# Positive characteristic Poincaré Lemma

Edileno de Almeida SANTOS  Instituto de Ciência, Engenharia e Tecnologia
(ICET) - Universidade Federal dos Vales do Jequitinhonha e Mucuri (UFVJM),
Campus do Mucuri, Teófilo Otoni - MG, Brazil<EMAIL_ADDRESS>and
Sergio RODRIGUES  Faculdade de Ciências Exatas e Tecnologia (FACET) -
Universidade Federal da Grande Dourados (UFGD), Rodovia Dourados - Itahum, Km
12 - Cidade Universitátia, Dourados - MS, Brazil<EMAIL_ADDRESS>
###### Abstract.

Let $K$ be a field of characteristic $p>0$ and $\omega$ be an $r$-form in
$K^{n}$. In this case, differently of fields of characteristic zero, the
Poincaré Lemma is not true because there are closed $r$-forms that are not
exact. We present here a definition of a $p$-closed $r$-forms and a version of
the Poincaré Lemma that is valid for $p$-closed polynomial or rational
$r$-forms on $K^{n}$ and, as a consequence, the de Rham cohomology modules of
$K^{n}$ are not trivial.

###### Key words and phrases:

Positive Characteristic, Differential forms, Cohomology

###### 1991 Mathematics Subject Classification:

37F75

##  1\. Introduction

The Poincaré Lemma ([25], 4.18, p. 155) asserts that if $\omega$ is a
differential $r$-form ($r>0$) on $\mathbb{R}^{n}$ or $\mathbb{C}^{n}$ (zero
characteristic), then: $\omega$ is closed ($d\omega=0)$ if and only if
$\omega$ is exact, that is, $\omega=d\eta$ for some $(r-1)$-form $\eta$. This
is true in the differentiable category, but also can be done in the algebraic
one if $K$ is of characteristic $0$.

If the field $K$ has characteristic $p>0$, there are closed $r$-forms on
$K^{n}$ that are not exact. M.E. Sweedler in [22] has defined the notion of
$p$-closed $1$-form on $K^{n}$, and he proves that $\omega$ is $p$-closed if
and if it is exact.

Finally, in Section 3, we extend the definition of a $p$-closed form. We make
the alternative definition of "closedness" for $r$-forms ("$p$-closedness",
see Definition 3.1), in such a way that it is possible to obtain the following
algebraic general version of Poincaré Lemma bellow.

######  Theorem A (Theorem 3.5).

Let $K$ be a field of positive characteristic $p>0$ and $R$ be $K[z]$ or
$K(z)$. An $r$-form $\omega\in\Omega^{r}_{R/K}$ is exact if and only if
$\omega$ is $p$-closed.

In Section 4, we deduce from Poincaré Lemma the general Inverse of Cartier
Operator.

######  Theorem B (Theorem 4.5).

$\gamma_{0}:\Omega_{K(z)/K}^{*}\rightarrow H_{K(z)}^{*}$ is an isomorphism.

In fact, we will see also how can be followed the opposite direction in order
to conclude the Poincaré Lemma from the Cartier Operator, as made in Section
5.

##  2\. Positive Characteristic Differential Forms

We introduce here the main notions about differential forms on $K^{n}$, where
$K$ is a positive characteristic field. The reader can see in [19] and [20]
the details of the exterior algebra of differential forms over $K^{n}$.

######  Definition 2.1.

Let $K$ be a field of arbitrary characteristic $p$ ($p=0$ or $p$ is a prime
integer). The ring of differential constants is the subring of
$K[z]:=K[z_{1},...,z_{n}]$ given by

$K[z^{p}]:=K[z_{1}^{p},...,z_{n}^{p}]$

and the field of differential constants is the sub-field of
$K(z):=K(z_{1},...,z_{n})$ given by

$K(z^{p}):=K(z_{1}^{p},...,z_{n}^{p})$

There are natural inclusions $K[z]\subset K(z)$ and $K[z^{p}]\subset
K(z^{p})$. Note that, in characteristic $0$, we have
$K(z^{p})=K(z^{0})=K[z^{p}]=K[z^{0}]=K$. Otherwise, in prime characteristic
$p>0$, we obtain $K[z^{p}]$ as a $K$-submodule of $K[z]$ generated by
$\\{G^{p}:G\in K[z]\\}$ and $K(z^{p})$ as the $K$-vector subspace of $K(z)$
generated by $\\{g^{p}:g\in K(z)\\}$. The elements of $K(z^{p})$ are called
$\partial$-constants, because $d(g^{p})=d(G^{p})=0$.

Here we will deal with $\Omega_{R/K}^{r}$, where $R$ is the ring $K[z]$ or the
field $K(z)$. The closedness or the exactness of an algebraic $r$-form
$\omega$ is not changed by multiplication by a constant $\lambda\in K$ or a
$\partial$-constant $\lambda\in K(z^{p})$ and, by (iii) of the next
proposition, one can always suppose that $\omega$ is a polynomial $r$-form.

In what follows we consider multi-indexes $I=(i_{1},...,i_{n})$. If
$J=(j_{1},...,j_{n})$ is another multi-index of length $n$, we write $J<I$
(resp. $J\leq I$) if $j_{1}<i_{1}$,…, $j_{n}<i_{n}$ (resp. $j_{1}\leq
i_{1}$,…, $j_{n}\leq i_{n}$). We indicate: $\overrightarrow{0}=(0,...,0)$ and
$\overrightarrow{p}=(p,...,p)$.

######  Proposition 2.2 (See also [1], 1.1.1 Lemma, p. 3).

Let $K$ be a field of characteristic $p>0$.

  * (i)

$K[z]$ is a finite $K[z^{p}]$-module and $K(z)$ is a finite dimensional
$K(z^{p})$-vector space, and $\dim_{K(z^{p})}(K(z))=p^{n}$.

  * (ii)

$K(z^{p})=\\{f:f\in K(z):df=0\\}$ and $K[z^{p}]=\\{F:F\in K[z]:dF=0\\}$

  * (iii)

If $\omega\in\Omega_{K(z)/K}^{r}$ is a rational $r$-form, then there is a
$\partial$-constant $\lambda\in K(z^{p})$ such that $\lambda\omega$ is a
polynomial $r$-form.

###### Proof.

A base for $K[z]$ over $K[z^{p}]$ is given by the monomials
$z_{1}^{i_{1}}...z_{n}^{i_{n}}$, where $0\leq i_{1},...,i_{n}\leq p-1$ and
this is also a base for $K(z)$ over $K(z^{p})$. In fact, if $f=\frac{P}{Q}\in
K(z)$, we can write

$f=\frac{P}{Q}=\left(\frac{1}{Q}\right)^{p}\cdot Q^{p-1}P$

and the polynomial $Q^{p-1}P$ can be expressed as a $K[z^{p}]$-linear
combination of the monomials $z_{1}^{i_{1}}...z_{n}^{i_{n}}$.

Let $f\in K(z)$ be a rational function. As we have show

$f(z)=\sum_{\overrightarrow{0}\leq I<\overrightarrow{p}}g_{I}(z^{p})z^{I}$

Hence, writing $I=(i_{(I,1)},...,i_{(I,n)})$,

$\frac{\partial f}{\partial
z_{j}}=\sum_{i_{(I,j)}>0}g_{I}(z^{p})i_{(I,j)}z^{I}z_{i}^{-1}$

Since the monomials $i_{(I,j)}z^{I}z_{i}^{-1}$ are $K(z^{p})$-linearly
independent, if $df=0$, then all $g_{I}$ with $\overrightarrow{0}<I$ must be
$0$, so we must have $f=g_{\overrightarrow{0}}(z^{p})\in K(z^{p})$.

If $\omega$ is a rational $r$-form, then there is a polynomial $Q\in K[z]$
that clears all the denominators of $\omega$, that is, $Q\omega$ is polynomial
and then $Q^{p}\omega$ is polynomial.

∎

By part (ii) of the above proposition, $\Omega_{K(z)/K}^{r}$ is isomorphic to
$\Omega_{K(z)/K(z^{p})}^{r}$. Similarly $\Omega_{K[z]/K[z^{p}]}^{r}$ is
isomorphic to $\Omega_{K[z]/K}^{r}$.

##  3\. Positive Characteristic Poincaré Lemma

If $K$ is of characteristic 0 (for example $K=\mathbb{R}$ or $K=\mathbb{C}$)
it is well known that a polynomial $r$-form is closed if and only if it is
exact, that is, $d\omega=0$ if and only if $\omega=d\eta$ (Poincaré’s Lemma).
If $K$ is of characteristic $p>0$, then there exist closed forms that are not
exact: for example, $\eta=z_{i}^{p-1}dz_{i}\in\Omega^{1}_{K(z^{p})/K}$ is
closed, but it is not exact because the "integration $\int z_{i}^{p-1}dz_{i}$"
does not results a polynomial. In this section we will develop a similar
result of Poincaré Lemma over fields of arbitrary characteristic.

The $d$ operator on $r$-forms over $R$, where $R=K[z]$ or $R=K(z)$, gives us a
chain sequence:

$0\longrightarrow K\hookrightarrow
R\overset{d}{\longrightarrow}\Omega^{1}_{R/K}\overset{d}{\longrightarrow}...\overset{d}{\longrightarrow}\Omega^{n}_{R/K}\longrightarrow
0$

If $K$ has characteristic $0$ and $R=K[z]$, this sequence is proved to be
exact (see [7], Proposition (7.1) (Poincaré lemma), page 53), that is, an
$r$-form is exact if and only if it is closed. In order to give a positive
characteristic version of this important theorem, we need more than simply the
closedness of forms. The following definition will provide a necessary and
sufficient condition for exactness.

######  Definition 3.1.

Let $K$ be a field of positive characteristic $p>0$. An $r$-form
$\omega=\sum_{I}a_{I}dz_{I}\in\Omega_{R/K}^{r}$ is $p$-closed if $\omega$ is
closed (that is, $d\omega=0$) and, for every multi-index
$I=(i_{1}<...<i_{r})$,

$\partial_{I}^{p-1}(a_{I}):=\left(\frac{\partial^{r}}{\partial
z_{I}}\right)^{p-1}(a_{I})=\frac{\partial^{r(p-1)}}{\partial
z_{I}^{p-1}}(a_{I})=\frac{\partial^{r(p-1)}}{\partial
z_{i_{1}}^{p-1}...\partial z_{i_{r}}^{p-1}}(a_{I})=0$

When $r=0$, a $0$-form is a rational function and $\omega\in R$ is $p$-closed
when $\omega\in K$.

If $\eta=\sum_{I}b_{i}dz_{I}\in\Omega^{r}_{R/K}$, we write

$\partial_{i}(\eta):=\sum_{I}\partial_{i}(b_{I})dz_{I}=\sum_{I}\frac{\partial
b_{I}}{\partial z_{i}}dz_{I}$

######  Definition 3.2.

Let $K$ be a field of positive characteristic $p>0$. A $p$-decomposable
$r$-form $\omega\in\Omega^{r}_{R/K}$ is defined inductively by the following
conditions.

  1. (1)

If $r=0$, that is, $\omega$ is a rational function, then we must have
$\omega\in K$.

  2. (2)

If $r>0$, then for every $i=1,...,n$, we require that

$\omega=dz_{i}\wedge(z_{i}^{p-1}\omega_{i}+\eta_{i})+\tau_{i}$

where $\omega_{i}$, $dz_{i}\wedge z_{i}^{p-1}\omega_{i}$,
$dz_{i}\wedge\eta_{i}+\tau_{i}$ are $p$-decomposable,
$\partial_{i}^{p-1}(\eta_{i})=0$ and $\tau_{i}$ is free of $dz_{i}$.

######  Theorem 3.3.

An $r$-form $\omega\in\Omega^{r}_{R/K}$, $r>0$, is $p$-closed if and only if,
for every $i=1,...,n$,

$\omega=dz_{i}\wedge(z_{i}^{p-1}\omega_{i}+\eta_{i})+\tau_{i}=z_{i}^{p-1}dz_{i}\wedge\omega_{i}+dz_{i}\wedge\eta_{i}+\tau_{i}$

where $\omega_{i}$, $z_{i}^{p-1}dz_{i}\wedge\omega_{i}$,
$dz_{i}\wedge\eta_{i}+\tau_{i}$ are $p$-closed,
$\partial_{i}^{p-1}(\eta_{i})=0$ and $\tau_{i}$ is free of $dz_{i}$.

###### Proof.

First suppose that $\omega$ is a closed $r$-form such that, for each
$i=1,...,n$, one can write

$\omega=dz_{i}\wedge(z_{i}^{p-1}\omega_{i}+\eta_{i})+\tau_{i}=z_{i}^{p-1}dz_{i}\wedge\omega_{i}+dz_{i}\wedge\eta_{i}+\tau_{i}$

where $\omega_{i}$, $z^{p-1}dz_{i}\wedge\omega_{i}$,
$dz_{i}\wedge\eta_{i}+\tau_{i}$ are $p$-closed,
$\partial_{i}^{p-1}(\eta_{i})=0$ and $\tau_{i}$ is free of $dz_{i}$. As sum of
$p$-closed forms, then $\omega$ is $p$-closed.

Reciprocally, we prove that if $\omega$ is $p$-closed, then $\omega$ has the
enunciated decomposition.

If $r=1$ and $\omega=\sum_{i=1}^{n}a_{i}dz_{i}$ is $p$-closed, then

$\omega=dz_{i}\wedge(z_{i}^{p-1}\omega_{i}+\eta_{i})+\tau_{i}$

where $\omega_{i}=0$, $\eta_{i}=a_{i}$, $\tau_{i}=\sum_{j\neq i}a_{j}dz_{j}$,
$z_{i}^{p-1}dz_{i}\wedge\omega_{i}=0$, $dz_{i}\wedge\eta_{i}+\tau_{i}=\omega$
are $p$-closed, $\partial_{i}^{p-1}(\eta_{i})=\partial_{i}^{p-1}(0)=0$ and
$\tau_{i}=0$ is free of $dz_{i}$.

Let $\omega$ be a $p$-closed $r$-form, $r>1$. For each $i=1,...,n$, $\omega$
has the decomposition

$\omega=dz_{i}\wedge(z_{i}^{p-1}\omega_{i}+\eta_{i})+\tau_{i}$

where ${\partial_{i}}(\omega_{i})={\partial_{i}}^{p-1}(\eta_{i})=0$ and
$\tau_{i}$ is free of $dz_{i}$. We will prove that $\omega_{i}$ is $p$-closed.

Since $d\omega=0$, we have

$0=d\omega=(-1)dz_{i}\wedge(z_{i}^{p-1}d\omega_{i}+d\eta_{i})+d\tau_{i}$

and $d\tau_{i}=\partial_{i}(dz_{i}\wedge\rho_{i})$ because $\tau_{i}$ has no
term in $dz_{i}$. By the above equation we can write

$z_{i}^{p-1}dz_{i}\wedge d\omega_{i}=-dz_{i}\wedge
d\eta_{i}+d\tau_{i}=-dz_{i}\wedge
d\eta_{i}+\partial_{i}(dz_{i}\wedge\rho_{i})$

As $\omega_{i}$ is free of $dz_{i}$, we can write
$d\omega_{i}=dz_{i}\wedge\partial_{i}(\omega_{i})+\xi_{i}$, where $\xi_{i}$ is
free of $dz_{i}$. Since $\partial_{i}(\omega_{i})=0$, we have that
$d\omega_{i}=\xi_{i}$ is free of $dz_{i}$. If $d\omega_{i}\neq 0$, then also
$dz_{i}\wedge d\omega_{i}=dz_{i}\wedge\xi_{i}\neq 0$ and, as
$(p-1)!\equiv(-1)\mod p$, then ${\partial_{i}}^{p-1}(z_{i}^{p-1}dz_{i}\wedge
d\omega_{i})=(-1)dz_{i}\wedge d\omega_{i}\neq 0$, but

$\partial_{i}^{p-1}(-dz_{i}\wedge d\eta_{i}+d\tau_{i})=-dz_{i}\wedge
d\partial_{i}^{p-1}(\eta_{i})+\partial_{i}^{p-1}\partial_{i}(dz_{i}\wedge\rho_{i})=0$

Therefore $d\omega_{i}=0$, that is, $\omega_{i}$ is closed. Now we will prove
that $\omega_{i}$ is $p$-closed. We can write
$\omega_{i}=\sum_{J}a_{J}dz_{J}$, $\eta_{i}=\sum_{J}b_{J}dz_{J}$, where
$J=(j_{1},\cdots,j_{r})$, and

$\omega=\sum_{I}a_{I}dz_{I}=dz_{i}\wedge(z_{i}^{p-1}\sum_{J}a_{J}dz_{J}+\sum_{J}b_{J}dz_{J})+\tau_{i}$

If $i=i_{m}\in I=(i_{1},...,i_{r})$ and
$J=I(i):=(i_{1},\cdots,\widehat{i},...,i_{r})$, then

$a_{I}=(-1)^{m+1}(z_{i}^{p-1}a_{I(i)}+b_{I(i)})$

and, as $\omega$ is $p$-closed,

$\begin{split}0&=(\partial_{I})^{p-1}(a_{I})\\\
&=(-1)^{m+1}(\partial_{I(i)})^{p-1}(\partial_{i})^{p-1}(z_{i}^{p-1}a_{I(i)})+(\partial_{I(i)})^{p-1}(\partial_{i})^{p-1}b_{I(i)}\\\
&=(-1)^{m}(\partial_{I(i)})^{p-1}(a_{I(i)})\end{split}$

and $\omega_{i}$ is $p$-closed. Hence $z_{i}^{p-1}dz_{i}\wedge\omega_{i}$ and
$dz_{i}\wedge\eta_{i}+\tau_{i}=\omega-z_{i}^{p-1}dz_{i}\wedge\omega_{i}$ are
also $p$-closed.

∎

######  Remark 3.4.

The set of $p$-closed $r$-forms is a $K(z^{p})$-vector subspace of
$\Omega^{r}_{K(z)/K}$.

The definition 3.1 generalizes the definition of $p$-closed $1$-form given in
[22], where a sketch of proof of the following theorem in the case $r=1$ can
be found.

######  Theorem 3.5 (Positive Characteristic Poincaré Lemma).

Let $K$ be a field of positive characteristic $p>0$ and $R$ be $K[z]$ or
$K(z)$. An $r$-form $\omega$ is exact if and only if $\omega$ is $p$-closed.

###### Proof.

By Proposition 2.2 there exists $\lambda\in K[z^{p}]$ such that
$\lambda\cdot\omega\in K[z]$ and hence we can suppose that $R=K[z]$.

$(\Rightarrow)$ Suppose first that $\omega\in\Omega_{R/K}^{r}$ is exact, that
is, $\omega=d\eta$. The case $r=0$ is trivial. If $r>0$ and
$\eta=\sum_{J}b_{J}dz_{J}$, then

$\omega=\sum_{I}a_{I}dz_{I}=\sum_{i=1}^{n}\sum_{J}\partial_{i}b_{J}dz_{i}\wedge
dz_{J}$

If $I=(i_{1}<...<i_{k}<...<i_{r})$ and
$I(k):=(i_{1}<...<\widehat{i_{k}}<...<i_{r})$, then

$a_{I}=\sum_{k=1}^{r}(-1)^{k}\partial_{i_{k}}b_{I(k)}$

Since $\partial_{i_{k}}^{p-1}\partial_{i_{k}}=\partial_{i_{k}}^{p}=0$ and
$|I|=r\geq 1$, then $\partial_{I}^{p-1}(a_{I})=0$.

$(\Leftarrow)$ Suppose that $\omega$ is $p$-closed. We proceed by induction on
$n$ and $r$. For $r=0$ we have that $\omega=\lambda\in K$ and there is nothing
to prove. Suppose the property true for $r$-forms, where $r\geq 0$. We now
prove by induction on $n$ that if $\omega$ is a $p$-closed $(r+1)$-form, then
$\omega$ is exact. The case $n=0$ is trivial. Suppose the result true for
$n\geq 0$ and let $\omega$ be a $(r+1)$-form over $K^{n+1}$. Since $\omega$ is
$p$-closed, we can write

$\omega=dz_{1}\wedge(z_{1}^{p-1}\omega_{1}+\eta_{1})+\tau_{1}=z_{1}^{p-1}dz_{1}\wedge\omega_{1}+dz_{1}\wedge\eta_{1}+\tau_{1}$

where $\omega_{1}$ is $p$-closed, $\partial_{1}^{p-1}(\eta_{1})=0$ and
$\tau_{1}$ is free of $dz_{1}$.

By induction hypothesis, $\omega_{1}$ is exact, that is, $\omega_{1}=d\alpha$.
Then $z_{1}^{p-1}dz_{1}\wedge\omega_{1}=d(-z_{1}^{p-1}dz_{1}\wedge\alpha)$ is
also exact, and hence $z_{1}^{p-1}dz_{1}\wedge\omega_{1}$ is $p$-closed, by
the first part of the proof. In this way, we have that $\omega-
z_{1}^{p-1}dz_{1}\wedge\omega_{1}$ is $p$-closed and then we can suppose
$\omega_{1}=0$. We define

$\theta=\int\eta_{1}dz_{1}$

(in the obvious algebraic "integration" sense, which is possible because
$(\frac{\partial}{\partial z_{1}})^{p-1}(\eta_{1})=0$). Then
$\theta\in\Omega^{r-1}_{R/K}$ and

$d\theta=dz_{1}\wedge\eta_{1}+\theta_{1}$

where $\theta_{1}$ does not involve $dz_{1}$.

Replacing $\omega$ by $\omega-d\theta=\tau_{1}-\theta_{1}$, we reduce to the
case where $\omega$ is free of $dz_{1}$ and we can see it in
$\Omega_{K(z_{1}^{p})[z_{2},...,z_{n+1}]/K(z_{1}^{p})}^{r+1}$. By our
induction hypothesis (over $n$) the $(r+1)$-form $\omega$ is $p$-closed as an
element of $\Omega_{K(z_{1}^{p})[z_{2},...,z_{n+1}]/K(z_{1}^{p})}^{r+1}$, and
hence it is also $p$-closed as an element of
$\Omega_{K[z_{1},z_{2},...,z_{n+1}]/K}^{r+1}$.

∎

######  Corollary 3.6.

Let $\omega=\sum_{I=(i_{1}<...<i_{r})}a_{I}(z)dz_{I}\in\Omega^{r}_{R/K}$ be a
closed $r$-form on $K^{n}$. If for each multi-index $I=(i_{1},...,i_{r})$
there is $s\in\\{1,...,r\\}$ such that

$\partial_{i_{s}}^{p-1}(a_{I})=0,$

then $\omega$ is exact.

###### Proof.

The enunciated property imply immediately the $p$-closedness condition.

∎

######  Remark 3.7.

The converse of the corollary is not true. In fact, if

$\omega=(x^{p-1}+y^{p-1})dx\wedge dy$

then

$\omega=d(xy^{p-1}dy-x^{p-1}ydx)$

but

${\partial_{x}}^{p-1}(x^{p-1}+y^{p-1})={\partial_{y}}^{p-1}(x^{p-1}+y^{p-1})=1$

##  4\. Decomposition of Closed Forms and the Inverse of Cartier Operator

Consider the $K(z)$-linear transformation

$\gamma_{0}:\Omega_{K(z)/K}^{*}\rightarrow H_{K(z)}^{*}$
$a_{I}(z)dz_{I}\mapsto a_{I}(z^{p})z_{I}^{p-1}dz_{I}\mod B_{K(z)}^{r}$

where $a(z^{p}):=a(z_{1}^{p},...,z_{n}^{p})$, and $K(z)$ acts on
$H_{K(z)}^{*}$ via $F_{0}:K(z)\rightarrow K(z)$, $a(z)\mapsto a(z^{p})$.

Using the Poincaré Lemma, we will prove in this section that $\gamma_{0}$ is
an isomorphism of $K(z)$-vector spaces. When $K$ is a perfect field (in
particular if $K$ is algebraically closed), we can take, instead of
$\gamma_{0}$, the $K(z)$-linear transformation

$\gamma:\Omega_{K(z)/K}^{*}\rightarrow H_{K(z)}^{*}$ $a_{I}dz_{I}\mapsto
a^{p}z_{I}^{p-1}dz_{I}\mod B_{K(z)}^{r}$

where $K(z)$ acts on $H_{K(z)}^{*}$ via the Frobenius endomorphism
$F:K(z)\rightarrow K(z)$, $a\mapsto a^{p}$. It will be also an isomorphism,
called the inverse of Cartier Operator, as mentioned in the next section,
where we show how the isomorphism $\gamma_{0}$ can be proved as a consequence
of the isomorphism $\gamma$, and how it can be used to prove the Poincaré
Lemma.

We need the following two operators
$\phi,\psi:\Omega_{K(z)/K}^{r}\rightarrow\Omega_{K(z)/K}^{r}$, given by

$\phi(\sum_{I}a_{I}dz_{I})=\sum_{I}\partial_{I}^{p-1}(a_{I})dz_{I}$

and

$\psi(\sum_{I}a_{I}dz_{I})=\sum_{I}\partial_{I}^{p-1}(a_{I})z_{I}^{p-1}dz_{I}$

We can now make the following definition.

######  Definition 4.1.

We call $\omega_{\mathcal{I}}:=\psi(\omega)$ the irrational part and
$\omega_{\mathcal{R}}:=\omega-\omega_{\mathcal{I}}$ the rational part of the
closed $r$-form $\omega$, giving the decomposition

$\omega=\omega_{\mathcal{I}}+\omega_{\mathcal{R}}$

Let $I=(i_{1}<...<i_{r})$ be an ordered multi-index. We write, by abuse of
notation, $i_{1}\in I$,…, $i_{r}\in I$, and $j\notin I$ if $j\neq i_{1}$,…,
$j\neq i_{r}$. Then

$I-i_{s}:=(i_{1}<...<\widehat{i_{s}}<...<i_{r})$

when $i_{s}\in I$, and

$I+j:=(i_{1}<...<i_{s}<j<i_{s+1}<...<i_{r})$

if $i_{s}<j<i_{s+1}$ (with the obvious adaptation to extremal situations).

######  Lemma 4.2.

If $\omega\in\Omega_{K(z)/K}^{r}$ is closed, then
$\partial_{j}\phi(\omega)=0$, for every $j=1$,…, $n$.

###### Proof.

Since $\omega=\sum_{I}a_{I}dz_{I}$ is closed, we obtain, for $j\notin I$,

$\partial_{j}(a_{I})=\sum_{i\in I}(-1)^{m_{I}(i,j)}\partial_{i}(a_{I-i+j})$

for suitable numbers $m_{I}(i,j)\in\mathbb{N}$. Therefore

$\partial_{j}\partial_{I}^{p-1}(a_{I})=\partial_{I}\partial_{j}(a_{I})=\sum_{i\in
I}(-1)^{m_{I}(i,j)}\partial_{I}^{p-1}\partial_{i}(a_{I-i+j})=0$

If $j\in I$, then
$\partial_{i}\partial_{I}^{p-1}(a_{I})=\partial_{j}^{p}\partial_{I-j}^{p-1}(a_{I})=0$.

∎

######  Proposition 4.3.

If $\omega=\sum_{I}a_{I}dz_{I}\in\Omega_{K(z)/K}^{r}$ is closed, then
$\phi(\omega)$ is $p$-closed (hence exact) and
$\omega_{\mathcal{I}}=\psi(\omega)$ is closed. More precisely, for every
multi-index $I$, $\partial_{I}^{p-1}(a_{I})\in K(z^{p})$.

###### Proof.

Let $\omega=\sum_{I}a_{I}dz_{I}$ be a closed $r$-form. By the above lemma,
$\partial_{i}\partial_{I}^{p-1}(a_{I})=0$ for every $i=1$,…, $n$, which
implies $\partial_{I}^{p-1}(a_{I})\in K(z^{p})$. Also $\phi(\omega)$ is closed
because

$d\phi(\omega)=\sum_{I}\sum_{j\notin
I}\partial_{j}\partial_{I}^{p-1}(a_{I})dz_{j}\wedge dz_{I}=0$

and hence $\phi(\omega)$ is $p$-closed because

$\partial_{I}^{p-1}\partial_{I}^{p-1}(a_{I})=0$

Finally $\omega_{\mathcal{I}}$ is closed because

$d\omega_{\mathcal{I}}=\sum_{I}\sum_{j\notin
I}\partial_{j}(\partial_{I}^{p-1}(a_{I})z_{I}^{p-1})dz_{j}\wedge
dz_{I}=\sum_{I}\sum_{j\notin
I}\partial_{I}^{p-1}(a_{I})\partial_{j}(z_{I}^{p-1})dz_{j}\wedge dz_{I}=0$

∎

######  Proposition 4.4.

If $\omega=\sum_{I}a_{I}dz_{I}$ is closed, then $\omega_{\mathcal{R}}$ is
$p$-closed (hence exact), while $\omega_{\mathcal{I}}$ is closed but it is
$p$-closed only in the case $\omega_{\mathcal{I}}=0$. So a closed $r$-form is
$p$-closed if and only if $\omega_{\mathcal{I}}=0$.

###### Proof.

By the above proposition, we have that

$\omega_{\mathcal{I}}=\sum_{I}\partial_{I}^{p-1}(a_{I})z_{I}^{p-1}dz_{I}=\sum_{I}b_{I}(z^{p})z_{I}^{p-1}dz_{I}$

is closed, with
$\partial_{I}^{p-1}(b_{I}(z^{p})z_{I}^{p-1})=b_{I}(z^{p})\partial_{I}^{p-1}(z_{I}^{p-1})=b_{I}(z^{p})$.
Also

$\omega_{\mathcal{R}}=\sum_{I}(a_{I}-\partial_{I}^{p-1}(a_{I})z_{I}^{p-1})dz_{I}=\sum_{I}(a_{I}-b_{I}(z^{p})z_{I}^{p-1})dz_{I}$

is closed, with
$\partial_{I}^{p-1}(a_{I}-b_{I}(z^{p})z_{I}^{p-1})=\partial_{I}^{p-1}(a_{I})-b_{I}(z^{p})=b_{I}(z^{p})-b_{I}(z^{p})=0$.
Therefore, $\omega_{\mathcal{R}}$ is $p$-closed.

∎

######  Theorem 4.5.

$\gamma_{0}:\Omega_{K(z)/K}^{*}\rightarrow H_{K(z)}^{*}$ is an isomorphism.

###### Proof.

Let $\omega=\sum_{I}a_{I}dz_{I}\in Z^{r}$ be a closed $r$-form. By Proposition
4.3, $\partial_{I}^{p-1}(a_{I})\in K(z^{p})$, then

$\omega=\gamma_{0}(\sum_{I}F_{0}^{-1}(\partial_{I}^{p-1}(a_{I}))dz_{I})$

and $\gamma_{0}$ is surjective.

We also have

$\gamma_{0}(\omega)=\sum_{I}a_{I}(z^{p})z_{I}^{p-1}dz_{I}\mod B^{r}$

If $\gamma_{0}(\omega)=0$, then $\sum_{I}a_{I}(z^{p})z_{I}^{p-1}dz_{I}$ is
$p$-closed, which implies $a_{I}=0$ for every $I$, so $\omega=0$ and hence
$\gamma_{0}$ is injective.

∎

##  5\. Deduction of Poincaré’s Lemma from the Inverse of Cartier Operator

In this section we follows closely the Chapter $1$ of book [1].

Let $A$ be a $K$-algebra (in general, $A=K[z]$ or $A=K(z)$). We have the
derivation operator

$d:\Omega_{A/K}^{*}\rightarrow\Omega_{A/K}^{*}$ $d(a_{1}da_{2}\wedge...\wedge
da_{r})=da_{1}\wedge da_{2}\wedge...\wedge da_{r}$

In particular, we have the derivation $d:A\rightarrow\Omega_{A/K}^{1}$. We
define de ring of $\partial$-constants

$A_{0}:=\\{a\in A;d(a)=0\\}$

Then obviously $K\subset A_{0}$. If $p$ is the characteristic of $K$, then

$A^{p}:=\\{a^{p};a\in A\\}\subset A_{0}$

(where if $p=0$, then $A^{p}=K$.)

The $d$-operator is $A_{0}$-linear and satisfies $d^{2}=0$, hence we obtain a
complex $(\Omega_{A}^{*},d)$, called the De Rham complex of $A$ (over $K$).

######  Lemma 5.1 ([1], 1.3.2 Lemma, p. 23).

Let $A$ be an algebra over an algebraically closed field $K$. The space
$Z_{A}^{*}:=\\{\alpha\in\Omega_{A/K};d\alpha=0\\}$ is a graded
$A^{p}$-subalgebra of $\Omega_{A}^{*}$, and the space
$B_{A}^{*}:=\\{d\alpha;\alpha\in\Omega_{A/K}^{*-1}\\}$ is a graded ideal of
$Z_{A}^{*}$. Hence, the quotient space

$H_{A}^{*}:=Z_{A}^{*}/B_{A}^{*}$

is a graded-commutative $A^{p}$-algebra. If also $A$ is a localization of a
finitely generated algebra, then the $A$-module $\Omega_{A/K}^{*}$ is finitely
generated, and the $A^{p}$-modules $Z_{A}^{*}$, $B_{A}^{*}$ and $H_{A}^{*}$
are finitely generated as well.

Consider the map $\gamma:A\rightarrow\Omega_{A/K}^{1}$, $\gamma(a)=a^{p-1}da$.

######  Lemma 5.2 ([1], 1.3.3 Lemma, p. 24).

With the above notations, for all $a,b\in A$, we have:

  1. (1)

$\gamma(ab)=a^{p}\gamma(b)+b^{p}\gamma(a)$.

  2. (2)

$d\gamma(a)=0$.

  3. (3)

$\gamma(a+b)-\gamma(a)-\gamma(b)\in B_{A}^{1}$.

Now consider the composition

$A\overset{\gamma}{\longrightarrow}Z_{A}^{1}\longrightarrow
Z_{A}^{1}/B_{A}^{1}=H_{A}^{1}$

which is a $K$-derivation and $A$ acts on itself by multiplication, and on
$H_{A}^{1}$ via the it Frobenius endomorphism $F:A\rightarrow A$,
$F(a)=a^{p}$. This derivation will be also denoted by $\gamma$. The universal
property of Kähler differentials yields an $A$-algebra homomorphism

$\gamma:\Omega_{A/K}^{*}\rightarrow H_{A}^{*}$ $a_{1}da_{2}\wedge...\wedge
da_{r}\mapsto a_{1}^{p}a_{2}^{p-1}...a_{r}^{p-1}da_{2}\wedge...\wedge
da_{r}\mod B_{A}^{*}$

where $A$ acts on $H_{A}^{*}$ via the Frobenius endomorphism.

######  Theorem 5.3 ([1], 1.3.4 Theorem, p. 24).

If $K$ is algebraically closed and $A$ is a regular ring, then
$\gamma:\Omega_{A/K}^{*}\rightarrow H_{A}^{*}$ is an isomorphism.

The inverse of the isomorphism $\gamma$ on the above theorem is called the
Cartier Operator. In particular, suppose $K$ algebraically closed and
$A=K(z)=K(z_{1},...,z_{n})$. Then

$\gamma(a(z)dz_{I})=a(z)^{p}\cdot z_{I}^{p-1}dz_{I}\mod B_{A}^{\mid I\mid}$

and every closed $r$-form $\omega$ can be write as

$\omega=\sum_{I}a_{I}dz_{I}=\sum_{I}b_{I}^{p}z_{I}^{p-1}dz_{I}+d\eta=\sum_{I}\left(b_{I}^{p}z_{I}^{p-1}+\sum_{i\in
I}(-1)^{m(I,i)}\frac{\partial c_{(I-i)}}{\partial z_{i}}\right)dz_{I}$

for suitable numbers $m(I,i)\in\mathbb{N}$. Hence

$a_{I}=b_{I}^{p}z_{I}^{p-1}+\sum_{i\in I}(-1)^{m(I,i)}\frac{\partial
c_{(I-i)}}{\partial z_{i}}$

so that

$\left(\frac{\partial^{\mid I\mid}}{\partial
z_{I}}\right)^{p-1}(a_{I})=b_{I}^{p}$

This imply that $\omega=d\eta$ if and only if $\omega$ is $p$-closed.

In the general case where $K$ is not necessarily algebraically closed, we can
consider the algebraic closure $\bar{K}$ of $K$ and then we obtain an
inclusion $K(z)\hookrightarrow\bar{K}(z)$, which enable us to make the same
argument to conclude the $p$-closedness condition for exactness. Also the
$K$-linear transformation

$\gamma_{0}:\Omega_{K(z)/K}^{*}\rightarrow H_{K(z)}^{*}$
$a(z)dz_{i_{1}}\wedge...\wedge dz_{i_{r}}\mapsto
a(z^{p})z_{i_{1}}^{p-1}...z_{i_{r}}^{p-1}dz_{i_{1}}\wedge...\wedge
dz_{i_{r}}\mod B_{K(z)}^{r}$

where $a(z^{p}):=a(z_{1}^{p},...,z_{n}^{p})$, is an isomorphism of
$K(z)$-vector spaces, if $K(z)$ acts on $H_{K(z)}^{*}$ via
$F_{0}:K(z)\rightarrow K(z^{p})$, $a(z)\mapsto a(z^{p})$.

##  6\. Rational and irrational parts of closed forms

In this section we describe how to split a polynomial or rational closed
$r$-form into a $p$-closed and a non $p$-closed part, and how these parts
generate the algebraic de Rham $r$-cohomology over the ring $R$ of the
polynomials $K[z]$ or over the field $K(z)$.

In this section we denote by $z=(z_{1},\cdots,z_{n})\in K^{n}$ and

$(z_{i}\partial_{i})^{p-1}\eta:=z_{i}^{p-1}\partial_{i}^{p-1}\eta$

The following proposition presented by Sweedler [22] is a particular case of
the splitting in the case of an 1-form.

######  Proposition 6.1.

Let $K$ be a field of characteristic $p>0$ and
$\omega=\sum_{i=1}^{n}a_{i}(z)dz_{i}$ be a polynomial or rational $1$-form on
$K^{n}$. Suppose that $\omega$ is closed and define the decomposition
$\omega=\omega_{\mathcal{R}}+\omega_{\mathcal{I}}$ such that

$\omega_{I}=\sum_{i=1}^{n}b_{i}dz_{i},\qquad
b_{i}=(-1)(z_{i}\partial_{i})^{p-1}a_{i},\qquad\omega_{\mathcal{R}}=\omega-\omega_{\mathcal{I}}$

Then this decomposition is such that

  * a)

$\omega_{\mathcal{R}}$ is $p$-closed (hence exact) and is called the rational
part of the integral of $\omega$.

  * b)

$\omega_{\mathcal{I}}$ is closed and
$\omega_{\mathcal{I}}=\sum_{i=1}^{n}(a_{i}z_{i}^{p})\frac{dz_{i}}{z_{i}}$ is
called the irrational part of the integral of $\omega$.

###### Proof.

As $\omega$ is closed and if $i\neq j=1,2,...,n$ then
$\partial_{i}a_{j}-\partial_{j}a_{i}=0$ and

$(-1)\partial_{i}b_{j}=\partial_{i}(z_{j}\partial_{j})^{p-1}a_{j}=(z_{j}\partial_{j})^{p-1}(\partial_{i}a_{j})=(z_{j}\partial_{j})^{p-1}(\partial_{j}a_{i})=z_{j}(\partial_{j})^{p}a_{i}=0$

and then $\partial_{i}b_{j}=\partial_{j}b_{i}=0$, that is, $\omega_{I}$ is
closed. As $\omega$ is closed then $\omega_{R}=\omega-\omega_{I}$ is also
closed. As $(p-1)!\equiv(-1)$ mod $p$ and $\partial_{i}^{p}=0$ therefore

$\partial_{i}^{p-1}b_{i}=(-1)\partial_{i}^{p-1}(z_{i}^{p-1}\partial_{i}^{p-1}a_{i})=(-1)(p-1)!\partial_{i}^{p-1}a_{i}=\partial_{i}^{p-1}a_{i}$

and then

$\partial_{i}^{p-1}(\omega_{R})_{i}=\partial_{i}^{p-1}(\omega-\omega_{I})_{i}=\partial_{i}^{p-1}a_{i}-\partial_{i}^{p-1}b_{i}=\partial_{i}^{p-1}a_{i}-\partial_{i}^{p-1}a_{i}=0$

By definition 3.1 $\omega_{R}$ is $p$-closed and one could conclude applying
the Poincaré Lemma LABEL:T:Main3 that $\omega$ is exact but we will follow the
reasoning sketched in [22].

If $\omega_{R}\in\Omega_{K(z)/K}$ is rational $r$-form then, by 2.2, there is
a $\partial$-constant $q^{p}$ such that $q^{p}\omega$ is a polynomial $r$-form
and one can suppose that $\omega_{R}=\sum_{i=1}^{n}c_{i}$ is a polynomial
1-form and $c_{i}$ is free of $z_{i}^{p-1}$ for each $i=1,\cdots,n$. As
$\omega_{R}$ is closed $\partial_{j}c_{i}=\partial_{i}c_{j}$ and then
$\partial_{j}\int c_{i}dz_{i}=c_{j}$ and then $\gamma=(1/n)\sum_{i=1}^{n}\int
c_{i}dz_{i}$ is such that $\partial_{i}\gamma=c_{i}$ for $i=1,\cdots,n$ and
then $d\gamma=\omega$, that is, $\omega_{R}$ is exact. ∎

######  Definition 6.2.

Let $K$ be a field of characteristic $p>0$, $r$ an integer $1\leq r\leq n$,
$J=\\{i_{1}<...<i_{r}\\}$ an ordered subset of $\\{1,2,...,n\\}$ and
$\mathcal{S}$ be the set of all ordered subsets of $\\{1,2,...,n\\}$ with $r$
elements. If $\omega\in\Omega^{r}_{R/K}$ is an algebraic $r$-form and
$\omega=\sum_{J}a_{J}dz_{J}$, where $J\in\mathcal{S}$, then we define:

  * a)

The operator $\mathcal{P}_{J}:R\rightarrow R$

$\mathcal{P}_{J}:=\prod_{i\in J}z_{i}^{p-1}\partial_{i}^{p-1}=(z_{i_{1}}\cdots
z_{i_{r}})^{p-1}\partial_{i_{1}}^{p-1}\cdots\partial_{i_{1}}^{p-1}$

With this notation
$\mathcal{P}_{(i)}=\mathcal{P}_{i}=z_{i}^{p-1}\partial_{i}^{p-1}$,
$\mathcal{P}_{J}=\prod_{i\in J}\mathcal{P}_{i}$ and
$\mathcal{P}_{i}\mathcal{P}_{j}=\mathcal{P}_{j}\mathcal{P}_{i}$.

  * b)

The operator
$\mathcal{Q}_{r}:\Omega^{r}_{R/K(z)}\rightarrow\Omega^{r}_{R/K(z)}$ such that,
for $\omega=\sum_{J}a_{J}dz_{J}$,

$\mathcal{Q}_{r}(\omega)=(-1)^{r}\sum_{J\in\mathcal{S}}\mathcal{P}_{J}(a_{J})dz_{J}$

  * c)

The decomposition $\omega=\omega_{\mathcal{R}}+\omega_{\mathcal{I}}$ such that
$\omega_{\mathcal{I}}=\mathcal{Q}_{r}(\omega)$ is the irrational part and
$\omega_{\mathcal{R}}=\omega-\omega_{\mathcal{I}}$ is the rational part of the
$r$-form $\omega$.

######  Proposition 6.3.

The operators $\mathcal{P}_{J}$ and $\mathcal{Q}_{r}$ have the following
properties.

  * (i)

If $i\in\\{1,2,\cdots,n\\}$, then
$\mathcal{P}_{i}\mathcal{P}_{i}=-\mathcal{P}_{i}$ and
$\mathcal{P}_{\\{i<j\\}}=\mathcal{P}_{i}\mathcal{P}_{j}$.

  * (ii)

If $J\in\mathcal{S}$ and $i\in J$, then $\mathcal{P}_{i}\partial_{i}=0$ and
$\mathcal{P}_{i}\mathcal{P}_{J}=-\mathcal{P}_{J}$ .

  * (iii)

If $J\in\mathcal{S}$ and $|J|=r$, then
$\mathcal{P}_{J}\mathcal{P}_{J}=(-1)^{r}\mathcal{P}_{J}$ and
$\mathcal{Q}_{r}\mathcal{Q}_{r}=\mathcal{Q}_{r}$ .

  * (iv)

$\mathcal{Q}_{r}(\omega_{\mathcal{R}})=0$ and if $\omega$ is closed then
$\omega_{\mathcal{I}}$ is closed and $\omega_{\mathcal{R}}$ is $p$-closed.

###### Proof.

As $(p-1)!\equiv-1\mod p$ and $\partial_{i}^{q}=0$ if $q\geq p$, then

$\begin{split}\mathcal{P}_{i}\mathcal{P}_{i}&=z_{i}^{p-1}\partial_{i}^{p-1}z_{i}^{p-1}\partial_{i}^{p-1}\\\
&=z_{i}^{p-1}(-1)\partial_{i}^{p-1}\\\ &=-\mathcal{P}_{i}\end{split}$

If $i<j$, we have

$\begin{split}\mathcal{P}_{i}\mathcal{P}_{j}&=z_{i}^{p-1}\partial_{i}^{p-1}z_{j}^{p-1}\partial_{j}^{p-1}\\\
&=z_{i}^{p-1}z_{j}^{p-1}\partial_{i}^{p-1}\partial_{j}^{p-1}\\\
&=\mathcal{P}_{\\{i<j\\}}\end{split}$

Without loss of generality one can suppose that $J=(1,2,...,r)$ and $i=r$,
then

$\mathcal{P}_{J}\partial_{r}=z_{1}^{p-1}\partial_{1}^{p}...z_{r}^{p-1}\partial_{r}^{p-1}\partial_{r}=z_{1}^{p-1}\partial_{1}^{p}...z_{r}^{p-1}\partial_{r}^{p}=0$

and, now supposing $i=1$,

$\begin{split}\mathcal{P}_{1}\mathcal{P}_{J}&=z_{1}^{p-1}\partial_{1}^{p-1}\mathcal{P}_{J}\\\
&=z_{1}^{p-1}\partial_{1}^{p-1}\prod_{i\in J}\mathcal{P}_{i}\\\
&=z_{1}^{p-1}\partial_{1}^{p-1}z_{1}^{p-1}\partial_{1}^{p-1}z_{2}^{p-2}\partial_{2}^{p-1}...z_{r}^{p-1}\partial_{r}^{p-1}\\\
&=z_{1}^{p-1}\partial_{1}^{p-1}(z_{1}^{p-1})\partial_{1}^{p-1}z_{2}^{p-2}\partial_{2}^{p-1}...z_{r}^{p-1}\partial_{r}^{p-1}+z_{1}^{2(p-1)}\partial_{1}^{2(p-1)}z_{2}^{p-2}\partial_{2}^{p-1}...z_{r}^{p-1}\partial_{r}^{p-1}\\\
&=z_{1}^{p-1}(-1)\partial_{1}^{p-1}z_{2}^{p-2}\partial_{2}^{p-1}...z_{r}^{p-1}\partial_{r}^{p-1}+0\\\
&=-\mathcal{P}_{J}\end{split}$

As $\partial_{i}\partial_{j}=\partial_{j}\partial_{i}$, then

$\mathcal{P}_{J}\mathcal{P}_{J}=(\prod_{i\in
J}\mathcal{P}_{i})\mathcal{P}_{J}=(-1)^{|J|}\mathcal{P}_{J}$

and

$\begin{split}\mathcal{Q}_{r}\mathcal{Q}_{r}(\omega)&=(-1)^{2r}\sum_{J\in\mathcal{S}}(\mathcal{P}_{J})^{2}(a_{J})dz_{J}\\\
&=(-1)^{|J|}\sum_{J\in\mathcal{S}}\mathcal{P}_{J}(a_{J})dz_{J}\\\
&=(-1)^{r}\sum_{J\in\mathcal{S}}\mathcal{P}_{J}(a_{J})dz_{J}\\\
&=\mathcal{Q}_{r}(\omega)\end{split}$

$\mathcal{Q}_{r}(\omega_{\mathcal{R}})=\mathcal{Q}_{r}(\omega)-\mathcal{Q}_{r}(\omega_{\mathcal{I}})=\mathcal{Q}_{r}(\omega)-\mathcal{Q}_{r}^{2}(\omega)=\mathcal{Q}_{r}(\omega)-\mathcal{Q}_{r}(\omega)=0$

Let $\mathcal{S}_{r}$ be all the ordered subsets of size $r$ of
$\\{1,2,...,n\\}$ and $\mathcal{S}_{r+1}$ be all the ordered subsets of size
$r+1$ of $\\{1,2,...,n\\}$. If $\omega=\sum_{J\in\mathcal{S}_{r}}a_{J}dz_{J}$,
then $d\omega=\sum_{I\in\mathcal{S}_{r+1}}b_{I}dz_{I}$. If
$I=\\{i_{1},i_{2},\cdots,i_{r+1}\\}$ and
$I(i_{k})=\\{i_{1},i_{2},\cdots,\widehat{i_{k}},\cdots,i_{r+1}\\}$, then

$b_{I}=\sum_{k=1}^{r+1}(-1)^{k-1}\partial_{i_{k}}a_{I(i_{k})}$

If $d\omega=0$, then $b_{I}=0$ and, for each $k=1,...,r+1$, one can isolate
$a_{I(i_{k})}$ on the left hand of $b_{I}=0$ such that

$(-1)^{k-1}\partial_{i_{k}}a_{I(i_{k})}=\sum_{j=1,j\neq
k}^{r+1}(-1)^{j}\partial_{i_{j}}a_{I(i_{j})}\qquad(*)$

Note that $\\{I(i_{k}):k=1,...,r+1,J\in\mathcal{S}_{r+1}\\}=\mathcal{S}_{r}$,
that is, each $I(i_{k})$ is such that $I(i_{k})=J$ for some
$J\in\mathcal{S}_{r}$ and reciprocally. Therefore

$\begin{split}d(\omega_{\mathcal{I}})&=d(\sum_{J\in\mathcal{I}}\mathcal{P}_{J}(a_{J})dz_{J})\\\
&=\sum_{J\in\mathcal{I}}\sum_{i\notin\mathcal{J}}\mathcal{P}_{J}\partial_{i}(a_{J})dz_{i}\wedge
dz_{J}\\\
&=\sum_{J\in\mathcal{I}}\sum_{i\notin\mathcal{J}}\sum_{j\in\mathcal{J}}(-1)^{R_{ij}}\mathcal{P}_{J}\partial_{j}(a_{(J-\\{j\\})\cup\\{i\\}})dz_{i}\wedge
dz_{J}\\\ &=0\end{split}$

and then $\omega_{\mathcal{I}}$ is closed and
$\omega_{\mathcal{R}}=\omega-\omega_{\mathcal{I}}$ is also closed.

If we suppose that $\omega_{\mathcal{R}}=\sum_{J\in\mathcal{I}}c_{J}dx_{J}$ is
not $p$-closed, then there is some $J\in\mathcal{I}$ such that
$\mathcal{P}_{J}c_{I}\neq 0$, and then
$\mathcal{Q}_{r}(\omega_{\mathcal{R}})\neq 0$, which is a contradiction. ∎

We summarize the previous discussions about $p$-closedness of an $r-$form on
$K^{n}$

######  Theorem 6.4.

If $\omega$ is an $r$-form and $\omega\in\Omega^{r}_{R/K}$, then the following
conditions are equivalent

  * (a)

$\omega$ is $p$-closed;

  * (b)

$\omega$ is $p$-decomposable;

  * (c)

$\omega$ is exact;

  * (d)

$\omega$ is closed and $\mathcal{Q}_{r}(\omega)=0$.

From Proposition 6.3 ( see [25]) follows the

######  Theorem 6.5.

Let $K$ be a field of characteristic $p>0$ and $R$ be the ring of polynomial
$K(z)$ or the field of rational functions $K(z)$, $\omega$ be a closed
polynomial or rational $r$-form with $1\leq r\leq n$. Then

  * a)

The algebraic de Rham cohomology $H^{r}(K^{n},R)$ over the ring $R$ is
generated by the class of the irrational forms $\omega_{I}$ with
$\omega\in\Omega^{r}(K^{n})$ and the chain sequence defined by exterior
derivation $d_{r-1}:\Omega^{r-1}(K^{n})\rightarrow\Omega^{r}(K^{n})$ is not
exact.

  * b)

The algebraic rational de Rham cohomology defined by the $p$-closed $r$-forms
$\Omega_{p}^{r}$ is such that $H_{p}^{r}(K^{n},R)=0$ and the chain sequence
defined by exterior derivation on closed and $p$-closed $r$-forms
$d_{r-1}:\Omega_{p}^{r-1}(K^{n})\rightarrow\Omega_{p}^{r}(K^{n})$ is exact.

##  7\. Completely integrable closed r-forms

A polynomial $p$-closed $r$-form in $K^{n}$ may present some powers
$x_{i}^{p-1}$ and in this case it is not possible to integrate $\int
z_{i}^{p-1}dz_{i}$ . In this section we define an operator that splits a
polynomial $r$-form $\omega$ in $K^{n}$ into two parts such that one of them
is free of powers $z_{i}^{p-1}$ and we prove that if $\omega$ is closed then
these parts of the decomposition are also closed.

######  Definition 7.1.

With the same hypothesis and notations of definition 6.2 we define

  * a)

the operator $\mathcal{O}_{J}:K[z]\rightarrow K[z]$

$\mathcal{O}_{J}=\prod_{i\in J}(1-(z_{i}\partial_{i})^{p-1})-1;$

where $1=id$ denotes the identity operator. With this notation
$\mathcal{O}_{\\{i\\}}=(z_{i}\partial_{i})^{p-1}$

  * b)

the operator $\mathcal{O}_{r}:\Omega^{r}(K^{n})\rightarrow\Omega^{r}(K^{n})$
such that

$\mathcal{O}_{r}(\omega)=(-1)\sum_{J\in\mathcal{I}}\mathcal{O}_{J}a_{J}dz_{J};$

  * c)

the decomposition $\omega=\omega_{C}+\omega_{T}$ such that
$\omega_{T}=\mathcal{O}_{r}(\omega)$ is the  restrict part, that is, with
integrability restriction in at least one variable, and
$\omega_{C}=\omega-\omega_{T}$ is the completely integrable part of the
$r$-form $\omega$.

######  Example 7.2.

  * a)

If $r=2$, $n=3$ and $\omega=a_{12}dz_{1}\wedge dz_{2}+a_{13}dz_{1}\wedge
dz_{3}+a_{23}dz_{2}\wedge dz_{3}$ then the restrict part of $\omega_{T}$ is
given by $\omega_{T}=b_{12}dz_{1}\wedge dz_{2}+b_{13}dx_{1}\wedge
dz_{3}+b_{23}dz_{2}\wedge dz_{3}$ such that

$(-1)b_{ij}=(z_{i}\partial_{i})^{p-1}a_{ij}+(z_{j}\partial_{j})^{p-1}a_{ij}+(z_{i}\partial_{i})^{p-1}(z_{j}\partial_{j})^{p-1}a_{ij}$

  * b)

If $r=3$, $n=3$ and $\omega=a_{123}dz_{1}\wedge dz_{2}\wedge dz_{3}$ then the
restrict part of $\omega$ is given by $\omega_{T}=b_{123}dz_{1}\wedge
dz_{2}\wedge dz_{3}$ such that

$(-1)b_{123}=\\{(z_{1}\partial_{1})^{p-1}+(z_{2}\partial_{2})^{p-1}+(z_{3}\partial_{3})^{p-1}+$
$(z_{1}z_{2}\partial_{1}\partial_{2})^{p-1}+(z_{1}z_{3}\partial_{1}\partial_{3})^{p-1}+(z_{2}z_{3}\partial_{2}\partial_{3})^{p-1}+$
$(z_{1}z_{2}z_{3}\partial_{1}\partial_{2}\partial_{3})^{p-1}\\}a_{123}$

In this case
$(-1)b_{123}=\\{\prod_{i=1}^{3}(1-(z_{i}\partial_{i})^{p-1})-1\\}a_{123}$.

The following two lemmas are a preparation to Proposition 7.5.

######  Lemma 7.3.

In the conditions and notations of the above definition :

  * a)

If $i\in I$ then

$\partial_{i}^{p-1}\mathcal{O}_{I}=(-1)\partial_{i}^{p-1};$

  * b)

If $k=1,\cdots,n$ then

$\partial_{k}\mathcal{O}_{J}=\mathcal{O}_{J}\partial_{k}$

  * c)

$\mathcal{O}_{J}\mathcal{O}_{J}=-\mathcal{O}_{J}\quad\mathrm{and}\quad\mathcal{O}_{r}\mathcal{O}_{r}=-\mathcal{O}_{r}$

  * d)

If we denote $\mathcal{O}_{\\{i\\}}=\mathcal{O}_{i}$ and if $\mathcal{J}$
denotes all non void subsets of $J=\\{1,2,\cdots,r\\}$ and $J_{q}$ is the set
of all sets of $\mathcal{J}$ with $q$ elements then

$\mathcal{O}_{J}=\sum_{q=1}^{r}\sum_{\quad\\{i_{1},\cdots,i_{q}\\}\in
J_{q}}\mathcal{O}_{i_{1}}\cdots\mathcal{O}_{i_{q}}$

and we call the above expression as the expanded polynomial form of the
operator $\mathcal{O}_{J}$

###### Proof.

a) The proof follows by induction on the number $r$ . If $r=1$ the assertion
is true as in the proof of 6.1, that is, if
$\omega=\sum_{i=1}^{n}a_{i}(z)dz_{i}$ then for any $i=1,\cdots,n$

$\partial_{i}^{p-1}\mathcal{O}_{\\{i\\}}a_{i}=\partial_{i}^{p-1}b_{i}=(-1)\partial_{i}^{p-1}a_{i}$

Suppose, by induction hypothesis, that if $r>1$ and $J$ is any ordered subset
of $\\{1,\cdots,n\\}$ with $r-1$ elements then for any $j\in J$

$\partial_{j}^{p-1}\mathcal{O}_{J}=(-1)\partial_{j}^{p-1}.$

If $I$ has $r$ elements and $i\in I$ then for any $k\in I$ and $k\neq i$

$\partial_{i}^{p-1}\mathcal{O}_{I}=\partial_{i}^{p-1})\\{(1-(z_{k}\partial_{k})^{p-1}\left(\prod_{j\in
I-k}(1-(z_{j}\partial_{j})^{p-1})-1+1\right)-1\\}=$
$(1-(z_{k}\partial_{k})^{p-1})\quad\partial_{i}^{p-1}\left(\mathcal{O}_{I-k}+1\right)-\partial_{i}^{p-1}$

by induction hypothesis of

$\partial_{i}^{p-1}\mathcal{O}_{I}=(1-(z_{k}\partial_{k})^{p-1})(\quad(-1)\partial_{i}^{p-1}+\partial_{i}^{p-1})-\partial_{i}^{p-1}=(-1)\partial_{i}^{p-1}$

b) As all the the partial derivation commutes then
$\partial_{k}\mathcal{O}_{J}=\mathcal{O}_{J}\partial_{k}$.

c) As in the proof of item a)

$\mathcal{O}_{\\{i\\}}\mathcal{O}_{\\{}i\\}=(z_{i}\partial_{i})^{p-1}\mathcal{O}_{\\{i\\}}=(-1)(z_{i}\partial_{i})^{p-1}=(-1)\mathcal{O}_{\\{i\\}}$

then

$\left((1+(z_{i}\partial_{i})^{p-1})\right)^{2}=(1+\mathcal{O}_{\\{i\\}})^{2}=(1+\mathcal{O}_{\\{i\\}})$

by the definition of $\mathcal{O}_{J}$

$\mathcal{O}_{J}^{2}=\left(\prod_{i\in
J}(1-\mathcal{O}_{\\{i\\}})-1\right)^{2}=(-1)\mathcal{O}_{J}$

and by the definition of $\mathcal{O}_{r}$ we have that
$\mathcal{O}_{r}\mathcal{O}_{r}=(-1)\mathcal{O}_{r}$.

d) If $\mathcal{O}_{i}=(x_{i}\partial_{i})^{p-1}$ then expanding the product
in the definition of $O_{J}$ we have

$\mathcal{O}_{J}=\prod_{i\in
J}(1-\mathcal{O}_{\\{i\\}})-1=\sum_{q=1}^{r}\sum_{\quad\\{i_{1},\cdots,i_{q}\\}\in
J_{q}}\mathcal{O}_{i_{1}}\cdots\mathcal{O}_{i_{q}}$

∎

Let $K$ be a field of characteristic $p>0$, $\omega$ a polynomial or rational
$r$-form in $\Omega^{r}(K^{n})$ and $1\leq r<n$. If
$J=\\{j_{1}<j_{2}<\cdots<j_{r}\\}$ is an ordered subset of
$\\{1,2,\cdots,n\\}$ and $\mathcal{I}$ is the set of all ordered subsets of
$\\{1,2,\cdots,n\\}$ then $\omega=\sum_{J\in\mathcal{I}}a_{J}dx_{J}$. If
$\mathcal{G}$ is the set o all ordered subsets of $\\{1,2,\cdots,n\\}$ with
$r+1$ elements, $G=\\{i_{1}<i_{2}<\cdots<i_{r+1}\\}$,
$dx_{G}=dx_{i_{1}}\wedge\cdots\wedge dx_{i_{r+1}}$,
$G(k)=i_{1},\cdots\widehat{i_{k}}\cdots i_{r+1}$,
$\sigma_{k}=(i_{k},i_{1},\cdots\widehat{i_{k}},\cdots i_{r+1})$ is a
permutation and $sign(\sigma)=k-1$ and $d\omega$ is written as

$d\omega=\sum_{G\in\mathcal{G}}\left(\sum_{k=1}^{r+1}(-1)^{k-1}\partial_{i_{k}}a_{G(k)}\right)dz_{G}$

######  Lemma 7.4.

  * a)

For every $i=1,2,\cdots,n$ $\mathcal{O}_{i}\partial_{i}=0$ and

$\mathcal{O}_{i_{1},\cdots{i_{k}}\cdots
i_{r+1}}\partial_{i_{k}}=\mathcal{O}_{i_{1},\cdots\widehat{i_{k}}\cdots
i_{r+1}}\partial_{i_{k}}$

.

  * b)

If $\omega=0$ then for every $G\in\mathcal{G}$ then
$\sum_{k=1}^{r+1}(-1)^{k-1}\partial_{k}a_{G(k)}=0$ and for each $1\leq k\leq
r+1$

$(-1)^{k-1}\partial_{k}a_{G(k)}=\sum_{i\neq
k,i=1}^{r+1}(-1)^{i}\partial_{j_{i}}a_{G(j_{i})}\qquad(*)$

.

  * c)

If the the irrational part of $\omega$ is
$\omega_{I}=\sum_{J\in\mathcal{I}}\mathcal{O}_{J}a_{J}dz_{J}$ then

$d(\omega_{I})=\sum_{G\in\mathcal{G}}b_{G}dz_{G}=\sum_{G\in\mathcal{G}}\left(\sum_{k=1}^{r+1}\mathcal{O}_{G(k)}(-1)^{k-1}\partial_{i_{k}}a_{G(k)}\right)dz_{G}\qquad(**)$

.

###### Proof.

a)
$\partial_{i}\mathcal{O}_{i}=\partial_{i}(z_{i}\partial_{i})^{p-1}=z_{i}\partial_{i}^{p}=0$
By definition 7.1 the partial derivation $\partial_{i_{k}}$ commutes with
$\partial_{i_{j}}$ and the result follows from distribution and elimination of
$(z_{i_{k}}\partial_{i_{k}})$.

b) If $d\omega=0$ then each coefficient of $dx_{G}$ is zero and in equation
(*) the term $(-1)^{k-1}\partial_{k}a_{G(k)}$ is isolated in the left hand of
the equation.

c)By the item (b) of 7.4 the operator $\mathcal{O}_{G}(k)$ commutes with
$\partial_{k}$. ∎

######  Proposition 7.5.

Let $K$ be a field of characteristic $p>0$ and $\omega$ be a polynomial or
rational $r$-form with $1\leq r<n$.

  * a)

If $\omega$ is a closed $r$-form then $\omega_{T}$ and $\omega_{C}$ are
closed.

  * b)

The decomposition $\omega=\omega_{C}+\omega_{T}$ is unique such that
$\mathcal{O}_{r}(\omega_{C})=0$ and $\mathcal{O}_{r}(\omega_{T})=-\omega_{T}$.

  * c)

If $d_{r-1}:\Omega^{r-1}(K^{n})\rightarrow\Omega^{r}(K^{n})$ is the exterior
derivation then

$\mathcal{O}_{r}d_{r-1}=0$

###### Proof.

a) The expression of $d(\omega_{i})$ is given in the item (c) of the previous
Lemma. Note that for each coefficient of $G\in\mathcal{G}$ of $d(\omega_{i})$
the index $i_{k}$ of $\partial_{i_{k}}$ determines uniquely the permutation
$\sigma_{k}=(i_{k},i_{1},\cdots\widehat{i_{k}},\cdots i_{r+1})$ and the set
$G(k)$, therefore there is no ambiguity in denoting only $\partial_{i_{k}}$
instead of $\partial_{i_{k}}a_{G(k)}$. We will prove that for each
$G\in\mathcal{G}$ the coefficient

$b_{G}=\sum_{k=1}^{r+1}\mathcal{O}_{G(k)}(-1)^{k-1}\partial_{i_{k}}a_{G(k)}=0$

.

Without loss of generality one can suppose that $G=\\{1,2,\cdots,r\\}$ and
then $\partial_{i_{k}}=\partial_{k}$,
$G(k)=\\{1,2,\cdots,\hat{k}\cdots,r+1\\}$ and

$b_{G}=\sum_{k=1}^{r+1}\mathcal{O}_{G(k)}(-1)^{k-1}\partial_{k}$

Substitute the term $(-1)^{k-1}\partial_{k}$ by the right hand of the equation
(*) of item (b) of Lemma 7.4, and so we have

$b_{G}=\sum_{k=1}^{n}\mathcal{O}_{G(k)}\sum_{i\neq
k,i=1}^{r+1}(-1)^{i}\partial_{i}$

By the item (a) of 7.4 if $i\neq k$ then
$\mathcal{O}_{G(k)}\partial_{i}=\mathcal{O}_{G(i,k)}\partial_{i}$ where
$G(i,k)=G-\\{i,k\\}$ and then

$b_{G}=\sum_{k=1}^{n}\sum_{i\neq
k,i=1}^{r+1}(-1)^{i}\mathcal{O}_{G(k,i)}\partial_{i}$

for each $k=1,\cdots,n$ collect the coefficients of $\partial_{k}$ such that

$b_{G}=\sum_{k=1}^{n}(\sum_{i\neq
k,i=1}^{r+1}\mathcal{O}_{G(i,k)})(-1)^{k}\partial_{k}$

Now we replace the operator $\mathcal{O}_{G(i,k)}$ by its expanded polynomial
expression as in item d) of the Lemma 7.3

$b_{G}=\sum_{k=1}^{n}\left(\sum_{i\neq
k,i=1}^{r+1}\left(\sum_{q=1}^{r-2}\sum_{\quad\\{j_{1},\cdots,j_{q}\\}\in
G(i,k)_{q}}\mathcal{O}_{j_{1}}\cdots\mathcal{O}_{j_{q}}\right)\right)(-1)^{k}\partial_{k}=$
$=\sum_{k=1}^{n}H(k)(-1)^{k}\partial_{k}$

where $G(i,k)_{q}$ denotes all the non void subsets of $G(i,k)$ with $q$
elements. Each coefficient $H(k)$ of $(-1)^{k}\partial_{k}$ is a sum of
product of operators where the operator $\mathcal{O}_{k}$ does not appear

$H(k)=(n-2)(\mathcal{O}_{1}+\cdots+\widehat{\mathcal{O}_{k}}\cdots+\mathcal{O}_{n})+(n-3)(\mathcal{O}_{1}\mathcal{O}_{2}+\cdots+\widehat{\mathcal{O}_{k}\mathcal{O}_{n}}+\cdots+\mathcal{O}_{n-1}\mathcal{O}_{n})+$
$+\cdots+\sum_{i\neq
k,i=1}^{r+1}\mathcal{O}_{1}\mathcal{O}_{2}\cdots\widehat{\mathcal{O}_{i}}\cdots\widehat{\mathcal{O}_{k}}\cdots\mathcal{O}_{r+1}$

We will conclude that $d(\omega_{I})$ is closed distributing the product of
$(-1)^{k}\partial_{k}$ to the above sum and factoring each operator that
appears in $H(k)$. Note that for each $G(k)$ with $k=1,2,\cdots n$, as in item
b) of the Lemmma 7.4, is such that

$\partial_{1}+\cdots+(-1)^{k-1}+\cdots+(-1)^{r}\partial_{r+1}=0$

. Then, for each $i=1,2,\dots,r+1$

$\mathcal{O}_{i}\left((-1)\partial_{1}+\cdots\widehat{\partial_{i}}\cdots+(-1)^{n+1}\partial_{r+1}\right)=\mathcal{O}_{i}(-1)^{i+1}\partial_{i}=0$

For each $i\neq j=1,2,\cdots,r+1$

$\mathcal{O}_{i}\mathcal{O}_{j}\left(\partial_{1}+\cdots+\widehat{(-1)^{i-1}\partial_{i}}\cdots+\widehat{(-1)^{j-1}\partial_{j}}+\cdots++(-1)^{n}\partial_{r+1}\right)=$

$=\mathcal{O}_{i}\mathcal{O}_{j}((-1)^{i}\partial_{i}+(-1)^{j}\partial_{j})=0$

and

$\mathcal{O}_{1}\mathcal{O}_{2}\cdots\widehat{\mathcal{O}_{i}}\cdots\widehat{\mathcal{O}_{k}}\cdots\mathcal{O}_{r+1}((-1)^{i}\partial_{i})=$
$=\mathcal{O}_{1}\mathcal{O}_{2}\cdots\widehat{\mathcal{O}_{i}}\cdots\widehat{\mathcal{O}_{k}}\cdots\mathcal{O}_{r+1}\left((\partial_{1}+\cdots+\widehat{(-1)^{i-1}\partial_{i}}+\cdots+(-1)^{n}\partial_{n+1}\right)=0$

and then $d(\omega_{I})=0$. As $\omega$ and $\omega_{I}$ are closed then the
rational part $\omega_{R}$ is also closed.

b) By definition 7.1
$\mathcal{O}_{r}(\omega)=(-1)\sum_{J\in\mathcal{I}}\mathcal{O}_{J}a_{J}dz_{J}$
and by Lemma 7.3 a) $\partial_{i}^{p-1}\mathcal{O}_{I}=(-1)\partial_{i}^{p-1}$
and if $\omega=\sum_{J\in\mathcal{I}}a_{J}dz_{J}$ and if $i\in J$ then
$(\partial_{i})^{p-1}a_{J}=(-1)O_{J}a_{J}$ and

$(\partial_{i})^{p-1}(\omega_{R})_{J}=(\partial_{i})^{p-1}(a_{J}-\mathcal{O}_{J}(a_{J}))=0$

By Corollary 3.6 $\omega_{R}$ is exact and then it is $p$-closed.

c) As $d_{r-1}\omega$ is an exact $r$-form then, by Poincaré Lemma
LABEL:T:Main3, $\omega$ is $p$-closed and
$(d_{r-1}\omega)_{T}=\mathcal{O}_{r}d_{r-1}=0$ ∎

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