f.g. over $R_{Y}$. Finally, apply Nakayama over $R_{Y}$ to the inclusion
$M_{X}\subseteq M_{Y}+(f)\cdot M_{X}$.

  2. 2.

One has $\mathfrak{a}\cdot M_{X}\in mod\text{-}R_{X}$ and
$\mathfrak{a}\cdot\mathfrak{b}\cdot M_{X}\subseteq\mathfrak{a}\cdot(f)\cdot
M_{X}+\mathfrak{a}\cdot\mathfrak{b}^{2}\cdot M_{X}$. Nakayama over $R_{X}$
gives $\mathfrak{a}\cdot\mathfrak{b}\cdot
M_{X}\\!\subseteq\\!\mathfrak{a}\cdot(f)\cdot M_{X}.$ The initial assumption
gives $\mathfrak{a}\cdot
M_{X}\\!\subseteq\\!M_{Y}\\!+\\!\mathfrak{a}\\!\cdot\\!(f)\\!\cdot\\!M_{X}$.
Apply part one. $\blacksquare$

######  Corollary 6.6.

Suppose the pair $(R_{X},R_{Y})$ satisfies Weierstraß finiteness condition,
e.g. it is one of
$({\footnotesize\left.\raisebox{1.37775pt}{$\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$}\right.},\Bbbk\langle y\rangle)$,
§2.1.ii. Fix some $j\geq-1$.

  1. 1.

If $T_{\mathscr{A}^{(j)}}f+(f)\cdot M\supseteq M$ then
$T_{\mathscr{A}^{(j)}}f\supseteq M$.

  2. 2.

Suppose $\mathfrak{a}_{\mathscr{A}}\subseteq{\mathfrak{m}}$. If
$T_{\mathscr{A}^{(j)}}f+(\mathfrak{a}_{\mathscr{A}}+(f))\cdot M\supseteq M$
then $T_{\mathscr{A}^{(j)}}f\supseteq M$.

Proof. We can assume $M\supseteq T_{\mathscr{R}^{(j)}}f$.

  1. 1.

Pass to the quotients,
${\footnotesize\left.\raisebox{1.37775pt}{$T_{\mathscr{A}^{(j)}}f$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}+(f)\cdot{\footnotesize\left.\raisebox{1.37775pt}{$M$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}\supseteq{\footnotesize\left.\raisebox{1.37775pt}{$M$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}$.
Here
${\footnotesize\left.\raisebox{1.37775pt}{$M$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}\in
mod\text{-}R_{X}$ and
${\footnotesize\left.\raisebox{1.37775pt}{$T_{\mathscr{A}^{(j)}}f$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}\in
mod\text{-}R_{Y}$. Therefore (by part one of lemma 6.5)
${\footnotesize\left.\raisebox{1.37775pt}{$T_{\mathscr{A}^{(j)}}f$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}\supseteq{\footnotesize\left.\raisebox{1.37775pt}{$M$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}$.

  2. 2.

Multiply the assumption by $\mathfrak{a}_{\mathscr{A}}$ to get
$\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{A}^{(j)}}f+\mathfrak{a}_{\mathscr{A}}\cdot(\mathfrak{a}_{\mathscr{A}}+(f))\cdot
M\\!\supseteq\\!\mathfrak{a}_{\mathscr{A}}\\!\cdot\\!M.$ Nakayama over $R_{X}$
gives
$\mathfrak{a}_{\mathscr{A}}\\!\cdot\\!T_{\mathscr{A}^{(j)}}f\\!\supseteq\\!\mathfrak{a}_{\mathscr{A}}\\!\cdot\\!M.$
Then $T_{\mathscr{A}^{(j)}}f\\!+\\!(f)\\!\cdot\\!M\\!\supseteq\\!M,$ by part
two of lemma 6.4. Now invoke part 1. $\blacksquare$

######  Example 6.7.

Let $(R_{X},R_{Y})$ be one of 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$}\right.},\Bbbk\langle y\rangle)$,
see §2.1.ii.
Let $I={\mathfrak{m}}$ and suppose
${\mathfrak{a}_{\mathscr{A}}}\subseteq{\mathfrak{m}}$. Take
$M={\mathfrak{m}}^{j}\cdot(f)\cdot R^{\oplus p}_{X}$ for some $j\geq 0$. Part
2 of the lemma gives:

(39) If
$T_{\mathscr{A}^{(j)}}f+(\mathfrak{a}_{\mathscr{A}}+(f))\cdot{\mathfrak{m}}^{j}\cdot(f)\cdot
R^{\oplus p}_{X}\supseteq{\mathfrak{m}}^{j}(f)R^{\oplus p}_{X}$ then
$T_{\mathscr{A}^{(j)}}f=T_{{\mathscr{K}}\\!^{(j)}}f$.

For $j=0$ this is well known, e.g. theorem 5.1 of [Ruas].

######  Remark 6.8.

One would like a stronger bound, “If
$T_{\mathscr{A}}f+(\sqrt{\mathfrak{a}_{\mathscr{A}}}+(f))\cdot M\supseteq M$
then $T_{\mathscr{A}}f\supseteq M$”. This fails already for the case
$\sqrt{\mathfrak{a}_{\mathscr{A}}}={\mathfrak{m}}$. For example, one always
has $T_{\mathscr{A}}f+{\mathfrak{m}}\cdot R^{\oplus p}_{X}=R^{\oplus p}_{X}$,
though of course $T_{\mathscr{A}}f\neq R^{\oplus p}_{X}$.

######  Corollary 6.9.

(The assumptions of corollary 6.6.) Suppose $\mathfrak{b}\cdot R^{\oplus
p}_{X}\subseteq
T_{\mathscr{A}^{(j)}}f+\mathfrak{b}\cdot(\mathfrak{b}(f):(f)^{j+1})R^{\oplus
p}_{X}$, for an ideal $\mathfrak{b}\subset R_{X}$. Then $\mathfrak{b}\cdot
R^{\oplus p}_{X}\subseteq T_{\mathscr{A}^{(j)}}f$.

Proof. Pass to the quotients,
$\mathfrak{b}\\!\cdot\\!{\footnotesize\left.\raisebox{1.37775pt}{$R^{\oplus
p}_{X}$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}\\!\subseteq\\!{\footnotesize\left.\raisebox{1.37775pt}{$T_{\mathscr{A}^{(j)}}f$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}+\mathfrak{b}\\!\cdot\\!(\mathfrak{b}(f):(f)^{j+1})\\!\cdot\\!{\footnotesize\left.\raisebox{1.37775pt}{$R^{\oplus
p}_{X}$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}.$ Note:
$(\mathfrak{b}(f):(f)^{j+1})\\!\cdot\\!{\footnotesize\left.\raisebox{1.37775pt}{$T_{\mathscr{A}^{(j)}}f$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}\\!\subseteq\\!\mathfrak{b}\\!\cdot\\!(f)\\!\cdot\\!{\footnotesize\left.\raisebox{1.37775pt}{$R^{\oplus
p}_{X}$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}.$ Then
(by part 2 of lemma 6.5):
$\mathfrak{b}\\!\cdot\\!{\footnotesize\left.\raisebox{1.37775pt}{$R^{\oplus
p}_{X}$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}\\!\subseteq\\!{\footnotesize\left.\raisebox{1.37775pt}{$T_{\mathscr{A}^{(j)}}f$}\\!/\\!\raisebox{-1.37775pt}{$T_{\mathscr{R}^{(j)}}f$}\right.}.$
$\blacksquare$

######  Example 6.10.

  1. i.

For $0\leq j<r$, $\mathfrak{b}=I^{r}$ and $(f)\subseteq I$ one gets the well-
known bound: if $T_{\mathscr{A}^{(j)}}f+I^{2r-j}\cdot R^{\oplus
p}_{X}\supseteq I^{r}\cdot R^{\oplus p}_{X}$ then
$T_{\mathscr{A}^{(j)}}f\supseteq I^{r}\cdot R^{\oplus p}_{X}$. (See e.g.
corollary 1.23, pg. 17, of [Ruas].)

  2. ii.

Moreover, for any map $g\in{\mathfrak{m}}\cdot R^{\oplus p}_{X}$ satisfying
$g-f\in I^{2r-2j}\cdot R^{\oplus p}_{X}$ one gets (by the direct check)
$T_{\mathscr{A}^{(j)}}g\supseteq I^{r}\cdot R^{\oplus p}_{X}$. For
$I={\mathfrak{m}}$ this is [du Plessis, corollary 2.4].

###  6.4. Artin-Rees type properties of $T_{\mathscr{A}}f$

The image tangent spaces $T_{\mathscr{R}}f$, $T_{\mathscr{K}}\\!f\subseteq
R^{\oplus p}_{X}$ are $R_{X}$-submodules. For them the condition
“$T_{\mathscr{G}^{(j)}}f\supseteq I^{d}\cdot R^{\oplus p}_{X}$” implies
“$T_{\mathscr{G}^{(j+\bullet)}}f\supseteq I^{d+\bullet}\cdot R^{\oplus
p}_{X}$”, for every $\bullet\geq 0$. This is heavily used in the study of
$\mathscr{R}$, ${\mathscr{K}}\\!$-orbits, theorems 4.5, 5.4.

The tangent space $T_{\mathscr{A}}f$ is a module over $R_{Y}$, not over
$R_{X}$. The condition $T_{\mathscr{A}^{(\bullet)}}f\supseteq I^{d}\cdot
R^{\oplus p}_{X}$ does not imply “$T_{\mathscr{A}^{(j+\bullet)}}f\supseteq
I^{d+\bullet}\cdot R^{\oplus p}_{X}$ for every $\bullet\geq 0$”. (For example,
in the infinitesimally stable case, i.e. $T_{\mathscr{A}}f=R^{\oplus p}_{X}$,
one does not have $T_{\mathscr{A}^{(0)}}f\supseteq{\mathfrak{m}}\cdot
R^{\oplus p}_{X}$.) Yet, some properties of Artin-Rees type do hold.

Take $f:X\to Y$ and the corresponding map
$f^{\\#}:R_{Y}\stackrel{{\scriptstyle}}{{\to}}R_{X}$.

######  Lemma 6.11.

For a pair $mod\text{-}R_{Y}\ni M_{Y}\subset M_{X}\in mod\text{-}R_{X}$ have
$M_{Y}\cap{\mathfrak{m}}^{d_{j}}\cdot
M_{X}\subseteq{\mathfrak{m}}^{j}_{Y}\cdot M_{Y}\cap{\mathfrak{m}}^{d}_{X}\cdot
M_{X}$ for every $j$ an a corresponding $d_{j}<\infty$.

Proof. The filtration ${\mathfrak{m}}^{\bullet}_{X}\cdot M_{X}$ is strictly
decreasing (by Nakayama over $R_{X}$). Take the induced filtration
${\footnotesize\left.\raisebox{1.37775pt}{$M_{Y}$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}^{j}_{Y}\cdot
M_{Y}$}\right.}\supset{\footnotesize\left.\raisebox{1.37775pt}{$M_{Y}\cap{\mathfrak{m}}^{\bullet}_{X}\cdot
M_{X}+{\mathfrak{m}}^{j}_{Y}\cdot
M_{Y}$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}^{j}_{Y}\cdot
M_{Y}$}\right.}$. Here
${\footnotesize\left.\raisebox{1.37775pt}{$M_{Y}$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}^{j}_{Y}\cdot
M_{Y}$}\right.}$ is an Artinian module over the ring
${\footnotesize\left.\raisebox{1.37775pt}{$R_{Y}$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}^{j}_{Y}$}\right.}$.
Thus the induced filtration stabilizes,
${\footnotesize\left.\raisebox{1.37775pt}{$M_{Y}\cap{\mathfrak{m}}^{d}_{X}\cdot
M_{X}+{\mathfrak{m}}^{j}_{Y}\cdot
M_{Y}$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}^{j}_{Y}\cdot
M_{Y}$}\right.}=\cap_{\bullet}{\footnotesize\left.\raisebox{1.37775pt}{$M_{Y}\cap{\mathfrak{m}}^{\bullet}_{X}\cdot
M_{X}+{\mathfrak{m}}^{j}_{Y}\cdot
M_{Y}$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}^{j}_{Y}\cdot
M_{Y}$}\right.}$ for $d\gg 1$.

Finally,
$\cap_{\bullet}{\footnotesize\left.\raisebox{1.37775pt}{$M_{Y}\cap{\mathfrak{m}}^{\bullet}_{X}\cdot
M_{X}+{\mathfrak{m}}^{j}_{Y}\cdot
M_{Y}$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}^{j}_{Y}\cdot
M_{Y}$}\right.}\subseteq\cap_{\bullet}{\footnotesize\left.\raisebox{1.37775pt}{${\mathfrak{m}}^{\bullet}_{X}\cdot
M_{X}+{\mathfrak{m}}^{j}_{Y}\cdot
M_{Y}$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}^{j}_{Y}\cdot
M_{Y}$}\right.}\twoheadleftarrow\cap_{\bullet}{\mathfrak{m}}^{\bullet}_{X}\cdot
M_{X}=0$. Thus
${\footnotesize\left.\raisebox{1.37775pt}{$M_{Y}\cap{\mathfrak{m}}^{d}\cdot
M_{X}+{\mathfrak{m}}_{Y}\cdot
M_{Y}$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}_{Y}\cdot
M_{Y}$}\right.}=0$ for $d\gg 1$. Hence the statement. $\blacksquare$

######  Lemma 6.12.

Let $R_{X}$ be a local henselian $\Bbbk$-algebra and assume $(f)\subseteq I$.

  1. 1.

Suppose $T_{\mathscr{A}^{(j_{l})}}f\supseteq{\mathfrak{c}}_{l}\cdot R^{\oplus
p}_{X}$ for $l=1,2$ and some ideals ${\mathfrak{c}}_{l}\subseteq I^{j_{l}+1}$,
with $j_{l}\geq 0$. Then
$T_{\mathscr{A}^{(j_{1}+j_{2}+1)}}f\supseteq{\mathfrak{c}}_{1}\cdot{\mathfrak{c}}_{2}\cdot
R^{\oplus p}_{X}$.

  2. 2.

Suppose $T_{\mathscr{A}^{(j)}}f\supseteq I^{d_{j}}\cdot R^{\oplus p}_{X}$ and
$T_{{\mathscr{K}}\\!^{(i)}}f\supseteq I^{d_{i}}\cdot R^{\oplus p}_{X}$ for
some $d_{j}\geq i\geq 0$ and $d_{j}>j\geq-1$. Then
$T_{\mathscr{A}^{(j+1)}}f\supseteq I^{d_{j}+d_{i}-i}\cdot R^{\oplus p}_{X}$.
By iterating this we get: $T_{\mathscr{A}^{(j+k)}}f\supseteq
I^{d_{j}+k(d_{i}-i)}\cdot R^{\oplus p}_{X}$ for each $k\geq 0$.

  3. 3.

Suppose an ideal $\mathfrak{a}\subset R_{X}$ and a subset
$\mathfrak{b}_{Y}\subseteq f^{\\#}(y)^{2}\cap\mathfrak{a}\subset R_{X}$
satisfy:
$\mathfrak{a}^{2}\subseteq\mathfrak{a}_{\mathscr{A}}\cdot\mathfrak{b}_{Y}$.
Then $\mathfrak{a}\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}^{(1)}}f\supseteq\mathfrak{b}_{Y}\cdot\mathfrak{a}_{\mathscr{A}}\cdot
R^{\oplus p}_{X}\supseteq\mathfrak{a}^{2}\cdot R^{\oplus p}_{X}$.

For a subset $\mathfrak{b}_{Y}\subseteq
f^{\\#}(y)^{2}\cap(\mathfrak{a}\cdot{\mathfrak{m}})\subset R_{X}$ satisfying
$\mathfrak{a}^{2}\subseteq\mathfrak{a}_{\mathscr{A}}\cdot C$ one has:
$\mathfrak{a}\cdot{\mathfrak{m}}\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}^{(1)}}f\supseteq\mathfrak{b}_{Y}\cdot\mathfrak{a}_{\mathscr{A}}\cdot
R^{\oplus p}_{X}\supseteq\mathfrak{a}^{2}\cdot R^{\oplus p}_{X}$.

  4. 4.

Take a subset $C\subseteq(y)\subset R_{Y}$. Then $C\cdot{\mathfrak{m}}\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}^{(1)}}f\supseteq
C\cdot\mathfrak{a}_{\mathscr{A}^{(o)}}\cdot R^{\oplus p}_{X}$.

In particular, $\mathfrak{a}_{\mathscr{A}^{(o)}}\cdot
T_{\mathscr{R}^{(0)}}f+T_{\mathscr{L}^{(1)}}f\supseteq(\mathfrak{a}_{\mathscr{A}^{(o)}}\cap
f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{R}^{(o)}})\mathfrak{a}_{\mathscr{A}^{(o)}}\cdot
R^{\oplus p}_{X}$.

And also: $(\mathfrak{a}_{\mathscr{A}}\cap
f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{R}^{(o)}})\cdot
T_{\mathscr{R}^{(0)}}f+(\mathfrak{a}_{\mathscr{A}}\cap y)^{2}\cdot
T_{\mathscr{L}}f\supseteq(\mathfrak{a}_{\mathscr{A}}\cap
f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{R}^{(o)}})^{2}\cdot R^{\oplus p}_{X}$.

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

  5. 5.

Suppose $(f)\subseteq I$ then
$\mathfrak{a}_{\mathscr{K}}\\!\cdot(\mathfrak{a}_{\mathscr{A}^{(j)}}\cap
I^{j+2})\subseteq\mathfrak{a}_{\mathscr{A}^{(j+1)}}$ for each $j\geq-1$.
If $(f)\subseteq I$ and $\mathfrak{a}_{\mathscr{A}^{(j)}}\cdot
T_{\mathscr{R}}f\subseteq T_{\mathscr{R}^{(j+1)}}f$ then
$\mathfrak{a}_{\mathscr{K}}\\!\cdot\mathfrak{a}_{\mathscr{A}^{(j)}}\subseteq\mathfrak{a}_{\mathscr{A}^{(j+1)}}$
for each $j\geq-1$.

We remark that in part 2 the sequence $\\{d_{i}-i\\}$ is non-increasing and
$d_{i}-i\geq ord(f)$.
Proof. All the proofs are direct verifications.

  1. 1.

${\mathfrak{c}}_{1}\cdot{\mathfrak{c}}_{2}\cdot R^{\oplus
p}_{X}\subseteq{\mathfrak{c}}_{1}\cdot\Big{(}T_{\mathscr{R}^{(j_{2})}}f+f^{\\#}(\mathfrak{b}_{j_{2}})\cdot
T_{\mathscr{L}}f\Big{)}\subseteq
T_{\mathscr{R}^{(j_{2}+j_{1}+1)}}f+f^{\\#}(\mathfrak{b}_{j_{2}})\cdot\Big{(}T_{\mathscr{R}^{(j_{1})}}f+$
$+f^{\\#}(\mathfrak{b}_{j_{1}})\cdot T_{\mathscr{L}}f\Big{)}\subseteq
T_{\mathscr{R}^{(j_{1}+j_{2}+1)}}f+f^{\\#}(\mathfrak{b}_{j_{1}+j_{2}+1})\cdot
T_{\mathscr{L}}f=T_{\mathscr{A}^{(j_{1}+j_{2}+1)}}f.$

  2. 2.

$I^{d_{j}+d_{i}-i}\cdot R^{\oplus p}_{X}\subseteq
I^{d_{j}-i}\cdot\left(T_{\mathscr{R}^{(i)}}f+I^{i}\cdot(f)R^{\oplus
p}_{X}\right)\subseteq T_{\mathscr{R}^{(d_{j})}}f+I^{d_{j}}\cdot(f)R^{\oplus
p}_{X}\subseteq T_{\mathscr{R}^{(d_{j})}}f+(f)\cdot
T_{\mathscr{R}^{(j)}}f+(y)\cdot T_{\mathscr{L}^{(j)}}f\subseteq
T_{\mathscr{A}^{(j+1)}}f$.

  3. 3.

Multiply the inclusion
$T_{\mathscr{A}}f\supseteq\mathfrak{a}_{\mathscr{A}}\cdot R^{\oplus p}_{X}$ by
$\mathfrak{b}_{Y}$.

  4. 4.

Multiply the inclusion
$T_{\mathscr{A}^{(o)}}f\supseteq\mathfrak{a}_{\mathscr{A}^{(o)}}\cdot
R^{\oplus p}_{X}$ by $C$.

  5. 5.

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

######  Example 6.13.

  1. i.

Part 1 gives:
$\mathfrak{a}^{2}_{\mathscr{A}^{(0)}}\subseteq\mathfrak{a}_{\mathscr{A}^{(1)}}$.

  2. ii.

Let $I={\mathfrak{m}}$ and $\mathfrak{a}\subseteq T_{\mathscr{A}^{(0)}}f$ and
$(f)\subseteq\mathfrak{a}$. Then $\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X}\subseteq I\cdot{\mathfrak{m}}\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}^{(1)}}f$.

###  6.5. Filtration closures, $\overline{T_{\mathscr{A}}f}$ vs
$T_{\mathscr{A}}f$.

We give the $\mathscr{A}$-version of lemma 3.6. Fix some $j\geq-1$.

######  Lemma 6.14.

  1. 1.

$\mathfrak{a}_{\mathscr{A}}\cdot\overline{T_{\mathscr{A}^{(j)}}f}=\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{A}^{(j)}}f$.

  2. 2.

If $f$ is ${\mathscr{K}}\\!$-finite and the Weierstraß finiteness (§2.1.vi)
holds for $(R_{X},R_{Y})$ then
$T_{\mathscr{A}^{(j)}}f=\overline{T_{\mathscr{A}^{(j)}}f}$.

  3. 3.

If $\mathfrak{a}_{\mathscr{A}}$ contains a non-zero divisor and
$\overline{T_{\mathscr{A}^{(j)}}f}\supseteq I^{d}\cdot R^{\oplus p}_{X}$ for
some $d\gg 1$, then
$T_{\mathscr{A}^{(j)}}f=\overline{T_{\mathscr{A}^{(j)}}f}$.

Proof.

  1. 1.

Observe the obvious inclusions: $\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{A}^{(j)}}f\subseteq\mathfrak{a}_{\mathscr{A}}\cdot\overline{T_{\mathscr{A}^{(j)}}f}\subseteq\overline{\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{A}^{(j)}}f}$. Note that $\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{A}^{(j)}}f\subseteq R^{\oplus p}_{X}$ is an $R_{X}$-module.
Therefore it is closed (lemma 3.6), $\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{A}^{(j)}}f=\overline{\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{A}^{(j)}}f}$. Hence $\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{A}^{(j)}}f=\mathfrak{a}_{\mathscr{A}}\cdot\overline{T_{\mathscr{A}^{(j)}}f}$.

  2. 2.

We have the $(f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}})$-submodule
$\overline{T_{\mathscr{A}}f}\subset R^{\oplus p}_{X}$. We have
$T_{\mathscr{A}}+I^{d}\cdot R^{\oplus
p}_{X}\supseteq\overline{T_{\mathscr{A}}f}\supseteq T_{\mathscr{A}}f$ for any
$d\gg 1$. Therefore $T_{\mathscr{A}}+(I^{d}\cdot R^{\oplus
p}_{X})\cap\overline{T_{\mathscr{A}}f}\supseteq\overline{T_{\mathscr{A}}f}$.

The ring $f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}}$ is local Noetherian
(lemma 6.4), therefore $\overline{T_{\mathscr{A}}f}$ is f.g. (§2.1.vii).

For $d\gg 1$ one has $I^{d}\subseteq(f)+\mathfrak{a}_{\mathscr{A}}$, by
property 6.2.iv. By lemma 6.11 one has:
$((f^{\\#}(y)+\mathfrak{a}_{\mathscr{A}})^{d}\cdot R^{\oplus
p}_{X})\cap\overline{T_{\mathscr{A}}f}\subseteq(f^{\\#}(y)+\mathfrak{a}_{\mathscr{A}})\cdot\overline{T_{\mathscr{A}}f}$.
Thus we get
$T_{\mathscr{A}}f+(f^{\\#}(y)+\mathfrak{a}_{\mathscr{A}})\cdot\overline{T_{\mathscr{A}}f}\supseteq\overline{T_{\mathscr{A}}f}$.
Apply Nakayama to $T_{\mathscr{A}}f$ as a f.g. module over
$f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}}$.

  3. 3.

As $T_{\mathscr{A}^{(j)}}f+I^{d}\cdot R^{\oplus
p}_{X}\supseteq\overline{T_{\mathscr{A}^{(j)}}f}$, it is enough to prove
$T_{\mathscr{A}^{(j)}}f\supseteq I^{d}\cdot R^{\oplus p}_{X}$ for $d\gg 1$.
Equivalently, we should prove: $\mathfrak{a}_{\mathscr{A}^{(j)}}\supseteq
I^{d}$ for $d\gg 1$. By part 1 of lemma 6.12 it is enough to prove:
$\mathfrak{a}_{\mathscr{A}}\supseteq I^{d}$ for $d\gg 1$.

By part 1 we have: $\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{A}}f\supseteq\mathfrak{a}_{\mathscr{A}}\cdot I^{d}\cdot R^{\oplus
p}_{X}$. This gives $T_{\mathscr{A}}f\supseteq I^{d}\cdot R^{\oplus p}_{X}$
for $d\gg 1$, see §2.1.viii. Therefore $\mathfrak{a}_{\mathscr{A}}\supseteq
I^{d}$ for $d\gg 1$. $\blacksquare$

######  Example 6.15.

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 $f$ is
${\mathscr{K}}\\!$-finite then
$T_{\mathscr{A}^{(j)}}f=\overline{T_{\mathscr{A}^{(j)}}f}$.

##  7\. Criteria for $\mathscr{A}$-orbits

###  7.1. The $\mathscr{A}$-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$}\right.}$, §2.1.ii. For $J\neq 0$
we assume the $jet_{0}$-condition of §2.3. Take a map $f:X\to(\Bbbk^{p},o)$.
If its critical locus, $Crit(f)$, is a point, i.e.
$\sqrt{\mathfrak{a}_{\mathscr{R}}}={\mathfrak{m}}$, then already theorem 4.1
gives a good control on the orbit, e.g.
$\mathscr{R}f\supseteq\\{f\\}+{\mathfrak{m}}\cdot\mathfrak{a}^{2}_{\mathscr{R}}\cdot
R^{\oplus p}_{X}$. For $p\geq 2$ (and $f$ not a submersion) the critical locus
is always of positive dimension. And “in most cases” the instability locus,
$V(\mathfrak{a}_{\mathscr{A}})$, is of smaller dimension than $Crit(f)$.

Take an ideal
${\mathfrak{m}}\cdot\mathfrak{a}_{\mathscr{R}}\subseteq\mathfrak{a}\subseteq{\mathfrak{m}}$.
When $J\neq 0$ we assume the condition $f^{-1}(f(V(\mathfrak{a})))\cap
Crit(f)=V(\mathfrak{a})$, i.e. $\sqrt{\mathfrak{a}\cap
f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{R}}}=\sqrt{\mathfrak{a}}$.

######  Theorem 7.1.

Suppose either $Crit(f)=V({\mathfrak{m}})\subset(\Bbbk^{n},o)$ or
$V(\mathfrak{a})\subset Crit(f)$ contains no irreducible component of
$Crit(f)\subset(\Bbbk^{n},o)$. Suppose $\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X}\subseteq\mathfrak{a}\cdot{\mathfrak{m}}\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}^{(1)}}f$.

  1. a.

Suppose $(\Bbbk,{\mathfrak{m}}_{\Bbbk})$ is a local ring. Then
${\mathscr{A}f}\supseteq\\{f\\}+({\mathfrak{m}}_{\Bbbk}+(f))\cdot\mathfrak{a}^{2}\cdot
R^{\oplus p}_{X}+T_{\mathscr{L}^{(1)}}f$.

  2. b.

Suppose $\Bbbk$ contains an infinite field. Then
${\mathscr{A}f}\supseteq\\{f\\}+\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X}+T_{\mathscr{L}^{(1)}}f$.

A trivial choice $\mathfrak{a}={\mathfrak{m}}\cdot\mathfrak{a}_{\mathscr{R}}$
satisfies the assumptions. Then the statement is implied by theorem 4.1.

We remark: the assumption $\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}^{(1)}}f$ is preserved by
$\mathscr{A}$-equivalence.
Proof. The proof is an essentially more complicated version of the proofs of
theorems 4.1, 5.1. The additional ingredients now are: the Weierstraß
division, an “arcwise” solution, passage to the finite jets, and the Zariski
closedness of orbits of unipotent algebraic groups.

  1. Step 1.

(Preparations)

    1. i.

Consider the $R_{X}$-module
$M:={\footnotesize\left.\raisebox{1.37775pt}{$\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X}+\mathfrak{a}\cdot{\mathfrak{m}}\cdot
T_{\mathscr{R}}f$}\\!/\\!\raisebox{-1.37775pt}{$\mathfrak{a}\cdot{\mathfrak{m}}\cdot
T_{\mathscr{R}}f$}\right.}$. Take the annihilator ideal
$\mathfrak{a}_{M}:=Ann(M)\subset R_{X}$. We have
$\mathfrak{a}_{\mathscr{R}}\cdot M=0$, thus $Supp(M)\subseteq
Crit(f)=V(\mathfrak{a}_{\mathscr{R}})\subset X$. Moreover, by the direct check
one has
$\mathfrak{a}_{M}\supseteq(\mathfrak{a}\cdot{\mathfrak{m}}\cdot\mathfrak{a}_{\mathscr{R}}):\mathfrak{a}^{2}$.

We claim: $\sqrt{\mathfrak{a}_{M}}=\sqrt{\mathfrak{a}_{\mathscr{R}}}$. This is
obvious if $Crit(f)=V({\mathfrak{m}})\in(\Bbbk^{n},o)$. Thus we assume:
$V(\mathfrak{a})$ contains no irreducible component of $Crit(f)$. Obviously
$Supp(M)\supseteq V(\mathfrak{a}_{\mathscr{R}})\setminus V(\mathfrak{a})$. In
addition $Supp(M)\subset(\Bbbk^{n},o)$ is closed. Therefore
$Supp(M)=V(\mathfrak{a}_{\mathscr{R}})$, i.e.
$\sqrt{\mathfrak{a}_{M}}=\sqrt{\mathfrak{a}_{\mathscr{R}}}$.

Consider $M$ as an $R_{Y}$-module. By our assumption $M$ is a submodule of
${\footnotesize\left.\raisebox{1.37775pt}{$(y)^{2}\cdot
T_{\mathscr{L}}f+\mathfrak{a}T_{\mathscr{R}}f$}\\!/\\!\raisebox{-1.37775pt}{$\mathfrak{a}T_{\mathscr{R}}f$}\right.}$.
The later module is finitely generated over $R_{Y}$. Therefore $M$ is f.g.
over $R_{Y}$, §2.1.vii.

Each $x_{i}\in R_{X}$ acts on $M$ as an $R_{Y}$-linear operator. And its
action is filtration-nilpotent, i.e. $x_{i}^{d}\cdot M\subseteq
f^{\\#}(y)\cdot M$ for $d\gg 1$. Therefore the characteristic polynomial of
$x_{i}$ is of the form

(40) $\chi_{x_{i}}(t):=det[t{1\rm I}-x_{i}]=t^{d_{i}}+y(\dots)\in R_{Y}[t].$

One has $\chi_{x_{i}}(x_{i})=0\circlearrowright M$, by Cayley-Hamilton
theorem. Therefore we can apply Weierstraß division,
$R_{X}=\mathfrak{a}_{M}+Span_{R_{Y}}\\{v_{\bullet}\\}$. Here
$\\{v_{\bullet}\\}$ is a finite tuple that generates the $R_{Y}$-module
${\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{$\mathfrak{a}_{M}$}\right.}$.
We take $v_{0}=1$ and $v_{l>0}\in{\mathfrak{m}}$.

    2. ii.

We claim: $\\{f\\}+J\cdot R^{\oplus
p}_{X}+T_{\mathscr{L}^{(1)}}f\subseteq\mathscr{L}(\\{f\\}+J\cdot R^{\oplus
p}_{X})$, for any ideal $J\subset R_{X}$. Indeed, given an element $Q(y)\in
T_{\mathscr{L}^{(1)}}f$, define the automorphism
$\Phi_{Y}\in\mathscr{L}^{(1)}$ by $\Phi_{Y}(y)=y+Q(y)$. Then
$\Phi^{-1}_{Y}(\\{f\\}+J\cdot R^{\oplus p}_{X}+Q(f))\subseteq\\{f\\}+J\cdot
R^{\oplus p}_{X}$.

In addition, by Theorem 4.1,
$\mathscr{R}f\supseteq\\{f\\}+{\mathfrak{m}}^{2}\cdot\mathfrak{a}^{2}_{\mathscr{R}}\cdot
R^{\oplus p}_{X}$. In view of the previous observations, it is enough to
prove:

(41) $\\{f\\}+\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X}\subseteq\mathscr{A}(\\{f\\}+\mathfrak{a}^{d}_{M}\cdot R^{\oplus
p}_{X}+T_{\mathscr{L}^{(1)}}f)\quad\text{ for }\quad d\gg 1.$

    3. iii.

We claim: $\\{f\\}+\mathfrak{a}\cdot T_{\mathscr{R}}f+\mathfrak{a}^{2}\cdot
R^{\oplus p}_{X}\subseteq\mathscr{R}(\\{f\\}+\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X})$. (This will be used in Step 4.) Indeed, take an element
$\xi(f)+q\in\mathfrak{a}\cdot T_{\mathscr{R}}f+\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X}$. For $J=0$ take the coordinate change $\Phi_{X}:x\to x+\xi(x)$. For
$J\neq 0$ use the $jet_{0}$-assumption to extend this to a coordinate change
$\Phi_{X}:x\to x+\xi(x)+h$, where $h\in\mathfrak{a}^{2}$. We get:
$\Phi_{X}(f)=f+\xi(f)+\tilde{q}$, where $\tilde{q}\in\mathfrak{a}^{2}\cdot
R^{\oplus p}_{X}$. Therefore
$f+\xi(f)+q=\Phi_{X}(f+\Phi_{X}^{-1}(q-\tilde{q}))$. Finally use
$\Phi_{X}(\mathfrak{a}^{2})=\mathfrak{a}^{2}$ to get
$\Phi_{X}^{-1}(q-\tilde{q})\in\mathfrak{a}^{2}\cdot R^{\oplus p}_{X}$.

  2. Step 2.

We transform the condition (41) into a system of implicit function equations.
As in the $\mathscr{R},{\mathscr{K}}\\!$-cases we fix some generators
$\\{a_{i}\\}$ of $\mathfrak{a}$, $\\{a^{M}_{m}\\}$ of $\mathfrak{a}_{M}$, and
$\\{\xi_{j}(f)\\}$ of $T_{\mathscr{R}}f$. Below we use the expansion
$R_{X}=\mathfrak{a}_{M}+Span_{R_{Y}}\\{v_{\bullet}\\}$.

From now and until Step 4 we assume $J=0$.

    1. i.

Take a “perturbation” $g\in\mathfrak{a}^{2}\cdot R^{\oplus p}_{X}$. Expand it,
$g=\sum a_{i}c^{g}_{ij}\xi_{j}(f)+T_{\mathscr{L}^{(1)}}f$, here
$c^{g}_{ij}\in{\mathfrak{m}}$. Accordingly define $\Phi_{X}\in\mathscr{R}$ by
$x\to x+\sum a_{i}c_{ij}\xi_{j}(x)$, here $\\{c_{ij}\\}\in R_{X}$ are
unknowns. As in the $\mathscr{R},{\mathscr{K}}\\!$-cases we have:

       * •

$\Phi_{X}(f)-f-\sum a_{i}\cdot
c_{ij}\cdot\xi_{j}(f)\in\mathfrak{a}^{2}\cdot\\{c_{**}\\}^{2}\cdot R^{\oplus
p}_{X}$.

       * •

$\Phi_{X}(g)-g\in\mathfrak{a}\cdot\\{c_{*}\\}\cdot(\mathfrak{a}^{2})^{\prime}\cdot
R^{\oplus p}_{X}+\mathfrak{a}^{2}\cdot H_{\geq 2}(\\{c_{*}\\})$.

Now expand
$c_{ij}=\sum_{l}c_{ij,l}v_{l}+\sum_{m}\tilde{c}_{ij,m}\mathfrak{a}^{M}_{m}$,
with the unknowns $c_{ij,l}\in R_{Y}$, $\tilde{c}_{ij,m}\in R_{X}$. Similarly
one expands $\\{c^{g}_{ij}\\}$. We get:

(42) $\Phi_{X}(f+g)=f+f^{\\#}(y)^{2}H(\\{c_{ij,l}\\})+\hskip 256.0748pt$
$\quad\quad\sum_{ij}\Big{[}\sum_{l}v_{l}(c_{ij,l}+c^{g}_{ij,l})+\sum_{m}a^{M}_{m}(\tilde{c}_{ij,m}+\tilde{c}^{g}_{ij,m})+Q^{Y}_{ij}(\\{c_{**}\\})+\mathcal{H}_{ij}(\\{c_{**}\\},\\{\tilde{c}_{**}\\})\Big{]}a_{i}\xi_{j}(f).$

Here $Q^{Y}_{ij}(z)\in(z)^{2}\cdot R_{Y}[[z]]$,
$\mathcal{H}_{ij}(z,w)\in(z,w)^{2}\cdot R_{X}[[z,w]]$ and $H\in
T_{\mathscr{L}}f$.

We would like to get rid of the term in the brackets, the coefficient of
$a_{i}\xi_{j}(f)$. This cannot be done directly. Instead we will force this
coefficient to belong to $\mathfrak{a}^{d}_{M}$ for $d\gg 1$. This will ensure
the condition (41).

    2. ii.

Iterating the expansion $R_{X}=\mathfrak{a}_{M}+Span_{R_{Y}}\\{v_{\bullet}\\}$
we present
$R_{X}=\mathfrak{a}^{d+1}_{M}+\sum^{d}_{k=0}\mathfrak{a}^{k}_{M}\cdot
Span_{R_{Y}}\\{v_{\bullet}\\}$, for any $d\geq 1$. Accordingly we expand the
unknown,
$c_{ij}=\sum^{d}_{k=0}\sum_{l,{\underline{m}}}a^{M}_{{\underline{m}}}v_{l}c_{ij,{\underline{m}},l}$.
Here ${\underline{m}}=(m_{1},\dots,m_{k})$ is the multi-index, thus
$a^{M}_{{\underline{m}}}=a^{M}_{m_{1}}\cdots a^{M}_{m_{k}}$, while
$c_{ij,{\underline{m}},l}\in R_{Y}$ are the new unknowns.

Below we work modulo $\mathfrak{a}^{d+1}_{M}$. Repeat the procedure of Step
2.i, then equation (42) becomes:

(43) $\Phi_{X}(f+g)\stackrel{{\scriptstyle mod\
a^{d+1}_{M}}}{{\equiv}}f+\sum_{ij}\sum^{d}_{k=0}\sum_{\begin{subarray}{c}l,{\underline{m}}\\\
|{\underline{m}}|=k\end{subarray}}v_{l}\cdot\mathfrak{a}^{M}_{\underline{m}}\big{[}c_{ij,{\underline{m}},l}+c^{g}_{ij,{\underline{m}},l}+Q^{Y}_{ij,{\underline{m}},l}(\\{c_{**}\\})\big{]}a_{i}\xi_{j}(f)+f^{\\#}(y)^{2}H(\\{c_{ij,0,l}\\}).$

Here $Q^{Y}_{ij,{\underline{m}},l}(z)\in(z)\cdot R_{Y}[[y\cdot z]][z]$, resp.
$(z)\cdot R_{Y}\\{y\cdot z\\}[z]$, resp. $(z)\cdot R_{Y}\langle y\cdot
z\rangle[z]$, rather than just $Q^{Y}_{ij,{\underline{m}},l}(z)\in(z)\cdot
R_{Y}[[z]]$. We get the (finite) system of implicit function equations over
the ring $R_{Y}$:

(44)
$c_{ij,{\underline{m}},l}+c^{g}_{ij,{\underline{m}},l}+Q_{ij,{\underline{m}},l}(\\{c_{**}\\})=0.$

As $v_{0}=1$ and $v_{l>0}\in{\mathfrak{m}}$ one has here:
$c^{g}_{ij,0,0}\in(y)$ and $Q_{ij,o,o}(z)\in(y)\cdot(z)$.

Unlike the $\mathscr{R},{\mathscr{K}}\\!$-cases we cannot immediately apply
the $IFT_{1\rm I}$ of §2.1.iii. In fact the coefficients
$c^{g}_{ij,{\underline{m}},l}$ do not necessarily belong to the ideal
$(y)+{\mathfrak{m}}_{\Bbbk}\subset R_{Y}$. Neither can one assume
$Q^{Y}_{ij,{\underline{m}},l}(z)\in(f^{\\#}(y)+{\mathfrak{m}}_{\Bbbk})\cdot(z)$.

As a cure we introduce below the indeterminate $t$, to get an “arc-solution”.
Then we extend this arc-solution to an ordinary solution.

  3. Step 3.

    1. i.

Extend the base ring $\Bbbk$ to $\Bbbk[[t]]$, resp. $\Bbbk\\{t\\}$, resp.
$\Bbbk\langle t\rangle$, for an indeterminate $t$. Instead of $f+g$ take
$f+tg$. Accordingly define $\Phi_{X}$ by $x\to x+t\sum a_{i}c_{ij}\xi_{j}(x)$.
Repeat the previous steps up to equation (43):

(45) $\Phi_{X}(f+tg)\stackrel{{\scriptstyle mod\
\mathfrak{a}^{d+1}_{M}}}{{\equiv}}f+t\sum_{ij}\sum^{d}_{k=0}\sum_{\begin{subarray}{c}l,{\underline{m}}\\\
|{\underline{m}}|=k\end{subarray}}v_{l}\cdot
a^{M}_{\underline{m}}\big{[}c_{ij,{\underline{m}},l}+c^{g}_{ij,{\underline{m}},l}+t\cdot
Q^{Y}_{ij,{\underline{m}},l}(\\{c_{**}\\})\big{]}a_{i}\xi_{j}(f)+f^{\\#}(y)^{2}H(\\{c_{ij,0,l}\\}).$

Then the system (44) becomes
$c_{ij,{\underline{m}},l}+c^{g}_{ij,{\underline{m}},l}+t\cdot
Q_{ij,{\underline{m}},l}(\\{c_{**}\\})=0$. Applying $IFT_{1\rm I}$ over the
ring $\Bbbk[[y,t]]$, resp. $\Bbbk\\{y,t\\}$, resp. $\Bbbk\langle y,t\rangle$,
we get the solution $\\{c_{ij,{\underline{m}},l}(y,t)\\}$.

For this solution we define the coordinate change $\Phi_{X}\in\mathscr{R}$ by
$x\to x+t\sum
a_{i}c_{ij,{\underline{m}},l}v_{l}a^{M}_{\underline{m}}\xi_{j}(x)$. This
coordinate change is unipotent, as $\mathfrak{a}\subseteq{\mathfrak{m}}$ and
$c_{ij,0,0}\in(y)$ and
$v_{l>0},\mathfrak{a}^{M}_{\underline{m}}\in{\mathfrak{m}}$ for
$|{\underline{m}}|>0$.

Altogether, we have proved:
$\Phi_{X}(f+tg)\in\\{f\\}+\mathfrak{a}^{d+1}_{M}\cdot R^{\oplus
p}_{X}+T_{\mathscr{L}^{(1)}}f$, for an indeterminate $t$ and any $d\gg 1$. By
part iii. of Step 1 we get:
$f+tg\stackrel{{\scriptstyle\mathscr{A}^{(1)}_{t}}}{{\sim}}f$.

At this moment we have proved part a. of the theorem. (One just takes
$t={\mathfrak{m}}_{\Bbbk}+(y)$.)

From now on we assume: $\Bbbk$ contains an infinite field. Moreover, we assume
$\\{Q^{Y}_{***}(z)\\}\not\in{\mathfrak{m}}_{k}+(y)$. (Otherwise the system
(44) is directly resolvable.)

    2. ii.

We prove: $f+tg\stackrel{{\scriptstyle\mathscr{A}}}{{\sim}}f$ for an infinite
set of values of $t\in\Bbbk$. We still have to resolve the system
$c_{ij,{\underline{m}},l}+c^{g}_{ij,{\underline{m}},l}+t\cdot
Q^{Y}_{ij,{\underline{m}},l}(\\{c_{**}\\})=0$, now for a fixed $t\in\Bbbk$.
Here $Q^{Y}_{ij,{\underline{m}},l}(z)$ are power series in $z$, not
polynomials. However, it is enough to resolve this system over the quotient
ring
${\footnotesize\left.\raisebox{1.37775pt}{$R_{Y}$}\\!/\\!\raisebox{-1.37775pt}{$(y)+{\mathfrak{m}}_{\Bbbk}$}\right.}$.
Indeed, $Q^{Y}_{ij,{\underline{m}},l}\in R_{Y}[[y\cdot z]][z]$ (resp.
analytic, resp. algebraic). And if $\\{c^{0}_{**}\\}$ is a solution
$mod(y+{\mathfrak{m}}_{\Bbbk})$, then by shifting the variables,
$c_{**}\rightsquigarrow c_{**}-c^{0}_{**}$, we get the system of $IFT_{1\rm
I}$-type, $c_{ij,{\underline{m}},l}=\tilde{Q}_{ij,{\underline{m}},l}(c_{**})$,
where
$\tilde{Q}^{Y}_{ij,{\underline{m}},l}(z)\in((y)+{\mathfrak{m}}_{\Bbbk})\cdot(z)$.

Over the ring
${\footnotesize\left.\raisebox{1.37775pt}{$R_{Y}$}\\!/\\!\raisebox{-1.37775pt}{$(y)+{\mathfrak{m}}_{\Bbbk}$}\right.}$
we get a polynomial system in variables $\\{c_{ij,{\underline{m}},l}\\}$, over
the (infinite) field
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}_{\Bbbk}$}\right.}$.
This system defines a closed algebraic subscheme of a finite dimensional
affine space
$\mathbb{A}^{N}_{{\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}_{\Bbbk}$}\right.}}$.
In Step 3.i we have constructed an arc on this subscheme,
$\\{c_{ij,{\underline{m}},l}(t)\\}$. This arc is non-trivial (i.e. not a
point) as $\\{Q^{Y}_{***}(z)\\}\not\in{\mathfrak{m}}_{k}+(y)$. Hence, as the
field
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}_{\Bbbk}$}\right.}$
is infinite, the system is solvable for an infinite number of $t$-values. Thus
the set of
${\footnotesize\left.\raisebox{1.37775pt}{$\Bbbk$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}_{\Bbbk}$}\right.}$-points
of this (affine, algebraic) subscheme is of positive dimension.

    3. iii.

Finally we deduce: $f+tg\stackrel{{\scriptstyle\mathscr{A}}}{{\sim}}f$ for any
$t\in\Bbbk$.

Through the whole proof we have used only the unipotent subgroup
$\mathscr{A}^{(1)}<\mathscr{A}$, for the filtration
${\mathfrak{m}}^{\bullet}\subset R_{X}$. Passing to the quotient
${\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}^{d}$}\right.}$
for $d\gg 1$, we get the action on the finite dimensional vector space,
$\mathscr{A}^{(1)}\circlearrowright({\mathfrak{m}}\cdot{\footnotesize\left.\raisebox{1.37775pt}{$R_{X}$}\\!/\\!\raisebox{-1.37775pt}{${\mathfrak{m}}^{d}$}\right.})^{\oplus
p}$. This action is algebraic and unipotent. Therefore its orbits are Zariski-
closed. (This is the Kostant-Rosenlicht theorem, see e.g. [F.-S. R., Theorem
2.11].) The intersection of such an orbit with a line,
$\mathscr{A}^{(1)}(f)\cap\\{f+tg\\}_{t}$ is either finite, or contains the
whole line. In our case this intersection is infinite, thus
$\mathscr{A}^{(1)}(f)\supseteq f+tg$ for each $t\in\Bbbk$.

We have proved the theorem for the case $J=0$.

  4. Step 4.

The general case is a reduction to the case $J=0$. Given
$g\in\mathfrak{a}^{2}\cdot R^{\oplus p}_{X}$ we have constructed the maps
$\Phi_{X},\Phi_{Y}$ satisfying $\Phi_{X}\circ\Phi_{Y}(f+g)=f$. For $J\neq 0$
the map $\Phi_{X}:x\to x+\sum a_{i}c_{ij}\xi_{j}(x)$ is not necessarily a
coordinate change, as it does not preserve the ideal $J$. Using the
$jet_{0}$-assumption extend $\Phi_{X}$ to an automorphism, $\Psi_{X}:x\to
x+\sum a_{i}c_{ij}\xi_{j}(x)+\mathfrak{a}^{2}\in\mathscr{R}$. We factorize
$\Psi_{X}=\tilde{\Phi}_{X}\circ\Phi_{X}$, where $\tilde{\Phi}_{X}:X\to
x+\mathfrak{a}^{2}$ is an automorphism of $\Bbbk[[x]]$, but not of $R_{X}$.
Working over $\Bbbk[[x]]$ we have
$\Psi_{X}\circ\Phi_{Y}(f+g)=\tilde{\Phi}_{X}(f)\in\\{f\\}+\mathfrak{a}^{2}\cdot
T_{\mathscr{R}}f+\mathfrak{a}^{4}\cdot R^{\oplus p}_{X}$. Apply step 1.iv. to
get $\\{f\\}+\mathfrak{a}^{2}\cdot T_{\mathscr{R}}f+\mathfrak{a}^{4}\cdot
R^{\oplus p}_{X}\subseteq\mathscr{R}(f+\mathfrak{a}^{4}\cdot R^{\oplus
p}_{X})$.

Thus for $J\neq 0$ we get: $\\{f\\}+\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X}\subseteq\mathscr{A}(\\{f\\}+\mathfrak{a}^{4}\cdot R^{\oplus p}_{X})$.
Observe: $\mathfrak{a}^{4}\cdot R^{\oplus
p}_{X}\subseteq(\mathfrak{a}^{3}+\mathfrak{a}\cdot
f^{\\#}(y)^{2}){\mathfrak{m}}\cdot T_{\mathscr{R}}f+T_{\mathscr{L}^{(3)}}f$
(by the direct check). Therefore we can repeat the previous steps to get:
$\\{f\\}+\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X}\subseteq\mathscr{A}(\\{f\\}+\mathfrak{a}^{d}\cdot R^{\oplus p}_{X})$
for any $d\gg 1$.

Finally, by the initial assumption $\mathfrak{a}^{d}\subseteq
f^{\\#}(y)^{2}+\mathfrak{a}^{3}_{\mathscr{R}}$ for $d\gg 1$. Then theorem 4.1
gives:

(46) $\\{f\\}+\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X}\subseteq\mathscr{A}(\\{f\\}+\mathfrak{a}^{3}_{\mathscr{R}}\cdot
R^{\oplus
p}_{X}+T_{\mathscr{L}^{1}}f)\subseteq\mathscr{A}(\\{f\\}+\mathfrak{a}^{3}_{\mathscr{R}}\cdot
R^{\oplus p}_{X})\subseteq\mathscr{A}f.\quad\quad\quad{\hfill\blacksquare}$

######  Remark 7.2.

The factor ${\mathfrak{m}}$ in the assumption $\mathfrak{a}^{2}\cdot R^{\oplus
p}_{X}\subseteq\mathfrak{a}\cdot{\mathfrak{m}}\cdot
T_{\mathscr{R}}f+y^{2}\cdot T_{\mathscr{L}}f$ is used only once in this proof,
in Step 3.i. It is needed to ensure that the coordinate change $\Phi_{X}:x\to
x+t\sum a_{i}c_{ij,{\underline{m}},l}v_{l}a^{M}_{\underline{m}}\xi_{j}(x)$ is
unipotent. But this unipotence holds trivially if one assumes
$\mathfrak{a}\subseteq{\mathfrak{m}}^{2}$. Thus (almost) the same proof leads
to the statement: If $\mathfrak{a}\subseteq{\mathfrak{m}}^{2}$ and
$\mathfrak{a}^{2}\cdot R^{\oplus p}_{X}\subseteq\mathfrak{a}\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}^{(1)}}f$ then $\\{f\\}+\mathfrak{a}^{2}\cdot
R^{\oplus p}_{X}+T_{\mathscr{L}^{(1)}}f\subseteq\mathscr{A}f$.

######  Example 7.3.

  1. i.

Suppose $\mathfrak{a}_{\mathscr{A}^{(o)}}\supseteq{\mathfrak{m}}(f)$,
equivalently $T_{\mathscr{A}^{(o)}}f=T_{{\mathscr{K}}\\!^{(o)}}f$,
equivalently
$\mathfrak{a}_{\mathscr{A}^{(o)}}=\mathfrak{a}_{{\mathscr{K}}\\!^{(o)}}$.
(This holds, e.g. for infinitesimally-stable maps, see Property 6.2.ii.) Then

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

Therefore we get: $\\{f\\}+\mathfrak{a}_{{\mathscr{K}}\\!^{(o)}}^{2}R^{\oplus
p}_{X}\subseteq\mathscr{A}f$. This is close to Mather’s statement “Stable maps
are determined by their local algebras”.

Note: it is much simpler to compute/control the ideal
$\mathfrak{a}_{{\mathscr{K}}\\!^{(o)}}$ than
$\mathfrak{a}_{\mathscr{A}^{(o)}}$. As a particular case take a stable map. It
is the unfolding of its genotype, i.e.
$f:(\Bbbk^{n+\tau-1},o)\to(\Bbbk^{p+\tau-1},o)$ with $f=(f_{o}+\sum
v_{j}u_{j},u)$, where $f_{o}:(\Bbbk^{n},o)\to(\Bbbk^{p},o)$ and
$Span_{\Bbbk}(\\{v_{\bullet}\\})={\mathfrak{m}}\cdot
T^{1}_{\mathscr{K}}\\!f_{o}$. Here
$\mathfrak{a}_{{\mathscr{K}}\\!^{(o)}}(F)=\mathfrak{a}_{{\mathscr{K}}\\!^{(o)}}(f_{o})+(u)$,
see example 3.12.vi. Thus we get
$\mathscr{A}f\supset\\{f\\}+(\mathfrak{a}_{{\mathscr{K}}\\!^{(o)}}\cdot(f_{o})+(u))^{2}\cdot
R^{\oplus p}_{X}$.

Compare this to the classical bound
$\mathscr{A}f\supset\\{f\\}+{\mathfrak{m}}^{p+\tau}\cdot R^{\oplus p}_{X}$.

As we have seen in examples 4.4, 5.2, in most cases the ideal
$(\mathfrak{a}_{{\mathscr{K}}\\!^{(o)}}(f_{o})+(u))^{2}$ is larger than
${\mathfrak{m}}^{r+\tau}$. (Even the asymptotics is better.)

  2. ii.

More generally, suppose
$\mathfrak{a}_{\mathscr{A}^{(j)}}\supset{\mathfrak{m}}(f)^{j+1}$ for some
$j\geq 0$. Then $\mathfrak{a}_{\mathscr{A}^{(j)}}^{2}R^{\oplus
p}_{X}\subseteq(\mathfrak{a}_{\mathscr{A}^{(j)}}+(f)^{j+1})T_{\mathscr{R}^{(j)}}f+\mathfrak{a}_{\mathscr{A}^{(j)}}T_{\mathscr{L}^{(1)}}f\subseteq\mathfrak{a}_{\mathscr{A}^{(j)}}T_{\mathscr{R}^{(j)}}f+T_{\mathscr{L}^{(1)}}f$.
Therefore $\\{f\\}+\mathfrak{a}_{\mathscr{A}^{(j)}}^{2}R^{\oplus
p}_{X}\subseteq\mathscr{A}f$.

Yet more generally, it is enough to assume
$f^{\sharp}(y)^{j+1}\cdot\mathfrak{a}_{\mathscr{A}^{(j)}}\supset\mathfrak{a}_{\mathscr{A}^{(j)}}T_{\mathscr{R}^{(j)}}f+T_{\mathscr{L}^{(1)}}f$
for some $j\geq 0$.

To apply the theorem efficiently we need the following bound.

######  Corollary 7.4.

Let $R_{X}$ be one of $\Bbbk[[x]]$, $\Bbbk\\{x\\}$, $\Bbbk\langle x\rangle$,
where $\Bbbk$ contains an infinite field.

  1. 1.

Suppose
$\mathfrak{b}^{2}_{Y}\subseteq\mathfrak{b}_{Y}\cdot\mathfrak{a}_{\mathscr{A}^{(o)}}$
for an ideal $\mathfrak{b}_{Y}\subseteq f^{\\#}(y)$. Then
$\\{f\\}+(\mathfrak{b}_{Y}+\mathfrak{a}_{\mathscr{R}^{(o)}})^{2}R^{\oplus
p}_{X}+T_{\mathscr{L}^{(1)}}f\subseteq\mathscr{A}f$

  2. 2.

Suppose
$\mathfrak{b}^{2}_{Y}\subseteq\mathfrak{b}_{Y}\cdot\mathfrak{a}_{\mathscr{A}}$
for an ideal $\mathfrak{b}_{Y}\subseteq f^{\\#}(y)^{2}$. Suppose
$\mathfrak{a}_{\mathscr{R}}\subseteq{\mathfrak{m}}^{2}$. Then
$\\{f\\}+(\mathfrak{b}_{Y}+\mathfrak{a}_{\mathscr{R}})^{2}R^{\oplus
p}_{X}+T_{\mathscr{L}^{(1)}}f\subseteq\mathscr{A}f$

Proof. (Part one.) It is enough to observe:
$(\mathfrak{b}_{Y}+\mathfrak{a}_{\mathscr{R}^{(o)}})^{2}R^{\oplus
p}_{X}\subseteq(\mathfrak{b}_{Y}+\mathfrak{a}_{\mathscr{R}^{(o)}})T_{\mathscr{R}^{(0)}}+\mathfrak{b}_{Y}^{2}\cdot
R^{\oplus
p}_{X}\subseteq(\mathfrak{b}_{Y}+\mathfrak{a}_{\mathscr{R}^{(o)}})T_{\mathscr{R}^{(0)}}+\mathfrak{b}_{Y}\cdot
T_{\mathscr{L}^{(0)}}$. Then we apply theorem 7.1.

For part two use remark 7.2. $\blacksquare$

######  Example 7.5.

  1. i.

Take $\mathfrak{b}_{Y}=\mathfrak{a}_{\mathscr{A}^{(o)}}\cap f^{\\#}(y)$ to get
$\mathscr{A}f\supset\\{f\\}+(\mathfrak{a}_{\mathscr{A}^{(o)}}\cap
f^{\\#}(y)+\mathfrak{a}_{\mathscr{R}^{(0)}})^{2}\cdot R^{\oplus
p}_{X}+T_{\mathscr{L}^{(1)}}f$.

Suppose $\mathfrak{a}_{\mathscr{R}}\subseteq{\mathfrak{m}}^{2}$. Using remark
7.2 we get also $\mathscr{A}f\supset\\{f\\}+(\mathfrak{a}_{\mathscr{A}}\cap
f^{\\#}(y)^{2}+\mathfrak{a}_{\mathscr{R}})^{2}\cdot R^{\oplus
p}_{X}+T_{\mathscr{L}^{(1)}}f$.

More generally, if
$\mathfrak{b}^{2}_{Y}\subseteq\mathfrak{b}_{Y}\cdot\mathfrak{a}_{\mathscr{A}^{(o)}}$
then the ideal $\mathfrak{b}_{Y}+\mathfrak{a}_{\mathscr{A}^{(o)}}\cap
f^{\\#}(y)$ satisfies this condition as well.

  2. ii.

Suppose $\mathfrak{a}_{\mathscr{A}}\supseteq{\mathfrak{m}}\cdot(f)^{d}$, then
we have $\mathfrak{a}_{\mathscr{A}}\cap f^{\\#}(y)\supseteq f^{\\#}(y)^{d}$.
Assuming $\mathfrak{a}_{\mathscr{R}}\subset{\mathfrak{m}}^{2}$ we get
$\\{f\\}+((f)^{d}+\mathfrak{a}_{\mathscr{R}})^{2}R^{\oplus
p}_{X}+T_{\mathscr{L}^{(1)}}f\subseteq\mathscr{A}f$.

We remark:
$\sqrt{(f)^{d}+\mathfrak{a}_{\mathscr{R}^{(o)}}}=\sqrt{\mathfrak{a}_{\mathscr{K}}\\!}$,
thus in this case the $\mathscr{A}$ and ${\mathscr{K}}\\!$-determinacy loci
coincide. In particular, if $f$ is $\mathscr{A}$-finite then it is
$\mathscr{A}$-finitely determined.

######  Remark 7.6.

Geometrically we have the locus $V(\mathfrak{a}_{\mathscr{A}^{(0)}})\subseteq
Crit(f)\subset X$. The ideal $R_{X}(\mathfrak{a}_{\mathscr{A}^{(0)}}\cap
f^{\\#}(y))+\mathfrak{a}_{\mathscr{R}^{(0)}}$ defines the preimage
$f^{-1}(f(V(\mathfrak{a}_{\mathscr{A}^{(0)}})))\cap Crit(f)$. This preimage
contains the instability locus, $V(\mathfrak{a}_{\mathscr{A}})$, and these two
often coincide.

If
$V(\mathfrak{a}_{\mathscr{A}^{(o)}})=f^{-1}(f(V(\mathfrak{a}_{\mathscr{A}^{(o)}})))\cap
Crti(f)$ as a scheme, then
$\mathscr{A}f\supseteq\\{f\\}+\mathfrak{a}^{d}_{\mathscr{A}^{(o)}}\cdot
R^{\oplus p}_{X}+T_{\mathscr{L}^{(1)}}f$. In this sense $f$ is
$\mathscr{A}$-determined by its finite jet on the ”locus of weakly
instability”.

In the case $\sqrt{\mathfrak{a}_{\mathscr{A}}}\subsetneq{\mathfrak{m}}$, i.e.
$dimV(\mathfrak{a}_{\mathscr{A}})>0$, this result seems to be new even for
$\Bbbk\in\mathbb{R},\mathbb{C}$.

######  Example 7.7.

(Computational examples.)

  1. i.

Define the map $(\Bbbk^{2},o)\stackrel{{\scriptstyle f}}{{\to}}(\Bbbk^{3},o)$
by $f(x_{1},x_{2})=(x_{1},x_{2}^{2},x_{2}^{3}+x_{1}^{k+1}x_{2})$. Suppose
$2,3,k+1\in\Bbbk^{\times}$. By direct check:
$\mathfrak{a}_{\mathscr{A}}=(x_{1}^{k},x_{2}^{2})\subset R_{X}$ and
$\mathfrak{a}^{2}_{\mathscr{A}}\cdot R^{\oplus
p}_{X}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}}f$. Thus
$\mathscr{A}f\supseteq\\{f\\}+(x_{1}^{k},x_{2}^{2})^{2}\cdot R^{\oplus
p}_{X}+T_{\mathscr{L}^{(1)}}f$. Compare this to the known criterion:
${\mathfrak{m}}^{k+1}\subseteq\mathfrak{a}_{\mathscr{A}}$ and therefore
$\mathscr{A}f\supset\\{f\\}+{\mathfrak{m}}^{2k+2}\cdot R^{\oplus p}_{X}$.

  2. ii.

Define the map $(\Bbbk^{2},o)\stackrel{{\scriptstyle f}}{{\to}}(\Bbbk^{3},o)$
by $f(x_{1},x_{2})=(x_{1},x_{2}^{2},x_{1}x_{2}^{k})$. Suppose
$2,k\in\Bbbk^{\times}$. Note that
$f^{\\#}(x_{2})\supset(x_{1}x_{2}^{k-1})\subset R_{X}$. By direct check:
$\mathfrak{a}_{\mathscr{A}}=(x_{2}^{k},x_{1}x_{2}^{k-1})\subset R_{X}$. This
is a non-isolated instability.

By direct check: $\mathfrak{a}^{2}_{\mathscr{A}}\cdot R^{\oplus
p}_{X}\subseteq{\mathfrak{m}}\cdot\mathfrak{a}_{\mathscr{A}}\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}}f$. Thus
$\mathscr{A}f\supseteq\\{f\\}+(x_{2}^{k},x_{1}x_{2}^{k-1})^{2}\cdot R^{\oplus
p}_{X}+T_{\mathscr{L}^{(1)}}f$.

###  7.2. The filtration criterion for $\mathscr{A}$-orbits

Let $R_{X}$ be a local henselian $\Bbbk$-algebra. Fix an ideal $I\subset
R_{X}$ and take the filtration $M_{\bullet}:=I^{\bullet}\cdot R^{\oplus
p}_{X}$.

######  Theorem 7.8.

Fix some integers $1\leq j<d$. Assume one of the following:

  1. i.

either the conditions $jet_{N}(Exp)$, $jet_{N}(Ln)$ of §2.3 hold for all
$N\geq 1$. (Thus in particular $\Bbbk\supseteq\mathbb{Q}$.)

  2. ii.

or $(f)\subseteq I$ and the conditions $jet_{N}(Exp)$, $jet_{N}(Ln)$ hold for
$N=\lceil\frac{2d-1-ord(f)}{j}\rceil+1$.

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

Proof. The part $\Lleftarrow$.

  1. The case $i$.

Let $w_{d}\in M_{d}$ then $w_{d}\equiv\xi_{Y}(y)|_{f}+\xi_{X}(f)\ mod\
M_{d^{\prime}}$ for some $(\xi_{Y},\xi_{X})\in T_{\mathscr{A}^{(j)}}f$. We can
assume $d^{\prime}\gg d$. Using the $jet_{N}(Exp)$-assumption, for
$N=\lceil\frac{d-ord(f)}{j}\rceil+1$, we take the corresponding automorphisms
$jet_{N}(e^{\xi_{X}})\in\mathscr{R}^{(j)}$,
$jet_{N}(e^{\xi_{Y}})\in\mathscr{L}^{(j)}$.

By remark 3.10, applied to $M_{d}$ and $\mathscr{A}^{(j)}$, we have:
$jet_{N}(e^{\xi_{X}})\cdot jet_{N}(e^{\xi_{Y}})f-f-w_{d}\in M_{d+j}$.
Therefore $\mathscr{A}^{(j)}f+M_{d+j}\supseteq\\{f\\}+M_{d}$. Now apply the
same argument to the pair $(M_{d+j},\mathscr{A}^{(j)})$, with
$jet_{N}(Exp)$-assumption for $N=\lceil\frac{d+j-ord(f)}{j}\rceil+1$. Iterate
this to get $\mathscr{A}^{(j)}f+{M_{d^{\prime}}}\supseteq\\{f\\}+M_{d}$ for
every $d^{\prime}\gg 1$. Hence
$\\{f\\}+M_{d}\subseteq\overline{\mathscr{A}^{(j)}f}$.

  2. The case $ii$.

It is enough to prove: $\mathscr{A}^{(j)}f+M_{k+1}\supset\\{f\\}+M_{k}$ for
each $k\geq d$. For this we prove:

(48) $\mathscr{A}^{(kj+k-1)}f+M_{kd+l+1}\supseteq\\{f\\}+M_{kd+l},\quad\text{
for each }\ k\geq 1\ \text{ and each }\ \ l\in 0,\dots,d-1.$

At each step we work modulo $M_{d^{\prime}}$ with $d^{\prime}\gg 1$, thus we
can replace $\overline{T_{\mathscr{A}^{(\bullet)}}}f$ by
$T_{\mathscr{A}^{(\bullet)}}f$.

By part one of lemma 6.12 we get $T_{\mathscr{A}^{(kj+k-1)}}f\supseteq M_{kd}$
for each $k\geq 1$.

Take any element $w\in M_{kd+l}$ and present it as
$w=\xi_{X}(f)+\xi_{Y}(y)|_{f}\in T_{\mathscr{A}^{(kj+k-1)}}f$. Take the
corresponding automorphisms $jet_{N}(e^{\xi_{X}})$, $jet_{N}(e^{\xi_{Y}})$,
ensured by the condition $jet_{N}(Exp)$.

Apply remark 3.10 to $M_{kd+l}$ and $\mathscr{A}^{(kj+k-1)}$ to get:
$jet_{N}(e^{\xi_{X}})\cdot jet_{N}(e^{\xi_{Y}})f-f-w\in M_{(kd+l)+(kj+k-1)}.$
This verifies the condition (48).

We should only justify the use of remark 3.10, i.e. to verify the condition:

(49)
$\frac{kd+l-ord(f)}{kj+k-1}+1\leq\lceil\frac{2d-1-ord(f)}{j}\rceil+1\quad\text{
for all $l=0,\dots,d-1$, and all $k,j\geq 1$.}$

Here the maximum of the left hand side is achieved for $l=d-1$ and $k=1$. And
then we get the equality.

The part $\Rrightarrow$.

  1. The case $i$.

Take $w_{d}\in M_{d}$ then $f+w_{d}\equiv(\Phi_{Y},\Phi_{X})f\ mod\
M_{d^{\prime}}$ for some $(\Phi_{Y},\Phi_{X})\in\mathscr{A}^{(j)}$. We can
assume $d^{\prime}\gg d$. Using the $jet_{N}$-assumptions we approximate
$(\Phi_{Y},\Phi_{X})$ by the elements $jet_{N}(e^{\xi_{X}})\in Aut^{(j)}_{X}$,
$jet_{N}(e^{\xi_{Y}})\in Aut^{(j)}_{Y}$ satisfying:

(50) $\Phi_{X}-jet_{N}(e^{\xi_{X}})\in
End^{(jN+1)}_{\Bbbk}(R_{X}),\quad\quad\quad\Phi_{Y}-jet_{N}(e^{\xi_{Y}})\in
End^{(jN+1)}_{\Bbbk}(R_{Y}).$

Therefore $(jet_{N}(e^{\xi_{X}}),jet_{N}(e^{\xi_{Y}}))f-f-w_{d}\in
M_{d^{\prime}}+M_{jN+1+ord(f)}$. Here we can assume $d^{\prime}\geq
jN+1+ord(f)\geq d+j$. Apply remark 3.10 to the pair
$(M_{d},\mathscr{A}^{(j)})$ to get: $(\xi_{Y}+\xi_{X})f-w_{d}\in M_{d+j}$.
Therefore $T_{\mathscr{A}^{(j)}}f+M_{d+j}\supseteq M_{d}$.

Now apply the same argument to the pair $(M_{d+j},\mathscr{A}^{(j)})$, and so
on. One gets $T_{\mathscr{A}^{(j)}}f+M_{d^{\prime}}\supseteq M_{d}$ for any
$d^{\prime}$. Hence the statement.

  2. The case $ii$.

Apply the argument of case i. to the pair $(M_{d},\mathscr{A}^{(j)})$ and
$N=\lceil\frac{d-ord(f)}{j}\rceil+1$ to get:
$T_{\mathscr{A}^{(j)}}f+M_{d+j}\supseteq M_{d}$. Iterate this argument for all
pairs $(M_{d+l},\mathscr{A}^{(j)})$ and
$N=\lceil\frac{d+l-ord(f)}{j}\rceil+1$, with $l=1,\dots,d-1$. We reach
$T_{\mathscr{A}^{(j)}}f+M_{2d}\supseteq M_{d}$, and for this we have used the
$jet_{N}$-assumptions with $N=\lceil\frac{d-ord(f)}{j}\rceil+1$, …,
$\lceil\frac{2d-1-ord(f)}{j}\rceil+1$.

And now observe:

$M_{2d}=I^{d}\cdot M_{d}\subseteq
I^{d}(T_{\mathscr{A}^{(j)}}f+M_{2d})\subseteq I^{d}\cdot
T_{\mathscr{R}^{(j)}}f+f^{\\#}(\mathfrak{b}_{j})T_{\mathscr{L}^{(j)}}f+M_{3d}\stackrel{{\scriptstyle(f)\subseteq
I}}{{\subseteq}}T_{\mathscr{A}^{(2j)}}f+M_{3d}.$

Iterate this to get $T_{\mathscr{A}^{(j)}}f+M_{d^{\prime}}\supseteq M_{d}$ for
all $d^{\prime}$. Hence the statement. $\blacksquare$

Let $R_{X}\\!\\!\in\\!\Bbbk[[x]],\Bbbk\\{x\\},\Bbbk\langle x\rangle$, where
$\Bbbk$ contains an infinite field. Fix some integers $1\leq j<d$. Suppose
either $char(\Bbbk)\\!\\!=\\!\\!0$ or
$char(\Bbbk)\\!\\!>\\!\\!\lceil\frac{2d-1-ord(f)}{j}\rceil\\!+\\!1$. Take an
ideal $(f)\subseteq I\subseteq{\mathfrak{m}}$. Suppose $V(I)\supseteq
f^{-1}(f(V(\mathfrak{a}_{\mathscr{A}})))\cap Crit(f)$.

######  Corollary 7.9.

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

For $\Bbbk=\mathbb{R},\mathbb{C}$ and $I={\mathfrak{m}}$ this is Theorem 2.5
of [B.P.W.].
Proof. W.l.o.g. we assume $I\neq 0$, thus
$V(\mathfrak{a}_{\mathscr{A}})\subsetneq(\Bbbk^{n},o)$.

  * •

The part $\Rrightarrow$. By theorem 7.8 we have
$\overline{T_{\mathscr{A}^{(j)}}}f\supseteq I^{d}\cdot R^{\oplus p}_{X}$. As
the instability locus is not the whole $(\Bbbk^{n},o)$ the annihilator
$\mathfrak{a}_{\mathscr{A}}$ contains a non-zero divisor. Then
${T_{\mathscr{A}^{(j)}}}f\supseteq I^{d}\cdot R^{\oplus p}_{X}$, by part three
of lemma 6.14.

  * •

The part $\Lleftarrow$. By theorem 7.8 it is enough to show:
$\mathscr{A}^{(j)}f\supseteq\\{f\\}+I^{N}\cdot R^{\oplus p}_{X}$ for some
$N\gg 1$. Using Example 7.5 we have: $\mathscr{A}f\supseteq\\{f\\}+I^{2N}\cdot
R^{\oplus
p}_{X}\subseteq\\{f\\}+(\mathfrak{a}_{\mathscr{R}}+\mathfrak{a}_{\mathscr{A}}\cap
f^{\sharp}(y))^{2}R^{\oplus p}_{X}\subset\mathscr{A}f$.

Moreover, in the proof of theorem 7.1 we get $\tilde{c}_{**}\in I^{j}$,
therefore the needed coordinate change is an $\mathscr{R}^{(j)}$-element.
Similarly, the involved $\mathscr{L}$-elements are in fact $\mathscr{L}^{(j)}$
elements. Therefore $\mathscr{A}^{(j)}f\supseteq\\{f\\}+I^{N}\cdot R^{\oplus
p}_{X}$. $\blacksquare$

######  Example 7.10.

Let $R_{X}$ be one of $\Bbbk[[x]]$, $\Bbbk\\{x\\}$, $\Bbbk\langle x\rangle$,
with $\Bbbk$ a field. Take the filtration $I^{\bullet}\cdot R^{\oplus p}_{X}$
and the corresponding pairs $(T_{\mathscr{A}^{(j)}},\mathscr{A}^{(j)})$ of
§3.4. Assume $(f)\subseteq I$.

  1. i.

Suppose $T_{\mathscr{A}^{(0)}}f\supseteq I^{d_{\mathscr{A}}}\cdot R^{\oplus
p}_{X}$ and $T_{{\mathscr{K}}\\!^{(0)}}f\supseteq I^{d_{\mathscr{K}}\\!}\cdot
R^{\oplus p}_{X}$. By part 2 of lemma 6.12 we get:
$T_{\mathscr{A}^{(1)}}f\supseteq I^{d_{\mathscr{A}}+d_{\mathscr{K}}\\!}\cdot
R^{\oplus p}_{X}$. Assume either $char(\Bbbk)=0$ or
$char(\Bbbk)>d_{\mathscr{A}}+d_{\mathscr{K}}\\!-ord(f)$. Then corollary 7.9
gives:
$\mathscr{A}^{(1)}f\supseteq\\{f\\}+I^{d_{\mathscr{A}}+d_{\mathscr{K}}\\!}\cdot
R^{\oplus p}_{X}$. For $I={\mathfrak{m}}$ and $\Bbbk=\mathbb{R},\mathbb{C}$
this is close to [Mo. N.B., Corollary 6.3]:

(51) If $T_{\mathscr{A}}f\supseteq{\mathfrak{m}}^{d_{\mathscr{A}}}\cdot
R^{\oplus p}_{X}$ and
$T_{{\mathscr{K}}\\!}f\supseteq{\mathfrak{m}}^{d_{\mathscr{K}}\\!}\cdot
R^{\oplus p}_{X}$ then
$\mathscr{A}^{(1)}f\supseteq\\{f\\}+{\mathfrak{m}}^{d_{\mathscr{A}}+d_{\mathscr{K}}\\!+1}\cdot
R^{\oplus p}_{X}$.

  2. ii.

We compare the conclusion
$\mathscr{A}^{(1)}f\supseteq\\{f\\}+I^{d_{\mathscr{A}}+d_{\mathscr{K}}\\!}\cdot
R^{\oplus p}_{X}$ to the classical Theorems 0.2 and 2.1 of [Gaffney-du
Plessis]: “Let $\Bbbk=\mathbb{R},\mathbb{C}$ and assume $f$ is
${\mathscr{K}}\\!$-finite. If ${\mathfrak{m}}\cdot T_{\mathscr{R}}f+(y)\cdot
T_{\mathscr{L}}\supseteq I^{d}\cdot R^{\oplus p}_{X}$ then
$\mathscr{A}f\supseteq\\{f\\}+I^{2d+2}\cdot R^{\oplus p}_{X}$”. Recall also
Theorem 6.2 of [Mo. N.B.]: “If ${\mathfrak{m}}\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}^{(0)}}f\supseteq{\mathfrak{m}}^{d}\cdot
R^{\oplus p}_{X}$ then $\mathscr{A}f\supseteq\\{f\\}+{\mathfrak{m}}^{2d}\cdot
R^{\oplus p}_{X}$”.

While we do not assume ${\mathscr{K}}\\!$-finiteness, we assume $V(I)\supseteq
f^{-1}(f(V(\mathfrak{a}_{\mathscr{A}})))\cap Crit(f)$. Moreover, the
assumption $T_{\mathscr{A}^{(0)}}f\supseteq I^{d}\cdot R^{\oplus p}_{X}$ is
slightly stronger than ${\mathfrak{m}}\cdot
T_{\mathscr{R}}f+T_{\mathscr{L}^{(0)}}f\supseteq I^{d}\cdot R^{\oplus p}_{X}$,
as $I\cdot T_{\mathscr{R}}f\subseteq
T_{\mathscr{R}^{(0)}}f\subseteq{\mathfrak{m}}\cdot T_{\mathscr{R}}f$.

But in the case $V(\mathfrak{a}_{\mathscr{A}})=o\in\Bbbk^{n}$ one can take
$\sqrt{I}={\mathfrak{m}}$, and our conclusion,
$\mathscr{A}^{(1)}f\supseteq\\{f\\}+I^{d_{\mathscr{A}}+d_{\mathscr{K}}\\!}\cdot
R^{\oplus p}_{X}$, strengthens Theorem 6.2 of [Mo. N.B.].

###  7.3. Geometric characterization of determinacy

Recall that the space $T^{1}_{\mathscr{A}}f$ is non-zero exactly at the locus
of instability of a map. This is translated into the determinacy as follows.

######  Corollary 7.11.

  1. 1.

Let $\Bbbk$ be an(y) infinite field. Suppose $f$ is unstable exactly over
$V(I_{Y})\subset(\Bbbk^{p},o)$, i.e. $I_{Y}=ann[f_{*}T^{1}_{\mathscr{A}}f]$,
i.e. $Supp(f_{*}T^{1}_{\mathscr{A}}f)=V(I_{Y})$. (Here $I_{Y}\subseteq(y)^{2}$
and $\mathfrak{a}_{\mathscr{R}}\subseteq{\mathfrak{m}}^{2}$.) Then
$\mathscr{A}f\supseteq\\{f\\}+(\mathfrak{a}_{\mathscr{R}}+f^{\\#}(I_{Y}))^{d}\cdot
R^{\oplus p}_{X}$ for $d\gg 1$.

  2. 2.

Suppose $char(\Bbbk)=0$ or $char(\Bbbk)\gg 1$. If
$\mathscr{A}^{(1)}f\supseteq\\{f\\}+f^{\\#}(I^{d}_{Y})\cdot R^{\oplus p}_{X}$
then
$T_{\mathscr{A}}f\supseteq(\mathfrak{a}_{\mathscr{R}}+f^{\\#}(I_{Y})^{d})\cdot
R^{\oplus p}_{X}$. In particular,
$Supp(f_{*}T^{1}_{\mathscr{A}}f)=V(I_{Y})\subset(\Bbbk^{p},o)$, i.e. $f$ is
infinitesimally stable over $(\Bbbk^{p},o)\setminus V(I_{Y})$.

For $I={\mathfrak{m}}$ and $R_{X}=\mathbb{C}\\{x\\}$ this gives the classical
criterion of Mather-Gaffney.
Proof.

  1. 1.

Take the ring extension
${\tilde{R}}_{Y}:=f^{\\#}(R_{Y})+\mathfrak{a}_{\mathscr{A}}$ and denote
$\tilde{I}_{Y}:={\tilde{R}}_{Y}\cdot I_{Y}$. Then $T^{1}_{\mathscr{A}}f\in
mod{\tilde{R}}_{Y}$, by lemma 6.4. The localizations vanish:
$(T^{1}_{\mathscr{A}}f)_{\tilde{{\mathfrak{p}\hskip 1.42271pt}}}=0$ for each
${\tilde{R}}_{Y}\supset{\tilde{{\mathfrak{p}\hskip
1.42271pt}}}\not\supseteq\tilde{I}_{Y}$. Thus
$Supp(T^{1}_{\mathscr{A}}f)\subseteq V(\tilde{I}_{Y})\subset Y$. By
Noetherianity we get: $f^{\\#}(I_{Y})^{d}\cdot T^{1}_{\mathscr{A}}f=0$ for
$d\gg 1$. Thus $f^{\\#}(I_{Y})^{d}\cdot R^{\oplus p}_{X}\subseteq
T_{\mathscr{A}}f$, hence $f^{\\#}(I_{Y})^{2d}\cdot R^{\oplus p}_{X}\subseteq
f^{\\#}(I_{Y})^{d}\cdot T_{\mathscr{R}}f+f^{\\#}(I_{Y})^{d}\cdot
T_{\mathscr{L}}f$. Therefore
$\mathscr{A}f\supseteq\\{f\\}+(\mathfrak{a}_{\mathscr{R}}+f^{\\#}(I^{d}_{Y}))^{2}\cdot
R^{\oplus p}_{X}$.

  2. 2.

By theorem 7.8 we get: $T_{\mathscr{A}}\supseteq I^{d}_{Y}\cdot R^{\oplus
p}_{X}$ for $d\gg 1$. Thus $\mathfrak{a}_{\mathscr{A}}\supset I^{d}_{Y}$.

$\blacksquare$

## References

  * [A.G.L.V.] V. I. Arnold, V.V. Goryunov, O.V. Lyashko, V.A. Vasil’ev, Singularity theory. I. Reprint of the original English edition from the series Encyclopaedia of Mathematical Sciences [Dynamical systems. VI, Encyclopaedia Math. Sci., 6, Springer, Berlin, 1993; MR1230637]. Springer-Verlag, Berlin, 1998. iv+245 pp
  * [Artin] M. Artin, Algebraic approximation of structures over complete local rings, Publ. Math. IHES, 36, (1969), 23-58. DOI: 10.1007/BF02684596
  * [B.K.16] G. Belitski, D. Kerner, Group actions on filtered modules and finite determinacy. Finding large submodules in the orbit by linearization, C. R. Math. Acad. Sci. Soc. R. Can. 38 (2016), no. 4, 113–153.
  * [B.G.K.22] A.-F. Boix, G.-M. Greuel, D. Kerner, Pairs of Lie-type and large orbits of group actions on filtered modules. (A characteristic-free approach to finite determinacy.), Math. Z. 301 (2022), no. 3, 2415–2463.
  * [Bou.Gr.Ma.11] Y. Boubakri, G.-M. Greuel, Th. Markwig, Normal forms of hypersurface singularities in positive characteristic. Mosc. Math. J. 11, No. 4, 657–683 (2011).
  * [Bou.Gr.Ma.12] Y. Boubakri, G.-M. Greuel, Th. Markwig, Invariants of hypersurface singularities in positive characteristic. Rev. Mat. Complut. 25 (2012), no. 1, 61–85.
  * [B.P.W.] J.W. Bruce, A.A. du Plessis, C.T.C. Wall, Determinacy and unipotency. Invent. Math. 88 (1987), no. 3, 521–554.
  * [Cu.Sri.1997] S. D. Cutkosky, H. Srinivasan, Equivalence and finite determinancy of mappings. J. Algebra 188 (1997), no. 1, 16–57.
  * [Damon84] J. Damon, The unfolding and determinacy theorems for subgroups of $\mathscr{A}$ and ${\mathscr{K}}\\!$. Mem. Amer. Math. Soc. 50 (1984), no. 306, x+88 pp.
  * [Damon91] J. Damon, $\mathscr{A}$-equivalence and the equivalence of sections of images and discriminants. Singularity theory and its applications. Pt. I: Geometric aspects of singularities, Proc. Symp., Warwick/UK 1988-89, Lect. Notes Math. 1462, 93–121 (1991).
  * [du Plessis] A. du Plessis, On the determinacy of smooth map-germs. Invent. Math. 58 (1980), no. 2, 107–160.
  * [Ebeling] W. Ebeling, Functions of several complex variables and their singularities. Translated from the 2001 German original by Philip G. Spain. Graduate Studies in Mathematics, 83. American Mathematical Society, Providence, RI, 2007. xviii+312 pp.
  * [Eisenbud] D. Eisenbud, Commutative algebra. With a view toward algebraic geometry. Graduate Texts in Mathematics, 150. Springer-Verlag, New York, 1995.
  * [F.-S. R.] W. R. Ferrer Santos, A. Rittatore, Actions and invariants of algebraic groups. Second edition. Monographs and Research Notes in Mathematics. CRC Press, Boca Raton, FL, 2017. xx+459 pp.
  * [Gabrielov] A. M. Gabrièlov, Formal relations among analytic functions. Izv. Akad. Nauk SSSR Ser. Mat. 37 (1973), 1056–1090.
  * [Gaffney] T. Gaffney, A note on the order of determination of a finitely determined germ. Invent. Math. 52 (1979), no. 2, 127–130.
  * [Gaffney-du Plessis] T. Gaffney, A. du Plessis, More on the determinacy of smooth map-germs. Invent. Math. 66 (1982), no. 1, 137–163.
  * [Greuel.18] G. M. Greuel, Singularities in positive characteristic: equisingularity, classification, determinacy. Singularities, algebraic geometry, commutative algebra, and related topics, 37–53, Springer, Cham, 2018
  * [Greuel-Kröning.90] G.-M. Greuel, H. Kröning, _Simple singularities in positive characteristic_ , Math. Z. 203 (1990), 339–354.
  * [Gr.Lo.Sh] G.-M. Greuel, C. Lossen, E. Shustin, Introduction to singularities and deformations. Springer Monographs in Mathematics. Springer, Berlin, 2007. xii+471 pp.
  * [Greuel-Nguyen.16] G.-M. Greuel, H. D. Nguyen, Right simple singularities in positive characteristic. J. Reine Angew. Math. 712 (2016), 81–106.
  * [Greuel-Pham.19] G.-M. Greuel, T.H. Pham, Finite determinacy of matrices and ideals in arbitrary characteristics, J. Algebra 530 (2019), 195–214.
  * [H.W.-A.] K. Houston, R. Wik Atique, $\mathscr{A}$-classification of map-germs via ${}_{V}\\!{\mathscr{K}}\\!$-equivalence, arXiv:1302.1104.
  * [Jacobson] N. Jacobson, Lie algebras. Republication of the 1962 original. Dover Publications, Inc., New York, 1979. ix+331 pp.
  * [Kerner] D. Kerner, Unfoldings of maps, the first results on stable maps, and results of Mather-Yau/Gaffney-Hauser type in arbitrary characteristic, Arxiv temporary submission identifier: submit/4489457.
  * [Kucharz] W. Kucharz, Power series and smooth functions equivalent to a polynomial. Proc. Amer. Math. Soc. 98 (1986), no. 3, 527–533.
  * [Lafon] J.-P. Lafon, Anneaux henséliens et théorème de préparation. C. R. Acad. Sci. Paris Sér. A-B 264 (1967), A1161-A1162.
  * [Martinet] J. Martinet, Singularities of smooth functions and maps, Lecture Note Seris 58, Cambridge University Press, 1982, 256 pp.
  * [Mather] J.N. Mather, Stability of $C^{\infty}$ mappings. I. The division theorem. Ann. of Math. (2) 87, 1968, 89–104.
Stability of $C^{\infty}$ mappings. II. Infinitesimal stability implies
stability. Ann. of Math. (2) 89, 1969, 254–291.
Stability of $C^{\infty}$ mappings. III. Finitely determined map-germs. Publ.
Inst. Hautes Études Sci. Publ. Math. No. 35, 1968, 279–308.
Stability of $C^{\infty}$ mappings. IV. Classification of stable germs by
R-algebras. Publ. Inst. Hautes Études Sci. Publ. Math. No. 37, 1969, 223–248.

  * [Mond-Montaldi] D. Mond, J. Montaldi, Deformations of maps on complete intersections, Damon’s ${\mathscr{K}}\\!_{V}$-equivalence and bifurcations. Brasselet, Jean-Paul (ed.), Singularities. Papers of the international congress ‘Singularities in geometry and topology’, held in Lille (France), 3-8 June, 1991. Cambridge: Cambridge University Press. Lond. Math. Soc. Lect. Note Ser. 201, 263-284 (1994).
  * [Mo. N.B.] D. Mond, J.-J. Nuño-Ballesteros, Singularities of Mappings The Local Behaviour of Smooth and Complex Analytic Mappings, 2020.
  * [Moret-Bailly] L. Moret-Bailly, A henselian preparation theorem, arXiv:2112.13552.
  * [Nguyen.19] H. D. Nguyen, Right unimodal and bimodal singularities in positive characteristic. Int. Math. Res. Not. IMRN 2019, no. 6, 1612–1641.
  * [Pham.2016] T. H. Pham, On Finite Determinacy of Hypersurface Singularities and Matrices in Arbitrary Characteristic, PhD thesis, 2016.
  * [Pfister-Popescu] G. Pfister, D. Popescu, Die strenge Approximationseigenschaft lokaler Ringe. Invent. Math. 30 (1975), no. 2, 145–174.
  * [Ruas] M. A. S. Ruas, On the finite $C^{l}$-determinacy and applications, PhD thesis, 1983, São-Carlos.
  * [R.W.] D. Rumynin, M. Westaway, Integration of modules II: Exponentials, Trans. Amer. Math. Soc., DOI: https://doi.org/10.1090/tran/8449.
  * [Tougeron] J.C. Tougeron, Idéaux de fonctions différentiables. I. Ann. Inst. Fourier (Grenoble) 18 1968 fasc. 1, 177–240.
  * [Wall.81] C.T.C. Wall, Finite determinacy of smooth map-germs. Bull. London Math. Soc. 13 (1981), no. 6, 481–539.
  * [Wall.95] C. T. C. Wall, Classification and stability of singularities of smooth maps. Singularity theory (Trieste, 1991), 920-952, World Sci. Publ., River Edge, NJ, 1995.