# Essential dimension of cohomology classes via valuation theory

Danny Ofek  and  Zinovy Reichstein  Department of Mathematics
University of British Columbia
Vancouver, BC V6T 1Z2
Canada

###### Abstract.

We give a formula for the essential dimension of a cohomology class $\alpha$
in $H^{d}(K,\mathbb{Q}_{p}/\mathbb{Z}_{p}(d))$ when $K$ is a strictly
Henselian field. This formula is particularly explicit in the case, where
$\alpha$ is a Brauer class (for $d=2$). As an application of our bound with
$d=3$, we study the essential dimension of exceptional groups by examining the
image of the Rost invariant.

###### Key words and phrases:

Essential dimension, Galois cohomology, Brauer group, valued field

###### 2020 Mathematics Subject Classification:

12G05, 12J20, 16K50, 19C30

Danny Ofek was partially supported by a graduate fellowship from the
University of British Columbia.

Zinovy Reichstein was partially supported by an Individual Discovery Grant
from the National Sciences and Engineering Research Council of Canada.

##  1\. Introduction

Let $k$ be a base field and $k\subset K$ be a field extension. Consider an
algebraic object $\alpha$ defined over $K$. That is, $\alpha\in\mathcal{F}(K)$
is an object of a covariant functor
$\mathcal{F}:\operatorname{Fields}_{k}\to\operatorname{Sets}$, where
$\operatorname{Fields}_{k}$ is the category of field extensions $K/k$ and
$\operatorname{Sets}$ is the category of sets. We think of the functor
$\mathcal{F}$ as specifying the type of object under consideration (a
quadratic form, an associative algebra, a Lie algebra, etc.) and
$\mathcal{F}(K)$ as the set of isomorphism classes of objects this type
defined over $K$. The essential dimension $\operatorname{ed}_{k}(\alpha)$ of
our object $\alpha\in\mathcal{F}(K)$ is the minimal transcendence degree
$\operatorname{trdeg}_{k}(K_{0})$ of an intermediate field $k\subset
K_{0}\subset K$ to which $\alpha$ descends. Informally speaking,
$\operatorname{ed}_{k}(\alpha)$ is the minimal number of parameters required
to define $\alpha$. The essential dimension
$\operatorname{ed}_{k}(\mathcal{F})$ of the functor $\mathcal{F}$ is then
defined as the maximal value of $\operatorname{ed}_{k}(\alpha)$ taken over all
$L\in\operatorname{Fields}_{k}$ and all $\alpha\in\mathcal{F}(L)$. These
definitions are recalled in a more formal way in Section 2. For surveys of
this research area, we refer the reader to [1, 29, 37].

Much of the work on essential dimension has centered on the functor

(1.1) $H^{1}(\ast,G)\colon K\mapsto H^{1}(K,G).$

Here $H^{1}(K,G)$ denotes the non-abelian cohomology set, whose elements are
isomorphism classes of (fppf) $G$-torsors over $\operatorname{Spec}(K)$. The
essential dimension of this functor is called the essential dimension of $G$
and is denoted by $\operatorname{ed}_{k}(G)$.

The main focus of this paper will be on a different functor,

$H_{p}^{d}\colon K\mapsto H^{d}(K,\mathbb{Q}_{p}/\mathbb{Z}_{p}(d)),$

where $p\neq\operatorname{char}k$ is a prime. We will study the essential
dimension of objects of this functor (i.e., cohomology classes) by valuation-
theoretic methods, in the spirit of [26, Section 3f] or [30].

Let $(F,\nu)$ be a valued field with value group $\mathbb{Z}^{r}$. Assume
$k\subset F$ and $\nu$ is trivial on $k$. There is a canonical homomorphism

$\wedge\nu:H^{d}_{p}(F)\longrightarrow\mathbb{Q}_{p}/\mathbb{Z}_{p}\otimes\bigwedge\mathbb{Z}^{r},$

where $\bigwedge\mathbb{Z}^{r}$ is the exterior algebra on $\mathbb{Z}^{r}$.
It is given, on symbols, by

$\wedge\nu(a_{1},\dots,a_{d})_{p^{n}}=\frac{1}{p^{n}}\otimes\nu(a_{1})\wedge\dots\wedge\nu(a_{d});$

for details see Section 3.

In Definition 6.1, we associate to any
$\omega\in\mathbb{Q}_{p}/\mathbb{Z}_{p}\otimes\bigwedge\mathbb{Z}^{r}$, a
finite abelian subgroup
$A_{\omega}\subset(\mathbb{Q}_{p}/\mathbb{Z}_{p})^{r}$. Given $\omega$, the
group $A_{\omega}$ is usually easy to compute. Our main result is the
following:

######  Theorem 1.1.

Let $(F,\nu)$ be a valued field with value group $\mathbb{Z}^{r}$. Assume
$\nu$ is trivial on a subfield $k\subset F$ with $\operatorname{char}k\neq p$.
Let $\alpha\in H^{d}_{p}(F)$ and $\omega=\wedge\nu(\alpha)$. Then

$\operatorname{ed}_{k}(\alpha)\geqslant\operatorname{ed}_{k}(\alpha;p)\geqslant\dim_{\mathbb{F}_{p}}(A_{\omega}/pA_{\omega}).$

Moreover, if $(F,\nu)$ is strictly Henselian, then equality holds:

$\operatorname{ed}_{k}(\alpha)=\operatorname{ed}_{k}(\alpha;p)=\dim_{\mathbb{F}_{p}}(A_{\omega}/pA_{\omega}).$

Here $\operatorname{ed}_{k}(\alpha;p)$ is _the essential dimension of $\alpha$
at $p$_; see Section 2.1. Note that first inequality in the statement of
Theorem 1.1,
$\operatorname{ed}_{k}(\alpha)\geqslant\operatorname{ed}_{k}(\alpha;p)$, is
immediate from the definition of $\operatorname{ed}_{k}(\alpha;p)$; see (2.1).
As a consequence of Theorem 1.1 we obtain the following:

######  Corollary 1.2.

Let $p$ be a prime and $k$ be a field of characteristic $\neq p$. Then

(1.2) $\operatorname{ed}_{k}(H^{d}_{p};p)=\infty$

for any $d\geqslant 2$.

For a proof, see Section 7.

### Central simple algebras and Brauer classes

Let us now consider the functor (1.1) in the special case, where $G$ is the
projective linear group $\operatorname{PGL}_{n}$. It is well known that the
functor $H^{1}(*,\operatorname{PGL}_{n})$ is isomorphic to

$\operatorname{CSA}_{n}\colon K\mapsto\\{\text{isomorphism classes of central
simple algebras of degree $n$ over $K$}\\}.$

Computing the essential dimension of $\operatorname{CSA}_{n}$ is a long-
standing problem, going back to Procesi [35, Section 2]. The deepest results
to date have been about $\operatorname{ed}_{k}(\operatorname{CSA}_{n};p)$.
Using primary decomposition it is easy to see that
$\operatorname{ed}(\operatorname{CSA}_{n};p)=\operatorname{ed}_{k}(\operatorname{CSA}_{p^{r}};p)$,
where $p^{r}$ is the highest power of $p$ dividing $n$. Thus we may assume
without loss of generality that $n=p^{r}$. When $r=1$, it is known that
$\operatorname{ed}(\operatorname{PGL}_{p};2)=2$; see, e.g., [39, Lemma 8.5.7].
For $r\geqslant 2$, we have

$(r-1)p^{r}+1\leqslant\operatorname{ed}(\operatorname{PGL}_{p^{r}};p)\leqslant
p^{2r-2}+1\,,$

where the lower bound is due to Merkurjev [28] and the upper bound is due to
Ruozzi [41]. In particular,

(1.3)
$\operatorname{ed}_{k}(\operatorname{PGL}_{p^{2}})=\operatorname{ed}_{k}(\operatorname{PGL}_{p^{2}};p)=p^{2}+1;$

see [27].

Now consider the morphism of functors
$\operatorname{CSA}_{n}\to\operatorname{Br}$ taking a central simple algebra
$A$ to its Brauer class $[A]$. If $A$ is not split and $k$ is algebraically
closed, then Tsen’s Theorem tells us that

(1.4) $\operatorname{ed}_{k}([A];p)\geqslant 2.$

We also have the obvious inequalities

(1.5)
$\operatorname{ed}_{k}([A])\leqslant\operatorname{ed}_{k}(A)\quad\text{and}\operatorname{ed}_{k}([A];p)\leqslant\operatorname{ed}_{k}(A;p)\,.$

The inequalities (1.5) may be sharp, because the essential dimension of
$\operatorname{M}_{s}(A)$ may be strictly smaller than the essential dimension
of $A$ for some $s>1$. In fact, this is exactly what happens when $A$ is a
universal division algebra of degree $4$. In this case (1.3) tells us that
$\operatorname{ed}_{k}(A)=\operatorname{ed}_{k}(A;2)=5$; on the other hand, by
[22, Corollary 1.4], $\operatorname{ed}_{k}(\operatorname{M}_{2}(A))=4$.

Using Theorem 1.1, we can give lower bounds on the essential dimension of some
Brauer classes that go beyond (1.4).

######  Theorem 1.3.

Let $p$ be a prime, $k$ be a field containing a primitive root of unity of
degree $p^{d}$ for every $d\geqslant 1$. Let $(F,\nu)$ be a valued field with
value group $\mathbb{Z}^{r}$. Assume that $k\subset F$ and $\nu_{|k}$ is
trivial. Let $\alpha\in\operatorname{Br}(F)$ is a sum of Brauer classes of
symbol algebras:

$\alpha=(a_{1},b_{1})_{p^{n}}+\dots+(a_{r},b_{r})_{p^{n}},$

for some integer $n\geqslant 1$. Consider the skew-symmetric matrix

$M=\sum_{i=1,\dots,k}\nu(a_{i})\nu(b_{i})^{t}-\nu(b_{i})\nu(a_{i})^{t}\in\operatorname{M}_{r}(\mathbb{Z}),$

where we view $\nu(a_{i})$ and $\nu(b_{i})\in\mathbb{Z}^{r}$ as $r\times 1$
matrices with integer entries and their transposes, $\nu(a_{i})^{t}$ and
$\nu(b_{i})^{t}$ as $1\times r$ matrices. Let $d_{1}\mid d_{2}\mid\dots\mid
d_{r}$ be the elementary divisors of $M$ and assume $i_{0}$ is the largest
subscript such that $p^{n}$ does not divide $d_{i_{0}}$. Then

(a) $\operatorname{ed}_{k}(\alpha;p)\geqslant i_{0}$, where $\alpha$ is viewed
as an object of the functor $\operatorname{Br}$. In particular, if $p^{r}$
does not divide $\det(M)$, then $\operatorname{ed}_{k}(\alpha)\geqslant r$.

(b) If $(F,\nu)$ is strictly Henselian, then
$\operatorname{ed}_{k}(\alpha)=\operatorname{ed}_{k}(\alpha;p)=i_{0}$.

The assumption that $\operatorname{char}k\neq p$ is crucial here. If
$\operatorname{char}(k)=p$, then (1.4) is tight; see Proposition 9.2.

As far as we know, Theorem 1.3 is the first known bound on
$\operatorname{ed}_{k}(\alpha)$ that is stronger than (1.4). There are some
lower bounds in the literature on the essential dimension of Brauer classes
considered as objects of $\operatorname{Br}_{p}$, where
$\operatorname{Br}_{p}(K)$ denotes the $p$-torsion subgroup of
$\operatorname{Br}(K)$ (see [23] and the last paragraph of Section 2 in [37]).
When a Brauer class $\alpha$ of exponent $p$ is considered as an object of
$\operatorname{Br}_{p}$, it is only allowed to descend to Brauer classes of
exponent $p$. When we view $\alpha$ as an object of $\operatorname{Br}$ (as we
do in the setting of Theorem 1.3), it is allowed to descend to a Brauer class
of higher exponent, so a priori the essential dimension may drop.

### Essential dimension of exceptional groups and the Rost invariant

A cohomological invariant of an algebraic group $G$ is a morphism of functors:

$\eta:H^{1}(\ast,G)\to H^{d}(\ast,M).$

Here $M$ is any discrete $\operatorname{Gal}(k)$-module. We refer the reader
to [44] for a detailed discussion of cohomological invariants. We will assume
that $\eta$ is normalized in the sense of [44, Chapter I, 4.5], i.e., takes
the trivial $G$-torsor to zero.

As an easy consequence of the definition of essential dimension, we see that

(1.6)
$\operatorname{ed}(G;p)\geqslant\operatorname{ed}(\gamma;p)\geqslant\operatorname{ed}(\eta(\gamma);p);$

cf. [37, Lemma 2.2]. Of particular interest to us will be the Rost invariant

$R_{G}\colon H^{1}(\ast,G)\to H^{3}(\ast,\mathbb{Q}_{p}/\mathbb{Z}_{p}(2)).$

Recall that the Rost invariant is defined for every semisimple simply
connected group $G$; see [19, Section 31B] or [25].

In Section 12 we will use the inequality (1.6) in combination with Theorem 1.1
to prove the following lower bounds:

(1.7) $\text{(i) \;}\operatorname{ed}_{k}(E_{7}^{\rm sc};2)\geqslant
7,\quad\quad\text{(ii) \;}\operatorname{ed}_{k}(E_{8};2)\geqslant
9,\quad\quad\text{(iii) \;}\operatorname{ed}_{k}(E_{8};3)\geqslant 5.$

Here the base field $k$ is assumed to be of characteristic different from $2$
in parts (i) and (ii) and different from $3$ in (iii). Specifically, we will
set $d=3$, $\eta=R_{G}$ to be the Rost invariant. The field $F$ will be
$F_{7}$, $F_{9}$, and $F_{5}$ in parts (i), (ii) and (iii), respectively,
where $F=F_{n}$ is the iterated power series field
$F_{n}=k((t_{0}))((t_{2}))\ldots((t_{n-1}))$ 111In part (iii), we will first
adjoin a primitive 3rd root of unity to $k$. This is harmless, as enlarging
$k$ does not increase the essential dimension..

Note that the inequalities (1.7) are known. In characteristic $0$ they were
first proved in [40]. In full generality (i) and (ii) were proved in [7] and
(iii) in [14]. Moreover, the arguments used in [7] and [14] show that there
exists an $E_{7}^{\rm sc}$-torsor $T\to\operatorname{Spec}(F_{7})$ such that
$\operatorname{ed}_{k}(T;2)=7$, and similarly in parts (ii) and (iii). What is
new here is that the lower bounds in (1.7) can, in fact, be extracted from the
Rost invariant of $T$ in the following sense.

######  Theorem 1.4.

Let $k$ be a base field of characteristic $\neq 2$ and $F_{n}$ be the iterated
Laurent series field $F_{n}=k((t_{0}))((t_{2}))\ldots((t_{n-1}))$. Then there
exist (i) an $E_{7}^{\rm sc}$-torsor $T_{1}\to\operatorname{Spec}(F_{7})$,
(ii) an $E_{8}$-torsor $T_{2}\to\operatorname{Spec}(F_{9})$, and (iii) an
$E_{8}$-torsor $T_{3}\to\operatorname{Spec}(F_{5})$ such that (i)
$\operatorname{ed}_{k}(R_{E_{7}^{\rm sc}}(T_{1});2)=7$, (ii)
$\operatorname{ed}_{k}(R_{E_{8}}(T_{2});2)=9$, and (iii)
$\operatorname{ed}_{k}(R_{E_{8}}(T_{3});3)=5$, respectively.

Here in (iii) we are assuming that $k$ contains a primitive $3$rd root of
unity.

Our proof of Theorem 1.4 in Section 12 relies on the formulas for the Rost
invariant, due to Chernousov [6] and Garibaldi [12]. We also note that the
exponential lower bounds on $\operatorname{ed}_{k}(\operatorname{Spin}_{n};2)$
from [4] cannot be recovered by this method; see Remark 12.1.

##  2\. Notation and preliminaries

###  2.1. Essential dimension

Let $k$ be a base field, $\operatorname{Fields}_{k}$ be the category of field
extensions $K/k$, $\operatorname{Sets}$ be the category of sets, and
$\mathcal{F}\colon\operatorname{Fields}_{k}\to\operatorname{Sets}$ be a
covariant functor. Given a field extension $K/k$, we will say that
$\alpha\in\mathcal{F}(K)$ _descends_ to an intermediate field $k\subseteq
K_{0}\subseteq K$ if $\alpha$ is in the image of the induced map
$\mathcal{F}(K_{0})\to\mathcal{F}(K)$. The essential dimension
$\operatorname{ed}_{k}(\alpha)$ is the smallest integer $d$ such that $\alpha$
descends to a field $k\subset K_{0}\subset K$ with
$\operatorname{trdeg}_{k}(K_{0})=d$. The essential dimension
$\operatorname{ed}_{k}(\alpha;p)$ of $\alpha$ at a prime $p$ is defined as the
minimal value of $\operatorname{ed}_{k}(\alpha_{L})$, where $L/K$ ranges over
field extensions whose degree $[L:K]$ is finite and prime to $p$.

The essential dimension $\operatorname{ed}_{k}(\mathcal{F})$ (respectively,
the essential $p$-dimension $\operatorname{ed}_{k}(\mathcal{F};p)$) of the
functor $\mathcal{F}$ is the supremum of $\operatorname{ed}_{k}(\alpha)$
(respectively, $\operatorname{ed}_{k}(\alpha;p)$) taken over all
$\alpha\in\mathcal{F}(K)$ with $K$ in $\operatorname{Fields}_{k}$. Clearly

(2.1)
$\operatorname{ed}_{k}(\alpha)\geqslant\operatorname{ed}_{k}(\alpha;p)\quad\text{and}\quad\operatorname{ed}_{k}(\mathcal{F})\geqslant\operatorname{ed}_{k}(\mathcal{F};p)$

for every object $\alpha$ of $\mathcal{F}$ and every prime $p$.

We also remark that if $f\colon\mathcal{F}\to\mathcal{G}$ is a morphism of
functors from $\operatorname{Fields}_{k}$ to $\operatorname{Sets}$, then

(2.2)
$\operatorname{ed}_{k}(\alpha)\geqslant\operatorname{ed}_{k}(f(\alpha))\quad\text{and}\quad\operatorname{ed}_{k}(\alpha;p)\geqslant\operatorname{ed}_{k}(f(\alpha);p)$

for every object $\alpha$ of $\mathcal{F}$; cf. [37, Lemma 2.2]. If $k\subset
k^{\prime}\subset L\subset L^{\prime}$ are fields with $k^{\prime}/k$
algebraic and $\alpha\in\mathcal{F}(L)$, then

(2.3)
$\operatorname{ed}_{k}(\alpha;p)\leqslant\operatorname{ed}_{k}(\alpha_{L^{\prime}};p)\quad\text{and}\quad\operatorname{ed}_{k}(\alpha;p)=\operatorname{ed}_{k^{\prime}}(\alpha;p).$

Here the inequality on the left follows from the fact that any prime-to-$p$
extension of $L$ embeds into a prime-to-$p$ extension of $L^{\prime}$, see
[26, Lemma 6.1]. The equality on the right follows from the fact that
$\operatorname{trdeg}_{k}(L_{0})=\operatorname{trdeg}_{k^{\prime}}(L_{0})$ for
any intermediate field $k\subset k^{\prime}\subset L_{0}\subset L^{\prime}$.

###  2.2. The norm residue isomorphism

Let $F$ be a field over $k$ and let $p$ be a prime different from
$\operatorname{char}k$. For any integer $d\in\mathbb{N}$ we set

$\mathbb{Q}_{p}/\mathbb{Z}_{p}(d):=\underset{r\in\mathbb{N}}{\operatorname{colim}}\
\mathbb{Z}/p^{r}(d);$

see [32, Definition 7.3.6]. Here the colimit is taken relative to the maps
$\mathbb{Z}/p^{r}\to\mathbb{Z}/p^{s}$ given by multiplication by ${p^{s-r}}$,
where $s\geqslant r\geqslant 0$. Note that $\mathbb{Q}_{p}/\mathbb{Z}_{p}(1)$
is isomorphic to the $p$-primary part of the group of roots of unity in
$k^{\operatorname{sep}}$. Set

$H_{p}(F):=\bigoplus_{d\in\mathbb{N}}H^{d}(F,\mathbb{Q}_{p}/\mathbb{Z}_{p}(d)).$

This abelian group is naturally graded and functorial in $F$. For any
$a_{1},\dots,a_{d}\in F^{*}$ and $n\in\mathbb{N}$, the Kummer map gives
cohomology classes $(a_{i})_{p^{n}}\in H^{1}(F,\mu_{p^{n}})\cong
F^{*}/F^{*p^{n}}$. The cup product of these classes defines a class
$(a_{1},\dots,a_{d})_{p^{n}}\in H^{d}_{p}(F)$ which is called a symbol of
degree $d$. The norm residue isomorphism theorem gives a simple presentation
for $H_{p}(F)$ in terms of symbols.

######  Theorem 2.1.

Let $K(F)$ denote the Milnor K-theory of a field $F$. For any prime $p$
different from $\operatorname{char}F$, there is an isomorphism of abelian
graded groups $\mathbb{Q}_{p}/\mathbb{Z}_{p}\otimes
K(F)\overset{h}{\to}H_{p}(F)$ given on generators by:

$h(\frac{1}{p^{n}}\otimes\\{a_{1},\dots,a_{d}\\})=(a_{1},\dots,a_{n})_{p^{n}}$

Theorem 2.1 is equivalent to the standard formulation of the norm residue
isomorphism theorem. We include an explanation of the equivalence in the
appendix.

######  Remark 2.2.

The norm residue isomorphism theorem is notoriously difficult to prove for
general fields. For strictly Henselian fields it is much simpler. In this
setting it is an easy consequence of [8, Theorem 2.6]; see [48, Corollary
3.13]. To simplify the exposition, we will appeal to the norm residue theorem
over general fields. Some readers may prefer to pass to the strict
Henselization of a field before applying the norm residue isomorphism theorem.
Doing this repeatedly will show that the proofs of our main results only
require the norm residue isomorphism theorem over strictly Henselian fields.

###  2.3. Notational conventions

Throughout this paper $k$ will denote a base field and $k^{\rm sep}$ will
denote the separable closure of $k$.

Let $A$ be an abelian group. For any positive integer $m$, we set $A/m:=A/mA$.
We will write elements of $A/m$ as $a\mod m$, where $a\in A$. If $p$ is a
prime, we will write

$A/p^{\infty}:=\mathbb{Q}_{p}/\mathbb{Z}_{p}\otimes_{\mathbb{Z}}A=\operatorname{colim}_{n\in\mathbb{N}}A/p^{n}.$

We denote the $n$-th exterior product of $A$ by $\bigwedge^{n}A$. When $A$ is
a finitely-generated free $\mathbb{Z}$ or $\mathbb{Z}/m$-module and
$e_{1},\ldots,e_{d}$ is a basis of $A$, $\bigwedge^{n}A$ is also a free module
generated by the “pure wedges” $e_{i_{1}}\wedge\ldots\wedge e_{i_{n}}$ for
some $1<i_{1}<\ldots<i_{n}\leqslant d$.

Note that there is a canonical isomorphism between $(\bigwedge^{n}A)/m$ and
$\bigwedge^{n}(A/m)$. We will identify these groups and simply write
$\bigwedge^{n}A/m$. The group $\bigwedge A/p^{\infty}$ is $(\bigwedge
A)/p^{\infty}$ and not $\bigwedge(A/p^{\infty})$ (the latter is the trivial
group).

We fix a compatible system of roots of unity $\zeta_{m}\in
k^{\operatorname{sep}}$ for all $m$ different from $\operatorname{char}(k)$.
That is, $\zeta_{m_{1}m_{2}}^{m_{1}}=\zeta_{m_{2}}$. This is equivalent to
fixing compatible isomorphisms $\mu_{m}\to\mathbb{Z}/m(1)$, where
$\mathbb{Z}/m(1)$ is the Tate twist of $\mathbb{Z}/m$; see [32, Definition
7.3.6].

A _valued field over_ $k$ is a field $F$ equipped with a valuation $\nu\colon
F^{*}\to\mathbb{Z}^{r}$ such that $k\subset F$, $\nu(k^{*})=0$. We will call
$(F,\nu)$ _strictly Henselian_ if it satisfies Hensel’s Lemma and its residue
field is separably closed. We will need to consider the _Henselization_ and
the _strict Henselization_ of a valued field. We refer the reader to [46,
Appendix A.3] for the definition and properties of the Henselization. The
strict Henselization of a valued field is the _inertial closure_ of its
Henselization (also called its maximal unramified extension) [46, Definition
A.20]. We will frequently use the fact that passing to the (strict)
Henselization of a valued field does not change the value group [46, Corollary
A.28].

An important example of a Henselian valuation is the
$(t_{0},\dots,t_{n-1})$-adic valuation $\nu:F_{n}^{*}\to\mathbb{Z}^{n}$ on the
field of iterated Laurent series $F_{n}=k((t_{0}))\dots((t_{n-1}))$, where $n$
is a positive integer; see [46, Example A.16]. Here $\mathbb{Z}^{n}$ is given
the reverse lexicographic ordering with respect to the standard basis
$e_{0},\dots,e_{n-1}$ and $\nu(t_{i})=e_{i}$ for any $0\leq i\leq n-1$. The
residue field of $(F_{n},\nu)$ is naturally identified with $k$. In
particular, $F_{n}$ is strictly Henselian if $k=k^{\operatorname{sep}}$.

We will denote the $d$-th Galois cohomology group of a
$\operatorname{Gal}(F)$-module $M$ by $H^{d}(F,M)$.

##  3\. The homomorphism $\wedge\nu$

Let $(F,\nu)$ be a valued field over $k$. That is, $F$ contains the base field
$k$ and $\nu_{|k}=0$ (remember that $\operatorname{char}k\neq p$). Denote the
value group $\nu F$ by $\Gamma$. Our first goal is to construct the
homomorphism $H_{p}(F)\to\bigwedge\Gamma/p^{\infty}$. Our starting point is
the following homomorphism.

######  Proposition 3.1.

There is a well-defined homomorphism of graded algebras:

$\wedge^{\prime}\nu:K(F)\to\bigwedge\Gamma,\ \
\\{a_{1},\dots,a_{d}\\}\mapsto\nu(a_{1})\wedge\dots\wedge\nu(a_{d}).$

Our proof will make use of the fact that

(3.1) If $\nu(a)>\nu(b)$, then $\nu(a+b)=\nu(b)$;

see, e.g., [21, p. 481].

###### Proof.

Since the function $(F^{*})^{d}\to\bigwedge^{d}\Gamma,\
(a_{1},\dots,a_{d})\mapsto\nu(a_{1})\wedge\dots\wedge\nu(a_{d})$ is
$\mathbb{Z}$-multilinear, it factors through the tensor product to give
homomorphisms:

$(F^{*})^{\otimes d}\to\bigwedge^{d}\Gamma,$

for all $d\in\mathbb{N}$. To check that these homomorphisms through $K(F)$ we
need to verify the Steinberg relation,

$\nu(a)\wedge\nu(1-a)=0,$

for every $a\in F^{*}$. Consider three cases.

Case 1. If $\nu(a)=0$, then $\nu(a)\wedge\nu(1-a)=0$ by bilinearity.

Case 2: If $\nu(a)>0$, then $\nu(1-a)=\nu(1)=0$; see (3.1). Hence,
$\nu(a)\wedge\nu(1-a)=0$.

Case 3. If $\nu(a)<0$ then $\nu(1-a)=\nu(a)$ by (3.1). In this case,
$\nu(a)\wedge\nu(1-a)=0$ by anti-symmetry. ∎

######  Remark 3.2.

It is clear from the definition of $\wedge^{\prime}\nu$ that it is functorial.
For any extension of valued fields $(F,\nu)\subset(\tilde{F},\tilde{\nu})$ and
any element $\alpha\in K(F)$ we have

$\wedge^{\prime}(\tilde{\nu})(\alpha_{\tilde{F}})=\wedge^{\prime}\nu(\alpha),$

under the natural identification of $\bigwedge\nu F$ with a subring of
$\bigwedge\tilde{\nu}\tilde{F}$.

Combining Proposition 3.1 with Theorem 2.1, we obtain:

######  Corollary 3.3.

There exists a well-defined homomorphism of graded abelian groups
$\wedge\nu:H_{p}(F)\to\bigwedge\Gamma/p^{\infty}$ given on symbols by:

$\wedge\nu(a_{1},\dots,a_{d})_{p^{n}}=\frac{1}{p^{n}}\otimes\nu(a_{1})\wedge\dots\wedge\nu(a_{d}).$

##  4\. First properties of $\wedge\nu$

Let $(F,\nu)$ be a valued field over $k$ as in the previous section. In this
section we will explore some basic properties of $\wedge\nu$. To this end, it
is convenient to choose a uniformizing parameter for $\nu$.

######  Definition 4.1.

A left inverse $\pi\colon\Gamma\to F^{*}$ to $\nu\colon F^{*}\to\Gamma$ will
be called a _uniformizing parameter_. Since the group operation in $\Gamma$ is
written additively while $F^{*}$ is written multiplicatively, it will be
convenient for us to use the exponential notation $\pi^{\gamma}$ in place of
$\pi(\gamma)$, for any $\gamma\in\Gamma$.

Note that a uniformizing parameter always exists because $\Gamma$ is a free
abelian group. From now on we will fix a uniformizing parameter $\pi$ for
$\nu$. Since elements in $H^{1}(F,\mathbb{Z}/{p^{n}})$ anti-commute for any
$n$, $\pi$ induces a left inverse $s_{\pi}:\bigwedge\Gamma/p^{\infty}\to
H_{p}(F)$ given on generators by:

$s_{\pi}(\frac{1}{p^{n}}\otimes\gamma_{1}\wedge\dots\wedge\gamma_{d})\mapsto(\pi^{\gamma_{1}},\dots,\pi^{\gamma_{d}})_{p^{n}}.$

We call the image of $s_{\pi}$ the group of $\pi$-monomial classes.

######  Definition 4.2.

An element $\alpha\in H_{p}(F)$ will be called _monomial_ if it is of the form
$\alpha=s_{\pi}(\omega)$ for some uniformizing parameter $\pi$ and class
$\omega\in\bigwedge\Gamma/p^{\infty}$.

It is important to keep in mind that any information captured by $\wedge\nu$
is contained in the subgroup of $\pi$-monomial classes, which is a small part
of $H_{p}(F)$. Moreover, $\wedge\nu$ allows us to split $H_{p}(F)$ as a direct
product of a divisible subgroup and the kernel of $\wedge\nu$. The following
lemma gives a convenient generating set for this kernel.

######  Lemma 4.3.

The kernel of $\wedge\nu:H_{p}^{d}(F)\to\bigwedge^{d}\Gamma/p^{\infty}$ is
generated by symbols $(a_{1},\dots,a_{d})_{p^{n}}$ with $\nu(a_{1})=0$.

###### Proof.

Let $U\subset H_{p}^{d}(F)$ be the subgroup generated by symbols
$(a_{1},\dots,a_{d})_{p^{n}}$ where $\nu(a_{1})=0$. It is clear that
$\wedge\nu(U)=0$. Note that any symbol $s=(a_{1},\dots,a_{d})_{p^{n}}\in
H_{p}^{d}(F)$ is equivalent to a $\pi$-monomial class modulo $U$. Indeed, if
we denote:

$u_{1}=(a_{1}\cdot\pi^{-\nu(a_{1})},a_{2},\dots,a_{d})_{p^{n}}\in U,$

then by multi-multiplicativity:

$s-u_{1}=(\pi^{\nu(a_{1})},a_{2},\dots,a_{d})_{p^{n}}\in U$

Proceeding iteratively and using anti-symmetry we find elements
$u_{2},\dots,u_{d}\in U$ such that:

$s-u_{1}-\dots-u_{d}=(\pi^{\nu(a_{1})},\dots,\pi^{\nu(a_{d})})_{p^{n}}.$

Thus for any element $\alpha\in H_{p}^{d}(F)$ mapped to $0$ under $\wedge\nu$
we can find an element $u\in U$ such that $\alpha-u$ is $\pi$-monomial. Since
$\wedge\nu(\alpha)=\wedge\nu(u)=0$ and $s_{\pi}$ is a section of $\wedge\nu$
this implies:

$\alpha-u=s_{\pi}(\wedge\nu(\alpha-u))=0.$

Therefore $\alpha\in U$, as we wanted to show. ∎

Let $(\hat{F},\nu)$ be the strict Henselization of $(F,\nu)$. The passage to
the Henselization may be viewed as a process of localization with respect to
$\nu$. The next corollary shows $\wedge\nu$ gives a local description of
$H_{p}(F)$. It can be easily deduced from a theorem of Wadsworth together with
the norm-residue isomorphism theorem for strictly Henselian fields [48,
Proposition 2.1]. In [5], Brussel used a similar description to study the
Brauer group of a strictly Henselian field; see also [46, Chapter 6].

######  Corollary 4.4.

Denote by $H_{p}(\hat{F}/F)\subset H_{p}(F)$ the subgroup of elements split by
$\hat{F}$. There is an exact sequence:

$0\to H_{p}(\hat{F}/F)\to
H_{p}(F)\overset{\wedge\nu}{\to}\bigwedge\Gamma/p^{\infty}\to 0.$

In particular, if $(F,\nu)$ is strictly Henselian, then $\wedge\nu$ is an
isomorphism.

###### Proof.

Since $(\hat{F},\nu)$ has the same value group as $(F,\nu)$, one has for all
$\alpha\in H(F)$:

$\wedge\nu(\alpha)=\wedge\nu(\alpha_{\hat{F}}).$

Therefore $H_{p}(\hat{F}/F)\subset\ker\wedge\nu$. For the reverse inclusion,
it suffices to check that symbols of the form $(a_{1},\dots,a_{d})_{p^{n}}$
with $\nu(a_{1})=0$ are split by $\hat{F}$ by Lemma 4.3. This follows from the
fact that if $a\in\hat{F}$ and $\nu(a)=0$, then $a$ has an ${p^{n}}$-th root
in $\hat{F}$. The residue class of $a$ has a ${p^{n}}$-th root in the residue
field of $\hat{F}$ because it is separably closed. The residue field has the
same characteristic as $k$ because $\nu(k^{*})=0$ and so this ${p^{n}}$-th
root may be lifted to $\hat{F}$ using Hensel’s lemma. ∎

##  5\. A lower bound on essential dimension

Now that we have constructed the invariant $\wedge\nu$, our goal is to show
that if $\wedge\nu(\alpha)$ is ”complicated enough” for $\alpha\in H_{p}(F)$,
then $\alpha$ cannot descend to a subfield $F_{0}\subset F$ of small
transcendence degree over $k$. The first step is to define a measure of
complexity for elements of $\bigwedge\Gamma/p^{\infty}$.

######  Definition 5.1.

The width $\rho(\omega)$ of an element $\omega\in\bigwedge\Gamma/p^{\infty}$
is defined as follows:

$\rho(\omega)=\min\bigg{\\{}\operatorname{rank}_{\mathbb{Z}}W\mid\
\begin{array}[]{cc}\text{Subgroups}\ W\subset\Gamma,\\\ \text{such that
}\omega\in\bigwedge W/p^{\infty}\end{array}\bigg{\\}}.$

This definition is clearly analogous to the definition of essential dimension.
Our next proposition makes the analogy precise. In the next section we will
explain how to compute $\rho(\omega)$ in general.

######  Proposition 5.2.

Let $(F,\nu)$ be a valued field with residue field $k$. Assume
$\operatorname{char}k\neq p$. Then

(a) $\operatorname{ed}_{k}(\beta;p)\geqslant\rho(\wedge\nu(\beta))$ for any
$\beta\in H_{p}(F)$

(b) Moreover, if $\beta$ is monomial (see Definition 4.2), then
$\operatorname{ed}_{k}(\beta)=\operatorname{ed}_{k}(\beta;p)=\rho(\wedge(\beta))$.

###### Proof.

(a) Let $F\subset L$ be a prime to $p$ extension such that
$\operatorname{ed}_{k}(\beta_{L})=\operatorname{ed}_{k}(\beta;p)$. By Lemma
13.3 we can choose an extension $\tilde{\nu}$ of $\nu$ to $L$ such that
$[\tilde{\nu}L:\nu E]$ is prime to $p$. By Lemma 13.4 for any subgroup
$W\subset\tilde{\nu}L$ we have:

(5.1) $\bigwedge W\cap\nu E/p^{\infty}=\bigwedge W/p^{\infty}.$

This implies:

(5.2) $\rho(\wedge\nu(\beta))=\rho(\wedge\tilde{\nu}(\beta_{L})).$

Assume that $\beta_{L}$ descends to $k\subset L_{0}\subset L$. Then
$\wedge\tilde{\nu}(\beta_{L})\in\bigwedge\tilde{\nu}L_{0}/p^{\infty}$, and
(5.2) gives

(5.3) $\operatorname{rank}(\nu
L_{0})\geqslant\rho(\wedge\tilde{\nu}(\beta_{L}))=\rho(\wedge\nu(\beta)).$

Now recall that by [17, Chapter XVII, Section 4, Theorem II],
$\operatorname{trdeg}_{k}(L_{0})\geqslant\operatorname{rank}(\nu L_{0})$; see
also [49, Chapter VI, Theorem 3, Corollary 1], [30, Theorem 3.1]. Combining
this with (5.3), we obtain
$\operatorname{trdeg}_{k}(L_{0})\geqslant\rho(\wedge\nu(\beta))$. Therefore,
$\operatorname{ed}_{k}(\beta;p)=\operatorname{ed}_{k}(\beta_{L})\geqslant\rho(\wedge\nu(\beta))$.

(b) Now assume that $\beta$ is monomial. Then there exists a uniformizer $\pi$
of $\nu$ and a class $\omega\in\bigwedge\Gamma/p^{\infty}$ such that
$s_{\pi}(\omega)=\beta$. By definition we can find a subgroup $W\subset\Gamma$
such that $\omega\in\bigwedge W/p^{\infty}$ and

$\operatorname{rank}(W)=\rho(\wedge\nu(\beta)).$

Choose a basis $e_{1},\dots,e_{r}$ of $W$ and set
$F_{0}=k(\pi^{e_{1}},\dots,\pi^{e_{r}})$. Since $\omega\in\bigwedge
W/p^{\infty}$ and $W=\tilde{\nu}F_{0}$, we have:

$\beta=s_{\pi}(\omega)\in\operatorname{im}(H_{p}(F_{0})\to H_{p}(F)).$

We conclude that if $\beta$ is monomial, then

$\operatorname{ed}_{k}(\beta)\leqslant\operatorname{trdeg}_{k}(F_{0})\leqslant
r=\rho(\wedge\nu(\beta)).$

Combining this inequality with part (a) and remembering (2.1), we deduce part
(b). ∎

##  6\. Computing $\rho$

In this section we show $\rho$ is relatively easy to compute using linear
algebra. Our main tools are the contraction maps of $\bigwedge\Gamma$.
Contraction is a way of “applying” an element of $\bigwedge\Gamma$ to an
element of the dual group
$\Gamma^{*}=\operatorname{Hom}_{\mathbb{Z}}(\Gamma,\mathbb{Z})$ that
generalizes multiplying a vector by a matrix (this special case will play a
key role in Section 8).

######  Definition 6.1.

For any $\eta\in\bigwedge\Gamma^{*}$, we will write
$\iota_{\eta}:\bigwedge\Gamma/p^{\infty}\to\bigwedge\Gamma/p^{\infty}$ for the
tensor product
$\operatorname{id}_{\mathbb{Q}_{p}/\mathbb{Z}_{p}}\otimes\iota^{\prime}_{\eta}$
where $\iota^{\prime}_{\eta}:\bigwedge\Gamma\to\bigwedge\Gamma$ is the usual
contraction map from linear algebra, see [3, Chapter III, p. 602]. We recall
the explicit formula for $\iota_{\eta}$ because it will be used often. For any
$f\in\Gamma^{*}$, $n\in\mathbb{N}$ and $\gamma_{1},\dots,\gamma_{d}\in\Gamma$
we have:

$\iota_{f}(\frac{1}{p^{n}}\otimes\gamma_{1}\wedge\dots\wedge\gamma_{d})=\sum_{i=1,\dots,d}\frac{f(\gamma_{i})(-1)^{i-1}}{p^{n}}\otimes(\gamma_{1}\wedge\dots\hat{\wedge\gamma_{i}\wedge}\dots\wedge\gamma_{d}).$

Here $\hat{\gamma_{i}}$ means ”omit $\gamma_{i}$”. This formula determines
$\iota_{f}$ uniquely by linearity together with the condition
$\iota_{f}(\frac{1}{p^{n}}\otimes 1_{\bigwedge\Gamma})=0$, for all
$n\in\mathbb{N}$. For general $\eta\in\bigwedge\Gamma^{*}$, $\iota_{\eta}$ is
defined by linearity in $\eta$ and by the formula:

$\iota_{f_{1}\wedge\dots\wedge
f_{d}}=\iota_{f_{1}}\circ\dots\circ\iota_{f_{d}}.$

We use the convention $\iota_{1}(\cdot)=\operatorname{id}_{\bigwedge\Gamma}$,
where $1=1_{\bigwedge\Gamma^{*}}$ is the unit of $\bigwedge\Gamma^{*}$. Note
that for $\eta\in\bigwedge^{d}\Gamma^{*}$ and
$\omega\in\bigwedge^{t}\Gamma/p^{\infty}$ we have
$\iota_{\eta}(\omega)\in\bigwedge^{t-d}\Gamma/p^{\infty}$. In particular, if
$d>t$, then $\iota_{\eta}(\omega)=0$. For any
$\omega\in\bigwedge\Gamma/p^{\infty}$, we denote by
$A_{\omega}\subset\Gamma/p^{\infty}$ the finite subgroup generated by degree
$1$ parts $[\iota_{\eta}(\omega)]_{1}$ as we vary over
$\eta\in\bigwedge\Gamma^{*}$.

The next proposition shows that we can compute $\rho(\omega)$ using
contractions.

######  Proposition 6.2.

Let $\omega\in\bigwedge\Gamma/p^{\infty}$. The following hold:

  1. (1)

If $W\subset\Gamma$ is a subgroup such that $\omega\in\bigwedge W/p^{\infty}$,
then $A_{\omega}\subset W/p^{\infty}$.

  2. (2)

The minimal number of generators of $A_{\omega}$ is $\rho(\omega)$.

  3. (3)

$\rho(\omega)=\dim_{\mathbb{F}_{p}}A_{\omega}/p$.

###### Proof.

(1) It is clear from the definition that for any $\eta\in\bigwedge\Gamma^{*}$
and subgroup $W\subset\Gamma$:

$\iota_{\eta}\bigwedge W/p^{\infty}\subset\bigwedge W/p^{\infty}.$

Therefore if $\omega\in\bigwedge W/p^{\infty}$, then
$\iota_{\eta}(\omega)\in\bigwedge W/p^{\infty}$. Letting $\eta$ vary over
$\bigwedge\Gamma^{*}$, we see that $A_{\omega}\subset W/p^{\infty}$.

(2) Assume that $\omega\in\bigwedge W/p^{\infty}$ for some subgroup
$W\subset\Gamma$. By part (1), $A_{\omega}\subset W/p^{\infty}$. For some $n$,
we have $p^{n}\omega=0$ and so:

$A_{\omega}\subset\frac{1}{p^{n}}\mathbb{Z}_{p}/\mathbb{Z}_{p}\otimes W\cong
W/p^{n}W.$

Therefore $A_{\omega}$ is generated by $\operatorname{rank}W$ elements [46,
Proposition A.36]. Taking $\operatorname{rank}W$ to be minimal we get that
$A_{\omega}$ is generated by $\rho(\omega)$ elements.

It remains to show that $A_{\omega}$ cannot be generated by fewer than
$\rho(\omega)$ elements. Suppose that $A_{\omega}$ is generated by $m$
elements. Our goal is to prove $\rho(\omega)\leqslant m$. Let
$u_{1},\dots,u_{m}\in\Gamma$ and $n\in\mathbb{N}$ be elements such that:

(6.1) $A_{\omega}=\langle\frac{1}{p^{n}}\otimes u_{i}\mid i=1,\dots,m\rangle.$

Let $U$ be the saturation of the subgroup generated by $u_{1},\dots,u_{m}$ in
$\Gamma$. Then

(6.2) $\operatorname{rank}U\leqslant m.$

Now, we use the fact that any saturated subgroup of $\Gamma$ is a direct
summand. Therefore we may assume that $\Gamma=\mathbb{Z}^{r}$ with basis
$e_{1},\dots,e_{r}$ and $U=\langle e_{1},\dots,e_{s}\rangle$, where
$s=\operatorname{rank}U$. There are integers $a_{i_{1},\dots,i_{d}}$ indexed
by increasing sequences such that:

$\omega=\sum_{1\leqslant i_{1}<\dots<i_{d}\leqslant
r}\frac{a_{i_{1},\dots,i_{d}}}{p^{n}}\otimes e_{i_{1}}\wedge\dots\wedge
e_{i_{d}}.$

Let $j_{1}<\dots<j_{d}$ be a sequence such $j_{d}>s$. It suffices to show
$a_{j_{1},\dots,j_{d}}$ is divisible by $p^{n}$ for any such sequence. Let
$e^{1},\dots,e^{r}\in\Gamma^{*}$ be the dual basis of $\Gamma^{*}$. We denote:

$\eta=e^{j_{1}}\wedge\dots\wedge\hat{e^{j_{t}}}\wedge\dots\wedge e^{j_{d}},$

where $\hat{e^{j_{t}}}$ means “omit $e^{j_{t}}$”, and compute the degree $1$
homogeneous part of $\iota_{\eta}(\omega)$:

$\displaystyle[\iota_{\eta}(\omega)]_{1}$ $\displaystyle=\sum_{1\leqslant
i_{1}<\dots<i_{d}\leqslant
r}\frac{a_{i_{1},\dots,i_{d}}}{p^{n}}\otimes[\iota_{\eta}(e_{i_{1}}\wedge\dots\wedge
e_{i_{d}})]_{1}$ $\displaystyle=\pm\frac{a_{j_{1},\dots,j_{d}}}{p^{n}}\otimes
e_{j_{d}}+\sum_{i\neq j_{d}}\frac{b_{i}}{p^{n}}\otimes e_{i}.$

for some $b_{i}\in\mathbb{Z}$. Since $[\iota_{\eta}(\omega)]_{1}\in
A_{\omega}$, (6.1) implies there exists $u\in U=\langle
e_{1},\dots,e_{s}\rangle$ such that:

$[\iota_{\eta}(\omega)]_{1}=\frac{1}{p^{n}}\otimes u.$

We conclude that $a_{j_{1},\dots,j_{d}}$ is divisible by $p^{n}$ because
$j_{d}>s$. This shows that $\omega\in\bigwedge U/p^{\infty}$. Together with
(6.2) this gives $\rho(\omega)\leqslant s=\operatorname{rank}U\leqslant m$, as
desired.

(3) Since $A_{\omega}$ is an abelian $p$-group, the minimal number of
generators of $A_{\omega}$ equals $\dim_{\mathbb{F}_{p}}(A_{\omega}/p)$. ∎

##  7\. Proof of Theorem 1.1 and Corollary 1.2

Putting together the results of the previous two sections we are now able to
prove our main result, Theorem 1.1. We restate it here in slightly greater
generality.

######  Theorem 7.1.

Let $(F,\nu)$ be a valued field over $k$ with value group $\Gamma$. Assume
$\operatorname{char}k\neq p$. Let $\alpha\in H_{p}(F)$ be a cohomology class,
$\omega=\wedge\nu(\alpha)\in\bigwedge\Gamma/p^{\infty}$, and
$A_{\omega}\subset\Gamma/p^{\infty}$ be the subgroup associated to $\omega$;
see Definition 6.1. Then

(7.1)
$\operatorname{ed}_{k}(\alpha;p)\geqslant\dim_{\mathbb{F}_{p}}(A_{\omega}/pA_{\omega}).$

Furthermore, if $\alpha$ is monomial (see Definition 4.2), then

(7.2)
$\operatorname{ed}_{k}(\alpha)=\operatorname{ed}_{k}(\alpha;p)=\dim_{\mathbb{F}_{p}}(A_{\omega}/pA_{\omega}).$

In particular, equality (7.2) holds whenever $(F,\nu)$ is strictly Henselian.

###### Proof.

Both (7.1) and (7.2) follow immediately from Proposition 5.2 and Proposition
6.2. If $(F,\nu)$ is strictly Henselian, then every class is monomial (see
Corollary 4.4) and so (7.2) holds in that case. ∎

We now turn to the proof of Corollary 1.2. We need to show that for any
$d\geqslant 2$ and any prime number $p$ there exist objects of $H^{d}_{p}(*)$
of arbitrarily large essential $p$-dimension. This follows from Proposition
7.2 below: just let $r\longrightarrow\infty$.

######  Proposition 7.2.

Let $r\geqslant 1$ and $d\geqslant 2$ be integers. Let

$F_{rd}=k((a_{11}))\ldots((a_{1d}))((a_{21}))\ldots((a_{2d}))\ldots((a_{rd})),$

be the field of iterated Laurent series in $rd$ variables $a_{ij}$ over $k$ .
Consider the class $\alpha\in H_{p}^{d}(K)$ given by

$\alpha=(a_{11},\ldots,a_{1d})_{p^{n}}+\ldots+(a_{r1},\ldots,a_{rd})_{p^{n}}.$

Then $\operatorname{ed}_{k}(\alpha)=\operatorname{ed}_{k}(\alpha;p)=rd$.

###### Proof.

Let $\nu:F_{rd}\to\mathbb{Z}^{rd}$ be the $(a_{11},\dots,a_{rd})$-adic
valuation on $F_{rd}$ (see Section 2.3) and recall that $\nu(a_{ij})=e_{ij}$
for all $i,j$, where $e_{ij}$ is the natural basis of $\mathbb{Z}^{rd}$. We
now evaluate $\omega=\wedge\nu(\alpha)$:

$\omega=\frac{1}{p^{n}}\otimes(e_{11}\wedge\ldots\wedge
e_{1d}+e_{21}\wedge\ldots\wedge e_{2d}+\ldots+e_{r1}\wedge\ldots\wedge
e_{rd}).$

Denote by $e^{ij}\in(\mathbb{Z}^{rd})^{*}$ the dual basis. That is, for all
$i,i^{\prime},j,j^{\prime}$:

$e^{ij}(e_{i^{\prime}j^{\prime}})=\begin{cases}1&\text{if }i=i^{\prime}\
\text{and }j=j^{\prime}\\\ 0&\text{otherwise}.\end{cases}$

For any $1\leqslant i,j\leqslant r$ let
$\eta=e^{1j}\wedge\dots\wedge\hat{e^{ij}}\wedge\dots\wedge
e^{dj}\in\bigwedge^{d-1}\Gamma^{*}$ be the wedge product where we omit
$e^{ij}$. We have:

$\iota_{\eta}(\omega)=\frac{1}{p^{n}}\otimes e_{ij}.$

Therefore $\mathbb{Z}^{rd}/p^{n}\subset A_{\omega}$. The inclusion
$A_{\omega}\subset\mathbb{Z}^{rd}/p^{n}$ is checked easily using the
definition of the contraction map. Therefore
$A_{\omega}=\mathbb{Z}^{rd}/p^{n}$ and $A_{\omega}/p\cong\mathbb{F}_{p}^{rd}$.
Since $\alpha$ is monomial, Theorem 7.1 gives
$\operatorname{ed}_{k}(\alpha)=\operatorname{ed}_{k}(\alpha;p)=rd$. ∎

##  8\. Proof of Theorem 1.3

In this section we will deduce Theorem 1.3 from Theorem 7.1. Let $(F,\nu)$ be
a valued field over $k$ with value group $\Gamma=\mathbb{Z}^{r}$. There is a
natural identification of $\bigwedge^{2}\mathbb{Z}^{r}/p^{\infty}$ with the
skew-symmetric matrices in $M_{r}(\mathbb{Q}_{p}/\mathbb{Z}_{p})$ by the
homomorphism:

$\frac{1}{p^{n}}\otimes u\wedge v\mapsto\frac{1}{p^{n}}(uv^{t}-vu^{t}).$

Here we view $u,v\in\mathbb{Z}^{r}$ as $r\times 1$ matrices (i.e. column
vectors) and their transposes, $u^{t}$ and $v^{t}$ as $1\times r$ matrices
(i.e., row vectors) with integer entries. The products $uv^{t}$ and $vu^{t}$
are $r\times r$ matrices.

Under the above identification, the contraction map is given by matrix
multiplication. That is, if $\omega\in\bigwedge^{2}\mathbb{Z}^{r}/p^{\infty}$
is identified with the matrix $M\in M_{r}(\mathbb{Q}_{p}/\mathbb{Z}_{p})$ and
we identify $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}^{r},\mathbb{Z})$ with
$\mathbb{Z}^{r}$ using the dot product, then:

$\iota_{(\cdot)}(\omega):\mathbb{Z}^{r}\to\bigwedge^{1}\Gamma/p^{\infty}=\mathbb{Z}^{r}/p^{\infty}=(\mathbb{Q}_{p}/\mathbb{Z}_{p})^{r}.$

is given by $\iota_{v}(\omega)=Mv$. To see this, assume that
$\omega=\frac{1}{p^{n}}u\wedge w$ for some $u,w\in\mathbb{Z}^{r}$ and
calculate using Definition 6.1:

$Mv=\frac{1}{p^{n}}(uw^{t}-wu^{t})v=\frac{1}{p^{n}}(w^{t}v)u-\frac{1}{p^{n}}(u^{t}v)w=\iota_{v}(\omega).$

We now proceed with the proof of Theorem 1.3. In fact, we will prove a
slightly stronger assertion, Theorem 8.1 below. If $k$ contains a primitive
root of unity of degree $p^{d}$ for every $d\geqslant 1$, then
$\mathbb{Q}_{p}/\mathbb{Z}_{p}(2)$ is isomorphic to
$\mathbb{Q}_{p}/\mathbb{Z}_{p}(1)$. For any field $F$ containing $k$,
$H_{p}^{2}(F)=H^{2}(F,\mathbb{Q}_{p}/\mathbb{Z}_{p}(2))$ is then isomorphic to
the $p$-primary part of $\operatorname{Br}(F)$. Moreover, under this
isomorphism the symbols $(a,b)_{p^{n}}$ correspond to Brauer classes of cyclic
algebras; see [15, Proposition 4.7.1].

######  Theorem 8.1.

Let $(F,\nu)$ be a valued field over $k$ with value group $\mathbb{Z}^{r}$ and
$\operatorname{char}k\neq p$. Assume $\alpha\in H_{p}^{2}(F)$ is a sum of
symbols,

$\alpha=(a_{1},b_{1})_{p^{n}}+\dots+(a_{k},b_{k})_{p^{n}}.$

Let $M\in M_{r}(\mathbb{Z})$ be the following skew-symmetric matrix:

$M=\sum_{i=1,\dots,k}\nu(a_{i})\nu(b_{i})^{t}-\nu(b_{i})\nu(a_{i})^{t}.$

Assume $d_{1}\mid d_{2}\mid\dots\mid d_{r}$ are the elementary divisors of $M$
and $i_{0}$ is the largest subscript such that ${p^{n}}$ does not divide
$d_{i_{0}}$. The following hold

(a) $\operatorname{ed}_{k}(\alpha;p)\geqslant i_{0}$. In particular, if
$p^{r}$ does not divide $\det(M)$, then
$\operatorname{ed}_{k}(\alpha)\geqslant r$.

(b) If $(F,\nu)$ is strictly Henselian, then
$\operatorname{ed}_{k}(\alpha)=\operatorname{ed}_{k}(\alpha;p)=i_{0}$.

###### Proof.

Denote $\omega=\wedge\nu(\alpha)$. By Corollary 3.3:

$\omega=\frac{1}{p^{n}}\otimes(\nu(a_{1})\wedge\nu(b_{1})+\dots+\nu(a_{k})\wedge\nu(b_{k})).$

Under the identification of $\bigwedge^{2}\mathbb{Z}^{r}/p^{\infty}$ with
skew-symmetric matrices in $M_{r}(\mathbb{Q}_{p}/\mathbb{Z}_{p})$ we have:

$\omega=\frac{1}{p^{n}}\sum_{i=1,\dots,k}\nu(a_{i})\nu(b_{i})^{t}-\nu(b_{i})\nu(a_{i})^{t}=\frac{1}{p^{n}}M.$

As explained above, we can identify
$\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}^{r},\mathbb{Z})$ with
$\mathbb{Z}^{r}$ so that $\iota_{v}(\omega)$ is given by matrix
multiplication:

(8.1)
$\iota_{v}(\omega)=\frac{1}{p^{n}}Mv\in(\mathbb{Q}_{p}/\mathbb{Z}_{p})^{r}.$

Set $A_{\omega}=\langle\iota_{v}(\omega)\mid
v\in\mathbb{Z}^{r}\rangle\subset(\mathbb{Q}_{p}/\mathbb{Z}_{p})^{r}$, as in
Definition 6.1. Let $d_{1}\mid d_{2}\mid\dots\mid d_{r}$ be the elementary
divisors of $M\in M_{r}(\mathbb{Z})$ and let $\displaystyle
k_{i}:=\frac{p^{n}}{\gcd({p^{n}},d_{i})}$ for each $i=1,\dots,r$. Using (8.1)
and elementary group theory one sees that

$A_{\omega}\simeq\mathbb{Z}/k_{1}\oplus\dots\oplus\mathbb{Z}/k_{r}.$

Therefore $\dim_{\mathbb{F}_{p}}(A_{\omega}/p)=i_{0}$, where $i_{0}$ is the
number of $k_{i}$’s that are different from $1$. Equivalently, $i_{0}$ is the
largest integer such that ${p^{n}}$ does not divide $d_{i_{0}}$.

Now parts $(a)$ and $(b)$ follow from Theorem 7.1. ∎

######  Example 8.2.

Let
$F_{2r}=k((a_{11}))((a_{12})),((a_{21})),(a_{22}))\ldots((a_{r1}))((a_{r2}))$,
be the field of iterated Laurent series in $2r$ variables $a_{ij}$. Assume
that $k$ contains a primitive root of unity of degree $p^{n}$. By Proposition
7.2, the Brauer class $[D]\in\operatorname{Br}(F)$ of the division algebra
$D=(a_{11},a_{12})_{p^{n}}\otimes\dots\otimes(a_{r1},a_{r2})_{p^{n}}$ has
essential dimension $2r$. This can also be seen from Theorem 1.3. Indeed,
carrying out the algorithm outlined there, we see that in this case

$M=\begin{pmatrix}B&0&\dots&0\\\ 0&B&\dots&0\\\ {4}\\\
0&0&\dots&B\end{pmatrix}.$

Here each entry represents a $2\times 2$ block: $0$ stands for the $2\times 2$
zero matrix, and $\displaystyle B=\begin{pmatrix}0&1\\\ -1&0\end{pmatrix}$.
The diagonal $2\times 2$ block $B$ appears $r$ times. Clearly,
$\det(M)=\det(B)^{r}=1$ and Theorem 1.3 tells us that
$\operatorname{ed}_{k}\,[D]=2r$.

######  Corollary 8.3.

Let $(F,\nu)$ be a strictly Henselian valued field over a field $k$ with
$\operatorname{char}k\neq p$. Then
$\operatorname{ed}_{k}(\alpha)=\operatorname{ed}_{k}(\alpha,p)$ is even for
any a Brauer class $\alpha\in\operatorname{Br}(F)$.

###### Proof.

This follows from Theorem 1.3 because non-zero elementary divisors of an anti-
symmetric matrix have even multiplicity; see [33, Chapter IV, Theorem 3]. ∎

##  9\. Some limitations of Theorem 1.3

The following proposition shows that Theorem 1.3 cannot be used to prove a
lower bound greater than $2n$ on the essential dimension of a Brauer class of
index $p^{n}$. This is because $\wedge\nu$ is unaffected by passing to the
strict Henselization; see Remark 3.2.

######  Proposition 9.1.

Let $(F,\nu)$ be a strictly Henselian valued field over $k$ with value group
$\Gamma\cong\mathbb{Z}^{r}$. If $\alpha\in H^{2}_{p}(F)$ is split by a field
extension $L/F$ of degree $p^{n}$, then

$\operatorname{ed}_{k}(\alpha)\leqslant 2n.$

###### Proof.

By Chevalley’s extension theorem [9, Theorem 3.1.1], we can extend $\nu$ to a
valuation $\tilde{\nu}\colon L^{*}\to\tilde{\Gamma}$. By Ostrowski’s theorem
[46, Theorem A.12],

$[\tilde{\Gamma}:\Gamma]\mid[L:F]=p^{n}.$

By the Elementary Divisors Theorem [21, Theorem III.7.8], we there exists a
basis $e_{1},\dots,e_{r}$ of $\tilde{\Gamma}$ and integers $d_{1},\dots,d_{r}$
such that $d_{1}e_{1},\dots,d_{r}e_{r}$ is a basis for $\Gamma$ and
$\prod_{i}d_{i}=[\tilde{\Gamma}:\Gamma]\mid p^{n}$. Thus each $d_{i}$ is $\pm$
a power of $p$, and we assume without loss of generality that $d_{i}=\pm 1$
for all $i>n$. For some $a_{ij}\in\mathbb{Q}_{p}/\mathbb{Z}_{p}$ we have:

$\wedge\nu(\alpha)=\sum_{1\leqslant i<j\leqslant
r}a_{ij}\otimes(d_{i}e_{i}\wedge d_{j}e_{j}).$

Since $\alpha_{L}=0$, $\wedge\nu(\alpha)$ is in the kernel of the
homomorphism:

$\phi:\bigwedge\Gamma/p^{\infty}\to\bigwedge\tilde{\Gamma}/p^{\infty}.$

Since $e_{1},\dots,e_{r}$ is a basis of $\tilde{\Gamma}$,
$\phi(\wedge\nu(\alpha))=0$ implies for all $i<j$:

$\phi(a_{ij}\otimes d_{i}e_{i}\wedge
d_{j}e_{j})=a_{ij}d_{i}d_{j}\otimes(e_{i}\wedge e_{j})=0$

We conclude that $a_{ij}=0$ for all $n<i<j\leqslant r$ because
$d_{i},d_{j}=\pm 1$. Assume that $\displaystyle a_{ij}=\frac{b_{ij}}{p^{N}}$
for some integers $b_{ij}$. Then

$\wedge\nu(\alpha)=\sum_{1\leqslant i\leqslant n}\frac{1}{p^{N}}\otimes
d_{i}e_{i}\wedge v_{i},$

where $v_{i}=\sum_{i<j\leqslant r}b_{ij}d_{j}e_{j}$. Let $W\subset\Gamma$ be
the subgroup generated by $d_{1}e_{1},\dots,d_{n}e_{n}$ and
$v_{1},\dots,v_{n}$. Since $\wedge\nu(\alpha)\in\bigwedge W/p^{\infty}$,
Theorem 7.1 tells us that
$\operatorname{ed}_{k}(\alpha)=\rho(\wedge\nu(\alpha))\leqslant 2n$. ∎

We will now show that the assumption that $\operatorname{char}(k)\neq p$ in
the statement of Theorem 8.1 cannot be dropped.

######  Proposition 9.2.

Let $k$ be a field of characteristic $p$. Then
$\operatorname{ed}_{k}(\operatorname{Br};p)=2$.

###### Proof.

First we will show that $\operatorname{ed}_{k}(\operatorname{Br};p)\leqslant
2$. In other words, $\operatorname{ed}_{k}(\alpha;p)\leqslant 2$ for every
field $K$ containing $k$ and every $\alpha\in\operatorname{Br}(K)$. By the
Primary Decomposition Theorem, we can write $\alpha=\alpha_{p}+\beta$, where
the index of $\alpha_{p}$ is a power of $p$, and the index of $\beta$ is prime
to $p$. Let $K^{\prime}/K$ be a finite prime-to-$p$ extension which splits
$\beta$. Then $\alpha_{K^{\prime}}=(\alpha_{p})_{K^{\prime}}$, and
$\operatorname{ed}_{k}(\alpha;p)=\operatorname{ed}_{k}(\alpha_{K^{\prime}};p)$.
Here the last equality is an easy consequence of the definition of
$\operatorname{ed}_{k}(\alpha;p)$; see, e.g., the proof of [26, Proposition
1.5(2)]. After replacing $K$ by $K^{\prime}$ and $\alpha$ by
$\alpha_{K^{\prime}}$, we may assume without loss of generality that the index
of $\alpha$ is a power of $p$. By a Theorem of Albert, $\alpha$ is represented
by a cyclic algebra over $K$; see [15, Theorem 9.1.8]. In other words, it lies
in the image of the natural (functorial) morphism

$j_{r}\colon H^{1}(K,\mathbb{Z}/p^{r}\mathbb{Z})\times
H^{0}(K,\mathbb{G}_{m})\to\operatorname{Br}(K)$

for some integer $r\geqslant 1$; see [15, p. 260]. Recall that
$H^{1}(K,\mathbb{Z}/p^{r}\mathbb{Z})$ is in a functorial bijective
correspondence with isomorphism classes of cyclic étale algebras $L/K$ of
dimension $p^{r}$ and $H^{0}(K,\mathbb{G}_{m})=K^{*}$. If $\tau$ is a
generator of the (cyclic) Galois group of $L/K$, then $j_{r}([L/K],[b])$ is
the cyclic algebra generated by $L$ and $u$, subject to relations
$ua=\tau(a)u$ and $u^{p^{r}}=b$. Here $[L/K]$ is the class of $L/K$ in
$H^{1}(K,\mathbb{Z}/p^{r}\mathbb{Z})$ and $[b]$ is the class of $b$ in
$H^{0}(K,\mathbb{G}_{m})$. Then

$\displaystyle\operatorname{ed}(\alpha;p)$
$\displaystyle\leqslant\operatorname{ed}_{k}(H^{1}(\ast,\mathbb{Z}/p^{r}\mathbb{Z})\times
H^{0}(K,\mathbb{G}_{m});\,p)$
$\displaystyle\leqslant\operatorname{ed}_{k}(H^{1}(\ast,\mathbb{Z}/p^{r}\mathbb{Z});\,p)+\operatorname{ed}_{k}(H^{0}(\ast,\mathbb{G}_{m});\,p)$
$\displaystyle=\operatorname{ed}_{k}(\mathbb{Z}/p^{r}\mathbb{Z};\,p)+\operatorname{ed}_{k}(H^{0}(\ast,\mathbb{G}_{m});\,p).$

Here the first inequality follows from the fact that $\alpha$ lies in the
image of $j_{r}$ and the second from [29, Proposition 2.2]. The equality on
the third line follows from the definition of
$\operatorname{ed}_{k}(\mathbb{Z}/p^{r}\mathbb{Z};\,p)$. Now observe that
$\operatorname{ed}_{k}(\mathbb{Z}/p^{r}\mathbb{Z};\,p)=1$ by [38, Theorem 1].
On the other hand,

$\operatorname{ed}(H^{0}(\ast,\mathbb{G}_{m});\,p)\leqslant\operatorname{ed}(H^{0}(\ast,\mathbb{G}_{m}))\leq
1.$

Here the first inequality is a special case of (2.1), and the second follows
from the fact that an object $b\in H^{0}(K,\mathbb{G}_{m})$ descends to
$k(b)$. We conclude that $\operatorname{ed}(\alpha;p)\leqslant 1+1=2$, as
desired.

It remains to show that $\operatorname{ed}_{k}(\operatorname{Br};p)\geqslant
2$. Assume the contrary. Then

$\operatorname{ed}_{\overline{k}}(\operatorname{Br};p)\leqslant\operatorname{ed}_{k}(\operatorname{Br};p)\leqslant
1,$

where $\overline{k}$ is the algebraic closure of $k$; see, e.g., [26,
Proposition 1.5(2)]. After replacing $k$ by $\overline{k}$, we may assume that
$k$ is algebraically closed. Choose a class $\alpha\in\operatorname{Br}(K)$ of
index $p$ for some field $K$ containing $k$ (for example, the class of a
universal division algebra of degree $p$). By our assumption
$\operatorname{ed}_{k}(\alpha;p)\leqslant 1$, i.e., there exists a prime-
to-$p$ field extension $L/K$ such that
$\operatorname{ed}_{k}(\alpha_{L})\leqslant 1$. In other words, $\alpha$
descends to $\alpha_{0}\in\operatorname{Br}(L_{0})$ for some intermediate
field $k\subset L_{0}\subset L$ such that
$\operatorname{trdeg}_{k}(L_{0})\leqslant 1$. By Tsen’s Theorem,
$\operatorname{Br}(L_{0})=0$; see [15, Lemma 6.2.3 and Proposition 6.2.8]. We
conclude that $\alpha_{L_{0}}=0$ and thus $\alpha_{L}=0$ in
$\operatorname{Br}(L)$, i.e., $L/K$ splits $\alpha$. This is a contradiction,
because $\alpha$ has index $p$ over $K$, and $[L:K]$ is prime to $p$. Thus
$\operatorname{ed}_{k}(\operatorname{Br};p)\geqslant 2$. ∎

##  10\. A further consequence of Theorem 1.1

######  Proposition 10.1.

Let $(F,\nu)$ be a strictly Henselian valued field over $k$. Let $\alpha\in
H^{d}_{p}(F)$ for some $p\neq\operatorname{char}k$. The following holds:

  1. (1)

$\alpha\neq 0$ if and only if $\operatorname{ed}_{k}(\alpha;p)\geqslant d$.

  2. (2)

$\alpha$ is not a symbol if and only $\operatorname{ed}_{k}(\alpha;p)\geqslant
d+2$.

In particular, $\operatorname{ed}_{k}(\alpha)$ can never be equal to $d+1$.

###### Proof.

Set $\omega:=\wedge\nu(\alpha)\in\bigwedge^{d}\nu F/p^{\infty}$. By Theorem
7.1,

(10.1)
$\rho(\omega)=\operatorname{ed}_{k}(\alpha)=\operatorname{ed}_{k}(\alpha;p).$

Choose a subgroup $W\subset\nu F$ of rank $\rho(\omega)$ such that
$\omega\in\bigwedge^{d}W/p^{\infty}$ .

$(1)$ If $\operatorname{ed}_{k}(\alpha;p)<d$ then
$\operatorname{rank}W=\rho(\omega)<d$ and thus
$\omega\in\bigwedge^{d}W/p^{\infty}=0$. Now Corollary 4.4 tells us that
$\alpha=0$.

$(2)$ If $\alpha=(a_{1},\dots,a_{d})_{p}$ is a symbol, then $\alpha$ descends
to $k(a_{1},\dots,a_{d})$ and so $d\geqslant\operatorname{ed}_{k}(\alpha;p)$.

Now assume that $d+1\geqslant\operatorname{ed}_{k}(\alpha;p)$. Our goal is to
show that $\alpha$ is a symbol. By Corollary 4.4, it suffices to show $\omega$
is a pure tensor. By (10.1), we may enlarge $W$ to assume
$\operatorname{rank}W=d+1$ and $\omega\in\bigwedge^{d}W/p^{\infty}$. Since
$\bigwedge^{d}W/p^{\infty}$ is generated by pure tensors it suffices to show
the sum of two pure tensors is a pure tensor. A pure tensor in
$\bigwedge^{d}W/p^{\infty}$ is of the form:

$\eta=\frac{a}{p^{n}}\otimes u_{1}\wedge\dots\wedge u_{d},$

for some $u_{1},\dots,u_{d}\in W$. Let $U=\langle u_{1},\dots,u_{d}\rangle$
and denote by $U^{\operatorname{sat}}$ its saturation in $W$. If
$u^{\prime}_{1},\dots,u^{\prime}_{d}$ are a basis of $U^{\operatorname{sat}}$,
then $\eta=\frac{a^{\prime}}{p^{n}}\otimes u^{\prime}_{1}\wedge\dots\wedge
u^{\prime}_{d}$ for some $a^{\prime}\in\mathbb{Z}$ (one can check this using
the Smith normal form). Therefore we may assume $U$ is saturated. Let $\theta$
be another pure tensor:

$\theta=\frac{b}{p^{n}}\otimes v_{1}\wedge\dots\wedge v_{d},$

for some $v_{1},\dots,v_{d}\in W$ such that $V=\langle
v_{1},\dots,v_{d}\rangle$ is saturated in $W$. We will show $\eta+\theta$ is a
pure tensor. Since $U$ and $V$ are saturated we have:

$(U\cap V)_{\mathbb{Q}}=U_{\mathbb{Q}}\cap V_{\mathbb{Q}},$

where $U_{\mathbb{Q}}=U\otimes\mathbb{Q}$ and similarly for $V_{\mathbb{Q}}$
and $(U\cap V)_{\mathbb{Q}}$. Therefore,

$\operatorname{rank}_{\mathbb{Z}}(U\cap
V)=\dim_{\mathbb{Q}}(U_{\mathbb{Q}}\cap
V_{\mathbb{Q}})\geqslant\dim_{\mathbb{Q}}(U_{\mathbb{Q}})+\dim_{\mathbb{Q}}(V_{\mathbb{Q}})-\dim(W_{\mathbb{Q}})=2d-\dim_{\mathbb{Q}}(W_{\mathbb{Q}})=d-1.$

Since $U\cap V$ is clearly saturated in both $U$ and $V$ we can complete a
basis of $U\cap V$ to a basis of $U$ and to a basis of $V$. In other words, we
may assume $u_{i}=v_{i}$ for all $1\leqslant i\leqslant d-1$. Then

$\displaystyle\eta+\theta$ $\displaystyle=\frac{a}{p^{n}}\otimes
u_{1}\wedge\dots\wedge u_{d}+\frac{b}{p^{n}}\otimes u_{1}\wedge\dots\wedge
u_{d-1}\wedge v_{d}$ $\displaystyle=\frac{1}{p^{n}}\otimes
u_{1}\wedge\dots\wedge(au_{d}+bv_{d}).$

The right hand side is a pure tensor. This shows that any element in
$\bigwedge^{d}W/p^{\infty}$ is a pure tensor and thus completes the proof of
part (b). ∎

In light of Corollary 8.3, one might guess that
$\operatorname{ed}_{k}(\alpha)$ is divisible by $d$ for any $\alpha\in
H^{d}_{p}(F)$. The following proposition shows that this is false for any
$d\geqslant 3$.

######  Proposition 10.2.

Let $F_{n}=k((t_{0}))\dots((t_{n-1}))$. Assume that $d\geqslant 3$. Then for
any $n\geqslant d+2$ and $p\neq\operatorname{char}k$, there exists $\alpha\in
H^{d}_{p}(F_{n})$ such that $\operatorname{ed}_{k}(\beta;p)=n$.

###### Proof.

We claim that

$\alpha=\sum_{i_{1}+\ldots+i_{d}\equiv
0\pmod{n}}(t_{i_{1}})_{p}\cup\ldots\cup(t_{i_{d}})_{p}$

has the desired property. Here the sum is taken over ordered $d$-tuples
$(i_{1},\ldots,i_{d})$ such that $0\leqslant i_{1}<\ldots<i_{d}\leqslant n-1$
and $i_{1}+\ldots+i_{d}\equiv 0\pmod{n}$. Let $\nu$ be the
$(t_{0},\dots,t_{n-1})$-adic valuation on $F_{n}$ and set

$\omega:=\wedge\nu(\alpha)=\frac{1}{p}\otimes\sum_{i_{1}+\ldots+i_{d}\equiv
0\pmod{n}}e_{i_{1}}\wedge\ldots\wedge e_{i_{d}},$

where $e_{0},e_{1},\ldots,e_{n-1}$ is the standard basis in $\mathbb{Z}^{n}$.
In view of Theorem 7.1 it suffices to show that
$A_{\omega}=\mathbb{Z}^{n}/p\subset\mathbb{Z}^{n}/p^{\infty}$. Let
$e^{1},\dots,e^{n}\in(\mathbb{Z}^{n})^{*}$ be the dual basis. From the
definition of $\omega$, we see that when we contract $\omega$ with a tensor of
the form $e^{j_{1}}\wedge\ldots\wedge e^{j_{d-1}}$, where $0\leqslant
j_{1}<\ldots<j_{d-1}$, we obtain $\pm e_{j}$, where $0\leqslant j\leqslant
n-1$ is uniquely determined by the congruence $j\equiv-j_{1}-\ldots-
j_{d-1}\pmod{n}$. By definition, every $e_{j}$ of this form lies in
$A_{\omega}$. To complete the proof, we need to show that every basis vector
$e_{0},\ldots,e_{n-1}$ can be obtained in this way. Equivalently, it remains
to prove the following.

Claim: Assume $d\geqslant 3$ and $n\geqslant d+2$ are integers. Then for every
$j\in\mathbb{Z}/n$, there exists a subset $S\subset\mathbb{Z}/n$ consisting of
$d$ distinct elements, $i_{1},\ldots,i_{d}$, such that $j\in S$ and
$i_{1}+\ldots+i_{d}=0$ in $\mathbb{Z}/n$.

To prove the Claim, let $\mathcal{P}$ be the set of pairs $\\{i,-i\\}$, where
$2i\neq 0$ in $\mathbb{Z}/n$. If $n$ is odd, then $2i=0$ is only possible in
$\mathbb{Z}/n$ if $i=0$, and thus $|\mathcal{P}|=(n-1)/2$. If $n=2m$ is even,
then the equation $2i=0$ has two solutions in $\mathbb{Z}/n$, $i=0$ or $i=m$.
In this case $|\mathcal{P}|=(n-2)/2=m-1$.

Case 1. $2j\neq 0$ in $\mathbb{Z}/n$, and $d$ is even. Take $S$ to be the
union of $d/2$ pairs from $\mathcal{P}$, one of these pairs being $\\{\pm
j\\}$. This is possible because we are assuming that $n\geqslant d+2$, and
thus $|\mathcal{P}|\geqslant(n-2)/2\geqslant d/2$.

Case 2. $2j\neq 0$ in $\mathbb{Z}/n$, and $d$ is odd. Take $S$ to be the union
of $0$ and $(d-1)/2$ pairs from $\mathcal{P}$, one of these pairs being
$\\{\pm j\\}$. Again, this is possible because $|\mathcal{P}|\geqslant d/2$.

Case 3. $2j=0$ and $d$ is odd. Take $S$ to be the union of $\\{1,j,j-1\\}$ and
$(d-3)/2$ pairs from $\mathcal{P}$, other than $\\{\pm 1\\}$ and
$\\{\pm(j-1)\\}$. Note that $1$, $j$ and $j-1$ are distinct in $\mathbb{Z}/n$;
this readily follows from our assumption that $n\geqslant d+2\geqslant 5$.
Note also that $(d-3)/2$ pairs, other than $\\{\pm 1\\}$ and $\\{\pm(j-1)\\}$,
exist in $\mathcal{P}$, because when $d$ is odd, the inequality
$|\mathcal{P}|\geqslant d/2$ can be strengthened to
$|\mathcal{P}|\geqslant(d+1)/2=(d-3)/2+2$.

Case 4a. $2j=0$, $d$ is even, and $n$ is odd. In this case $j=0$ in
$\mathbb{Z}/n$. Moreover, for even $d$, the inequality $d\geqslant 3$ can be
strengthened to $d\geqslant 4$, and for odd $n$, the inequality $n\geqslant
d+2\geqslant 6$ can be strengthened to $n\geqslant 7$. Consequently, the four
elements $0$, $1$, $2$ and $-3$ are distinct in $\mathbb{Z}/n$, and we define
$S$ to be the union of $\\{0,1,2,-3\\}$ and $(d-4)/2$ pairs from
$\mathcal{P}$, other than $\\{\pm 1\\}$, $\\{\pm 2\\}$ and $\\{\pm 3\\}$. Note
that $(d-4)/2$ pairs with this property exist because for $n$ odd
$|\mathcal{P}|=(n-1)/2\geqslant(d+1)/2$. Since $d$ is even, this translates to
$|\mathcal{P}|\geqslant(d+2)/2=(d-4)/2+3$.

Case 4b. $2j=0$, $d$ is even, and $n=2m$ is even. Here $j=0$ or $m$, and we
define $S$ as the union of $\\{0,1,m-1,m\\}$ and $(d-4)/2$ pairs from
$\mathcal{P}$, other than $\\{\pm 1\\}$ and $\\{m-1,m+1\\}$. The elements $0$,
$1$, $m$, $m+1$ are distinct in $\mathbb{Z}/n$, because $n\geqslant
d+2\geqslant 6$, and $\mathcal{P}$ has $(d-4)/2$ pairs, other than $\\{\pm
1\\}$ and $\\{m-1,m+1\\}$, because $|\mathcal{P}|\geqslant d/2=(d-4)/2+2$. ∎

######  Remark 10.3.

Similarly, in light of Proposition 9.1, one might guess that if $\alpha\in
H^{d}_{p}(F)$ is split by a field extension $E/F$ of degree $n$, then
$\operatorname{ed}_{k}(\alpha)\leqslant dn$. This is also false. For example,
let $d=3$, $F=k((t_{0}))\dots((t_{2r}))$, and

$\alpha=(t_{0})_{p}\cup\big{(}(t_{1},t_{2})_{p}+\ldots+(t_{2r-1},t_{2r})_{p}\big{)}.$

Then $\alpha$ is split by $E=F(\sqrt[p]{t_{0}})$, where $[E:F]=p$. On the
other hand, contracting

$\omega:=\wedge\nu(\alpha)=\frac{1}{p}\otimes e_{0}\wedge[e_{1}\wedge
e_{2}+\ldots+e_{2r-1}\wedge e_{2r}]$

with tensors of the form $e^{j_{1}}\wedge e^{j_{2}}$, as we did in the proof
of Proposition 10.2, we see that
$A_{\omega}=\mathbb{Z}^{2r+1}/p\subset\mathbb{Z}^{2r+1}/p^{\infty}$. By
Theorem 7.1,
$\operatorname{ed}_{k}(\alpha)=\operatorname{ed}_{k}(\alpha;p)=2r+1$. Since
$r$ can be taken to be an arbitrary positive integer, the ratio
$\displaystyle\frac{\operatorname{ed}_{k}(\alpha)}{[E:F]}=\frac{2r+1}{p}$ can
be arbitrarily large.

##  11\. Anisotropic torsors

In this section take a slight detour to prove a structural result about
anisotropic $E_{8}$-torsors, Proposition 11.3 below. This proposition will be
used in the proof of Theorem 1.4 in the next section.

Let $G$ be a connected quasi-split reductive group and $T\in H^{1}(F,G)$ a
$G$-torsor over a field $F\in\operatorname{Fields}_{k}$. For any subgroup
$H\subset G$, we will say that $T$ _admits reduction of structure to_ $H$, if
it is in the image of $H^{1}(F,H)\to H^{1}(F,G)$. The torsor $T$ is called
_isotropic_ if it admits reduction of structure to a proper parabolic subgroup
of $G$ (or equivalently, if the twisted group ${}_{T}G$ is isotropic as an
algebraic group [34, Lemma 2.2]). Otherwise, $T$ is said to be _anisotropic_.
Recall that if $H$ is a finite constant group, $H^{1}(F,H)$ classifies Galois
etale algebras of $F$ whose Galois group is a homomorphic image of $H$. The
next lemma is inspired by [14, Proposition 7.2].

######  Lemma 11.1.

Let $(F,\nu)$ be a strictly Henselian field with residue field $k$ and
valuation ring $\mathcal{O}$. Let $G$ be a connected quasi-split reductive
group over a field $F_{0}\subset\mathcal{O}$. Assume that $G_{\mathcal{O}}$
contains a constant finite subgroup $\iota:H\hookrightarrow G_{\mathcal{O}}$
such that $H_{\overline{k}}$ is not contained in any proper parabolic subgroup
of $G_{\overline{k}}$. Then for any torsor $T\in H^{1}(F,H)$ corresponding to
an $H$-Galois field extension $L/F$, the push-forward $\iota_{*}(T)\in
H^{1}(F,G)$ is anisotropic.

###### Proof.

By assumption, there exists an isomorphism $c:\operatorname{Gal}(L/F)\to H$
and $\iota_{*}(T)$ is represented by the cocycle $\iota\circ c$. Assume that
$\iota_{*}(T)$ admits reduction of structure to a parabolic subgroup $P\subset
G$. Our goal is to show $P=G$. Since $\iota_{*}(T)$ admits reduction of
structure to $P$, there exists $g\in G(L)$ such that for all
$\sigma\in\operatorname{Gal}(L/F)$:

(11.1) $g^{-1}c_{\sigma}{}^{\sigma}g\in P(L).$

Setting $x\in(G/P)(L)$ to be the coset represented by $g$, we see that for all
$\sigma\in\operatorname{Gal}(L/F)$:

(11.2) $c(\sigma){}^{\sigma}x=x.$

Since $\nu$ is Henselian, it has a unique extension to $L$. By abuse of
notation, we will continue to denote this extended valuation by $\nu$. We will
write $\mathcal{O}_{L}$ for the valuation ring of $\nu$ in $L$ and $l$ be the
residue field. Since $k$ is separably closed, $l$ is a purely inseparable
extension of $k$. Since $G/P$ is projective, $x\in(G/P)(L)$ can be lifted to a
point in $(G/P)(\mathcal{O}_{L})$. By abuse of notation, we will simply say
that $x\in(G/P)(\mathcal{O}_{L})$. Note that $\operatorname{Gal}(L/F)$
preserves $\nu$ because $L$ is Henselian. Therefore $\operatorname{Gal}(L/F)$
acts on $\mathcal{O}_{L}$. The reduction homomorphism $\mathcal{O}_{L}\to l$
is $\operatorname{Gal}(L/F)$-equivariant with respect to the trivial
$\operatorname{Gal}(L/F)$-action on $l$ because $l/k$ is purely inseparable.
Let $\pi:G(\mathcal{O}_{L})\to G(l)$ be the induced homomorphism. Applying
$\pi$ to (11.2) gives:

$\pi(x)=\pi(c(\sigma){}^{\sigma}x)=\pi(c(\sigma))\pi(x).$

Since $H\subset G_{\mathcal{O}}$ is constant, $H(\mathcal{O}_{L})=H(l)$ and
the restriction of $\pi$ to $H$ is the identity. Therefore we get:

$\pi(x)=\pi(c(\sigma){}^{\sigma}x)=c(\sigma)\pi(x).$

Since $l$ is separably closed, we can find $g\in G(l)$ such that
$\pi(x)=gP(l)$. Since $c$ is an isomorphism we get $HgP(l)=gP(l)$ and
therefore:

$H_{l}\subset gP_{l}g^{-1}.$

Since we are assuming that $H_{\overline{k}}$ is not contained in any proper
parabolic of $G_{\overline{k}}$, we conclude that $gP_{l}g^{-1}=G_{l}$ and
thus $P=G$. This finishes the proof. ∎

The next lemma gives a convenient sufficient condition for a subgroup
$H\subset G$ to not be contained in any proper parabolic subgroup $P\subset
G$.

######  Lemma 11.2.

Let $G$ be a connected semisimple group over a field $k$ and let $H\subset G$
be a finite subgroup. If the centralizer $C_{G}(H)$ is finite (i.e.,
0-dimensional) and $|H|$ is coprime to $\operatorname{char}k$, then $H$ is not
contained in any proper parabolic subgroup of $G$.

###### Proof.

Assume that $H$ is contained in a parabolic subgroup $P\subset G$. Since $H$
is of order prime to $p$ it is linearly reductive over $k$. Therefore $H$ is
contained in a Levi subgroup $L\subset P$ (see [18, Lemma 11.24]). If $P\neq
G$, then the center $Z_{L}$ of $L$ contains a split torus [2, Proposition
20.6]. This contradicts our assumption that $C_{G}(H)$ is finite because
clearly $Z_{L}\subset C_{G}(H)$. Therefore we must have $P=G$. ∎

We are now ready to proceed with the main result of this section.

######  Proposition 11.3.

Let $G$ be a split group of type $E_{8}$ over a field $k$ and assume
$\operatorname{char}k\neq 2,3$. Let $T\in H^{1}(F,G)$ be a torsor over a
strictly Henselian field $F\in\operatorname{Fields}_{k}$ over $k$. The
following are equivalent:

  1. (1)

$\operatorname{ed}_{k}(T;3)\geqslant 5$.

  2. (2)

$\operatorname{ed}_{k}(R_{G}(T);3)\geqslant 5$.

  3. (3)

$T_{L}$ is anisotropic for any prime-to-$3$ extension $F\subset L$.

###### Proof.

Passing from $F$ to its prime-to-$3$ extension $F^{(3)}$ does not affect the
validity of assertions (1), (2) and (3). Thus we may assume without loss of
generality that $F$ is a $3$-special field. In particular, since
$\operatorname{char}F\neq 3$, this implies that $F$ is perfect. By assumption,
$F$ is equipped with a Henselian valuation $\nu$ that is trivial on $k$.
Therefore the residue field of $\nu$ has the same characteristic as $k$.
Hensel’s lemma implies that $F$ contains all primitive roots of unity of order
$3^{n}$ for $n\in\mathbb{N}$. Let $k\subset k^{\prime}$ be the field generated
by all roots of unity in $F$. We replace $k$ by $k^{\prime}$ without effecting
assertions $(1)$ and $(2)$ by (2.3). This allows us to view the Rost invariant
as a natural homomorphism $H^{1}(\ast,G)\to H^{3}_{3}(\ast)$ because
$\mathbb{Q}_{3}/\mathbb{Z}_{3}(2)\cong\mathbb{Q}_{3}/\mathbb{Z}_{3}(3)$ as
Galois modules over $k$.

(3) $\iff$ (2) By Proposition 10.1,
$\operatorname{ed}_{k}(R_{G}(T);3)\geqslant 5$ if and only if $R_{G}(T)$ is
not a symbol. Since $F$ is $3$-special, [11, Theorem 10.24] implies that
$R_{G}(T)$ is a symbol if and only if $T$ is isotropic. Therefore $T$ is
anisotropic if and only if $\operatorname{ed}_{k}(R_{G}(T);3)\geqslant 5$

(2)$\implies$ (1) follows directly from (1.6).

(1)$\implies$ (3) We assume $T$ is isotropic and show
$\operatorname{ed}_{k}(T;3)\leqslant 4$. Since $T$ is isotropic, it admits
reduction of structure to some proper parabolic subgroup $P\subset G$. Assume
without loss of generality that $P$ is maximal amongst the proper parabolic
subgroups. Let $L\subset P$ be a Levi subgroup of $P$ and let
$L^{\prime}\subset L$ be the corresponding derived subgroup (it exists by [31,
Theorem 25.6]). Since $F$ is perfect, $H^{1}(F,L)\to H^{1}(F,P)$ is bijective
[42, Lemme 1.13]. Note that $Z(L)^{\circ}$ is a split torus because it is
contained in a maximal torus of $G$. Therefore, $L/L^{\prime}$ is a split
torus as a homomorphic image of $Z(L)^{\circ}$ [2, Chapter V, Theorem 15.4],
and $H^{1}(F,L^{\prime})\to H^{1}(F,L)$ is surjective by Hilbert $90$.
Therefore the composition $H^{1}(F,L^{\prime})\to H^{1}(F,P)$ is surjective
and $T$ admits reduction of structure to $L^{\prime}$. Therefore
$\operatorname{ed}_{k}(T;3)\leqslant\operatorname{ed}_{k}(L^{\prime};3)$ and
it suffices to prove:

(11.3) $\operatorname{ed}_{k}(L^{\prime};3)\leqslant 4.$

Since $P$ is a maximal proper parabolic of $G$, the Dynkin diagram of
$L^{\prime}$ is obtained from the Dynkin diagram of $E_{8}$ by removing one
vertex [47]. By [45, Corollary 5.4(b)], $L^{\prime}$ is simply connected.
Hence, $L^{\prime}$ is a product of simple simply connected groups. Removing
each of the eight vertices in the Dynkin diagram of $E_{8}$, we obtain the
following possibilities for $L^{\prime}$.

(i) $D_{7}$, (ii) $A_{1}\times A_{6}$, (iii) $A_{2}\times A_{1}\times A_{4}$,
(iv) $A_{7}$, (v) $A_{4}\times A_{3}$, (vi) $D_{5}\times A_{2}$,

(vii) $E_{6}\times A_{1}$, (viii) $E_{7}$.

That is, in case (i), $L^{\prime}$ is isomorphic to
$\operatorname{Spin}_{14}$, it case (ii) to
$\operatorname{SL}_{2}\times\operatorname{SL}_{7}$, etc. We will now check
that the the inequality (11.3) holds in each of these eight cases. Recall that
if $G=H_{1}\times\dots\times H_{r}$, then

$\operatorname{ed}_{k}(G;3)\leqslant\operatorname{ed}_{k}(H_{1};3)+\dots+\operatorname{ed}_{k}(H_{r};3);$

see, e.g., [29, Proposition 3.3]. The simply connected simple group of type
$A_{n}$ is $\operatorname{SL}_{n}$. This group has essential dimension $0$. In
cases (ii) - (v), $L^{\prime}$ is a direct product of simply connected simple
groups of type $A$ and thus
$\operatorname{ed}_{k}(L^{\prime})=\operatorname{ed}_{k}(L^{\prime};3)=0$. The
simply connected group of type simple group of type $D_{n}$ is
$\operatorname{Spin}_{2n}$. Since $3$ is not a torsion prime for these groups,
we have $\operatorname{ed}_{k}(\operatorname{Spin}_{2n};3)=0$; see [29,
Proposition 3.16]. We conclude that $\operatorname{ed}_{k}(L^{\prime};3)=0$ in
cases (i) and (vi) as well. In case (vii), we have

$\operatorname{ed}_{k}(L^{\prime};3)=\operatorname{ed}_{k}(E_{6}^{\rm
sc}\times\operatorname{SL}_{2};3)\leqslant\operatorname{ed}_{k}(E_{6}^{\rm
sc};3)+\operatorname{ed}_{k}(\operatorname{SL}_{2};3)=\operatorname{ed}_{k}(E_{6}^{\rm
sc};3)=3.$

For the last equality, see [29, Section 3h]. In case (viii),

$\operatorname{ed}_{k}(L^{\prime};3)=\operatorname{ed}_{k}(E_{7}^{\rm
sc};3)=4,$

where the last equality is again taken from [29, Section 3h]. We conclude that
(11.3) holds in each of the cases (i) - (viii). ∎

##  12\. Proof of Theorem 1.4 and a remark on spin groups

For the reader’s convenience, we begin by reproducing the statement of Theorem
1.4.

Theorem 1.4. Let $k$ be a base field of characteristic $\neq 2$ and $F_{n}$ be
the iterated Laurent series field
$F_{n}=k((t_{0}))((t_{2}))\ldots((t_{n-1}))$. Then there exist (i) an
$E_{7}^{\rm sc}$-torsor $T_{1}\to\operatorname{Spec}(F_{7})$, (ii) an
$E_{8}$-torsor $T_{2}\to\operatorname{Spec}(F_{9})$, and (iii) an
$E_{8}$-torsor $T_{3}\to\operatorname{Spec}(F_{5})$ such that (i)
$\operatorname{ed}_{k}(R_{E_{7}^{\rm sc}}(T_{1});2)=7$, (ii)
$\operatorname{ed}_{k}(R_{E_{8}}(T_{2});2)=9$, and (iii)
$\operatorname{ed}_{k}(R_{E_{8}}(T_{3});3)=5$, respectively.

Here in (iii) we are assuming that $k$ contains a primitive $3$rd root of
unity.

###### Proof.

We assume at first that $k$ is separably closed. In this case, The functors
$H^{3}(\ast,\mathbb{Q}_{p}/\mathbb{Z}_{p}(2))$ and $H^{3}_{p}(\ast)$ are
naturally isomorphic because $\mathbb{Q}_{p}/\mathbb{Z}_{p}(2)$ and
$\mathbb{Q}_{p}/\mathbb{Z}_{p}(3)$ are isomorphic as Galois modules for all
$p\neq\operatorname{char}k$. Therefore we can and shall see $R_{G}$ as taking
values in $H^{3}_{p}$ where $p=2$ in parts (i),(ii) and $p=3$ in part (iii).

Part (i): Let $G$ be a simply connected split group of type $E_{7}$ over $k$.
Let $L=k(s_{0},\dots,s_{6})$ be a field of rational functions in $7$
independent variables and $H$ a quasi-split simply connected group of type
$E_{6}$ over $L$ that is split by the extension $L(\sqrt{s_{0}})$. In [6,
Section 6.2.3], Chernousov constructed a torsor $S\in H^{1}(L,H)$ such that:

$R_{H}(S)=(s_{0})_{2}\cup[(s_{1},s_{3})_{2}+(s_{2}s_{3}s_{5},s_{4})_{2}+(s_{5},s_{6})_{2}].$

Moreover, there is an embedding $\iota:H\to G$ such that
$R_{G}(\iota_{*}(S))=R_{H}(S)$ [10, Proposition 3.6] (see also [13, Example
A.3]). We make a rational change of variables $t_{0}=s_{0}$,
$t_{1}=s_{1},t_{2}=s_{3},t_{3}=s_{2}s_{3}s_{4},t_{4}=s_{4},t_{5}=s_{5},t_{6}=s_{6}$.
Clearly $L=k(t_{0},\dots,t_{6})$ and we have:

$R_{G}(\iota_{*}(S))=(t_{0})_{2}\cup[(t_{1},t_{2})_{2}+(t_{3},t_{4})_{2}+(t_{5},t_{6})_{2}].$

We set $T_{1}=\iota_{*}(T_{0})_{F_{7}}$ and observe that
$\operatorname{ed}_{k}(R_{G}(T_{1});2)=7$ by Remark 10.3.

Part (ii): Let $G$ be a split group of type $E_{8}$ over $k$. In [12,
Corollary 11.3] Garibaldi proves there exists a torsor $T_{2}\in
H^{1}(F_{9},G)$ such that:

$R_{G}(T_{2})=(t_{0})_{2}\cup[(t_{1},t_{2})_{2}+\dots+(t_{7},t_{8})_{2}].$

We have $\operatorname{ed}_{k}(R_{G}(T_{2});2)=9$ by Remark 10.3 again.

Part (iii): Let $G=E_{8}$ as in the previous part. Denote
$H:=\mathbb{Z}^{5}/3$. By [16, Lemma 11.5], there is an embedding
$\iota:H\hookrightarrow G$ such that $\iota(H)$ is self-centralizing. Let
$K_{5}=k(t_{0},\dots,t_{4})$ be a rational field in $5$ variables. We denote
by $S\in H^{1}(K_{5},H)$ the torsor corresponding to the $H$-Galois extension
$L_{5}=K_{5}(t_{0}^{1/3},\dots,t_{4}^{1/3})$ of $K_{5}$. Let
$T_{3}=\iota_{*}(S)_{F_{5}}$. Inequality (2.2) implies:

$\operatorname{ed}_{k}(R_{G}(T_{3});3)\leqslant\operatorname{ed}_{k}(R_{G}(T_{3}))\leqslant\operatorname{ed}_{k}(T_{3})\leqslant\operatorname{trdeg}_{k}(K_{5})=5.$

Let $\nu$ be the $(t_{0},\dots,t_{4})$-adic valuation on $F_{5}$. Since $k$ is
separably closed, $(F_{5},\nu)$ is strictly Henselian. Let $E/F_{5}$ be an
arbitrary prime-to-$3$ extension. There is a unique extension of $\nu$ to $E$
which we will again denote $\nu$ by abuse of notation. Note that $(E,\nu)$ is
strictly Henselian [46, Proposition A.30]. The torsor $S_{E}$ corresponds to
the $H$-Galois field extension
$E_{5}:=E(t_{0}^{1/3},\dots,t_{4}^{1/3})=E\otimes_{F_{5}}F_{5}(t_{0}^{1/3},\dots,t_{4}^{1/3})$
of $E$. The inclusion $H\subset G_{k}$ extends to an inclusion $H\subset
G_{\mathcal{O}}$ where $k\subset\mathcal{O}\subset E$ is the valuation ring of
$\nu$. Moreover, $C_{G}(H)=H$ implies that $H$ is not contained in any proper
parabolic of $G_{\overline{k}}$ by Lemma 11.2. Therefore
$T_{3,E}=\iota_{*}(S_{E})$ is anisotropic by Lemma 11.1. This shows that
$T_{3}$ satisfies the condition of Proposition 11.3 and therefore
$\operatorname{ed}_{k}(R_{G}(T_{3});3)\geqslant 5$.

We now explain how to generalize to the case where $k$ is not separably
closed. We will focus on part (iii), which is the most complicated. Parts (i)
and (ii) are handled in an analogous way. Denote
$F^{\prime}_{5}=k^{\operatorname{sep}}((t_{0}))\dots((t_{4}))$,
$K_{5}=k(t_{0},\dots,t_{4})$ and let $G=E_{8}$. The above proof gives us a
$G$-torsor $T^{\prime}_{3}$ over $F^{\prime}_{5}$ whose Rost invariant
satisfies
$\operatorname{ed}_{k^{\operatorname{sep}}}(R_{G}(T^{\prime}_{3},3))=5$. To
check that $T^{\prime}_{5}$ descends to $K_{5}$, we note that the embedding
$\iota:H\hookrightarrow G$ is defined over $k$ because $k$ contains a
primitive $3$-rd root of unity. We have
$T^{\prime}_{3}=\iota_{*}(S)_{F^{\prime}_{5}}$, where $S\in H^{1}(K_{5},H)$ is
the torsor corresponding to the $H$-Galois extension
$L_{5}=K_{5}(t_{0}^{1/3},\dots,t_{4}^{1/3})$ of $K_{5}$. We set
$T_{3}=\iota_{*}(S)_{F_{5}}$. The inequality
$\operatorname{ed}_{k}(R_{G}(T_{3});3)\leqslant 5$ follows from (2.2) because
$\operatorname{ed}_{k}(T_{3})\leqslant\operatorname{trdeg}_{k}(K_{5})=5$.
Moreover, $T_{3,F^{\prime}_{5}}=T^{\prime}_{3}$ and so (2.3) gives us:

$\operatorname{ed}_{k}(R_{G}(T_{3});3)\geqslant\operatorname{ed}_{k}(R_{G}(T_{3})_{F^{\prime}_{5}};3)=\operatorname{ed}_{k^{\operatorname{sep}}}(R_{G}(T^{\prime}_{3});3)=5.$

We conclude that $\operatorname{ed}_{k}(R_{G}(T_{3});3)=5$. ∎

######  Remark 12.1.

One may wonder if similar techniques can be used to recover the exponential
lower bounds

(12.1)
$\operatorname{ed}(\operatorname{Spin}_{n};2)\geqslant\begin{cases}2^{(n-1)/2}-\frac{(n-1)n}{2}&\text{if
$n$ is odd, and}\\\ 2^{(n-2)/2}-\frac{(n-1)n}{2}&\text{if $n$ is
even}\end{cases}$

from [4, Theorem 3.3]. The answer is “no”. (Here we assume that $k$ is of
characteristic $0$ and $n\geqslant 15$. In the case, where $n$ is divisible by
$4$, the lower bound in (12.1) is not optimal, but that shall not concern us
here.)

Indeed, recall that the Rost invariant for $R_{G}\colon
H^{1}(F,\operatorname{Spin}_{n})\to H^{3}(F,\mu_{2})\subset
H^{3}(F,\mathbb{Q}/\mathbb{Z}(2))$ for $G=\operatorname{Spin}_{n}$ may be
computed as follows. Let $T\in H^{1}(F,\operatorname{Spin}_{n})$ be a torsor.
The image of $T$ under the natural map $H^{1}(F,\operatorname{Spin}_{n})\to
H^{1}(F,\operatorname{SO}_{n})$ is represented by a quadratic form $q$ of rank
$n$, with trivial discriminant and trivial Hasse-Witt invariant. Then
$R_{G}(T)={\rm ar}(q)$ is the Arason invariant of $q$; see [20, p. 437]. In
other words, if we represent $q$ as a sum of 3-fold Pfister forms

(12.2) $q=\langle\langle a_{1},a_{2},a_{3}\rangle\rangle+\ldots+\langle\langle
a_{3r-2},a_{3r-1},a_{3r}\rangle\rangle$

in the Witt ring $W(F)$, then

${\rm
ar}(q)=(a_{1})_{2}\cup(a_{2})_{2}\cup(a_{3})_{2}+\ldots+(a_{3r-2})_{2}\cup(a_{3r-1})_{2}\cup(a_{3r})_{2}\in
H^{3}(K,\mu_{2}).$

Note that $q$ can be represented in the form (12.2) by a theorem of Merkurjev
[24].

Now assume $F$ is equipped with a strictly Henselian valuation. By a theorem
of Raczek [36, Theorem 1.13], we can assume $r\leqslant n^{2}/8$ in (12.2).
Since $R_{G}(T)={\rm ar}(q)$ descends to $k(a_{1},\dots,a_{3r})$ we conclude:

$\operatorname{ed}_{k}(R_{G}(T);2)\leqslant\operatorname{ed}_{k}(R_{G}(T))\leqslant
3r\leqslant 3n^{2}/8.$

For large $n$ this is smaller than the lower bound on
$\operatorname{ed}_{k}(\operatorname{Spin}_{n};2)$ in (12.1).

##  13\. Appendix

###  13.1. The norm residue isomorphism

Let $F$ be a field. Denote the separable closure of $F$ by
$F^{\operatorname{sep}}$. Choose compatible a compatible system of roots of
primitive $m$th unity $\zeta_{m}\in F^{\operatorname{sep}}$, as $m$ ranges
over the positive integers prime to $\operatorname{char}(F)$. That is,
$\zeta_{m_{1}m_{2}}^{m_{1}}=\zeta_{m_{2}}$, as in Section 2.3.

Now fix one particular integer $m$ prime to $\operatorname{char}(F)$. Recall
that the Kummer map is the connecting homomorphism :

(13.1) $\partial_{m}:F^{*}\to H^{1}(F,\mu_{m})$

induced by the short exact sequence
$1\to\mu_{m}\to(F^{\operatorname{sep}})^{*}\overset{\cdot
m}{\to}(F^{\operatorname{sep}})^{*}\to 1$ [43, Chapter 2, Section 1.2]. The
norm residue isomorphism theorem states that for all $m$ coprime to
$\operatorname{char}k$ and $d\in\mathbb{N}$ there is an isomorphism:

$K_{d}(F)/m\to H^{d}(F,\mu^{\otimes d}_{m}),\ \
\\{a_{1},\dots,a_{d}\\}\mapsto\partial_{m}(a_{1})\cup\dots\cup\partial_{m}(a_{d}).$

Here $K_{d}$ is Milnor $K$-theory. We will denote by $h^{d}_{m}:K_{d}(F)\to
H^{d}(F,\mathbb{Z}/{m}(d))$ the Galois symbol, given by the composition:

$K_{d}(F)\to H^{d}(F,\mu^{\otimes d}_{m})\to
H^{d}(F,(\mathbb{Z}/m(1))^{\otimes d})\to H^{d}(F,\mathbb{Z}/m(d)).$

For $d=0$, we set $h^{0}_{m}:K_{0}(F)=\mathbb{Z}\to\mathbb{Z}/m$ to be the
reduction modulo $m$ homomorphism. Note that $h^{d}_{m}$ depends on our choice
of the root of unity $\zeta_{m}$, which induces an isomorphism
$\mu_{m}\to\mathbb{Z}/m(1)$.

We will use the following notation for the symbol defined by
$a_{1},\dots,a_{d}\in F^{*}$:

$(a_{1},\dots,a_{d})_{m}:=h^{d}_{m}(\\{a_{1},\dots,a_{d}\\}).$

Putting these together for all $d\geqslant 0$ yields a homomorphism of graded
groups

$h_{m}:=\oplus
h^{d}_{m}:K(F)\overset{}{\to}\bigoplus_{d}H^{d}(F,\mathbb{Z}/m(d)).$

As we vary $m$, the homomorphisms $h_{m}$ transform like our chosen roots of
unity $\zeta_{m}$.

######  Lemma 13.1.

For all $m\mid n$ coprime to $\operatorname{char}F$ we have:

$i_{m,n}\circ h_{m}=\frac{n}{m}h_{n}$

where
$i_{m,n}:\bigoplus_{d}H^{d}(F,\mathbb{Z}/m(d))\to\bigoplus_{d}H^{d}(F,\mathbb{Z}/n(d))$
is the natural inclusion.

###### Proof.

Let $\iota:\mathbb{Z}/m(d)\to\mathbb{Z}/n(d)$ be the inclusion given by
multiplication by ${\frac{n}{m}}$ and let
$\iota_{*}:H^{d}(F,\mathbb{Z}/m(d))\to H^{d}(F,\mathbb{Z}/n(d))$ be the push-
forward on Galois cohomology. It suffices to prove for all
$a_{1},\dots,a_{d}\in F^{*}$:

(13.2) $\iota_{*}(a_{1},\dots,a_{d})_{m}=\frac{n}{m}(a_{1},\dots,a_{d})_{n}.$

We do this on the level of cocycles. Let $b_{i}\in{F^{\operatorname{sep}}}$ be
an $n$-th root of $a_{i}$. By definition of the Kummer map, the function

$c^{\prime}_{i}(\sigma):=b_{i}\sigma(b_{i})^{-1}\in\mu_{n}$

is a cocycle representing the class of $\partial_{n}(a_{i})\in
H^{1}(F,\mu_{n})$; see (13.1). Similarly, the cocycle

$c_{i}(\sigma)=b_{i}^{\frac{n}{m}}\sigma(b_{i}^{-\frac{n}{m}})=(b_{i}\sigma(b_{i}^{-1}))^{\frac{n}{m}}=c^{\prime}_{i}(\sigma)^{\frac{n}{m}}$

represents the class of $\partial_{m}(a_{i})\in H^{1}(F,\mu_{m})$. Using our
choice of isomorphisms $\mu_{n}\cong\mathbb{Z}/n(1)$, we can find a cocycle
$\chi_{i}:\operatorname{Gal}(F)\to\mathbb{Z}/n(1)$ such that

(13.3) $\text{for all
}\sigma\in\operatorname{Gal}(F):c^{\prime}_{i}(\sigma)=\zeta_{n}^{\chi_{i}(\sigma)}$

We calculate:

$c_{i}(\sigma)=c^{\prime}_{i}(\sigma)^{\frac{n}{m}}=(\zeta^{\chi_{i}(\sigma)}_{n})^{\frac{n}{m}}=\zeta^{\chi_{i}(\sigma)}_{m}.$

This shows that under the isomorphism
$H^{1}(F,\mu_{m})\tilde{\to}H^{1}(F,\mathbb{Z}/m(1))$ the cohomology class
$\partial_{m}(a_{i})$ is sent to the class represented by the cocycle:

$\operatorname{Gal}(F)\to\mathbb{Z}/m(1),\ \ \sigma\mapsto\chi_{i}(\sigma)\mod
m.$

Therefore, $(a_{1},\dots,a_{d})_{m}$ is represented by the $n$-cocycle:

$f:\operatorname{Gal}(F)^{n}\to\mathbb{Z}/m(n),\ \
f(\sigma_{1},\dots,\sigma_{d})=\prod_{1\leqslant i\leqslant
n}\chi_{i}(\sigma_{i})\mod m.$

Pushing $f$ forward using $\iota$, we get a cocycle representing
$\iota_{*}(a_{1},\dots,a_{d})_{m}$:

$\iota_{*}(f)(\sigma_{1},\dots,\sigma_{d})=\frac{n}{m}\prod_{1\leqslant
i\leqslant n}\chi_{i}(\sigma_{i})\mod n.$

This cocycle represents $\displaystyle\frac{n}{m}(a_{1},\dots,a_{d})_{n}$
because the cocycle $\chi_{i}$ represents the image of $\partial_{n}(a_{i})$
under the isomorphism $H^{1}(F,\mu_{n})\tilde{\to}H^{1}(F,\mathbb{Z}/n(1))$ by
(13.3). This implies (13.2) and finishes the proof. ∎

We can now deduce the ”global” form of the norm residue isomorphism stated in
Theorem 2.1 from the norm residue isomorphism theorem and Lemma 13.1. Recall
that $K(F)/p^{\infty}=\mathbb{Q}_{p}/\mathbb{Z}_{p}\otimes_{\mathbb{Z}}K(F)$
by definition.

######  Theorem 13.2.

For any prime $p$ different from $\operatorname{char}k$, there is an
isomorphism of abelian graded groups $K(F)/p^{\infty}\overset{h}{\to}H_{p}(F)$
given on generators by:

$h(\frac{1}{p^{n}}\otimes\\{a_{1},\dots,a_{d}\\})=h^{d}_{p^{n}}(\\{a_{1},\dots,a_{d}\\})=(a_{1},\dots,a_{n})_{p^{n}}$

###### Proof.

The homomorphism $h$ is well-defined by Lemma 13.1 and because $h_{1}=0$ by
definition. Moreover $h$ is surjective by the norm residue isomorphism
theorem. Assume $h(\alpha)=0$ for some $\alpha\in\mathbb{T}K(F)$. We can write
$\alpha\cong\frac{1}{p^{n}}x$ for some $x\in K(F)$ and $n\in\mathbb{N}$. Since
$h(\alpha)=h_{p^{n}}(x)=0$, the norm residue isomorphism theorem implies
$x=p^{n}x^{\prime}$ for some $x^{\prime}\in K(F)$. Therefore $\alpha=0$. This
shows $h$ is indeed an isomorphism. ∎

###  13.2. Extensions of valuations

The following lemmas will be used to handle extensions of degree prime to $p$.

######  Lemma 13.3.

Let $(F,\nu)$ be a valued field over $k$ and let $F\subset L$ be a finite
extension of degree prime to $p$. There exists an extension $w$ of $\nu$ to
$L$ such that the ramification degree $[wL:\nu F]$ is prime to $p$.

###### Proof.

Let $w_{1},\dots,w_{r}$ be all the extensions of $\nu$ to $L$. Denote by
$F^{h}$ the Henselization of $(F,\nu)$, and choose Henselizations
$F^{h}\subset L^{h}_{i}$ of $(L,w_{i})$ (these exist by [46, Theorem A.30]).
Using [46, Theorem A.32], we get an $F^{h}$-linear isomorphism:

$L\otimes_{F}F^{h}\cong L^{h}_{1}\times\dots\times L^{h}_{r}.$

Comparing dimensions over $F^{h}$ we find:

(13.4) $[L:F]=[L^{h}_{1}:F^{h}]+\dots+[L^{h}_{r}:F^{h}].$

Let $e_{i}=[w_{i}L:\nu F]$ be the ramification degree of $(L,w_{i})$ over
$(F,\nu)$ and note that it is also the ramification degree of
$(L^{h}_{i},w_{i})$ over $(F^{h},\nu)$. Ostrowski’s theorem implies $e_{i}$
divides $[L^{h}_{i}:F^{h}]$ for all $i$ [46, Theorem A.12]. Since the left
hand side of (13.4) is prime to $p$, we conclude that $e_{i}$ is prime to $p$
for some $i$. ∎

######  Lemma 13.4.

Let $\Gamma\subset\Gamma^{\prime}$ be finitely generated free abelian groups
such that $[\Gamma^{\prime}:\Gamma]$ is prime to $p$. The canonical inclusion
$\bigwedge\Gamma\subset\bigwedge\Gamma^{\prime}$ induces an isomorphism:

(13.5) $\bigwedge\Gamma/p^{\infty}\cong\bigwedge\Gamma^{\prime}/p^{\infty}.$

###### Proof.

By the Elementary Divisors Theorem (as in the proof of Proposition 9.1),
$\Gamma^{\prime}$ has a basis $e_{1},\dots,e_{r}$ such that
$d_{1}e_{1},\dots,d_{r}e_{r}$ is a basis of $\Gamma$ for some positive
integers $d_{1},\ldots,d_{r}$. Then $[\Gamma^{\prime}:\Gamma]=\prod d_{i}$; in
particular, $d_{i}$ is prime to $p$ for every $i$. The group
$\bigwedge\Gamma^{\prime}$ has a basis $\\{e_{i_{1}}\wedge\dots\wedge
e_{i_{k}}\\}$ indexed by increasing sequences $1\leqslant
i_{1}<\dots<i_{k}\leqslant r$. Similarly, the pure tensors
$\\{d_{i_{1}}e_{i_{1}}\wedge\dots\wedge d_{i_{k}}e_{i_{k}}\\}$ form a basis
for $\bigwedge\Gamma$. Therefore the index
$[\bigwedge\Gamma^{\prime}:\bigwedge\Gamma]$ is prime to $p$. Consequently,
the inclusion $\bigwedge\Gamma\hookrightarrow\bigwedge\Gamma^{\prime}$ induces
an isomorphism
$(\bigwedge\Gamma)/p^{n}\stackrel{{\scriptstyle\sim}}{{\to}}(\bigwedge\Gamma^{\prime})/p^{n}$
for every $n\geqslant 1$. Taking the colimit as $n\to\infty$, we obtain the
isomorphism (13.5). ∎

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