# Detecting modules inside the group-orbits $\mathscr{R}$, ${\mathscr{K}}\\!,$
$\mathscr{A}\circlearrowright Maps(X,(\Bbbk^{p},o))$.
The case of arbitrary characteristic

Dmitry Kerner  Department of Mathematics, Ben Gurion University of the Negev,
P.O.B. 653, Be’er Sheva 84105, Israel<EMAIL_ADDRESS>
(Date: filename: main.tex)

###### Abstract.

The germs of maps $(\Bbbk^{n},o)\stackrel{{\scriptstyle
f}}{{\to}}(\Bbbk^{p},o)$ are traditionally studied up to the right
($\mathscr{R}$), left-right ($\mathscr{A}$) or contact (${\mathscr{K}}\\!$)
equivalence. Various questions about the group-orbits $\mathscr{G}f$ are
transformed to the tangent spaces, $T_{\mathscr{G}}f$. Classically this
passage was done by vector fields integration, hence it was bound to the
$\mathbb{R}/\mathbb{C}$-analytic or $C^{r}$-category.

The characteristic-free approach to the groups $\mathscr{R},{\mathscr{K}}\\!$
has been developed during the last decade. The results were still restricted.
E.g. one could not address the (essentially more complicated)
$\mathscr{A}$-equivalence. Moreover, the characteristic-free criteria were
weaker than those in characteristic zero, because of the (inevitable)
pathologies of positive characteristic.

In this paper we close these omissions.

  * •

We establish the general (characteristic-free) passage
$T_{\mathscr{G}}f\rightsquigarrow\mathscr{G}f$ for the groups $\mathscr{R}$,
${\mathscr{K}}\\!,$ $\mathscr{A}$. Submodules of $T_{\mathscr{G}}f$ ensure
submodules of $\mathscr{G}f$. For the $\mathscr{A}$-equivalence this extends
(and strengthens) various classical results by Mather, Gaffney, du Plessis,
and others.

  * •

Given a filtration on the space of maps one has the filtration on the group,
$\mathscr{G}^{(\bullet)}$, and on the tangent space,
$T_{\mathscr{G}^{(\bullet)}}$. We establish the criteria of type
“$T_{\mathscr{G}^{(j)}}f$ vs $\mathscr{G}^{(j)}f$” in their strongest form,
for arbitrary base field/ring, provided the characteristic is zero or high for
a given $f$.

As an auxiliary step, important on its own, we study the mixed module
structure of the tangent space $T_{\mathscr{A}}f$ and establish several
properties of the annihilator ideal
$\mathfrak{a}_{\mathscr{A}}:=Ann[T^{1}_{\mathscr{A}}f]$.

I was supported by the Israel Science Foundation (grant No. 1910/18)

##  1\. Introduction

### 1.1.

Consider $\Bbbk$-analytic map-germs, $(\Bbbk^{n},o)\stackrel{{\scriptstyle
f}}{{\to}}(\Bbbk^{p},o)$, here $\Bbbk$ is $\mathbb{R}$ or $\mathbb{C}$.
Whitney’s initial results on stable maps were greatly developed and extended
by Thom, Mather and many others. (See e.g. [A.G.L.V.], [Martinet], [Mo.
N.B.].) The maps are traditionally studied up to the left-right
($\mathscr{A}$) equivalence, with the auxiliary right ($\mathscr{R}$) and
contact (${\mathscr{K}}\\!$) equivalences. The first questions are about the
group orbits, $\mathscr{G}f$, in particular the finite determinacy.

The classical methods relied heavily on vector fields integration. This
chained the whole theory to the $\mathbb{R},\mathbb{C}$-analytic (or
$C^{k}$-differentiable) setting.

The study of $\mathscr{R},{\mathscr{K}}\\!$-equivalences in zero and positive
characteristic (by purely algebraic methods) has begun in 80’s, e.g. [Greuel-
Kröning.90]. By now numerous results are available, [Bou.Gr.Ma.12],
[Bou.Gr.Ma.11], [Greuel-Nguyen.16], [Nguyen.19], [Greuel-Pham.19], [B.K.16],
[B.G.K.22]. The results in positive characteristic are (naturally) weaker than
those in zero characteristic.

The $\mathscr{A}$-case is essentially more complicated than the
$\mathscr{R},{\mathscr{K}}\\!$-cases, the case of arbitrary characteristic was
untouched. The $\mathscr{A}$-orbits cannot be studied by the (algebraic)
methods of these papers, e.g. because of the following difficulties.

  * •

The group-action
$\mathscr{A}\circlearrowright\rm{Maps}\big{(}(\Bbbk^{n},o),\\!(\Bbbk^{p},o)\big{)}$
is neither additive, nor $\Bbbk$-multiplicative, unlike the actions
$\mathscr{R},{\mathscr{K}}\\!^{lin}\circlearrowright\rm{Maps}\big{(}(\Bbbk^{n},o),\\!(\Bbbk^{p},o)\big{)}$.

  * •

The tangent space $T_{\mathscr{A}}f$ is not an
$\mathcal{O}_{(\Bbbk^{n},o)}$-module, it is a “mixed
$(\mathcal{O}_{(\Bbbk^{n},o)},\mathcal{O}_{(\Bbbk^{p},o)})$-module”.

  * •

The traditional Artin approximation (so helpful for
$\mathscr{R},{\mathscr{K}}\\!$-equivalences) is absent for the left-right
equivalence. (The “inverse Artin question” has negative answer, [Gabrielov].)

In the current paper we study the $\mathscr{A}$-equivalence in arbitrary
characteristic, in particular developing the needed algebraic methods.
Somewhat surprisingly, our results are stronger than the known ones, even in
the classical case of $\mathscr{R},{\mathscr{K}}\\!$-equivalences over
$\mathbb{R},\mathbb{C}$. (Even the asymptotic is stronger.)

In [Kerner] we use these methods to develop the theory of unfoldings, the
theory of stable maps, and to establish Mather-Yau/Gaffney-Hauser results in
arbitrary characteristic.

###  1.2. The structure and contents of the paper

We study maps of henselian germs (of any characteristic)
$Maps(X,(\Bbbk^{p},o))$. Here $X=Spec(R_{X})$ and $(\Bbbk^{p},o)=Spec(R_{Y})$,
the main examples of $(R_{X},R_{Y})$ are:
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk[[y]])$,
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk\\{y\\})$,
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk\langle y\rangle)$,
see §2.1.ii.

  1. §2

is preparatory. In §2.1 we fix the notations and recall numerous auxiliary
results (e.g. Weierstraß finiteness, Baker-Campbell-Hausdorff formula, Artin
approximation).

In §2.2 we recall the “local coordinate changes”, i.e. the ring automorphisms
$Aut_{X}:=Aut_{\Bbbk}(R_{X})$, the germs of vector fields (i.e. derivations),
$Der_{X}:=Der_{\Bbbk}(R_{X})$, and the corresponding filtrations, on the
group, $\\{Aut^{\bullet}_{X}\\}$, and on the module,
$\\{Der^{\bullet}_{X}\\}$.

In §2.3 we recall the truncated exp/log maps. In zero or high characteristic
they substitute the integration of vector fields.

  2. §3

We define the group actions, $\mathscr{G}\circlearrowright
Maps(X,(\Bbbk^{p},o))\cong{\mathfrak{m}}\cdot R^{\oplus p}_{X}$, for
$\mathscr{G}\in\mathscr{R},{\mathscr{K}}\\!,\mathscr{A}$, and the
corresponding (extended) tangent spaces $T_{\mathscr{G}}f$. The filtration
$I^{\bullet}\cdot R^{\oplus p}_{X}$ induces the filtrations of the group,
$\mathscr{G}\rhd\mathscr{G}^{(\bullet)}$, and of the module,
$T_{\mathscr{G}}\supseteq T_{\mathscr{G}^{(\bullet)}}$, see §3.4. For
$\mathscr{G}\in\mathscr{R},{\mathscr{K}}\\!$ this is known, [B.K.16],
[B.G.K.22], but the $\mathscr{A}$-case is new. In §3.4.2 we settle “little
issues” of $\mathscr{A}^{\bullet}$, $T_{\mathscr{A}^{\bullet}}$, when $\Bbbk$
is an infinite field.

This filtration induces the filtration topology, §3.5. The tangent space
$T_{\mathscr{G}}f$, and the orbit $\mathscr{G}f$, are closed in this topology
for $\mathscr{G}\in\mathscr{R},{\mathscr{K}}\\!$. The $\mathscr{A}$-case is
more complicated, see §6.

In §3.6 we establish the algebraic version of Thom-Levine’s “naturality of
vector fields integration”. This is used through the paper (when the
characteristic is zero or high enough) in the same way as the original Thom-
Levine’s statement is used in the $\mathbb{R},\mathbb{C}$-analytic case.

In §3.7 we introduce the annihilator ideal of the tangent space to
$\mathscr{G}$-miniversal unfoldings,
$\mathfrak{a}_{\mathscr{G}}:=Ann[T^{1}_{\mathscr{G}}f]$. For functions on
smooth germs, i.e. $(\Bbbk^{n},o)\to(\Bbbk^{1},o)$, these are the classical
Jacobian/Tjurina ideals, $\mathfrak{a}_{\mathscr{R}}=Jac(f)$,
$\mathfrak{a}_{\mathscr{K}}\\!=Jac(f)+(f)$. More generally, the ideal
$\mathfrak{a}_{\mathscr{R}}$ defines the critical locus of a map,
$\mathfrak{a}_{\mathscr{K}}\\!$ defines the singular locus,
$\mathfrak{a}_{\mathscr{A}}$ defines the instability locus. We rederive the
classical fact: the restriction $f|_{Crit(f)}:X\to(\Bbbk^{p},o)$ is a finite
morphism iff $f$ is ${\mathscr{K}}\\!$-finite (and other characterizations),
lemma 3.14.

  3. §4

is a warmup, addressing the $\mathscr{R}$-equivalence.

    1. §4.1

contains the “$\mathscr{R}$-implicit function theorem”, ensuring that a large
part of the tangent space sits inside the orbit:
$\mathscr{R}f\supseteq\\{f\\}+\mathfrak{a}\cdot T_{\mathscr{R}}f$. This
statement (and its proof) is characteristic-free.

    2. §4.2

Numerous examples show that this bound is much stronger than the known bounds
(even in the $\mathbb{R},\mathbb{C}$-cases). As a trivial corollary we get the
Morse lemma over an arbitrary field. A geometric corollary: the map $f$ is
$\mathscr{R}$-determined by its “2-jet” on the critical locus,
$\mathscr{R}f\supseteq\\{f\\}+\mathfrak{a}_{\mathscr{R}}^{2}\cdot R^{\oplus
p}_{X}$.

    3. §4.3

We prove the filtration criterion:
“$\mathscr{R}^{(j)}f\supseteq\\{f\\}+I^{d}\cdot R^{\oplus p}_{X}$ iff
$T_{\mathscr{R}^{(j)}}f\supseteq I^{d}\cdot R^{\oplus p}_{X}$”. (In positive
characteristic the integers $j,d$ are restricted by the condition
$\frac{d-ord(f)}{j}<char(\Bbbk)$, i.e. for a given $j,d$ the characteristic
must be large enough.) The proof of this theorem 4.5 follows closely those of
[B.K.16], [B.G.K.22], using the truncated maps $jet_{N}(Exp)$, $jet(Ln)$.

In zero characteristic this criterion is well known. In positive
characteristic it strengthens the known criteria, [Greuel-Pham.19],
[B.G.K.22], all being of type “If $T_{\mathscr{R}^{(j)}}f\supseteq I^{d}\cdot
R^{\oplus p}_{X}$ then
$\mathscr{R}^{(j+d)}f\supseteq\\{f\\}+I^{2d-ord(f)+1}\cdot R^{\oplus p}_{X}$”.
(We remark that the new statement is of “iff” type.)

  4. §5

is the second warmup, addressing the ${\mathscr{K}}\\!$-equivalence.

    1. §5.1

contains the “${\mathscr{K}}\\!$-implicit function theorem”, ensuring that a
large part of the tangent space sits inside the orbit:
${\mathscr{K}}\\!f\supseteq\\{f\\}+(\mathfrak{a}\cdot\mathfrak{a}_{\mathscr{R}}+{\mathfrak{m}}(f))\cdot
R^{\oplus p}_{X}$. As in the $\mathscr{R}$-case, this is characteristic-free
and much stronger than the known bounds (as is shown by several examples). A
geometric corollary: the map $f$ is ${\mathscr{K}}\\!$-determined by its
“2-jet” on the singular locus,
${\mathscr{K}}\\!f\supseteq\\{f\\}+\mathfrak{a}^{2}_{\mathscr{K}}\\!\cdot
R^{\oplus p}_{X}$.

    2. §5.2

is a simple application, we study the ${\mathscr{K}}\\!$-determinacy of a
reduced complete intersection curve germ $(C,o)\subset(\Bbbk^{n},o)$.

    3. §5.3

we prove the filtration criterion:
“${\mathscr{K}}\\!^{(j)}f\supseteq\\{f\\}+I^{d}\cdot R^{\oplus p}_{X}$ iff
$T_{{\mathscr{K}}\\!^{(j)}}f\supseteq I^{d}\cdot R^{\oplus p}_{X}$”. (As in
the $\mathscr{R}$-case, for $char(\Bbbk)>0$ the integers $j,d$ are restricted
by the characteristic, $\frac{d-ord(f)}{j}<char(\Bbbk)$.) As in the
$\mathscr{R}$-case, this statement is known in zero characteristic, but
strengthens the known criteria in positive (high enough) characteristic.

The proof uses (besides the truncated maps $jet_{N}(Exp)$, $jet(Ln)$) the
analysis of the full Baker-Campbell-Hausdorff expansion. We bound the prime
numbers appearing in the denominators of this formula.

The $\mathscr{A}$-equivalence is much more complicated. (Recall that the
${\mathscr{K}}\\!$-equivalence was initially introduced as a simpler version
of $\mathscr{A}$-equivalence.) The next two sections are the core of the
paper.

  5. §6

is “The theory of $T_{\mathscr{A}}f$”.

    1. §6.1.

The annihilator ideal of the tangent to the $\mathscr{A}$-miniversal
unfolding, $\mathfrak{a}_{\mathscr{A}}\\!:=\\!Ann[T^{1}_{\mathscr{A}}f]$ is
much more delicate than the corresponding annihilators
$\mathfrak{a}_{\mathscr{R}},\\!\mathfrak{a}_{\mathscr{K}}\\!$. We establish
the basic properties of $\mathfrak{a}_{\mathscr{A}}.$

    2. §6.2.

The (extended) tangent space $T_{\mathscr{A}}f$ is not an $R_{X}$-module.
Considered as an $R_{Y}$-module it is often not finitely-generated (e.g. when
$dim(X)>p$). We prove that $T_{\mathscr{A}}f$ is a module over the extended
ring $f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}}$. Moreover, $T_{\mathscr{A}}f$
is finitely-generated over $f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}}$ in the
most important case, “$f$ is of finite singularity type” (i.e. $f$ is
${\mathscr{K}}\\!$-finite).

    3. §6.3.

As $T_{\mathscr{A}}f$ is not an $R_{X}$-module, the usual Nakayama statement
“$M\subseteq T_{\mathscr{A}}f+{\mathfrak{m}}\cdot M$ implies $M\subseteq
T_{\mathscr{A}}f$” does not hold. We prove weaker versions.

    4. §6.4.

Similarly the assumption $T_{\mathscr{A}^{(j)}}f\supseteq I^{d}\cdot R^{\oplus
p}_{X}$ does not imply $T_{\mathscr{A}^{(j+1)}}f\supseteq I^{d+1}\cdot
R^{\oplus p}_{X}$. Yet, we establish weaker statements of Artin-Rees type.

    5. §6.5.

Unlike the tangent spaces $T_{\mathscr{R}}f,T_{\mathscr{K}}\\!f$, the space
$T_{\mathscr{A}}f$ is not “obviously filtration-closed”. One cannot deduce
$\overline{T_{\mathscr{A}}f}=T_{\mathscr{A}}f$ just from the
faithfulness/exactness of the completion functor
$\lim\limits_{\leftarrow}{\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{$I^{\bullet}$}\right.}\otimes$.
Yet we prove: $T_{\mathscr{A}}f=\overline{T_{\mathscr{A}}f}$ under weak
assumptions on $f$ and $R_{X}$.

  6. §7

is about large modules inside $\mathscr{A}$-orbits.

    1. §7.1

contains the $\mathscr{A}$-implicit function theorem,
$\mathscr{A}f\supseteq\\{f\\}+\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X}+T_{\mathscr{L}^{(1)}}f$. This is the main result of the paper. The
proof is essentially more complicated than those in the
$\mathscr{R},{\mathscr{K}}\\!$-cases. The additional ingredients are: the
Weierstraß division, an arc solution, the passage to the finite jets, and
Zariski-closedness of the orbits of unipotent algebraic groups.

The first immediate corollary is for the stable maps:
$\mathscr{A}f\supseteq\\{f\\}+\mathfrak{a}^{2}_{{\mathscr{K}}\\!^{(o)}}\cdot
R^{\oplus p}_{X}$. This bound is completely new even in the classical case
$R_{X}=\mathbb{C}\\{x\\},\mathbb{R}\\{x\\}$, and is much stronger than the
known bounds.

Then we give other examples showing how to apply this $\mathscr{A}$-$IFT$
theorem and to extend numerous classical results to arbitrary characteristic.
(Any $\mathscr{A}$-finite map is $\mathscr{A}$-finitely determined, various
explicit determinacy bounds, and so on.)

    2. §7.2

contains the filtration criterion,
“$\mathscr{A}^{(j)}f\supseteq\\{f\\}+I^{d}\cdot R^{\oplus p}_{X}$ iff
$T_{\mathscr{A}^{(j)}}f\supseteq I^{d}\cdot R^{\oplus p}_{X}$”. (As in the
$\mathscr{R},{\mathscr{K}}\\!$-cases, for $char(\Bbbk)>0$ the integers $j,d$
are restricted by the characteristic, $\frac{2d-1-ord(f)}{j}<char(\Bbbk)$.).)
This statement is known for $R_{X}=\Bbbk\\{x\\}$,
$\Bbbk\in\mathbb{R},\mathbb{C}$, but is new in zero/positive characteristic.
The essential ingredient in the proof is the Thom-Levine lemma of §3.6.

    3. §7.3

contains the geometric criterion of determinacy, $f$ is determined by its
values over the image of the instability locus,
$f(V(\mathfrak{a}_{\mathscr{A}}))$. For $\mathscr{A}$-finite maps this extends
the classical Mather-Gaffney criterion.

###  1.3. Acknowledgement

Thanks are to G.-M. Greuel, D. Mond, M. A. S. Ruas for important advices. I
was heavily influenced by their papers and also by those of T. Gaffney, A. du
Plessis and L. Wilson.

##  2\. Notations, conventions, preliminaries

###  2.1. Local rings/germs of schemes

All the rings in this paper are commutative, Noetherian, local
$\Bbbk$-algebras, where $\Bbbk$ is a local ring. In the simplest case $\Bbbk$
is a field, classically $\Bbbk=\mathbb{R},\mathbb{C}$. The case when $\Bbbk$
is not a field is important for deformations/unfoldings, e.g.
$\Bbbk=\tilde{\Bbbk}[[t]]$, see [Kerner]. We always assume that the algebra
structure map is injective, $\Bbbk\hookrightarrow R$, and identify $\Bbbk$
with its image.

  1. i.

Denote by $\Bbbk^{\times}$ the group of invertible elements of $\Bbbk$.
Sometimes we impose the condition “$2,\dots,d\in\Bbbk^{\times}$”. This means:
either $\Bbbk\supseteq\mathbb{Q}$, or $\Bbbk$ contains a field of
characteristic bigger than $d$.

  2. ii.

Let $(R,{\mathfrak{m}})$ be a local henselian $\Bbbk$-algebra. Geometrically
we have the germ of a $\Bbbk$-scheme $X:=Spec(R)$. The most important examples
are:

     * •

The formal power series $R=\Bbbk[[x]]$, for $x=(x_{1},\dots,x_{n})$. In this
case we take as $\Bbbk$ either a(ny) field or a complete local ring. Then $R$
is complete with respect to the filtrations ${\mathfrak{m}}^{\bullet}$ and
$(x)^{\bullet}$.

     * •

The locally convergent power series $R=\Bbbk\\{x\\}$. In this case we always
assume $\Bbbk$ to be a normed ring, complete with respect to its norm. The
simplest case is a normed field, e.g. $\mathbb{R},\mathbb{C},\mathbb{Q}_{p}$.

     * •

The algebraic power series $R=\Bbbk\langle x\rangle$. In this case we assume:

       * –

either $\Bbbk$ is a(ny) field. Then $R=\Bbbk\langle x\rangle$ is the ring of
power series satisfying polynomial equations. This is a local henselian ring.

       * –

or $\Bbbk$ is a local henselian ring (with the maximal ideal
${\mathfrak{m}}_{\Bbbk}$). Then $R=\Bbbk\langle x\rangle$ denotes the
henselization of the ring $\Bbbk[x]$ at the ideal
${\mathfrak{m}}_{\Bbbk}+(x)$. See [Moret-Bailly] for more detail.

     * •

The quotient of these rings by an ideal,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$.

Geometrically $X=Spec(R)\subseteq(\Bbbk^{n},o)$ is a
formal/$\Bbbk$-analytic/henselian germ, and $x$ denotes the local coordinates
on the ambient space $(\Bbbk^{n},o)$.

The assumptions on $\Bbbk$ (local complete, resp. local henselian, resp.
normed and complete for its norm) are needed to ensure the implicit function
theorem, §2.1.iv.

  3. iii.

Take the ${\mathfrak{m}}$-completion, $R\to{\widehat{R}}$, it is injective as
$R$ is Noetherian. If $\Bbbk$ is a field then, by Cohen’s structure theorem,
we identify
${\widehat{R}}={\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$.
If $\Bbbk$ is a local ring then we take the
${\mathfrak{m}}\cap\Bbbk$-completion, $\Bbbk\hookrightarrow\hat{\Bbbk}$, and
then identify
${\widehat{R}}={\footnotesize\left.\raisebox{1.37775pt}{$\hat{\Bbbk}[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$.
In both cases the elements of $R$ can be presented by power series,
occasionally we write $f(x)$. We have $(x)\subseteq{\mathfrak{m}}\subset R$.
If $\Bbbk$ is a field then $(x)={\mathfrak{m}}$.

For the general element $f(z)\in R[[z]]$ the composition $f\circ
g\in{\widehat{R}}$ is not defined unless $g\in{\mathfrak{m}}\subset R$. In the
proof of theorems 4.1, 5.1, 7.1 the series $f$ are of special form, $f(z)\in
R[[{\mathfrak{m}}\cdot z]][z]$, i.e. $f$ is a polynomial in $z$ whose
coefficients are power series in ${\mathfrak{m}}\cdot z$. In this case the
composition $f\circ g$ is well defined for any $g\in R$.

For an element $f\in R$ and an ideal $I\subset R$ denote the $I$-order of $f$
by $ord(f)$. Thus $(f)\subseteq I^{ord(f)}$ and $(f)\not\subseteq
I^{ord(f)+1}$. Sometimes we use the ideal $I^{ord(f)-2}$. If $ord(f)\leq 2$ or
$I=R_{X}$ we take $I^{ord(f)-2}:=R_{X}$.

  4. iv.

Implicit function theorem with unit linear part, $\mathbf{IFT_{1\rm I}}$. We
often have to resolve implicit function equations of type $z=f(z)$ for a
vector of power series $f(z)\in R[[z]]^{\oplus p}$, here
$z=(z_{1},\dots,z_{p})$. To specify that the equations belong to the same
category as $R$ (e.g. analytic/henselian) we impose the valuative condition
$f({\mathfrak{m}}\cdot R^{\oplus p})\subseteq R^{\oplus p}$. For
$R={\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$I$}\right.}$
(with $\Bbbk$ any ring) this condition holds trivially for any power series
$f$. For many rings this condition is ensured by composition property, e.g.
for the analytic ring,
$R={\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
any analytic power series $f\in R\\{z\\}$ is allowed. Similarly for
$R={\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$ any algebraic series
$f\in R\langle z\rangle$ is allowed.

######  Definition 2.1.

[B.K.16, §3.4] We say that $IFT_{1\rm I}$ holds in $R$ if the system of
equations $z=f(z)$ has a solution (in $R$) for any vector
$f(z)\in({\mathfrak{m}}+(z)^{2})\cdot R[[z]]^{\oplus p}$ satisfying
$f({\mathfrak{m}})\subseteq{\mathfrak{m}}\cdot R^{\oplus p}$.

This system has the (obvious) unique formal solution. The definition says:
this solution belongs to $R$.

Sometimes the system to resolve has the form $z=h(z)+c$, for a vector $c\in
R^{\oplus p}$, where the entries of the vector $h(z)$ belong to the subset
${\mathfrak{m}}\cdot R[[{\mathfrak{m}}\cdot z]][z]\subset R[[z]]$. This case
is converted to that of definition 2.1 by the substitution ${\tilde{z}}=z-c$.

We remark that the involved equations are non-polynomial. Therefore the
$IFT_{1\rm I}$ condition cannot be replaced by Hensel’s lemma, i.e. $IFT_{1\rm
I}$ does not necessarily hold in all henselian rings.

######  Example 2.2.

$IFT_{1\rm I}$ holds in many rings.

     * •

It holds in the rings
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$, for any field $\Bbbk$,
see example 2.2 of [B.G.K.22].

     * •

In the ring
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
with $\Bbbk$ any complete local ring, $IFT_{1\rm I}$ holds trivially. Indeed,
the solution of $z=f(z)$ can be obtained by an order-by-order procedure.

     * •

$IFT_{1\rm I}$ holds in the ring
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$, for $\Bbbk$ a local
henselian ring, [Lafon]. Indeed, first one gets a formal solution,
$z(x)\in{\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$.
This power series satisfies the equation $z=f(z)$. As $f$ is an algebraic
power series, it satisfies a polynomial equation,
$f^{d}+a_{d-1}f^{d-1}+\cdots+a_{0}=0$. But then $z(x)$ satisfies the same
equation. And therefore
$z(x)\in{\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$.

######  Lemma 2.3.

$IFT_{1\rm I}$ holds in the ring
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$
for $\Bbbk$ a normed ring complete with respect to its norm.

Proof. Fix a system of equations
$z={\tilde{f}}(z)\in\big{(}{\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x,z\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}\big{)}^{\oplus
p}$. Let $f\in\Bbbk\\{x,z\\}^{\oplus p}$ be a representative of $f$. Let
$\\{C_{\bullet}\in\Bbbk\\}$ be the (countable) collection of the non-zero
coefficients of ${\tilde{f}}$. Associate to this collection a set of
algebraically independent positive reals,
$\\{C_{\mathbb{R},\bullet}\in\mathbb{R}_{>0}\\}$, satisfying the norm
condition: $\|C_{j}\|_{\Bbbk}\leq\|C_{\mathbb{R},j}\|_{\mathbb{R}}\leq
2\|C_{j}\|_{\Bbbk}$. Thus we associate to $f$ the power series
$f_{\mathbb{R}}\in\mathbb{R}_{>0}[[{\tilde{x}},{\tilde{z}}]]^{\oplus p}$. By
our construction $f_{\mathbb{R}}$ is the tuple of coverging power series, i.e.
$f_{\mathbb{R}}\in\mathbb{R}_{>0}\\{{\tilde{x}},{\tilde{z}}\\}^{\oplus p}$.
Thus to the system of equations $z=f(z)$ over
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$
we associate the system of equations ${\tilde{z}}=f_{\mathbb{R}}({\tilde{z}})$
over $\mathbb{R}\\{x\\}$. And one still has
$f_{\mathbb{R}}({\tilde{z}})\in(\tilde{\mathfrak{m}}+{\tilde{z}}^{2})\cdot\mathbb{R}\\{x,z\\}^{\oplus
p}$.

The unique solution $z_{\mathbb{R}}({\tilde{x}})$ of the equation
${\tilde{z}}=f_{\mathbb{R}}({\tilde{z}})$ is locally convergent. The
coefficients in the expansion $z_{\mathbb{R}}({\tilde{x}})$ are polynomials in
$\\{C_{\mathbb{R},\bullet}\\}$. Moreover, the coefficients of these
polynomials are positive.

Finally we go back, substituting
$\\{\tilde{C}_{\bullet}\\}\rightsquigarrow\\{C_{\bullet}\\}$ in
$z_{\mathbb{R}}({\tilde{x}})$. We get a power series
$z(x)\in\big{(}{\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}\big{)}^{\oplus
p}$ that solves the equation $z=f(z)$. Moreover, the coefficients of $z(x)$
are norm-bounded by those of $z_{\mathbb{R}}({\tilde{x}})$. Thus $z(x)$ is
locally convergent, i.e.
$z(x)\in({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.})^{\oplus
p}$. $\blacksquare$

  5. v.

Weierstraß division with remainder. Take a local subring
$R\subseteq\Bbbk[[x]]$, here $\Bbbk$ is a local ring (e.g. a field). The
Weierstraß division holds in $R$ if for any $x_{n}$-regular element $f\in R$
of $x_{n}$-order $d_{n}$, and any $g\in R$, one can present $g=q\cdot f+r$,
where $q\in R$ and $r\in R\cap\Bbbk[[x_{1},\dots,x_{n-1}]][x_{n}]$, with
$deg_{x_{n}}r<d_{n}$. This division with remainder holds in many rings, e.g.
$R=\Bbbk[[x]]$, $R=\Bbbk\\{x\\}$ and $R=\Bbbk\langle x\rangle$, see §2.1.ii.
(For the case when $\Bbbk$ is not a field and $\Bbbk\langle x\rangle$ is the
henselization see Theorem 1.1 of [Moret-Bailly].)

  6. vi.

Weierstraß finiteness condition holds for a pair of local $\Bbbk$-algebras
$R_{X},R_{Y}$ if for any homomorphism
$R_{Y}\stackrel{{\scriptstyle\phi}}{{\to}}R_{X}$ and any finitely generated
$R_{X}$ module $M$, with
${\footnotesize\left.\raisebox{1.37775pt}{$M$}\\!/\\!\raisebox{-1.37775pt}{$\phi^{*}({\mathfrak{m}}_{Y})\cdot
M$}\right.}$ f.g. over $\Bbbk$, the module $M$ is f.g. over $R_{Y}$.

This condition is implied by the Weierstraß division condition, in particular
it holds for the pairs
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk[[y]])$,
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk\\{y\\})$,
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk\langle y\rangle)$.

A remark: the $C^{\infty}$-version of Weierstraß finiteness is a deep
preparation theorem of Malgrange.

  7. vii.

Recall the general fact: any $R_{X}$-submodule $M\subseteq R^{\oplus p}_{X}$
is finitely generated. Proof: $R^{\oplus p}_{X}$ itself is a f.g. module over
the Noetherian ring $R_{X}$, therefore $R^{\oplus p}_{X}$ is a Noetherian
module. Hence $M$ is f.g.

  8. viii.

Let $R$ be a local Noetherian ring, $M\subseteq R^{\oplus p}$ a submodule.
Suppose an ideal $I\subset R$ contains a non-zero divisor.

######  Lemma 2.4.

If $I\cdot M\supseteq I\cdot J\cdot R^{\oplus p}$ then $M\supseteq J^{d}\cdot
R^{\oplus p}$ for $d\gg 1$.

Proof.

    1. The case

$\mathbf{J=R}$. Here $I\cdot M\supseteq I\cdot R^{\oplus p}$. If
$M\subseteq{\mathfrak{m}}\cdot R^{\oplus p}$ then we get ${\mathfrak{m}}\cdot
I\cdot R^{\oplus p}\supseteq I\cdot R^{\oplus p}$. Thus (by Nakayama) $I=0$,
contradicting the non-zero divisor in $I$.

If $M\not\subseteq{\mathfrak{m}}\cdot R^{\oplus p}$ then, by reshuffling the
generators of $R^{\oplus p}$, we have $\phi:\ R^{\oplus
p}\xrightarrow{\,\smash{\raisebox{-2.79857pt}{$\scriptstyle\sim$}}\,}R^{\oplus
p_{1}}\oplus R^{\oplus p_{2}}$, with $\phi(M)=R^{\oplus
p_{1}}\oplus(\phi(M)\cap{\mathfrak{m}}\cdot R^{\oplus p_{2}})$. Then by the
previous argument we get $\phi(M)\cap{\mathfrak{m}}\cdot R^{\oplus p_{2}}=0$.
Hence $p_{2}=0$, i.e. $M=R^{\oplus p}$.

    2. The case

$\mathbf{J\subseteq{\mathfrak{m}}\subset R}$. Take a prime ideal not
associated to $J$, thus ${\mathfrak{q}\hskip 1.42271pt}\not\supseteq J$.
Localizing at ${\mathfrak{q}\hskip 1.42271pt}$ we have: $(I\cdot
M)_{\mathfrak{q}\hskip 1.42271pt}\supseteq I_{\mathfrak{q}\hskip
1.42271pt}\cdot R^{\oplus p}_{\mathfrak{q}\hskip 1.42271pt}=R^{\oplus
p}_{\mathfrak{q}\hskip 1.42271pt}$. By the previous step we get:
$M_{\mathfrak{q}\hskip 1.42271pt}=R^{\oplus p}_{\mathfrak{q}\hskip
1.42271pt}$. Therefore $({\footnotesize\left.\raisebox{1.37775pt}{$R^{\oplus
p}$}\\!/\\!\raisebox{-1.37775pt}{$M$}\right.})_{\mathfrak{q}\hskip
1.42271pt}=0$.

Hence this quotient module
${\footnotesize\left.\raisebox{1.37775pt}{$R^{\oplus
p}$}\\!/\\!\raisebox{-1.37775pt}{$M$}\right.}$ is supported only inside the
locus $V(J)\subset Spec(R)$. But then (as $R$ is Noetherian) we get:
$J^{d}\cdot{\footnotesize\left.\raisebox{1.37775pt}{$R^{\oplus
p}$}\\!/\\!\raisebox{-1.37775pt}{$M$}\right.}=0$ for $d\gg 1$. Hence
$J^{d}\cdot R^{\oplus p}\subseteq M$. $\blacksquare$

  9. ix.

Baker-Campbell-Hausdorff formula. Suppose $\Bbbk\supseteq\mathbb{Q}$. Let $M$
be a $\Bbbk$-module (possibly not finitely generated), with a filtration
$M_{\bullet}$. Suppose some operators (not necessarily $\Bbbk$-linear)
$\xi,\eta\circlearrowright M$ are filtration-nilpotent, i.e.
$\xi,\eta(M_{\bullet})\subseteq M_{\bullet+1}$. Then the operators $exp(\xi)$,
$\exp(\eta)$, $\exp(\xi+\eta)$ are defined on the $M_{\bullet}$-completion of
$M$. One has the classical relation
$exp(\xi)exp(\eta)=exp(\sum^{\infty}_{l=1}p_{l}(\xi,\eta))$. Here
$p_{l}(\xi,\eta)$ is a homogeneous polynomial of degree $l$ in non-commuting
variables $\xi,\eta$. The expansion begins as follows: $exp(\xi)\cdot
exp(\eta)=exp\big{(}\xi+\eta+\frac{[\xi,\eta]}{2}+\frac{[\xi,[\xi,\eta]]-[\eta,[\xi,\eta]]}{12}+\cdots\big{)}$.
Below we use the full expansion, see e.g. §5 of chapter 5 of [Jacobson]:

(1)
$\sum^{\infty}_{l=1}p_{l}(\xi,\eta)=\sum^{\infty}_{i=1}\frac{(-1)^{i}}{i}\sum_{\begin{subarray}{c}r_{1}+s_{1}>0\\\
\dots\\\
r_{i}+s_{i}>0\end{subarray}}\frac{[\xi^{r_{1}}\eta^{s_{1}}\cdots\xi^{r_{i}}\eta^{s_{i}}]}{\Big{(}\sum\limits^{i}_{\bullet=1}(r_{\bullet}+s_{\bullet})\Big{)}\Big{(}\mathop{\prod}\limits^{i}_{\bullet=1}r_{\bullet}!\Big{)}\Big{(}\mathop{\prod}\limits^{i}_{\bullet=1}s_{\bullet}!\Big{)}}.$

Here $[\xi^{r_{1}}\eta^{s_{1}}\cdots\xi^{r_{i}}\eta^{s_{i}}]$ is a certain
combination of commutators with $\mathbb{Z}$ coefficients. Comparing the two
sides we get $\sum^{i}_{\bullet=1}(r_{\bullet}+s_{\bullet})=l$ and therefore:

(2)
$p_{l}(\xi,\eta)=\frac{1}{l}\sum^{l}_{i=1}\frac{(-1)^{i}}{i}\sum_{\begin{subarray}{c}\\{r_{\bullet}+s_{\bullet}>0\\}_{\bullet}\\\
\sum^{i}_{\bullet=1}(r_{\bullet}+s_{\bullet})=l\end{subarray}}\frac{[\xi^{r_{1}}\eta^{s_{1}}\cdots\xi^{r_{i}}\eta^{s_{i}}]}{\Big{(}\prod^{i}_{\bullet=1}r_{\bullet}!\Big{)}\Big{(}\prod^{i}_{\bullet=1}s_{\bullet}!\Big{)}}.$

In particular, the rescaled polynomial $(l!)^{4}\cdot p_{l}(\xi,\eta)$ has
only integer coefficients.

We emphasize that the relation $exp(\xi)\cdot exp(\eta)=\dots$ is purely
formal, and no linearity/algebraicity is assumed on the action
$\xi,\eta\circlearrowright M$.

Finally we recall that in the analytic case, i.e.
$R={\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
$\Bbbk\supseteq\mathbb{Q}$, the series $\sum^{\infty}_{l=1}p_{l}(\xi,\eta)$
can be non-convergent for $\xi,\eta\in Der_{\Bbbk}(R)$. However this series is
convergent, i.e. the resulting derivation is analytic,
$\sum^{\infty}_{l=1}p_{l}(\xi,\eta)\in Der_{\Bbbk}(R)$, if one assumes the
filtration-nilpotence, $\xi,\eta(M_{\bullet})\subseteq M_{\bullet+1}$. Indeed,
in this case we get the (filtration-unipotent) analytic automorphisms,
$exp(\xi),exp(\eta)\in Aut_{\Bbbk}(R)$. (See e.g. the proof of part 2 of Lemma
3.17 in [B.K.16].) Their product is a filtration-unipotent automorphism as
well, $exp(\sum^{\infty}_{l=1}p_{l}(\xi,\eta))\in Aut_{\Bbbk}(R)$. And
therefore $\sum^{\infty}_{l=1}p_{l}(\xi,\eta)\in Der_{\Bbbk}(R)$.

  10. x.

Artin approximation. Let $R_{X}$ be one of
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$, see §2.1.ii. Take a
(finite) system of implicit function equations, $F(y)=0$, for $F(y)\in
R_{X}[[y]]^{\oplus N}$, resp. $R_{X}\\{y\\}^{\oplus N}$, resp. $R_{X}\langle
y\rangle^{\oplus N}$. Suppose we have an order-by-order solution, i.e. a
sequence ${\tilde{y}}_{\bullet}(x)\in R^{\oplus q}_{X}$ satisfying:
$F({\tilde{y}}_{\bullet}(x))\in{\mathfrak{m}}^{\bullet}\cdot R^{\oplus
N}_{X}$. The Pfister-Popescu theorem, [Pfister-Popescu] (see also §2.4.1 of
[B.G.K.22]), ensures a formal solution, i.e.
$\hat{y}(x)\in{\widehat{R}}_{X}^{\oplus q}$ satisfying $F(\hat{y}(x))=0$.

Suppose $R_{X}$ is one of
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$. By the Artin
approximation theorem, [Artin], this formal solution is approximated by
$R_{X}$-solutions. Namely, for each $d\geq 1$ there exists $y_{d}(x)\in
R_{X}^{\oplus q}$ satisfying: $F(y_{d}(x))=0$ and
$\hat{y}(x)-y_{d}(x)\in(x)^{d}\cdot{\widehat{R}}_{X}^{\oplus q}$.

A remark: the general Artin approximation holds for polynomial equations over
any henselian local ring. In our case the equations $F(y)=0$ are non-
polynomial, they are (algebraic/analytic) power series. Therefore we restrict
to the particular rings
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$. See [B.G.K.22, §2.4.2]
for more detail.

###  2.2. Germs of coordinate changes (ring automorphisms) and germs of vector
fields (ring derivations)

  1. i.

Denote by $Aut_{\Bbbk}(R)$ the group of $\Bbbk$-linear automorphisms of $R$.
Presenting the elements of $R$ as power series, see §2.1.iii, the
automorphisms act by substitution: $\phi(f(x))=f(\phi(x))$, here
$\phi(x)\in(x)$. Geometrically these are “the coordinate changes” of the germ
$X=Spec(R)$. See [B.G.K.22, §2.3.2] for more detail.

Because of this identification we abbreviate $Aut_{\Bbbk}(R)$ to $Aut_{X}$.

Classically, for $\Bbbk=\mathbb{R},\mathbb{C}$, this group is exactly the
group of right equivalences that preserve the origin, $\mathscr{R}$.

  2. ii.

Denote by $Der_{\Bbbk}(R)\subset End_{\Bbbk}(R)$ the set of $\Bbbk$-linear
derivations. This is an $R$-module. Geometrically this is the module of germs
of vector fields on $X=Spec(R)$, classically denoted by $\Theta_{X}$.

For $R$ a regular ring this module is often generated by the partial
derivatives, $Der_{\Bbbk}(R)=R\langle\frac{\partial}{\partial
x_{1}},\dots,\frac{\partial}{\partial x_{n}}\rangle$. This holds e.g. for
$R=\Bbbk[[x]]$, $\Bbbk\\{x\\}$, $\Bbbk\langle x\rangle$.

More generally, for
$R={\footnotesize\left.\raisebox{1.37775pt}{$S$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$
with $S=\Bbbk[[x]],\Bbbk\\{x\\},\Bbbk\langle x\rangle$, we have the
presentation

(3) $J\cdot Der_{\Bbbk}(S)\to Der(log(J))\to Der_{\Bbbk}(R)\to 0.$

Here $Der(log(J))=\\{\sum c_{j}\frac{\partial}{\partial x_{j}}|\ \sum
c_{j}\frac{\partial}{\partial x_{j}}(J)\subseteq J\\}\subset Der_{\Bbbk}(S)$
is the module of log-derivations. Its elements correspond to the vector fields
tangent to the subgerm $V(J)\subset(\Bbbk^{n},o)$.

Denoting $X=Spec(R)$ we abbreviate $Der_{\Bbbk}(R)$ to $Der_{X}$ and
$End_{\Bbbk}(R)$ to $End_{X}$.

Applying the derivations to an element $f\in R^{\oplus p}$ we get the
submodule $Der_{X}(f)\subseteq R^{\oplus p}$. While the module $Der_{X}$ is
not always finitely generated, the module $Der_{X}(f)$ is f.g., see §2.1.vii.

  3. iii.

In $\mathbb{R}/\mathbb{C}$-analytic (or $C^{\infty}$-) geometry one integrates
vector fields to flows. Taking the flows at time $t=1$ one gets the map
$\Theta_{X}\to Aut_{X}$. Over an arbitrary ring $\Bbbk$ (without any
topology/integration) the maps $Der_{X}\rightleftarrows Aut_{X}$ are
constructed in §2.3, using the following filtration-topology.

Fix a descending filtration by ideals, $R=I_{0}\supset I_{1}\supset\cdots$,
with $\cap I_{j}=(0)$. This defines the filtration topology on $R$.
Accordingly, for each $j\geq 0$, one considers the coordinate changes that are
identity modulo higher order terms:

(4) $Aut^{(j)}_{X}:=\\{\Phi\in Aut_{X}|\ \Phi(I_{d})=I_{d}\ \text{ and
}\quad\Phi|_{{\footnotesize\left.\raisebox{1.37775pt}{$I_{d}$}\\!/\\!\raisebox{-1.37775pt}{$I_{d+j}$}\right.}}=Id|_{{\footnotesize\left.\raisebox{1.37775pt}{$I_{d}$}\\!/\\!\raisebox{-1.37775pt}{$I_{d+j}$}\right.}},\
\forall\ d\geq 1\\}.$

Thus $Aut^{(0)}_{X}$ consists of automorphisms that preserve the filtration,
while $Aut^{(1)}_{X}$ consists of “topologically unipotent” automorphism.

The sets $Aut^{(j)}_{X}$ are normal subgroups, and we get the filtration
$Aut_{X}\trianglerighteq Aut^{(0)}_{X}\trianglerighteq
Aut^{(1)}_{X}\triangleright\cdots$.

Similarly, for every $j\in\mathbb{Z}$, we consider the “topologically
nilpotent” derivations,

(5) $Der^{(j)}_{X}:=\\{\xi\in Der_{X}|\ \xi(I_{d})\subseteq I_{j+d},\ \forall\
d\geq 1\\}=Der_{X}\cap End^{(j)}_{\Bbbk}(R).$

We get the filtration by submodules, $Der_{X}\supseteq\cdots\supseteq
Der^{(-1)}_{X}\supseteq Der^{(0)}_{X}\supseteq Der^{(1)}_{X}\supset\cdots$.

######  Example 2.5.

For the filtration by powers of ideals, $I^{\bullet}\subset R$, we have
$Der_{X}=Der^{(-1)}_{X}$ and $Der^{(j)}_{X}\supseteq I^{j+1}\cdot Der_{X}$. In
the particular case $I={\mathfrak{m}}$ one has $Aut_{X}=Aut^{(0)}_{X}$.

For more detail see [B.G.K.22, §2.2.3].

###  2.3. Exponential/logarithmic maps and their approximations. Integrating
vector fields

Let $(R,{\mathfrak{m}},I_{\bullet})$ be a local filtered $\Bbbk$-algebra.
Assume $\Bbbk\supseteq\mathbb{Q}$ and $R$ is complete with respect to the
filtration. Then the derivations and automorphisms admit the exponential and
logarithmic maps, see Theorem 3.6 of [B.G.K.22]:

(6) $\forall\ j\geq 1:\quad\quad Der^{(j)}_{X}\stackrel{{\scriptstyle
exp}}{{\to}}Aut^{(j)}_{X},\quad
exp(\xi):=e^{\xi}=\sum\nolimits^{\infty}_{i=0}\frac{\xi^{i}}{i!},$ $\forall\
j\geq 1:\quad\quad Aut^{(j)}_{X}\stackrel{{\scriptstyle
ln}}{{\to}}Der^{(j)}_{X},\quad
ln(\Phi):=-\sum\nolimits^{\infty}_{i=1}\frac{({1\rm I}-\Phi)^{i}}{i}.$

When $\Bbbk\not\supseteq\mathbb{Q}$ or $R$ is not $I_{\bullet}$-complete one
cannot use the classical exponential/logarithmic maps. But in many cases one
can use the following “$N$’th jet approximations” to these maps. Fix $j\geq 1$
and $1\leq N\leq\infty$. For the filtration criteria, §4.3,§5.3,S7.2, we
impose the conditions:

  * •

$\boldsymbol{jet_{N}(Exp):}$ for any derivation $\xi\in Der^{(j)}_{X}$ there
exists an endomorphism $\phi_{\xi}\in End^{(jN+1)}_{X}$ that gives the
automorphism $jet_{N}(e^{\xi}):=\sum^{N}_{i=0}\frac{\xi^{i}}{i!}+\phi_{\xi}\in
Aut^{(j)}_{X}$.

  * •

$\boldsymbol{jet_{N}(Ln):}$ for any automorphism $\phi\in Aut^{(j)}_{X}$ there
exists an endomorphism $\xi_{\phi}\in End^{(jN+1)}_{X}$ that gives the
derivation
$jet_{N}(ln(\phi)):=-\sum^{N}_{i=1}\frac{(1-\phi)^{i}}{i}+\xi_{\phi}\in
Der^{(j)}_{X}$.

Here $End_{X}:=End_{\Bbbk}(R)$ is the $\Bbbk$-module of $\Bbbk$-linear
endomorphisms of $R$, and $End^{(jN+1)}_{X}\subset End_{X}$ is the
corresponding submodule of topologically nilpotent endomorphisms.

These conditions imply, in particular: $2,\dots,N\in\Bbbk^{\times}$, see
§2.1.i.

For the “$\mathscr{G}$-implicit function theorems” we impose the following
condition. Take the completion (§2.1.iii)
$R_{X}\hookrightarrow{\footnotesize\left.\raisebox{1.37775pt}{$\hat{\Bbbk}[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$.
Present $\xi\in Der_{X}$ by the derivation $\sum c_{i}(x)\partial_{x_{i}}$. If
$J=0$ then the map $x\to x+\sum c_{i}\partial_{x_{i}}(x)$ with
$c_{\bullet}\in{\mathfrak{m}}^{2}$ defines an automorphism $\Phi\in Aut_{X}$.
For $J\neq 0$ this is not a self-map of $R_{X}$, as it does not preserve $J$.
Therefore we assume:

$\bullet\boldsymbol{jet_{0}:}$ Any map $x\to x+\sum c_{i}\partial_{x_{i}}(x)$
with $c_{\bullet}\in{\mathfrak{m}}^{2}$ and $\sum c_{i}\partial_{x_{i}}\in
Der_{X}$ extends to an automorphism $\Phi\in Aut_{X}$ that satisfies:
$\Phi(x)-x-\sum c_{i}\partial_{x_{i}}(x)\in(\\{c_{\bullet}\\})^{2}$.

######  Example 2.6.

The condition $jet_{1}(Exp)$ means: “any $\xi\in Der^{(j)}_{X}$ extends to a
coordinate change $\Phi_{X}\in Aut^{(j)}_{X}$.” This condition holds for any
field if $J=(0)$, and holds for any $J$ if $\Bbbk\supseteq\mathbb{Q}$. For
$char(\Bbbk)>0$ and $J\neq(0)$ this is a non-trivial condition, see §3.2 of
[B.G.K.22], and also §2.4 below. The condition $jet_{0}$ is of the same nature
and is usually weaker than $jet_{1}$.

######  Example 2.7.

For $N\geq 2$ the conditions $jet_{N}(Exp),jet_{N}(Ln)$ hold, e.g. in the
following cases.

  * •

$R={\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
with $\Bbbk\supseteq\mathbb{Q}$, and any $N\leq\infty$. Here one takes just
the full power series in (6).

  * •

$R={\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$
or $R={\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$, with
$\Bbbk\supseteq\mathbb{Q}$, and any $N<\infty$. Here one shows that such
$\phi_{\xi}$, $\xi_{\phi}$ do exists, under very weak assumptions on the
filtration:

$Der^{(1)}_{X}({\mathfrak{m}})\subseteq{\mathfrak{m}}^{2}\quad\text{ and
}\quad I_{d}\cdot Der_{X}\subseteq Der^{(1)}_{X}\quad\text{ for }\quad d\gg
1.$

For more detail see [B.G.K.22, Theorem 3.6].

  * •

$R=\Bbbk[[x]]$, $\Bbbk\\{x\\}$ or $\Bbbk\langle x\rangle$, with
$2,\dots,N\in\Bbbk^{\times}$, see §3.2.2 of [B.G.K.22].

We remark that the condition $jet_{N}(Exp)$, resp. $jet_{N}(Ln)$, implies
$jet_{l}(Exp)$, resp. $jet_{l}(Ln)$, for $l<N$.

Geometrically the condition $jet_{\infty}(Exp)$ means: every vector field
$\xi\in T_{X}$ integrates to the flow $\Phi_{t}\circlearrowright X$.

###  2.4. Pathologies of non-regular rings

Already for the ring
$R={\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$
with $0\neq J\subseteq(x)^{2}$ many of the previous properties do not hold.

  1. i.

The composition of elements of $R$ is not well defined, unlike the case of
§2.1.iii. For example, suppose $\sqrt{J}\subsetneq(x)$, and take a linear form
$l(x)\not\in\sqrt{J}$. Then for any $0\neq q\in J$ one has $q=0\in R$, but
$q(l(x))\neq 0$.

  2. ii.

If $\Bbbk\not\supseteq\mathbb{Q}$ then even the conditions $jet_{1}(Exp)$,
$jet_{1}(Ln)$ of §2.3 do not always hold. For example, let $\Bbbk$ be a field
of characteristic $p$ and take $f(x,y)=x^{p}+y^{3p}\in\Bbbk[[x,y]]$. Take the
ring
$R={\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x,y]]$}\\!/\\!\raisebox{-1.37775pt}{$(f)$}\right.}$
filtered by the ideals $(x,y)^{\bullet}$. Take
$\xi:=y^{2}\frac{\partial}{\partial x}\in Der^{(1)}_{X}$. The condition
$jet_{1}(Exp)$ here means: there exist $\phi_{x},\phi_{y}\in(x,y)^{3}$ such
that the map $(x,y)\to(x+y^{2}+\phi_{x},y+\phi_{y})$ defines an automorphism
of $R$. But then we must have

(7) $f(x,y)\to
f(x+y^{2}+\phi_{x},y+\phi_{y})=x^{p}+y^{2p}+(\phi_{x})^{p}+y^{3p}+(\phi_{y})^{3p}\in(f(x,y))=(x^{p}+y^{3p}).$

And this condition is clearly non-resolvable because of the term $y^{2p}$.

##  3\. Maps of germs, group actions $\mathscr{G}\circlearrowright Maps(X,Y)$,
and the corresponding tangent spaces

Classically one studies maps of germs
$(\Bbbk^{n},o)\stackrel{{\scriptstyle}}{{\to}}(\Bbbk^{p},o)$, for
$\Bbbk=\mathbb{R},\mathbb{C}$, up to the right ($\mathscr{R}$), left
($\mathscr{L}$), contact (${\mathscr{K}}\\!$) or left-right ($\mathscr{A}$)
equivalences. We introduce the corresponding general notions.

###  3.1. The identification $\mathbf{Maps(X,Y)\approx{\mathfrak{m}}\cdot
R^{\oplus p}_{X}}$

Let $\Bbbk$ be a local ring (e.g. a field) and $(R_{X},{\mathfrak{m}})$ a
local $\Bbbk$-algebra. Here ${\mathfrak{m}}$ is generated by the ideal $(x)$
and the maximal ideal of $\Bbbk$. Let $R_{Y}$ be a regular local henselian
$\Bbbk$-algebra. Typically $R_{Y}$ will be one of $\Bbbk[[y]]$, $\Bbbk\langle
y\rangle$, $\Bbbk\\{y\\}$, for $y=(y_{1},\dots,y_{p})$, see §2.1.ii.
Geometrically $X:=Spec(R_{X})$ is the (germ of) $\Bbbk$-scheme, while
$Y:=Spec(R_{Y})=(\Bbbk^{p},o)$ is the (formal/henselian/analytic) germ of the
affine space over $\Bbbk$.

We identify the maps of germs of $\Bbbk$-schemes and the homomorphisms of
local $\Bbbk$-algebras, $f\in Maps(X,Y)\approx Hom_{\Bbbk}(R_{Y},R_{X})\ni
f^{\\#}$. Any homomorphism $R_{Y}\to R_{X}$ is determined by the image of
$y=(y_{1},\dots,y_{p})$ in ${\mathfrak{m}}$. Thus we have the injection
$Hom_{\Bbbk}(R_{Y},R_{X})\hookrightarrow{\mathfrak{m}}\cdot R^{\oplus p}_{X}$.
In this paper we always assume:

  1. i.

The injection $Hom_{\Bbbk}(R_{Y},R_{X})\hookrightarrow{\mathfrak{m}}\cdot
R^{\oplus p}_{X}$ is an isomorphism. Namely, any assignment
$(y_{1},\dots,y_{p})\to(f_{1},\dots,f_{p})\in{\mathfrak{m}}\cdot R^{\oplus
p}_{X}$ extends (uniquely) to a homomorphism $f^{\\#}:\ R_{Y}\to R_{X}$.

  2. ii.

The pair $(X,Y)$, i.e. $(R_{X},R_{Y})$, satisfies the Weierstraß finiteness
condition, §2.1.vi. Thus any quasi-finite morphism is in fact finite.

These conditions mean: $X$, $Y$ belong to “the same category”, e.g. $Y$ is a
formal/analytic/henselian germ iff so is $X$. More precisely:

  * •

The pairs
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk[[y]])$,
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk\\{y\\})$,
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk\langle y\rangle)$
satisfy these conditions.

  * •

The pairs
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk\\{y\\})$,
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk\langle
y\rangle)$,
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk\langle
y\rangle)$ satisfy i. but not ii.

  * •

The pairs
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk[[y]])$,
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk\\{y\\})$,
$({\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.},\Bbbk[[y]])$ satisfy ii.
(trivially) but not i.

Accordingly we identify $Maps(X,Y)\approx
Hom_{\Bbbk}(R_{Y},R_{X})\cong{\mathfrak{m}}\cdot R^{\oplus p}_{X}$. Though
initially $Maps(X,Y)$ is just a set, this identification equips $Maps(X,Y)$
with the $R_{X}$-module structure.

If $\Bbbk$ is a field then ${\mathfrak{m}}=(x)$ and the maps fix the origin,
$X\ni o\to o\in(\Bbbk^{p},o)$. If $\Bbbk$ is a local ring then
${\mathfrak{m}}\supsetneq(x)$, and $Maps(X,Y)$ can be interpreted as families
of maps, possibly displacing the origin.

Given a map $f\in Hom(R_{Y},R_{X})\cong{\mathfrak{m}}\cdot R^{\oplus p}_{X}$
the space $R^{\oplus p}_{X}$ becomes a (not finitely-generated) $R_{Y}$-module
via the composition,

(8) $q(y)\cdot R^{\oplus p}_{X}:=f^{\\#}(q(y))\cdot R^{\oplus
p}_{X}:=q(f_{1},\dots,f_{p})\cdot R^{\oplus p}_{X}.$

We have the $R_{Y}$-submodule $f^{\\#}(R^{\oplus p}_{Y})\subseteq R^{\oplus
p}_{X}$ and the $R_{X}$-submodule $f^{*}(R^{\oplus p}_{Y})=R^{\oplus p}_{X}$.
One has $f^{\\#}(y)\subseteq f^{*}(y)=(f)\subset R_{X}$, and the first
inclusion is usually proper.

Recall the classical notations, $\omega f({\mathfrak{m}}_{p})$ for
$f^{\\#}(y)$, and $f^{*}({\mathfrak{m}}_{p})$ for $f^{*}(y)$, [Mo. N.B.].

###  3.2. The group-actions
$\mathbf{\mathscr{R},\mathscr{L},\mathscr{C},{\mathscr{K}}\\!,\mathscr{A}\circlearrowright
Maps(X,Y)}$

Consider the elements of $R$ as power series in $x$, see §2.1.iii.

  1. i.

The right action $\mathscr{R}\circlearrowright Maps(X,Y)$ is defined by the
action $Aut_{X}\circlearrowright{\mathfrak{m}}\cdot R^{\oplus p}_{X}$, which
is

$\Phi_{X}(f)=\Phi(f_{1},\dots,f_{p})=f\circ\Phi^{-1}_{X}=(f_{1}\circ\Phi^{-1}_{X},\dots,f_{p}\circ\Phi^{-1}_{X}).$

The action $Aut_{X}\circlearrowright{\mathfrak{m}}\cdot R^{\oplus p}_{X}$ is
$\Bbbk$-linear.

  2. ii.

Present the automorphisms $Aut_{Y}\circlearrowright R_{Y}$ explicitly,
$(y_{1},\dots,y_{p})=y\to\Phi_{Y}(y)=\big{(}\Phi_{1}(y),\dots,\Phi_{p}(y)\big{)}$.
Accordingly define the left action $\mathscr{L}\circlearrowright Maps(X,Y)$,
i.e. $Aut_{Y}\circlearrowright{\mathfrak{m}}\cdot R^{\oplus p}_{X}$, by
$f\to\Phi_{Y}(y)|_{f}$.

We remark that the group $\mathscr{L}$ does not act on the whole module
$R^{\oplus p}_{X}$, as the power series $\Phi_{Y}(y)|_{f}$ are not defined for
$f\not\in{\mathfrak{m}}\cdot R^{\oplus p}_{X}$. (See §2.iii.)

The action $Aut_{Y}\circlearrowright R_{Y}$ is $\Bbbk$-linear, but the action
$Aut_{Y}\circlearrowright{\mathfrak{m}}\cdot R^{\oplus p}_{X}$ is neither
additive nor multiplicative.

  3. iii.

The contact group ${\mathscr{K}}\\!\circlearrowright X\times Y$ acts by
$(x,y)\to(\Phi(x),\Psi(x,y))$. Here $\Phi\in Aut_{X}$ and $\Psi\in
Maps(X\times Y,Y)$ with $\Psi(o,-)\in Aut_{Y}$. Algebraically:
$\Psi(x,y)\in(y)\cdot R_{X}[[y]]^{\oplus p}$, with
$\Psi(o,y)\in\Bbbk[[y]]^{\oplus p}$, and
$\Psi({\mathfrak{m}},{\mathfrak{m}})\in R^{\oplus p}_{X}$. The later condition
ensures in particular that we “remain in the same category”, §2.1.iv., e.g. if
$R_{X}$ is analytic/algebraic then so is $\Psi$.

Accordingly we have the action ${\mathscr{K}}\\!\circlearrowright
Maps(X,Y)\approx{\mathfrak{m}}\cdot R^{\oplus p}_{X}$ by
$f(x)\to\Psi(x,f(\Phi^{-1}(x))$.

Taking $\Phi_{X}=Id$ we get the subgroup $\mathscr{C}\subset{\mathscr{K}}\\!$.
It acts on the fibres of the projection $X\times Y\to X$. This presents the
contact group as the semi-direct product
${\mathscr{K}}\\!=\mathscr{C}\rtimes\mathscr{R}$.

The action ${\mathscr{K}}\\!\circlearrowright{\mathfrak{m}}\cdot R^{\oplus
p}_{X}\cong Maps(X,Y)$ is neither additive nor $\Bbbk$-multiplicative. But the
${\mathscr{K}}\\!$-orbits coincide with the orbits of the $\Bbbk$-linear
action ${\mathscr{K}}\\!^{lin}:=GL(p,R_{X})\rtimes Aut_{X}\circlearrowright
X\times Y$. For $R\\!=\\!\Bbbk\\{x\\}$, $\Bbbk\in\mathbb{R},\mathbb{C}$ this
is well known, see e.g.[Mo. N.B., pg.110]. For the general case see [B.K.16,
pg.123].

Therefore the maps $f,{\tilde{f}}\in{\mathfrak{m}}\cdot R^{\oplus p}_{X}$ are
${\mathscr{K}}\\!$-equivalent iff the corresponding local $\Bbbk$-algebras are
isomorphic,
${\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{$(f)$}\right.}\cong{\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{$({\tilde{f}})$}\right.}$.

  4. iv.

The action of the left-right group,
$\mathscr{A}:=\mathscr{L}\times\mathscr{R}=Aut_{Y}\times
Aut_{X}\circlearrowright Maps(X,Y)$, is defined by

(9) $Aut_{Y}\times Aut_{X}\circlearrowright{\mathfrak{m}}\cdot R^{\oplus
p}_{X},\quad\quad\quad(\Phi_{Y},\Phi_{X})(f):=\Phi_{Y}\circ
f\circ\Phi_{X}^{-1}:=\Phi_{Y}(y)|_{f\circ\Phi_{X}^{-1}}.$

This action is compatible with the product structure:
$(\Phi_{Y},\Phi_{X})\circ(\tilde{\Phi}_{Y},\tilde{\Phi}_{X})=(\Phi_{Y}\circ\tilde{\Phi}_{Y},\Phi_{X}\circ\tilde{\Phi}_{X})$.
In particular,
$(\Phi_{Y},\Phi_{X})=(\Phi_{Y},Id_{X})\circ(Id_{Y},\Phi_{X})=(Id_{Y},\Phi_{X})\circ(\Phi_{Y},Id_{X})$.

While the actions
$\mathscr{R},{\mathscr{K}}\\!^{lin}\circlearrowright{\mathfrak{m}}\cdot
R^{\oplus p}_{X}$ are $\Bbbk$-linear, the actions
$\mathscr{L},\mathscr{A}\circlearrowright{\mathfrak{m}}\cdot R^{\oplus p}_{X}$
are neither additive nor $\Bbbk$-multiplicative. To my knowledge it is not
known whether/how the $\mathscr{A}$-action can be “reduced” to a
$\Bbbk$-linear one. See [Damon91], [Mond-Montaldi], [H.W.-A.] for the related
works.

###  3.3. The (extended) tangent spaces
$\mathbf{T_{\mathscr{R}},T_{\mathscr{L}},T_{\mathscr{A}},T_{\mathscr{K}}\\!}$

The natural extended tangent spaces to the groups
$\mathscr{R},\mathscr{L},{\mathscr{K}}\\!,\mathscr{A}$ are defined via the
modules of derivations and endomorphisms, §2.2. We have:

(10) $T_{\mathscr{R}}:=Der_{X},\quad\quad T_{\mathscr{L}}:=Der_{Y},\quad\quad
T_{\mathscr{K}}\\!:=End_{R_{X}}(R^{\oplus p}_{X})\oplus
T_{\mathscr{R}},\quad\quad T_{\mathscr{A}}:=T_{\mathscr{L}}\oplus
T_{\mathscr{R}}.$

If $R_{Y}$ is one of $\Bbbk[[y]]$, $\Bbbk\\{y\\}$, $\Bbbk\langle y\rangle$,
then one can write explicitly $Der_{Y}=R_{Y}\langle\frac{\partial}{\partial
y_{1}},\dots,\frac{\partial}{\partial y_{p}}\rangle$, i.e.
$\xi_{Y}=\sum\xi_{Y,i}\frac{\partial}{\partial y_{i}}$, here $\xi_{Y,i}\in
R_{Y}$.

The name “tangent space” is justified by the relation
$T_{\mathscr{G}}\rightleftarrows\mathscr{G}$, see §2.3, and [B.G.K.22, §3].

These tangent spaces act on the space of maps as follows:

(11) $T_{\mathscr{R}}\circlearrowright R^{\oplus p}_{X}\quad\quad{\rm
by}\quad\quad\xi_{X}(f):=\xi_{X}(f_{1},\dots,f_{p})=(\xi_{X}(f_{1}),\dots,\xi_{X}(f_{1}))\in
R^{\oplus p}_{X},$ $T_{\mathscr{L}}\\!\\!:\ {\mathfrak{m}}\cdot R^{\oplus
p}_{X}\to R^{\oplus p}_{X}\quad\quad{\rm
by}\quad\quad\xi_{Y}(f):=\xi_{Y}(y)|_{f}=(\xi_{Y,1}|_{f},\dots,\xi_{Y,p}|_{f})\in
R^{\oplus p}_{X}.$

The action $T_{\mathscr{R}}\circlearrowright R^{\oplus p}_{X}$ is
$\Bbbk$-linear. This embeds $Der_{X}\subset End_{\Bbbk}(R^{\oplus p}_{X})$ as
an $R_{X}$-submodule. The map $T_{\mathscr{L}}:{\mathfrak{m}}\cdot R^{\oplus
p}_{X}\to R^{\oplus p}_{X}$ is a power series, neither additive nor
$\Bbbk$-multiplicative.

For a map $f\in{\mathfrak{m}}\cdot R^{\oplus p}_{X}$ one gets the (extended)
image tangent space $T_{\mathscr{G}}f\subseteq R^{\oplus p}_{X}$.

(12) $T_{\mathscr{R}}f=Der_{X}(f),\quad\quad
T_{\mathscr{L}}f=f^{\\#}(R_{Y})^{\oplus p},\quad\quad
T_{\mathscr{A}}f=T_{\mathscr{R}}f+T_{\mathscr{L}}f,\quad\quad
T_{\mathscr{K}}\\!f=T_{\mathscr{R}}f+(f)\cdot R^{\oplus p}_{X}.$

######  Remark 3.1.

  1. i.

Classically these are called “the extended tangent spaces”, with the notations
from differential geometry, [Mo. N.B., pg.72]:

(13) $R_{X}^{\oplus p}=\theta(f),\quad\quad\quad
T_{\mathscr{R}^{e}}f=tf(\theta_{n}),\quad\quad\quad
T_{\mathscr{L}^{e}}f=\omega f(\theta_{p}),\quad\quad\quad
T_{{\mathscr{K}}\\!^{e}}f=tf(\theta_{n})+f^{*}{\mathfrak{m}}_{p}\cdot\theta(f).$

We avoid the notation $T_{\mathscr{G}^{e}}$ to prevent any confusion with the
filtered tangent spaces, $T_{\mathscr{G}^{(j)}}$.

  2. ii.

In the case of a field $\Bbbk\supseteq\mathbb{Q}$ these “image tangent spaces”
coincide with the “tangent spaces to the orbit”,
$T_{\mathscr{G}}f=T_{\mathscr{G}f}$. In positive characteristic the two
notions differ, see §1.5.2 of [B.G.K.22].

  3. iii.

Note that $T_{\mathscr{R}}f,T_{\mathscr{K}}\\!f\subset R^{\oplus p}_{X}$ are
$R_{X}$-submodules, while $T_{\mathscr{L}}f,T_{\mathscr{A}}f\subset R^{\oplus
p}_{X}$ are only $R_{Y}$-submodules, §3.1. Here $R_{Y}$ acts by substitution,
$q(y_{1},\dots,y_{p})\cdot T_{\mathscr{A}}f:=q(f_{1},\dots,f_{p})\cdot
T_{\mathscr{A}}f\subset R^{\oplus p}_{X}$.

  4. iv.

The automorphisms $\Phi_{X}$, $\Phi_{Y}$ commute, see (9). The derivations do
not commute in a naive way,

(14)
$\xi_{X}(\xi_{Y}(y)|_{f})=\xi_{X}(\xi_{Y,1}|_{f},\dots,\xi_{Y,p}|_{f})=(\sum\nolimits_{i}\partial_{y_{i}}\xi_{Y,1}|_{f}\cdot\xi_{X}(f_{i}),\dots,\sum\nolimits_{i}\partial_{y_{i}}\xi_{Y,p}|_{f}\cdot\xi_{X}(f_{i}))=\\\
=\sum\nolimits_{i}\partial_{y_{i}}\xi_{Y}(y)|_{f}\cdot\xi_{X}(f_{i})\neq\xi_{Y}(y)|_{\xi_{X}(f)}.$

To pass from $\xi_{X},\xi_{Y}$ to $\xi_{Y},\xi_{X}$ we aply lemma 3.8. This is
used in §7.2.

###  3.4. Filtrations on the group $\mathscr{G}$ and on the tangent space
$T_{\mathscr{G}}$

Fix an ideal $I\subseteq{\mathfrak{m}}\subset R_{X}$ and take the filtration
by powers of this ideal, $M_{\bullet}:=I^{\bullet}\cdot R^{\oplus
p}_{X}\subseteq R^{\oplus p}_{X}$. Thus $M_{\bullet}\subset Maps(X,Y)$
consists of maps that ‘vanish to a given order” on the locus $V(I)$.

#### 3.4.1.

For $\mathscr{G}\in\mathscr{R},\mathscr{L}$ we get the corresponding
filtrations on the groups/tangent spaces, as in §2.2.iii:

(15) $\mathscr{G}^{(j)}:=\\{g\in\mathscr{G}|\ g\cdot M_{d}=M_{d}\ \text{ and
}\
g|_{{\footnotesize\left.\raisebox{1.37775pt}{$M_{d}$}\\!/\\!\raisebox{-1.37775pt}{$M_{d+j}$}\right.}}=Id|_{{\footnotesize\left.\raisebox{1.37775pt}{$M_{d}$}\\!/\\!\raisebox{-1.37775pt}{$M_{d+j}$}\right.}},\
\forall d\geq 1\\}.$ $T_{\mathscr{G}^{(j)}}:=\\{\xi\in
T_{\mathscr{G}}|\quad\xi(M_{d})\subseteq M_{d+j},\quad\forall d\geq 1\\}.$

For the right equivalence we have $T_{\mathscr{R}^{(j)}}=T_{\mathscr{R}}\cap
End^{(j)}_{\Bbbk}(R^{\oplus p}_{X})$ and
$T_{\mathscr{R}}=T_{\mathscr{R}^{(-1)}}\supsetneq T_{\mathscr{R}^{(0)}}$, cf.
(5).

For the left equivalence we have
$\mathscr{L}^{(j)}=\\{\Phi_{Y}\in\mathscr{L}|\ \Phi_{Y}(y)-y|_{I^{d}\cdot
R^{\oplus p}_{X}}\in I^{d+j}\cdot R^{\oplus p}_{X}\\}$, i.e.
$\Phi_{Y}(y)=y+\phi(y)$, where $\phi(I^{d})\subseteq I^{d+j}$. In particular
$\mathscr{L}^{(0)}=\mathscr{L}$.

The tangent space filtration is $T_{\mathscr{L}^{(j)}}=\\{\xi_{Y}\in Der_{Y}|\
\xi_{Y}(y)|_{I^{d}}\subseteq I^{d+j}\cdot R^{\oplus p}_{X},\ \forall\ d\geq
1\\}$. In particular:

(16) $T_{\mathscr{L}}\supsetneq T_{\mathscr{L}^{(0)}}=(y)\cdot
T_{\mathscr{L}}\supsetneq T_{\mathscr{L}^{(1)}}=(y)^{2}\cdot T_{\mathscr{L}}.$

More generally, $T_{\mathscr{L}^{(j)}}=\mathfrak{b}_{j}\cdot T_{\mathscr{L}}$,
for an ideal $\mathfrak{b}_{j}\subset R_{Y}$ satisfying:
$f^{\\#}(\mathfrak{b}_{j})\subseteq I^{d+j}$ for all $d\in\mathbb{N}$ and all
$f\in I^{d}\cdot R^{\oplus p}_{X}$. In particular, the tangent image is
$T_{\mathscr{L}^{(j)}}f=f^{\\#}(\mathfrak{b}_{j}\cdot R^{\oplus
p}_{Y})\subseteq I^{d+j}\cdot R^{\oplus p}_{X}$.

We have the obvious inclusions: $(y)^{j+1}\subseteq\mathfrak{b}_{j}\subset
R_{Y}$, and $(f)^{j+1}\subseteq f^{\\#}(\mathfrak{b}_{j})\cdot R_{X}$ when
$(f)\subseteq I$. These are equalities when $\Bbbk\supseteq\mathbb{Q}$ is a
field, but can be proper if $\Bbbk$ is a finite field or a ring.

######  Example 3.2.

  1. i.

Suppose $R_{X}$ is one of $\Bbbk[[x]]$, $\Bbbk\\{x\\}$, $\Bbbk\langle
x\rangle$, where $\Bbbk$ is an infinite field. Suppose
$I={\mathfrak{m}}=(x)\subset R_{X}$. Take a map $f\in(x)\cdot R^{\oplus
p}_{X}$. Then:

$T_{\mathscr{R}^{(j)}}=(x)^{j+1}\cdot T_{\mathscr{R}},\quad\quad
T_{\mathscr{R}^{(j)}}f=(x)^{j+1}\cdot Der_{X}(f),\quad\quad
T_{\mathscr{L}^{(j)}}=(y)^{j+1}\cdot T_{\mathscr{L}},\quad\quad
T_{\mathscr{L}^{(j)}}f=f^{\\#}(y)^{j+1}\cdot T_{\mathscr{L}}f\subset R^{\oplus
p}_{X}.$

In the classical notations, cf. (13):
$T_{\mathscr{R}^{(j)}}f=tf({\mathfrak{m}}^{j+1}_{n}\cdot\Theta_{n})$ and
$T_{\mathscr{L}^{(j)}}f=\omega f({\mathfrak{m}}^{j+1}_{p}\cdot\Theta_{p})$.

  2. ii.

Consider $Maps((\Bbbk,o),(\Bbbk^{2},o))$, where $\Bbbk$ is a field with two
elements. Take the filtration $(x)^{\bullet}\cdot R^{\oplus 2}_{X}$. Take the
derivation $\xi_{Y}=q(y)\partial_{y_{1}}+0\cdot\partial_{y_{2}}\in Der_{Y}$,
where $q(y)=y_{1}y_{2}(y_{1}+y_{2})\in\Bbbk[[y_{1},y_{2}]]$. Then
$q((x)^{i},(x)^{i})\subseteq(x)^{3i+1}$. Thus $\xi_{Y}\in
T_{\mathscr{L}^{(3)}}$, even though $q(y)\not\in(y)^{4}$.

  3. iii.

Let $\Bbbk=\tilde{\Bbbk}[[t]]$ for a field $\tilde{\Bbbk}\supseteq\mathbb{Q}$.
Take $I=(t,{\mathfrak{m}})$, then
$T_{\mathscr{L}^{(j)}}=(t,y)^{j+1}T_{\mathscr{L}}\supsetneq(y)^{j+1}T_{\mathscr{L}}$.

######  Remark 3.3.

  1. i.

In the definition (15) it is important to assume $d\geq 1$, rather than $d\geq
0$. In fact the group $\mathscr{L}$ does not act on the whole $R^{\oplus
p}_{X}$, thus the condition
“$g|_{{\footnotesize\left.\raisebox{1.37775pt}{$M_{0}$}\\!/\\!\raisebox{-1.37775pt}{$M_{j}$}\right.}}=Id|_{{\footnotesize\left.\raisebox{1.37775pt}{$M_{0}$}\\!/\\!\raisebox{-1.37775pt}{$M_{j}$}\right.}}$”
makes no sense. One could restrict to
${\footnotesize\left.\raisebox{1.37775pt}{${\mathfrak{m}}\cdot
M_{0}$}\\!/\\!\raisebox{-1.37775pt}{$M_{j}$}\right.}$, but then, for
$\sqrt{I}\subsetneq{\mathfrak{m}}$, the condition
$g|_{{\footnotesize\left.\raisebox{1.37775pt}{${\mathfrak{m}}\cdot
M_{0}$}\\!/\\!\raisebox{-1.37775pt}{$M_{0+j}$}\right.}}=Id|_{{\footnotesize\left.\raisebox{1.37775pt}{${\mathfrak{m}}\cdot
M_{0}$}\\!/\\!\raisebox{-1.37775pt}{$M_{0+j}$}\right.}}$ would imply
$\mathscr{L}^{(j)}=\\{e\\}$. Similarly, the condition $\xi({\mathfrak{m}}\cdot
M_{0})\subseteq M_{j}$ would imply $T_{\mathscr{L}^{(j)}}=(0)$.

  2. ii.

The relation $(y)\cdot T_{\mathscr{L}}=T_{\mathscr{L}^{(0)}}$ motivates the
notation $T_{\mathscr{L}^{(-1)}}:=T_{\mathscr{L}}$. This should be used
carefully, as $T_{\mathscr{L}}(I^{d})\not\subseteq I$ for any $d$.

#### 3.4.2.

For $\mathscr{G}\in\mathscr{A},{\mathscr{K}}\\!$ we define the filtrations in
the special way:

  * •

$\mathscr{A}^{(j)}:=\mathscr{L}^{(j)}\times\mathscr{R}^{(j)}$ and
$T_{\mathscr{A}^{(j)}}:=T_{\mathscr{L}^{(j)}}\oplus T_{\mathscr{R}^{(j)}}$.

  * •

${\mathscr{K}}\\!^{(j)}:=\mathscr{C}^{(j)}\rtimes\mathscr{R}^{(j)}$ and
$T_{{\mathscr{K}}\\!^{(j)}}:=T_{\mathscr{C}^{(j)}}\oplus
T_{\mathscr{R}^{(j)}}$. Here $T_{\mathscr{C}^{(j)}}:=Mat_{p\times p}(I^{j})$.

In particular $T_{\mathscr{K}}\\!=T_{{\mathscr{K}}\\!^{(-1)}}$. As in remark
3.3 we put $T_{\mathscr{A}^{(-1)}}:=T_{\mathscr{A}}$. Again care is needed, as
$T_{\mathscr{A}}I^{d}\not\subseteq I$ for any $d$.

One would like to use the general definition (15), for the groups
$\mathscr{A},{\mathscr{K}}\\!$, i.e. to define

(17) $\mathscr{A}^{(j)}=(\mathscr{L}\times\mathscr{R})^{(j)}:=\mathscr{A}\cap
Maps^{(j)}((x)\cdot R^{\oplus p}_{X}),\text{ and }\
T_{\mathscr{A}^{(j)}}=(T_{\mathscr{R}}\oplus
T_{\mathscr{L}})^{(j)},\quad\text{ and so on.}$

In this definition $\mathscr{A}^{(j)},T_{\mathscr{A}^{(j)}}$ become non-
explicit/complicated. But the two versions often coincide.

######  Lemma 3.4.

Suppose $j\geq 1$, and $\Bbbk$ is an infinite field, and $I^{\bullet}\neq(0)$
for each $\bullet$. Then:

  1. 1.

$(\mathscr{L}\times\mathscr{R})^{(j)}=\mathscr{L}^{(j)}\times\mathscr{R}^{(j)}$,
i.e. $(Aut_{Y}\times Aut_{X})^{(j)}=Aut^{(j)}_{Y}\times Aut^{(j)}_{X}$.

  2. 2.

$(T_{\mathscr{L}}\oplus T_{\mathscr{R}})^{(j)}=T_{\mathscr{L}^{(j)}}\oplus
T_{\mathscr{R}^{(j)}}$, i.e. $(Der_{Y}\oplus
Der_{X})^{(j)}=Der^{(j)}_{Y}\oplus Der^{(j)}_{X}$.

Proof. The inclusions
$(\mathscr{L}\times\mathscr{R})^{(j)}\geq\mathscr{L}^{(j)}\times\mathscr{R}^{(j)}$
and $(T_{\mathscr{L}}\oplus T_{\mathscr{R}})^{(j)}\supseteq
T_{\mathscr{L}^{(j)}}\oplus T_{\mathscr{R}^{(j)}}$ are obvious. We prove the
parts $\leq$ and $\subseteq$.

Below we use the property: there exists $f\in I$ such that $f^{d}\in
I^{d}\setminus I^{d+1}$ for each $d\geq 1$. Indeed, otherwise we would get
$I^{d}\subseteq I^{d+1}$ for $d\gg 1$, as $I$ is finitely generated. But then
$I^{d}=0$ by Nakayama.

  1. 1.

Suppose $(\Phi_{Y},\Phi_{X})\in(\mathscr{L}\times\mathscr{R})^{(j)}$. As
$\Phi_{Y}\in\mathscr{L}=\mathscr{L}^{(0)}$, we get:
$\Phi_{X}\in\mathscr{R}^{(0)}$. Suppose
$\Phi_{Y}\in\mathscr{L}^{({\tilde{l}})}\setminus\mathscr{L}^{({\tilde{l}}+1)}$,
then after a $\Bbbk$-linear reshuffling of $y$-coordinates we can assume:
$\Phi_{Y}(y_{1},0,\dots,0)=(q(y_{1}),\dots)$, where $q(y_{1})=y_{1}+a\cdot
y^{l+1}_{1}+h.o.t.$, for some $l\leq{\tilde{l}}$. Here $0\neq a\in\Bbbk$ and
$h.o.t.\in(y_{1})^{l+2}$. Given $f\in I$ define the difference

(18)
$\Delta_{f}:=[\Phi_{Y}(y_{1},0,\dots,0)|_{f}]_{1}-\Phi_{X}(f)=f+q(f)-\Phi_{X}(f)\in
I^{ord(f)+j}.$

     * •

The case $l=0$. Then $\Phi_{Y}\in\mathscr{L}^{(0)}\setminus\mathscr{L}^{(1)}$
and $\Phi_{X}\in\mathscr{R}=\mathscr{R}^{(0)}\setminus\mathscr{R}^{(1)}$.
Expand $\Phi_{X}(x)=L(x)+h.o.t$, where $L(x)$ is $\Bbbk$-linear, while
$h.o.t\in(x)^{2}$. Then

$\Delta_{f}=(1+a)\cdot f-f(L(x)+h.o.t.)+h.o.t\in I^{ord(f)+j}.$

As this should hold for all $f\in I$ we get: $a=0$ and $L(x)=x$, in
contradiction to $\Phi_{Y}\not\in\mathscr{L}^{(1)}$,
$\Phi_{X}\not\in\mathscr{R}^{(1)}$.

     * •

The case $l>0$. For any $f\in I$, $c\in\Bbbk^{\times}$ we have:

(19) $\Delta_{c\cdot f}-c\cdot\Delta_{f}=q(c\cdot f)-c\cdot
q(f)=a(c^{l+1}-c)f^{l+1}+h.o.t.\in I^{ord(f)+j}.$

As $\Bbbk$ is infinite there exists $c\in\Bbbk^{\times}$ for which
$c^{l+1}-c\neq 0$. Thus for each $f\in I$ and $l>0$ we get: $f^{l+1}\in
I^{ord(f)+j}$. Therefore $l\geq j$, hence $\Phi_{Y}\in\mathscr{L}^{(j)}$, and
thus $\Phi_{X}\in\mathscr{R}^{(j)}$.

  2. 2.

Suppose $(\xi_{Y},\xi_{X})\in(T_{\mathscr{L}}\oplus T_{\mathscr{R}})^{(j)}$,
thus $(\xi_{Y},\xi_{X})(I\cdot R^{\oplus p}_{X})\subseteq I^{1+j}\cdot
R^{\oplus p}_{X}$. If $\xi_{Y}\in T_{\mathscr{L}^{(-1)}}\setminus
T_{\mathscr{L}^{(0)}}$ then $\xi_{Y}(I)\not\subseteq{\mathfrak{m}}$. Therefore
$\xi_{Y}\in T_{\mathscr{L}^{(0)}}$ and then $\xi_{X}\in
T_{\mathscr{R}^{(0)}}$. Suppose $\xi_{Y}\in
T_{\mathscr{L}^{({\tilde{l}})}}\setminus T_{\mathscr{L}^{({\tilde{l}}+1)}}$,
then after a $\Bbbk$-linear reshuffling of $y$-coordinates we can assume
$\xi_{Y}(y_{1},0,\dots,0)=(q(y_{1}),\dots,)$, where $q(y_{1})=a\cdot
y^{l+1}_{1}+h.o.t.$ for some $l\leq{\tilde{l}}$. Here $0\neq a\in\Bbbk$ and
$h.o.t.\in(y_{1})^{l+2}$.

     * •

The case $l=0$. For any $f\in I$ we have

(20) $[(\xi_{Y},\xi_{X})|_{f^{d},0,\dots,0}]_{1}-d\cdot
f^{d-1}\cdot[(\xi_{Y},\xi_{X})|_{f,0,\dots,0}]_{1}=q(f^{d})-d\cdot
f^{d-1}\cdot q(f)=a(1-d)f^{d}+h.o.t\in I^{ord(f^{d})+j}.$

Choose $f\in I\setminus I^{2}$ satisfying $f^{2}\in I^{2}\setminus I^{3}$ to
get the contradiction.

     * •

The case $l>0$. For any $f\in I$ and $c\in\Bbbk$ we have

(21) $[(\xi_{Y},\xi_{X})|_{c\cdot
f,0,\dots,0}]_{1}-c\cdot[(\xi_{Y},\xi_{X})|_{f,0,\dots,0}]_{1}=q(c\cdot
f)-c\cdot q(f)=a(c^{l+1}-c)f^{l+1}+h.o.t\in I^{ord(f)+j}.$

As $\Bbbk$ is infinite there exists $c\in\Bbbk$ such that $c^{l+1}-c\neq 0$.
Therefore we get $l\geq j$, hence $\xi_{Y}\in T_{\mathscr{L}^{(j)}}$, and thus
$\xi_{X}\in T_{\mathscr{R}^{(j)}}$.

######  Remark 3.5.

By the similar arguments one can prove for
${\mathscr{K}}\\!=\mathscr{C}\rtimes\mathscr{R}$: if $\Bbbk$ is an infinite
field then

$(\mathscr{C}\rtimes\mathscr{R})^{(j)}=\mathscr{C}^{(j)}\rtimes\mathscr{R}^{(j)},\quad(GL(p,R_{X})\rtimes
Aut_{X})^{(j)}=GL^{(j)}(p,R_{X})\rtimes
Aut^{(j)}_{X},\quad(T_{\mathscr{C}}\oplus
T_{\mathscr{R}})^{(j)}=T_{\mathscr{C}^{(j)}}\oplus T_{\mathscr{R}^{(j)}}.$

###  3.5. The closures in the filtration topology

For our groups $\mathscr{G}\in\mathscr{R},{\mathscr{K}}\\!,\mathscr{A}$ we
have the closures of the orbits in the filtration topology (for
$M_{\bullet}=I^{\bullet}\cdot R^{\oplus p}_{X}$):

(22) $\overline{T_{\mathscr{G}^{(j)}}f}=\cap_{\bullet\geq
1}(T_{\mathscr{G}^{(j)}}f+M_{\bullet}),\quad\quad\quad\quad\overline{\mathscr{G}^{(j)}f}=\cap_{\bullet\geq
1}(\mathscr{G}^{(j)}f+M_{\bullet}).$

We often use the following observation: to verify the inclusion
$\overline{\mathscr{G}^{(j)}f}\supseteq\\{f\\}+M_{d}$ it is enough to verify
$\mathscr{G}^{(j)}f+M_{\bullet+1}\supseteq\\{f\\}+M_{\bullet}$ for every
$\bullet\geq d$. And similarly for $\overline{T_{\mathscr{G}^{(j)}}f}\supseteq
M_{d}$.

######  Lemma 3.6.

  1. 1.

($R_{X}$ is local) Any $R_{X}$-submodule of $R^{\oplus p}_{X}$ is closed in
$I^{\bullet}$-topology.
In particular, the tangent spaces $T_{\mathscr{R}^{(j)}}f$,
$T_{{\mathscr{K}}\\!^{(j)}}f$ are closed.

  2. 2.

($R_{X}$ is one of
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$) The orbits
$\mathscr{R}^{(j)}f$, ${\mathscr{K}}\\!^{(j)}f$ are closed.

(The case of $T_{\mathscr{A}}$ is more delicate, see lemma 6.14.)
Proof.

  1. 1.

Let $M\subseteq R^{\oplus p}_{X}$ then $M+I^{d}\cdot R^{\oplus
p}_{X}\supseteq\overline{M}$ for any $d\geq 1$. Therefore $M+(I^{d}\cdot
R^{\oplus p}_{X})\cap\overline{M}\supseteq\overline{M}$. Applying the Artin-
Rees lemma to the (finitely generated, §2.1.vii) submodule
$\overline{M}\subset R^{\oplus p}_{X}$ we get:
$M+I\cdot\overline{M}\supseteq\overline{M}$. Then Nakayama gives
$M=\overline{M}$.

  2. 2.

(The ${\mathscr{K}}\\!$-case.) Take a perturbation $g\in{\mathfrak{m}}\cdot
R^{\oplus p}_{X}$ satisfying: $f+g\in\overline{{\mathscr{K}}\\!f}$. Thus the
system of implicit function equations $f(\Phi(x))=U\cdot(f+g)$ has an order-
by-order solution, see §2.1.x. By Popescu-Pfister, §2.1.x., it has a formal
solutions, $(\hat{\Phi},\hat{U})\in\widehat{({\mathscr{K}}\\!^{lin})^{(j)}}$.
Apply Artin approximation to get an ordinary solution,
$(\Phi,{U})\in({\mathscr{K}}\\!^{lin})^{(j)}$. $\blacksquare$

###  3.6. Related derivations and related automorphisms

Recall the classical “naturality” result on vector fields integration, e.g.
[Mo. N.B., §2.5 ]. Given a map of ($C^{\infty}$ or complex/real-analytic)
manifolds, $X\stackrel{{\scriptstyle f}}{{\to}}{Y}$, take some vector fields
$\xi_{X}\in\Theta_{X}$, $\xi_{Y}\in\Theta_{Y}$, and the corresponding time-one
flows, i.e. the automorphisms $\Phi_{X}\circlearrowright X$,
$\Phi_{Y}\circlearrowright Y$.

######  Lemma 3.7.

[Thom-Levine, e.g. [Mo. N.B., Lemma 2.6]] If the vector fields are related,
i.e. $\xi_{Y}\circ f=df\circ\xi_{X}$, then so are the corresponding flows,
$\Phi_{Y}\circ f=f\circ\Phi_{X}$.

We need the algebraic version of this and other results on vector fields.
Below we use the exponential and logarithmic maps of §2.3.

Let $(R,{\mathfrak{m}},I_{\bullet})$ be a local filtered $\Bbbk$-algebra,
assume $\Bbbk\supseteq\mathbb{Q}$ and $R$ is complete with respect to the
filtration. Take the filtration of $R^{\oplus p}_{X}$ by
$M_{\bullet}:=I_{\bullet}\cdot R^{\oplus p}_{X}$. The map $exp$ of (6) is well
defined.

######  Lemma 3.8.

Fix some elements $(\xi_{Y},\xi_{X})\in T_{\mathscr{A}^{(l)}}$, for $l\geq 1$,
and $f\in(x)\cdot R^{\oplus p}_{X}$, some $1\leq d\leq\infty$, and $w_{d}\in
M_{d}$. (Here $M_{\infty}=0$.)

  1. 1.

$\xi_{Y}(e^{\xi_{X}}f)=e^{\xi_{X}}\cdot\xi_{Y}(y)|_{f}$.

  2. 2.

If $\xi_{X}(f)\stackrel{{\scriptstyle mod\ M_{d}}}{{\equiv}}\xi_{Y}(y)|_{f}$
then $\xi^{i}_{X}(\xi^{j}_{Y}(y)|_{f})\stackrel{{\scriptstyle mod\
M_{d+l}}}{{\equiv}}\xi^{i+j}_{Y}(y)|_{f}$ for any $i+j\geq 2$.
In particular, $e^{\xi_{X}}f-e^{\xi_{Y}}f\in M_{d}$.

  3. 3.

(algebraic “naturality”) $\xi_{Y}(y)|_{f}-\xi_{X}(f)\stackrel{{\scriptstyle
mod\ M_{d+l}}}{{\equiv}}w_{d}$ if and only if
$e^{\xi_{Y}}f-e^{\xi_{X}}f\stackrel{{\scriptstyle mod\
M_{d+l}}}{{\equiv}}w_{d}$.

Proof.

  1. 1.

The direct check:
$\xi_{Y}(e^{\xi_{X}}f)=\xi_{Y}(y)|_{e^{\xi_{X}}f}=\xi_{Y}(y)|_{f\circ
e^{\xi_{X}}}=e^{\xi_{X}}\cdot\xi_{Y}(y)|_{f}$.

  2. 2.

Induction on $i$. The case $i=0$ is trivial. Assuming the statement for $i$ we
verify it for $i+1$:

(23)
$\xi^{i+1}_{X}(\xi^{j}_{Y}(y)|_{f})=\xi_{X}\Big{(}\xi^{i}_{X}(\xi^{j}_{Y}(y)|_{f})\Big{)}=\xi_{X}\Big{(}\xi^{i+j}_{Y}(y)|_{f}\
mod\ M_{d}\Big{)}\stackrel{{\scriptstyle mod\
M_{d+l}}}{{\equiv}}\xi_{X}\Big{(}\xi^{i+j}_{Y}(y)|_{f}\Big{)}=\\\
\sum\nolimits_{q}\partial_{y_{q}}\Big{(}\xi^{i+j}_{Y}(y)|_{f}\Big{)}\cdot\xi_{X}(f_{q})\stackrel{{\scriptstyle
mod\
M_{d+l}}}{{\equiv}}\sum\nolimits_{q}\partial_{y_{q}}\Big{(}\xi^{i+j}_{Y}(y)|_{f}\Big{)}\cdot\xi_{Y}(y)_{q}|_{f}=\xi^{i+j+1}_{Y}(y)|_{f}.$

The statement $e^{\xi_{X}}f-e^{\xi_{Y}}f\in M_{d}$ is now obtained by Taylor
expansion of exponents.

  3. 3.

The part $"\Rrightarrow"$. $e^{\xi_{Y}}f-e^{\xi_{X}}f=w_{d}+\sum_{j\geq
2}\frac{\xi_{Y}^{j}(y)|_{f}-\xi_{X}^{j}(f)}{j!}\equiv w_{d}\ mod\ M_{d+k}$.
For the last equality we use part 2.
The part $"\Lleftarrow"$.
$\xi_{Y}(y)|_{f}-\xi_{X}(f)=ln(e^{\xi_{Y}})(y)|_{f}-ln(e^{\xi_{X}})f\stackrel{{\scriptstyle!}}{{=}}ln(e^{-\xi_{X}}e^{\xi_{Y}})(y)|_{f}=-\sum_{j\geq
1}\frac{(1-e^{-\xi_{X}}e^{\xi_{Y}})^{j}}{j}f=\sum_{j\geq
1}\frac{(1-e^{-\xi_{X}}e^{\xi_{Y}})^{j-1}}{j}e^{-\xi_{X}}(w_{d}\ mod\
M_{d+k})\stackrel{{\scriptstyle mod\ M_{d+k}}}{{\equiv}}w_{d}$. For the
transition ”!” we use the commutativity
$e^{\xi_{X}}e^{\xi_{Y}}=e^{\xi_{Y}}e^{\xi_{X}}$.

######  Remark 3.9.

For the subsequent applications one would like the strongest Thom-Levine
statement. In Part 3 the conclusion cannot be strengthened to statement
“$(mod\ M_{d+l+1})$” . For example, define the map
$(\Bbbk^{1},o)\to(\Bbbk^{1},o)$ by $f(x)=x+x^{d-1}\in\Bbbk[[x]]$. Here
$\Bbbk\supseteq\mathbb{Q}$, the ring $R_{X}=\Bbbk[[x]]$ is filtered by
$(x)^{\bullet}$, the ring $R_{Y}=\Bbbk[[y]]$ is filtered by $(y)^{\bullet}$.
Take $\xi_{X}=x^{2}\partial_{x}$, $\xi_{Y}=y^{2}\partial_{y}$. One has
$\xi_{X}(f)-\xi_{Y}(y)|_{f}\stackrel{{\scriptstyle mod\
x^{2d-2}}}{{\equiv}}(d-3)x^{d}$. But $e^{\xi_{X}}(x)=\frac{x}{1-x}$ and
therefore
$e^{\xi_{X}}(f)-e^{\xi_{Y}}(f)=f(\frac{x}{1-x})-\frac{f(x)}{1-f(x)}=(d-3)x^{d}\big{(}1+\frac{3d-2}{2}x+\dots\big{)}$.
Thus $e^{\xi_{X}}(f)-e^{\xi_{Y}}(f)\not\equiv(d-3)x^{d}\ mod\ (x)^{d+2}$.

######  Remark 3.10.

In the proof of part 3 we needed the exponential/logarithmic expansions only
up to order $\lceil\frac{d-ord(f)}{l}\rceil+1$. Therefore this lemma can be
used even when the full exp/log maps do not exist, e.g. when
$\Bbbk\not\supseteq\mathbb{Q}$ or $R^{\oplus p}_{X}$ is not
$M_{\bullet}$-complete. Namely, fix some $(\xi_{Y},\xi_{X})\in
T_{\mathscr{A}^{(l)}}$, $w_{d}\in M_{d}$, and suppose the assumptions
$jet_{N}Exp$, $jet_{N}Ln$ of §2.3 hold for
$N=\lceil\frac{d-ord(f)}{l}\rceil+1$. Then we get the statement:

(24) $\xi_{Y}(y)|_{f}-\xi_{X}(f)\stackrel{{\scriptstyle mod\
M_{d+l}}}{{\equiv}}w_{d}\quad\text{ if and only if }\quad
jet_{N}(e^{-\xi_{X}})jet_{N}(e^{\xi_{Y}})f\stackrel{{\scriptstyle mod\
M_{d+l}}}{{\equiv}}f+w_{d}.$

###  3.7. The annihilator of $T^{1}_{\mathscr{G}}f$

Let $\mathscr{G}\in\mathscr{R},\mathscr{A},{\mathscr{K}}\\!$ and fix a map
$f\in{\mathfrak{m}}\cdot R^{\oplus p}_{X}\cong Maps(X,Y)$. Take the image
tangent space $T_{\mathscr{G}}f\subseteq R^{\oplus p}_{X}$. Take the quotient
module
$T^{1}_{\mathscr{G}}f:={\footnotesize\left.\raisebox{1.37775pt}{$R^{\oplus
p}_{X}$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{G}}f$}\right.}$. (In
[Kerner] we show that this is the tangent space to the
$\mathscr{G}$-miniversal unfolding.) The standard way to measure how large is
the tangent space goes via the support of $T^{1}_{\mathscr{G}}f$, i.e. the
annihilator of the quotient module

(25) $\mathfrak{a}_{\mathscr{G}}:=Ann[T^{1}_{\mathscr{G}}f]=\\{q\in R_{X}|\
q\cdot R^{\oplus p}_{X}\subseteq T_{\mathscr{G}}f\\}\subseteq R_{X}.$

This is the largest ideal satisfying:
$T_{\mathscr{G}}f\supseteq\mathfrak{a}_{\mathscr{G}}\cdot R^{\oplus p}_{X}$.

Fix a filtration $I^{\bullet}\subset R_{X}$ and the corresponding filtrations
$\mathscr{G}^{(\bullet)}$, $T_{\mathscr{G}^{(\bullet)}}$, see §3.4. One gets
the filtered annihilators
$\mathfrak{a}_{\mathscr{G}^{(j)}}:=Ann[T^{1}_{\mathscr{G}^{(j)}}f]\subseteq
R_{X}$ for $j\geq 0$.

######  Definition 3.11.

  1. 1.

For $\mathfrak{a}_{\mathscr{G}}\subset R_{X}$ the germ
$V(\mathfrak{a}_{\mathscr{G}})\subseteq X$ is called the critical ($Crit(f)$)
resp. the singular ($Sing(f)$) resp. the instability locus of $f$.

  2. 2.

The map $f$ is called $\mathscr{G}$-finite if
$\sqrt{\mathfrak{a}_{\mathscr{G}}}\supseteq{\mathfrak{m}}$.

The ${\mathscr{K}}\\!$-finite maps are called also “of finite singularity
type”, [Mo. N.B., pg.224].

If $X$ is smooth and $f_{1},\dots,f_{p}\in R_{X}$ is a regular sequence then
$Sing(f)=Sing(V(f))$, the singular locus of the complete intersection.
Otherwise the two objects differ.

Below we discuss the annihilators
$\mathfrak{a}_{\mathscr{R}},\mathfrak{a}_{\mathscr{K}}\\!$. The annihilator
$\mathfrak{a}_{\mathscr{A}}$ is much more delicate, see §6.

######  Example 3.12.

Let $R$ be one of
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$, see §2.1.ii.

  1. i.

For $p\\!=\\!1$ and $R$ one of $\Bbbk[[x]],\Bbbk\\{x\\},\Bbbk\langle x\rangle$
the annihilators are just the Jacobian and the Tjurina ideals,

$T^{1}_{\mathscr{R}}f={\footnotesize\left.\raisebox{1.37775pt}{$R$}\\!/\\!\raisebox{-1.37775pt}{$Jac(f)$}\right.},\quad\quad\mathfrak{a}_{\mathscr{R}}=Jac(f),\quad\quad
T^{1}_{\mathscr{K}}\\!f={\footnotesize\left.\raisebox{1.37775pt}{$R$}\\!/\\!\raisebox{-1.37775pt}{$(f)+Jac(f)$}\right.},\quad\quad\mathfrak{a}_{\mathscr{K}}\\!=Jac(f)+(f).$

Here (when $\Bbbk$ is not a field) the derivatives are taken with respect to
$x$-variables only.

  2. ii.

For $p\geq 2$ the annihilators are not computed easily, but one has immediate
bounds via the determinantal ideals of the generating matrix
$[\xi_{i}(f_{j})]$ of $T_{\mathscr{R}}f$. (For regular rings this is just the
matrix of derivatives, $[\partial_{x_{i}}f_{j}]$.) One has (see e.g.
[Eisenbud, §20]):

(26)
$I_{p}[\xi_{i}(f_{j})]\subseteq\mathfrak{a}_{\mathscr{R}}\subseteq\sqrt{I_{p}[\xi_{i}(f_{j})]},\quad\quad\quad\quad
I_{p}[\xi_{i}(f_{j})]+(f)\subseteq\mathfrak{a}_{\mathscr{K}}\\!\subseteq\sqrt{I_{p}[\xi_{i}(f_{j})]+(f)}.$

  3. iii.

In particular one gets:
$\mathfrak{a}_{\mathscr{R}}+(f)\subseteq\mathfrak{a}_{\mathscr{K}}\\!\subseteq\sqrt{\mathfrak{a}_{\mathscr{R}}+(f)}$.
Hence $Sing(f)=Crit(f)\cap V(f)$.

  4. iv.

Let $p\geq 2$. In many cases $\mathfrak{a}_{\mathscr{R}}\cdot R^{\oplus
p}_{X}\subseteq{\mathfrak{m}}\cdot T_{\mathscr{R}}f$. For example, this holds
if the determinantal ideal $I_{p}[\xi_{i}(f_{j})]\subset R_{X}$ is radical, as
then $\mathfrak{a}_{\mathscr{R}}=I_{p}[\xi_{i}(f_{j})]$.

  5. v.

Take the filtration $I^{\bullet}\subset R_{X}$. One has
$I^{j+1}\cdot\mathfrak{a}_{\mathscr{G}}\subseteq\mathfrak{a}_{\mathscr{G}^{(j)}}\subseteq\mathfrak{a}_{\mathscr{G}}$.

  6. vi.

(The case of unfolding) Take a map
$(X\times\Bbbk^{r}_{u},o)\to(\Bbbk^{p}_{y}\times\Bbbk^{r}_{u},o)$,
$(x,u)\to(f_{o}(x)+g(x,u),u)$, where $(g(x,u))\subseteq(x)\cdot(u)$. Then
$\mathfrak{a}_{\mathscr{K}}\\!(f)=\mathfrak{a}_{\mathscr{K}}\\!(f_{o})+(u)$
and
$\mathfrak{a}_{{\mathscr{K}}\\!^{(0)}}(f)=\mathfrak{a}_{{\mathscr{K}}\\!^{(0)}}(f_{o})+(u)$.

######  Remark 3.13.

Classically the Fitting (determinantal) ideals of $T^{1}_{\mathscr{R}}f$,
$T^{1}_{\mathscr{K}}\\!f$ were used instead of the annihilators. But
$\\{Fitt_{j}(T^{1}_{\mathscr{A}}f)\\}_{j}$ are not immediately defined, as
$T^{1}_{\mathscr{A}}f$ is not an $R_{X}$ module, while as $R_{Y}$-module it is
not finitely generated.

The following criterion is well known for $\Bbbk=\mathbb{R},\mathbb{C}$.

######  Lemma 3.14.

Suppose the Weierstraß-finiteness holds for the pair $(R_{X},R_{Y})$, §3.1.
The following conditions are equivalent for a map $f:X\to(\Bbbk^{p},o)$.

  1. 1.

The restriction of $f$ to $Crit(f)\to(\Bbbk^{p},o)$ is a finite morphism (i.e.
the $R_{X}$-module
${\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{$\mathfrak{a}_{\mathscr{R}}$}\right.}$
is a f.g. $R_{Y}$-module);

  2. 2.

The $R_{X}$-module $T^{1}_{\mathscr{R}}f$ is a f.g. $R_{Y}$-module;

  3. 3.

$f$ is ${\mathscr{K}}\\!$-finite;

  4. 4.

$f^{-1}(o)\cap Crit(f)$ is a (fat) point.

Moreover, if $T^{1}_{\mathscr{K}}\\!f\in mod\text{-}\Bbbk$ is generated by
$\\{v_{\bullet}\\}\subset R^{\oplus p}_{X}$, then $T^{1}_{\mathscr{R}}f\in
mod(R_{Y})$ is generated by $\\{v_{\bullet}\\}$.

Proof. $\bf 1\Rrightarrow 2.$ $T^{1}_{\mathscr{R}}f$ is a f.g. module over
${\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{$\mathfrak{a}_{\mathscr{R}}$}\right.}$,
and thus a f.g. module over $R_{Y}$.

$\bf 2\Rrightarrow 3.$ If $T^{1}_{\mathscr{R}}f$ is a f.g. module over $R_{Y}$
then
$T^{1}_{\mathscr{R}}f\otimes{\footnotesize\left.\raisebox{1.37775pt}{$R_{Y}$}\\!/\\!\raisebox{-1.37775pt}{$(y)$}\right.}$
is a finite-dimensional $\Bbbk$-vector space. Observe:
$T^{1}_{\mathscr{R}}f\otimes{\footnotesize\left.\raisebox{1.37775pt}{$R_{Y}$}\\!/\\!\raisebox{-1.37775pt}{$(y)$}\right.}=T^{1}_{\mathscr{K}}\\!f$.
Thus $f$ is ${\mathscr{K}}\\!$-finite.

$\bf 3\Rrightarrow 4.$ ${\mathscr{K}}\\!$-finiteness means
$\sqrt{\mathfrak{a}_{\mathscr{K}}\\!}=\sqrt{\mathfrak{a}_{\mathscr{R}}+(f)}={\mathfrak{m}}$.
Thus $f^{-1}(o)\cap Crit(f)$ is zero dimensional.

$\bf 4\Rrightarrow 1$ follows by the Weierstraß-finiteness. $\blacksquare$

##  4\. Criteria for $\mathscr{R}$-orbits

###  4.1. The $\mathscr{R}$-implicit function theorem

Let $R_{X}$ be one of
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$, see §2.1.ii. If $J\neq
0$ and $\Bbbk\not\supseteq\mathbb{Q}$ we assume the $jet_{0}$-condition of
§2.3.

Take two ideals $\mathfrak{a}\subseteq I\subseteq R_{X}$, with
$\mathfrak{a}\subseteq{\mathfrak{m}}^{2}$, and an element $f\in I\cdot
R^{\oplus p}_{X}$. (One often takes $I=R_{X}$ or $I={\mathfrak{m}}$ or
$I=\sqrt{(f)}$.) For the notation $ord(f)$ see §2.1.iii.

######  Theorem 4.1.

If the ideal $\mathfrak{a}$ satisfies $\mathfrak{a}^{2}\cdot
I^{ord(f)-2}\\!\cdot R^{\oplus
p}_{X}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}\cdot T_{\mathscr{R}}f$ then
$\\{f\\}\\!\\!+\mathfrak{a}\cdot T_{\mathscr{R}}f\subseteq\mathscr{R}f$.

We remark: the assumption $\mathfrak{a}^{2}\cdot I^{ord(f)-2}\\!\cdot
R^{\oplus p}_{X}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}\cdot
T_{\mathscr{R}}f$ is preserved by $\mathscr{R}$-equivalence.
Proof. Fix an element $g\in\mathfrak{a}\cdot T_{\mathscr{R}}f$. We want to
resolve the condition $\Phi_{X}(f)=f+g$, where $\Phi_{X}\in Aut_{X}$. Fix some
(finite sets of) generators, $\\{a_{i}\\}$ of $\mathfrak{a}$, and
$\\{\xi_{j}(f)\\}$ of $T_{\mathscr{R}}f=Der_{X}(f)$. Present an element of
$\mathfrak{a}\cdot T_{\mathscr{R}}f$ in the form $\sum
a_{i}c^{g}_{ij}\xi_{j}(f)$, here $c^{g}_{ij}\in R_{X}$.

  * •

The case of regular $R_{X}$, i.e. $J=0$. We look for the coordinate change
$\Phi_{X}\in\mathscr{R}$ in the form $x\to x+\sum a_{i}c_{ij}\xi_{j}(x)$. Thus
we want to resolve the equation

(27) $f(x+\sum a_{i}c_{ij}\xi_{j}(x))=f(x)+\sum
a_{i}c^{g}_{ij}\xi_{j}(f),\text{ with the unknowns $\\{c_{ij}\\}$}.$

Taylor-expand the left hand side: $f(x)+\sum a_{i}c_{ij}\xi_{j}(f)+H_{\geq
2}(\\{a_{i}c_{ij}\xi_{j}(x)\\})$. Here $H_{\geq 2}(z)\in\sum^{ord(f)}_{l\geq
2}I^{ord(f)-l}\cdot(z)^{l}\cdot R^{\oplus p}_{X}$. As $\mathfrak{a}\subseteq
I$ one has: $H_{\geq 2}(\mathfrak{a}\cdot z)\in
I^{ord(f)-2}\cdot(\mathfrak{a}\cdot z)^{2}\cdot R_{X}[[\mathfrak{a}\cdot
z]][z]$. These are polynomials in $z$ whose coefficients are
(formal/analytic/algebraic) power series in $\mathfrak{a}\cdot z$, see
§2.1.iii. Therefore:

(28) $H_{\geq
2}(\\{a_{i}c_{ij}\xi_{j}(x)\\})\in(\\{c_{ij}\\})^{2}\cdot\mathfrak{a}^{2}\cdot
I^{ord(f)-2}\cdot R^{\oplus
p}_{X}\subseteq(c_{ij})^{2}\cdot{\mathfrak{m}}\cdot\mathfrak{a}\cdot
T_{\mathscr{R}}f.$

Thus we can present $H_{\geq 2}(a_{i}c_{ij}\xi_{j}(x))=\sum
a_{i}h_{ij}(c_{*})\xi_{j}(f)$, for some (fixed) power series
$h_{ij}(z)\in(z)^{2}\cdot{\mathfrak{m}}\cdot R_{X}[[\mathfrak{a}\cdot z]][z]$.
These (formal/analytic/algebraic) series $h_{ij}$ satisfy:
$h_{ij}(R_{X})\subset{\mathfrak{m}}$.

Thus to resolve the equation (27) it is enough to resolve the following
(finite) system of equations: $\\{c_{ij}+h_{ij}(c_{*})=c^{g}_{ij}\\}_{ij}$.
And now apply the $IFT_{1\rm I}$, see §2.1.iv, to get the solutions
$c_{ij}(x)\in R_{X}$.

Finally we take the coordinate change $\Phi_{X}:x\to x+\sum
a_{i}c_{ij}(x)\xi_{j}(x)$. It is invertible (as
$\mathfrak{a}\subseteq{\mathfrak{m}}^{2}$), and defines the needed
automorphism $\Phi_{X}\in Aut_{X}$.

  * •

The case $J\neq 0$. Given an element $g=\sum
a_{i}c^{g}_{ij}\xi_{j}(f)\in\mathfrak{a}\cdot T_{\mathscr{R}}f$, we have
constructed the map $\Phi_{X}:x\to x+\sum a_{i}c_{ij}(x)\xi_{j}(x)$ satisfying
$\Phi_{X}(f)=f+g$. For $J\neq 0$ this $\Phi_{X}$ is not necessarily a
coordinate change. By the assumption $jet_{0}$ we can extend it to a
coordinate change $\Psi_{X}\in Aut_{X}$, of the form $x\to x+\sum
a_{i}c_{ij}(x)\xi_{j}(x)+\mathfrak{a}^{2}$. Then
$\Psi_{X}(f)-f-g\in\mathfrak{a}\cdot{\mathfrak{m}}\cdot T_{\mathscr{R}}f$.

Thus we can apply the previous step iteratively, to get
$\Psi_{X,d}^{-1}(f+g)-f\in\mathfrak{a}\cdot{\mathfrak{m}}^{d}\cdot
T_{\mathscr{R}}f$ for all $d\gg 1$. Hence $f+g\in\overline{\mathscr{R}f}$ (the
closure in ${\mathfrak{m}}^{\bullet}$-filtration topology). Now apply lemma
3.6. $\blacksquare$

######  Example 4.2.

Below $p\geq 2$ and $R_{X}$ is one of $\Bbbk[[x]]$, $\Bbbk\\{x\\}$,
$\Bbbk\langle x\rangle$, see §2.1.ii. Take $I=R_{X}$.

  1. i.

Suppose $f$ is a submersion, i.e. $T_{\mathscr{R}}f=R^{\oplus p}_{X}$. (Thus
$p\leq n$.) Then for $\mathfrak{a}={\mathfrak{m}}^{2}$ one gets
$\mathscr{R}f\supseteq\\{f\\}+{\mathfrak{m}}^{2}\cdot R^{\oplus p}_{X}$. In
particular one gets the normal form of submersion,
$f\stackrel{{\scriptstyle\mathscr{R}}}{{\sim}}(x_{1},\dots,x_{p})$.

  2. ii.

The theorem gives: If
$\mathfrak{a}^{2}\\!\subseteq{\mathfrak{m}}\cdot\mathfrak{a}\cdot\mathfrak{a}_{\mathscr{R}}$
then $\mathscr{R}f\supseteq\\{f\\}\\!+\mathfrak{a}\cdot T_{\mathscr{R}}f$.
(Here $\mathfrak{a}_{\mathscr{R}}$ is the annihilator of the critical locus,
§3.7.) For example, one has
$\mathscr{R}f\supseteq\\{f\\}\\!+{\mathfrak{m}}\cdot\mathfrak{a}^{2}_{\mathscr{R}}\cdot
R^{\oplus p}_{X}$.

  3. iii.

In many cases one has $\mathfrak{a}_{\mathscr{R}}\cdot R^{\oplus
p}_{X}\subseteq{\mathfrak{m}}\cdot T_{\mathscr{R}}f$, see example 3.12.iv.
Then theorem 4.1 gives:
$\mathscr{R}f\supseteq\\{f\\}+\mathfrak{a}^{2}_{\mathscr{R}}\cdot R^{\oplus
p}_{X}$. Geometrically: $f$ is determined up to $\mathscr{R}$-equivalence by
its 2-jet on the critical locus $Crit(f)=V(\mathfrak{a}_{\mathscr{R}})$.

###  4.2. More examples for the case of a power series, i.e.
$\boldsymbol{p}=1$

Let $R_{X}$ be one of $\Bbbk[[x]]$, $\Bbbk\\{x\\}$, $\Bbbk\langle x\rangle$.
Take $f\in{\mathfrak{m}}^{2}$, $I={\mathfrak{m}}$ and
$\mathfrak{a}\subseteq{\mathfrak{m}}^{2}$ . Theorem 4.1 reads:

(29) If
$\mathfrak{a}^{2}\cdot{\mathfrak{m}}^{ord(f)-2}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}\cdot
Jac(f)$ then $\\{f\\}+\mathfrak{a}\cdot Jac(f)\subseteq\mathscr{R}f$.

Below we use a simpler though more restrictive condition:
$\mathfrak{a}\cdot{\mathfrak{m}}^{ord(f)-2}\subseteq{\mathfrak{m}}\cdot
Jac(f)$. Then we get:

######  Corollary 4.3.

  1. 1.

$\mathscr{R}f\supseteq\\{f\\}+Jac(f)\cdot\Big{(}({\mathfrak{m}}\cdot
Jac(f)):{\mathfrak{m}}^{ord(f)-2}\Big{)}$.

  2. 2.

Take the smallest $d\in\mathbb{N}$ satisfying
${\mathfrak{m}}^{2d+ord(f)-2}\subseteq{\mathfrak{m}}^{d+1}\cdot Jac(f)$. Then
$\mathscr{R}f\supseteq\\{f\\}+{\mathfrak{m}}^{d}\cdot Jac(f)$.

######  Example 4.4.

  1. i.

Obviously $({\mathfrak{m}}\cdot
Jac(f)):{\mathfrak{m}}^{ord(f)-2}\supseteq{\mathfrak{m}}\cdot Jac(f)$.
Therefore: $\\{f\\}+{\mathfrak{m}}\cdot Jac(f)^{2}\subseteq\mathscr{R}f$. In
particular, if $\sqrt{Jac(f)}={\mathfrak{m}}$ then $f$ is finitely-determined.

  2. ii.

(Morse lemma) Let $\Bbbk$ be a field, $f\in{\mathfrak{m}}^{2}$, and
$rank[f^{\prime\prime}|_{o}]=r$. Then
$f\stackrel{{\scriptstyle\mathscr{R}}}{{\sim}}Q_{2}(x)+{\tilde{f}}(y)$, where
$x=(x_{1},\dots,x_{r})$, $y=(y_{1},\dots,y_{n-r})$, ${\tilde{f}}(y)\in(y)^{3}$
(is independent of $x$) and $Q_{2}(x)$ is a homogeneous quadratic polynomial.

Moreover, for $\Bbbk=\bar{\Bbbk}$ one can diagonalize the quadratic form
$Q_{2}$ to get: $f\stackrel{{\scriptstyle\mathscr{R}}}{{\sim}}\sum
x^{2}_{i}+{\tilde{f}}(y)$.
Proof. Apply a $GL(n,\Bbbk)$ transformation to get
$f\\!-\\!Q_{2}(x)\\!\in\\!(x,y)^{3}$. Then
$Jac(f)\\!+\\!(x,y)^{2}\\!\supset\\!(x_{1},\dots,x_{r})$. And thus we have
$f-Q_{2}(x)\in(x,y)\cdot Jac(f)^{2}+f^{\\#}(y)^{3}$. The statement follows now
by part i. $\blacksquare$

  3. iii.

Assuming $ord(f)\geq 3$ one has ${\mathfrak{m}}\cdot
Jac(f):{\mathfrak{m}}^{ord(f)-2}\supseteq Jac(f)$. Then we get:
$\\{f\\}+Jac(f)^{2}\subseteq\mathscr{R}f$. For $R_{X}=\mathbb{C}\\{x\\}$ this
is Lemma 3.20 in [Ebeling]. Its proof is based on vector field integration,
thus cannot be used when $\Bbbk$ is an arbitrary field.

  4. iv.

If $ord(f)=2$ then by Morse lemma we can assume $f=Q_{2}(x)+{\tilde{f}}(y)$,
with ${\tilde{f}}(y)\in(y)^{3}$. Then we get:
$\mathscr{R}f\supseteq\\{f\\}+(x)^{2}\cdot(x,y)+(x)\cdot
Jac({\tilde{f}}(y))+Jac({\tilde{f}}(y))^{2}$.

  5. v.

To show how this corollary strengthens the classical bounds, let $\Bbbk$ be a
field of zero characteristic, and take $f$ with an ordinary multiple point at
$o\in(\Bbbk^{n},o)$. This means: the partials
$\partial_{1}f,\dots,\partial_{n}f$ form a regular sequence, and their leading
terms (of order $ord(f)-1$) form a regular sequence as well. Let $q$ be the
smallest integer satisfying
${\mathfrak{m}}^{q}\subseteq{\mathfrak{m}}^{2}\cdot Jac(f)$. Then
$q=n\cdot(ord(f)-2)+1$, see exercise 2.2.7 in [Gr.Lo.Sh]. The known
determinacy bound is: $\\{f\\}+{\mathfrak{m}}^{q}\subset\mathscr{R}f$, see §5
of [B.G.K.22].

As $f$ has an ordinary multiple point we have:
${\mathfrak{m}}^{q}=Jac(f)\cdot{\mathfrak{m}}^{q-ord(f)+1}$. Assume
$ord(f)\geq 2+\frac{3}{n-1}$ and denote $d:=\lceil\frac{q-ord(f)}{2}\rceil+1$.
Take $\mathfrak{a}={\mathfrak{m}}^{d}\subset I={\mathfrak{m}}$. We claim:

(30)
$\mathfrak{a}^{2}\cdot{\mathfrak{m}}^{ord(f)-2}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}\cdot
Jac(f),\quad\quad\text{ i.e.
}\quad\quad{\mathfrak{m}}^{2d+ord(f)-2}\subseteq{\mathfrak{m}}^{d+1}\cdot
Jac(f).$

Indeed, by our assumption:

(31) $2d+ord(f)-2\geq q\quad\text{ and
}\quad(d+1)+(ord(f)-1)=\lceil\frac{q-ord(f)}{2}\rceil+1\leq q.$

Therefore
$\mathscr{R}f\supseteq\\{f\\}+{\mathfrak{m}}^{\lceil\frac{q-ord(f)}{2}\rceil}\cdot
Jac(f)$. This improves the bound
$\\{f\\}+{\mathfrak{m}}^{q}\subset\mathscr{R}f$ even asymptotically.

  6. vi.

Take the $E_{6}$ singularity, $f(x,y)=x^{3}+y^{4}$. Assume $\Bbbk$ is a field,
$char(\Bbbk)\neq 2,3$. Denote ${\mathfrak{m}}=(x,y)$,
$\mathfrak{a}={\mathfrak{m}}^{2}\subset I={\mathfrak{m}}$. We get:
$\mathfrak{a}^{2}\cdot{\mathfrak{m}}^{ord(f)-1}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}\cdot
Jac(f)$. Therefore
$\mathscr{R}f\supseteq\\{f\\}+{\mathfrak{m}}^{2}\cdot(x^{2},y^{3})$. The ideal
${\mathfrak{m}}^{2}\cdot(x^{2},y^{3})$ contains all the monomials lying above
the Newton diagram of $x^{3}+y^{4}$. Compare this result to the standard
determinacy statement: $\mathscr{R}f\supseteq\\{f\\}+{\mathfrak{m}}^{5}$.

###  4.3. The filtration criterion for $\mathscr{R}$-orbits

Let $R_{X}$ be a local henselian $\Bbbk$-algebra, §2.1.ii. Take the filtration
$M_{\bullet}:=I^{\bullet}\cdot R^{\oplus p}_{X}$. Fix some integers $1\leq
j<d$.

######  Theorem 4.5.

Suppose the conditions $jet_{N}(Exp)$, $jet_{N}(Ln)$ of §2.3 hold for
$N=\lceil\frac{d-ord(f)}{j}\rceil$. Then:

$\overline{\mathscr{R}^{(j)}f}\supseteq\\{f\\}+I^{d}\cdot R^{\oplus p}_{X}$ if
and only if  ${T_{\mathscr{R}^{(j)}}f}\supseteq I^{d}\cdot R^{\oplus p}_{X}$.

Proof. The part $\boldsymbol{``\Rrightarrow"}$. First we prove:
$T_{\mathscr{R}^{(j)}}f+M_{d+1}\supseteq M_{d}$. Let $w\in M_{d}$, then
$gf\equiv f+w\ mod\ M_{d+1}$ for some $g\in\mathscr{R}^{(j)}$. Use the
assumption $jet_{N}(Ln)$ to get the derivation $-\sum^{N}_{i=1}\frac{({1\rm
I}-g)^{i}}{i}+\xi_{g}\in T_{\mathscr{R}^{(j)}}$, for some $\xi_{g}\in
End^{(Nj+1)}_{\Bbbk}(R)$. And then

(32) $T_{\mathscr{R}^{(j)}}f\ni\Big{(}-\sum^{N}_{i=1}\frac{({1\rm
I}-g)^{i}}{i}+\xi_{g}\Big{)}f\equiv w\ mod(M_{d+1}+M_{Nj+1+ord(f)})\equiv w\
mod\ M_{d+1}.$

Here $Nj+ord(f)\geq d$ because of the assumption
$N=\lceil\frac{d-ord(f)}{j}\rceil$.

Therefore we have $T_{\mathscr{R}^{(j)}}f+M_{d+1}\supseteq M_{d}$. As
$M_{d+1}=I\cdot M_{d}$, we get (by Nakayama) $T_{\mathscr{R}^{(j)}}f\supseteq
M_{d}$.

The part $\boldsymbol{``\Lleftarrow"}$. First we prove:
$\mathscr{R}^{(j+k)}f+M_{d+k+1}\supseteq\\{f\\}+M_{d+k}$ for every $k\geq 0$.
Indeed, $T_{\mathscr{R}^{(j)}}f\supseteq M_{d}$ implies
$T_{\mathscr{R}^{(j+k)}}f\supseteq M_{d+k}$ for every $k\geq 0$. (Using
$M_{\bullet}=I^{\bullet}\cdot R^{\oplus p}_{X}$.) Accordingly we present an
element $w\in M_{d+k}$ as $w=\xi(f)$, for some $\xi\in
T_{\mathscr{R}^{(j+k)}}$. The condition $jet_{N}(Exp)$ ensures an element
$\Phi:=\sum^{N}_{i=0}\frac{\xi^{i}}{i!}+\phi_{\xi}\in\mathscr{R}^{(j+k)}$,
where $\phi_{\xi}\in End^{(N(j+k)+1)}_{\Bbbk}(R_{X})$. And then

$\Phi(f)-f-w\in M_{d+k+1}+M_{(j+k)N+ord(f)+1}.$

Finally use $(j+k)N+ord(f)\geq d+k$ to get $\Phi(f)-f-w\in M_{d+k+1}$.

Iterating $\mathscr{R}^{(j+k)}f+M_{d+k+1}\supseteq\\{f\\}+M_{d+k}$ for $k\geq
0$ we get: $\overline{\mathscr{R}^{(j)}f}\supseteq\\{f\\}+M_{d}$.
$\blacksquare$

######  Corollary 4.6.

Let $R_{X}\in{\Bbbk[[x]]}{}$, ${\Bbbk\\{x\\}}{}$, ${\Bbbk\langle x\rangle}{}$,
with $\Bbbk$ a field, see §2.1.ii. Take $I\subseteq(x)$ and $f\in(x)\cdot
R^{\oplus p}_{X}$.

  1. 1.

Suppose either $char(\Bbbk)=0$ or
$char(\Bbbk)>\lceil\frac{d-ord(f)}{j}\rceil$. Then:

$\mathscr{R}^{(j)}f\supseteq\\{f\\}+I^{d}\cdot R^{\oplus p}_{X}$ if and only
if $T_{\mathscr{R}^{(j)}}f\supseteq I^{d}\cdot R^{\oplus p}_{X}$.

  2. 2.

Suppose $char(\Bbbk)\neq 2$. If $T_{\mathscr{R}^{(j)}}f\supseteq I^{d}\cdot
R^{\oplus p}_{X}$ then $\mathscr{R}^{(d-j-
ord(f))}f\supseteq\\{f\\}+I^{2d-2j-ord(f)}\cdot R^{\oplus p}_{X}$.

Proof. 1. Combine theorem 4.5 with lemma 3.6. The $jet_{N}$-condition holds by
example 2.7.

2. We get $T_{\mathscr{R}^{(d-j-ord(f))}}f\supseteq I^{2d-2j-ord(f)}\cdot R^{\oplus p}_{X}$. Now apply part one with $N:=\lceil\frac{2d-2j-2ord(f)}{d-j-ord(f)}\rceil=2$. $\blacksquare$

######  Example 4.7.

  1. i.

Take $p=1$, $I={\mathfrak{m}}$, $j=1$ and suppose either $char(\Bbbk)=0$ or
$char(\Bbbk)>d-ord(f)$. Then:
$\mathscr{R}^{(1)}f\supseteq\\{f\\}+{\mathfrak{m}}^{d}$ if and only if
${\mathfrak{m}}^{2}\cdot Jac(f)\supseteq{\mathfrak{m}}^{d}$.

In particular, the $\mathscr{R}$-order of determinacy of $f$ is
$\leq\mu(f)+1$. In $char(\Bbbk)=0$ this is well known, e.g. [Gr.Lo.Sh,
Corollary 2.24]. For $char(\Bbbk)>d-ord(f)$ this improves the known bound,
[Bou.Gr.Ma.12], [Greuel-Pham.19]: the $\mathscr{R}$-order of determinacy of
$f$ is $\leq 2\mu(f)-ord(f)+2$.

  2. ii.

In part two of corollary 4.6 the conclusion is weaker, but also the assumption
is weaker. For $j=1$ this part two is known, e.g. see corollary 5.2 of
[B.G.K.22].

##  5\. Criteria for ${\mathscr{K}}\\!$-orbits

###  5.1. The ${\mathscr{K}}\\!$-implicit function theorem

Let $R_{X}$ be one of
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk[[x]]$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\\{x\\}$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$,
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk\langle
x\rangle$}\\!/\\!\raisebox{-1.37775pt}{$J$}\right.}$, see §2.1.ii. If $J\neq
0$ and $\Bbbk\not\supseteq\mathbb{Q}$ we assume the $jet_{0}$-condition of
§2.3.

Take two ideals $\mathfrak{a}\subseteq I\subseteq R_{X}$, with
$\mathfrak{a}\subseteq{\mathfrak{m}}^{2}$, and an element $f\in I\cdot
R^{\oplus p}_{X}$. (One often takes $I=R_{X}$ or $I={\mathfrak{m}}$ or
$I=\sqrt{(f)}$.) For the notation $ord(f)$ see §2.1.iii.

######  Theorem 5.1.

If $\mathfrak{a}^{2}\cdot I^{ord(f)-2}\cdot R^{\oplus
p}_{X}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}\cdot
T_{\mathscr{R}}f+{\mathfrak{m}}\cdot(f)\cdot R^{\oplus p}_{X}$ then
$\\{f\\}+(\mathfrak{a}\cdot\mathfrak{a}_{\mathscr{R}}+{\mathfrak{m}}\cdot(f))\cdot
R^{\oplus p}_{X}\subseteq{\mathscr{K}}\\!f$.

We remark: the assumption $\mathfrak{a}^{2}\cdot I^{ord(f)-2}\cdot R^{\oplus
p}_{X}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}\cdot
T_{\mathscr{R}}f+{\mathfrak{m}}\cdot(f)\cdot R^{\oplus p}_{X}$ is preserved by
${\mathscr{K}}\\!$-equivalence.
Proof. Fix an element
$g\in(\mathfrak{a}\cdot\mathfrak{a}_{\mathscr{R}}+{\mathfrak{m}}\cdot(f))\cdot
R^{\oplus p}_{X}$. We want to resolve the condition $\Phi_{X}(f)=({1\rm
I}+U)\cdot(f+g)$, where $\Phi_{X}\in Aut_{X}$ and $U\in Mat_{p\times
p}({\mathfrak{m}})$. As in the proof of Theorem 4.1 we fix some (finite) sets
of generators, $\\{a_{i}\\}$ of $\mathfrak{a}$, $\\{a^{\mathscr{R}}_{i}\\}$ of
$\mathfrak{a}_{\mathscr{R}}$, and $\\{\xi_{j}(f)\\}$ of $T_{\mathscr{R}}f$.

  * •

The case of regular $R_{X}$, i.e. $J=0$. Present the coordinate change in the
form $\Phi_{X}:x\to x+\sum a_{i}c_{ij}\xi_{j}(x)$, here $\\{c_{ij}\\}$ are
unknowns.

    * –

As in the $\mathscr{R}$-case we get: $\Phi_{X}(f)-f-\sum
a_{i}c_{ij}\xi_{j}(f)\in(c_{*})^{2}\cdot\mathfrak{a}^{2}\cdot
I^{ord(f)-2}\cdot R^{\oplus p}_{X}$.

Therefore, as in the $\mathscr{R}$-case, we can present

(33) $\Phi_{X}(f)=f+\sum
a_{i}\big{(}c_{ij}+H^{(1)}_{ij}(\\{c_{*}\\})\big{)}\xi_{j}(f)+H^{(2)}(\\{c_{*}\\})\cdot
f.$

Here $H^{(1)}_{ij}(z)\in{\mathfrak{m}}\cdot R_{X}[[{\mathfrak{m}}\cdot z]][z]$
and $H^{(1)}({\mathfrak{m}})\subseteq{\mathfrak{m}}\subset R_{X}$. Similarly,
$H^{(2)}(z)\in Mat_{p\times p}({\mathfrak{m}}\cdot R_{X}[[{\mathfrak{m}}\cdot
z]][z])$ and $H^{(2)}({\mathfrak{m}})\subset Mat_{p\times p}({\mathfrak{m}})$.

    * –

Expand $g=\sum a_{i}a^{\mathscr{R}}_{k}c^{g}_{ik}+U^{g}\cdot f$, here
$c^{g}_{ik}\in R^{\oplus p}_{X}$ and $U^{g}\in Mat_{p\times
p}({\mathfrak{m}})$. Thus $({1\rm I}+U)(f+g)=f+g+(U+U^{g}+U\cdot U^{g})f+\sum
a_{i}a^{\mathscr{R}}_{k}\cdot U\cdot c^{g}_{ik}$. We can present
$\sum_{k}a^{\mathscr{R}}_{k}\cdot({1\rm I}+U)\cdot
c^{g}_{ik}=\sum_{j}l_{ij}(U)\cdot\xi_{j}(f)$, where $l_{ij}(U)$ are linear
functions in the entries of $U$.

Altogether the condition $\Phi_{X}(f)=({1\rm I}+U)\cdot(f+g)$ becomes:

(34) $f+\sum
a_{i}\big{(}c_{ij}+H^{(1)}_{ij}(\\{c_{*}\\})\big{)}\xi_{j}(f)+H^{(2)}(\\{c_{*}\\})\cdot
f=f+\sum_{ij}a_{i}l_{ij}(U)\cdot\xi_{j}(f)+(U+U^{g}+U\cdot U^{g})\cdot f.$

It remains to resolve the system of equations:

(35) $c_{ij}+H^{(1)}_{ij}(\\{c_{*}\\})=a_{i}l_{ij}(U),\quad
H^{(2)}(\\{c_{*}\\})=U+U^{g}+U\cdot U^{g},\quad\text{with the unknowns
$\\{c_{ij}\\},U$}.$

Apply the $IFT_{1\rm I}$, §2.1.iv, to determine $c_{ij},U$. We get the
transformation $(\Phi_{X},U)\in\mathscr{R}\ltimes GL(p,R_{X})$ that satisfies
the condition $\Phi_{X}(f)=({1\rm I}+U)\cdot(f+g)$.

  * •

The case $J\neq 0$. Given an element
$g\in(\mathfrak{a}\cdot\mathfrak{a}_{\mathscr{R}}+{\mathfrak{m}}\cdot(f))\cdot
R^{\oplus p}_{X}$ apply the previous argument to construct the map
$\Phi_{X}:x\to x+\sum a_{i}c_{ij}\xi_{j}(x)$ satisfying $\Phi_{X}(f)=({1\rm
I}+U)(f+g)$. This $\Phi_{X}$ is not necessarily a coordinate change. By the
$jet_{0}$-assumption we can extend it to a coordinate change $\Psi_{X}:x\to
x+\sum a_{i}c_{ij}\xi_{j}(x)+\mathfrak{a}^{2}$. Then
$\Psi_{X}^{-1}\big{(}({1\rm
I}+U)(f+g)\big{)}-f\in\mathfrak{a}^{2}I^{ord(f)-1}R^{\oplus p}_{X}$.

Thus we can iterate the previous step to get
$\Psi_{X,d}^{-1}(f+g)-f\in\mathfrak{a}^{2}\cdot{\mathfrak{m}}^{d}\cdot
I^{ord(f)-1}R^{\oplus p}_{X}$ for all $d\gg 1$. Thus
$f+g\in\overline{{\mathscr{K}}\\!f}$ (the closure in the
${\mathfrak{m}}^{\bullet}$-filtration topology). Now apply lemma 3.6.
$\blacksquare$

######  Example 5.2.

Let $R_{X}$ be one of $\Bbbk[[x]]$, $\Bbbk\\{x\\}$, $\Bbbk\langle x\rangle$,
see §2.1.ii. Take $I={\mathfrak{m}}$ and assume
$(f)\subseteq{\mathfrak{m}}^{2}$. We get the following corollaries.

  1. i.

${\mathscr{K}}\\!f\supseteq\\{f\\}+({\mathfrak{m}}^{2}\cdot\mathfrak{a}^{2}_{\mathscr{R}}+{\mathfrak{m}}\cdot(f))\cdot
R^{\oplus p}_{X}$. For $p=1$ this reads:
${\mathscr{K}}\\!f\supseteq{\mathfrak{m}}^{2}\cdot
Jac(f)^{2}+f\cdot{\mathfrak{m}}$.

     * •

In particular, suppose $f$ has an isolated singularity, i.e. $Crit(f)\cap
V(f)=\\{o\\}\in\Bbbk^{n}$. Then $f$ is finitely-${\mathscr{K}}\\!$-determined.

     * •

The inclusion
${\mathscr{K}}\\!f\supseteq\\{f\\}+({\mathfrak{m}}^{2}\cdot\mathfrak{a}^{2}_{\mathscr{R}}+{\mathfrak{m}}\cdot(f))\cdot
R^{\oplus p}_{X}$ strengthens Theorem 2.1 of [Cu.Sri.1997], which reads: If
$R_{X}=\Bbbk[[x]]$ and $\mathfrak{a}=(f)+Fitt_{0}(T^{1}_{\mathscr{R}}f)$ then
${\mathscr{K}}\\!f\supseteq\\{f\\}+\mathfrak{a}^{3}\cdot R^{\oplus p}_{X}$.

(Note that $\mathfrak{a}_{\mathscr{R}}\supseteq
Fitt_{0}(T^{1}_{\mathscr{R}}f)$, and for $p\geq 2$ one often has
${\mathfrak{m}}\cdot\mathfrak{a}_{\mathscr{R}}\supseteq
Fitt_{0}(T^{1}_{\mathscr{R}}f)$.)

  2. ii.

If $(f)\subseteq{\mathfrak{m}}^{3}$ then
${\mathscr{K}}\\!f\supseteq\\{f\\}+(\mathfrak{a}^{2}_{\mathscr{R}}+{\mathfrak{m}}\cdot(f))\cdot
R^{\oplus p}_{X}$. For $p=1$ this reads:
${\mathscr{K}}\\!f\supseteq\\{f\\}+Jac(f)^{2}+f\cdot{\mathfrak{m}}$.

  3. iii.

For $p=1$ and $f\in{\mathfrak{m}}^{2}$ use Morse lemma, example 4.4.ii, to
present $f=Q_{2}(x)+{\tilde{f}}(y)$, with ${\tilde{f}}(y)\in(y)^{3}$. Then we
get:
${\mathscr{K}}\\!f\supseteq\\{f\\}+{\mathfrak{m}}\cdot(x)\cdot((x)+Jac_{y}({\tilde{f}}(y)))+(Jac_{y}({\tilde{f}}(y)))^{2}+{\mathfrak{m}}\cdot(f)$.

  4. iv.

For $p>1$ the assumption on modules $\mathfrak{a}^{2}\cdot I^{ord(f)-2}\cdot
R^{\oplus p}_{X}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}\cdot
T_{\mathscr{R}}f+{\mathfrak{m}}\cdot(f)\cdot R^{\oplus p}_{X}$ is weaker than
the assumption on the ideals $\mathfrak{a}^{2}\cdot
I^{ord(f)-2}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}\cdot\mathfrak{a}_{\mathscr{R}}+{\mathfrak{m}}\cdot(f)$,
cf. §2.1.viii. But the later assumption is often simpler to verify.

  5. v.

Below we assume $p=1$.

     * •

Suppose
$\mathfrak{a}^{2}\cdot{\mathfrak{m}}^{ord(f)-2}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}\cdot
Jac(f)+f\cdot{\mathfrak{m}}$. Then
${\mathscr{K}}\\!f\supseteq\\{f\\}+\mathfrak{a}\cdot
Jac(f)+f\cdot{\mathfrak{m}}$

     * •

Suppose
${\mathfrak{m}}^{e}\cdot\mathfrak{a}_{\mathscr{R}}+(f){\mathfrak{m}}\supseteq{\mathfrak{m}}^{d}$
for $e=\lceil\frac{d-ord(f)}{2}\rceil+1$. Take
$\mathfrak{a}={\mathfrak{m}}^{e}$ to get:
${\mathscr{K}}\\!f\supset\\{f\\}+({\mathfrak{m}}^{e}\cdot\mathfrak{a}_{\mathscr{K}}\\!+{\mathfrak{m}}(f))\cdot
R^{\oplus p}_{X}$.

     * •

Take the smallest $d\in\mathbb{N}$ satisfying
${\mathfrak{m}}^{2d+ord(f)-2}\subseteq{\mathfrak{m}}^{d+1}\cdot
Jac(f)+{\mathfrak{m}}\cdot(f)$. Then
${\mathscr{K}}\\!f\supseteq\\{f\\}+{\mathfrak{m}}^{d}\cdot
Jac(f)+{\mathfrak{m}}\cdot f$.

###  5.2. The ${\mathscr{K}}\\!$-determinacy of a reduced complete
intersection curve germ via the valuation semigroup

Let $R_{X}$ be one of $\Bbbk[[x]]$, $\Bbbk\\{x\\}$, $\Bbbk\langle x\rangle$,
where $\Bbbk=\bar{\Bbbk}$ is a field, see §2.1.ii. Suppose
$(C,o):=V(f)\subset(\Bbbk^{n},o)$ is a reduced complete intersection curve
germ. Take its branch decomposition, $(C,o)=\cup^{r}_{i=1}(C_{i},o)$. We have
the normalization morphism
$\mathcal{O}_{(C,o)}{\hookrightarrow}\prod\mathcal{O}_{(\tilde{C}_{i},o)}$.
Here each ring $\mathcal{O}_{(\tilde{C}_{i},o)}$ is a principal ideal domain.
It has the valuation map
$val_{i}:\mathcal{O}_{(\tilde{C}_{i},o)}\stackrel{{\scriptstyle}}{{\to}}{\bar{\mathbb{N}}}:=\mathbb{N}\cup\\{\infty\\}$.
The “full” valuation map is the product of morphisms,
$val:R_{X}\to\mathcal{O}_{(C,o)}{\hookrightarrow}\prod\mathcal{O}_{\tilde{C}_{i},o}\to\oplus_{i}{\bar{\mathbb{N}}}$.
Accordingly we have two sub-semigroups:
$val(\mathcal{O}_{(C,o)})\subseteq\oplus_{i}{\bar{\mathbb{N}}}$ and
$val(\mathfrak{a}^{2}_{\mathscr{R}})\subseteq\oplus_{i}{\bar{\mathbb{N}}}$.
Note that
$val(\mathfrak{a}^{2}_{\mathscr{R}})=val(\mathfrak{a}^{2}_{\mathscr{R}}+(f))$.

Below we consider the $\mathbb{N}^{r}$ submodules, i.e. subsets
$S\subset\mathbb{N}^{r}$ satisfying: $S+\mathbb{N}^{r}=S$.

######  Proposition 5.3.

Suppose a submodule ${\mathcal{S}}\subset{\bar{\mathbb{N}}}^{r}$ satisfies:
${\mathcal{S}}\cap val(R_{X})={\mathcal{S}}\cap
val({\mathfrak{m}}\cdot\mathfrak{a}^{2}_{\mathscr{R}})$. Then
${\mathscr{K}}\\!f\supseteq\\{f\\}+{\mathfrak{m}}\cdot
val^{-1}({\mathcal{S}})\cdot R^{\oplus p}_{X}$.

As was mentioned before, if $(C,o)$ is not a plane curve singularities, then
in many cases $\mathfrak{a}_{\mathscr{R}}R^{\oplus
p}_{X}\subseteq{\mathfrak{m}}\cdot T_{\mathscr{R}}f$. Therefore the bound can
be strengthened:
${\mathscr{K}}\\!f\supseteq\\{f\\}+val^{-1}({\mathcal{S}})\cdot R^{\oplus
p}_{X}$.

Proof. Let $g\in val^{-1}({\mathcal{S}})\subset R_{X}$. Then $val(g)=val(q)$
for an element $q\in\mathfrak{a}^{2}_{\mathscr{R}}$. Thus exists $c\in\Bbbk$
satisfying $val(g-c\cdot q)>val(g)$. Namely, $val_{i}(g-c\cdot q)\geq
val_{i}(g)$ for each $i=1,\dots,r$, and this inequality is strict at least for
one $i$.

Apply this procedure for each coordinate of the vector
$val(g)\in{\bar{\mathbb{N}}}^{r}$ to get an element
$q\in\mathfrak{a}^{2}_{\mathscr{R}}$ that satisfies: $val(g-c\cdot q)\geq
val(g)+(1,\dots,1)$. Iterating this we get
$val^{-1}({\mathcal{S}})\subset\mathfrak{a}^{2}_{\mathscr{R}}+(f)+val^{-1}({\mathcal{S}}+d(1,\dots,1))$
for each $d\geq 1$.

As $(C,o)$ is a reduced germ, the ideal $\mathfrak{a}^{2}_{\mathscr{R}}+(f)$
is ${\mathfrak{m}}$-primary. Therefore we get:
$val^{-1}({\mathcal{S}})\subset\mathfrak{a}^{2}_{\mathscr{R}}+(f)$. And hence
$\\{f\\}+{\mathfrak{m}}\cdot val^{-1}({\mathcal{S}})\cdot R^{\oplus
p}_{X}\subset\\{f\\}+{\mathfrak{m}}\cdot(\mathfrak{a}^{2}_{\mathscr{R}}+(f))\cdot
R^{\oplus p}_{X}$. The statement follows now by Theorem 5.1. $\blacksquare$

###  5.3. The filtration criterion for ${\mathscr{K}}\\!$-orbits

Let $R_{X}$ be a local henselian $\Bbbk$-algebra, §2.1.ii. Take the filtration
$M_{\bullet}:=I^{\bullet}\cdot R^{\oplus p}_{X}$. Fix some integers $1\leq
j<d$.

######  Theorem 5.4.

Suppose the conditions $jet_{N}(Exp)$, $jet_{N}(Ln)$ of §2.3 hold for
$N=\lceil\frac{d-ord(f)}{j}\rceil$. Then:

$\overline{{\mathscr{K}}\\!^{(j)}f}\supseteq\\{f\\}+I^{d}\cdot R^{\oplus
p}_{X}$ if and only if  ${T_{{\mathscr{K}}\\!^{(j)}}f}\supseteq I^{d}\cdot
R^{\oplus p}_{X}$.

Proof.

  1. Step 1.

In the proof of $\mathscr{R}$-case, Theorem 4.5, we have used the transitions
$T_{\mathscr{R}^{(j)}}\rightleftarrows\mathscr{R}^{(j)}$, ensured by the
$jet_{N}$ assumptions. Now we need the transitions
$T_{{\mathscr{K}}\\!^{(j)}}\rightleftarrows{\mathscr{K}}\\!^{(j)}$. These are
ensured by lemma 3.22 of [B.G.K.22]:
$\bullet$ For any element $(u,\xi)\in T_{{\mathscr{K}}\\!^{(j)}}$ there exists
an element $(U,\Phi)\in{\mathscr{K}}\\!^{(j)}$ satisfying:

$ord_{I}[(U,\Phi)f-f-(u,\xi)f]>ord_{I}[(u,\xi)f].$

$\bullet$ For any element $(U,\Phi)\in{\mathscr{K}}\\!^{(j)}$ there exists an
element $(u,\xi)\in T_{{\mathscr{K}}\\!^{(j)}}$ satisfying:

$ord_{I}[(U,\Phi)f-f-(u,\xi)f]>ord_{I}[(U,\Phi)f-f].$

The statement of that lemma in [B.G.K.22] assumes the conditions
$jet_{N}(Exp)$, $jet_{N}(Ln)$ for each $N$. In particular this implies:
$\Bbbk\supseteq\mathbb{Q}$. However, the proof in [B.G.K.22] uses these
conditions only up to $N=\lceil\frac{d-ord(f)}{j}\rceil$, as we explain now.

The only part of the proof involving the conditions $jet_{N}(Exp)$,
$jet_{N}(Ln)$ is the Baker-Campbell-Hausdorff formula, 2.1.ix. For the
transitions $T_{{\mathscr{K}}\\!^{(j)}}\rightleftarrows{\mathscr{K}}\\!^{(j)}$
we need the expansion only up to order $l$ that satisfies: $l\cdot
j+ord(f)\geq d$. Thus it suffices to expand up to order
$l=\lceil\frac{d-ord(f)}{j}\rceil$, which coincides with $N$. Our
$jet_{N}(\dots)$ assumptions imply $2,\dots,N\in\Bbbk^{\times}$. Therefore the
factor $(l!)^{4}$ is invertible. Therefore the Baker-Campbell-Hausdorff
formula, (1), holds in $R_{X}$ up to order $l=N$. Thus lemma 3.22 of
[B.G.K.22] is applicable to our case.

  2. Step 2.

We restate lemma 3.22 of [B.G.K.22] in the form:

(36) $w_{d}\in T_{{\mathscr{K}}\\!^{(j)}}f+M_{d+1}\quad\quad\text{ iff
}\quad\quad w_{d}\in({\mathscr{K}}\\!^{(j)}f-f)+M_{d+1}.$

Using this equivalence the proof goes as follows.

     * •

If $\overline{{\mathscr{K}}\\!^{(j)}}f\supseteq\\{f\\}+M_{d}$ then
$T_{{\mathscr{K}}\\!^{(j)}}f+M_{d+1}\supseteq M_{d}$. As $M_{d+1}=I\cdot
M_{d}$ we get (by Nakayama): $T_{{\mathscr{K}}\\!^{(j)}}f\supseteq M_{d}$.

     * •

If $T_{{\mathscr{K}}\\!^{(j)}}f\supseteq M_{d}$ then
$T_{{\mathscr{K}}\\!^{(j+l)}}f\supseteq M_{d+l}$ for every $l\geq 0$. For each
$l$ we use (36) with $N=\frac{d+l-ord(f)}{j+l}\leq\frac{d-ord(f)}{j}$. And
then ${\mathscr{K}}\\!^{(j+l)}f+M_{d+l+1}\supseteq\\{f\\}+M_{d+l}$ for every
$l\geq 0$. Thus $\overline{{\mathscr{K}}\\!^{(j)}}f\supseteq\\{f\\}+M_{d}$.
$\blacksquare$

######  Corollary 5.5.

Let $R_{X}\in{\Bbbk[[x]]}{}$, ${\Bbbk\\{x\\}}{}$, ${\Bbbk\langle x\rangle}{}$,
see §2.1.ii. Take $I\subseteq(x)$ and $f\in(x)\cdot R^{\oplus p}_{X}$.

  1. 1.

Suppose either $char(\Bbbk)=0$ or
$char(\Bbbk)>\lceil\frac{d-ord(f)}{j}\rceil$. Then:
${\mathscr{K}}\\!^{(j)}f\supseteq\\{f\\}+I^{d}\cdot R^{\oplus p}_{X}$ if and
only if ${T_{{\mathscr{K}}\\!^{(j)}}f}\supseteq I^{d}\cdot R^{\oplus p}_{X}$.

  2. 2.

Suppose $char(\Bbbk)\neq 2$. If $T_{{\mathscr{K}}\\!^{(j)}}f\supseteq
I^{d}\cdot R^{\oplus p}_{X}$ then ${\mathscr{K}}\\!^{(d-j-
ord(f))}f\supseteq\\{f\\}+I^{2d-2j-ord(f)}\cdot R^{\oplus p}_{X}$.

The proofs of both parts are the same as in Corollary 4.6.

######  Example 5.6.

  1. i.

Take $p=1$, $I={\mathfrak{m}}$, $j=1$ and suppose either $char(\Bbbk)=0$ or
$char(\Bbbk)>d-ord(f)$. Then:

(37) ${\mathscr{K}}\\!^{1}f\supseteq\\{f\\}+{\mathfrak{m}}^{d}$ if and only if
${\mathfrak{m}}^{2}\cdot
Jac(f)+{\mathfrak{m}}\cdot(f)\supseteq{\mathfrak{m}}^{d}$.

In particular, the ${\mathscr{K}}\\!$-order of determinacy of $f$ is
$\leq\tau(f)+1$. In $char(\Bbbk)=0$ this is well known, e.g. [Gr.Lo.Sh,
Corollary 2.24]. For $char(\Bbbk)>d-ord(f)$ this strengthens the bound
$(2\tau(f)-ord(f)+2)$ of [Bou.Gr.Ma.11], [Greuel-Pham.19].

  2. ii.

For $j=1$ part two of Corollary 5.5 is known, e.g. see Corollary 5.10 of
[B.G.K.22].

##  6\. Properties of the tangent space $T_{\mathscr{A}}f$

The tangent space $T_{\mathscr{A}}f$ is not an $R_{X}$-module. And
$T_{\mathscr{A}}f$ is not finitely generated as a module over $R_{Y}$.
Numerous simple/immediate properties of $T_{\mathscr{R}}f,T_{\mathscr{K}}\\!f$
become cumbersome (and non-trivial) for $T_{\mathscr{A}}f$. Below we study the
annihilator $\mathfrak{a}_{\mathscr{A}}$, the extended module structure of
$T_{\mathscr{A}}f$, the Nakayama&Artin-Rees properties, and the topological
closure $\overline{T_{\mathscr{A}}f}$.

Let $R_{X}$ be a local henselian $\Bbbk$-algebra, §2.1.ii. Take the filtration
$I^{\bullet}\cdot R^{\oplus p}_{X}$, we always assume $(f)\subseteq I$.
Accordingly one has the filtrations $\mathscr{A}^{(\bullet)}$ and
$T_{\mathscr{A}^{(\bullet)}}$, with $T_{\mathscr{A}^{(-1)}}:=T_{\mathscr{A}}$,
see §3.4.2. Fix a map $f:X\to(\Bbbk^{p},o)$, i.e. $f\in{\mathfrak{m}}\cdot
R^{\oplus p}_{X}$. Take the annihilator ideals
$\mathfrak{a}_{\mathscr{A}},\mathfrak{a}_{\mathscr{A}^{(j)}}\subset R_{X}$
defining the instability locus, see §3.7. In general
$\mathfrak{a}_{\mathscr{A}},\mathfrak{a}_{\mathscr{A}^{(j)}}\not\subset
f^{\\#}(R_{Y})$.

The map $f$ is called infinitesimally $\mathscr{A}$-stable if
$T_{\mathscr{A}}f=R^{\oplus p}_{X}$, i.e. $\mathfrak{a}_{\mathscr{A}}=R_{X}$.

###  6.1. The annihilator $\mathfrak{a}_{\mathscr{A}}$

The case $p=1$ is simple and (probably) known.

######  Lemma 6.1.

Assuming $f\in{\mathfrak{m}}^{2}$ one has
$\mathfrak{a}_{\mathscr{R}}\subseteq\mathfrak{a}_{\mathscr{A}}\subseteq\mathfrak{a}_{\mathscr{R}}:{\mathfrak{m}}$
and
$\mathfrak{a}_{\mathscr{A}}=\mathfrak{a}_{\mathscr{R}}+\mathfrak{a}_{\mathscr{A}}\cap
f^{\\#}(R_{Y})$.

In particular, for $R_{X}$ one of $\Bbbk[[x]]$, $\Bbbk\\{x\\}$, $\Bbbk\langle
x\rangle$, one has: $Jac(f)\subseteq\mathfrak{a}_{\mathscr{A}}\subseteq
Jac(f):{\mathfrak{m}}$.
Proof. (The part
$\mathfrak{a}_{\mathscr{R}}\subseteq\mathfrak{a}_{\mathscr{A}}$ is obvious.)
Take any $g\in\mathfrak{a}_{\mathscr{A}}$, then
$g\in\mathfrak{a}_{\mathscr{R}}+f^{\\#}(y)^{d}$. For any $q\in{\mathfrak{m}}$
one has $q\cdot g\in\mathfrak{a}_{\mathscr{R}}+f^{\\#}(y)^{d^{\prime}}$. (Here
$d,d^{\prime}\leq\infty$ are the largest possible.)

  * •

If $d^{\prime}\leq d$ then $(1-\tilde{q}\cdot
q)g\in\mathfrak{a}_{\mathscr{R}}$ for some $\tilde{q}\in R_{Y}\subset R_{X}$.
And therefore $g\in\mathfrak{a}_{\mathscr{R}}$.

  * •

Suppose $d^{\prime}>d$ for each $q\in{\mathfrak{m}}$. Then $q\cdot
g\in\mathfrak{a}_{\mathscr{R}}+g\cdot(f)$. Thus ${\mathfrak{m}}\cdot
g\subseteq\mathfrak{a}_{\mathscr{R}}+g\cdot{\mathfrak{m}}^{2}$. Nakayama
gives: ${\mathfrak{m}}\cdot g\subseteq\mathfrak{a}_{\mathscr{R}}$.

In both cases we get
${\mathfrak{m}}\\!\cdot\\!\mathfrak{a}_{\mathscr{A}}\\!\subseteq\\!\mathfrak{a}_{\mathscr{R}}$.
Hence
$\mathfrak{a}_{\mathscr{A}}\\!\subseteq\\!\mathfrak{a}_{\mathscr{R}}\\!:\\!{\mathfrak{m}}$.
The statement
$\mathfrak{a}_{\mathscr{A}}=\mathfrak{a}_{\mathscr{R}}+\mathfrak{a}_{\mathscr{A}}\cap
f^{\\#}(R_{Y})$ follows. $\blacksquare$

For $p\geq 2$ the annihilator $\mathfrak{a}_{\mathscr{A}}$ is more
complicated. First we list several basic properties. Most of them are
immediate/well known.

######  Properties 6.2.

Let the pair $(R_{X},R_{Y})$ be as in §3.1.

  1. i.

$\mathfrak{a}_{\mathscr{A}}\cdot R^{\oplus
p}_{X}\subseteq\mathfrak{a}_{\mathscr{K}}\\!\cdot R^{\oplus
p}_{X}+\Bbbk^{\oplus p}$.

  2. ii.

If $(f)\subseteq\mathfrak{a}_{\mathscr{A}}$ then
$\mathfrak{a}_{\mathscr{K}}\\!\subseteq\mathfrak{a}_{\mathscr{A}}$.
If $f$ is infinitesimally stable, i.e. $\mathfrak{a}_{\mathscr{A}}=R_{X}$,
then $T_{\mathscr{A}^{(0)}}f=T_{{\mathscr{K}}\\!^{(0)}}f$.
(In Lemma 7.1 of [Mo. N.B.] this is stated for $I={\mathfrak{m}}$, but the
proof is the same.)
More generally, $T_{\mathscr{A}^{(0)}}f=T_{{\mathscr{K}}\\!^{(0)}}f$ iff
$(f)\cdot R^{\oplus p}_{X}\subseteq T_{\mathscr{A}^{(0)}}f$ iff
$\mathfrak{a}_{\mathscr{A}^{(0)}}=\mathfrak{a}_{{\mathscr{K}}\\!^{(0)}}$.

  3. iii.

If $(f)\subseteq{\mathfrak{m}}^{2}$ and
$\mathfrak{a}_{\mathscr{A}}\subseteq{\mathfrak{m}}$ then
$\mathfrak{a}_{\mathscr{A}}+(f)\subseteq\mathfrak{a}_{\mathscr{K}}\\!$.

  4. iv.

If $\mathfrak{a}_{\mathscr{A}}\subseteq{\mathfrak{m}}$ then
$\sqrt{\mathfrak{a}_{\mathscr{A}}+(f)}=\sqrt{\mathfrak{a}_{\mathscr{K}}\\!}\subset
R_{X}$. Moreover,
$\sqrt{\mathfrak{a}_{\mathscr{A}^{(j)}}+(f)}=\sqrt{\mathfrak{a}_{{\mathscr{K}}\\!^{(j)}}}$
for $j\geq 0$.
(Geometrically: if $f$ is $\mathscr{A}$-unstable then the singular locus
$Sing(f)$ coincides with the instability locus restricted to the germ $V(f)$.)
Proof. The part $\mathbf{\subseteq}$. For $d\\!\gg\\!1$ we have:
$(\mathfrak{a}_{\mathscr{A}}+(f))^{d}\\!\cdot\\!R^{\oplus
p}_{X}\\!\subseteq\\!T_{\mathscr{R}}f+T_{\mathscr{L}^{(0)}}f\\!\subseteq\\!T_{\mathscr{R}}f+(f)\\!\cdot\\!R^{\oplus
p}_{X}=T_{\mathscr{K}}\\!f.$

The case $j\geq 0$ is immediate.

The part $\mathbf{\supseteq}$. Take the filtration
${\mathfrak{m}}^{\bullet}\subset R_{X}$. Then
$\sqrt{\mathfrak{a}_{\mathscr{A}}+(f)}\supseteq\sqrt{\mathfrak{a}_{\mathscr{A}^{(0)}}+(f)}\supseteq\sqrt{\mathfrak{a}_{\mathscr{R}^{(0)}}+(f)}=\sqrt{\mathfrak{a}_{\mathscr{R}}+(f)}=\sqrt{\mathfrak{a}_{\mathscr{K}}\\!}$.
The last transition is by example 3.12.

The statement with $j\geq 0$ is similar:
$\sqrt{\mathfrak{a}_{\mathscr{A}^{(j)}}+(f)}\supseteq\sqrt{\mathfrak{a}_{\mathscr{R}^{(j)}}+(f)}=\sqrt{\mathfrak{a}_{{\mathscr{K}}\\!^{(j)}}}$.
$\blacksquare$

  5. v.

If $\mathfrak{a}_{\mathscr{K}}\\!\subseteq\sqrt{I}$ then
$\sqrt{\mathfrak{a}_{\mathscr{A}^{(j)}}+(f)}=\sqrt{\mathfrak{a}_{\mathscr{A}}+(f)}$
for each $j\geq 0$.
Proof.
$\sqrt{\mathfrak{a}_{\mathscr{A}^{(j)}}+(f)}=\sqrt{\mathfrak{a}_{{\mathscr{K}}\\!^{(j)}}}\supseteq\sqrt{I\cdot\mathfrak{a}_{{\mathscr{K}}\\!}}=\sqrt{\mathfrak{a}_{\mathscr{K}}\\!}=\sqrt{\mathfrak{a}_{\mathscr{A}}+(f)}$.
$\blacksquare$

  6. vi.

$\mathfrak{a}_{\mathscr{A}}\cdot\mathfrak{a}_{\mathscr{K}}\\!\cdot R^{\oplus
p}_{X}\subseteq(\mathfrak{a}_{\mathscr{A}}+(f))\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}^{(0)}}f$.
If $(f)\subseteq{\mathfrak{m}}^{2}$ and
$\mathfrak{a}_{\mathscr{A}}\subseteq{\mathfrak{m}}$ then
$\mathfrak{a}_{\mathscr{A}}\cdot\mathfrak{a}_{\mathscr{K}}\\!\cdot R^{\oplus
p}_{X}\subseteq(\mathfrak{a}_{\mathscr{A}}+(f))\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}^{(1)}}f$.

  7. vii.

$\mathfrak{a}_{\mathscr{A}^{(0)}}\cdot\mathfrak{a}_{{\mathscr{K}}\\!}\cdot
R^{\oplus p}_{X}\subseteq\mathfrak{a}_{\mathscr{A}^{(0)}}\cdot
T_{\mathscr{R}}f+(f)\cdot T_{\mathscr{R}^{(0)}}f+T_{\mathscr{L}^{(1)}}f$.
$\mathfrak{a}_{\mathscr{A}^{(0)}}\cdot\mathfrak{a}_{{\mathscr{K}}\\!^{(0)}}\cdot
R^{\oplus p}_{X}\subseteq(\mathfrak{a}_{\mathscr{A}^{(0)}}+(f))\cdot
T_{\mathscr{R}^{(0)}}f+T_{\mathscr{L}^{(1)}}f$.
$\mathfrak{a}_{\mathscr{A}}\cdot\mathfrak{a}^{2}_{\mathscr{K}}\\!\cdot
R^{\oplus
p}_{X}\subseteq(\mathfrak{a}_{\mathscr{A}}\cdot\mathfrak{a}_{\mathscr{K}}\\!+(f)^{2})\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}^{(1)}}f$.

  8. viii.

$\sqrt{\mathfrak{a}_{\mathscr{A}^{(j)}}}=\sqrt{I\cdot\mathfrak{a}_{\mathscr{A}}}$
for every $j\geq 0$.

Proof. The part $\subseteq$. By the assumption $(f)\subseteq I$. Therefore
$T_{\mathscr{A}^{(j)}}f\subseteq I\cdot R^{\oplus p}_{X}$. Thus
$\mathfrak{a}_{\mathscr{A}^{(j)}}\subseteq I\cap\mathfrak{a}_{\mathscr{A}}$.
Therefore (by Artin-Rees or via the primary decomposition)
$\mathfrak{a}_{\mathscr{A}^{(j)}}\subseteq\sqrt{I\cdot\mathfrak{a}_{\mathscr{A}}}$.

The part $\supseteq$. The quotient
${\footnotesize\left.\raisebox{1.37775pt}{$T_{\mathscr{A}}f$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}}f$}\right.}$
is a finitely-generated module over $R_{Y}$. Thus the submodule
${\footnotesize\left.\raisebox{1.37775pt}{$\mathfrak{a}_{\mathscr{A}}\cdot
R^{\oplus
p}_{X}+T_{\mathscr{R}}f$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}}f$}\right.}\subseteq{\footnotesize\left.\raisebox{1.37775pt}{$T_{\mathscr{A}}f$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}}f$}\right.}$
is finitely generated, see §2.1.vii. Take the filtration
$(I\cdot\mathfrak{a}_{\mathscr{A}})^{\bullet}\cdot{\footnotesize\left.\raisebox{1.37775pt}{$\mathfrak{a}_{\mathscr{A}}\cdot
R^{\oplus
p}_{X}+T_{\mathscr{R}}f$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}}f$}\right.}$.
It is strictly decreasing. Therefore for each $i\geq 1$ there exists
$d_{i}<\infty$ satisfying:

$(I\cdot\mathfrak{a}_{\mathscr{A}})^{d_{i}}\cdot{\footnotesize\left.\raisebox{4.13332pt}{$\mathfrak{a}_{\mathscr{A}}\cdot
R^{\oplus
p}_{X}+T_{\mathscr{R}}f$}\\!\\!\diagup\\!\\!\raisebox{-4.13332pt}{$T_{\mathscr{R}}f$}\right.}\subset
f^{\\#}(y)^{i+1}{\footnotesize\left.\raisebox{1.37775pt}{$T_{\mathscr{A}}f$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}}f$}\right.}.$

Hence $(I\cdot\mathfrak{a}_{\mathscr{A}})^{d_{i}+1}\cdot R^{\oplus
p}_{X}\subset T_{\mathscr{R}}f+T_{\mathscr{L}^{(i)}}f$. Then
$(I\cdot\mathfrak{a}_{\mathscr{A}})^{d_{i}+1}\cdot R^{\oplus p}_{X}\subset
T_{\mathscr{R}}f\cap I^{i+1}\cdot R^{\oplus p}_{X}+T_{\mathscr{L}^{(i)}}f$,
because $(f)\subseteq I$. Artin-Rees over $R_{X}$ gives:
$(I\cdot\mathfrak{a}_{\mathscr{A}})^{d}\cdot R^{\oplus p}_{X}\subset
T_{\mathscr{R}^{(j)}}f+T_{\mathscr{L}^{(j)}}f$ for each $j\geq 0$ and $d\gg
j$. Hence $\sqrt{\mathfrak{a}_{\mathscr{A}^{(j)}}}\supseteq
I\cdot\mathfrak{a}_{\mathscr{A}}$. $\blacksquare$

The following technical lemma is used to apply theorem 7.1 in particular
cases.

######  Lemma 6.3.

  1. 1.

$(\mathfrak{a}_{\mathscr{R}}+\mathfrak{a}_{\mathscr{A}}\cap
f^{\\#}(y))^{2}\cdot R^{\oplus
p}_{X}\subseteq(\mathfrak{a}_{\mathscr{R}}+\mathfrak{a}_{\mathscr{A}}\cap
f^{\\#}(y))\cdot T_{\mathscr{R}}f+(\mathfrak{a}_{\mathscr{A}}\cap
f^{\\#}(y))\cdot T_{\mathscr{L}}f$.

  2. 2.

$(\mathfrak{a}_{\mathscr{R}^{(0)}}+\mathfrak{a}_{\mathscr{A}^{(0)}}\cap
f^{\\#}(y))^{2}\cdot R^{\oplus
p}_{X}\subseteq(\mathfrak{a}_{\mathscr{R}^{(0)}}+\mathfrak{a}_{\mathscr{A}^{(0)}}\cap
f^{\\#}(y))\cdot T_{\mathscr{R}^{(0)}}f+(\mathfrak{a}_{\mathscr{A}^{(0)}}\cap
f^{\\#}(y))\cdot T_{\mathscr{L}^{(0)}}f$.

Proof. We prove the second statement. Take the largest $R_{Y}$-submodule
$M\subseteq T_{\mathscr{A}^{(0)}}f$ such that $M\subseteq{\mathfrak{m}}\cdot
R^{\oplus p}_{X}$ is an $R_{X}$-submodule. (Thus $M$ is uniquely defined.) In
particular $M\supseteq
T_{\mathscr{R}^{(0)}}f+\mathfrak{a}_{\mathscr{A}^{(0)}}\cdot R^{\oplus
p}_{X}$. Therefore we can present $M=T_{\mathscr{R}^{(0)}}f+\Lambda$ for some
$R_{Y}$-submodule $\Lambda\subseteq T_{\mathscr{L}^{(0)}}f$. We take $\Lambda$
the largest possible, thus
$\Lambda\supseteq\mathfrak{a}_{\mathscr{A}^{(0)}}R^{\oplus p}_{X}\cap
T_{\mathscr{L}^{(0)}}f$.

In fact we claim: $Ann({\footnotesize\left.\raisebox{1.37775pt}{$R^{\oplus
p}_{Y}$}\\!/\\!\raisebox{-1.37775pt}{$\Lambda$}\right.})=\mathfrak{a}_{\mathscr{A}^{(0)}}\cap(y)$.
Indeed, suppose $\Lambda\supseteq\mathfrak{b}_{Y}\cdot R^{\oplus p}_{Y}$ for
an ideal $\mathfrak{b}_{Y}\subset f^{\\#}(R_{Y})$. Then $M\supseteq
R_{X}\cdot(\mathfrak{b}_{Y}\cdot R^{\oplus p}_{Y})$. And thus
$\mathfrak{a}_{\mathscr{A}^{(0)}}\supseteq R_{X}\cdot\mathfrak{b}_{y}$. Hence
$\mathfrak{b}_{Y}\subseteq\mathfrak{a}_{\mathscr{A}^{(0)}}\cap
f^{\\#}(R_{Y})$.

Altogether: $(\mathfrak{a}_{\mathscr{A}^{(0)}}\cap(y))\cdot R^{\oplus
p}_{X}\subseteq M=T_{\mathscr{R}^{(0)}}f+\Lambda$. Therefore

(38)
$(\mathfrak{a}_{\mathscr{R}^{(0)}}+\mathfrak{a}_{\mathscr{A}^{(0)}}\cap(y))^{2}\cdot
R^{\oplus
p}_{X}=(\mathfrak{a}_{\mathscr{R}^{(0)}}+\mathfrak{a}_{\mathscr{A}^{(0)}}\cap(y))\cdot\mathfrak{a}_{\mathscr{R}^{(0)}}\cdot
R^{\oplus p}_{X}+(\mathfrak{a}_{\mathscr{A}^{(0)}}\cap(y))^{2}\cdot R^{\oplus
p}_{X}\subseteq$ $\hskip
256.0748pt\subseteq(\mathfrak{a}_{\mathscr{R}^{(0)}}+\mathfrak{a}_{\mathscr{A}^{(0)}}\cap(y))\cdot
T_{\mathscr{R}^{(0)}}+(\mathfrak{a}_{\mathscr{A}^{(0)}}\cap(y))\cdot\Lambda.\hskip
28.45274pt{\hfill\blacksquare}$

###  6.2. Extending the module structure of $T_{\mathscr{A}^{(\bullet)}}f$

The tangent spaces $T_{\mathscr{A}^{(\bullet)}}f\subset R^{\oplus p}_{X}$ are
$R_{Y}$-submodules. In fact they are modules over a larger ring.

######  Lemma 6.4.

  1. 1.

$\mathfrak{a}_{\mathscr{A}}\cdot T_{\mathscr{A}}f\subseteq T_{\mathscr{A}}f$.

  2. 2.

If $(f)\subseteq I$ then $\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{A}^{(j)}}f\subseteq T_{\mathscr{A}^{(j)}}f$ for each $j\geq 0$.
Therefore all the spaces $T_{\mathscr{A}^{(j)}}f$ are modules over the subring
$f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}}\subseteq R_{X}$.

  3. 3.

The map $f$ is ${\mathscr{K}}\\!$-finite (§3.7) iff the $\Bbbk$-module
${\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{$(f)+\mathfrak{a}_{\mathscr{A}}$}\right.}$
is finitely generated.
Moreover, if the Weierstraß finiteness holds (§2.1.vi) then any
$(f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}})$-submodule of $R^{\oplus p}_{X}$
is finitely generated. In particular $R^{\oplus p}_{X}$ and $T_{\mathscr{A}}f$
are f.g. over the local Noetherian ring
$f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}}$.

Proof.

  1. 1.

Immediate verification: $\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{A}}f\subseteq\mathfrak{a}_{\mathscr{A}}\cdot R^{\oplus
p}_{X}\subseteq T_{\mathscr{A}}f$.

  2. 2.

We have: $\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{A}^{(j)}}f\stackrel{{\scriptstyle\S\ref{Sec.3.Filtrations.on.G.TG}}}{{=}}\mathfrak{a}_{\mathscr{A}}\big{(}T_{\mathscr{R}^{(j)}}f+f^{\\#}(\mathfrak{b}_{j})\cdot
T_{\mathscr{L}}f\big{)}\subseteq
T_{\mathscr{R}^{(j)}}f+f^{\\#}(\mathfrak{b}_{j})\cdot T_{\mathscr{A}}f$. Note
that $f^{\\#}(\mathfrak{b}_{j})\subseteq I^{j+1}$, as $(f)\\!\subseteq\\!I$.
Thus
$f^{\\#}(\mathfrak{b}_{j})\\!\cdot\\!T_{\mathscr{R}}f\\!\subseteq\\!T_{\mathscr{R}^{(j)}}f.$
In addition
$f^{\\#}(\mathfrak{b}_{j})\\!\cdot\\!T_{\mathscr{L}}f\\!\subseteq\\!T_{\mathscr{L}^{(j)}}f.$
Therefore
$f^{\\#}(\mathfrak{b}_{j})\\!\cdot\\!T_{\mathscr{A}}f\\!\subseteq\\!T_{\mathscr{A}^{(j)}}f.$

  3. 3.

The subring $f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}}\subseteq R_{X}$ is
local as the rings $R_{Y}$, $R_{X}$ are local and
$\mathfrak{a}_{\mathscr{A}}\subseteq R_{X}$ is an ideal.

We can assume $\mathfrak{a}_{\mathscr{A}}\subseteq{\mathfrak{m}}$. Then, by
property 6.2.iv we have
$\sqrt{(f)+\mathfrak{a}_{\mathscr{A}}}=\sqrt{\mathfrak{a}_{\mathscr{K}}\\!}={\mathfrak{m}}$.
Hence
${\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{$(f)+\mathfrak{a}_{\mathscr{A}}$}\right.}$
is $\Bbbk$-finite iff $\sqrt{(f)+\mathfrak{a}_{\mathscr{A}}}={\mathfrak{m}}$
iff $f$ is ${\mathscr{K}}\\!$-finite.

Applying Weierstraß finiteness to
${\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{$\mathfrak{a}_{\mathscr{A}}$}\right.}\otimes{\footnotesize\left.\raisebox{1.37775pt}{$R_{Y}$}\\!/\\!\raisebox{-1.37775pt}{$(y)$}\right.}$
we get:
${\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{$\mathfrak{a}_{\mathscr{A}}$}\right.}$
is f.g. over $R_{Y}$.

And then $R_{X}$ is a f.g. module over
$f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}}$. (Fix a finite set of
$R_{X}$-preimages of the generators of
${\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{$\mathfrak{a}_{\mathscr{A}}$}\right.}\in
mod\text{-}f^{\\#}(R_{Y})$. They generate $R_{X}$ over
$f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}}$.)

Therefore the subring $f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}}\subset R_{X}$
is Noetherian. (This is the Eakin-Nagata theorem, [Eisenbud, Exercise A.3.7,
pg.625].)

Finally, apply the remark of §2.1.vii to $R^{\oplus p}_{X}$ as a f.g. module
over $f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}}$. $\blacksquare$

###  6.3. Nakayama-type results for $T_{\mathscr{A}}f$

Suppose a submodule $M\subseteq R^{\oplus p}_{X}$ satisfies
$T_{\mathscr{G}}f+{\mathfrak{m}}\cdot M\supseteq M$. For
$\mathscr{G}\in\mathscr{R},{\mathscr{K}}\\!$ one concludes (by Nakayama):
$T_{\mathscr{G}}f\supseteq M$. But $T_{\mathscr{A}}f$ is not an
$R_{X}$-module. If considered as $R_{Y}$-modules, $T_{\mathscr{A}}f,M$ are not
always finitely generated. Recall the “adjusted Nakayama lemma”.

######  Lemma 6.5.

[Wall.81, lemma 1.6] Take a homomorphism of (local, Noetherian) rings
$(R_{Y},(y))\stackrel{{\scriptstyle f}}{{\to}}(R_{X},{\mathfrak{m}})$, with
Weierstraß finiteness condition. Take f.g. modules $M_{X}\in
mod\text{-}R_{X}$, $M_{Y}\in mod\text{-}R_{Y}$.

  1. 1.

If $M_{X}\subseteq M_{Y}+(f)\cdot M_{X}$ then $M_{X}\subseteq M_{Y}$.

  2. 2.

Suppose the inclusion $\mathfrak{a}\cdot M_{X}\subseteq
M_{Y}+\mathfrak{a}\cdot\mathfrak{b}\cdot M_{X}$ holds for some ideals
$\mathfrak{b}\subseteq{\mathfrak{m}}$, $\mathfrak{a}\subseteq R_{X}$
satisfying: $\mathfrak{b}\cdot M_{Y}\subseteq\mathfrak{a}\cdot(f)\cdot M_{X}$.
Then $\mathfrak{a}\cdot M_{X}\subseteq M_{Y}$.

Proof.

  1. 1.

One has
${\footnotesize\left.\raisebox{1.37775pt}{$R_{Y}$}\\!/\\!\raisebox{-1.37775pt}{$(y)$}\right.}\otimes
M_{X}\cong{\footnotesize\left.\raisebox{1.37775pt}{$M_{X}$}\\!/\\!\raisebox{-1.37775pt}{$(f)M_{X}$}\right.}\subseteq{\footnotesize\left.\raisebox{1.37775pt}{$M_{Y}+(f)M_{X}$}\\!/\\!\raisebox{-1.37775pt}{$(f)M_{X}$}\right.}$.
Here
${\footnotesize\left.\raisebox{1.37775pt}{$M_{Y}+(f)M_{X}$}\\!/\\!\raisebox{-1.37775pt}{$(f)M_{X}$}\right.}$
is f.g. over $R_{Y}$. Thus
${\footnotesize\left.\raisebox{1.37775pt}{$R_{Y}$}\\!/\\!\raisebox{-1.37775pt}{$(y)$}\right.}\otimes
M_{X}$ is f.g. over $R_{Y}$, see §2.1.vii. By Weierstraß finiteness $M_{X}$ is