# THE BEHAVIOR OF DIFFERENTIAL FORMS UNDER PURELY INSEPARABLE EXTENSIONS

Ahmed Laghribi, Roberto Aravire Flores, Manuel O’Ryan

► To cite this version:

Ahmed Laghribi, Roberto Aravire Flores, Manuel O’Ryan. THE BEHAVIOR OF DIFFERENTIAL FORMS UNDER PURELY INSEPARABLE EXTENSIONS. Journal of Pure and Applied Algebra, 2019, 223 (1), pp.439-457. hal-04125066

HAL Id: hal-04125066

https://univ-artois.hal.science/hal-04125066

Submitted on 11 Jun 2023

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

<figure><img src="image_1.png" /></figure>

Distributed under a Creative Commons Attribution 4.0 International License
THE BEHAVIOR OF DIFFERENTIAL FORMS UNDER PURELY
INSEPARABLE EXTENSIONS

ROBERTO ARAVIRE, AHMED LAGHRIBI, AND MANUEL O'RYAN

ABSTRACT. Let $F$ be a field of characteristic 2. In this paper we give a complete computation of the kernel of the homomorphism $H_2^{m+1}(F) \to H_2^{m+1}(L)$ induced by scalar extension, where $L/F$ is a purely inseparable extension (of any degree), $H_2^{m+1}(F)$ is the cokernel of the Artin-Schreier operator $\wp : \Omega_F^m \to \Omega_F^m/d\Omega_F^{m-1}$ given by $x \frac{dx_1}{x_1} \wedge \cdots \wedge \frac{dx_m}{x_m} \mapsto (x^2-x) \frac{dx_1}{x_1} \wedge \cdots \wedge \frac{dx_m}{x_m} + d\Omega_F^{m-1}$, where $\Omega_F^m$ is the space of absolute $m$-differential forms over $F$ and $d$ is the differential operator. Other related results are included.

# 1. INTRODUCTION

Let $F$ be a field of characteristic 2. For any integer $m \ge 1$, let $\Omega_F^m = \wedge^m \Omega_F^1$ denote the space of absolute $m$-differential forms $(\Omega_F^0 = F)$, where $\Omega_F^1$ is the $F$-vector space generated by the symbols $dx$, $x \in F$, subject to the relations: $d(x+y) = dx+dy$ and $d(xy) = xdy+ydx$ for $x,y \in F$. In particular, there is an $F^2$-linear map $F \to \Omega_F^1$, given by: $x \mapsto dx$. This map extends to the differential operator $d: \Omega_F^m \to \Omega_F^{m+1}$ given by: $d(xdx_1 \wedge \cdots \wedge dx_m) = dx \wedge dx_1 \wedge \cdots \wedge dx_m$. The Artin-Schreier operator $\wp: \Omega_F^m \to \Omega_F^m/d\Omega_F^{m-1}$ is defined by: $x \frac{dx_1}{x_1} \wedge \cdots \wedge \frac{dx_m}{x_m} \mapsto (x^2-x) \frac{dx_1}{x_1} \wedge \cdots \wedge \frac{dx_m}{x_m} + d\Omega_F^{m-1}$, and the cokernel of this operator is denoted by $H_2^{m+1}(F)$.

For any field extension $L/F$, we have a group homomorphism $H_2^{m+1}(F) \to H_2^{m+1}(L)$ induced by the inclusion $F \subset L$. An important problem consists in computing the kernel $H_2^{m+1}(L/F)$ of this homomorphism. The motivation of considering this problem is the computation of the (graded-)Witt kernel for the extension $L/F$ which reduces to the computation of $H_2^{m+1}(L/F)$ by a celebrated result of Kato [11]. Up to now, the kernel $H_2^{m+1}(L/F)$ is known in the following cases:

(a) $L$ is the function field of a projective $F$-quadric given by a bilinear Pfister form of arbitrary dimension, or a quadratic Pfister form of dimension $2^k$ such that $m \le k$ [3], [7, See after Question 8.1].

<footnote>Date: November 28, 2017.</footnote>

<footnote n="2010">2010 Mathematics Subject Classification. 11E04, 11E81.</footnote>

<footnote>Key words and phrases. Differential forms, purely inseparable extensions.</footnote>

<footnote>The authors were supported by Fondecyt #1130796. The second author acknowledges the support of the French Agence Nationale de la Recherche (ANR) under reference ANR-12-BL01-0005.</footnote>
(b) $L$ is a quadratic extension of $F$ ([2] and [3] for the separable and the inseparable case, respectively).

(c) $L$ is a separable biquadratic extension of $F$ [4]. This result has been recently extended to other quartic extensions in [5].

(d) $L$ is a multiquadratic extension of $F$ of separability degree $\le 2$ [6].

(e) $L$ is the compositum of a multiquadratic extension of $F$ of separability degree $\le 2$ with the function field of a bilinear Pfister form over $F$ [7].

The main result of this paper is Theorem 4.1 giving a complete computation of $H_2^{m+1}(L/F)$ where $L/F$ is an arbitrary purely inseparable extension of finite degree. This is a generalization of the kernel (d) above concerning purely inseparable multiquadratic extensions. We note that Theorem 4.1 has been independently proved by Sobiech [13] in the case of a modular purely inseparable extension, using a different approach.

The proof of Theorem 4.1 will be done in two steps. The first step is developed in Section 3 and is devoted to the case of a simple purely inseparable extension. This result can be viewed as a sort of generalization of Ahmad's result [1, Theorem 3]. As mentioned before, the kernel of quadratic inseparable extensions is done in [3]. To see how our arguments work, we start with the details in the case of purely inseparable quartic extensions, and then we prove Theorem 3.1 concerning the kernel of simple purely inseparable extensions of arbitrary degree. After that, we give in Section 4 our proof of Theorem 4.1 using some arguments from the proof of Theorem 3.1, an induction on the exponents of the radicals composing the extension $L/F$ and some preliminary facts that we give in Section 2. In Section 5 we complete Theorem 4.1 by giving the kernel $H_2^{m+1}(L/F)$ where $L/F$ is a purely inseparable extension of infinite degree (Theorem 5.1). We finish with some computations of the kernel $\text{Ker}(\Omega_F^m \to \Omega_L^m)$ and a couple of examples.

## 2. FORMULA FOR DIFFERENTIAL 2$^k$ POWER AND OTHERS

Let $\mathcal{B} = \{e_i | i \in I\}$ be a 2-basis of $F$. We take an ordering on $I$, and let us consider the set of multi-indices $\sum_{m,F} := \{\sigma : \{1, \dots, m\} \to I \mid \sigma(i) < \sigma(j) \text{ whenever } i < j\}$, which is equipped with the lexicographic ordering. Thus, the set $\{\frac{de_\sigma}{e_\sigma} \mid \sigma \in \sum_{m,F}\}$ is an $F$-basis of $\Omega_F^m$, where $\frac{de_\sigma}{e_\sigma} = \frac{de_{\sigma(1)}}{e_{\sigma(1)}} \wedge \dots \wedge \frac{de_{\sigma(m)}}{e_{\sigma(m)}}$ for each $\sigma \in \sum_{m,F}$.

Using the 2-basis $\mathcal{B}$, we get the Artin-Schreier operator $\wp : \Omega_F^m \to \Omega_F^m$ given by:

$$ (1) \qquad \wp\left(\sum_{\text{finite}} c_{\sigma} \frac{de_{\sigma}}{e_{\sigma}}\right) = \sum_{\text{finite}} (c_{\sigma}^2 - c_{\sigma}) \frac{de_{\sigma}}{e_{\sigma}}. $$

This operator depends on the 2-basis $\mathcal{B}$, but it is well-defined modulo $d\Omega_F^{m-1}$. In our proofs we will use the operator $\wp$ as defined in (1), but not the one defined in the introduction modulo $d\Omega_F^{m-1}$. We will also take $H_2^{m+1}(F)$ as the quotient $\Omega_F^m/(d\Omega_F^{m-1} + \wp(\Omega_F^m))$.
The square operator $s : \Omega_F^m \to \Omega_F^m$ is given by:

$$s(\sum_{\text{finite}} c_{\sigma} \frac{\mathrm{d}e_{\sigma}}{e_{\sigma}}) = \sum_{\text{finite}} c_{\sigma}^2 \frac{\mathrm{d}e_{\sigma}}{e_{\sigma}}.$$

Since $\wp(w) = s(w) + w$ for any $w \in \Omega_F^m$, it follows that the operator $s$ also depends on the choice of the 2-basis, and it is well-defined modulo $d\Omega_F^{m-1}$. We simply write $w^{[2]}$ instead of $s(w)$ for any $w \in \Omega_F^m$.

Let $c \in F \setminus F^2$ and consider the inseparable quadratic extension $M = F(\sqrt{c})$. In this section we establish some facts needed in our computations. One of these facts concerns the computation of the group $(d\Omega_F^{m-1})^{[2^k]}$, $k \ge 1$ which is defined to be $s^k(d\Omega_F^{m-1})$ where $s^k = s \circ \dots \circ s$ (k factors).

The first fact is easy and well-known.

**Fact 2.1.** For any integer $m \ge 1$, we have:

$$\Omega_M^m = \Omega_F^m \oplus \sqrt{c} \Omega_F^m \oplus \Omega_F^{m-1} \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}} \oplus \sqrt{c} \Omega_F^{m-1} \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}}.$$

In particular, for any integer $m \ge 1$, we deduce:

$$\begin{align*} (2) \qquad \Omega_M^m \wedge \mathrm{d}\sqrt{c} &= (\Omega_F^m \oplus \sqrt{c}\Omega_F^m) \wedge \mathrm{d}\sqrt{c} \\ &= \Omega_F^m \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}} \oplus \sqrt{c}\Omega_F^m \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}}. \end{align*}$$

Now let $\lambda \in \Omega_M^m$. Writing $\lambda = \lambda_0 + \sqrt{c}\lambda_1 + \lambda_2 \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}} + \sqrt{c}\lambda_3 \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}}$ with $\lambda_0, \lambda_1 \in \Omega_F^m$ and $\lambda_2, \lambda_3 \in \Omega_F^{m-1}$. Then $d\lambda = d\lambda_0 + \sqrt{c}d\lambda_1 + \sqrt{c}\lambda_1 \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}} + d\lambda_2 \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}} + \sqrt{c}d\lambda_3 \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}}$.

From this:

$$(d\lambda)^{[2]} = (d\lambda_0)^{[2]} + c(d\lambda_1)^{[2]} + c\lambda_1^{[2]} \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}} + (d\lambda_2)^{[2]} \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}} + c(d\lambda_3)^{[2]} \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}}.$$

Hence, for any integer $k \ge 1$, we have:

$$\begin{align*} (3) \qquad (d\lambda)^{[2^k]} &= (d\lambda_0)^{[2^k]} + c^{2^{k-1}} (d\lambda_1)^{[2^k]} + c^{2^{k-1}} \lambda_1^{[2^k]} \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}} + \\ & \qquad (d\lambda_2)^{[2^k]} \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}} + c^{2^{k-1}} (d\lambda_3)^{[2^k]} \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}} \\ &= (d\lambda_0)^{[2^k]} + c^{2^{k-1}} (d\lambda_1)^{[2^k]} + \\ & \qquad \left( c^{2^{k-1}} \lambda_1^{[2^k]} + (d\lambda_2)^{[2^k]} + c^{2^{k-1}} (d\lambda_3)^{[2^k]} \right) \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}}. \end{align*}$$

A couple of identities we will use several times are the following:

**Fact 2.2.** Let $u = \alpha_0 + \sqrt{c}\alpha_1 + \alpha_2 \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}} + \sqrt{c}\alpha_3 \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}}$ in $\Omega_M^m$, with $\alpha_0, \alpha_1 \in \Omega_F^m$ and $\alpha_2, \alpha_3 \in \Omega_F^{m-1}$, and similarly let $v = \beta_0 + \sqrt{c}\beta_1 + \sqrt{c}\beta_2 \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}} + \beta_3 \wedge \frac{\mathrm{d}\sqrt{c}}{\sqrt{c}}$ in
$\Omega_M^{m-1}$. Then, we have for the $\wp$ operator:

$$ \begin{aligned} \wp(u) &= \wp(\alpha_0) + \sqrt{c}\alpha_1 + c\alpha_1^{[2]} + \wp(\alpha_2) \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \sqrt{c}\alpha_3 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + c\alpha_3^{[2]} \wedge \frac{d\sqrt{c}}{\sqrt{c}}, \\ &= \wp(\alpha_0) + c\alpha_1^{[2]} + \sqrt{c}\alpha_1 + (\wp(\alpha_2) + c\alpha_3^{[2]}) \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \sqrt{c}\alpha_3 \wedge \frac{d\sqrt{c}}{\sqrt{c}}. \end{aligned} $$

And for d:

$$ \begin{aligned} dv &= d\beta_0 + \sqrt{c}d\beta_1 + \beta_1 \wedge d\sqrt{c} + \sqrt{c}d\beta_2 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + d\beta_3 \wedge \frac{d\sqrt{c}}{\sqrt{c}}, \\ &= d\beta_0 + \sqrt{c}d\beta_1 + d\beta_3 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \sqrt{c}(\beta_1 + d\beta_2) \wedge \frac{d\sqrt{c}}{\sqrt{c}}. \end{aligned} $$

Also we will need:

**Fact 2.3.** Let $K$ be a field extension of $F$. For $0 \neq x \in F$ consider $x^2(d\Omega_K^m)^{[2^k]}$ for integers $k, m \ge 1$. Then, we have

$$ x^2(d\Omega_K^m)^{[2^k]} \equiv x(d\Omega_K^m)^{[2^{k-1}]} \mod (\wp + d). $$

*Proof.* It is clear from $\wp(x(d\Omega_K^m)^{[2^{k-1}]}) = x^2(d\Omega_K^m)^{[2^k]} + x(d\Omega_K^m)^{[2^{k-1}]}$. $\square$

3. THE KERNEL OF A SIMPLE EXTENSION $L = F(\sqrt[2^n]{c})$.

Let us consider the simple purely inseparable extension $L = F(\sqrt[2^n]{c})$ for $c \in F$ and $n$ a positive integer.

Our aim in this section is to compute $H_2^{m+1}(L/F)$. We already know it when $n=1$ (and $c \notin F^2$), see [3]. As a matter of fact

$$ (4) \qquad H_2^{m+1}(F(\sqrt{c})/F) = \Omega_F^{m-1} \wedge dc + \wp(\Omega_F^m) + d\Omega_F^{m-1}. $$

Before stating the general result we will work with more detail in the 4-th degree extension $L = F(\sqrt[4]{c})$ of $F$ in order to get a better idea of the following computations. We notice that the Witt kernel $W_q(F(\sqrt[4]{c})/F)$ was computed by Ahmad in [1, Theorem 3]. Our result takes care of the graded Witt group.

Now let $w \in \Omega_F^m$ be such that $\bar{w} \in H_2^{m+1}(F(\sqrt[4]{c})/F)$ and consider the intermediate quadratic field $M = F(\sqrt{c})$. Then, it is clear that $\bar{w}_M \in H_2^{m+1}(F(\sqrt[4]{c})/M)$.

We can express this last kernel using (4) above obtaining:

$$ H_2^{m+1}(F(\sqrt[4]{c})/M) = \Omega_M^{m-1} \wedge d\sqrt{c} + \wp(\Omega_M^m) + d\Omega_M^{m-1}. $$

Using relation (2), we get $\Omega_M^{m-1} \wedge d\sqrt{c} = \Omega_F^{m-1} \wedge \frac{d\sqrt{c}}{\sqrt{c}} \oplus \sqrt{c}\Omega_F^{m-1} \wedge \frac{d\sqrt{c}}{\sqrt{c}}$.

Thus, $w_M \in \Omega_F^{m-1} \wedge \frac{d\sqrt{c}}{\sqrt{c}} \oplus \sqrt{c}\Omega_F^{m-1} \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \wp(\Omega_M^m) + d\Omega_M^{m-1}$. Hence, there are $\lambda_0, \lambda_1 \in \Omega_F^{m-1}$, $u \in \Omega_M^m$ and $v \in \Omega_M^{m-1}$ such that:

$$ w_M = \lambda_0 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \sqrt{c}\lambda_1 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \wp(u) + d(v). $$

Using the expressions for $\wp(u)$ and $d(v)$ from Fact 2.2 we can write:
$$
\begin{align*}
w_M &= \lambda_0 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \sqrt{c}\lambda_1 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \wp(u) + d(v) \\
&= \lambda_0 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \sqrt{c}\lambda_1 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \wp(\alpha_0) + c\alpha_1^{[2]} + \sqrt{c}\alpha_1 + \\
&\qquad (\wp(\alpha_2) + c\alpha_3^{[2]}) \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \sqrt{c}\alpha_3 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + d\beta_0 + \sqrt{c}d\beta_1 + \\
&\qquad d\beta_3 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \sqrt{c}(\beta_1 + d\beta_2) \wedge \frac{d\sqrt{c}}{\sqrt{c}}.
\end{align*}
$$

Since $w$ is defined over $F$, comparing coefficients we have:

$$
\begin{align*}
\text{(Integral part:)} \qquad w_M &= c\alpha_1^{[2]} + \wp(\alpha_0) + d\beta_0 \\
\text{(factor of } \sqrt{c} \text{:)} \qquad 0 &= \alpha_1 + d\beta_1 \\
\text{(factor of } d\sqrt{c} \text{:)} \qquad 0 &= \lambda_1 + \alpha_3 + \beta_1 + d\beta_2 \\
\text{(factor of } d\sqrt{c}/\sqrt{c} \text{:)} \qquad 0 &= \lambda_0 + \wp(\alpha_2) + c\alpha_3^{[2]} + d\beta_3.
\end{align*}
$$

Replacing $\alpha_1$ by $d\beta_1$ in the first equation, we get: $w_M = c(d\beta_1)^{[2]} + \wp(\alpha_0) + d\beta_0$.

Here we should notice that solving for $\beta_1$ from the third equation does not add any
more useful information and also that the terms involved in the last equation don't
show up in the first two.

Now $w$, $\alpha_0$, $\beta_0$ and $\beta_1$ are defined over $F$, then we have $w = c(d\beta_1)^{[2]} + \wp(\alpha_0) + d\beta_0 + \gamma \wedge dc$, where $\gamma \in \Omega_F^{m-1}$, showing that the kernel $H_2^{m+1}(F(\sqrt[4]{c})/F)$ is contained in $c(d\Omega_F^{m-1})^{[2]} + \Omega_F^{m-1} \wedge dc + \wp(\Omega_F^m) + d\Omega_F^{m-1}$. Conversely, it is clear that $\Omega_F^{m-1} \wedge dc + \wp(\Omega_F^m) + d\Omega_F^{m-1}$ is contained in $H_2^{m+1}(F(\sqrt[4]{c})/F)$ as $c$ is a square in $F(\sqrt[4]{c})$. Moreover, working over the field $L = F(\sqrt[4]{c})$ and using fact 2.3, we deduce

$$
\begin{align*}
c (d\Omega_F^{m-1})^{[2]} &\equiv \sqrt{c} d\Omega_F^{m-1} \pmod{\wp(\Omega_L^m) + d\Omega_L^{m-1}} \\
&\equiv (\sqrt[4]{c})^2 (d\Omega_F^{m-1}) \pmod{\wp(\Omega_L^m) + d\Omega_L^{m-1}} \\
&\equiv 0 \pmod{\wp(\Omega_L^m) + d\Omega_L^{m-1}},
\end{align*}
$$

and thus $c (d\Omega_F^{m-1})^{[2]} + \wp(\Omega_F^m) + d\Omega_F^{m-1}$ is also contained in $H_2^{m+1}(F(\sqrt[4]{c})/F)$.
Hence, we get:

$$
H_2^{m+1}(F(\sqrt[4]{c})/F) = c(d\Omega_F^{m-1})^{[2]} + \Omega_F^{m-1} \wedge dc + \wp(\Omega_F^m) + d\Omega_F^{m-1}.
$$

For the general case of a purely inseparable simple extension we have the following:

**Theorem 3.1.** Let $c \in F$ and $L = F(\sqrt[2]{c})$ for $n \ge 2$. Then

$$
H_2^{m+1}(L/F) = \left( \sum_{k=1}^{n-1} \sum_{j=1}^{2^{k-1}} c^{2j-1} (d\Omega_F^{m-1})^{[2^k]} \right) + \Omega_F^{m-1} \wedge dc + \wp(\Omega_F^m) + d\Omega_F^{m-1}.
$$
In this theorem we do not suppose that $[L : F] = 2^n$. In fact, we will be confronted by this kind of extension in the proof of Theorem 4.1. So we start by giving a lemma that helps to clarify the situation.

**Lemma 3.2.** Let $n \ge 2$ be an integer, $\alpha \in F$ and $c = \alpha^2$. Then, modulo the group $d\Omega_F^{m-1} + \wp(\Omega_F^m)$, the group

$$G_1 = \sum_{k=1}^{n-1} \sum_{j=1}^{2^{k-1}} c^{2j-1} (d\Omega_F^{m-1})^{[2^k]} + \Omega_F^{m-1} \wedge dc$$

coincides with the group

$$G_2 = \sum_{k=1}^{n-2} \sum_{j=1}^{2^{k-1}} \alpha^{2j-1} (d\Omega_F^{m-1})^{[2^k]} + \Omega_F^{m-1} \wedge d\alpha.$$

*Proof.* (a) Let $A \in G_1$. Note that $\Omega_F^{m-1} \wedge dc = 0$ as $c$ is a square in $F$. Let us write $A = \sum_{k=1}^{n-1} \sum_{j=1}^{2^{k-1}} c^{2j-1}(d\lambda_{k,j})^{[2^k]}$ for $\lambda_{k,j} \in \Omega_F^{m-1}$. Then, modulo $d\Omega_F^{m-1} + \wp(\Omega_F^m)$, $A = \sum_{k=1}^{n-1} \sum_{j=1}^{2^{k-1}} \alpha^{2j-1}(d\lambda_{k,j})^{[2^{k-1}]}$. Hence, modulo $d\Omega_F^{m-1} + \wp(\Omega_F^m)$, we get:

$$A = \sum_{k=0}^{n-2} \sum_{j=1}^{2^k} \alpha^{2j-1} (d\lambda_{k+1,j})^{[2^k]} \\ = \underbrace{\sum_{k=1}^{n-2} \sum_{j=1}^{2^{k-1}} \alpha^{2j-1} (d\lambda_{k+1,j})^{[2^k]}}_{A_2} + \underbrace{\alpha \wedge d\lambda_{1,1} + \sum_{k=1}^{n-2} \sum_{j=2^{k-1}+1}^{2^k} \alpha^{2j-1} (d\lambda_{k+1,j})^{[2^k]}}_{A_3}$$

The element $A_1$ is in the group $G_2$. Modulo the group $d\Omega_F^{m-1}$, the element $A_2$ is equal to $\lambda_{1,1} \wedge d\alpha \in G_2$. Finally, each element $\alpha^{2j-1}(d\lambda_{k+1,j})^{[2^k]}$ of $A_3$ can be written as $\alpha^{2^k+(2t-1)}(d\lambda_{k+1,j})^{[2^k]}$ for $1 \le t \le 2^{k-1}$, and thus it is equal to the element $\alpha^{2t-1}(\lambda_{k+1,j}d\alpha)^{[2^k]} + \alpha^{2t-1}(d(\alpha\lambda_{k+1,j}))^{[2^k]}$, which belongs to $\sum_{j=1}^{2^{k-1}} \alpha^{2j-1}(d\Omega_F^{m-1})^{[2^k]} + \Omega_F^{m-1} \wedge d\alpha \subset G_2$.

(b) Conversely, one can prove by similar arguments that $G_2 \subset G_1$. $\square$

*Proof of Theorem 3.1.* We proceed by induction on $n$.

Suppose that $n = 2$. There is nothing to prove if $c \in F^4$. If $c \notin F^2$, then the theorem was proved at the beginning of this section. If $c \in F^2 \setminus F^4$, then we combine the quadratic case with Lemma 3.2.

Now suppose $n > 2$ and the result is valid for all fields $k$ (of characteristic 2) and extensions $K = k(\sqrt[2]{c})$ with $c \in k$ and $l < n$.
Let $L = F(\sqrt[2^n]{c})$ for $c \in F$. If $c \in F^2$, say $c = \alpha^2$ for $\alpha \in F$, then the theorem follows from Lemma 3.2 and induction hypothesis applied to the field $L = F(\sqrt[2^{n-1}]{\alpha})$. So suppose that $c \notin F^2$. As we did before, define $M = F(\sqrt{c})$ so that $L/M$ is a simple purely inseparable extension of degree $2^{n-1}$. Thus, by induction hypothesis,

$$H_2^{m+1}(L/M) = \sum_{k=1}^{n-2} \sum_{j=1}^{2^{k-1}} (\sqrt{c})^{2j-1} (d\Omega_M^{m-1})^{[2^k]} + \Omega_M^{m-1} \wedge d\sqrt{c} + \wp(\Omega_M^m) + d\Omega_M^{m-1}.$$

Consider $w \in \Omega_F^m$ such that $\bar{w} \in H_2^{m+1}(L/F)$. Then, $\bar{w}_M \in H_2^{m+1}(L/M)$ and so for every $k$ there are $(d\Lambda_j)^{[2^k]} \in (d\Omega_M^{m-1})^{[2^k]}$ and $\alpha \in \Omega_M^m, \beta \in \Omega_M^{m-1}, \gamma \in \Omega_M^{m-1}$ such that

$$ \begin{aligned} w_M &= \sum_{k=1}^{n-2} \sum_{j=1}^{2^{k-1}} (\sqrt{c})^{2j-1} (d\Lambda_j)^{[2^k]} + \gamma \wedge d\sqrt{c} + \wp(\alpha) + d\beta \\ &= \sqrt{c} \sum_{k=1}^{n-2} \sum_{j=1}^{2^{k-1}} c^{j-1} (d\Lambda_j)^{[2^k]} + \gamma \wedge d\sqrt{c} + \wp(\alpha) + d\beta. \end{aligned} $$

Using relation (3) for each $j$ we can write

$$(d\Lambda_j)^{[2^k]} = (d\lambda_{0j})^{[2^k]} + c^{2^{k-1}} (d\lambda_{1j})^{[2^k]} + (c^{2^{k-1}} \lambda_{1j}^{[2^k]} + (d\lambda_{2j})^{[2^k]} + c^{2^{k-1}} (d\lambda_{3j})^{[2^k]}) \wedge \frac{d\sqrt{c}}{\sqrt{c}}$$

Inserting this into the expression for $w_M$, and using fact 2.2, we get:

$$ \begin{aligned} w_M = & \sqrt{c} \sum_{k=1}^{n-2} \sum_{j=1}^{2^{k-1}} c^{j-1} \left( (d\lambda_{0j})^{[2^k]} + c^{2^{k-1}} (d\lambda_{1j})^{[2^k]} \right) + \\ & \sqrt{c} \left( \sum_{k=1}^{n-2} \sum_{j=1}^{2^{k-1}} c^{j-1} \left( c^{2^{k-1}} \lambda_{1j}^{[2^k]} + (d\lambda_{2j})^{[2^k]} + c^{2^{k-1}} (d\lambda_{3j})^{[2^k]} \right) \right) \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \\ & \gamma_0 \wedge d\sqrt{c} + \gamma_1 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \\ & \wp(\alpha_0) + c\alpha_1^{[2]} + \sqrt{c}\alpha_1 + (\wp(\alpha_2) + c\alpha_3^{[2]}) \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \sqrt{c}\alpha_3 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \\ & d\beta_0 + \sqrt{c}d\beta_1 + d\beta_3 \wedge \frac{d\sqrt{c}}{\sqrt{c}} + \sqrt{c}(\beta_1 + d\beta_2) \wedge \frac{d\sqrt{c}}{\sqrt{c}}. \end{aligned} $$

As in the quartic case, comparing factors we get:
$$ (\text{Integral part:}) \qquad w_M = c\alpha_1^{[2]} + \wp(\alpha_0) + d\beta_0 $$

$$ (\text{factor of } \sqrt{c}: ) \qquad 0 = \alpha_1 + \sum_{k=1}^{n-2} \sum_{j=1}^{2^{k-1}} c^{j-1} \left( (d\lambda_{0j})^{[2^k]} + c^{2^{k-1}} (d\lambda_{1j})^{[2^k]} \right) + d\beta_1 $$

$$ (\text{factor of } d\sqrt{c}: ) \qquad 0 = \sum_{k=1}^{n-2} \sum_{j=1}^{2^{k-1}} c^{j-1} \left( c^{2^{k-1}} \lambda_{1j}^{[2^k]} + (d\lambda_{2j})^{[2^k]} + c^{2^{k-1}} (d\lambda_{3j})^{[2^k]} \right) + \\ \gamma_0 + \alpha_3 + \beta_1 + d\beta_2 $$

$$ (\text{factor of } d\sqrt{c}/\sqrt{c}: ) \qquad 0 = \gamma_1 + \wp(\alpha_2) + c\alpha_3^{[2]} + d\beta_3. $$

Replacing the value of $\alpha_1$ from the second equation into the first one, we obtain in $\Omega_M^m$:

$$ \begin{aligned} w_M &= (c \sum_{k=1}^{n-2} \sum_{j=1}^{2^{k-1}} c^{2j-2} \left( (d\lambda_{0j})^{[2^{k+1}]} + c^{2^k} (d\lambda_{1j})^{[2^{k+1}]} \right)) + c(d\beta_1)^{[2]} + \\ & \wp(\alpha_0) + d\beta_0 \\ &= \sum_{k=2}^{n-1} \sum_{j=1}^{2^{k-1}} c^{2j-1} (d\mu_{kj})^{[2^k]} + c(d\beta_1)^{[2]} + \wp(\alpha_0) + d\beta_0 \\ &= \sum_{k=1}^{n-1} \sum_{j=1}^{2^{k-1}} c^{2j-1} (d\mu_{kj})^{[2^k]} + \wp(\alpha_0) + d\beta_0. \end{aligned} $$

Now as $w$ is defined over $F$, we have

$$ w = \sum_{k=1}^{n-1} \sum_{j=1}^{2^{k-1}} c^{2j-1} (d\mu_{kj})^{[2^k]} + \gamma \wedge dc + \wp(\alpha_0) + d\beta_0, $$

for a suitable $\gamma \in \Omega_F^{m-1}$.

Therefore the kernel $H_2^{m+1}(L/F)$ is contained in

$$ \sum_{k=1}^{n-1} \sum_{j=1}^{2^{k-1}} c^{2j-1} (d\Omega_F^{m-1})^{[2^k]} + \Omega_F^{m-1} \wedge dc + \wp(\Omega_F^m) + d(\Omega_F^{m-1}), $$

and since every term in the summation is in the kernel we have the equality.

### 4. GENERAL CASE

The aim of this section is to extend the previous Theorem 3.1 by giving a description of the kernel $H_2^{m+1}(L/F)$ where $L = F(\sqrt[2^n]{c_1}, \dots, \sqrt[2^n]{c_s})$ for $s > 1$ and $c_1, \dots, c_s \in F$.
Before giving an exact description of the kernel we consider some of the ingredients that appear in it.

Since $F \subseteq F(\sqrt[2^{n_1}]{c_i}) \subseteq L$ we see that every kernel $H_2^{m+1}(F(\sqrt[2^{n_1}]{c_i})/F)$ must be contained in $H_2^{m+1}(L/F)$. More generally every intermediate field $K, F \subseteq K \subseteq L$ must contribute to this kernel.

The description of intermediate fields of a purely inseparable extension started with the paper [10] of Jacobson where a Galois theory for purely inseparable extensions of exponent 1 was introduced. Later the case of extensions of finite exponent $> 1$ was treated in [9] by Gerstenhaber and Zaromp. The main tool is the use of (higher) derivations. The easiest intermediate fields one can think of are subextensions generated by a single element of the type $\sqrt[2^{n_1}]{c_1^{c_1}} \cdots \sqrt[2^{n_s}]{c_s^{c_s}} \in L$, and keeping in mind the expression for the kernel from Theorem 3.1 we will have in the kernel elements of the kind $c_1^{l_1} \cdots c_s^{l_s} (d\Omega_F^{m-1})^{[2^k]}$ for some exponents $l_1, \dots, l_s$ and some $k$. More precisely, the description of the kernel of $H_2^{m+1}(L/F)$ is as follows:

**Theorem 4.1.** Let $c_1, \dots, c_s \in F$ and $L$ be the finitely generated purely inseparable extension $L = F(\sqrt[2^{n_1}]{c_1}, \dots, \sqrt[2^{n_s}]{c_s})$ of $F$. If $r = \max\{n_1, \dots, n_s\}$ then

$$ H_2^{m+1}(L/F) = \sum_{i=1}^{s} H_2^{m+1}(F(\sqrt[2^{n_i}]{c_i})/F) \\ (5) \qquad + \sum_{k=1}^{r-1} \sum_{\substack{l_i 2^{n_i-k-1} \in \mathbb{N} \\ 0 \le l_i \le 2^k-1}} c_1^{l_1} \cdots c_s^{l_s} (d\Omega_F^{m-1})^{[2^k]} + \wp(\Omega_F^m) + d\Omega_F^{m-1}. $$

**Remark 4.2.** It is known that in general it is impossible to express a finite purely inseparable extension $L$ of $F$ as $L = F(\sqrt[2^{n_1}]{c_1}, \dots, \sqrt[2^{n_s}]{c_s})$, $c_1, \dots, c_s \in F$ with degree $[L : F] = 2^{n_1+\cdots+n_s}$, see for instance [8, page 198]. In order to deal with this behavior, we don't exclude in Theorem 4.1 the case where some scalars among $c_1, \dots, c_s$ are squares as we assumed in Theorem 3.1. To treat this case we need the following lemma whose proof proceeds as for that of Lemma 3.2 using successive equivalences modulo the group $\wp(\Omega_F^m) + d\Omega_F^{m-1}$.

**Lemma 4.3.** Let $n_1, \dots, n_s$ be positive integers, and $c_i = \alpha_i^2$ where $\alpha_i \in F$ for $1 \le i \le s$. Let $r = \max\{n_1, \dots, n_s\}$. Then, modulo $d\Omega_F^{m-1} + \wp(\Omega_F^m)$, the group

$$ \sum_{k=1}^{r-1} \sum_{\substack{l_i 2^{n_i-k-1} \in \mathbb{N} \\ 0 \le l_i \le 2^k-1}} c_1^{l_1} \cdots c_s^{l_s} (d\Omega_F^{m-1})^{[2^k]} $$

coincides with the group

$$ \sum_{k=1}^{r-2} \sum_{\substack{l_i 2^{n_i-k-2} \in \mathbb{N} \\ 0 \le l_i \le 2^k-1}} \alpha_1^{l_1} \cdots \alpha_s^{l_s} (d\Omega_F^{m-1})^{[2^k]}. $$
*Proof of Theorem 4.1.* Without loss of generality, by the results of the previous section we assume $s \ge 2$.

First, we claim that every term on right hand side of the above formula (5) belongs to the kernel $H_2^{m+1}(L/F)$.

Since $F(\sqrt[2^n]{c_i}) \subseteq L$ we have $H_2^{m+1}(F(\sqrt[2^n]{c_i})/F) \subseteq H_2^{m+1}(L/F)$ for all $i \ge 1$.

Now consider $\Gamma = c_1^{l_1} \cdots c_s^{l_s} (d\lambda)^{[2^k]} \in \Omega_F^m$, for $1 \le k \le r-1$ where $r = \max\{n_1, n_2, \dots, n_s\}$ and $0 \le l_i \le 2^k - 1$ satisfying $l_i 2^{n_i-k-1} \in \mathbb{N}, 1 \le i \le s$.

For each $i$, there is $q_i \in \mathbb{N}$ such that $l_i 2^{n_i} = 2^{k+1}q_i$.

Let $\zeta_i = \sqrt[2^n]{c_i} \in L$ then $c_i^{l_i} = \zeta_i^{l_i 2^{n_i}}$. Hence considering $\Gamma$ over $L$ we have

$$ \begin{align*} \Gamma_L &= \zeta_1^{l_1 2^{n_1}} \cdots \zeta_s^{l_s 2^{n_s}} (d\lambda)^{[2^k]} \\ (6) \qquad &= \zeta_1^{q_1 2^{k+1}} \cdots \zeta_s^{q_s 2^{k+1}} (d\lambda)^{[2^k]} \\ &= (\zeta_1^{q_1} \cdots \zeta_s^{q_s})^{2^{k+1}} (d\lambda)^{[2^k]} \\ &\equiv (\zeta_1^{q_1} \cdots \zeta_s^{q_s})^2 d\lambda \quad \operatorname{mod} \wp(\Omega_L^m) + d\Omega_L^{m-1}. \end{align*} $$

Note that $(\zeta_1^{q_1} \cdots \zeta_s^{q_s})^2 d\lambda = d((\zeta_1^{q_1} \cdots \zeta_s^{q_s})^2 \lambda) \in d\Omega_L^{m-1}$. Therefore, $\Gamma + \wp(\Omega_F^m) + d\Omega_F^{m-1} \subset H_2^{m+1}(L/F)$. This proves our claim.

For the reverse inclusion we proceed by induction on $\sum_{i=1}^s n_i$.

The case $n_1 = n_2 = \cdots = n_s = 1$ is just [6, Prop. 2]. So without loss of generality, we may suppose that $n_1 \ge 2$.

If $c_1, \dots, c_s \in F^2$, then clearly $L = F(\sqrt[2^{n_1-1}]{\alpha_1}, \dots, \sqrt[2^{n_s-1}]{\alpha_s})$, where $\alpha_i^2 = c_i$ for $1 \le i \le s$, and the theorem follows by combining the induction hypothesis with Lemma 4.3. Without loss of generality, we may suppose that $c_1 \notin F^2$. Consider the intermediate extension $M = F(\sqrt{c_1})$. It is clear that

$$ L = M( \sqrt[2^{n_1-1}]{c_1}, \sqrt[2^{n_2}]{c_2}, \dots, \sqrt[2^{n_s}]{c_s} ), $$

and it follows from induction hypothesis that

$$ H_2^{m+1}(L/M) = H_2^{m+1}(M(\sqrt[2^{n_1-1}]{\sqrt{c_1}})/M) + \sum_{i=2}^{s} H_2^{m+1}(M(\sqrt[2^{n_i}]{c_i})/M) + \\ (7) \qquad \sum_{\substack{k=1 \\ \ell_0 \le l_1 \le 2^k-1 \\ l_1 2^{n_1-k-2} \in \mathbb{N}}}^{r'-1} \sum_{\sqrt{c_1}^{l_1} c_{(k)}^{\ell_1} (d\Omega_M^{m-1})^{[2^k]} + \wp(\Omega_M^m) + d(\Omega_M^{m-1}), $$

where $r' = \max\{n_1-1, n_2, \dots, n_s\}$ and $c_{(k)}^{\ell_1} = c_2^{l_2} \cdots c_s^{l_s}$, for $0 \le l_i \le 2^k - 1$ such that $l_i 2^{n_i-k-1} \in \mathbb{N}$.
Let $w \in \Omega_F^m$ be such that $\bar{w} \in H_2^{m+1}(L/F)$. Since $\bar{w}_M \in H_2^{m+1}(L/M)$, it follows from (7) that $w_M$ can be expressed as:

$$ (8) \quad w_M = \Delta_1 + \sum_{i=2}^{s} \Delta_i + \sum_{k=1}^{r'-1} \sum_{\substack{l, 0 \le l_1 \le 2^{k-1} \\ l_1 2^{n_1-k-2} \in \mathbb{N}}} (\sqrt{c_1})^{l_1} c_{(k)}^{l_1} (d\Lambda_j)^{[2^k]} + \\ + \wp(\alpha) + d\beta, $$

where $\Delta_1 \in H_2^{m+1}(M(\sqrt[2^{n_1-1}]{\sqrt{c_1}})/M)$, $\Delta_i \in H_2^{m+1}(M(\sqrt[2^{n_i}]{c_i})/M)$, $2 \le i \le s$, $d\Lambda_j^{[2^k]} \in (d\Omega_M^{m-1})^{[2^k]}$ and $\alpha \in \Omega_M^m, \beta \in \Omega_M^{m-1}$.

Now we analyze each term of (8) separately, showing that each part belongs to either some of the kernels $H_2^{m+1}(F(\sqrt[2^{n_i}]{c_i})/F)$ or to

$$ \sum_{k=1}^{r-1} \sum_{\substack{l_i 2^{n_i-k-1} \in \mathbb{N} \\ 0 \le l_i \le 2^{k-1}}} c_1^{l_1} \cdots c_s^{l_s} (d\Omega_F^{m-1})^{[2^k]}. $$

I) By Theorem 3.1 we see that $\Delta_1$ can be expressed as:

$$ \Delta_1 = \sum_{k=1}^{n_1-2} \sum_{j=1}^{2^{k-1}} (\sqrt{c_1})^{2j-1} (d\Lambda_j)^{[2^k]} + \gamma \wedge d\sqrt{c_1}. $$

Since we will get, in the final expression, many elements of the shape $\wp(\alpha) + d(\beta)$ with $\alpha \in \Omega_M^m$ and $\beta \in \Omega_M^{m-1}$, we omit them at this stage and add them at the final step in (15).

As in the proof of Theorem 3.1 we use identity (3) to get:

$$ (9) \quad \Delta_1 = \sqrt{c_1} \sum_{k=1}^{n_1-2} \sum_{j=1}^{2^{k-1}} c_1^{j-1} \left( (d\lambda_{0j})^{[2^k]} + c_1^{2^{k-1}} (d\lambda_{1j})^{[2^k]} \right) + \\ \sqrt{c_1} \left( \sum_{k=1}^{n_1-2} \sum_{j=1}^{2^{k-1}} c_1^{j-1} \left( c_1^{2^{k-1}} \lambda_{1j}^{[2^k]} + (d\lambda_{2j})^{[2^k]} + c_1^{2^{k-1}} (d\lambda_{3j})^{[2^k]} \right) \right) \wedge \frac{d\sqrt{c_1}}{\sqrt{c_1}} + \\ \gamma_{10} \wedge d\sqrt{c_1} + \gamma_{11} \wedge \frac{d\sqrt{c_1}}{\sqrt{c_1}}, $$

where $d\lambda_{rj} \in d\Omega_F^{m-1}$ for $0 \le r \le 1$, $d\lambda_{sj} \in d\Omega_F^{m-2}$ for $2 \le s \le 3$ and $\gamma_{10}, \gamma_{11} \in \Omega_F^{m-1}$.
II) Similarly we can write every $\Delta_i$ for $i > 1$ as

$$ (10) \quad \Delta_i = \sum_{k=1}^{n_i-1} \sum_{j=1}^{2^{k-1}} c_i^{2j-1} \left( (d\delta_{i0,j})^{[2^k]} + c_1^{2^{k-1}} (d\delta_{i1,j})^{[2^k]} + \right. \\ \qquad \left. + [c_1^{2^{k-1}} \delta_{i1,j}^{[2^k]} + (d\delta_{i2,j})^{[2^k]} + c_1^{2^{k-1}} (d\delta_{i3,j})^{[2^k]}] \wedge \frac{d\sqrt{c_1}}{\sqrt{c_1}} \right) + \\ \qquad + \gamma_{i0} \wedge dc_i + \sqrt{c_1} \gamma_{i1} \wedge dc_i + \\ \qquad + \gamma_{i2} \wedge dc_i \wedge \frac{d\sqrt{c_1}}{\sqrt{c_1}} + \sqrt{c_1} \gamma_{i3} \wedge dc_i \wedge \frac{d\sqrt{c_1}}{\sqrt{c_1}} $$

for suitable $\delta_{il,j} \in \Omega_F^{m-1}$ (for $l = 0, 1$), $\delta_{il,j} \in \Omega_F^{m-2}$ (for $l = 2, 3$), and $\gamma_{i0}, \gamma_{i1} \in \Omega_F^{m-1}$ and $\gamma_{i2}, \gamma_{i3} \in \Omega_F^{m-2}$. From each $\Delta_i$, for $i > 1$, we consider two parts:

$$ (11) \quad \Delta_{i1} = \sum_{k=1}^{n_i-1} \sum_{j=1}^{2^{k-1}} c_i^{2j-1} (d\delta_{i0,j})^{[2^k]}, $$

and

$$ (12) \quad \Delta_{i2} = \sum_{k=1}^{n_i-1} \sum_{j=1}^{2^{k-1}} c_i^{2j-1} c_1^{2^{k-1}} (d\delta_{i1,j})^{[2^k]}. $$

Arguing as in the beginning of the proof, it is easy to see that $\Delta_{i1} \in H_2^{m+1}(F(\sqrt[2^{n_i}]{c_i})/F)$ and $\Delta_{i2} \in H_2^{m+1}(L/F)$. Notice that $\Delta_{i2}$ is included in the mixed terms of Theorem 4.1.

III) The third term of (8), that is $\sum_{k=1}^{r'-1} \sum_{\substack{l_1, 0 \le l_1 \le 2^k-1 \\ l_1 2^{n_1-k-2} \in \mathbb{N}}} (\sqrt{c_1})^{l_1} c_{(k)}^l (d\Lambda_j)^{[2^k]}$, can be split as:

$$ (13) \quad \sum_{k=1}^{r'-1} \sum_{\substack{l_1, 0 \le l_1 \le 2^k-1 \\ l_1 2^{n_1-k-2} \in \mathbb{N}}} (\sqrt{c_1})^{l_1} c_{(k)}^l (d\Lambda_j)^{[2^k]} = \\ \qquad \sum_{k=1}^{r'-1} \sum_{\substack{l_1, 0 \le l_1 \le 2^k-1 \\ l_1 2^{n_1-k-2} \in \mathbb{N} \\ l_1 \text{ even}}} c_1^{l_1/2} c_{(k)}^l (d\Lambda_j)^{[2^k]} + \\ \qquad \sqrt{c_1} \sum_{k=1}^{r'-1} \sum_{\substack{l_1, 0 \le l_1 \le 2^k-1 \\ l_1 2^{n_1-k-2} \in \mathbb{N} \\ l_1 \text{ odd}}} c_1^{(l_1-1)/2} c_{(k)}^l (d\Lambda_j)^{[2^k]} $$
Now we use the expressions for $(d\Lambda_j)^{[2^k]}$ from (3) getting:

$$
\begin{align*}
(14) \quad & \sum_{k=1}^{r'-1} \sum_{\substack{l,0 \le l_1 \le 2^k-1 \\ l_1 2^{n_1-k-2} \in \mathbb{N}}} (\sqrt{c_1})^{l_1} c_{(k)}^{l_1} (d\Lambda_j)^{[2^k]} = \\
& \qquad \sum_{k=1}^{r'-1} \sum_{\substack{l,0 \le l_1 \le 2^k-1 \\ l_1 2^{n_1-k-2} \in \mathbb{N} \\ l_1 \text{ even}}} c_1^{l_1/2} c_{(k)}^{l_1} \left( (d\epsilon_{0j})^{[2^k]} + c_1^{2^{k-1}} (d\epsilon_{1j})^{[2^k]} \right) + \\
& \qquad \sqrt{c_1} \sum_{k=1}^{r'-1} \sum_{\substack{l,0 \le l_1 \le 2^k-1 \\ l_1 2^{n_1-k-2} \in \mathbb{N} \\ l_1 \text{ odd}}} c_1^{(l_1-1)/2} c_{(k)}^{l_1} \left( (d\epsilon_{0j})^{[2^k]} + c_1^{2^{k-1}} (d\epsilon_{1j})^{[2^k]} \right) + \\
& \qquad A \wedge d\sqrt{c_1} + B \wedge \frac{d\sqrt{c_1}}{\sqrt{c_1}},
\end{align*}
$$

for some $A$ and $B$ in $\Omega_F^{m-1}$.

As we did before we compare factors as $w$ is defined over $F$. We only record the integral part of the equations and the $\sqrt{c_1}$ parts of equations as they are the only ones involved in the computation of $w$.

For $\wp(\alpha)$ and $d\beta$ we have the exact same expressions as in the proof of Theorem 3.1: $\wp(\alpha_0) + c_1\alpha_1^{[2]} + \sqrt{c_1}\alpha_1 + (\wp(\alpha_2) + c_1\alpha_3^{[2]})\wedge\frac{d\sqrt{c_1}}{\sqrt{c_1}} + \sqrt{c_1}\alpha_3\wedge\frac{d\sqrt{c_1}}{\sqrt{c_1}}$ and $d\beta_0 + \sqrt{c_1}d\beta_1 + d\beta_3\wedge\frac{d\sqrt{c_1}}{\sqrt{c_1}} + \sqrt{c_1}(\beta_1 + d\beta_2)\wedge\frac{d\sqrt{c_1}}{\sqrt{c_1}}$.

Collecting terms we have:

For the integral part, we have terms from equations (11), (12), (14) and $\wp(\alpha)$ and $d\beta$:

$$
\begin{align*}
(15) \quad w_M = & \sum_{i=2}^{s} (\Delta_{i1} + \gamma_{i0} \wedge d c_i) + \sum_{i=2}^{s} \Delta_{i2} + \\
& \sum_{k=1}^{r'-1} \sum_{\substack{l,0 \le l_1 \le 2^k-1 \\ l_1 2^{n_1-k-2} \in \mathbb{N} \\ l_1 \text{ even}}} c_1^{l_1/2} c_{(k)}^{l_1} \left( (d\epsilon_{0j})^{[2^k]} + c_1^{2^{k-1}} (d\epsilon_{1j})^{[2^k]} \right) + \\
& c_1 \alpha_1^{[2]} + \wp(\alpha_0) + d\beta_0.
\end{align*}
$$
For the $\sqrt{c_1}$ part, we have terms from equations (9), (10), (14) and $\wp(\alpha)$ and $d\beta$:

$$ (16) \quad 0 = \sum_{k=1}^{n_1-2} \sum_{j=1}^{2^{k-1}} c_1^{j-1} \left( (d\lambda_{0j})^{[2^k]} + c_1^{2^{k-1}} (d\lambda_{1j})^{[2^k]} \right) + \\ \sum_{k=1}^{r'-1} \sum_{\substack{l, 0 \le l_1 \le 2^{k-1} \\ l_1 2^{n_1-k-2} \in \mathbb{N} \\ l_1 \text{ odd}}} c_1^{(l_1-1)/2} c_{(k)}^{l} \left( (d\epsilon_{0j})^{[2^k]} + c_1^{2^{k-1}} (d\epsilon_{1j})^{[2^k]} \right) + \\ \sum_{i=2}^{s} \gamma_{i1} \wedge dc_i + \alpha_1 + d\beta_1. $$

As we did before, we compute $c_1\alpha_1^{[2]}$ using the expression of $\alpha_1$ from the above equation.

Squaring the first term of (16) and multiplying by $c_1$, we get:

$$ \sum_{k=1}^{n_1-2} \sum_{j=1}^{2^{k-1}} c_1^{2j-1} \left( (d\lambda_{0j})^{[2^{k+1}]} + c_1^{2k} (d\lambda_{1j})^{[2^{k+1}]} \right) = \\ \sum_{k=2}^{n_1-1} \sum_{j=1}^{2^{k-1}} c_1^{2j-1} (d\mu_{jk})^{[2^k]}, $$

for some $\mu_{jk}$. We notice that this term is contained in $H_2^{m+1}(F(\sqrt[2^{n_1}]{c_1})/F)$.

Squaring the second term of (16) and multiplying by $c_1$, we get:

$$ (17) \quad \sum_{k=1}^{r'-1} \sum_{\substack{l, 0 \le l_1 \le 2^{k-1} \\ l_1 2^{n_1-k-2} \in \mathbb{N} \\ l_1 \text{ odd}}} c_1^{l_1} c_{(k)}^{2l} \left( (d\epsilon_{0j})^{[2^{k+1}]} + c_1^{2k} (d\epsilon_{1j})^{[2^{k+1}]} \right) = \\ \sum_{k=1}^{r'-1} \sum_{\substack{l, 0 \le l_1 \le 2^{k-1} \\ l_1 2^{n_1-k-2} \in \mathbb{N} \\ l_1 \text{ odd}}} (c_1^{l_1} c_{(k)}^{2l} (d\epsilon_{0j})^{[2^{k+1}]} + c_1^{l_1+2k} c_{(k)}^{2l} (d\epsilon_{1j})^{[2^{k+1}]}). $$

Squaring the third term of (16) and multiplying by $c_1$, we get:

$$ \sum_{i=2}^{s} c_1(\gamma_{i1} \wedge dc_i)^{[2]} + c_1(d\beta_1)^{[2]}. $$

Note that for each $i, 2 \le i \le s$, the term $c_1(\gamma_{i1} \wedge dc_i)^{[2]} = c_1 c_i^2 (\gamma_{i1})^{[2]} \wedge dc_i \in \Omega_F^{m-1} \wedge dc_i$ is included in $H_2^{m+1}(F(\sqrt[2^{n_i}]{c_i})/F)$. This observation also applies to the other term.
So, from the expression for $c_1\alpha_1^{[2]}$ we only need to consider the terms from equation (17).

For $k$ a non-negative integer, let us define:

$$ \mathcal{M}_k = \langle c_1^{l_1} \cdots c_s^{l_s} (d\eta)^{[2^k]} \mid \eta \in \Omega_F^{m-1}, 0 \le l_i \le 2^k - 1, \text{ such that } l_i 2^{n_i-k-1} \in \mathbb{N}, 1 \le i \le s \rangle. $$

With this notation we see that the last term of kernel in Theorem 4.1 can be written as

$$ \sum_{k=1}^{r-1} \sum_{\substack{l_i 2^{n_i-k-1} \in \mathbb{N} \\ 0 \le l_i \le 2^k - 1}} c_1^{l_1} \cdots c_s^{l_s} (d\Omega_F^{m-1})^{[2^k]} = \sum_{k=1}^{r-1} \mathcal{M}_k $$

Elements of the shape of those from $\mathcal{M}_k$ belong to the kernel $H_2^{m+1}(L/F)$ as we explained in (6). Now it remains to check that each element from the third term of the RHS of (15), and (17) belongs to some $\mathcal{M}_k$.

Let us fix $k_0$ with $1 \le k_0 \le r' - 1$.

From the third term of the RHS of (15), we have two kinds of elements of the type $(d\pi)^{[2^{k_0}]}$:

(a) $c_1^{l_1/2} c_{(k_0)}^{l_0} (d\epsilon)^{[2^{k_0}]}$ for some $\epsilon$, where $l_1 2^{n_1-k_0-2} \in \mathbb{N}$, $l_1$ is even and $l$ satisfies the conditions for $k_0$.

(b) $c_1^{l_1/2+2^{k_0-1}} c_{(k_0)}^{l_0} (d\epsilon')^{[2^{k_0}]}$ for some $\epsilon'$, where $l_1 2^{n_1-k_0-2} \in \mathbb{N}$, $l_1$ is even and $l$ satisfies the conditions for $k_0$.

From equation (17) we have another two kinds of elements:

(c) $c_1^{l_1} c_{(k_0-1)}^{2l_0} (d\tau)^{[2^{k_0}]}$ for some $\tau$, where $l_1 2^{n_1-k_0-1} \in \mathbb{N}$, $l_1$ is odd and $l$ satisfies the conditions for $k_0 - 1$.

(d) $c_1^{l_1+2^{k_0-1}} c_{(k_0-1)}^{2l_0} (d\tau)^{[2^{k_0}]}$ for some $\tau$, where $l_1 2^{n_1-k_0-1} \in \mathbb{N}$, $l_1$ is odd and $l$ satisfies the conditions for $k_0 - 1$.

Let's check that each type belongs to $\mathcal{M}_{k_0}$.

(a) We know that $l_1 2^{n_1} \in 2^{k_0+2} \mathbb{N}$ so $(l_1/2) 2^{n_1} \in 2^{k_0+1} \mathbb{N}$ and since $l_i, i > 1$ satisfy conditions of level $k_0$ we see that this type of elements belong to $\mathcal{M}_{k_0}$.

(b) As in the previous case we have $(l_1/2) 2^{n_1} \in 2^{k_0+1} \mathbb{N}$ and then clearly $(l_1/2 + 2^{k_0-1}) 2^{n_1} \in 2^{k_0+1} \mathbb{N}$ because $n_1 \ge 2$. It is also clear that $l_i, i > 1$ satisfy conditions of level $k_0$. Since $0 \le l_1/2 < 2^{k_0-1}$ it follows that $l_1/2 + 2^{k_0-1} \le 2^{k_0} - 1$, we conclude that this type of elements are in $\mathcal{M}_{k_0}$.

(c) We know that $l_1 2^{n_1-k_0-1} \in \mathbb{N}$ so $l_1 2^{n_1} \in 2^{k_0+1} \mathbb{N}$. Since $l_i, i > 1$ satisfy conditions of level $k_0 - 1$, that is $l_i 2^{n_i} \in 2^{k_0} \mathbb{N}$, we see that $(2l_i) 2^{n_i} \in 2^{k_0+1} \mathbb{N}$ and because $0 \le 2l_i \le 2^{k_0} - 1$ we conclude that this type of elements are in $\mathcal{M}_{k_0}$.
(d) From type (c) we know that $l_1 2^{n_1} \in 2^{k_0+1}\mathbb{N}$, therefore $(l_1 + 2^{k_0-1}) 2^{n_1} \in 2^{k_0+1}\mathbb{N}$ because $n_1 \ge 2$. As in (c) we see that $\ell$ also satisfies conditions of level $k_0$. As in case (b) $l_1 + 2^{k_0-1} \le 2^{k_0} - 1$. Hence this type of elements are in $M_{k_0}$.

This completes the proof of the theorem.

## 5. INFINITE DIMENSIONAL EXTENSIONS

**5.1. Reduction to finite purely inseparable extensions.** We now consider the case of purely inseparable extensions of $F$ of infinite degree. Our main result in this case is the following theorem:

**Theorem 5.1.** Let $L$ be an algebraic purely inseparable extension of $F$. Then

$$H_2^{m+1}(L/F) = \bigcup_{\substack{F \subseteq K \subseteq L \\ K/F \text{ finite}}} H_2^{m+1}(K/F).$$

Clearly, Theorem 4.1 combined with Theorem 5.1 gives a complete computation of the kernel $H_2^{m+1}(L/F)$, where $L/F$ is a purely inseparable extension of infinite degree. The proof of Theorem 5.1 is based on the kernel $\text{Ker}(\Omega_F^m \to \Omega_L^m)$ which we describe in the following proposition:

**Proposition 5.2.** Let $L$ be a purely inseparable algebraic extension of $F$ (of any degree). Then for any integer $m \ge 1$

$$\text{Ker}(\Omega_F^m \to \Omega_L^m) = \bigcup_{\substack{F \subseteq K \subseteq L \\ K/F \text{ finite}}} \text{Ker}(\Omega_F^m \to \Omega_K^m)$$

*Proof.* It is clear that the RHS is contained in the LHS.

For the reverse inclusion, consider $w \in \Omega_F^m$ such that $w$ becomes 0 as an element in $\Omega_L^m$. Let's write $w_K$ when we consider $w \in \Omega_K^m$, for $K$ a field containing $F$. Thus we have $w_L = 0$.

Let's write

$$(18) \qquad w = \sum_{\text{finite}} a_{\alpha} db_{\alpha(1)} \wedge \cdots \wedge db_{\alpha(m)}$$

where $a_\alpha \in F$ and $B = \cup_\alpha \{b_{\alpha(1)}, \dots, b_{\alpha(m)}\}$ is a finite 2-independent set over $F$. Since $w_L = 0$, the set $B$ can not be 2-independent over $L$. Thus, there are scalars $\lambda_j \in L, j \in J$, for $J$ finite such that at least one of the $b_{\alpha(j)}$ can be expressed as a 2-combination in $L$ of the others. Assume without loss of generality that $b_{\alpha_1(1)}$ is a 2-combination of the others.

Let's define $F_1 = F(\lambda_j, j \in J)$ Since $L$ is algebraic over $F$ we conclude that $F_1$ is of finite degree over $F$.

Now consider $w$ as an element of $F_1$. Replace $b_{\alpha_1(1)}$ in each of terms of $w$ in (18) to get a similar expression for $w_{F_1}$ that does not involve $b_{\alpha_1(1)}$. Since $w_{F_1} \in$
Ker($\Omega_F^m \to \Omega_L^m$). Now arguing as before using $F_1$ instead of $F$ we see that we can eliminate another one of the $b$'s, say $b_{\alpha_2(1)}$, and build a finite extension $F_2$ of $F$, and as before $w_{F_2}$ can be expressed a sum like in (18) that does not contain neither $b_{\alpha_1(1)}$ nor $b_{\alpha_2(1)}$.

Given that $\mathcal{B}$ is a finite set, this process stops after a finite number of steps, where we have constructed $K \subseteq L$ a finite extension of $F$ so that $w_K = 0$, that is $w \in \text{Ker}(\Omega_F^m \to \Omega_K^m)$. $\square$

*Proof of Theorem 5.1.* We keep the same notations as in Theorem 5.1. Clearly

$$ \bigcup_{\substack{F \subseteq K \subseteq L \\ K/F \text{ finite}}} H_2^{m+1}(K/F) \subset H_2^{m+1}(L/F). $$

For the reverse inclusion, consider $w \in \Omega_F^m$ such that $\bar{w} \in H_2^{m+1}(L/F)$. Then there exist $u \in \Omega_L^m$ and $v \in \Omega_L^{m-1}$ such that $w = \wp(u) + d(v)$ in $\Omega_L^m$.

Since $L$ is an algebraic extension of $F$, we can find a finite extension of $F$, say $K$, where $u$ and $v$ of the above identity are defined, that is $u \in \Omega_K^m$ and $v \in \Omega_K^{m-1}$.

This means that $\delta := w_K + \wp(u) + d(v) \in \text{Ker}(\Omega_K^m \to \Omega_L^m)$. By Proposition 5.2 there exists a finite extension $M/K$ such that $M \subset L$ and $\delta_M = 0$. Hence, $\bar{w} \in H_2^{m+1}(M/F)$, and thus we are done since the extension $M/F$ is finite. $\square$

**5.2. Some computations of the kernel** $\text{Ker}(\Omega_F^m \to \Omega_L^m)$. We compute the kernel $\text{Ker}(\Omega_F^m \to \Omega_L^m)$, in two cases: $L/F$ is a modular purely inseparable extension, i.e., $L = F(\sqrt[2^n]{c_1}, \dots, \sqrt[2^n]{c_s})$ such that $[L : F] = 2^{n_1+\dots+n_s}$ and for $L = F(\sqrt[2^n]{c_1}, \dots, \sqrt[2^{n_s-1}]{c_{s-1}})(\sqrt[2^n]{c_s})$ with $[L : F] < 2^{n_1+\dots+n_s}$ and $[F(\sqrt[2^n]{c_1}, \dots, \sqrt[2^{n_s-1}]{c_{s-1}}) : F] = 2^{n_1+\dots+n_{s-1}}$.

**Proposition 5.3.** Let $c_1, \dots, c_s \in F$ and $L$ be the finitely generated purely inseparable extension $L = F(\sqrt[2^n]{c_1}, \dots, \sqrt[2^{n_s}]{c_s})$ of $F$ such that $[L : F] = 2^{n_1+\dots+n_s}$. Then

$$ \text{Ker}(\Omega_F^m \to \Omega_L^m) = \sum_{i=1}^{s} \Omega_F^{m-1} \wedge \text{d}c_i. $$

*Proof.* Clearly $\sum_{i=1}^s \Omega_F^{m-1} \wedge \text{d}c_i$ is contained in $\text{Ker}(\Omega_F^m \to \Omega_L^m)$ as $c_1, \dots, c_s$ are squares in $L$.

For the reverse inclusion we proceed by induction on $s$. The case $s = 1$ can be done as in [3, Lemma 2.4]. Suppose $s > 1$ and the result is true for any modular purely inseparable extension $L'$ of $F$ generated by less than $s$ radicals. Let $M = F(\sqrt[2^n]{c_1})$ and $r = \sqrt[2^n]{c_1}$.
Let $w \in \text{Ker}(\Omega_F^m \to \Omega_L^m)$. Then, $w_M \in \text{Ker}(\Omega_M^m \to \Omega_L^m)$. Since $L/M$ is modular, we deduce by induction hypothesis that

$$ (19) \qquad w_M = \sum_{i=2}^{s} \lambda_i \wedge dc_i, $$

where $\lambda_i \in \Omega_M^{m-1}$ for $2 \le i \le s$. As $c_1 \notin F^2$, there exists $\{e_i | i \in I\} \cup \{c_1\}$ a 2-basis of $F$, and thus $\{e_i | i \in I\} \cup \{r\}$ is a 2-basis of $M$. Now we may write $w = w_1 + w_2 \wedge dc_1$, where $w_1 \in \Omega_F^m$ and $w_2 \in \Omega_F^{m-1}$ are free from $dc_1$. Note that $w_M = (w_1)_M$. Similarly, we may write each $\lambda_i$, $2 \le i \le s$, as follows:

$$ (20) \qquad \lambda_i = \left( \sum_{j=0}^{2^{n_1}-1} r^j \lambda_{i,j}^1 \right) + \left( \sum_{j=0}^{2^{n_1}-1} r^j \lambda_{i,j}^2 \right) \wedge dr, $$

where $\lambda_{i,j}^1 \in \Omega_F^{m-1}$ and $\lambda_{i,j}^2 \in \Omega_F^{m-2}$. Now substituting (20) in (19) and taking the form that belongs to $\Omega_F^m$ (i.e., the integral part), we get $w_M = (w_1)_M = \sum_{i=2}^s \lambda_{i,0}^1 \wedge dc_i$. Using the kernel $\text{Ker}(\Omega_F^m \to \Omega_M^m) = \Omega_F^{m-1} \wedge dc_1$, we conclude that $w = \delta \wedge dc_1 + \sum_{i=2}^s \lambda_{i,0}^1 \wedge dc_i$ for a suitable $\delta \in \Omega_F^{m-1}$. $\square$

Now we consider the case of a non modular extension. Let $L = F(\sqrt[2^{n_1}]{c_1}, \dots, \sqrt[2^{n_s}]{c_s})$ be such that $c_s$ is not 2-independent of $\{c_1, \dots, c_{s-1}\}$ over the base field $F$, and $\{c_1, \dots, c_{s-1}\}$ is a 2-independent set over $F$. (This implies $c_i \notin F^2$ for $1 \le i \le s-1$). We also assume $c_s \notin F^2$.

Let $\rho_i = \sqrt[2^{n_i}]{c_i}$, $1 \le i \le s$. So we have a chain of fields

$$ F \subset F_1 = F(\rho_1) \subset F_2 = F_1(\rho_2) \subset \dots \subset F_s = F_{s-1}(\rho_s) = L. $$

Since $L/F$ is non modular and $F_{s-1}/F$ is modular, we get $[L : F_{s-1}] > 1$. Put $2^k = [L : F_{s-1}]$.

Our objective is to build a 2-basis of $L = F_s$ that allows us to compute the desired kernel.

Since $\{c_1, \dots, c_{s-1}\}$ is 2-independent we extend this set to a 2-basis of $F$: $B = \{c_1, \dots, c_{s-1}, b_1, b_2, \dots\}$. Then a 2-basis of $F_{s-1} = F(\rho_1, \dots, \rho_{s-1})$ is given by $B_{s-1} = \{\rho_1, \dots, \rho_{s-1}, b_1, b_2, \dots\}$.

We know that $c_s = \sum_{\epsilon} a_{\epsilon}^2 \underline{\rho}^{\epsilon}$, where $a_{\epsilon} \in F$, $\underline{c}^{\epsilon} = c_1^{\epsilon_1} \cdots c_{s-1}^{\epsilon_{s-1}}$ and $\epsilon$ is the multindex $\epsilon = (\epsilon_1, \dots, \epsilon_{s-1})$ with $\epsilon_j \in \{0, 1\}$ and since $c_i = \rho_i^{2^{n_i}}$ for $1 \le i \le s-1$, we have $c_s = \sum_{\epsilon} a_{\epsilon}^2 \underline{\rho}^{\epsilon}$ where $\epsilon$ is as above and $\underline{\rho} = \rho_1^{2^{n_1}} \cdots \rho_{s-1}^{2^{n_{s-1}}}$.

We can re-write the last expression as:

$$ c_s = \sum_{\epsilon} a_{\epsilon}^2 \underline{\rho}^{\epsilon} = \left( \sum_{\epsilon} a_{\epsilon} \underline{\rho}'^{\epsilon} \right)^2, $$

where $\epsilon$ is as above and $\underline{\rho}' = \rho_1^{2^{n_1}-1} \cdots \rho_{s-1}^{2^{n_{s-1}}}-1$. By the way this implies that in order for $F_s$ to be a proper extension of $F_{s-1}$ we must have $n_s > 1$.
We notice that $\rho_s^{2^{n_s-1}} = \sum_{\epsilon} a_{\epsilon} \underline{\rho}^{\epsilon}$. But this expression could be a square in $F_{s-1}$ or even a higher power of 2. Say that:

$$ \rho_s^{2^{n_s-1}} = \sum_{\epsilon} a_{\epsilon} \underline{\rho}^{\epsilon} = g^{2^r}, \text{ with } g \in F_{s-1} \setminus F_{s-1}^2. $$

Therefore $2^k = 2^{n_s-r-1}$, $k > 0$ and so $\rho_s^{2^k} = g \in F_{s-1}$.

Writing $g$ in terms of the 2-basis $B_{s-1}$ this can be expressed as:

$$ (21) \qquad \rho_s^{2^k} = g = \sum_{\mu=(\mu_1, \mu_2)} \lambda_\mu^2 \rho^{\mu_1} b^{\mu_2}, $$

with $\lambda_\mu \in F_{s-1}$. Here $\rho^{\mu_1} = \rho_1^{\mu_1,1} \cdots \rho_{s-1}^{\mu_1,s-1}$ and $b^{\mu_2} = \prod_j b_j^{\mu_2,j}$ with $\mu_{i,j} \in \{0,1\}$ and almost all $\mu_{2,j} = 0$.

Equation (21) gives a 2-dependence relation (over $F_s$) among

$$ \mathcal{D} = \{\rho_1, \dots, \rho_s, b_1, \dots, b_l\}. $$

Therefore in order to get a 2-basis of $F_s$ we choose one element of $\mathcal{D} \setminus \{\rho_s\}$ and replace $\rho_s$ in $B_s$. Here we have two choices: pick a $\rho_j$ or a $b_i$.

We consider the two cases separately:

**Case I** There is $\mu_1 \neq (0, \dots, 0)$.

In this case there is $\rho_j$, $1 \le j \le s-1$ in relation (21) and thus we clearly see that

$$ B_s = \{\rho_1, \dots, \rho_{j-1}, \rho_s, \rho_{j+1}, \dots, \rho_{s-1}, b_1, \dots\} $$

is a 2-basis of $F_s$.

**Case II** All $\mu_1 = (0, \dots, 0)$.

Since $g$ is not a square in $F_{s-1}$ at least one $\mu_2$ must be non zero. Let $b_{j_0}$ be the element with nonzero exponent in relation (21), whose subindex $j_0$ is the smallest. Then

$$ B_s = \{\rho_1, \dots, \rho_{s-1}, b_1, \dots, b_{j_0-1}, \rho_s, b_{j_0+1}, \dots\} $$

is a 2-basis of $F_s$.

Thus we have proved the following

**Lemma 5.4.** Let $c_1, \dots, c_s \in F \setminus F^2$ and $F_{s-1} = F(\sqrt[2^{n_1}]{c_1}, \dots, \sqrt[2^{n_{s-1}}]{c_{s-1}})$ be the finitely generated purely inseparable extension such that $[F_{s-1} : F] = 2^{n_1+\dots+n_{s-1}}$. Let $F_s = F_{s-1}(\sqrt[2^{n_s}]{c_s})$ be an extension of $F_{s-1}$ of degree $[F_s : F_{s-1}] = 2^k < 2^{n_s}$. Let $\rho_i = \sqrt[2^{n_i}]{c_i}$ for $1 \le i \le s$.

Given the 2-basis $B_{s-1} = \{\rho_1, \dots, \rho_{s-1}, b_1, \dots\}$ of $F_{s-1}$ where $b_i \in F$, then a 2-basis of $F_s$ is given by:

(1) $B_s = \{\rho_1, \dots, \rho_{j-1}, \rho_s, \rho_{j+1}, \dots, \rho_{s-1}, b_1, \dots\}$, in case I.

(2) $B_s = \{\rho_1, \dots, \rho_{s-1}, b_1, \dots, b_{j_0-1}, \rho_s, b_{j_0+1}, \dots\}$, in case II.
Now that we know a 2-basis of $F_s$ we can compute the kernel $W = \text{Ker}(\Omega_F^m \to \Omega_{F_s}^m)$

Let $w \in W$, then

$$w = \omega_1 \wedge \mathrm{d}c_1 + \cdots + \omega_{s-1} \wedge \mathrm{d}c_{s-1} + \sum_{\sigma \in \Sigma_{m,F}} \beta_{\sigma} \mathrm{d}b_{\sigma} \text{ with } \omega_i \in \Omega_F^{m-1}, \beta_{\sigma} \in F.$$

Since $c_1, \dots, c_{s-1}$ are squares in $F_{s-1} \subset F_s$, we may assume $w = \sum_{\sigma \in \Sigma_{m,F}} \beta_{\sigma} \mathrm{d}b_{\sigma}$ (here $\Sigma_{m,F}$ is the set of multi-indices as defined at the beginning of section 2).

**In Case I.** We know that $\{b_1, \dots\}$ is part of a 2-basis of $F_s$. Then $\{\mathrm{d}b_{\sigma}, \dots\}$ is part of a basis of $\Omega_F^m$ therefore $w = 0$ in $\Omega_{F_s}^m$ implies $\beta_{\sigma} = 0$ all $\sigma$, hence $w = 0$ in $\Omega_F^m$, and thus $W = \Omega_F^{m-1} \wedge \mathrm{d}c_1 + \cdots + \Omega_F^{m-1} \wedge \mathrm{d}c_{s-1}$.

**In Case II.** Here we can write $\rho_s^{2k} = g = \sum_{\mu_2} \lambda_{\mu_2}^2 \mathbf{b}^{\mu_2}$, with $\lambda_{\mu_2} \in F_{s-1}$.

Differentiating $g$ we obtain the following relation in $F_{s-1}$:

$$\mathrm{d}g = \sum_{J} \mathrm{D}_{b_{j}}(g)\mathrm{d}b_{j}, \text{ with } \mathrm{D}_{b_{j_{0}}}(g) \neq 0.$$

where $\mathrm{D}_{b_i}$ denotes the partial derivative with respect to $b_i$.

Since $g$ is a power of 2 in $F_s$ we have

$$(22) \qquad \sum_{J} \mathrm{D}_{b_j}(g) \mathrm{d}b_j = 0, \text{ in } \Omega_{F_s}.$$

Let $w \in \Omega_F^{m-1} \wedge (\mathrm{d}b_{j_0} + \gamma_{j_1} \mathrm{d}b_{j_1} + \cdots + \gamma_{j_t} \mathrm{d}b_{j_t})$ be such that $\gamma_{j_i} \in F$ satisfies $\mathrm{D}_{b_{j_i}}(g) = \gamma_{j_i} \mathrm{D}_{b_{j_0}}(g)$ for all $1 \le i \le t$. Then, $w \in W$. To this end, it suffices to prove that $w_0 := \mathrm{d}b_{j_0} + \gamma_{j_1} \mathrm{d}b_{j_1} + \cdots + \gamma_{j_t} \mathrm{d}b_{j_t} = 0 \in \Omega_{F_s}^1$. In fact, we have

$$\begin{align*} w_0 &= \mathrm{d}b_{j_0} + \gamma_{j_1} \mathrm{d}b_{j_1} + \cdots + \gamma_{j_t} \mathrm{d}b_{j_t} \\ &= \mathrm{d}b_1 + \frac{\mathrm{D}_{b_2}(g)}{\mathrm{D}_{b_1}(g)} \mathrm{d}b_2 + \cdots + \frac{\mathrm{D}_{b_{j_t}}(g)}{\mathrm{D}_{b_1}(g)} \mathrm{d}b_{j_t} \\ &= \left( \mathrm{D}_{b_1}(g) \mathrm{d}b_1 + \mathrm{D}_{b_2}(g) \mathrm{d}b_2 + \cdots + \mathrm{D}_{b_{j_t}}(g) \mathrm{d}b_{j_t} \right) / \mathrm{D}_{b_1}(g) \\ &= 0 \text{ in } \Omega_{F_s} \end{align*}$$

Conversely, assume that $0 \ne w \in W$. As before it is enough to consider $w = \sum_{\sigma \in \Sigma_{m,F}} \beta_{\sigma} \mathrm{d}b_{\sigma}$.

Since each $b_j, b_j \ne b_{j_0}$ is part of a 2-basis of $F_s$, in the above expression for $w$ we must have some $\sigma$ for which $j_0 \in \text{Im}(\sigma)$ (here $\text{Im}(\sigma)$ denotes the image of the multi-index $\sigma$).

Using (22) we see that $\mathrm{D}_{b_{j_0}}(g) \ne 0$ and so we can write

$$\mathrm{d}b_{j_0} = \sum_{j \ne j_0} \left( \frac{\mathrm{D}_{b_j}(g)}{\mathrm{D}_{b_{j_0}}(g)} \right) \mathrm{d}b_j$$
Inserting this into $w$ we have:

$$ 
\begin{aligned}
0 &= w = \sum_{\sigma \in \Sigma_{m,F}} \beta_{\sigma} db_{\sigma} \\
&= \sum_{\substack{\sigma \in \Sigma_{m,F} \\ j_0 \in \mathrm{Im}(\sigma)}} \beta_{\sigma} db_{\sigma} + \sum_{\substack{\sigma \in \Sigma_{m,F} \\ j_0 \notin \mathrm{Im}(\sigma)}} \beta_{\sigma} db_{\sigma}
\end{aligned}
$$

$$ 
\begin{aligned}
(23) \quad &= \sum_{\substack{\tau \in \Sigma_{m-1,F} \\ j_0 \notin \mathrm{Im}(\tau)}} \beta_{\tau} db_{\tau} \wedge db_{j_0} + \sum_{\substack{\sigma \in \Sigma_{m,F} \\ j_0 \notin \mathrm{Im}(\sigma)}} \beta_{\sigma} db_{\sigma} \\
&= \sum_{\substack{\tau \in \Sigma_{m-1,F} \\ j_0 \notin \mathrm{Im}(\tau)}} \beta_{\tau} db_{\tau} \wedge \left( \sum_{j \neq j_0} \frac{D_{b_j}(g)}{D_{b_{j_0}}(g)} db_j \right) + \sum_{\substack{\sigma \in \Sigma_{m,F} \\ j_0 \notin \mathrm{Im}(\sigma)}} \beta_{\sigma} db_{\sigma} \\
&= \sum_{\substack{\tau \in \Sigma_{m-1,F} \\ j_0 \notin \mathrm{Im}(\tau), j \neq j_0}} \beta_{\tau} \frac{D_{b_j}(g)}{D_{b_{j_0}}(g)} db_{\tau} \wedge db_j + \sum_{\substack{\sigma \in \Sigma_{m,F} \\ j_0 \notin \mathrm{Im}(\sigma)}} \beta_{\sigma} db_{\sigma}
\end{aligned}
$$

Using that $db_{\sigma} \in \Omega_{F_s}^m, j_0 \notin \mathrm{Im}(\sigma)$, are linearly independent we conclude that:

$$ 
\left\{ 
\begin{array}{l}
\beta_{\tau} D_{b_j}(g) = \beta_{\sigma} D_{b_{j_0}}(g), \\
\sigma = \tau \cup \{j\}, \quad j \neq j_0 \text{ and } j_0 \notin \mathrm{Im}(\tau).
\end{array}
\right.
$$

Using the expression of $w$ from the third line in (23), we get

$$ 
\begin{aligned}
w &= \sum_{\substack{\tau \in \Sigma_{m-1,F} \\ j_0 \notin \mathrm{Im}(\tau)}} \beta_{\tau} db_{\tau} \wedge db_{j_0} + \sum_{\substack{\sigma \in \Sigma_{m,F} \\ j_0 \notin \mathrm{Im}(\sigma)}} \beta_{\sigma} db_{\sigma} \\
&= \sum_{\substack{\tau \in \Sigma_{m-1,F} \\ j_0 \notin \mathrm{Im}(\tau)}} \beta_{\tau} db_{\tau} \wedge db_{j_0} + \sum_{\substack{\tau \in \Sigma_{m-1,F} \\ j_0 \notin \mathrm{Im}(\tau), j \neq j_0}} \beta_{\tau} \frac{D_{b_j}(g)}{D_{b_{j_0}}(g)} db_{\tau} \wedge db_j \\
&= \sum_{\substack{\tau \in \Sigma_{m-1,F} \\ j_0 \notin \mathrm{Im}(\tau)}} \beta_{\tau} db_{\tau} \wedge db_{j_0} + \sum_{\substack{\tau \in \Sigma_{m-1,F} \\ j_0 \notin \mathrm{Im}(\tau)}} \beta_{\tau} db_{\tau} \wedge \left( \sum_{j \neq j_0} \frac{D_{b_j}(g)}{D_{b_{j_0}}(g)} db_j \right) \\
&= \sum_{\substack{\tau \in \Sigma_{m-1,F} \\ j_0 \notin \mathrm{Im}(\tau)}} \beta_{\tau} db_{\tau} \wedge \left( db_{j_0} + \sum_{j \neq j_0} \frac{D_{b_j}(g)}{D_{b_{j_0}}(g)} db_j \right)
\end{aligned}
$$

Thus we have proved,

**Theorem 5.5.** Let $L = F_s$ be a purely inseparable extension of $F$ as above and $g$ defined by equation (21). Then using the 2-basis defined in Lemma 5.4, the kernel
W is given by:

$$ \mathrm{Ker}(\Omega_F^m \to \Omega_L^m) = \sum_{i=1}^{s-1} \Omega_F^{m-1} \wedge \mathrm{d}c_i, $$

in case I.

For case II,

$$ \mathrm{Ker}(\Omega_F^m \to \Omega_L^m) = \sum_{i=1}^{s-1} \Omega_F^{m-1} \wedge \mathrm{d}c_i + \Omega_F^{m-1} \wedge \left( \mathrm{d}b_{j_0} + \sum_{j \neq j_0} \frac{D_{b_j}(g)}{D_{b_{j_0}}(g)} \mathrm{d}b_j \right). $$

5.3. **Some examples.** We finish our paper with some examples illustrating the computations done in the previous subsection.

Let $K$ be a field of characteristic 2 with $\mathcal{U} = \{u_1, \dots\}$ a 2-basis. Consider $x, y, z$ indeterminates over $K$ and the field $F = K(x, y, z)$ of rational functions over $K$. A 2-basis of $F$ is given by $\mathcal{B} = \{x, y, z\} \cup \mathcal{U}$.

For the various cases of kernels we have the following examples:

**Example 5.6.** Consider $L = F(\sqrt{x}, \sqrt[4]{y}, \sqrt[8]{xyz^2})$, then $W_1 = \mathrm{Ker}(\Omega_F \to \Omega_L) = Fdx + Fdy$ and

$$ \mathrm{Ker}(\Omega_F^m \to \Omega_L^m) = \Omega_F^{m-1} \wedge \mathrm{d}x + \Omega_F^{m-1} \wedge \mathrm{d}y $$

*Proof.* We just compute $W_1$, the general case follows easily.

Let $\rho_1 = \sqrt{x}, \rho_2 = \sqrt[4]{y}$ and $\rho_3 = \sqrt[8]{xyz^2}$, also $F_1 = F(\rho_1), F_2 = F_1(\rho_2), L = F_2(\rho_3)$.

In $L$ we have $\rho_3^8 = xyz^2 = \rho_1^2\rho_2^4z^2$. If $g = \rho_1\rho_2^2z \in F_2 \setminus F_2^2$ then $\rho_3^4 = g$ and $[L:F] = 32$. A 2-basis of $F_3$ is given by $\mathcal{B}_3 = \{\rho_3, \rho_2, z\} \cup \mathcal{U}$.

$$ \text{Therefore } W_1 = \mathrm{Ker}(\Omega_F \to \Omega_L) = Fdx + Fdy. \quad \square $$

**Example 5.7.** Let $L = F(\sqrt[4]{x}, \sqrt[4]{y^2+xz^2})$, then $W_1 = \mathrm{Ker}(\Omega_F \to \Omega_L) = Fdx$ and

$$ \mathrm{Ker}(\Omega_F^m \to \Omega_L^m) = \Omega_F^{m-1} \wedge \mathrm{d}x $$

*Proof.* We just compute $W_1$.

As before, define $\rho_1 = \sqrt[4]{x}, \rho_2 = \sqrt[4]{y^2+xz^2}, F_1 = F(\rho_1)$, and $L = F_1(\rho_2)$. If $g = y + \rho_1^2 z \in F_1$, then $g \notin F_1^2$ and $\rho_2^4 = g^2 = y^2 + \rho_1^4 z^2$. Thus $\rho_2^2 = g = y + \rho_1^2 z$, hence $[L:F] = 8$.

$$ \text{Over } L \text{ we have } 0 = d\rho_2^2 = dg = dy + \rho_1^2 dz = D_y(g)dy + D_z(g)dz $$

$$ \text{Since } D_z(g)/D_y(g) = \sqrt{x} \notin F, \text{ the result follows.} \quad \square $$

**Example 5.8.** Now consider $L = F(\sqrt[4]{x}, \sqrt[8]{x(y^2+z^2)})$, then $W_1 = \mathrm{Ker}(\Omega_F \to \Omega_L) = Fdx + Fd(y+z)$, and

$$ \mathrm{Ker}(\Omega_F^m \to \Omega_L^m) = \Omega_F^{m-1} \wedge \mathrm{d}x + \Omega_F^{m-1} \wedge \mathrm{d}(y+z). $$
Proof. We just compute $W_1$, since rest is straight forward.

We set $\rho_1 = \sqrt[4]{x}$, $\rho_2 = \sqrt[8]{x(y^2 + z^2)}$, and so $\rho_2^8 = x(y^2 + z^2) = \rho_1^4(y + z)^2$. If $g = \rho_1^2(y + z) \in F_1 = F(\rho_1)$ then $\rho_2^4 = g$, hence $[L : F] = 16$.

Computing over $L$ we see that $0 = d(y + z)$ since it is a quotient of two squares. Therefore $d(y + z) \in W_1$.

We note that in this case $D_z(g)/D_y(g) = 1 \in F$, also notice that the class $w \wedge d(y+z)$ in $H_2^{m+1}(F)$ of $w \wedge d(y+z) \in \Omega_F^m$, an element that is in $H_2^{m+1}(L/F)$ and somehow hidden in the mixed term

$$ \sum_{k=1}^{r-1} \sum_{\substack{l_i 2^{n_i-k-1} \in \mathbb{N} \\ 0 \le l_i \le 2^k-1}} c_1^{l_1} \cdots c_s^{l_s} (d\Omega_F^{m-1})^{[2^k]} $$

of Theorem 4.1. To see this, take $k = 1, c_1 = x, c_2 = x(y^2 + z^2)$ and use the reduction (fact 2.3). $\square$

## REFERENCES

[1] H. Ahmad, *The Witt kernels of purely inseparable quartic extensions*, L. Algebra and its Appl. **395** (2005), 265-273.

[2] R. Aravire, R. Baeza, *Milnor's K-Theory and quadratic forms over fields of characteristic two*, Communications in Algebra **20** (1992), 1087-1107.

[3] R. Aravire, R. Baeza, *The behavior of quadratic and differential forms under function field extensions in characteristic two*, J. Algebra **259** (2003), 361-414.

[4] R. Aravire, B. Jacob, *The graded Witt group kernel of biquadratic extensions in characteristic two*, J. Algebra **370** (2012), 297-319.

[5] R. Aravire, B. Jacob, *Cohomology and graded Witt group kernels for extensions of degree four in characteristic two*, J. Algebra **461** (2016), 314-350.

[6] R. Aravire, A. Laghribi, *Some results on Witt kernels of quadratic forms for multiquadratic extensions*, Proc. Amer. Math. Soc. **141** (2013) 4191-4197.

[7] R. Aravire, A. Laghribi, M. O'Ryan, *Graded Witt kernels of the compositum of multiquadratic extensions with the function fields of Pfister forms*, J. Algebra **449** (2016), 635-659.

[8] J. K. Deveney, J. Mordeson, *Higher derivation Galois theory of inseparable field extensions*, Handbook of algebra, Vol. 1, 187-220, North-Holland, Amsterdam, 1996.

[9] M. Gerstenhaber, A. Zaromp, *On the Galois theory of purely inseparable field extensions*, Bull. Amer. Math. Soc. **76** (1970), 1011-1014.

[10] N. Jacobson, *Galois theory of purely inseparable fields of exponent one*, Amer. J. Math. **66** (1944), 645-648.

[11] K. Kato, *Symmetric bilinear forms, quadratic forms and Milnor K-theory in characteristic 2*, Inventiones Math. **66** (1982), 493-510.

[12] M. Sweedler, *Structure of Inseparable Extensions*, Annals of Mathematics, Second Series, Vol. **87**, No. 3 (1968), 401-410.

[13] M. Sobiech, *The behavior of differential and quadratic forms under purely inseparable field extensions*, preprint 2015.
UNIVERSIDAD ARTURO PRAT, FACULTAD DE INGENIERÍA Y ARQUITECTURA, INSTITUTO DE
CIENCIAS EXACTAS Y NATURALES, CASILLA 121, IQUIQUE, CHILE

E-mail address: raravire@unap.cl

UNIVERSITÉ D'ARTOIS, FACULTÉ DES SCIENCES JEAN PERRIN, LABORATOIRE DE MATHÉMATIQUES
DE LENS EA 2462, RUE JEAN SOUVRAZ - SP18, 62307 LENS, FRANCE

E-mail address: ahmed.laghribi@univ-artois.fr

UNIVERSIDAD DE TALCA, INSTITUTO DE MATEMÁTICA Y FÍSICA, CASILLA 721, TALCA, CHILE

E-mail address: moryan@inst-mat.utalca.cl