On the behavior of modules of $m$-integrable derivations in the sense of Hasse-Schmidt under base change

TL;DR

This paper investigates how modules of $m$-integrable derivations, as defined by Hasse-Schmidt, behave under base change, especially in the context of separable extensions and polynomial rings over fields of positive characteristic.

Contribution

It provides new insights into the behavior of $m$-integrable derivations under base change in specific algebraic settings, including separable extensions and polynomial rings.

Findings

Modules of $m$-integrable derivations exhibit specific behaviors under base change.

Results apply to separable extensions over fields of positive characteristic.

Analysis includes polynomial rings in multiple variables.

Abstract

We study the behavior of modules of $m$-integrable derivations of a commutative finitely generated algebra in the sense of Hasse-Schmidt under base change. We focus on the case of separable ring extensions over a field of positive characteristic and on the case where the extension is a polynomial ring in an arbitrary number of variables.

Equations

$$\begin{array}[]{ccc}D_{0}=\operatorname{\rm Id}_{A},&\displaystyle D_{n}(xy)=\sum_{a+b=n}D_{a}(x)D_{b}(y)\end{array}$$

$$\begin{array}[]{ccc}D_{0}=\operatorname{\rm Id}_{A},&\displaystyle D_{n}(xy)=\sum_{i+j=n}D_{i}(x)D_{j}(y)\end{array}$$

$${\phi }_D:a\in A⟼\sum _{i\ge 0}^m{D}_i(a){\mu }^i\in A[|\mu |{]}_m$$

$${\mathrm{Hom}}_{k-\text{alg}}^{\circ }(A,A[|\mu |{]}_m):=\{f\in {\mathrm{Hom}}_{k-\text{alg}}(A,A[|\mu |{]}_m)|f(a)\equiv a\mathrm{mod}\mu \text{mbox}\forall a\in A\},$$

$$D\in {\mathrm{HS}}_k(A;m)\mapsto {\phi }_D\in {\mathrm{Hom}}_{k-\text{alg}}^{\circ }\left(A,A[|\mu |{]}_m\right).$$

$${\mathrm{Aut}}_{k[|\mu |{]}_m-\text{alg}}^{\circ }(A[|\mu |{]}_m):=\{f\in {\mathrm{Aut}}_{k[|\mu |{]}_m-\text{alg}}(A[|\mu |{]}_m)|f(a)\equiv a_0\mathrm{mod}\mu \text{mbox}\forall a\in A[|\mu |{]}_m\},$$

$$\ell (D):=\min \{h\ge 1\text{mbox}|\text{mbox}{D}_h\ne 0\}$$

$$\psi \bullet D\equiv \psi \bullet \left(\sum _{i=0}^m{D}_i{\mu }^i\right)=\sum _{i=0}^m\psi (\mu {)}^i{D}_i.$$

$$D[n]_{i}=\left\{\begin{array}[]{ll}D_{i/n}&\mbox{ if }i=0\mod n\\ 0&\mbox{ otherwise.}\end{array}\right.$$

$$0⟶I{\mathrm{Der}}_k(R)⟶{\mathrm{Der}}_k(\log I)⟶{\mathrm{Der}}_k(A)⟶0$$

$$\delta \in {\mathrm{Der}}_k(\log I)\mapsto \overline{\delta }\in {\mathrm{Der}}_k(A)\text{mbox}with\overline{\delta }(r+I)=\delta (r)+I\text{mbox}forallr\in R.$$

$$A$$

$${\psi }_B\bullet \left({\Pi }_{\mathrm{HS},m}^I(D)\right)={\Pi }_{\mathrm{HS},n}^I\left(\psi \bullet D\right).$$

$$\delta \in {\mathrm{IDer}}_k(\log I;m)⟼{\Pi }_m^I(\delta )=\overline{\delta }\in {\mathrm{IDer}}_k(A;m)$$

$$0⟶I({\mathrm{Der}}_k(R))⟶{\mathrm{IDer}}_k(\log I;m)\stackrel{{\Pi }_m^I}{\to }{\mathrm{IDer}}_k(R/I;m)⟶0.$$

$$\mathcal{J}(l,D):=\{j\in \mathbb{N}|\ell (D)\le j\le p^l,\text{mbox}p\nmid j\}.$$

$${\circ }_{j\in \mathcal{J}(l,D)}{F}^j=F^{p^l-1}\circ \cdots \circ F^{\ell (D)}$$

$$D=T[p]\circ \left({\circ }_{j\in \mathcal{J}(l,D)}\left({\psi }^j\bullet F^j\right)\right),$$

$$D^{\prime }=T^{\prime }[p]\circ \left({\circ }_{j\in \mathcal{J}(l,D^{\prime })}\left({\psi }^j\bullet F^j\right)\right)$$

$$D=F[p^t]\circ T^{\prime }[p]\circ \left({\circ }_{j\in \mathcal{J}(l,D)}\left({\psi }^j\bullet F^j\right)\right)=\left(F\left[p^{t-1}\right]\circ T^{\prime }\right)[p]\circ \left({\circ }_{j\in \mathcal{J}(l,D)}\left({\psi }^j\bullet F^j\right)\right).$$

$$D^{\prime }=T[p]\circ \left({\circ }_{j\in \mathcal{J}(l,D^{\prime })}\left({\psi }^j\bullet F^j\right)\right)\Rightarrow D=T[p]\circ \left({\circ }_{j\in \mathcal{J}(l,D^{\prime })}\left({\psi }^j\bullet F^j\right)\right)\circ ({\psi }^i\bullet F)$$

$$D=\left({\psi }^i\bullet F\right)\circ T^{\prime }[p]\circ \left({\circ }_{j\in \mathcal{J}(l,D^{\prime })}\left({\psi }^j\bullet F^j\right)\right)$$

$${\psi }^i\bullet F={\tau }_{i{p}^{s_i+1},p^l}(F[i])={\tau }_{i{p}^{s_i+1},p^l}(F[i/p][p])={\tau }_{i{p}^{s_i},p^{l-1}}(F[i/p])[p]$$

$$D=T[p]\circ \left({\circ }_{i\in \mathcal{J}(l,D)}\left({\psi }^i\bullet F^i\right)\right)$$

$$\begin{array}[]{rccc}\Phi^{L,A}:&L\otimes_{k}\operatorname{\rm Der}_{k}(A)=A_{L}\otimes_{A}\operatorname{\rm Der}_{k}(A)&\longrightarrow&\operatorname{\rm Der}_{L}(A_{L})\\ &c\otimes\delta&\longmapsto&c\widetilde{\delta}.\end{array}$$

$$L{\otimes }_k{\mathrm{Der}}_k(A)\simeq L{\otimes }_k{\mathrm{Hom}}_A({\Omega }_{A/k},A)⟶{\mathrm{Hom}}_{A_L}({\Omega }_{A/k},A_L)\simeq {\mathrm{Hom}}_{A_L}(A_L{\otimes }_A{\Omega }_{A/k},A_L)\simeq {\mathrm{Der}}_L(A_L).$$

$$\begin{array}[]{ccc}\Phi^{L,A}_{\text{\rm ind}}:L\otimes_{k}(I(\operatorname{\rm Der}_{k}(A)))\to I^{e}\operatorname{\rm Der}_{L}(A_{L})&\mbox{ and }&\Phi^{L,A}_{\text{\rm ind}}:L\otimes_{k}\operatorname{\rm Der}_{k}(\log I)\to\operatorname{\rm Der}_{L}(\log I^{e}).\end{array}$$

$$L{\otimes }_k\left(I({\mathrm{Der}}_k(R))\right)$$

$$I/I^2\to A{\otimes }_R{\Omega }_{R/k}\to {\Omega }_{A/k}\to 0$$

$$L{\otimes }_k{\mathrm{Der}}_k(A)\simeq A_L{\otimes }_A{\mathrm{Der}}_k(A)\simeq A_L{\otimes }_A{\mathrm{Hom}}_A({\Omega }_{A/k},A)\simeq {\mathrm{Hom}}_A({\Omega }_{A/k},A_L)\simeq$$

Full text

Keywords: Hasse-Schmidt derivation, Integrability, Base change, Separable algebras.

MSC 2010: 13N15.

INTRODUCTION

Let $k$ be a commutative ring and $A$ a commutative $k$-algebra. A Hasse-Schmidt derivation of $A$ over $k$ of length $m\in\mathbb{N}$ or $m=\infty$ is a sequence $D=(D_{n})_{n=0}^{m}$ such that:

[TABLE]

for all $x,y\in A$ and for all $n$. For $n\geq 1$, the component $D_{n}$ of a Hasse-Schmidt derivation is a differential operator of order $\leq n$ vanishing at 1, in particular $D_{1}$ is a $k$-derivation. Hasse-Schmidt derivations of length $m$, also called higher derivations of order $m$ (see [Ma2]), were introduced by H. Hasse and F.K. Schmidt ([Ha-Sh]) and they have been used by several authors in different contexts (see [Se], [Tr] or [Vo]).

An important notion related with Hasse-Schmidt derivations is integrability. Let $m\in\mathbb{N}$ or $m=\infty$, then we say that $\delta\in\operatorname{\rm Der}_{k}(A)$ is $m$-integrable if there exists a Hasse-Schmidt derivation $D$ of length $m$ such that $\delta=D_{1}$. The set of all $m$-integrable $k$-derivations is an $A$-submodule of $\operatorname{\rm Der}_{k}(A)$ for all $m$, which is denoted by $\operatorname{\rm IDer}_{k}(A;m)$. If $k$ has characteristic 0 or $A$ is [math]-smooth over $k$, then any $k$-derivation is $\infty$-integrable ([Ma2]), i.e. $\operatorname{\rm Der}_{k}(A)=\operatorname{\rm IDer}_{k}(A;\infty)$. However, if we consider $k$ a ring of positive characteristic, then we do not have the same property in general. Nonetheless, the modules $\operatorname{\rm IDer}_{k}(A;m)$ have better properties in some way than $\operatorname{\rm Der}_{k}(A)$ (see [Mo], [Fe-Na]) and so their exploration could be interesting for understanding singularities in positive characteristic.

In this paper we study the behavior of modules of $m$-integrable $k$-derivations under base change. Namely, if $k\to L$ is a ring extension and $A$ is a $k$-algebra, the well-known base change map $L\otimes_{k}\operatorname{\rm Der}_{k}(A)\to\operatorname{\rm Der}_{k}(L\otimes_{k}A)$ induces, for each $m\geq 1$ or $m=\infty$, a base change map $\Phi^{L,A}_{m}:L\otimes_{k}\operatorname{\rm IDer}_{k}(A;m)\to\operatorname{\rm IDer}_{L}(L\otimes_{k}A;m)$. We prove that if $A$ is finitely generated and $L$ is a polynomial ring over $k$ in an arbitrary number of variables or a separable $k$-algebra over a field $k$ of positive characteristic then $\Phi^{L,A}_{m}$ is an isomorphism for all $m\geq 1$.

This paper is organized as follows: In section 1 we recall the definition of Hasse-Schmidt derivations and give some basic results. In section 2 we prove that an $I$-logarithmic Hasse-Schmidt derivation of a polynomial ring $R=k[x_{1},\ldots,x_{d}]$ over a ring $k$ of positive characteristic (where $I\subseteq R$ is an ideal) can be decomposed into two special Hasse-Schmidt derivations if its 1-component is zero.

In section 3 we recall some classical results of base change maps for $k$-derivations and we study the induced maps $\Phi^{L,A}_{m}:L\otimes_{k}\operatorname{\rm IDer}_{k}(A;m)\to\operatorname{\rm IDer}_{L}(L\otimes_{k}A;m)$. We see that $\Phi^{L,A}_{m}$ is not surjective in general by giving a counterexample and we prove that if $L$ is a polynomial ring over $k$ in an arbitrary number of variables or if $L$ is separable algebra over a field $k$ of positive characteristic then $\Phi^{L,A}_{m}$ is bijective for any finitely generated $k$-algebra $A$ and for all integers $m$.

Throughout this paper, all rings (and algebras) are assumed to be commutative.

1 Hasse-Schmidt derivations

In this section, we recall the main definitions of the theory of Hasse-Schmidt derivations and give some basic results. From now on, $k$ will be a commutative ring and $A$ a commutative $k$-algebra. We denote $\overline{\mathbb{N}}:=\mathbb{N}\cup\{\infty\}$ and, for each integer $m\geq 1$, we will write $A[|\mu|]_{m}:=A[|\mu|]/\langle\mu^{m+1}\rangle$ and $A[|\mu|]_{\infty}:=A[|\mu|]$. General references for the definitions and results in this section are [Ma2, §27], [Na1] and [Na2].

Definition 1.1

A Hasse-Schmidt derivation (HS-derivation for short) of $A$ (over $k$) of length $m\geq 1$ (resp. of length $\infty$) is a sequence $D:=(D_{0},D_{1},\ldots,D_{m})$ (resp. $D=(D_{0},D_{1},\ldots)$) of $k$-linear maps $D_{i}:A\rightarrow A$, satisfying the conditions:

[TABLE]

for all $x,y\in A$ and for all $n$. We write $\operatorname{\rm HS}_{k}(A;m)$ (resp. $\operatorname{\rm HS}_{k}(A;\infty)=\operatorname{\rm HS}_{k}(A)$) for the set of HS-derivations of $A$ (over $k$) of length $m$ (resp. $\infty$).

For $i\geq 1$, the $D_{i}$ component of a HS-derivation $D\in\operatorname{\rm HS}_{k}(A;m)$ is a $k$-linear differential operator of order $\leq i$ vanishing at $1$. In particular, $D_{1}$ is a $k$-derivation.

The set $\operatorname{\rm HS}_{k}(A;m)$ has a natural group structure with identity $\mathbb{I}=(\operatorname{\rm Id},0,\ldots)$ and $D\circ D^{\prime}=D^{\prime\prime}\in\operatorname{\rm HS}_{k}(A;m)$ with $D^{\prime\prime}_{n}=\sum_{i+j=n}D_{i}\circ D_{j}^{\prime}$ for all $n$. We denote by $D^{\ast}\in\operatorname{\rm HS}_{k}(A;m)$ the inverse of $D\in\operatorname{\rm HS}_{k}(A;m)$. Observe that $D_{1}^{\ast}=-D_{1}$ and that the map: $(\operatorname{\rm Id},D_{1})\in\operatorname{\rm HS}_{k}(A;1)\mapsto D_{1}\in\operatorname{\rm Der}_{k}(A)$ is a group isomorphism.

Any HS-derivation $D\in\operatorname{\rm HS}_{k}(A;m)$ is determined by the $k$-algebra homomorphism

[TABLE]

satisfying $\varphi_{D}(a)\equiv a\mod\mu$. If we denote

[TABLE]

we have a bijection

[TABLE]

The map $\varphi_{D}$ can be uniquely extended to a $k[|\mu|]_{m}$-algebra automorphism $\widetilde{\varphi_{D}}:A[|\mu|]_{m}\rightarrow A[|\mu|]_{m}$ with $\widetilde{\varphi_{D}}(a)\equiv a_{0}$ for all $a=\sum_{i}a_{i}\mu^{i}\in A[|\mu|]_{m}$. If we denote

[TABLE]

we have a group isomorphism $D\in\operatorname{\rm HS}_{k}(A;m)\longmapsto\widetilde{\varphi_{D}}\in\operatorname{\rm Aut}_{k[|\mu|]_{m}-\text{alg}}^{\circ}(A[|\mu|]_{m}),$ and for $D,D^{\prime}\in\operatorname{\rm HS}_{k}(A;m)$ we have $\varphi_{D\circ D^{\prime}}:=\widetilde{\varphi_{D}}\circ\varphi_{D^{\prime}}$.

A HS-derivation $D$ of $A$ over $k$ of length $m$ can be understood as a power series $\sum_{i=1}^{m}D_{i}\mu^{i}$ with coefficients in $\operatorname{\rm End}_{k}(A)$ and so we can consider $\operatorname{\rm HS}_{k}(A;m)$ as a subgroup of the group of units of $\operatorname{\rm End}_{k}(A)[|\mu|]_{m}$.

Additional details for the above material can be found in [Na2, §5].

Definition 1.2

For each HS-derivation $D\in\operatorname{\rm HS}_{k}(A;m)$ such that $D\neq\mathbb{I}$, we denote

[TABLE]

and for $D=\mathbb{I}$, $\ell(D)=\infty$, i.e. $\ell(D)=\operatorname{\rm ord}(D-\mathbb{I})$.

The following lemma is clear (see [Na2, §5]).

Lemma 1.3

If $D,E\in\operatorname{\rm HS}_{k}(A;m)$, then $\ell(D\circ E)\geq\min\{\ell(D),\ell(E)\}$. In particular, if $\ell(D),\ell(E)\geq n$, then $\ell(D\circ E)\geq n$ and $(D\circ E)_{n}=D_{n}+E_{n}$.

1.1 The action of substitution maps

In this section, we recall some notions and results of [Na2, §6].

Definition 1.4

An $A$-algebra map $\psi:A[|\mu|]_{m}\rightarrow A[|\mu|]_{n}$ will be called a substitution map if $\psi(\mu)\in\langle\mu\rangle$. We say that a substitution map $\psi:A[|\mu|]_{m}\rightarrow A[|\mu|]_{n}$ has constant coefficients if $\psi(\mu)=\sum_{i\geq 1}a_{i}\mu^{i}$ with $a_{i}\in k$ for all $i$.

Compositions of substitution maps (with constant coefficients) are also substitution maps (with constant coefficients).

It is clear that for any $f\in\operatorname{\rm Hom}_{k-\text{\rm alg}}^{\circ}(A,A[|\mu|]_{m})$ and any substitution map $\psi:A[|\mu|]_{m}\rightarrow A[|\mu|]_{n}$, we have that $\psi\circ f\in\operatorname{\rm Hom}_{k-\text{\rm alg}}^{\circ}(A,A[|\mu|]_{n})$.

Notation 1.5

Let $\psi:A[|\mu|]_{m}\rightarrow A[|\mu|]_{n}$ be a substitution map and $D\in\operatorname{\rm HS}_{k}(A;m)$ a HS-derivation. We denote by $\psi\bullet D\in\operatorname{\rm HS}_{k}(A;n)$ the HS-derivation determined by $\varphi_{\psi\bullet D}=\psi\circ\varphi_{D}$. In terms of power series, we have:

[TABLE]

Example 1.6

In this paper we mainly use three types of substitution maps. Let $D\in\operatorname{\rm HS}_{k}(A;m)$ be a HS-derivation of length $m\in\overline{\mathbb{N}}$.

For each $a\in A$, we define $a\bullet D:=\psi\bullet D\in\operatorname{\rm HS}_{k}(A;m)$ where $\psi:A[|\mu|]_{m}\to A[|\mu|]_{m}$ is given by $\psi(\mu)=a\mu$. Namely: $a\bullet D=(a^{i}D_{i})_{i}$. 2. 2.

Let $1\leq n\leq m$ with $n\in\overline{\mathbb{N}}$ and let us consider the projection $\pi_{mn}:A[|\mu|]_{m}\to A[|\mu|]_{n}$ ($\pi_{mn}(\mu)=\mu$). The truncation $\tau_{mn}(D)$ is given by $\tau_{mn}(D)=\pi_{mn}\bullet D$, i.e. $\tau_{mn}(D)=(\operatorname{\rm Id},D_{1},\ldots,D_{n})\in\operatorname{\rm HS}_{k}(A;n)$. 3. 3.

For each integer $n\geq 1$, we define $D[n]=\psi\bullet D\in\operatorname{\rm HS}_{k}(A;nm)$ where $\psi:A[|\mu|]_{m}\to A[|\mu|]_{nm}$ is the substitution map given by $\psi(\mu)=\mu^{n}$. Namely:

[TABLE]

Substitution maps of type 2. and 3. of Example 1.6 have constant coefficients. Moreover, if $a\in k$, the substitution map $a\bullet(-)$ of type 1. has constant coefficients too.

The following lemma comes from 8. and Prop. 11 of [Na2, §6].

Lemma 1.7

Let $\phi:A[|\mu|]_{m}\rightarrow A[|\mu|]_{n}$ and $\psi:A[|\mu|]_{n}\rightarrow A[|\mu|]_{s}$ be substitution maps and $D,D^{\prime}\in\operatorname{\rm HS}_{k}(A;m)$ HS-derivations. We have the following properties:

If $\phi$ has constant coefficients, then $\phi\bullet(D\circ D^{\prime})=(\phi\bullet D)\circ(\phi\bullet D^{\prime})$.

• 2.

$\psi\bullet(\phi\bullet D)=(\psi\circ\phi)\bullet D$.

As a straightforward consequence we obtain the following corollary.

Corollary 1.8

Let $D,D^{1},\ldots,D^{t}\in\operatorname{\rm HS}_{k}(A,m)$ be HS-derivations of length $m\in\overline{\mathbb{N}}$. The following properties hold:

For each $a\in k$, we have $a\bullet(D^{1}\circ\cdots\circ D^{t})=(a\bullet D^{1})\circ\cdots\circ(a\bullet D^{t})$.

• 2.

$\tau_{mn}(D^{1}\circ\cdots\circ D^{t})=\tau_{mn}(D^{1})\circ\cdots\circ\tau_{mn}(D^{t})$* for any $n\leq m$.*

• 3.

$(D^{1}\circ\cdots\circ D^{t})[n]=D^{1}[n]\circ\cdots\circ D^{t}[n]$* for any $n\geq 1$.*

• 4.

$D[nn^{\prime}]=(D[n])[n^{\prime}]$* for any $n,n^{\prime}\geq 1$.*

The proof of the following lemma is easy and it is left up to the reader (see [Na1, §1.2]).

Lemma 1.9

Let $D\in\operatorname{\rm HS}_{k}(A;m)$ be a HS-derivation of length $m\in\overline{\mathbb{N}}$, $n\geq 1$ and $q\leq m$. The following properties hold:

$\left(a^{n}\bullet D\right)[n]=a\bullet(D[n])$* for all $a\in A$.*

• 2.

$\tau_{mn,m^{\prime}n}(D[n])=(\tau_{mm^{\prime}}(D))[n]$* for all $1\leq m^{\prime}\leq m$.*

• 3.

$\tau_{mq}(a\bullet D)=a\bullet(\tau_{mq}(D))$* for all $a\in A$.*

The following proposition is proved in [Na2, Prop. 11].

Proposition 1.10

Let $\psi:A[|\mu|]_{m}\rightarrow A[|\mu|]_{n}$ be a substitution map with constant coefficients. Then, $(\psi\bullet D)^{\ast}=\psi\bullet D^{\ast}$ for each $D\in\operatorname{\rm HS}_{k}(A;m)$.

The following two lemmas are clear.

Lemma 1.11

Let $D,E\in\operatorname{\rm HS}_{k}(A;m)$ be two HS-derivations of length $m\in\mathbb{N}$ such that $\tau_{m,m-1}(D)=\tau_{m,m-1}(E)$. Then, there exists $\delta\in\operatorname{\rm Der}_{k}(A)$ such that $D=E\circ(\operatorname{\rm Id},\delta)[m]$.

Lemma 1.12

Let $D\in\operatorname{\rm HS}_{k}(A;m)$ be a HS-derivation of length $m\in\mathbb{N}$ and $\delta\in\operatorname{\rm Der}_{k}(A)$, then $D\circ(\operatorname{\rm Id},\delta)[m]=(\operatorname{\rm Id},\delta)[m]\circ D$.

1.2 Integrable derivations

In this section, we recall the notion of $n$-integrable derivation. This notion was introduced in [Ma1] for $n=\infty$. The case of finite $n$ has been studied in [Na1]. We also recall the “logarithmic point of view” developed in loc. cit.. From now on, $k$ will be a commutative ring, $A$ a commutative $k$-algebra and $I\subseteq A$ an ideal.

Remember that a $k$-derivation $\delta:A\to A$ is called $I$-logarithmic if $\delta(I)\subset I$. The set of $I$-logarithmic $k$-derivations is an $A$-submodule of $\operatorname{\rm Der}_{k}(A)$ and will be denoted by $\operatorname{\rm Der}_{k}(\log I)$.

If $A$ is a finitely generated $k$-algebra, we may assume that $A$ is the quotient of $R=k[x_{1},\dots,x_{d}]$ by some ideal $I$. There is an exact sequence of $R$-modules:

[TABLE]

where the last map is given by:

[TABLE]

Definition 1.13

Let $D\in\operatorname{\rm HS}_{k}(A;m)$ with $m\in\overline{\mathbb{N}}$, $I\subset A$ an ideal and $n\geq m$.

• •

$D$* is $I$-logarithmic if $D_{i}(I)\subseteq I$ for all $i$. The set of $I$-logarithmic HS-derivations is denoted by $\operatorname{\rm HS}_{k}(\log I;m)$ and $\operatorname{\rm HS}_{k}(\log I):=\operatorname{\rm HS}_{k}(\log I;\infty)$. In particular we have $\operatorname{\rm Der}_{k}(\log I)\equiv\operatorname{\rm HS}_{k}(\log I;1)$.*

• •

More generally, for $r\leq m$, $D$ is $r-I$-logarithmic if $\tau_{mr}(D)\in\operatorname{\rm HS}_{k}(\log I;r)$.

• •

$D$* is $n$-integrable if there exists $E\in\operatorname{\rm HS}_{k}(A,n)$ such that $\tau_{nm}(E)=D$. Any such $E$ will be called an $n$-integral of $D$. If $D$ is $\infty$-integrable we simply say that $D$ is integrable. If $m=1$, we write $\operatorname{\rm IDer}_{k}(A;n)$ for the set of $n$-integrable derivations and $\operatorname{\rm IDer}_{k}(A):=\operatorname{\rm IDer}_{k}(A;\infty)$.*

• •

If $D\in\operatorname{\rm HS}_{k}(\log I;m)$, we say that $D$ is $I$-logarithmically $n$-integrable if there exists $E\in\operatorname{\rm HS}_{k}(\log I;n)$ such that $E$ is an $n$-integral of $D$. We denote $\operatorname{\rm IDer}_{k}(\log I;n)$ the set of $I$-logarithmically $n$-integrable derivations (i.e. for $m=1$) and $\operatorname{\rm IDer}_{k}(\log I):=\operatorname{\rm IDer}_{k}(\log I,\infty)$.

The following lemma is clear.

Lemma 1.14

Under the above conditions, the following properties hold:

$\operatorname{\rm HS}_{k}(\log I;m)$* is a subgroup of $\operatorname{\rm HS}_{k}(A;m)$ for all $m\in\overline{\mathbb{N}}$.* 2. 2.

$I$-logarithmicity and $I$-logarithmically $n$-integrability are stable by the action of substitution maps. 3. 3.

$\operatorname{\rm IDer}_{k}(A;n)$* and $\operatorname{\rm IDer}_{k}(\log I;n)$ are $A$-submodules of $\operatorname{\rm Der}_{k}(A)$ for all $n\in\overline{\mathbb{N}}$.*

Definition 1.15

Let $s>1$ be an integer. We say that the $k$-algebra $A$ has a leap at $s>1$ if the inclusion $\operatorname{\rm IDer}_{k}(A;s-1)\supsetneq\operatorname{\rm IDer}_{k}(A;s)$ is proper. The set of leaps of $A$ over $k$ is denoted by $\operatorname{\rm Leaps}_{k}(A)$.

If $k\supset\mathbb{Q}$, it is well-known that $\operatorname{\rm IDer}_{k}(A;m)=\operatorname{\rm Der}_{k}(A)$ for all $m\in\overline{\mathbb{N}}$ and so $A$ has no leaps (see [Ma1]). Let us recall Theorem 27.1 of [Ma2].

Theorem 1.16

If $A$ is [math]-smooth over $k$, then any HS-derivation of length $m<\infty$ over $k$ is $\infty$-integrable.

Let us recall Theorem 4.1 of [Ti2].

Theorem 1.17

Let $k$ be a ring of $\operatorname{\rm char}(k)=p>0$ (i.e. $\mathbb{F}_{p}\subset k$) and $A$ a $k$-algebra. Then, $\operatorname{\rm Leaps}_{k}(A)\subseteq\{p^{\tau}\ |\ \tau\geq 1\}$.

Definition 1.18

Let $I\subset A$ be an ideal. An $I$-differential operator is a ($k$-linear) differential operator $H:A\to A$ such that $H(I)\subset I$.

In the following lemma we collect some easy results that will be used later. Its proof is left up to the reader.

Lemma 1.19

Let $D\in\operatorname{\rm HS}_{k}(A;m)$ be a HS-derivation of length $m\in\overline{\mathbb{N}}$, $n,s\geq 1$ positive integers such that $n\leq m$ and $I\subset A$ an ideal. The following properties hold:

1. (a)

If $D$ is $(n-1)-I$-logarithmic, then $D[s]\in\operatorname{\rm HS}_{k}(A;ms)$ is $(ns-1)-I$-logarithmic. 2. (b)

If $D$ is $n-I$-logarithmic, then $D^{\ast}$ is $n-I$-logarithmic too. 3. (c)

If $D$ is $(n-1)-I$-logarithmic, then $D^{\ast}[s]\in\operatorname{\rm HS}_{k}(A;ms)$ is $(ns-1)-I$-logarithmic. 4. (d)

Let $D^{1},\ldots,D^{t}\in\operatorname{\rm HS}_{k}(A;m)$ be an ordered family of $(n-1)-I$-logarithmic HS-derivations and let us denote $D:=D^{1}\circ\cdots\circ D^{t}\in\operatorname{\rm HS}_{k}(A;m)$. Then, $D$ is $(n-1)-I$-logarithmic and $D_{n}=\sum_{i=1}^{t}D_{n}^{i}+F_{n}$ where $F_{n}$ is an $I$-differential operator of order $\leq n$. 5. (e)

If $D$ is $(n-1)-I$-logarithmic, then, $D_{n}^{\ast}=-D_{n}+F_{n}$ where $F_{n}$ is an $I$-differential operator of order $\leq n$. 6. (f)

If $D$ is $(n-1)-I$-logarithmic and $E\in\operatorname{\rm HS}_{k}(\log I;m)$, then $D\circ E\in\operatorname{\rm HS}_{k}(A,m)$ is $(n-1)-I$-logarithmic and $(D\circ E)_{n}=D_{n}+F_{n}$ where $F_{n}$ is an $I$-differential operator of order $\leq n$.

Let us consider a $k$-algebra $A$, $I\subseteq A$ an ideal and $m\in\overline{\mathbb{N}}$ and we denote $\pi:A\to A/I$ the natural projection. An $I$-logarithmic HS-derivation $D\in\operatorname{\rm HS}_{k}(\log I;m)$ gives rise to a unique $\overline{D}\in\operatorname{\rm HS}_{k}(A/I;m)$ such that $\overline{D}_{r}\circ\pi=\pi\circ D_{r}$ for all $r\geq 0$, and the following diagram is commutative:

[TABLE]

where $\pi_{m}:A[|\mu|]_{m}\to(A/I)[|\mu|]_{m}$ is the natural projection. The map $\Pi_{\operatorname{\rm HS},m}^{I}:D\in\operatorname{\rm HS}_{k}(\log I;m)\to\overline{D}\in\operatorname{\rm HS}_{k}(A/I)$ is clearly a homomorphism of groups.

Lemma 1.20

Let $I\subseteq A$ an ideal and $\psi:A[|\mu|]_{m}\to A[|\mu|]_{n}$ a substitution map. Let us denote $B=A/I$ and $\psi_{B}:B[|\mu|]_{m}\to B[|\mu|]_{n}$ the substitution map induced by $\psi$. Then, for each $D\in\operatorname{\rm HS}_{k}(\log I;m)$ we have that

[TABLE]

2 Hasse-Schmidt derivations on polynomial rings

In this section $k$ will be an arbitrary commutative ring and $R=k[x_{1},\ldots,x_{d}]$ a polynomial ring.

The following result is a straightforward consequence of Theorem 1.16.

Proposition 2.1

Any HS-derivation of $R$ (over $k$) of length $m\geq 1$ is integrable.

Proposition 2.2

[Na1, Prop. 1.3.4]* If $I\subseteq R$ is an ideal, the map $\Pi_{\operatorname{\rm HS},m}^{I}:\operatorname{\rm HS}_{k}(\log I;m)\to\operatorname{\rm HS}_{k}(R/I;m)$ is a surjective group homomorphism for all $m\in\overline{\mathbb{N}}$.*

Let $I\subset R$ be an ideal, $A=R/I$, $m\in\overline{\mathbb{N}}$ and let us denote by $\Pi^{I}_{m}:\operatorname{\rm IDer}_{k}(\log I;m)\rightarrow\operatorname{\rm IDer}_{k}(A;m)$ the map defined as:

[TABLE]

where $\overline{\delta}$ has been defined in (1).

The following proposition is clear thanks to [Na1, Corollary 2.1.9].

Proposition 2.3

Under the above conditions, the following short sequence of $R$-modules is exact:

[TABLE]

Corollary 2.4

Under the above conditions, $A$ has a leap at $s>1$ if and only if the inclusion $\operatorname{\rm IDer}_{k}(\log I;s-1)\supsetneq\operatorname{\rm IDer}_{k}(\log I;s)$ is proper.

2.1 A decomposition of logarithmic HS-derivations in characteristic $\bf{p>0}$

In this section, $k$ will be a ring of characteristic $p>0$, i.e. $\mathbb{F}_{p}\subset k$, $R=k[x_{1},\ldots,x_{d}]$ and $I\subset R$ an ideal.

We first recall the following theorem.

Theorem 2.5

[Ti2, Th. 3.14] Let $e,s\geq 1$ be two integers and $D\in\operatorname{\rm HS}_{k}(R;ep^{s})$ an $(ep^{s}-1)-I$-logarithmic HS-derivation with $\ell(D)\geq e$. Then, there exist an integral $D^{\prime}\in\operatorname{\rm HS}_{k}(R;p^{s})$ of $D_{e}$ and an $I$-differential operator $H$ of order $\leq p^{s}$ such that $D^{\prime}$ is $(p^{s}-1)-I$-logarithmic and $D_{p^{s}}^{\prime}=D_{ep^{s}}+H$.

Notation 2.6

Let $i,l\geq 1$ be two positive integers such that $i<p^{l}$. We define $C_{i,l}^{p}:=\{j\in\mathbb{N}|\mbox{ }ip^{j}<p^{l}\}.$

The proof of the following lemma is clear.

Lemma 2.7

Let $i,l\geq 1$ be two positive integers such that $i<p^{l}$ and $i$ is not a power of $p$ and denote $s=\max C_{i,l}^{p}$. Then $ip^{s+1}>p^{l}$.

Notation 2.8

Let $l\geq 1$ be an integer and $D\in\operatorname{\rm HS}_{k}(R;p^{l})$. We define:

[TABLE]

Note that if $E\in\operatorname{\rm HS}_{k}(R;p^{l})$ such that $\ell(E)\leq\ell(D)$, then $\mathcal{J}(l,D)\subseteq\mathcal{J}(l,E)$ and $\mathcal{J}(l,E)\setminus\mathcal{J}(l,D)=\{j\in\mathbb{N}\ |\ \ell(E)\leq j<\ell(D),\mbox{ }p\nmid j\}$. For each family $F^{j}\in\operatorname{\rm HS}_{k}(R;m)$, $j\in\mathcal{J}(l,D)$, we will write:

[TABLE]

(observe that we have chosen the decreasing ordering), where $F^{j}=\mathbb{I}$ if $j\not\in\mathcal{J}(l,D)$.

Proposition 2.9

Let $l>0$ be an integer and let us denote $s_{j}:=\max C_{j,l}^{p}$ for each integer $j$ with $1\leq j\leq p^{l}$. Then, for any $\left(p^{l}-1\right)-I$-logarithmic HS-derivation $D\in\operatorname{\rm HS}_{k}(R;p^{l})$ with $\ell(D)>1$, there exist:

• •

a $(p^{l-1}-1)-I$-logarithmic HS-derivation $T\in\operatorname{\rm HS}_{k}(R;p^{l-1})$,

• •

a $(p^{s_{j}+1}-1)-I$-logarithmic HS-derivation $F^{j}\in\operatorname{\rm HS}_{k}(R;p^{s_{j}+1})$, for each $j\in\mathcal{J}(l,D)$, and

• •

an $I$-differential operator $H$ of order $\leq p^{l}$

such that $T_{p^{l-1}}=D_{p^{l}}+H$ and

[TABLE]

where $\psi^{j}:R[|\mu|]_{p^{s_{j}+1}}\rightarrow R[|\mu|]_{p^{l}}$ is the substitution map given by $\psi^{j}(\mu)=\mu^{j}$.

Proof.

First, note that $\psi^{j}$ is well-defined for all $j\in\mathcal{J}(l,D)$ because $jp^{s_{j}+1}\geq p^{l}$ by definition of $s_{j}$. Moreover, observe that $\psi^{j}\bullet E=\tau_{jp^{s_{j}+1},p^{l}}(E[j])$ for any $E\in\operatorname{\rm HS}_{k}(R;p^{s_{j}+1})$. If $\ell(D)=\infty$, then $D=\mathbb{I}$, $\mathcal{J}(l,D)=\emptyset$ and we may take $T=\mathbb{I}$ to obtain the result. Let us suppose that $\ell(D)$ is finite, i.e. $1<\ell(D)\leq p^{l}$. We proceed by decreasing induction on $\ell(D)$.

Assume that $\ell(D)=p^{l}$. Then, $\mathcal{J}(l,D)=\emptyset$ and $D=(\operatorname{\rm Id},\delta)[p^{l}]=(\operatorname{\rm Id},\delta)\left[p^{l-1}\right][p]$ (see Corollary 1.8). So, if we put $T:=(\operatorname{\rm Id},\delta)\left[p^{l-1}\right]$, we have the result. Let us suppose that the proposition is true for all HS-derivations such that $\ell(\ast)>i$ and let us take a $(p^{l}-1)-I$-logarithmic HS-derivation $D\in\operatorname{\rm HS}_{k}(R;p^{l})$ with $1<\ell(D)=i<p^{l}$. We divide the proof in two cases:

If $i$ is a power of $p$.

Let us write $i=p^{t}$ where $t<l$. Since $\ell(D)>1$, then $t\geq 1$ and we can see $D\in\operatorname{\rm HS}_{k}(R;p^{t}p^{l-t})$. By Theorem 2.5, there exist an integral $F\in\operatorname{\rm HS}_{k}(R;p^{l-t})$ of $D_{p^{t}}$ and an $I$-differential operator $H$ of order $\leq p^{l-t}$ such that $F$ is $(p^{l-t}-1)-I$-logarithmic and $F_{p^{l-t}}=D_{p^{l}}+H$. Then, by Lemma 1.19, (c), and Proposition 1.10, $F^{\ast}[p^{t}]=(F[p^{t}])^{\ast}\in\operatorname{\rm HS}_{k}(R;p^{l})$ is $(p^{l}-1)-I$-logarithmic. Moreover, $(F[p^{t}])^{\ast}_{p^{t}}=F^{\ast}_{1}=-D_{p^{t}}$ and, by Lemma 1.19, (e), $(F[p^{t}])^{\ast}_{p^{l}}=F^{\ast}_{p^{l-t}}=-D_{p^{l}}-H+E$ where $E$ is an $I$-differential operator of order $\leq p^{l-t}$. We define $D^{\prime}:=(F[p^{t}])^{\ast}\circ D$. By Lemma 1.3, $\ell(D^{\prime})>i=p^{t}$ and, by Lemma 1.19, (d), $D^{\prime}$ is $(p^{l}-1)-I$-logarithmic and $D^{\prime}_{p^{l}}=D_{p^{l}}-D_{p^{l}}+H^{\prime}=H^{\prime}$ where $H^{\prime}$ is an $I$-differential operator of order $\leq p^{l}$. So, $D^{\prime}\in\operatorname{\rm HS}_{k}(\log I;p^{l})$. We apply the induction hypothesis to $D^{\prime}$ and we obtain that

[TABLE]

where $F^{j}\in\operatorname{\rm HS}_{k}(R;p^{s_{j}+1})$ is $(p^{s_{j}+1})-I$-logarithmic for $j\in\mathcal{J}(l,D^{\prime})$ and $T^{\prime}\in\operatorname{\rm HS}_{k}\left(R;p^{l-1}\right)$ is $\left(p^{l-1}-1\right)-I$-logarithmic with T^{\prime}_{p^{l-1}}=D^{\prime}_{p^{l}}+\mbox{some I-diff. op. of order}\leq p^{l}. Since $D^{\prime}\in\operatorname{\rm HS}_{k}(\log I;p^{l})$, we have that $T^{\prime}\in\operatorname{\rm HS}_{k}(\log I;p^{l-1})$. We put $F^{j}=\operatorname{\rm\mathbb{I}}\in\operatorname{\rm HS}_{k}(\log I;p^{s_{j}+1})$ for all $j\in\mathcal{J}(l,D)\setminus\mathcal{J}(l,D^{\prime})$. By Corollary 1.8,

[TABLE]

By Lemma 1.19, (a), $F[p^{t-1}]$ is $(p^{l-1}-1)-I$-logarithmic. Moreover, $F[p^{t-1}]_{p^{l-1}}=F_{p^{l-t}}=D_{p^{l}}+H$. So, by Lemma 1.19, (f), $T:=F\left[p^{t-1}\right]\circ T^{\prime}\in\operatorname{\rm HS}_{k}(R;p^{l-1})$ is $(p^{l-1}-1)-I$-logarithmic and T_{p^{l-1}}=F[p^{t-1}]_{p^{l-1}}+\mbox{ some }I\mbox{-diff. op. of order }\leq p^{l}=D_{p^{l}}+\mbox{some I-diff. op. of order }\leq p^{l} and we have the proposition in this case.

If $i$ is not a power of $p$.

Since $i$ is not a power of $p$, by Lemma 2.7, $ip^{s_{i}+1}>p^{l}$ where $s_{i}=\max C^{p}_{i,l}$. We consider $\tau_{p^{l},ip^{s_{i}}}(D)\in\operatorname{\rm HS}_{k}(\log I;ip^{s_{i}})$. If $s_{i}\geq 1$, then $D_{i}$ is $I$-logarithmically $p^{s_{i}}$-integrable by Theorem 2.5. If $s_{i}=0$, then $D_{i}\in\operatorname{\rm Der}_{k}(\log I)$. In both cases, since leaps only occur at powers of $p$ (Theorem 1.17 and Corollary 2.4), we have that $D_{i}$ is $I$-logarithmically $(p^{s_{i}+1}-1)$-integrable. Thanks to Proposition 2.1, we can integrate any $I$-logarithmic $(p^{s_{i}+1}-1)$-integral of $D_{i}$ so, there exists $F\in\operatorname{\rm HS}_{k}(R;p^{s_{i}+1})$ a $(p^{s_{i}+1}-1)-I$-logarithmic integral of $D_{i}$. Then, by Lemma 1.19 (c), $F^{\ast}[i]\in\operatorname{\rm HS}_{k}(R;ip^{s_{i}+1})$ is $ip^{s_{i}+1}-I$-logarithmic. By Proposition 1.10, $\psi^{i}\bullet F^{\ast}=\left(\psi^{i}\bullet F\right)^{\ast}\in\operatorname{\rm HS}_{k}(\log I;p^{l})$ and $(\psi^{i}\bullet F)^{\ast}_{i}=F[i]^{\ast}_{i}=-D_{i}$.

• a.

If $i\not\equiv 0\mod p$, then by Lemma 1.19 (f), and Lemma 1.3, $D^{\prime}:=D\circ(\psi^{i}\bullet F)^{\ast}$ is $(p^{l}-1)-I$-logarithmic with $\ell(D^{\prime})>i$ where $D^{\prime}_{p^{l}}=D_{p^{l}}+H$ with $H$ an $I$-differential operator of order $\leq p^{l}$. We apply the induction hypothesis to $D^{\prime}$ and we obtain that

[TABLE]

where $T\in\operatorname{\rm HS}_{k}(R;p^{l-1})$ is $(p^{l-1}-1)-I$-logarithmic with $T_{p^{l}-1}=D^{\prime}_{p^{l}}+\mbox{some }I\mbox{-diff. op. of order }\leq p^{l}=D_{p^{l}}+H^{\prime}$ where $H^{\prime}$ is an $I$-differential operator of order $\leq p^{l}$. Then, we put $F^{i}=F\in\operatorname{\rm HS}_{k}(R;p^{s_{i}+1})$ and $F^{j}=\operatorname{\rm\mathbb{I}}\in\operatorname{\rm HS}_{k}(\log I;p^{s_{j}+1})$ for $j\in\mathcal{J}(l,D)\setminus(\mathcal{J}(l,D^{\prime})\cup\{i\})$ and we have the result.

• b.

If $i$ is a multiple of $p$, then by Lemma 1.19 (d), and Lemma 1.3, $D^{\prime}:=\left(\psi^{i}\bullet F\right)^{\ast}\circ D$ is $(p^{l}-1)-I$-logarithmic with $\ell(D^{\prime})>i$ and $D^{\prime}_{p^{l}}=D_{p^{l}}+H$ where $H$ is an $I$-differential operator of order $\leq p^{l}$. Then, we apply the induction hypothesis to $D^{\prime}$ and we have that

[TABLE]

where $T^{\prime}\in\operatorname{\rm HS}_{k}(R;p^{l-1})$ is $(p^{l-1}-1)-I$-logarithmic with T^{\prime}_{p^{l-1}}=D^{\prime}_{p^{l}}+\mbox{some I-diff. op. of order }\leq p^{l}=D_{p^{l}}+H^{\prime} where $H^{\prime}$ is an $I$-differential operator of order $\leq p^{l}$. We put $F^{j}=\operatorname{\rm\mathbb{I}}$ for all $j\in\mathcal{J}(l,D)\setminus\mathcal{J}(l,D^{\prime})$. On the other hand, by Corollary 1.8 and Lemma 1.9,

[TABLE]

Since $F$ is $(p^{s_{i}+1}-1)-I$-logarithmic, $F[i/p]$ is $(ip^{s_{i}}-1)-I$-logarithmic by Lemma 1.19 (a), and, since $ip^{s_{i}}>p^{l-1}$, $\tau_{{ip^{s}_{i}},p^{l-1}}(F[i/p])\in\operatorname{\rm HS}_{k}(\log I,p^{l-1})$. By Lemma 1.19 (f), $T:=\tau_{ip^{s_{i}},p^{l-1}}(F[i/p])\circ T^{\prime}$ is $(p^{l-1}-1)-I$-logarithmic and T_{p^{l-1}}=T^{\prime}_{p^{l-1}}+\mbox{ some I-diff. op. of order }\leq p^{l}=D_{p^{l}}+H^{\prime\prime} where $H^{\prime\prime}$ is an $I$-differential operator of order $\leq p^{l}$. Since $D=T[p]\circ\left(\circ_{j\in\mathcal{J}(l,D)}\left(\psi^{j}\bullet F^{j}\right)\right)$, we have the proposition.

∎

Corollary 2.10

Let $l\geq 1$ be an integer and $D\in\operatorname{\rm HS}_{k}(R;p^{l})$ a $(p^{l}-1)-I$-logarithmic HS-derivation with $\ell(D)>1$. Then, there exist $F\in\operatorname{\rm HS}_{k}(\log I;p^{l})$ with $\ell(F)>1$ and a $(p^{l-1}-1)-I$-logarithmic HS-derivation $T\in\operatorname{\rm HS}_{k}(R;p^{l-1})$ such that $D=T[p]\circ F$.

Proof.

From Proposition 2.9, we have that

[TABLE]

for some $(p^{l-1}-1)-I$-logarithmic HS-derivation $T\in\operatorname{\rm HS}_{k}(R;p^{l-1})$ and some $(p^{s_{i}+1}-1)-I$-logarithmic HS-derivations $F^{i}\in\operatorname{\rm HS}_{k}(R;p^{s_{i}+1})$, for $i\in\mathcal{J}(l,D)$ and $s_{i}=\max C^{p}_{i,l}$. Since $\psi^{i}\bullet F^{i}=\tau_{ip^{s_{i}+1},p^{l}}(F^{i}[i])$ and $F^{i}[i]$ is $(ip^{s_{i}+1}-1)-I$-logarithmic by Lemma 1.19 (a), then $\psi^{i}\bullet F^{i}\in\operatorname{\rm HS}_{k}(\log I;p^{l})$ because $i\not\equiv 0\mod p$ and, by Lemma 2.7, $ip^{s_{i}+1}>p^{l}$. Hence, $F:=\circ_{i\in\mathcal{J}(l,D)}\left(\psi^{i}\bullet F^{i}\right)\in\operatorname{\rm HS}_{k}(\log I;p^{l})$. Moreover, $\ell(F^{i}[i])>1$ for all $j\in\mathcal{J}(l,D)$, so $\ell(\psi^{i}\bullet F^{i})>1$ and $\ell(F)>1$ by Lemma 1.3. ∎

3 Base change

Let $k$ be a commutative ring, $k\to L$ a ring extension and $A$ a commutative finitely generated $k$-algebra. We denote $A_{L}=L\otimes_{k}A$. In this section, we study the relationship between $\operatorname{\rm IDer}_{k}(A;m)$ and $\operatorname{\rm IDer}_{L}(A_{L};m)$ under suitable hypotheses on the ring extension $k\to L$.

3.1 Base change for derivations

For any commutative $k$-algebra $A$, let us denote $A_{L}:=L\otimes_{k}A$. For each $k$-derivation $\delta:A\to A$ let us denote by $\widetilde{\delta}:A_{L}\to A_{L}$ the natural $L$-linear extension given by $\widetilde{\delta}(c\otimes a)=c\otimes\delta(a)$ for all $c\in L$ and all $a\in A$, which is a $L$-derivation. The map $\delta\in\operatorname{\rm Der}_{k}(A)\mapsto\widetilde{\delta}\in\operatorname{\rm Der}_{L}(A_{L})$, being $A$-linear, gives rise to an $A_{L}$-linear base change map:

[TABLE]

The above map can be also described through the base change isomorphism for the module of differential forms $A_{L}\otimes_{A}\Omega_{A/k}=L\otimes_{k}\Omega_{A/k}\stackrel{{\scriptstyle\sim}}{{\to}}\Omega_{A_{L}/L}$, namely:

[TABLE]

If $I\subseteq A$ is an ideal, the map $\Phi^{L,A}:L\otimes_{k}\operatorname{\rm Der}_{k}(A)\to\operatorname{\rm Der}_{L}(A_{L})$ induces new $A_{L}$-linear maps:

[TABLE]

When $A=R=k[x_{1},\dots,x_{d}]$ is a polynomial ring, then $R_{L}=L[x_{1},\dots,x_{d}]$ is also a polynomial ring and since the module of derivations of a polynomial ring in a finite number of variables is free with basis the partial derivatives, we deduce that the map $\Phi^{L,R}$ is an isomorphism for an arbitrary ring extension $k\to L$.

We denote $I^{e}=IR_{L}=IL[x_{1},\ldots,x_{d}]$ the extended ideal of $I$ in $R_{L}$. It is clear that the following diagram is commutative:

[TABLE]

Moreover, it has exact rows and the left vertical arrow is surjective, and if $L$ is flat over $k$, then the top row is also left exact, the left vertical arrow is bijective and the middle vertical arrow is injective.

Proposition 3.1

Under the above hypotheses, if $k\to L$ is a flat ring extension, then the following properties are equivalent:

1. (a)

The map $\Phi^{L,R}_{\text{\rm ind}}:L\otimes_{k}\operatorname{\rm Der}_{k}(\log I)\to\operatorname{\rm Der}_{L}(\log I^{e})$ is an isomorphism. 2. (b)

The map $\Phi^{L,A}:L\otimes_{k}\operatorname{\rm Der}_{k}(A)\to\operatorname{\rm Der}_{L}(A_{L})$ is an isomorphism.

Moreover, both properties hold if $I$ is finitely generated (i.e. if $A$ is finitely presented over $k$).

Proof.

The equivalence (a) $\Leftrightarrow$ (b) comes from the five lemma. The last statement is well known (cf. [Gr, Prop. (16.5.11)]) but for the sake of completeness we recall its proof: from the second fundamental exact sequence

[TABLE]

we deduce that, if $I$ is finitely generated, then $\Omega_{A/k}$ is a finitely presented $A$-module and so

[TABLE]

∎

We also have the following result for any ideal $I\subset R=k[x_{1},\ldots,x_{d}]$ and for any finitely generated $k$-algebra $A=R/I$.

Proposition 3.2

Under the above hypotheses, if $k\to L$ is a free ring extension ($L$ is a free $k$-module) and $A=R/I$ is a finitely generated $k$-algebra, then properties (a) and (b) in Proposition 3.1 hold.

Proof.

Since $L$ is a (faithfully) flat extension of $k$, after Proposition 3.1 we only need to prove that the map $\Phi^{L,R}_{\text{\rm ind}}:L\otimes_{k}\operatorname{\rm Der}_{k}(\log I)\to\operatorname{\rm Der}_{L}(\log I^{e})$ is surjective. Let $\mathcal{B}=\{a_{i},i\in\mathcal{I}\}$ be a $k$-basis of $L$ and $\varepsilon:R_{L}\to R_{L}$ an $I^{e}$-logarithmic derivation. Since $\mathcal{B}$ is also a $R$-basis of $R_{L}$, there is a finite subset $\mathcal{I}_{0}\subseteq\mathcal{I}$ and unique elements $r_{ji}\in R$, $1\leq j\leq d$, $i\in\mathcal{I}_{0}$ such that $\varepsilon(x_{j})=\sum_{i\in\mathcal{I}_{0}}r_{ji}a_{i}$ for all $j=1,\dots,d$. We have

[TABLE]

and for each $f\in I$ we have $\varepsilon(f)=\sum_{i\in\mathcal{I}_{0}}a_{i}\delta_{i}(f)\in I^{e}$ and so $\delta_{i}(f)\in I$. We deduce that each $\delta_{i}$ is $I$-logarithmic and so $\varepsilon$ belongs to the image of $\Phi^{L,R}_{\text{\rm ind}}$. ∎

3.2 Base change for integrable derivations

Proposition 3.3

Let $A$ be a $k$-algebra, $I\subset A$ an ideal, $k\to L$ a ring extension, $I^{e}=IA_{L}$ the extended ideal and $m\in\overline{\mathbb{N}}$. For any HS-derivation $D\in\operatorname{\rm HS}_{k}(A;m)$, there is a unique HS-derivation $\widetilde{D}\in\operatorname{\rm HS}_{L}(A_{L};m)$ such that the following diagram is commutative:

[TABLE]

Moreover, if $D$ is $I$-logarithmic, then $\widetilde{D}$ is $I^{e}$-logarithmic.

Observe that for $m=1$, we know that $\operatorname{\rm Der}_{k}(R)\equiv\operatorname{\rm HS}_{k}(R;1)$ and the extension process $D\mapsto\widetilde{D}$ described in Proposition 3.3 coincides with the usual extension $\delta\mapsto\widetilde{\delta}$ of derivations.

Lemma 3.4

Let $A$ be a $k$-algebra, $I\subset A$ an ideal, $k\to L$ a ring extension, $m\in\overline{\mathbb{N}}$, $n\leq m$, $D\in\operatorname{\rm HS}_{k}(A;m)$ a HS-derivation and $\psi:A[|\mu|]_{m}\rightarrow A[|\mu|]_{n}$ a substitution map. The following properties hold:

1. (1)

The map $D\in\operatorname{\rm HS}_{k}(A;m)\mapsto\widetilde{D}\in\operatorname{\rm HS}_{L}(A_{L};m)$ is a group homomorphism. 2. (2)

$\widetilde{\psi\bullet D}=\widetilde{\psi}\bullet\widetilde{D}$, where $\widetilde{\psi}:A_{L}[|\mu|]_{m}\rightarrow A_{L}[|\mu|]_{n}$ is the substitution map induced by $\psi$. 3. (3)

If $D$ is $n-I$-logarithmic, then $\widetilde{D}$ is $n-I^{e}$-logarithmic.

Lemma 3.5

Let $I\subseteq A$ be an ideal, $B=A/I$ and $I^{e}=IA_{L}$ the extended ideal. Then, for each $D\in\operatorname{\rm HS}_{k}(\log I;m)$,

[TABLE]

(where $\widetilde{\Pi^{I}_{\operatorname{\rm HS},m}(D)}$ is the extension of $\Pi^{I}_{\operatorname{\rm HS},m}(D)\in\operatorname{\rm HS}_{k}(B;m)$ to $B_{L}=A_{L}/I^{e}$ and $\widetilde{D}\in\operatorname{\rm HS}_{L}(\log I^{e};m)\subseteq\operatorname{\rm HS}_{L}(A_{L};m)$).

Corollary 3.6

Under the hypotheses of Lemma 3.4, let $\delta:A\to A$ be a $k$-derivation (resp. an $I$-logarithmic $k$-derivation). If $\delta$ is $m$-integrable (resp. $I$-logarithmically $m$-integrable), then $\widetilde{\delta}$ is also $m$-integrable (resp. $I^{e}$-logarithmically $m$-integrable).

Proof.

Let us suppose that $\delta\in\operatorname{\rm IDer}_{k}(A;m)$ and let us consider an $m$-integral $D\in\operatorname{\rm HS}_{k}(A;m)$ of $\delta$, i.e. $D_{1}=\delta$. From Proposition 3.3, $\widetilde{D}\in\operatorname{\rm HS}_{L}(A_{L};m)$ is an $m$-integral of $\widetilde{D}_{1}=\widetilde{\delta}$, i.e. $\widetilde{\delta}\in\operatorname{\rm IDer}_{k}(A;m)$. Moreover, if $\delta\in\operatorname{\rm IDer}_{k}(\log I;m)$, then we can consider $D\in\operatorname{\rm HS}_{k}(\log I;m)$ and, by Proposition 3.3, $\widetilde{D}\in\operatorname{\rm HS}_{L}(\log I^{e};m)$. Hence, $\widetilde{\delta}\in\operatorname{\rm IDer}_{L}(\log I^{e};m)$. ∎

As a consequence of the above corollary, base change maps $\Phi^{L,A}:L\otimes_{k}\operatorname{\rm Der}_{k}(A)\to\operatorname{\rm Der}_{L}(A_{L})$ and $\Phi^{L,A}_{\text{\rm ind}}:L\otimes_{k}\operatorname{\rm Der}_{k}(\log I)\to\operatorname{\rm Der}_{L}(\log I^{e})$ induce, for each $m\in\overline{\mathbb{N}}$, new $A_{L}$-linear base change maps:

[TABLE]

From now on, we assume that $L$ is flat over $k$ and $A$ a finitely generated $k$-algebra. Then, we can put $A=R/I$ where $R=k[x_{1},\ldots,x_{d}]$ is a polynomial ring and $I\subset R$ an ideal.

From the exact sequence in Proposition 2.3, we obtain for each $m\in\overline{\mathbb{N}}$ a commutative diagram with exact rows (compare with (3)):

[TABLE]

Moreover the left vertical arrow is bijective and the middle vertical arrow is injective.

The proof of the following lemma is clear.

Lemma 3.7

Under the above hypotheses, the following properties hold:

$\Phi^{L,A}_{m}$* is injective.*

• 2.

$\Phi^{L,R}_{\text{\rm ind},m}$* is surjective if and only if $\Phi^{L,A}_{m}$ is surjective.*

Moreover, we have the following result about leaps.

Lemma 3.8

Assume that $L$ is faithfully flat over $k$ and $A$ a finitely generated $k$-algebra. If $\Phi^{L,A}_{m}$ is surjective for all $m\geq 1$ then,

[TABLE]

Proof.

Since $L$ is flat over $k$, we have that $\Phi^{L,A}_{m}$ is bijective so, $\operatorname{\rm IDer}_{L}(A_{L};m)=\operatorname{\rm IDer}_{L}(A_{L};m-1)$ if and only if

[TABLE]

Since $L$ is faithfully flat over $k$, the last equality holds if and only if $\operatorname{\rm IDer}_{k}(A;m-1)/\operatorname{\rm IDer}_{k}(A;m)=0$ and we have the result.

∎

In the rest of this section, we will study the surjectivity of $\Phi_{m}^{L,A}$. Let us start by giving a counterexample.

Counterexample 3.9

Let us consider $k=\mathbb{F}_{2}(s,t)$ the quotient field of $\mathbb{F}_{2}[s,t]$ and $L=\overline{k}$ the perfect closure of $k$. We denote $A:=k[x,y]/\langle h\rangle$ where $h\in k[x,y]$ is the irreducible polynomial $x^{2}+y^{2}+tx^{4}+sy^{4}$. Then, $\Phi^{L,A}_{4}$ is not surjective.

To prove this counterexample, we need to calculate the 4-integrable derivations of $A$ (resp. $A_{L}$) over $k$ (resp. over $L$). To do this, we use two general results:

Proposition 3.10

[Ti1, Prop. 2.10]* Let $k$ be a unique factorization domain of characteristic $p>0$, $R=k[x_{1},\ldots,x_{d}]$ the polynomial ring over $k$ and $h$ a polynomial of $R$. For all $n\in\mathbb{N}$, we have:*

[TABLE]

Proposition 3.11

[Na1, Prop. 2.2.4]* Let $h\in R=k[x_{1},\ldots,x_{d}]$, $I=\langle h\rangle$ and $J^{0}=\langle\partial_{1}(h),\ldots,\partial_{d}(h)\rangle$ the gradient ideal. If $\delta:R\rightarrow R$ is an $I$-logarithmic $k$-derivation with $\delta\in J^{0}\operatorname{\rm Der}_{k}(R)$, then $\delta$ admits an $I$-logarithmic integral $D\in\operatorname{\rm HS}_{k}(\log I)$ with $D_{i}(h)=0$ for all $i>i$. In particular, if $\delta(h)=0$, the integral $D$ can be taken with $\varphi_{D}(h)=h$.*

Proof of Counterexample 3.9. To calculate the $m$-integrable derivations of $A$, we will follow the same step of Example 7 of [Ma1]. Let us suppose that there exists $\delta\in\operatorname{\rm IDer}_{k}(A;m)$ and $D\in\operatorname{\rm HS}_{k}(A;m)$ an integral of $\delta$. Let us consider

[TABLE]

where $u_{i},v_{i}\in A$. To $\varphi_{D}$ be well defined, we need $\varphi_{D}(h)=0$, i.e.

[TABLE]

By looking at the coefficient of $\mu^{2}$ in the previous equation, we deduce $u_{1}^{2}+v_{1}^{2}=(u_{1}+v_{1})^{2}=0$, and since $A$ is a domain, $u_{1}=v_{1}$. By looking at the coefficient of $\mu^{4}$, we deduce $u_{2}^{2}+v_{2}^{2}+tu^{4}_{1}+sv^{4}_{1}=0$. We can write $w=u_{2}+v_{2}$ and $u=u_{1}=v_{1}$, and we obtain the equation:

[TABLE]

Let $W$ and $U$ be elements of $k[x,y]$ such that $W+\langle h\rangle=w$ and $U+\langle h\rangle=u$. Then, thanks to the previous equation:

[TABLE]

for some $G\in k[x,y]$. By applying the partial derivatives $\partial_{s}$ and $\partial_{t}$ to (5), we obtain:

[TABLE]

Then, if $g:=G+\langle h\rangle$, we have the following equalities in $A$:

[TABLE]

Since $A$ is a domain and $x^{4}\neq y^{4}$, $g=0$, so $u=u_{1}=v_{1}=0$. Then, we can not integrate any non-zero $k$-derivation until length 4, i.e. $\operatorname{\rm IDer}_{k}(A;4)=0$ and $L\otimes_{k}\operatorname{\rm IDer}_{k}(A;4)=0$.

To prove that $\operatorname{\rm IDer}_{L}(A_{L};4)$ is not zero, we calculate $\operatorname{\rm IDer}_{L}(\log\langle h\rangle^{e};4)$. Thanks to Proposition 3.10, it is enough to calculate $\operatorname{\rm IDer}_{L}(\log H;2)$ where $H=x+y+t^{1/2}x^{2}+s^{1/2}y^{2}$. Note that $J^{0}=\langle 1\rangle$ so, by Proposition 3.11, any $I$-logarithmic $k$-derivation is integrable. It is easy to see that $\operatorname{\rm Der}_{L}(\log H)=\langle\partial_{x}+\partial_{y},H\partial_{x}\rangle$. Hence, thanks to Proposition 2.3, $\operatorname{\rm IDer}_{L}(A_{L};4)=\langle\overline{\delta}_{1},\overline{\delta}_{2}\rangle\neq 0$ where $\overline{\delta}_{1}$ (resp. $\overline{\delta}_{2}$) is the derivation induced by $\partial_{x}+\partial_{y}$ (resp. $H\partial_{x}$) in the quotient. Therefore, $\Phi_{4}^{L,A}$ is not surjective.

$\square$

We have seen that $\Phi_{m}^{L,A}$ is not surjective in general, however, if we assume that $L$ is not only flat, but satisfies some additional conditions, then $\Phi^{L,A}_{m}$ will be surjective for all $m\geq 1$ and any finitely generated $k$-algebra $A$.

3.2.1 The extension is a polynomial ring

In this section, we assume that $k$ is a commutative ring and $L:=k[t_{i}\ |\ i\in\mathcal{I}]$ is a polynomial ring in an arbitrary number of variables. We define $\mathbb{N}^{(\mathcal{I})}=\{\alpha:=(\alpha_{i})_{i\in\mathcal{I}}\ |\ \alpha_{i}=0\mbox{ \it except for a finite number of }i\in\mathcal{I}\}$ and, for $\alpha\in\mathbb{N}^{(\mathcal{I})}$, we put $t^{\alpha}=\prod_{i\in\mathcal{I}}t_{i}^{\alpha_{i}}$. We start with some numerical results.

The following lemma is clear.

Lemma 3.12

Let $n\leq m$ be two positive integers. We have the following properties.

• a.

$\left(\lfloor m/n\rfloor+1\right)n-1\geq m$.

• b.

If $m\neq 0\mod n$, then $\lfloor m/n\rfloor=\lfloor(m-1)/n\rfloor$. Otherwise, $\lfloor m/n\rfloor=\lfloor(m-1)/n\rfloor+1$.

• c.

If $n<m$ such that $m=0\mod n$. Then, there exists a prime factor of $m$ which divides $m/n$.

Definition 3.13

Let $n$ be a positive integer. We define

[TABLE]

Lemma 3.14

Let $n,s$ be two positive integers such that $n\neq s$. Then, there do not exist $\alpha\in\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{n}$ and $\eta\in\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{s}$ such that $\alpha s=\eta n$.

Proof.

Suppose that there exist $\alpha\in\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{n}$ and $\eta\in\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{s}$ such that $\alpha s=\eta n$. If there were such a prime that divides $n$ and $s$, then we could simplify it. So, we can assume that $s$ and $n$ do not have prime factors in common. Now, as $s$ and $n$ are not the same, one of them, we say $s$, has a prime factor $q$ such that does not divide to the another one, in this case $n$. Since $\alpha s=\eta n$, we have that $\alpha_{i}s=\eta_{i}n$ for all $i\in\mathcal{I}$. So, $q$ divide $\eta_{i}$ for all $i\in\mathcal{I}$. Then $\eta=q\eta^{\prime}\in\mathcal{P}_{s}$ and we have a contradiction. ∎

Fix $m>1$ an integer and consider $m=q_{1}^{a_{1}}\cdots q_{s}^{a_{s}}$ its prime factorization, i.e. for all $j=1,\ldots s$, $q_{j}$ is a prime, $a_{j}>0$ and $q_{j}\neq q_{i}$ if $i\neq j$. Let us consider $\beta\in\mathcal{P}_{m}$. Then, we can write $\beta=q_{1}^{b_{1}}\cdots q_{s}^{b_{s}}\eta$ where $b_{j}\geq 1$ for some $j\in\{1,\ldots,s\}$ and $\eta\in\mathbb{N}^{(\mathcal{I})}$ such that $q_{j}\nmid\eta$ for any $j=1,\ldots,s$, i.e. for all $j$ there exists $\eta_{i_{j}}$ with $i_{j}\in\mathcal{I}$ such that $q_{j}\nmid\eta_{i_{j}}$. We can assume, without loss of generality, that there exists an integer $l_{\beta}$ such that $0\leq l_{\beta}\leq s$ and $a_{j}>b_{j}$ for all $j\leq l_{\beta}$ and $a_{j}\leq b_{j}$ for all $j>l_{\beta}$. Then, we define

[TABLE]

Lemma 3.15

For each $\beta\in\mathcal{P}_{m}$, there exists a unique $n\in\mathbb{N}$ with $1\leq n<m$ such that $m=0\mod n$ and $\beta n/m\not\in\mathcal{P}_{n}$.

Proof.

We write $\beta=q_{1}^{b_{1}}\cdots q_{s}^{b_{s}}\eta$, where $\eta\in\mathbb{N}^{(\mathcal{I})}$ such that $q_{j}\nmid\eta$ for any $j=1,\ldots,s$ and $b_{j}\geq 1$ for some $j\in\{1,\ldots,s\}$. We take $n=n_{\beta}$. It is obvious that $n$ divides $m$ and $1\leq n<m$. We denote $l:=l_{\beta}$ to simplify the notation. We put

[TABLE]

If $l=0$, then $n=1$ and $\mathcal{P}_{1}=\emptyset$ so, $\alpha\not\in\mathcal{P}_{n}$ (note that $\alpha\in\mathbb{N}^{(\mathcal{I})}$ because if $l=0$, then $b_{j}\geq a_{j}$ for all $j=1,\ldots,s$). If $l\geq 1$, then

[TABLE]

Note that set of primes which divide $n$ is $\{q_{1},\ldots,q_{l}\}$. Hence, $q_{j}\nmid\alpha$ for all $j=1,\ldots,l$ (recall that $q_{j}\nmid\eta$). So, $\alpha\not\in\mathcal{P}_{n}$.

Now, let us suppose that there exists another $n^{\prime}\in\mathbb{N}$ holding the lemma, in particular $\alpha^{\prime}:=\beta n^{\prime}/m\not\in\mathcal{P}_{n^{\prime}}$. Then, $\alpha n^{\prime}=\alpha^{\prime}n$ and we have a contradiction by Lemma 3.14. ∎

Theorem 3.16

Let $m\geq 1$ be an integer and $L=k[t_{i}\ |\mbox{ }i\in\mathcal{I}]$ a polynomial ring. Let us consider $D\in\operatorname{\rm HS}_{L}(R_{L};m)$. Then, for all $n=1,\ldots,m$ there exist a finite subset $L_{n}$ of $\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{n}$ and $N^{n,\alpha}\in\operatorname{\rm HS}_{k}(R)$ for each $\alpha\in L_{n}$ such that

[TABLE]

where $\psi^{n,m}_{\alpha}:R_{L}[|\mu|]\rightarrow R_{L}[|\mu|]_{m}$ is the substitution map of constant coefficients given by $\psi_{\alpha}^{n,m}(\mu)=t^{\alpha}\mu^{n}$.

Proof.

First, observe that, if $E\in\operatorname{\rm HS}_{L}(R_{L};m)$ then, $\psi_{\alpha}^{n,m}\bullet E=\tau_{\infty,m}\left(\left(t^{\alpha}\bullet E\right)[n]\right)$. We prove this theorem by induction on $m$. Assume that $m=1$ then, $D=(\operatorname{\rm Id},D_{1})\in\operatorname{\rm HS}_{L}(R_{L};1)$. Since $L$ is free over $k$ and $\{t^{\alpha}\mbox{ }|\mbox{ }\alpha\in\mathbb{N}^{(\mathcal{I})}\}$ is a $k$-basis of $L$, from the proof of Proposition 3.2, $D_{1}\in\operatorname{\rm Der}_{L}(R_{L})$ can be written as $D_{1}=\sum_{\alpha\in L_{1}}t^{\alpha}\widetilde{\delta_{\alpha}}$ where $L_{1}$ is a finite subset of $\mathbb{N}^{(\mathcal{I})}$ and $\delta_{\alpha}\in\operatorname{\rm Der}_{k}(R)$ for each $\alpha\in L_{1}$. Let us consider $N^{1,\alpha}$ an integral of $\delta_{\alpha}$ for $\alpha\in L_{1}$. Then, $\widetilde{N^{1,\alpha}}\in\operatorname{\rm HS}_{L}(R_{L})$ is an integral of $\widetilde{\delta}_{\alpha}$ (see Corollary 3.6). Hence,

[TABLE]

(note that the order of the composition in this equality is not important because $\operatorname{\rm HS}_{L}(R_{L};1)\equiv\operatorname{\rm Der}_{L}(R)$ is a commutative group) and we have the result when $m=1$. Let us assume that the theorem is true for any HS-derivation of length $m-1$ and we will prove it for $D\in\operatorname{\rm HS}_{L}(R_{L};m)$. By induction hypothesis, for all $n=1,\ldots,m-1$, there exist a finite subset $L_{n}^{\prime}$ of $\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{n}$ and $N^{n,\alpha}\in\operatorname{\rm HS}_{k}(R)$ for all $\alpha\in L^{\prime}_{n}$ such that

[TABLE]

We define

[TABLE]

where the composition is taken in the same order that in (6). Note that $\psi^{n,m-1}_{\alpha}=\tau_{m,m-1}\circ\psi^{n,m}_{\alpha}$, and thanks to Lemma 1.7 and Corollary 1.8, we have that:

[TABLE]

Then, by Lemma 1.11, $D=E\circ(\operatorname{\rm Id},\delta)[m]$ where $\delta\in\operatorname{\rm Der}_{L}(R_{L})$. From the proof of Proposition 3.2, $\delta=\sum_{\beta\in\mathcal{J}}t^{\beta}\widetilde{\delta_{\beta}}$ where $\mathcal{J}$ is a finite subset of $\mathbb{N}^{(\mathcal{I})}$ and $\delta_{\beta}\in\operatorname{\rm Der}_{k}(R)$ for all $\beta\in\mathcal{J}$. We denote $\Gamma=\{n\in\mathbb{N}\mbox{ }|\mbox{ }1\leq n\leq m-1,\mbox{ }m=0\mod n\}$. For all $n\in\Gamma$, we define

[TABLE]

Claim 1. For all $n,s\in\Gamma$ such that $n\neq s$, then $\mathcal{J}_{n}\cap\mathcal{J}_{s}=\emptyset$.

Let us suppose that there exists $\beta\in\mathcal{J}_{n}\cap\mathcal{J}_{s}$. In this case, there exist $\alpha\in L^{\prime}_{n}\subseteq\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{n}$ and $\eta\in L^{\prime}_{s}\subseteq\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{s}$ such that $\beta=\alpha(m/n)=\eta(m/s)$, i.e. $\alpha s=\eta n$ and this can not happen by Lemma 3.14.

Claim 2. $\mathcal{L}_{m}\cap\mathcal{J}_{n}=\emptyset$* for all $n\in\Gamma$.*

By Lemma 3.12 c., there exists a prime factor, $q$, of $m$ that divides $m/n$. Assume that $\beta\in\mathcal{J}_{n}$. Then, we have that $\beta=\alpha(m/n)$ for some $\alpha\in L_{n}^{\prime}$. Then, $q|\beta$ so, $\beta\in\mathcal{P}_{m}$.

Let us write $\mathcal{J}=\sqcup_{n\in\Gamma}\mathcal{J}_{n}\sqcup\mathcal{L}_{m}\sqcup\overline{\mathcal{J}}$ where $\overline{\mathcal{J}}=\mathcal{J}\setminus\left(\sqcup_{n\in\Gamma}\mathcal{J}_{n}\sqcup\mathcal{L}_{m}\right)$. Observe that $\overline{\mathcal{J}}\subseteq\mathcal{P}_{m}$ so, from Lemma 3.15, for all $\beta\in\overline{\mathcal{J}}$, there exists a unique $n_{\beta}\in\Gamma$ such that $(\beta n_{\beta})/m\not\in\mathcal{P}_{n_{\beta}}$. Therefore, if we denote $\overline{\mathcal{J}}_{n}=\{\beta\in\overline{\mathcal{J}}\mbox{ }|\mbox{ }n_{\beta}=n\}$ for all $n\in\Gamma$, we can write

[TABLE]

Now, for each $n\in\Gamma$ we can define

[TABLE]

Note that $\mathcal{L}^{\prime}_{n}\cap\overline{\mathcal{L}}_{n}=\emptyset$. Let us denote $\mathcal{L}_{n}=\mathcal{L}^{\prime}_{n}\cup\overline{\mathcal{L}}_{n}$. Hence, we can express

[TABLE]

By Corollary 1.8 and Lemma 1.9, for each $n\in\Gamma\cup\{m\}$ and $\alpha\in\mathcal{L}_{n}$, we have that:

[TABLE]

For each $n\in\Gamma\cup\{m\}$ and $\alpha\in\mathcal{L}_{n}$, let us consider $M^{n,\alpha}\in\operatorname{\rm HS}_{k}(R)$ an integral of $\delta_{\alpha(m/n)}$. We know that $\widetilde{M^{n,\alpha}}$ is an integral of $\widetilde{\delta_{\alpha(m/n)}}$, so $\widetilde{M^{n,\alpha}}[m/n]$ is an integral of $\left(\operatorname{\rm Id},\widetilde{\delta_{\alpha(m/n)}}\right)[m/n]$. Hence, by Lemma 1.9, we have

[TABLE]

To simplify the following expression, we put $\overline{\mathcal{L}}_{n}=\mathcal{L}^{\prime}_{n}=\emptyset$ for all $n\in\{1,\ldots,m-1\}\setminus\Gamma$. Moreover, for all $n\in\{1,\ldots,m-1\}$, if $\alpha\in L_{n}^{\prime}\setminus\mathcal{L}^{\prime}_{n}$ then we consider $\delta_{\alpha(m/n)}=0$ and $M^{n,\alpha}=\operatorname{\rm\mathbb{I}}\in\operatorname{\rm HS}_{k}(R)$ an integral of $\delta_{\alpha(m/n)}$. Thanks to Lemmas 1.12 and 1.7 and the previous equation, we can write:

[TABLE]

Thanks to Lemma 3.4 (2), $\widetilde{M^{n,\alpha}}[m/n]$ is the extension of the HS-derivation $M^{n,\alpha}[m/n]$ and, by Lemma 3.4 (1), $\widetilde{N^{n,\alpha}}\circ\widetilde{M^{n,\alpha}}[m/n]$ is the extension of $N^{n,\alpha}\circ M^{n,\alpha}[m/n]$. Therefore, if we denote $L_{n}=L_{n}^{\prime}\cup\overline{\mathcal{L}}_{n}\subseteq\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{n}$ and $L_{m}=\mathcal{L}_{m}$, we have the theorem. ∎

Theorem 3.17

Let $m\geq 1$ be an integer, $L=k[t_{i}\mbox{ }|\mbox{ }i\in\mathcal{I}]$ a polynomial ring, $I\subseteq R$ an ideal and $D\in\operatorname{\rm HS}_{L}(\log I^{e};m)$. For all $n=1,\ldots,m$, let $L_{n}$ be a finite subset of $\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{n}$ and $N^{n,\alpha}\in\operatorname{\rm HS}_{k}(R)$ for all $\alpha\in L_{n}$ such that

[TABLE]

where $\psi_{\alpha}^{n,m}:R_{L}[|\mu|]\to R_{L}[|\mu|]_{m}$ is the substitution map given by $\psi^{n,m}_{\alpha}(\mu)=t^{\alpha}\mu^{n}$. Then, for all $n=1,\ldots,m$ and $\alpha\in L_{n}$, $N^{n,\alpha}\in\operatorname{\rm HS}_{k}(R)$ is an $\lfloor m/n\rfloor-I$-logarithmic HS-derivation.

Proof.

We prove this result by induction on $m$. If $m=1$, we have to prove that $N^{1,\alpha}$ is $1-I$-logarithmic for all $\alpha\in L_{1}$, i.e. $N^{1,\alpha}_{1}\in\operatorname{\rm Der}_{k}(\log I)$ for all $\alpha\in L_{1}$. In this case,

[TABLE]

Since $D_{1}$ is $I^{e}$-logarithmic, doing the same process of the proof of Proposition 3.2, we have that $N^{n,\alpha}$ is $1-I$-logarithmic. Assume that the theorem is true for all $I^{e}$-logarithmic HS-derivation of length $m-1$ and let us take $D\in\operatorname{\rm HS}_{L}(\log I^{e};m)$ such that

[TABLE]

where $L_{n}\subseteq\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{n}$ is a finite set and $N^{n,\alpha}\in\operatorname{\rm HS}_{k}(R)$ for all $\alpha\in L_{n}$ and $n=1,\ldots,m$. By Corollary 1.8, we have that

[TABLE]

From Lemma 1.7, for any $E\in\operatorname{\rm HS}_{L}(R_{L})$, $\tau_{m,m-1}\bullet(\psi^{n,m}_{\alpha}\bullet(E))=(\tau_{m,m-1}\circ\psi^{n,m}_{\alpha})\bullet E=\psi^{n,m-1}_{\alpha}\bullet E$. Moreover, $\psi^{m,m-1}_{\alpha}\bullet E=\operatorname{\rm\mathbb{I}}$. So,

[TABLE]

Hence, since $\tau_{m,m-1}(D)\in\operatorname{\rm HS}_{L}(\log I^{e};m-1)$, we can apply the induction hypothesis and we deduce that $N^{n,\alpha}\in\operatorname{\rm HS}_{k}(R)$ is $\lfloor(m-1)/n\rfloor-I$-logarithmic for all $\alpha\in L_{n}$ and $n=1,\ldots,m-1$. We define

[TABLE]

where the order of the composition in $E^{n}$ is the same that in $D$.

Claim. $E^{n}$* is $(m-1)-I^{e}$-logarithmic:* Since $N^{n,\alpha}$ is $\lfloor(m-1)/n\rfloor-I$-logarithmic, by Lemma 3.4 (3), $t^{\alpha}\bullet\widetilde{N^{n,\alpha}}$ is $\lfloor(m-1)/n\rfloor-I^{e}$-logarithmic. From Lemma 1.19 (a), $\left(t^{\alpha}\bullet\widetilde{N^{n,\alpha}}\right)[n]$ is $\left((\lfloor(m-1)/n\rfloor+1)n-1\right)-I^{e}$-logarithmic. By Lemma 3.12 a., $m-1<(\lfloor(m-1)/n\rfloor+1)n-1$, so $\psi^{n,m}_{\alpha}\bullet\widetilde{N^{n,\alpha}}$ is $(m-1)-I^{e}$-logarithmic because $\psi^{n,m}_{\alpha}\bullet\ast=\tau_{\infty,m}\left((t^{\alpha}\bullet\ast)[n]\right)$. Hence, by Lemma 1.19 (d), $E^{n}$ is $(m-1)-I^{e}$-logarithmic for all $n$.

Let us consider $n\in\{1,\ldots,m\}$ such that $n\nmid m$. Then, by Corollary 1.8,

[TABLE]

Hence, $E^{n}_{m}=0$. Moreover, by Lemma 3.12 b., $\lfloor(m-1)/n\rfloor=\lfloor m/n\rfloor$, so $N^{n,\alpha}$ is $\lfloor m/n\rfloor-I$-logarithmic. Therefore, to prove the theorem we have to show that $N^{n,\alpha}$ is $(m/n)-I$-logarithmic for $n|m$. By Lemma 3.12 b., $m/n=\lfloor(m-1)/n\rfloor+1$ and, since $N^{n,\alpha}$ is $\lfloor(m-1)/n\rfloor-I$-logarithmic, it is enough to prove that $N^{n,\alpha}_{m/n}(I)\subseteq I$. Note that

[TABLE]

where ${\left(\widetilde{N^{n,\alpha}}\right)_{m/n}}_{|R}=N^{n,\alpha}_{m/n}$. Therefore, by Lemma 1.19 (d),

[TABLE]

where $F_{n}$ is an $I^{e}$-differential operator. Hence, again by Lemma 1.19 (d),

[TABLE]

where $F$ is an $I^{e}$-differential operator. Since $D_{m}$ is also an $I^{e}$-differential operator, we have that

[TABLE]

Observe that $\alpha(m/n)\neq\eta(m/s)$ for all $\alpha\in L_{n}$ and $\eta\in L_{s}$ by Lemma 3.14 because $L_{n}\subseteq\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{n}$ and $L_{s}\in\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{s}$. Doing the same process than in the proof of Proposition 3.2, we can deduce that $N^{n,\alpha}\in\operatorname{\rm HS}_{k}(R)$ is $\lfloor m/n\rfloor-I$-logarithmic for all $\alpha\in L_{n}$ and $n=1,\ldots,m$. ∎

Theorem 3.18

Let $m\geq 1$ be an integer, $L=k[t_{i}\mbox{ }|\mbox{ }i\in\mathcal{I}]$ a polynomial ring, $A$ a finitely generated $k$-algebra and $E\in\operatorname{\rm HS}_{L}(A_{L};m)$. Then, for all $n=1,\ldots,m$ there exist a finite subset $L_{n}\subseteq\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{n}$ and $M^{n,\alpha}\in\operatorname{\rm HS}_{k}(A;\lfloor m/n\rfloor)$ for each $\alpha\in L_{n}$ such that

[TABLE]

where $\phi^{n,m}_{\alpha}:A_{L}[|\mu|]_{\lfloor m/n\rfloor}\to A_{L}[|\mu|]_{m}$ is the substitution map of constant coefficients given by $\phi^{n,m}_{\alpha}(\mu)=t^{\alpha}\mu^{n}$.

Proof.

Since $A$ is a finitely generated $k$-algebra, we can take $A=R/I$ where $R=k[x_{1},\ldots,x_{d}]$ and $I\subseteq R$ an ideal. By Proposition 2.2, there exists $D\in\operatorname{\rm HS}_{k}(\log I^{e};m)$ such that $\Pi_{\operatorname{\rm HS};m}^{I^{e}}(D)=E$. By theorems 3.16 and 3.17, for all $n=1,\ldots,m$ there exist a finite subset $L_{n}$ of $\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{n}$ and an $\lfloor m/n\rfloor-I$-logarithmic HS-derivation $N^{n,\alpha}\in\operatorname{\rm HS}_{k}(R)$ such that

[TABLE]

where $\psi^{n,m}_{\alpha}:R_{L}[\mu|]\to R_{L}[|\mu|]_{m}$ is the substitution map given by $\psi^{n,m}_{\alpha}(\mu)=t^{\alpha}\mu^{n}$.

Let us consider $\theta^{n,m}_{\alpha}:R_{L}[|\mu|]_{\lfloor m/n\rfloor}\to R_{L}[|\mu|]_{m}$ the substitution map given by $\theta^{n,m}_{\alpha}(\mu)=t^{\alpha}\mu^{n}$. Then, $\psi^{n,m}_{\alpha}=\theta^{n,m}_{\alpha}\circ\tau_{\infty,\lfloor m/n\rfloor}$. So, let us rewrite $N^{n,\alpha}=\tau_{\infty,\lfloor m/n\rfloor}(N^{n,\alpha})\in\operatorname{\rm HS}_{k}(\log I;\lfloor m/n\rfloor)$ and we have that

[TABLE]

(note that $\widetilde{\tau_{\infty,s}(N})=\tau_{\infty,s}(\widetilde{N})$ for any $N\in\operatorname{\rm HS}_{k}(R;m)$ and $s\geq 1$ by Lemma 3.4). Moreover $\phi^{n,m}_{\alpha}$ is the induced map by $\theta^{n,m}_{\alpha}$ in $A_{L}$. Therefore, by Lemmas 1.20 and 3.5,

[TABLE]

where $\widetilde{M^{n,\alpha}}\in\operatorname{\rm HS}_{L}(A_{L};m)$ is the extension of $\Pi^{I}_{\operatorname{\rm HS},\lfloor m/n\rfloor}(N^{n,\alpha})\in\operatorname{\rm HS}_{k}(A;\lfloor m/n\rfloor)$ and the theorem is proved.

∎

Corollary 3.19

Let $k$ be a ring, $L=k[t_{i}\ |\mbox{ }i\in\mathcal{I}]$ and $A$ a finitely generated $k$-algebra. We denote $A_{L}=A\otimes_{k}L$. Then, $\Phi_{m}^{L,A}:L\otimes\operatorname{\rm IDer}_{k}(A;m)\rightarrow\operatorname{\rm IDer}_{L}(A_{L};m)$ is an isomorphism of $A_{L}$-modules for all $m\in\mathbb{N}$. Moreover, $\operatorname{\rm Leaps}_{k}(A)=\operatorname{\rm Leaps}_{L}(A_{L})$.

Proof.

Since $L$ is flat over $k$, from Lemma 3.7, $\Phi_{m}^{L,A}$ is injective. To prove the surjectivity, we take $\delta\in\operatorname{\rm IDer}_{L}(A_{L};m)$. By definition of integrability, there exists $E\in\operatorname{\rm HS}_{L}(A_{L};m)$ such that $E_{1}=\delta$. By the previous theorem, we can write $E$ as

[TABLE]

where, for all $n=1,\ldots,m$, $L_{n}$ is a finite subset of $\mathbb{N}^{(\mathcal{I})}\setminus\mathcal{P}_{n}$ and, for all $\alpha\in L_{n}$, $M^{n,\alpha}\in\operatorname{\rm HS}_{k}(A;\lfloor m/n\rfloor)$ and $\phi^{n,m}_{\alpha}:A_{L}[|\mu|]_{\lfloor m/n\rfloor}\to A_{L}[|\mu|]_{m}$ is the substitution map given by $\phi^{n,m}_{\alpha}(\mu)=t^{\alpha}\mu^{n}$. If $n>1$, then $\ell\left(\phi^{n,m}_{\alpha}\bullet N\right)>1$ for all $N\in\operatorname{\rm HS}_{L}(A_{L};m)$ and if $n=1$, then $M^{1,\alpha}_{1}\in\operatorname{\rm IDer}_{k}(A;m)$. Hence,

[TABLE]

So, $\Phi^{L,A}_{m}$ is surjective. Moreover, since $L$ is faithfully flat over $k$, $\operatorname{\rm Leaps}_{k}(A)=\operatorname{\rm Leaps}_{L}(A_{L})$ by Lemma 3.8. ∎

Let $L\supseteq k$ a pure transcendental field extension. Then, we can express $L=T^{-1}L^{\prime}$ where $L^{\prime}=k[t_{i}\ |\mbox{ }i\in\mathcal{I}]$ and $T=L^{\prime}\setminus\{0\}$. Hence, for any finitely generated $k$-algebra $A$, we have that

[TABLE]

Now, let us recall the following proposition:

Proposition 3.20

[Na1, Corollary 2.3.5]* Assume that $B$ is a finitely presented $C$-algebra, where $C$ is a commutative ring, and let $T\subseteq B$ be a multiplicative set. Then, for any integer $m\geq 1$, the canonical map*

[TABLE]

is an isomorphism of $(T^{-1}B)$-modules.

Hence, if $A$ is finitely presented $k$-algebra, $T^{-1}L^{\prime}\otimes_{L^{\prime}}\operatorname{\rm IDer}_{L^{\prime}}(A_{L^{\prime}};m)\cong\operatorname{\rm IDer}_{L^{\prime}}(T^{-1}L^{\prime}\otimes_{L^{\prime}}A_{L^{\prime}};m)=\operatorname{\rm IDer}_{L^{\prime}}(A_{L};m)$. Moreover, it is easy to prove that if $T\subseteq L^{\prime}$, then any Hasse-Schmidt derivation over $L^{\prime}$ is $T^{-1}L^{\prime}$-linear, so $\operatorname{\rm HS}_{L^{\prime}}(A_{L};m)=\operatorname{\rm HS}_{T^{-1}L^{\prime}}(A_{L};m)$. Therefore,

[TABLE]

and we have proved the following corollary:

Corollary 3.21

Let $k$ be a field and $L$ a pure transcendental field extension of $k$. Assume that $A$ is a finitely presented $k$-algebra. Then, $\Phi^{L,A}_{m}:L\otimes_{k}\operatorname{\rm IDer}_{k}(A;m)\rightarrow\operatorname{\rm IDer}_{L}(A_{L};m)$ is an isomorphism of $A_{L}$-modules for all $m\in\mathbb{N}$. Moreover, $\operatorname{\rm Leaps}_{k}(A)=\operatorname{\rm Leaps}_{L}(A_{L})$.

3.2.2 Separable extensions

Let us consider a field $k$ of characteristic $p>0$ and $L$ a $k$-algebra containing $k$. Recall that $L$ is separable over $k$ if $L_{K}:=K\otimes_{k}L$ is reduced for every possible extension $K$ of $k$. In this section, we prove that $\Phi_{m}^{L,A}$ is bijective when $L$ is a separable algebra over a field $k$ and $A$ a finitely generated $k$-algebra.

Hypothesis 3.22

Let $k$ be a ring of characteristic $p>0$ and $k\to L$ a free ring extension. Then, we assume that the following conditions hold.

For every k-linearly independent subset $\{a_{i},\ i\in\mathcal{I}\}$ of $L$, the subset $\left\{a_{i}^{p},\mbox{ }i\in\mathcal{I}\right\}$ of $L$ continues to be $k$-linearly independent.

• 2.

For every $k$-basis $\{a_{i},\mbox{ }i\in\mathcal{I}\}$ of $L$ and every $k$-linearly independent set $\{b_{1},\ldots,b_{s}\}$ of $L$, there exists $\mathcal{L}\subseteq\mathcal{I}$ such that $\{b_{1},\ldots,b_{s}\}\cup\{a_{i},\mbox{ }i\in\mathcal{L}\}$ is a $k$-basis of $L$.

Remark 3.23

If $k$ is a field, then the second condition always holds and the first one is equivalent to $L$ being a separable $k$-algebra (see [Bo, §15.4. Th. 2]). Then, if $L$ is a separable $k$-algebra, $L$ satisfies Hypothesis 3.22. Unfortunately, we do not know another type of extension that satisfies Hypothesis 3.22.

From now on, we put $R=k[x_{1},\ldots,x_{d}]$.

Hypothesis 3.24

Let $l\geq 1$ be an integer. We say that $I\subseteq R$ satisfies $S_{l}$ if $\Phi_{\text{ind},m}^{L,R}:L\otimes_{k}\operatorname{\rm IDer}_{k}(\log I;m)\to\operatorname{\rm IDer}_{L}(\log I^{e};m)$ is surjective of all $m<p^{l}$.

Note that if $k\to L$ is a flat ring extension where $k$ is a ring of characteristic $p>0$, $S_{1}$ is satisfied for all $I\subseteq R$ thanks to $\Phi_{\text{ind}}^{L,R}:L\otimes_{k}\operatorname{\rm Der}_{k}(\log I)\to\operatorname{\rm Der}_{k}(\log I^{e})$ is bijective and leaps only occur at powers of $p$.

Lemma 3.25

Let $l\geq 1$ be an integer and $k$ a ring of characteristic $p>0$. Assume that $k\to L$ is a free ring extension and $I\subseteq R$ satisfies $S_{l}$. Let us consider a $\left(p^{l}-1\right)-I$-logarithmic HS-derivation $D\in\operatorname{\rm HS}_{L}\left(R_{L};p^{l}\right)$. Then, for each $k$-basis $\{a_{i},\mbox{ }i\in\mathcal{I}\}$ of $L$, there exist a finite subset $\mathcal{I}_{0}\subseteq\mathcal{I}$ and a $\left(p^{l}-1\right)-I$-logarithmic HS-derivation $N^{i}\in\operatorname{\rm HS}_{k}\left(R;p^{l}\right)$ for each $i\in\mathcal{I}_{0}$ such that if

[TABLE]

(where we choose any order of composition) there exist a $\left(p^{l-1}-1\right)-I^{e}$-logarithmic HS-derivation $T\in\operatorname{\rm HS}_{L}\left(R_{L};p^{l-1}\right)$ and an $I^{e}$-logarithmic HS-derivation $F\in\operatorname{\rm HS}_{L}\left(\log I^{e};p^{l}\right)$ with $\ell(F)>1$ such that

[TABLE]

Proof.

Since $\Phi_{\text{ind},p^{l}-1}^{L,R}:L\otimes_{k}\operatorname{\rm IDer}_{k}(\log I;p^{l}-1)\to\operatorname{\rm IDer}_{L}(\log I^{e};p^{l}-1)$ is surjective and $D_{1}\in\operatorname{\rm IDer}_{L}(\log I^{e};p^{l}-1)$, there exist a subset $\mathcal{I}_{0}\subset\mathcal{I}$ and $\delta_{i}\in\operatorname{\rm IDer}_{k}(\log I;p^{l}-1)$ for each $i\in\mathcal{I}_{0}$ such that

[TABLE]

Let us consider a $(p^{l}-1)-I$-logarithmic integral $N^{i}\in\operatorname{\rm HS}_{k}(R;p^{l})$ of $\delta_{i}$ for all $i\in\mathcal{I}_{0}$. Then, $E:=\circ_{i\in\mathcal{I}_{0}}\left(a_{i}\bullet\widetilde{N^{i}}\right)$ is a $p^{l}$-integral of $D_{1}$ (note that the order of the composition is not important, $E$ is always an integral of $D_{1}$). Since $N^{i}$ is $(p^{l}-1)-I$-logarithmic for all $i\in\mathcal{I}_{0}$, we have that $\widetilde{N^{i}}$ is $(p^{l}-1)-I^{e}$-logarithmic (see Lemma 3.4 (3)). Hence, by Lemma 1.19 (b) and (d), $E^{\ast}$ is a $(p^{l}-1)-I^{e}$-logarithmic integral of $-D_{1}$. Therefore, $E^{\ast}\circ D\in\operatorname{\rm HS}_{L}(R_{L};p^{l})$ is a $(p^{l}-1)-I^{e}$-logarithmic HS-derivation such that $\ell(E^{\ast}\circ D)>1$. So, we can apply Corollary 2.10 to this HS-derivation. Then, there exist a $(p^{l-1}-1)-I^{e}$-logarithmic HS-derivation $T\in\operatorname{\rm HS}_{L}(R_{L};p^{l-1})$ and $F\in\operatorname{\rm HS}_{L}(\log I^{e};p^{l})$ with $\ell(F)>1$ such that

[TABLE]

and the result is proved. ∎

Theorem 3.26

Let $l\geq 1$ be an integer and assume that $k\to L$ satisfies Hypothesis 3.22 and the ideal $I\subseteq R$ satisfies $S_{l}$. Let us consider a $\left(p^{l}-1\right)-I^{e}$-logarithmic HS-derivation $D\in\operatorname{\rm HS}_{L}(R_{L};p^{l})$. Then, for every $k$-basis $\{a_{i},\mbox{ }i\in\mathcal{I}\}$ of $L$, there exist, for all $j=0,\ldots,l$,

• •

a finite subset $\mathcal{I}_{j}$ of $\mathcal{I}$ and

• •

a $\left(p^{l-j}-1\right)-I$-logarithmic HS-derivation $N^{j,n,i,j-n}\in\operatorname{\rm HS}_{k}\left(R;p^{l-j}\right)$ for each $i\in\mathcal{I}_{j-n}$, $0\leq n\leq j$

such that for all $j=0,\ldots,l$

[TABLE]

and, if we take

[TABLE]

for all $j=0,\ldots,l$ then, there exists $F\in\operatorname{\rm HS}_{L}(\log I^{e};p^{l})$ with $\ell(F)>1$ such that

[TABLE]

Proof.

By Lemma 3.25, there exist a finite subset $\mathcal{I}_{0}\subseteq\mathcal{I}$ and a $(p^{l}-1)-I$-logarithmic HS-derivation $N^{0,0,i,0}\in\operatorname{\rm HS}_{k}(R;p^{l})$ for each $i\in\mathcal{I}_{0}$ such that, if we take $E^{0}=\circ_{i\in\mathcal{I}_{0}}\left(a_{i}\bullet N^{0,0,i,0}\right)$, there exist a $(p^{l-1}-1)-I^{e}$-logarithmic HS-derivation $T^{1}\in\operatorname{\rm HS}_{L}(R_{L};p^{l-1})$ and $F\in\operatorname{\rm HS}_{L}(\log I^{e};p^{l})$ with $\ell(F)>1$ such that

[TABLE]

Observe that the set $\mathcal{C}_{0}:=\{a_{i},\mbox{ }i\in\mathcal{I}_{0}\}$ of $L$ is $k$-linearly independent so, by Hypothesis 3.22, we have that the set $\mathcal{C}_{0}^{p}:=\{a_{i}^{p},\mbox{ }i\in\mathcal{I}_{0}\}$ of $L$ is also $k$-linearly independent and from the point 2 in Hypothesis 3.22 (taking $\{a_{i},\mbox{ }i\in\mathcal{I}\}$ as $k$-basis) we obtain a subset $\mathcal{L}_{1}\subseteq\mathcal{I}$ such that $\mathcal{B}_{1}=\mathcal{C}_{0}^{p}\cup\{a_{i},\mbox{ }i\in\mathcal{L}_{1}\}$ is a $k$-basis of $L$. Note that if $l\neq 1$, we can apply the previous lemma to $T^{1}$ using the $k$-basis $\mathcal{B}_{1}$ of $L$.

Assumption.  Let us suppose that doing this process recursively we obtain that, for some integer $j$ such that $0\leq j\leq l$, there exist for all $s=0,\ldots,j-1$,

• •

a finite subset $\mathcal{I}_{s}$ of $\mathcal{I}$,

• •

a $(p^{l-s}-1)-I$-logarithmic HS-derivation $N^{s,n,i,s-n}\in\operatorname{\rm HS}_{k}(R;p^{l-s})$ for all $i\in\mathcal{I}_{s-n}$ and $0\leq n\leq s$

such that for all $s=0,\ldots,j-1$,

[TABLE]

and if we take

[TABLE]

for all $s=0,\ldots,j-1$ then, there exist

• •

$F\in\operatorname{\rm HS}_{L}(\log I^{e};p^{l})$ with $\ell(F)>1$ and

• •

a $(p^{l-j}-1)-I^{e}$-logarithmic HS-derivation $T^{j}\in\operatorname{\rm HS}_{L}(R_{L};p^{l-j})$

such that

[TABLE]

$\diamond$

Observe that since $\mathcal{C}_{j-1}$ is $k$-linearly independent, then $\mathcal{C}_{j-1}^{p}=\bigcup_{m=0}^{j-1}\left\{a_{i}^{p^{j-m}},\mbox{ }i\in\mathcal{I}_{m}\right\}$ is also a $k$-linearly independent finite set of $L$. So, there exists a subset $\mathcal{L}_{j}\subseteq\mathcal{I}$ such that $\mathcal{B}_{j}:=\mathcal{C}_{j-1}^{p}\cup\{a_{i},\mbox{ }i\in\mathcal{L}_{j}\}$ is a $k$-basis of $L$ (see Hypothesis 3.22).

Let us suppose that $j\neq l$, i.e. $l-j\geq 1$. Then, we can apply Lemma 3.25 to $T^{j}$ using the $k$-basis $\mathcal{B}_{j}$ of $L$. Hence, there exists a finite subset $\mathcal{I}^{\prime}_{m}$ of $\mathcal{I}_{m}$ for all $m=0,\ldots,j-1$, a finite set $\mathcal{I}^{\prime}_{j}$ of $\mathcal{L}_{j}$ and a $(p^{l-j}-1)-I$-logarithmic HS-derivation $N^{j,n,i,j-n}\in\operatorname{\rm HS}_{k}(R;p^{l-j})$ for each $0\leq n\leq j$ and $i\in\mathcal{I}^{\prime}_{j-n}$ such that, if we take

[TABLE]

then, there exist $F^{\prime}\in\operatorname{\rm HS}_{L}\left(\log I^{e};p^{l-j}\right)$ with $\ell(F^{\prime})>1$ and a $(p^{l-(j+1)}-1)-I^{e}$-logarithmic HS-derivation $T^{j+1}\in\operatorname{\rm HS}_{L}\left(R_{L};p^{l-(j+1)}\right)$ such that

[TABLE]

Note that we can take $\mathcal{I}_{m}^{\prime}=\mathcal{I}_{m}$ for all $0\leq n\leq j-1$ (it is enough to take $N^{j,n,i,j-n}=\operatorname{\rm\mathbb{I}}$ for all $i\in\mathcal{I}_{m}\setminus\mathcal{I}^{\prime}_{m}$) and let us rewrite $\mathcal{I}_{j}=\mathcal{I}_{j}^{\prime}$. Moreover, the subset $C_{j}=\bigcup_{m=0}^{j}\left\{a_{i}^{p^{j-m}},\mbox{ }i\in\mathcal{I}_{m}\right\}$ of $L$ is $k$-linearly independent and, if we replace $T^{j}$ in (7), we obtain that

[TABLE]

Observe that $F[p^{j}]\in\operatorname{\rm HS}_{L}(\log I^{e};p^{l})$ so, $F:=F^{\prime}[p^{j}]\circ F\in\operatorname{\rm HS}_{L}(\log I^{e};p^{l})$ with $\ell(F)>1$. Therefore, we have the same condition that Assumption for $j+1$. So that, we can apply this process until $j=l$.

Let us suppose that $j=l$ in Assumption. Then, $T^{l}\in\operatorname{\rm HS}_{L}(R_{L};1)\equiv\operatorname{\rm Der}_{L}(R_{L})$ and, by the proof of Proposition 3.2 with the $k$-basis $\mathcal{B}_{j}=\mathcal{B}_{l}$, there exists a finite subset $\mathcal{I}_{l}\subseteq\mathcal{L}_{l}\subseteq\mathcal{I}$ such that

[TABLE]

where $N^{l,n,i,l-n}\in\operatorname{\rm HS}_{k}(R;1)$ for each $i\in\mathcal{I}_{l-n}$ and $0\leq n\leq l$. It is obvious that $\bigcup_{m=0}^{l}\left\{a_{i}^{p^{j-m}},\mbox{ }i\in\mathcal{I}_{j-m}\right\}$ is a $k$-linearly independent set of $L$ and since $D=E_{0}\circ E_{1}[p]\circ\cdots\circ E^{l-1}[p^{l-1}]\circ T^{l}[p^{l}]\circ F$, we have the result. ∎

Theorem 3.27

Let $k\to L$ be a ring extension satisfying Hypothesis 3.22 and $A$ a commutative finitely generated $k$-algebra. Then, $\Phi_{m}^{L,A}:L\otimes_{k}\operatorname{\rm IDer}_{k}(A;m)\to\operatorname{\rm IDer}_{L}(A_{L};m)$ is an isomorphism of $A_{L}$-modules for all $m\in\mathbb{N}$. Moreover, $\operatorname{\rm Leaps}_{k}(A)=\operatorname{\rm Leaps}_{L}(A_{L})$.

Proof.

If $\Phi_{m}^{L,A}$ is bijective, since $L$ is faithfully flat over $k$, we have that $\operatorname{\rm Leaps}_{k}(A)=\operatorname{\rm Leaps}_{L}(A_{L})$ by Lemma 3.8. Moveover, by Lemma 3.7 1., $\Phi_{m}^{L,A}$ is injective for all $m\in\mathbb{N}$. So, we only need to prove that $\Phi^{L,A}_{m}$ is surjective.

Recall that we consider $A=R/I$ where $R=k[x_{1},\ldots,x_{d}]$ is a polynomial ring in a finite number of variable and $I\subseteq R$ an ideal. By Lemma 3.7, $\Phi_{m}^{L,A}$ is surjective if and only if $\Phi^{L,R}_{\text{ind},m}:L\otimes_{k}\operatorname{\rm IDer}_{k}(\log I;m)\to\operatorname{\rm IDer}_{L}(\log I^{e};m)$ is surjective. So, we will prove that $\Phi_{\text{ind},m}^{L,R}$ is surjective for all $m\in\mathbb{N}$. Moreover, since leaps only occur at powers of $p$, it is enough to see that $\Phi_{\text{ind},m}^{L,R}$ is surjective when $m=p^{l}$ for $l\geq 0$. We proceed by induction on $l\geq 0$.

If $l=0$, Proposition 3.2 gives us the result in this case. Now, let us assume that $\Phi_{\text{ind},m}^{L,R}$ is surjective for all $m<p^{l}$ with $l\geq 1$, i.e. $I$ satisfies $S_{l}$, and we prove the theorem for $\Phi_{\text{ind},p^{l}}^{L,R}$ with $l\geq 1$.

Let $\delta\in\operatorname{\rm IDer}_{L}(\log I^{e},p^{l})$ be an $L$-derivation of $R_{L}$, then there exists $D\in\operatorname{\rm HS}_{k}(\log I^{e};p^{l})$ an integral of $\delta$. In particular, $D$ is $(p^{l}-1)-I^{e}$-logarithmic and we can apply Theorem 3.26 to $D$. Let us consider a $k$-basis $\{a_{i},\mbox{ }i\in\mathcal{I}\}$ of $L$. Then, for all $j=0,\ldots,l$, there exist

• •

a finite subset $\mathcal{I}_{j}$ of $\mathcal{I}$ and

• •

a $(p^{l-j}-1)-I$-logarithmic HS-derivation $N^{j,n,i,j-n}\in\operatorname{\rm HS}_{k}(R;p^{l-j})$ for each $i\in\mathcal{I}_{j-n}$ and $0\leq n\leq j$

such that, for all $j=0,\ldots,l$ the subset

[TABLE]

and, if we take

[TABLE]

for all $j=0,\ldots,l$, there exists $F\in\operatorname{\rm HS}_{L}(\log I^{e};p^{l})$ with $\ell(F)>1$ such that

[TABLE]

For each $j=0,\ldots,l$, $N^{j,n,i,j-n}$ is $\left(p^{l-j}-1\right)-I$-logarithmic for all $0\leq n\leq j$ and $i\in\mathcal{I}_{j-n}$. So, $\widetilde{N^{j,n,i,j-n}}$ is $\left(p^{l-j}-1\right)-I^{e}$-logarithmic for all $0\leq n\leq j$ and $i\in\mathcal{I}_{j-n}$ (see Lemma 3.4 (3)). Therefore, by Lemma 1.19 (d), $E^{j}\in\operatorname{\rm HS}_{L}(R_{L};p^{l-j})$ is $\left(p^{l-j}-1\right)-I^{e}$-logarithmic and

[TABLE]

Hence, from Lemma 1.19 (a), $E^{j}\left[p^{j}\right]\in\operatorname{\rm HS}_{L}\left(R_{L};p^{l}\right)$ is $\left(p^{l}-1\right)-I^{e}$-logarithmic for all $j$ and

[TABLE]

So, by Lemma 1.19 (d),

[TABLE]

Since $D_{p^{l}}$ is an $I^{e}$-differential operator,

[TABLE]

is an $I^{e}$-differential operator.

Since $\mathcal{C}:=\bigcup_{k=0}^{l}\left\{a_{i}^{p^{l-k}},\mbox{ }i\in\mathcal{I}_{k}\right\}$ is a $k$-linearly independent finite set of $L$ and $\{a_{i},\mbox{ }i\in\mathcal{I}\}$ is a $k$-basis of $L$, by Hypothesis 3.22, there exists $\mathcal{L}\subseteq\mathcal{I}$ such that $\mathcal{C}\cup\{a_{i},\mbox{ }i\in\mathcal{L}\}$ is a $k$-basis of $L$. Hence, we can deduce, in the same way that in the proof of Proposition 3.2, that

[TABLE]

(note that $\widetilde{N_{p^{l-j}}^{j,j,i,0}}_{|R}=N_{p^{l-j}}^{j,j,i,0}$).

For all $i\in\mathcal{I}_{0}$, let us consider $D^{i}=N^{0,0,i,0}\circ N^{1,1,i,0}[p]\circ\cdots\circ N^{l,l,i,0}\left[p^{l}\right]\in\operatorname{\rm HS}_{k}(R;p^{l})$ an integral of $N^{0,0,i,0}_{1}$. Since $N^{j,j,i,0}\in\operatorname{\rm HS}_{k}(R;p^{l-j})$ is $(p^{l-j}-1)-I$-logarithmic for all $j=0,\ldots,l$, $N^{j,j,i,0}[p^{j}]\in\operatorname{\rm HS}_{k}(R;p^{l})$ is $(p^{l}-1)-I$-logarithmic and by Lemma 1.19 (d), $D^{i}\in\operatorname{\rm HS}_{k}(R;p^{l})$ is $(p^{l}-1)-I$-logarithmic and

[TABLE]

So, $D^{i}\in\operatorname{\rm HS}_{k}\left(\log I;p^{l}\right)$ and we can deduce that $N^{0,0,i,0}_{1}\in\operatorname{\rm IDer}_{k}\left(\log I;p^{l}\right)$. On the other hand, we recall that

[TABLE]

where $\ell(F)>1$. Then, $D_{1}=E^{0}_{1}$ and, since $E^{0}=\circ_{i\in\mathcal{I}_{0}}\left(a_{i}\bullet\widetilde{N^{0,0,i,0}}\right)$, we have that

[TABLE]

Therefore, $\Phi^{L,R}_{\text{ind},m}$ is bijective. ∎

Remark 3.28

If we change the condition 2. in Hypothesis 3.22 for

• 2’.

There exists a $k$-basis $\{a_{i},\mbox{ }i\in\mathcal{I}\}$ of $L$ such that $\left\{a^{p^{r}},\mbox{ }i\in\mathcal{I}\right\}\subseteq\{a_{i},\mbox{ }i\in\mathcal{I}\}$ for all $r\geq 1$.

then, Theorems 3.26 and 3.27 are true for that basis. For example, if we take $L=k[t_{i}\ |\ i\in\mathcal{I}]$, we can apply these theorems and we obtain that $\Phi_{m}^{L,A}$ is an isomorphism.

Corollary 3.29

Let $k$ be a field of characteristic $p>0$, $k\to L$ a separable extension and $A$ a commutative finitely generated $k$-algebra. Then, $\Phi^{L,A}_{m}:L\otimes_{k}\operatorname{\rm IDer}_{k}(\log I;m)\rightarrow\operatorname{\rm IDer}_{k}(\log I^{e};m)$ is an isomorphism of $A_{L}$-modules for all $m\geq 1$. Moreover, $\operatorname{\rm Leaps}_{k}(A)=\operatorname{\rm Leaps}_{L}(A_{L})$.

Acknowledgment. The author thanks Professor Luis Narváez Macarro for his careful reading of this paper with numerous useful comments.