid a_{i}\neq 0\\}.$

Then $4\leq T\leq t.$

2.2.2.1) $b_{2}\neq 0$: By the proof of Proposition 2.5,
$g\sim(x^{2}+y^{r},xy^{2})=H_{r+3}$ for some $r\geq 3$.

2.2.2.2) $b_{2}=0$: Set

$Q=\min\\{j\mid b_{j}\neq 0\\}.$

Then $Q\geq 3$. We consider the following cases:

2.2.2.2.1) $T\leq Q$: by Lemma 2.3, $g\sim(x^{2}+y^{3},y^{T})=I^{0}_{2T-1}$.

2.2.2.2.2) $T=Q+1$: if $p\nmid T$, by Lemma 2.3, $g\sim I^{0}_{2T-1}$. If
$p\mid T$, by the proof of Proposition 2.5, $g\sim I^{1}_{2T-1}$.

2.2.2.2.3) $T\geq Q+3$: by the proof of Proposition 2.5, $g\sim I^{0}_{2Q+2}$.

2.2.2.2.4) $T=Q+2$: by the proof of Proposition 2.5, $g\sim I^{0}_{2Q+2}$ if
$p\nmid 2Q+3$, and $g\sim I^{1}_{2Q+2}$ if $p\mid 2Q+3$ .

6\. a) We prove $I_{2q+2}^{0}=(x^{2}+y^{3},xy^{q})$, $q\geq 3$, is simple.
The proofs for the cases $p\mid 2q+3$ and $p\nmid 2q+3$ are put together as
below.
Consider the unfolding

$g=(x^{2}+y^{3}+cy^{2},a_{2}y^{2}+\ldots+a_{q+2}y^{q+2}+b_{1}xy+\ldots+b_{q-1}xy^{q-1}+b_{q}xy^{q}),$

where $b_{q}=1$, and $a_{q+2}=0$ if $p\nmid 2q+3.$

i) $b_{1}\neq 0:$ By Proposition 2.1, $g$ is of type $F$.

ii) $b_{1}=0:$

1) $c\neq 0:$ By Proposition 2.1, $g$ is of type $F$.

2) $c=0:$

2.1) $a_{2}\neq 0:$ by Lemma 2.3, $g\sim F_{3}^{2,2}$.

2.2) $a_{2}=0:$ we consider the following cases:

2.2.1) $a_{3}\neq 0:$

2.2.1.1) $b_{2}\neq 0:$ by the proof of Proposition 2.5, $g\sim(x^{2},y^{3})$
if $p\geq 5$,

and $g\sim(x^{2},y^{3}+xy^{2})$ if $p=3$.

2.2.1.2) $b_{2}=0:$ by Lemma 2.3, $g\sim(x^{2},y^{3})$.

2.2.2) $a_{3}=0:$

2.2.2.1) $b_{2}\neq 0$: by Lemma 2.3 and the proof of Proposition 2.5, $g\sim
H_{r+3}=(x^{2}+y^{r},xy^{2})$ for some $r\geq 3$.

2.2.2.2) $b_{2}=0$: Set $Q=\min\\{j\mid b_{j}\neq 0\\}$. Then $3\leq Q\leq q.$

If there is $i\geq 4$ such that $a_{i}\neq 0$ then we set

$T=\min\\{i\mid a_{i}\neq 0\\}.$

Then $T\geq 4$.

2.2.2.2.1) $T\leq Q$: by Lemma 2.3, $g\sim I^{0}_{2T-1}=(x^{2}+y^{3},y^{T})$.

2.2.2.2.2) $T=Q+1$: by the proof of Proposition 2.5, $g\sim I^{0}_{2T-1}$ if
$p\nmid T$, and $g\sim(x^{2}+y^{3},x^{T}+xy^{T-1})=I^{1}_{2T-1}$ if $p\mid T.$

2.2.2.2.3) $T\geq Q+3$: by the proof of Proposition 2.5, $g\sim I^{0}_{2Q+2}$.

2.2.2.2.4) $T=Q+2$: by the proof of Proposition 2.5, $g\sim
I^{0}_{2Q+2}=(x^{2}+y^{3},xy^{Q})$ if $p\nmid 2Q+3$, and $g\sim
I^{1}_{2Q+2}=(x^{2}+y^{3},xy^{Q}+y^{Q+2})$ if $p\mid 2Q+3$.

If $a_{i}=0$ for all $i$ then by Lemma 2.3, we obtain $g\sim I^{0}_{2Q+2}$.

6\. b) We prove $I_{2q+2}^{1}=(x^{2}+y^{3},xy^{q}+y^{q+2})$, $q\geq 3$, $p\mid
2q+3$ is simple.
Consider the unfolding

$g=(x^{2}+y^{3}+cy^{2},a_{2}y^{2}+\ldots+a_{q+2}y^{q+2}+b_{1}xy+\ldots+b_{q-1}xy^{q-1}+b_{q}xy^{q}),$

where $a_{q+2}=b_{q}=1$.

i) $b_{1}\neq 0:$ By Proposition 2.1, $g$ is of type $F$.

ii) $b_{1}=0:$

1) $c\neq 0:$ By Proposition 2.1, $g$ is of type $F$.

2) $c=0:$

2.1) $a_{2}\neq 0:$ by Lemma 2.3, $g\sim(x^{2},y^{2})\sim F_{3}^{2,2}$.

2.2) $a_{2}=0:$ we consider the following cases:

2.2.1) $a_{3}\neq 0:$

2.2.1.1) $b_{2}\neq 0:$ by the proof of Proposition 2.5, $g\sim
G^{0}_{5}=(x^{2},y^{3})$ if $p\geq 5$,

and $g\sim G^{1}_{5}=(x^{2},y^{3}+xy^{2})$ if $p=3$.

2.2.1.2) $b_{2}=0:$ by Lemma 2.3, $g\sim(x^{2},y^{3})$.

2.2.2) $a_{3}=0:$ Set

$T=\min\\{i\mid a_{i}\neq 0\\}.$

Then $T\geq 4$.

2.2.2.1) $b_{2}\neq 0$: by the proof of Proposition 2.5, $g\sim
H_{r+3}=(x^{2}+y^{r},xy^{2})$ for some $r\geq 3$.

2.2.2.2) $b_{2}=0$: Set $Q=\min\\{j\mid b_{j}\neq 0\\}$. Then $3\leq Q\leq q.$

2.2.2.2.1) $T\leq Q$: by Lemma 2.3, $g\sim I^{0}_{2T-1}=(x^{2}+y^{3},y^{T})$.

2.2.2.2.2) $T=Q+1$: by the proof of Proposition 2.5, $g\sim I^{0}_{2T-1}$ if
$p\nmid T$, and $g\sim(x^{2}+y^{3},x^{T}+xy^{T-1})=I^{1}_{2T-1}$ if $p\mid T.$

2.2.2.2.3) $T\geq Q+3$: by the proof of Proposition 2.5, $g\sim I^{0}_{2Q+2}$.

2.2.2.2.4) $T=Q+2$: by the proof of Proposition 2.5, $g\sim
I^{0}_{2Q+2}=(x^{2}+y^{3},xy^{Q})$ if $p\nmid 2Q+3$, and $g\sim
I^{1}_{2Q+2}=(x^{2}+y^{3},xy^{Q}+y^{Q+2})$ if $p\mid 2Q+3$. ∎

##  3 Simple isolated complete intersection singularities of $I_{2,2}$ in
characteristic 2

In this section, we assume that the field $K$ has characteristic $p=2$.

######  Lemma 3.1.

Let $f\in I_{2,2}$ be such that $\operatorname{ord}(f)=2$. Then either there
is a $g_{2}\in\mathfrak{m}^{2}$ such that

$f\sim(xy,g_{2})$

or there are $h_{1}\in\mathfrak{m}^{3}$ and $g_{2}\in\mathfrak{m}^{2}$ such
that

$f\sim(x^{2}+h_{1},g_{2}).$

If $g_{2}\in\mathfrak{m}^{3}$ then $f$ is not simple.

###### Proof.

Let $f=(f_{1},f_{2})$. We may assume that $\operatorname{ord}(f_{1})=2$. By
Lemma 1.18, either there is $\phi\in Aut(R)$ such that $\phi(f_{1})=xy$ or
there is $\varphi\in Aut(R)$ such that $\varphi(f_{1})=x^{2}+h_{1}$ for some
$h_{1}\in\mathfrak{m}^{3}$. Therefore, either $f\sim(xy,\phi(f_{2}))$ or
$f\sim(x^{2}+h_{1},\varphi(f_{2}))$. Proposition 2.4 implies that $f$ is not
simple if $g_{2}\in\mathfrak{m}^{3}$ . This proves the lemma. ∎

######  Proposition 3.2.

Let $g=(xy,g_{2})\in I_{2,2}$, where $g_{2}\in\mathfrak{m}^{2}$. Then there
are $m,n\geq 2$ such that $g\sim(xy,x^{n}+y^{m})=F^{n,m}_{n+m-1}$.

###### Proof.

The proof is similar to a part of the proof of Proposition 2.1, which does not
depend on characteristic. ∎

######  Proposition 3.3.

Let $g=(x^{2}+h_{1},g_{2})\in I_{2,2}$, where $h_{1}\in\mathfrak{m}^{3}$ and
$j_{2}(g_{2})=ax^{2}+bxy+cy^{2}\neq 0$, where $a,b,c\in K$.

  1. a)

If $b=0$ and $c\neq 0$ then $g\sim(x^{2},y^{2})$.

  2. b)

If $b\neq 0$ and $c\neq 0$ then $g\sim(x^{2},xy+y^{2})$.

###### Proof.

The proposed normal forms are of quasi homogeneous type $({\bf d};{\bf
a})=(2,2;1,1)$.

a) If $b=0$ and $c\neq 0$ then $g\sim(x^{2}+h_{1},y^{2}+h_{2})$,
$h_{i}\in\mathfrak{m}^{3}$. Setting

$f=(x^{2},y^{2}),\hskip 5.0pth=(h_{1},h_{2}),$

we have $v_{{\bf d},{\bf a}}(h)\geq 1$ and $T^{1}(f)=R^{2}/\langle
x^{2},y^{2}\rangle R^{2}$. It is easy to check $T^{1}_{\nu}(f)=0$ for all
$\nu\geq 1$. This implies $\sup\\{i\mid T^{1}_{i}(H)\neq 0\\}\leq 0<v_{{\bf
d},{\bf a}}(h)$ and by Proposition 1.8 we deduce $f\sim f+h\sim g$.

b) If $b\neq 0$ and $c\neq 0$ then $g\sim(x^{2}+h_{1},xy+y^{2}+h_{2})$,
$h_{i}\in\mathfrak{m}^{3}$. Setting

$f=(x^{2},xy+y^{2}),\hskip 5.0pth=(h_{1},h_{2}).$

we have $v_{{\bf d},{\bf a}}(h)\geq 1$. Moreover, $T^{1}(f)=R^{2}/D,$ with

$D:=\langle(0,y),(0,x)\rangle+\langle x^{2},xy+y^{2}\rangle R^{2}.$

Then $(x^{2},0)$, $(xy^{2},0)$, $(y^{3},0)$, $(0,xy^{2})$, $(0,y^{3})\in D.$
This implies $T^{1}_{\nu}(f)=0$ for all $\nu\geq 1$. By Proposition 1.8,
$f\sim f+h\sim g.$ ∎

######  Theorem 3.4.

All ICIS of the table 3 are simple and these are the only ones if $p=$
char$(K)=2$.

Type |  Equation of $f$ |
---|---|---
$F^{m,n}_{m+n-1}$ $m,n\geq 2$ | $(xy,x^{m}+y^{n})$ |
$F_{3}^{2,2;0}$ | $(x^{2},y^{2})$ |
$F_{3}^{2,2;1}$ | $(x^{2},xy+y^{2})$ |

Table 3

###### Proof.

To prove that the $f$ from table 3 are simple, we proceed as in the proof of
Theorem 2.6. We have the following bases of $T^{1,sec}(f)$:

  1. 1.

$F^{m,n}_{m+n-1}$, $m,n\geq 2$
$\bullet~{}m~{}\textit{or }n\text{ odd:}~{}(x,0),(y,0),$
$(0,x),\ldots,(0,x^{m-1}),(0,y),\ldots,(0,y^{n-1}).$
$\bullet~{}m,n\textit{ even},{m>n}:~{}(x,0),(y,0),$
$(0,x),\ldots,(0,x^{m}),(0,y),\ldots,(0,y^{n-1}).$
$\bullet~{}m=n\textit{ even:}~{}(x,0),(y,0),$
$(0,x),\ldots,(0,x^{m-1}),(0,y),\ldots,(0,y^{n}).$

$F^{m,n}_{m+n-1}$ is of quasi-homogeneous type $(m+n,mn;m,n)$.

  2. 2.

$F^{2,2;0}_{3}$
$(x,0),(y,0),(xy,0),(0,x),(0,y),(0,xy).$

$F^{2,2;1}_{3}$
$(x,0),(y,0),(y^{2},0).$

$F^{2,2}_{3}$ is of quasi-homogeneous type $(2,2;1,1)$.

1\. The proof is similar to the proof of Proposition 2.1, which works also for
$p=2$.

2\. We prove that $F^{2,2;0}_{3}=(x^{2},y^{2})$ is simple.

Consider $g=(x^{2}+axy,y^{2}+bxy)$ with $a,b\in K.$

i) $a=0$: $g=(x^{2},y^{2}+bxy)$

1) $b=0$: $g=(x^{2},y^{2})$.

2) $b\neq 0$: Using the automorphism $\phi:$ $x\mapsto b^{-1}x$, $y\mapsto y$
we get

$\phi(g)=(b^{-2}x^{2},xy+y^{2})\sim(x^{2},xy+y^{2}).$

ii) $a\neq 0$: Using the automorphism $\phi:$ $x\mapsto x$,
$y\mapsto-a^{-1}x+a^{-1}y$, we get

$\phi(g)=(xy,(a^{-2}-a^{-1}b)x^{2}+a^{-1}bxy+a^{-2}y^{2})\sim(xy,(a^{-2}-a^{-1}b)x^{2}+a^{-2}y^{2}).$

1) $a^{-1}\neq b$: $g\sim(xy,x^{2}+y^{2})=F^{2,2}_{3}.$

2) $a^{-1}=b$: $g\sim(xy,y^{2})$ is not a complete intersection.

We prove that $F^{2,2;1}_{3}=(x^{2},xy+y^{2})$ is simple.

Consider $g=(x^{2}+ay^{2},xy+y^{2})$, where $a\in K$. Using $\phi:x\mapsto
x-\sqrt{a}y$, $y\mapsto y$

we get $\phi(g)=(x^{2},(1-\sqrt{a})y^{2}+xy)$.

i) $a=1$: $g$ is not a complete intersection.

ii) $a\neq 1$: Let $c=1-\sqrt{a}$. Using the automorphism
$\varphi:x\mapsto\sqrt{c}x,$ $y\mapsto c^{-\frac{1}{2}}y$

we get $g\sim\varphi(\phi(g))=(cx^{2},xy+y^{2})\sim(x^{2},xy+y^{2}).$ ∎

Acknowledgements: This research is funded by Vietnam Ministry of Education and
Training (MOET) under grant number B2024-DQN-02. The latter author would like
to thank the Mathematical Institute of the University of Freiburg for its
hospitality.

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Thuy Huong Pham
Department of Mathematics and Statistics, Quy Nhon University, 170 An Duong
Vuong, Quy Nhon, Vietnam
Email address<EMAIL_ADDRESS>
Gerhard Pfister
Department of Mathematics, University of Kaiserslautern, Germany
Email address<EMAIL_ADDRESS>
Gert-Martin Greuel
Department of Mathematics, University of Kaiserslautern, Germany
Email address<EMAIL_ADDRESS>