# Sums of squares in function fields over henselian discretely valued fields

Gonzalo Manzano-Flores  Universidad de Santiago de Chile, Facultad de
Ciencias, Avenida Libertador Bernardo O’Higgins nº 3363, Estación Central,
Santiago, Chile<EMAIL_ADDRESS>
(Date: 22.02.2023)

###### Abstract.

Let $n\in\mathbb{N}$ and let $K$ be a field with a henselian discrete
valuation of rank $n$ with hereditarily euclidean residue field. Let $F/K$ be
a function field in one variable in one variable. It is known that every sum
of squares is a sum of $3$ squares. We determine the order of the group of
nonzero sums of $3$ squares modulo sums of $2$ squares in $F$ in terms of
equivalence classes of certain discrete valuations of $F$ of rank at most $n.$
In the case of function fields of hyperelliptic curves of genus $g,$ K.J.
Becher and J. Van Geel showed that the order of this quotient group is bounded
by $2^{n(g+1)}.$ We show that this bound is optimal. Moreover, in the case
where $n=1,$ we show that if $F/K$ is a hyperelliptic function field such that
the order of this quotient group is $2^{g+1},$ then $F$ is nonreal.

Keywords: Sums of squares, valuation, function field in one variable,
quadratic form, local-global principle, hereditarily pythagorean field.

Classification (MSC 2020): 11E81, 12D15, 12J10

##  1\. Introduction

Let $K$ be a field. For $d\in\mathbb{N},$ let $\mathsf{S}_{d}(K)$ denote the
set of nonzero sums of $d$ squares in $K.$ Let
$\mathsf{S}(K)=\displaystyle\bigcup_{n\in\mathbb{N}}\mathsf{S}_{n}(K),$ the
multiplicative group of nonzero sums of squares in $K.$ The Pythagoras number
$p(K)$ of a field $K$ is the smallest integer $d$ such that
$\mathsf{S}(K)=\mathsf{S}_{d}(K)$ if such an integer $d$ exists, otherwise it
is infinite. We call a finitely generated field extension of transcendence
degree one a function field in one variable, for short. In [20] it was shown
that $p(F)=2$ for every function field in one variable $F$ over $\mathbb{R}.$
However, the same does not hold for function field in one variables over
$\mathbb{R}(\\!(t_{1})\\!)\ldots(\\!(t_{n})\\!)$ when $n\geqslant 1.$ For
example, S. Tikhonov in [17, Example 3.8] showed that $tX$ is a sum of three
squares but not a sum of two squares in the function field $F$ of the curve
$Y^{2}=(tX-1)(X^{2}+1)$ over $\mathbb{R}(\\!(t)\\!),$ which shows that
$p(F)>2.$ However, in that example it was not known whether $tX$ was the only
nontrivial element of the quotient group of nonzero sums of squares in $F$
modulo the subgroup of nonzero sums of two squares in $F.$

Let $n\in\mathbb{N}.$ Consider a function field in one variable $F$ over the
base field $\mathbb{R}(\\!(t_{1})\\!)\ldots(\\!(t_{n})\\!).$ It was shown in
[3, Theorem 6.12] that $p(F)\leqslant 3$ and that the quotient
$\mathsf{S}(F)/\mathsf{S}_{2}(F),$ which measures the failure of every sum of
squares to be a sum of two squares, is a finite group. In this article, we
determine the order of $\mathsf{S}(F)/\mathsf{S}_{2}(F)$ in terms of the
cardinality of a certain finite set of valuations on $F.$ In fact, we consider
situations with slightly more general base fields (than
$\mathbb{R}(\\!(t_{1})\\!)\ldots(\\!(t_{n})\\!)$) namely fields carrying a
nontrivial henselian valuation with value group $\mathbb{Z}^{n}$ endowed with
the lexicographic order, and such that the Pythagoras number of the rational
function field over its residue field is bounded by $2^{r},$ for some
$r\in\mathbb{N}^{+}.$ Concretely, for a function field in one variable
$F/\mathbb{R}(\\!(t_{1})\\!)\ldots(\\!(t_{n})\\!),$ we show that
$|\mathsf{S}(F)/\mathsf{S}_{2}(F)|=2^{m},$ where $m$ is the number of
equivalence classes of discrete valuations of rank at most $n$ and with
nonreal residue field not containing $\sqrt{-1}.$ In this case, this
generalizes [3, Theorem 6.12]. It is natural to wonder whether we can bound
the order of $\mathsf{S}(F)/\mathsf{S}_{2}(F)$ in terms of the genus of $F/K.$
In [2], reduction theory of arithmetic curves over $\mathbb{R}[\\![t]\\!]$ is
applied to obtain the bound $|\mathsf{S}(F)/\mathsf{S}_{2}(F)|\leqslant
2^{g+1}$ for any function field in one variable $F$ of genus $g$ over
$\mathbb{R}(\\!(t)\\!).$ In the case of a function field $F$ of a
hyperelliptic curve of genus $g$ over
$\mathbb{R}(\\!(t_{1})\\!)\ldots(\\!(t_{n})\\!),$ it is a consequence of [5,
Theorem 3.10] that

$|\mathsf{S}(F)/\mathsf{S}_{2}(F)|\leqslant 2^{n(g+1)}.$

It is known that in the case where $n=1$ and $g=1,$ the bound is optimal. In
fact, for the nonreal function field $F$ of the curve
$Y^{2}=-(X^{2}+1)(X^{2}+t^{2})$ over $\mathbb{R}(\\!(t)\\!),$ it was shown in
[5, Example 5.12] that $|\mathsf{S}(F)/\mathsf{S}_{2}(F)|=2^{2}.$ In this
article, for any $g\in\mathbb{N},$ we will construct a hyperelliptic curve of
genus $g$ over $K$ such that $|\mathsf{S}(F)/\mathsf{S}_{2}(F)|=2^{n(g+1)}.$
This shows the optimality of the previously mentioned bounds for function
field in one variables over $\mathbb{R}(\\!(t)\\!)$ or for hyperelliptic
function fields over $\mathbb{R}(\\!(t_{1})\\!)\ldots(\\!(t_{n})\\!).$ We ask
in Section 4 whether our bounds in terms of valuations may equally serve to
find a bound in terms of $g$ and $n$ for function field in one variables $F/K$
of genus $g.$

##  2\. Preliminaries

For a ring $R,$ let $R^{\times}$ denote its group of invertible elements. Let
$K$ be a field. For a valuation $v$ on $K,$ we denote by $\mathcal{O}_{v}$ the
valuation ring associated to $v,$ by $\Gamma_{v}$ the value group
$v(K^{\times}),$ which is an ordered abelian group, by $\mathfrak{m}_{v}$ the
maximal ideal of $\mathcal{O}_{v}$ and by $\kappa_{v}$ the residue field
$\mathcal{O}_{v}/\mathfrak{m}_{v}.$ The residue in $\kappa_{v}$ of an element
$a\in\mathcal{O}_{v}$ will be denoted by $\overline{a}.$ For a valuation ring
$\mathcal{O}$ of $K,$ the corresponding valuation on $K$ with valuation ring
$\mathcal{O}$ and value group $K^{\times}/\mathcal{O}^{\times}$ is denoted by
$v_{\mathcal{O}}.$ We call two valuations $v$ and $w$ on $K$ equivalent if
$\mathcal{O}_{v}=\mathcal{O}_{w},$ equivalently, if and only if there exists
an order-isomorphism $\gamma:\Gamma_{w}\to\Gamma_{v}$ such that $v=\gamma\circ
w.$ See [8, Proposition 2.1.3].

The rank of an ordered abelian group $\Gamma$ is defined as the number of its
proper convex subgroups. For a valuation $v$ on $K,$ we denote by
$\mathsf{rk}(v)\in\mathbb{N}$ the rank of $\Gamma_{v},$ and we say that
$\mathsf{rk}(v)$ is the rank of $v.$ Let $n\in\mathbb{N}.$ We call a valuation
on $K$ having value group $(\mathbb{Z}^{n},\leqslant_{\mathsf{leq}}),$ where
$\leqslant_{\mathsf{leq}}$ is the lexicographic order, a
$\mathbb{Z}^{n}$-_valuation_ for short. Note that
$(\mathbb{Z}^{0},\leqslant_{\mathsf{leq}})$ is the trivial additive group and
that for a $\mathbb{Z}^{n}$-valuation $v$ on $K$ we have that
$\mathsf{rk}(v)=n.$

For $i\in\mathbb{N},$ let $V_{i}(K)$ be the set of $\mathbb{Z}^{i}$-valuations
on $K,$ where $V_{0}(K)$ is the trivial valuation on $K.$ We set

$V(K)=\bigcup_{i\in\mathbb{N}}V_{i}(K).$

Let $\Omega_{n}(K)=\\{\mathcal{O}_{v}\mid v\in V_{n}(K)\\},$ where
$\Omega_{0}(K)=K.$ We set

$\Omega(K)=\bigcup_{i\in\mathbb{N}}\Omega_{i}(K).$

We observe that every $\mathcal{O}\in\Omega_{n}(K)$ is a valuation ring of
Krull dimension $n$ by [8, Lemma 2.3.1]. For valuation rings
$\mathcal{O},\mathcal{O}^{\prime}$ of a field $K$ and valuations
$w,w^{\prime}$ on $K$ corresponding to $\mathcal{O}$ and
$\mathcal{O}^{\prime},$ respectively, we say that $\mathcal{O}^{\prime}$ is a
coarsening of $\mathcal{O}$ or that $\mathcal{O}$ is a refinement of
$\mathcal{O}^{\prime},$ respectively, $w^{\prime}$ is a coarsening of $w$ or
$w$ is a refinement of $w,$ if $\mathcal{O}\subseteq\mathcal{O}^{\prime}.$
Note that a field $K$ is trivially a coarsening of every valuation ring of
itself. We denote by $\mathcal{O}\mathcal{O}^{\prime}$ the smallest subring of
$K$ containing both $\mathcal{O}$ and $\mathcal{O}^{\prime}.$

Let $n,d\in\mathbb{N}$ with $d\leqslant n.$ We denote by
$\pi_{d}:\mathbb{Z}^{n}\to\mathbb{Z}^{d}$ the projection on the first $d$
components. Note that $\pi_{d}$ is a homomorphism of ordered abelian groups
with respect to the lexicographic orders on $\mathbb{Z}^{n}$ and
$\mathbb{Z}^{d}.$ Dually, we denote by
$\pi^{d}:\mathbb{Z}^{n}\to\mathbb{Z}^{d}$ the projection on the last $d$
components of $\mathbb{Z}^{n}.$ Note that $\pi^{d}$ is a group homomorphism,
but it is not order-preserving when $d<n.$

######  2.1 Proposition.

Let $n\in\mathbb{N}.$ Let $v$ be a $\mathbb{Z}^{n}$-valuation on $K.$ Let
$\mathcal{O}^{\prime}$ be a coarsening of $\mathcal{O}_{v}.$ Then
$\mathcal{O}^{\prime}=\mathcal{O}_{\pi_{r}\circ v},$ for a unique $r\leqslant
n.$

###### Proof.

First, note that $K^{\times}/\mathcal{O}_{v}^{\times}$ is an abelian group,
which has an order $\leqslant,$ given by $x\mathcal{O}_{v}^{\times}\leqslant
y\mathcal{O}_{v}^{\times}$ if and only if $yx^{-1}\in\mathcal{O}_{v}.$ Thus
$\Gamma_{v}$ is order-isomorphic to $K^{\times}/\mathcal{O}_{v}^{\times}.$ Let
$\mathcal{O}=\mathcal{O}_{v}.$ Let $\mathfrak{p}$ be the maximal ideal of
$\mathcal{O}^{\prime}.$ Then $\mathfrak{p}$ is a prime ideal of $\mathcal{O}$
and localizing $\mathcal{O}$ at $\mathfrak{p}$ we obtain
$\mathcal{O}^{\prime}=\mathcal{O}_{\mathfrak{p}}.$ Thus, we have a surjective
order-preserving group homomorphism of value groups
$\varphi:K^{\times}/\mathcal{O}^{\times}\to
K^{\times}/\mathcal{O}_{\mathfrak{p}}^{\times}$ sending
$x\mathcal{O}^{\times}$ to $x\mathcal{O}_{\mathfrak{p}}^{\times}.$ In fact, if
$a\mathcal{O}^{\times}\leqslant b\mathcal{O}^{\times},$ then
$ab^{-1}\in\mathcal{O}\subseteq\mathcal{O}_{\mathfrak{p}}$ and hence
$a\mathcal{O}_{\mathfrak{p}}\leqslant b\mathcal{O}_{\mathfrak{p}},$ for
$a,b\in K^{\times}.$ Since
$K^{\times}/\mathcal{O}^{\times}\simeq\mathbb{Z}^{n},$ then the quotient group
$(K^{\times}/\mathcal{O}^{\times})/\ker\varphi$ is order-isomorphic to
$\mathbb{Z}^{r},$ for some $r\in\mathbb{N}$ such that $r\leqslant n.$ Hence
$K^{\times}/\mathcal{O}_{\mathfrak{p}}\simeq\mathbb{Z}^{r}$ by the first
isomorphism theorem. Let $v^{\prime}=\pi_{r}\circ v.$ Since $\pi_{r}$ is
order-preserving, we have that $\mathcal{O}_{v^{\prime}}\in\Omega_{r}(K),$
which is a coarsening of $\mathcal{O}_{v}.$ By [8, Lemma 2.3.1] and since
$\mathbb{Z}^{n}$ has only $n$ proper convex subgroups, we have that
$\mathcal{O}^{\prime}=\mathcal{O}_{v^{\prime}}.$ ∎

Let $v\in V(K)$ and set $n=\mathsf{rk}(v).$ For $1\leqslant i\leqslant n,$ we
denote by $e_{i}^{n}$ the $n$-tuple $(e_{1},\ldots,e_{n})$ such that

$e_{j}=\left\\{\begin{array}[]{ll}1&\mbox{ if $j=(n+1)-i$ ,}\\\ 0&\mbox{ if
$j\neq(n+1)-i$ .}\end{array}\right.$

Note that $e_{1}^{n}$ is the minimal positive element of $\mathbb{Z}^{n}.$ An
$n$-tuple $(t_{1},\ldots,t_{n})\in K^{n}$ is called a parametrical system of
$v$ if $v(t_{i})=e_{i}^{n}$ for all $1\leqslant i\leqslant n.$ For a
$\mathbb{Z}$-valuation $v$ we call $t\in K$ a uniformizer of $v$ if $v(t)=1.$

######  Definition.

Let $n\in\mathbb{N}.$ Let $v$ be a $\mathbb{Z}^{n}$-valuation on $K$ and let
$\mathcal{O}^{\prime}$ be a coarsening of $\mathcal{O}_{v}.$ By 2.1 there
exists a unique $r\leqslant n,$ such that
$\mathcal{O}^{\prime}=\mathcal{O}_{\pi_{r}\circ v}.$ Let
$v^{\prime}=\pi_{r}\circ v.$ Let $x,y\in\mathcal{O}_{v^{\prime}}$ be such that
$x-y\in\mathfrak{m}_{v^{\prime}}$ and $x,y\notin\mathfrak{m}_{v^{\prime}}.$
Since $y\in\mathcal{O}_{v^{\prime}}^{\times}$ and
$\mathfrak{m}_{v^{\prime}}\subseteq\mathfrak{m}_{v},$ we have that
$(x-y)y^{-1}=x/y-1\in\mathfrak{m}_{v},$ and hence $v(x)=v(y).$ We denote by
$\overline{v}:\kappa_{v^{\prime}}\to\mathbb{Z}^{n-r}\cup\\{\infty\\},\overline{x}\mapsto(\pi^{n-r}\circ
v)(x),$ where $x$ is any lifting of $\overline{x},$ the residual valuation of
$v$ modulo $v^{\prime},$ which is well-defined by the above.

Let $r,d\in\mathbb{N}.$ On the other hand, let $v^{\prime}$ be a
$\mathbb{Z}^{r}$-valuation on $K$ and let $\overline{v}$ be a
$\mathbb{Z}^{d}$-valuation on $\kappa_{v^{\prime}}.$ Let
$(t_{1},\ldots,t_{r})\in K^{r}$ be a parametrical system of $v^{\prime}.$ For
$a\in K^{\times},$ there exists $a_{1},\ldots,a_{r}\in\mathbb{Z}$ such that
$a=t_{1}^{a_{1}}\ldots t_{r}^{a_{r}}u,$ for some
$u\in\mathcal{O}_{v^{\prime}}^{\times}.$ We can define a valuation
$v:K\to\mathbb{Z}^{r}\times\mathbb{Z}^{d}\cup\\{\infty\\},$ by
$v(a)=(v^{\prime}(a),\overline{v}(\overline{u})).$ The valuation $v$ is called
the composition of $v^{\prime}$ and $\overline{v}$ with respect to
$(t_{1},\ldots,t_{r}).$ Note that $\mathcal{O}_{v}$ is a refinement of
$\mathcal{O}_{v^{\prime}}.$

######  2.2 Example.

Let $n\in\mathbb{N}^{+}.$ We consider
$K_{n}=k(\\!(t_{1})\\!)\ldots(\\!(t_{n})\\!)$ the field of iterated Laurent
series over a field $k.$ By induction on $n,$ we show that there exists a
$\mathbb{Z}^{n}$-valuation $v$ on $K_{n}$ such that $(t_{1},\ldots,t_{n})\in
K_{n}^{n}$ is a parametrical system of $v.$ For $n=1,$ the valuation
$v_{t_{1}}$ on $K_{1},$ given by
$v_{t_{1}}(\displaystyle\operatorname{\mathsf{\Sigma}}_{i=m}^{\infty}a_{i}t_{1}^{i})=m,$
when $a_{m}\in k^{\times},$ is a $\mathbb{Z}$-valuation on $K_{1}$ such that
$t_{1}$ is a parametrical system $v_{t_{1}}.$ Let $n>1.$ Let $v_{t_{n}}$ be
the $t_{n}$-adic valuation on $K_{n},$ that is, the $\mathbb{Z}$-valuation
corresponding to the valuation ring $K_{n-1}[\\![t_{n}]\\!].$ By the induction
hypothesis, since $\kappa_{v_{t_{n}}}=K_{n-1},$ we may consider a
$\mathbb{Z}^{n-1}$-valuation $w$ on $K_{n-1}$ such that
$(t_{1},\ldots,t_{n-1})\in\kappa_{v_{t_{n}}}^{n-1}$ is a parametrical system
of $w.$ Let $v$ be the composition of $v_{t_{n}}$ and $w$ with respect to
$t_{n}.$ Hence $v$ is a $\mathbb{Z}^{n}$-valuation on $K_{n},$ and since
$v(t_{n})=(v_{t_{n}}(t_{n}),w(\overline{1}))=(1,0,\ldots,0)=e_{n}^{n},$ we
have that $(t_{1},\ldots,t_{n})\in K_{n}^{n}$ is a parametrical system of $v.$
Note that different choices of uniformizers of $v_{t_{n}}$ can lead to
distinct $\mathbb{Z}^{n}$-valuations on $K_{n}$ (e.g. $t_{n}t_{1}$ instead of
$t_{n}$).

######  2.3 Remark.

Let $r\in\mathbb{N}.$ Fixing a $\mathbb{Z}^{r}$-valuation $v^{\prime}$ on $K,$
a parametrical system $(t_{1},\ldots,t_{r})\in K^{r}$ for $v^{\prime},$ and
$d\in\mathbb{N},$ the above defined assignations define mutually inverse
bijections between equivalence classes of $\mathbb{Z}^{r+d}$-valuations $v,$
where $\mathcal{O}_{v}$ are refinements of $\mathcal{O}_{v^{\prime}}$ and
equivalence classes of $\mathbb{Z}^{d}$-valuations $\overline{v}$ on
$\kappa_{v},$ such that $\kappa_{v}=\kappa_{\overline{v}}.$ See [8, Section
2.3].

We call $W\subseteq\Omega(K)\smallsetminus\Omega_{0}(K)$ saturated if for all
$\mathcal{O}\in W$ we have $\mathcal{O}^{\prime}\in W$ for every nontrivial
coarsening $\mathcal{O}^{\prime}$ of $\mathcal{O}.$ Let $S,S^{\prime}\subseteq
V(K).$ We say that $S$ and $S^{\prime}$ are equivalent if
$\\{\mathcal{O}_{v}\mid v\in S\\}=\\{\mathcal{O}_{v}\mid v\in S^{\prime}\\}.$
We say that $S\subseteq V(K)\smallsetminus V_{0}(K)$ is coherent if

  * •

$S$ is a set of pairwise non-equivalent valuations,

  * •

and, if $v\in S$ is a $\mathbb{Z}^{n}$-valuation, then $\pi_{i}\circ v\in S,$
for all $1\leqslant i\leqslant n.$

######  2.4 Proposition.

Let $S\subseteq V(K)\smallsetminus V_{0}(K)$ be such that
$W=\\{\mathcal{O}_{v}\mid v\in S\\}$ is finite and saturated. Then there
exists a coherent set $S^{\prime}\subseteq V(K)$ equivalent to $S.$

###### Proof.

Let $n=\mathsf{max}\\{\dim\mathcal{O}_{v}\mid v\in S\\}.$ We prove the
statement by induction on $n.$ If $n=1,$ there is nothing to show, since every
subset of $V_{1}(K)$ is a set of pairwise of non-equivalent
$\mathbb{Z}$-valuations, which is trivially coherent. Assume now that $n>1.$
By the induction hypothesis, we have that for any
$S\subseteq\bigcup_{1\leqslant i\leqslant n-1}V_{i}(K)$ such that
$W=\\{\mathcal{O}_{v}\mid v\in S\\}$ is finite and saturated, there exists a
coherent set $S^{\prime}\subseteq V(K)$ with $W=\\{\mathcal{O}_{v}\mid v\in
S^{\prime}\\}.$ Let $S\subseteq\bigcup_{1\leqslant i\leqslant n}V_{i}(K)$ be
such that $W=\\{\mathcal{O}_{v}\mid v\in S\\}$ is finite and saturated.
Without loss of generality, we may assume that $S$ consist of a set of non-
equivalent valuations. Let $S_{n-1}=S\smallsetminus V_{n}(K).$ Then, by the
induction hypothesis, $S_{n-1}$ is equivalent to a coherent subset
$S_{n-1}^{\prime}$ of $\bigcup_{1\leqslant i\leqslant n-1}V_{i}(K).$ Let $v\in
S_{n-1}^{\prime}.$ Let $V_{v}=\\{w\in
S\mid\mathcal{O}_{w}\subsetneqq\mathcal{O}_{v}\\}.$ Let
$(t_{1},\ldots,t_{n-1})\in K^{n-1}$ be a parametrical system of $v.$ For $w\in
V_{v},$ let $\overline{w}$ be the residual valuation of $w$ modulo $v.$ Let
$w^{\prime}$ be the composition of $v$ and $\overline{w},$ with respect to
$(t_{1},\ldots,t_{n-1}).$ Thus, for every $w\in V_{v},$ we can define a
$\mathbb{Z}^{n}$-valuation $w^{\prime}$ on $K,$ which is equivalent to $w,$ by
Remark 2.3. Let $S(v)=\\{w^{\prime}\mid w\in V_{v}\\}.$ Then $S(v)\cup
S_{n-1}^{\prime}$ is a coherent set equivalent to $V_{v}\cup
S_{n-1}^{\prime}.$ We obtain that

$S^{\prime}=\displaystyle\bigcup_{v\in S_{n-1}^{\prime}\cap
V_{n-1}(K)}S(v)\cup S_{n-1}^{\prime},$

is a coherent finite set equivalent to $S.$ ∎

Let $v,w\in V(K).$ We call
$(\gamma_{v},\gamma_{w})\in\Gamma_{v}\times\Gamma_{w}$ compatible with respect
to $(v,w)$ if $\pi_{r}(\gamma_{v})=\pi_{r}(\gamma_{w}),$ where
$r\in\mathbb{N}$ is such that
$\mathcal{O}_{v}\mathcal{O}_{w}\in\Omega_{r}(K).$

Let $n\in\mathbb{N}.$ We recall that $\pi^{1}$ is the projection on the last
component of $\mathbb{Z}^{n}.$ Let $S\subseteq V(K)\smallsetminus V_{0}(K)$ be
a finite coherent set. We define the group homomorphism

(2.5) $\displaystyle\Phi_{S}:K^{\times}$ $\displaystyle\to\prod_{v\in
S}\mathbb{Z},a\mapsto(\pi^{1}(v(a)))_{v\in S}.$

######  2.6 Proposition.

Let $K$ be a field. Let $S\subseteq V(K)\smallsetminus V_{0}(K)$ be a finite
coherent set. Then $\Phi_{S}$ is surjective.

###### Proof.

Let $(e_{v})_{v\in S}$ be the canonical basis of $\prod_{v\in S}\mathbb{Z}$ as
a $\mathbb{Z}$-module. For every $v\in S,$ we show that there exists $x_{v}\in
K^{\times}$ such that $\Phi_{S}(x_{v})=e_{v}.$ Consider $v\in S$ and let
$\gamma_{v}=e_{1}^{\mathsf{rk}(v)}.$ For $w\in S$ a refinement of $v,$ let
$\gamma_{w}=e_{\mathsf{rk}(w)-\mathsf{rk}(v)+1}^{\mathsf{rk}(w)}$ in
$\Gamma_{w},$ otherwise $\gamma_{w}=0.$ We claim that for every
$w,w^{\prime}\in S,$ the pair $(\gamma_{w},\gamma_{w^{\prime}})$ in
$\Gamma_{w}\times\Gamma_{w^{\prime}},$ is compatible with respect to
$(w,w^{\prime})$. Let $w,w^{\prime}\in S.$ If $w,w^{\prime}$ are both not
refinements of $v,$ then $(\gamma_{w},\gamma_{w^{\prime}})$ are trivially
compatible. Assume $w$ is a refinement of $v$ and $w^{\prime}$ is not a
refinement of $v.$ It follows from 2.1 that
$\mathcal{O}_{w}\subseteq\mathcal{O}_{v}\subsetneqq\mathcal{O}_{w}\mathcal{O}_{w^{\prime}}\subseteq
K,$ because every valuation ring has a unique coarsening of a fixed rank. Let
$d\in\mathbb{N}$ be such that
$\mathcal{O}_{w}\mathcal{O}_{w^{\prime}}\in\Omega_{d}(K).$ Then
$d<\mathsf{rk}(v)\leqslant\mathsf{rk}(w)$ and hence
$\pi_{d}(\gamma_{w})=0=\pi_{d}(\gamma_{w^{\prime}})$ in $\mathbb{Z}^{d}.$
Finally, we assume that $w$ and $w^{\prime}$ are refinements of $v.$ Let
$d\in\mathbb{N}$ be such that
$\mathcal{O}_{w}\mathcal{O}_{w^{\prime}}\in\Omega_{d}(K).$ Then
$\mathcal{O}_{w}\mathcal{O}_{w^{\prime}}\subseteq\mathcal{O}_{v}$ and hence we
have that $\mathsf{rk}(v)\leqslant d<\mathsf{rk}(w),\mathsf{rk}(v)\leqslant
d<\mathsf{rk}(w^{\prime}).$ Thus, we have
$\pi_{d}(\gamma_{w})=\pi_{d}(\gamma_{w^{\prime}}),$ that is,
$(\gamma_{w},\gamma_{w^{\prime}})$ are compatible. Therefore, for every
$w,w^{\prime}\in S,$ the pair
$(\gamma_{w},\gamma_{w^{\prime}})\in\Gamma_{w}\times\Gamma_{w^{\prime}}$ is
compatible. By [16, Theorem 5] there exists $x_{v}\in K^{\times}$ such that
$w(x_{v})=\gamma_{w}$ for all $w\in W.$ Hence
$\pi^{1}(v(x_{v}))=\pi^{1}(\gamma_{v})=1,$ and $\pi^{1}(w(x_{v}))=0$ for all
$w\in S\smallsetminus\\{v\\}.$ Therefore $\Phi_{S}(x_{v})=e_{v}.$ Since $v\in
S$ was arbitrarily chosen, we conclude that $\Phi_{S}$ is surjective. ∎

######  2.7 Proposition.

Let $n\in\mathbb{N}.$ Assume that $K$ carries a henselian
$\mathbb{Z}^{n}$-valuation. Then $|\Omega_{i}(K)|=1$ for all $1\leqslant
i\leqslant n.$

###### Proof.

We prove this by induction on $n.$ For $n=1,$ follows from [3, Proposition
2.2]. We assume that $n>1.$ Let $v_{n-1}=\pi_{n-1}\circ v.$ Then $v_{n-1}$ is
a henselian $\mathbb{Z}^{n-1}$-valuation on $K,$ which is a coarsening of $v,$
by [8, Corollary 4.1.4] and 2.1. By the induction hypothesis
$|\Omega_{i}(K)|=1,$ for all $1\leqslant i\leqslant n-1.$ Let $\overline{v}$
be the residual valuation of $v$ modulo $v_{n-1}.$ Then $\overline{v}$ is a
henselian $\mathbb{Z}$-valuation on $\kappa_{v_{n-1}},$ by [8, Corollary
4.1.4], and it is the unique $\mathbb{Z}$-valuation on $\kappa_{v_{n-1}},$ by
[3, Proposition 2.2]. Therefore, by 2.3, we have that
$\Omega_{n}(K)=\\{\mathcal{O}_{v}\\}$ and hence $|\Omega_{n}(K)|=1.$ ∎

By a function field in one variable $F/K$ we mean a finitely generated field
extension of transcendence degree one.

######  2.8 Lemma.

Let $F/K$ be a function field in one variable and $r\in\mathbb{N}.$ Let $w$ be
a valuation on $F$ such that $\kappa_{w}/\kappa_{w|_{K}}$ is transcendental.
Then $\mathsf{rk}(w)=\mathsf{rk}(w|_{K})$ and $\kappa_{w}/\kappa_{w|_{K}}$ is
a function field in one variable.

###### Proof.

Let $v=w|_{K}.$ It follows from [8, Proposition 3.4.1, Theorem 3.4.3] that
$\mathsf{rk}(w)=\mathsf{rk}(v).$ We claim that $\kappa_{w}/\kappa_{v}$ is a
function field in one variable. Let $\alpha\in\kappa_{w}$ be a transcendental
element over $\kappa_{v}$ and let $\theta\in\mathcal{O}^{\times}$ be such that
$\overline{\theta}=\alpha.$ It follows by [8, Theorem 3.2.4] that
$K(\theta)/K$ is transcendental. Let $w^{\prime}:=w|_{K(\theta)}.$ It follows
from [8, Corollary 2.2.2] that $w^{\prime}$ is the Gauss extension of $v$ with
respect to $\theta.$ In particular $\kappa_{w^{\prime}}=\kappa_{v}(\alpha).$
Since $F/K$ is a function field in one variable, we have that $F/K(\theta)$ is
finite, then $\kappa_{w}/\kappa_{v}(\alpha)$ is a finite field extension and
hence $\kappa_{w}/\kappa_{v}$ is a function field in one variable. ∎

We fix a valued field $(K,v)$ and $F/K$ a function field in one variable. An
extension $w$ of $v$ to $F$ is called _residually transcendental_ if
$\kappa_{w}/\kappa_{v}$ is transcendental. We say that $F/K$ is _ruled_ if $F$
is a rational function field over some finite extension of $K.$

######  2.9 Proposition.

Let $v$ be a valuation on $K.$ Let $F/K$ be a ruled extension. Let $w$ be a
residually transcendental extension of $v.$ Then $\kappa_{w}/\kappa_{v}$ is
ruled.

###### Proof.

See [15, Theorem 3.3]. ∎

We define a set of equivalence classes of valuation extensions of coarsenings
of $v$ to $F$ in the following way. For $i\in\mathbb{N},$ let
$\Omega_{i}(F/v)$ denote the set of valuation rings
$\mathcal{O}\in\Omega_{i}(F)$ such that $v_{\mathcal{O}}$ is a residually
transcendental extension of a coarsening of $v.$ We set

$\Omega(F/v)=\bigcup_{i\in\mathbb{N}}\Omega_{i}(F/v).$

######  2.10 Proposition.

Let $n\in\mathbb{N}$ and let $v\in V_{n}(K).$ Let $F/K$ be a function field in
one variable. Let $\mathcal{O}\in\Omega(F/v).$ Then $v_{\mathcal{O}}$ is a
$\mathbb{Z}^{r}$-valuation, for some $r\leqslant n.$

###### Proof.

This follows directly from 2.1 and Lemma 2.8. ∎

For $i\in\mathbb{N}^{+},$ we set

$\Omega_{i}^{*}(F/v)=\\{\mathcal{O}\in\Omega_{i}(F/v)\mid\>\kappa_{\mathcal{O}}/\kappa_{\mathcal{O}\cap
K}\>\text{is nonruled}\>\\}.$

We set

$\Omega(F/v)=\bigcup_{i\in\mathbb{N}}\Omega_{i}^{*}(F/v).$

######  2.11 Lemma.

Let $F/K$ be a function field in one variable. Then $\Omega^{*}(F/v)$ is
saturated.

###### Proof.

Let $w$ be a valuation on $F$ such that $\mathcal{O}_{w}\in\Omega^{*}(F/v).$
Let $w^{\prime}$ be a coarsening of $w.$ We claim that
$\mathcal{O}_{w^{\prime}}\in\Omega^{*}(F/v).$ It follows by [8, Theorem 3.2.4]
that $\kappa_{w^{\prime}}/\kappa_{w^{\prime}|_{K}}$ is transcendental, and
hence is a function field in one variable, by Lemma 2.8. Let $\overline{w}$ be
the residual valuation of $w$ modulo $w^{\prime}.$ Let $\nu=w|_{K}.$ Note that
$w^{\prime}|_{K}$ is a coarsening of $\nu.$ Let $\overline{\nu}$ be the
residual valuation of $\nu$ modulo $w^{\prime}|_{K}.$ If
$\kappa_{w^{\prime}}/\kappa_{w^{\prime}|_{K}}$ were ruled, then
$\kappa_{\overline{w}}/\kappa_{\overline{\nu}}$ would be ruled, by Proposition
2.9, and hence $\kappa_{w}/\kappa_{\nu}$ would be ruled because
$\kappa_{w}=\kappa_{\overline{w}},\kappa_{\nu}=\kappa_{\overline{\nu}},$ which
is a contradiction. Therefore $\mathcal{O}_{w^{\prime}}\in\Omega^{*}(F/v).$ ∎

Let $F/K$ be a function field in one variable. It is natural to wonder whether
the set $\Omega^{*}(F/v)$ is finite. We shall give a positive answer in 2.12
under the assumption that $v\in V(K).$ Assuming that $v$ is a
$\mathbb{Z}$-valuation on $K,$ it was shown in [2, Theorem 5.3] by K. Becher
and D. Grimm $|\Omega^{*}(F/v)|\leqslant g+1,$ where $g$ is the genus of
$F/K.$ We will show in 4.5 that the above bound is optimal.

######  2.12 Theorem.

Let $n$ be a positive integer. Assume that $K$ carries a
$\mathbb{Z}^{n}$-valuation $v$ such that $\kappa_{v}$ is perfect. Let $F/K$ be
a function field in one variable. Then $\bigcup_{1\leqslant i\leqslant
n}\Omega_{i}^{*}(F/v)$ is finite.

###### Proof.

We prove the statement by induction on $n.$ If $n=1,$ then $v$ is a
$\mathbb{Z}$-valuation on $K,$ and it follows by [2, Theorem 5.3] that
$\Omega_{1}^{*}(F/v)$ is finite. Assume now that $n>1.$ By the induction
hypothesis, for any positive integer $s<n,$ for any field $L$ carrying a
$\mathbb{Z}^{s}$-valuation $v^{\prime}$ and for any function field in one
variable $E/L,$ the set $\bigcup_{1\leqslant i\leqslant
s}\Omega_{i}^{*}(E/v^{\prime})$ is finite. Let $v_{1}=\pi_{1}\circ v$ and let
$\overline{v}$ be the residual valuation of $v$ modulo $v_{1}.$ Then
$\overline{v}$ is a $\mathbb{Z}^{n-1}$-valuation on $\kappa_{v_{1}},$ by 4.1.
Let $r\in\\{1,\ldots,n\\}.$ We claim that $\Omega_{r}^{*}(F/v)$ is finite. If
$r=1,$ then this follows by [2, Theorem 5.3]. Assume $r>1.$ Let $w$ be a
valuation on $F$ such that $\mathcal{O}_{w}\in\Omega_{r}^{*}(F/v).$ Let
$w_{1}=\pi_{1}\circ w.$ It follows by Lemma 2.11 and by Lemma 2.8 that
$\mathcal{O}_{w_{1}}\in\Omega_{1}^{*}(F/v)$ and that $\mathcal{O}_{w_{1}}$ is
an extension of $\mathcal{O}_{v_{1}}.$ Since $\kappa_{w_{1}}/\kappa_{v_{1}}$
is a function field in one variable and $\overline{v}$ is a
$\mathbb{Z}^{n-1}$-valuation on $\kappa_{v_{1}},$ we have that
$\Omega_{r-1}^{*}(\kappa_{w_{1}}/\overline{v})$ is finite. Furthermore, we
have that $\mathcal{O}_{w}$ is determined by the induce valuation ring
$\mathcal{O}_{\overline{w}}\in\Omega_{r-1}^{*}(\kappa_{w_{1}}/\overline{v}),$
where $\overline{w}$ is the residual valuation of $w$ modulo $w_{1},$ by
Remark 2.3. Hence, we have that
$|\Omega_{r}^{*}(F/v)|=\operatorname{\mathsf{\Sigma}}_{\mathcal{O}\in\Omega_{1}^{*}(F/v)}|\Omega_{r-1}^{*}(\kappa_{\mathcal{O}}/\overline{v})|,$
where $\overline{v}$ is the residual valuation of $v$ modulo
$v_{\mathcal{O}}.$ Since for every $\mathbb{Z}$-valuation $\nu$ on $F$ the set
$\Omega_{r-1}(\kappa_{\nu}/\overline{v})$ is finite, we have that the set
$\Omega_{r}^{*}(F/v)$ is finite, and since $r$ was arbitrarily taken, we
obtain the statement. ∎

Let $v$ be a $\mathbb{Z}$-valuation on $K.$ We denote by $(K^{v},\hat{v})$ the
completion of $(K,v).$

######  2.13 Lemma.

Let $v$ be a henselian $\mathbb{Z}^{n}$-valuation on $K,$ and let
$v_{1}=\pi_{1}\circ v.$ Let $(K^{v_{1}},\hat{v}_{1})$ be the completion of
$(K,v_{1}).$ Then $v$ has a unique unramified extension $v^{\prime}$ to
$K^{v_{1}},$ and this extension is a henselian $\mathbb{Z}^{n}$-valuation with
$\kappa_{v^{\prime}}=\kappa_{v}.$

###### Proof.

We observe that $\kappa_{v_{1}}=\kappa_{\hat{v}_{1}}$ by [8, Theorem 1.3.4].
Let $\overline{v}$ be the residual valuation of $v$ modulo $v_{1},$ and let
$t$ be an uniformizer of $\hat{v}_{1}.$ We can consider the composition
$v^{\prime}$ of $\hat{v}_{1}$ and $\overline{v}$ with respect to $t.$ Note
that $\mathcal{O}_{v^{\prime}}\cap K=\mathcal{O}_{v},$ and using the fact that
$\overline{v}$ and $\hat{v}_{1}$ are henselian $\mathbb{Z}^{n-1}$ and
$\mathbb{Z}$-valuation on $\kappa_{v_{1}}$ and $K^{v_{1}},$ respectively, we
have that $v^{\prime}$ is a henselian $\mathbb{Z}^{n}$-valuation on
$K^{v_{1}}$ by [8, Corollary 4.1.4], and with
$\kappa_{v^{\prime}}=\kappa_{\overline{v}}=\kappa_{v},$ by 2.3. ∎

Let $F/K,K^{\prime}/K$ be two field extensions such that $K$ is separably
algebraically closed in $K^{\prime}.$ Then $F\otimes_{K}K^{\prime}$ is a
domain, see [11, Corollary 1, pag. 203]. We denote by $FK^{\prime}$ the
fraction field of $F\otimes_{K}K^{\prime}$ and we call it the compositum of
$F$ and $K^{\prime}$ over $K.$ Note that $FK^{\prime}$ is an extension of $F$
and of $K^{\prime}.$

A field $K$ that carries a henselian $\mathbb{Z}$-valuation $v$ is separably
algebraically closed in its completion $K^{v},$ by [19, Theorem 32.19], hence
the compositum $FK^{v}$ over $K$ exists for any field extension $F/K.$ In
particular, if $F/K$ is a function field in one variable, then $FK^{v}/K^{v}$
is a function field in one variable.

######  2.14 Proposition (D. Harbater, J. Hartmann, D. Krashen).

Assume that $K$ carries a henselian $\mathbb{Z}$-valuation $v$ such that
$\mathcal{O}_{v}$ is excellent. Let $F/K$ be a function field in one variable,
and let $E$ be the compositum of $F$ and $K^{v}$ over $K.$ Let $\varphi$ be a
quadratic form over $F$ of $\dim\varphi\geqslant 3.$ If $\varphi$ is isotropic
over $E,$ then $\varphi$ is isotropic over $F.$

###### Proof.

See [10, Lemma 4.11]. ∎

######  2.15 Corollary.

Assume that $K$ carries a henselian $\mathbb{Z}$-valuation $v$ such that
$\mathcal{O}_{v}$ is excellent. Let $F/K$ be a function field in one variable,
and let $E$ be the compositum of $F$ and $K^{v}$ over $K.$ Then

$F^{\times}\cap\mathsf{S}_{k}(E)=\mathsf{S}_{k}(F),$

for any $k\geqslant 2.$ In particular $p(F)\leqslant p(E).$

###### Proof.

Let $k\geqslant 2.$ Let $\sigma\in F^{\times}\cap\mathsf{S}_{k}(E).$ Let
$\varphi=k\times\langle 1\rangle\perp\langle-\sigma\rangle.$ Since $\varphi$
is isotropic over $E,$ then $\varphi$ is isotropic over $F$ by 2.14, hence
$\sigma\in\mathsf{S}_{k}(F).$ The second statement follows trivially. ∎

##  3\. The group of nonzero sums of squares

Let $K$ be a field. We define

$s(K)=\mathsf{inf}\\{d\in\mathbb{N}\mid-1\in
S_{d}(K)\\}\in\mathbb{N}\cup\\{\infty\\},$

the level of $K.$ If $s(K)<\infty,$ we say that $K$ is _nonreal_ , otherwise
we say that $K$ is real.

For a valued field $(K,v),$ a function field in one variable $F/K,$ and
$r\in\mathbb{N},$ we define

$\mathcal{X}^{r}(F/v)=\\{\mathcal{O}\in\Omega^{*}(F/v)\mid 2^{r}\leqslant
s(\kappa_{\mathcal{O}})<\infty\>\\},$

and

$\begin{split}\mathcal{E}^{r}(F/v)&=\\{x\in F^{\times}\mid
x\in\mathcal{O}^{\times}F^{\times 2}\>\mbox{for
all}\>\mathcal{O}\in\mathcal{X}^{r}(F/v)\\}\\\ &=\\{x\in F^{\times}\mid
v_{\mathcal{O}}(x)\in 2\Gamma_{v_{\mathcal{O}}}\>\mbox{for
all}\>\mathcal{O}\in\mathcal{X}^{r}(F/v)\\}.\end{split}$

######  3.1 Lemma.

Let $K$ be a field and $v\in V(K).$ Let $F/K$ be a function field in one
variable. Let $\mathcal{O}\in\Omega_{1}^{*}(F/v).$ Let $a\in F^{\times}$ and
$r\in\mathbb{N}.$ If $a\in\mathcal{E}^{r}(F/v)\cap\mathcal{O}^{\times},$ then
$\overline{a}\in\mathcal{E}^{r}(\kappa_{\mathcal{O}}/\overline{v}),$ where
$\overline{v}$ is the residual valuation of $v$ modulo $v_{\mathcal{O}}|_{K}.$

###### Proof.

Let $v_{1}=v_{\mathcal{O}}|_{K}.$ It follows by Lemma 2.8 and Proposition 2.10
that $v_{1}$ is equivalent to a $\mathbb{Z}$-valuation. Let $\overline{v}$ be
the residual valuation of $v$ modulo $v_{1}.$ Let
$a\in\mathcal{E}^{r}(F/v)\cap\mathcal{O}^{\times}.$ Let
$\mathcal{O}^{\prime}\in\mathcal{X}^{r}(\kappa_{\mathcal{O}}/\overline{v}).$
We need to show that
$\overline{a}\in\mathcal{O}^{\prime}\kappa_{\mathcal{O}}^{\times 2}.$ Let
$v^{\prime}=v_{\mathcal{O}^{\prime}}|_{\kappa_{v_{1}}}.$ Let
$\pi,\pi^{\prime}$ be uniformizers of $v_{\mathcal{O}}$ and $v_{1}$
respectively. Let $w$ be a composition of $v_{\mathcal{O}}$ and
$v_{\mathcal{O}^{\prime}}$ with respect to $\pi,$ and let $\nu$ be a
composition of $v_{1}$ and $v^{\prime}$ with respect to $\pi^{\prime}.$ Since
$\kappa_{\mathcal{O}^{\prime}}=\kappa_{w}$ and
$\kappa_{v^{\prime}}=\kappa_{\nu}$ by Remark 2.3, we have that
$\kappa_{w}/\kappa_{\nu}$ is a function field in one variable with
$2^{r}\leqslant s(\kappa_{w})<\infty$ and
$\mathcal{O}_{v}\subseteq\mathcal{O}_{w|_{K}}=\mathcal{O}_{\nu},$ that is
$\mathcal{O}_{w}\in\mathcal{X}^{r}(F/v).$ Then
$a\in\mathcal{O}_{w}^{\times}F^{\times 2}\cap\mathcal{O}^{\times},$ because
$a\in\mathcal{E}^{r}(F/v).$ This implies that
$\overline{a}\in\mathcal{O}^{\prime\times}\kappa_{\mathcal{O}}^{\times 2},$
because $(0,v_{\mathcal{O^{\prime}}}(\overline{a}))=w(a)\in 2\Gamma_{w}.$ ∎

We say that a valuation $v$ on $K$ is real or nonreal, respectively, if
$\kappa_{v}$ has the corresponding property. We say that a valuation $v$ on
$K$ is nondyadic if $\mathsf{char}(\kappa_{v})\neq 2.$ We recall that an
ordered abelian group that admits a minimal positive element is called
_discrete_.

######  3.2 Lemma.

Let $v$ be a nonreal and nondyadic valuation on $K$ such that $\Gamma_{v}$ is
discrete. Then $s(\kappa_{v})<p(K).$

###### Proof.

Let $d\in\mathbb{N}$ be such that $d=s(\kappa_{v}).$ There exist
$f\in\mathfrak{m}_{v},x_{1},\ldots,x_{d}\in K$ such that
$f=1+x_{1}^{2}+\cdots+x_{d}^{2}.$ Let
$b=(1-\frac{f}{2})^{2}+x_{1}^{2}+\cdots+x_{d}^{2}=\frac{f^{2}}{4}.$ Let
$\gamma\in\Gamma_{v}$ be the minimal positive element and let $z\in K$ be such
that $v(z)=\gamma.$ Note that $\gamma\notin 2\Gamma_{v}.$ Hence
$0<v(z)<2v(f)=v(b).$ Let
$\sigma=(z-(1-\frac{f}{2}))^{2}+x_{1}^{2}+\cdots+x_{d}^{2}.$ Since
$\sigma=z(z-2(1-\frac{f}{2}))+b,\>0<v(z)<v(b),$ and since $v(2-f)=0<v(z),$ we
have that $v(\sigma)=\gamma.$ It follows by Lemma [3, Lemma 4.1] that
$\sigma\notin\mathsf{S}_{d}(K),$ and hence $s(\kappa_{v})<p(K).$ ∎

######  3.3 Lemma.

Let $r\in\mathbb{N}^{+}.$ Let $K$ be a field and $v$ a real henselian
valuation in $V(K)$ such that $p(\kappa_{v}(X))\leqslant 2^{r}.$ Let
$v_{1}=\pi_{1}\circ v.$ Let $\overline{v}$ be the residual valuation of $v$
modulo $v_{1}.$ Let $(K^{v_{1}},\hat{v}_{1})$ be the completion of
$(K,v_{1}).$ Let $v^{\prime}$ be a composition of $\hat{v}_{1}$ and
$\overline{v}$ with respect to a uniformizer of $\hat{v}_{1}.$ Let $F/K$ be a
regular function field in one variable. Let $E$ be the compositum of $F$ and
$K^{v_{1}}$ over $K.$ Then

$\mathsf{S}(F)\cap\mathcal{E}^{r}(F/v)\subseteq\mathsf{S}(E)\cap\mathcal{E}^{r}(E/v^{\prime}).$

###### Proof.

Let $\mathcal{O}\in\mathcal{X}^{r}(E/v^{\prime}).$ Let
$\mathcal{O}_{F}=\mathcal{O}\cap F$ and $\mathcal{O}_{K}=\mathcal{O}\cap K.$
We claim that $\mathcal{O}_{F}\in\mathcal{X}^{r}(F/v)$ or
$\kappa_{\mathcal{O}_{F}}$ is real. By definition we have that
$\mathcal{O}_{v^{\prime}}\subseteq\mathcal{O}\cap K^{v_{1}},$ and since $v$ is
henselian, we have that $\mathcal{O}\cap
K^{v_{1}}\subseteq\mathcal{O}_{\hat{v}_{1}}\subseteq K^{v_{1}},$ by 2.7. Hence
$\mathcal{O}_{v}\subseteq\mathcal{O}_{K}\subseteq\mathcal{O}_{v_{1}}\subseteq
K.$ We denote by $\nu$ the residual valuation of $v$ modulo
$v_{\mathcal{O}_{K}}.$ We recall that $\nu$ is a henselian valuation on
$\kappa_{\mathcal{O}_{K}}$ such that $\kappa_{\nu}=\kappa_{v},$ by [8,
Corollary 4.1.4]. Since $p(\kappa_{v}(X))\leqslant 2^{r},$ it follows by [5,
Theorem 3.5] that $s(L)\leqslant 2^{r-1}$ for every finite nonreal extension
$L/\kappa_{v}.$ Let $L^{\prime}/\kappa_{\mathcal{O}_{K}}$ be a finite nonreal
extension. Since $\nu$ is henselian, any extension of $\nu$ to $L^{\prime}$ is
again henselian, and hence we obtain that $s(L^{\prime})\leqslant 2^{r-1}.$
Since $\kappa_{\mathcal{O}_{F}}\subseteq\kappa_{\mathcal{O}},$ we have that
$s(\kappa_{\mathcal{O}_{F}})\geqslant s(\kappa_{\mathcal{O}})\geqslant 2^{r}$
and hence the extension $\kappa_{\mathcal{O}_{F}}/\kappa_{\mathcal{O}_{K}}$
cannot be algebraic. Hence $\kappa_{\mathcal{O}_{F}}/\kappa_{\mathcal{O}_{K}}$
is transcendental. If $\kappa_{\mathcal{O}_{F}}$ is nonreal, since
$s(\kappa_{\mathcal{O}_{F}})\geqslant 2^{r},$ then
$2^{r}<p(\kappa_{\mathcal{O}_{F}}),$ by 3.2, because $v_{\mathcal{O}|_{F}}$ is
equivalent to a valuation in $V(F),$ by Proposition 2.10. This implies that
$\kappa_{\mathcal{O}_{F}}/\kappa_{\mathcal{O}_{K}}$ cannot be ruled, by [5,
Theorem 3.5], whenever $\kappa_{\mathcal{O}_{F}}$ is nonreal. Hence
$\mathcal{O}_{F}\in\mathcal{X}^{r}(F/v)$ because
$\mathcal{O}_{v}\subseteq\mathcal{O}_{K}.$ We conclude that
$s(\kappa_{\mathcal{O}_{F}})\geqslant 2^{r}.$

Let $a\in\mathsf{S}(F)\cap\mathcal{E}^{r}(F/v).$ Hence
$a\in\mathcal{O}_{F}^{\times}F^{\times 2}$ because
$\mathcal{O}_{F}\in\mathcal{X}^{r}(F/v)$ or $\kappa_{\mathcal{O}_{F}}$ is
real, and the latter follows by [3, Lemma 4.1]. Therefore
$a\in\mathcal{O}_{F}^{\times}F^{\times
2}\subseteq\mathcal{O}^{\times}E^{\times 2}.$ Since $\mathcal{O}$ was
arbitrarily chosen, we conclude that
$a\in\mathsf{S}(E)\cap\mathcal{E}^{r}(E/v^{\prime}).$ ∎

For a field $K,$ we set

$p^{\prime}(K)=\left\\{\begin{array}[]{ll}p(K)&\mbox{ if $K$ is real,}\\\
s(K)+1&\mbox{ if $K$ is nonreal.}\end{array}\right.$

######  3.4 Lemma.

Let $K$ be a perfect field. Let $L/K$ be an algebraic extension. Then

$p^{\prime}(L)\leqslant p(K(X)).$

###### Proof.

If the extension $L/K$ is finite, then the result follows from [3, Lemma 6.3].
Assume that $L/K$ is algebraic. If $p(K(X))=\infty,$ then the inequality is
trivially satisfied. Assume $p(K(X))<\infty.$ Let $r=p(K(X)),$ and let
$\sigma\in\mathsf{S}_{r+1}(L).$ Let $x_{0},\ldots,x_{r}\in L$ be such that
$\sigma=x_{0}^{2}+\cdots+x_{r}^{2},$ and let
$K^{\prime}=K(x_{0},\ldots,x_{r}).$ Hence $K^{\prime}/K$ is finite. Then
$p(K^{\prime})\leqslant r,$ by [3, Lemma 6.3], and since $K^{\prime}\subseteq
L,$ we have that $\sigma\in\mathsf{S}_{r}(L)$ because
$\mathsf{S}_{r}(K^{\prime})\subseteq\mathsf{S}_{r}(L).$ Therefore
$p(L)\leqslant r.$ ∎

######  3.5 Proposition.

Let $r\in\mathbb{N}^{+}.$ Let $K$ be a field and $v$ a real henselian
valuation in $V(K)$ such that $p(\kappa_{v}(X))\leqslant 2^{r}.$ Let $F/K$ be
a regular function field in one variable. Then

$\mathsf{S}(F)\cap\mathcal{E}_{r}(F/v)=\mathsf{S}_{2^{r}}(F).$

###### Proof.

It follows from [3, Lemma 4.1] that
$\mathsf{S}_{2^{r}}(F)\subseteq\mathsf{S}(F)\cap\mathcal{E}^{r}(F/v).$ Let
$n=\mathsf{rk}(v).$ Let us now show by induction over $n$ that
$\mathsf{S}(F)\cap\mathcal{E}_{r}(F/v)\subseteq\mathsf{S}_{2^{r}}(F).$ For
$n=0,$ it follows trivially from the assumption.

Let now $n>0.$ Let $v_{1}=\pi_{1}\circ v.$ Let $(K^{v_{1}},\hat{v}_{1})$ be
the completion of $(K,v_{1}).$ Let $v^{\prime}$ be a composition of
$\hat{v}_{1}$ and $\overline{v}$ with respect to a uniformizer of
$\hat{v}_{1}.$ It follows by Lemma 2.13 that $v^{\prime}$ is a henselian
valuation of rank $n$ on $K^{v_{1}}$ such that
$\kappa_{v}=\kappa_{v^{\prime}}$ and $\mathcal{O}_{v^{\prime}}\cap
K=\mathcal{O}_{v}.$

Let $E$ be the compositum of $F$ and $K^{v_{1}}$ over $K.$ Let
$\sigma\in\mathsf{S}(F)\cap\mathcal{E}^{r}(F/v).$ Then
$\sigma\in\mathsf{S}(E)\cap\mathcal{E}^{r}(E/v^{\prime}),$ by Lemma 3.3. Let
$\varphi$ be the quadratic form
$2^{r}\times\left<1\right>\perp\left<-\sigma\right>$ over $F.$ We first show
that $\varphi$ is isotropic over $E.$

Let $w\in V_{1}(E).$ We claim that $p(E^{w})\leqslant 2^{r}.$ If
$w|_{K^{v_{1}}}$ is trivial, then $\kappa_{w}/K^{v_{1}}$ is a finite field
extension, and it follows that $p(E^{w})\leqslant p^{\prime}(\kappa_{w}),$ by
[3, Proposition 4.3] and that $p^{\prime}(\kappa_{w})\leqslant 2^{r}$ by [5,
Theorem 3.5]. Let us assume now that $\mathcal{O}_{w}\cap
K^{v_{1}}=\mathcal{O}_{\hat{v}_{1}}$ and $\kappa_{w}/\kappa_{\hat{v}_{1}}$ is
algebraic. Since the residual valuation $\overline{v^{\prime}}$ of
$v^{\prime}$ modulo $\hat{v}_{1}$ is a henselian valuation on
$\kappa_{\hat{v}_{1}}$ such that
$p(\kappa_{\overline{v^{\prime}}}(X))\leqslant 2^{r},$ it follows by [5,
Theorem 3.5] and 3.4 that $p^{\prime}(\kappa_{w})\leqslant
p(\kappa_{\hat{v}_{1}}(X))\leqslant 2^{r}.$ Since $p(E^{w})\leqslant
p^{\prime}(\kappa_{w})\leqslant 2^{r}$ by Lemma 3.4 and by [3, Proposition
4.3], we have $p(E^{w})\leqslant 2^{r}.$ Therefore $\varphi$ is isotropic over
$E^{w}$ in both cases.

Let us assume now that $\mathcal{O}_{w}\cap
K^{v_{1}}=\mathcal{O}_{\hat{v}_{1}}$ and $\kappa_{w}/\kappa_{\hat{v}_{1}}$ is
a function field in one variable. We claim that $\varphi$ is isotropic over
$E^{w}.$ If $s(\kappa_{w})\leqslant 2^{r-1},$ since $s(E^{w})=s(\kappa_{w}),$
we have that $\varphi$ is isotropic over $E^{w}.$ Let
$d\in\mathbb{N},x_{1},\ldots,x_{d}\in F$ be such that
$\sigma=x_{1}^{2}+\cdots+x_{d}^{2}.$ If $\kappa_{w}$ is real, then
$w(\sigma)=\mathsf{min}\\{2w(x_{1}),\ldots,2w(x_{d})\\},$ by [3, Lemma 4.1].
If $2^{r}\leqslant s(\kappa_{w})<\infty,$ then $2^{r}<p(\kappa_{w})$ by 3.2.
Hence $\kappa_{w}/\kappa_{\hat{v}_{1}}$ cannot be ruled by [5, Theorem 3.5],
which implies that $\mathcal{O}_{w}\in\mathcal{X}^{r}(E/v^{\prime}).$ In any
case $w(\sigma)\in 2\Gamma_{w},$ because
$\sigma\in\mathcal{E}^{r}(E/v^{\prime}),$ and thus $w(\sigma)=2w(y)$ for some
$y\in E.$ Let $\tau=\sigma y^{-2}.$ Then
$\overline{\tau}\in\mathsf{S}(\kappa_{w})$ and, if
$\varphi^{\prime}=2^{r}\times\left<1\right>\perp\left<-\tau\right>$ is
isotropic over $E^{w},$ then $\varphi$ is isotropic over $E^{w}.$ By the
induction hypothesis, we have that
$\mathsf{S}_{2^{r}}(\kappa_{w})=\mathsf{S}(\kappa_{w})\cap\mathcal{E}^{r}(\kappa_{w}/\overline{v^{\prime}}).$
Since $\tau\in\mathcal{E}^{r}(E/v^{\prime}),$ it follows by Lemma 3.1 that
$\overline{\tau}\in\mathcal{E}^{r}(\kappa_{w}/\overline{v^{\prime}}),$ and
hence $\overline{\tau}\in\mathsf{S}_{2^{r}}(\kappa_{w}).$ Hence
$\overline{\varphi_{r}^{\prime}}$ is isotropic over $\kappa_{w},$ and then
$\varphi^{\prime}$ is isotropic over $E^{w},$ because the extension $\hat{w}$
of $w$ to $E^{w}$ is henselian, whereby $\varphi$ is isotropic over $E^{w}.$

By [9, Theorem 6.1], $\varphi$ is isotropic over $E$ if and only if $\varphi$
is isotropic over $E^{w}$ for every rank one valuation $w$ on $E$ such that
$w|_{K^{v_{1}}}$ is trivial or $\mathcal{O}_{w}\cap
K^{v_{1}}=\mathcal{O}_{\hat{v}_{1}}.$ If $\mathcal{O}_{w}\cap
K^{v_{1}}=\mathcal{O}_{\hat{v}_{1}}$ then it follows by Lemma 2.8 that, either
$\kappa_{w}/\kappa_{\hat{v}_{1}}$ is algebraic or
$\kappa_{w}/\kappa_{\hat{v}_{1}}$ is a function field in one variable. By the
above, in any case $\varphi$ is isotropic over $E,$ and it follows from 2.15
that $\varphi$ is isotropic over $F,$ whereby
$\sigma\in\mathsf{S}_{2^{r}}(F).$ ∎

For a field $K$ and $r\in\mathbb{N},$ we set
$G_{r}(K)=\mathsf{S}(K)/\mathsf{S}_{2^{r}}(K).$

######  3.6 Theorem.

Let $r\in\mathbb{N}^{+}$ and $n\in\mathbb{N}.$ Let $K$ be a field carrying a
real henselian $\mathbb{Z}^{n}$-valuation $v$ such that
$p(\kappa_{v}(X))\leqslant 2^{r}.$ Let $F/K$ be a function field in one
variable. Then

$|G_{r}(F)|=2^{|\mathcal{X}^{r}(F/v)|}.$

In particular $|G_{r}(F)|$ is finite.

###### Proof.

It follows by 2.12 that we may choose a finite and saturated set
$W\subseteq\Omega(F)$ such that $\mathcal{X}^{r}(F/v)\subseteq W,$ and it
follows by Proposition 2.4 that we can choose a coherent subset
$S^{\prime}\subseteq V(F)$ such that $W=\\{\mathcal{O}_{w}\mid w\in
S^{\prime}\\}.$ We consider $S=\\{w\in
S^{\prime}\mid\>\mathcal{O}_{w}\in\mathcal{X}^{r}(F/v)\\}.$ Let
$\Phi:S(F)\to\prod_{w\in S}\mathbb{Z}/2\mathbb{Z}$ be the map given by the
composition of $\Phi_{S^{\prime}}|_{\mathsf{S}(F)},$ where
$\Phi_{S^{\prime}}:F^{\times}\to\prod_{w\in S^{\prime}}\mathbb{Z}$ is defined
in Equation 2.5, and the natural surjective map

$\prod_{w\in S^{\prime}}\mathbb{Z}\to\prod_{w\in S}\mathbb{Z}\to\prod_{w\in
S}\mathbb{Z}/2\mathbb{Z}.$

We claim that $\Phi$ is a surjective group homomorphism with
$\ker(\Phi)=\mathsf{S}_{2^{r}}(F).$

First, we observe that $\Phi$ is a group homomorphism simply because
valuations and projections are group homomorphisms. The inclusion
$\mathsf{S}_{2^{r}}(F)\subseteq\ker(\Phi)$ follows directly from Proposition
3.5. Let us show that $\ker(\Phi)\subseteq\mathsf{S}_{2^{r}}(F).$ Let
$\sigma\in\ker(\Phi)$ and we assume that $\sigma\notin\mathsf{S}_{2^{r}}(F).$
By Proposition 3.5, $\sigma\notin\mathcal{O}^{\times}F^{\times 2},$ for some
$\mathcal{O}\in\mathcal{X}^{r}(F/v).$ Let $w\in S$ be such that
$\mathcal{O}_{w}=\mathcal{O},$ and let
$a_{1},\ldots,a_{\mathsf{rk}(w)}\in\mathbb{Z}$ be such that
$w(\sigma)=(a_{1},\ldots,a_{\mathsf{rk}(w)}).$ Since $w(\sigma)\notin
2\mathbb{Z},$ there exists $d\leqslant\mathsf{rk}(w)$ such that $a_{d}\notin
2\mathbb{Z}.$ Let $w^{\prime}=\pi_{d}\circ w.$ Since $w^{\prime}(\sigma)\notin
2\mathbb{Z},$ the residue field $\kappa_{w^{\prime}}$ cannot be real by [3,
Lemma 4.1]. Let $\overline{w}$ be the residual valuation of $w$ modulo
$w^{\prime}.$ Then $2^{r}\leqslant
s(\kappa_{w})=s(\kappa_{\overline{w}})\leqslant
s(\kappa_{w^{\prime}})<\infty,$ which implies that
$\mathcal{O}_{w^{\prime}}\in\mathcal{X}^{r}(F/v).$ Since $S^{\prime}$ is
coherent, we obtain that $w^{\prime}\in S^{\prime},$ and since
$\mathcal{O}_{w^{\prime}}\in\mathcal{X}^{r}(F/v),$ we have $w^{\prime}\in S.$
But $\pi^{1}(w^{\prime}(\sigma))=a_{d}\notin 2\mathbb{Z},$ which contradicts
the fact that $\sigma\in\ker(\Phi).$ This shows that
$\ker(\Phi)=\mathsf{S}_{2^{r}}(F).$

We show now that $\Phi$ is surjective. Let $(e_{w})_{w\in S}$ be the canonical
basis of $\prod_{w\in S}\mathbb{Z}/2\mathbb{Z}$ as a
$\mathbb{Z}/2\mathbb{Z}$-module. Consider $w\in S,$ and let
$d=\mathsf{rk}(w).$ We claim that there exists $\sigma\in\mathsf{S}(F)$ with
$\Phi(\sigma)=e_{w}.$ Since $2^{r}\leqslant s(\kappa_{w})<\infty,$ there
exists $f\in\mathfrak{m}_{w}$ and
$x_{1},\ldots,x_{m}\in\mathcal{O}_{w}^{\times},$ for some $m\in\mathbb{N},$
such that $f=1+x_{1}^{2}+\cdots+x_{m}^{2}.$ Let
$b=(1-\frac{f}{2})^{2}+x_{1}^{2}+\cdots+x_{m}^{2}=\frac{f^{2}}{4}.$ Note that
$w(b)=2w(f)>e_{1}^{d},$ where $e_{1}^{d}$ is the minimal positive element of
$\mathbb{Z}^{d}.$ By 2.6, there exists $z\in F$ such that for all $\nu\in S$
we have that $\nu(z)=e_{\mathsf{rk}(\nu)+1-d}^{\mathsf{rk}(\nu)}$ if $\nu$ is
a refinement of $w$ and $\nu(z)<\mathsf{min}\\{0,\nu(f)\\}$ otherwise. Let
$\sigma=(z-(1-\frac{f}{2}))^{2}+x_{1}^{2}+\cdots+x_{m}^{2}$ in
$\mathsf{S}(F).$ Let $\nu\in S$ be a refinement of $w.$ Since
$\sigma=z(z-2(1-\frac{f}{2}))+b$ and $\nu(z)<\nu(b)$ we have that
$\nu(\sigma)=\nu(z).$ Thus $\pi^{1}(\nu(\sigma))=0$ if $\mathcal{O}_{\nu}$ is
a proper refinement of $\mathcal{O}_{w}$ and
$\pi^{1}(w(\sigma))=\pi^{1}(e_{1}^{d})=1.$ Consider $\nu\in S$ not a
refinement of $w.$ Then $\nu(\sigma)=\nu(z(z-2(1-\frac{f}{2})))=2\nu(z).$ Thus
$\pi^{1}(\nu(\sigma))\in 2\mathbb{Z}.$ Hence $\Phi(\sigma)=e_{w}.$ Therefore
$\Phi$ is surjective and $|G_{r}(F)|=2^{|\mathcal{X}^{r}(F/v)|}.$ ∎

A field $K$ is called _hereditarily pythagorean_ if $K$ is real and $p(L)=1$
for every finite real extension $L$ of $K.$

######  3.7 Corollary.

Let $r\in\mathbb{N}^{+}$ and $n\in\mathbb{N}.$ Assume that $K$ carries a
henselian $\mathbb{Z}^{n}$-valuation $v$ such that $\kappa_{v}$ is
hereditarily pythagorean. Let $F/K$ be a function field in one variable. If
$|\kappa_{v}^{\times}/\kappa_{v}^{\times 2}|=2^{r},$ then

$|G_{r}(F)|=2^{|\mathcal{X}_{r}(F/v)|}.$

###### Proof.

This follows directly from [1, Corollary 3.3] and 3.6. ∎

##  4\. Hyperelliptic function fields

A hyperelliptic function field $F/K$ is the function field generated by two
variables $X,Y$ subject to the relation $Y^{2}=f(X),$ for some square-free
polynnomial $f$ over $K.$

In this section, we will apply the valuation-theoric description of the order
of $G(F):=G_{1}(F)$ from the previous section in the case where $F/K$ is a
hyperelliptic function field where $K$ carries a henselian
$\mathbb{Z}^{n}$-valuation $v$ such that $\kappa_{v}$ is hereditarily
pythagorean with $\kappa_{v}^{\times}=\kappa_{v}^{\times
2}\cup-\kappa_{v}^{\times 2},$ or equivalently, such that $\kappa_{v}$ is
hereditarily euclidean. Under this assumptions, it is a consequence of [1,
Theorem 5.8.2] and [5, Corollary 4.6] that $p(F)\leqslant 3$ (even if $F/K$ is
not hyperelliptic). The particular case where
$K=\mathbb{R}(\\!(t_{1})\\!)\ldots(\\!(t_{n})\\!)$ was showed earlier in [3,
Theorem 6.12]. Thus, in this case we have that

$G(F)=\mathsf{S}_{3}(F)/\mathsf{S}_{2}(F).$

Let $F/K$ be a hyperelliptic function field. Let $f\in K[X]$ be a square-free
polynomial such that $F=K(X)(\sqrt{f}).$ Let $g=\lfloor{\frac{\deg
f-1}{2}}\rfloor.$ The integer $g$ is the genus of the function field $F.$ See
[14, Proposition 7.4.24]. In [5, Theorem 3.10], an upper bound for $|G(F)|$ in
terms of the order of the square class group of the root fields of the
distinct nonreal irreducible factors of $f,$ is given in the general case
where $K$ is an arbitrary hereditarily pythagorean field. In the case where
$K$ is a field carrying a henselian $\mathbb{Z}^{n}$-valuation $v$ with
hereditarily euclidean residue field the bound becomes

$|G(F)|\leqslant 2^{n(g+1)},$

see 4.1. We will then proceed to construct hyperelliptic function fields $F/K$
such that the set $\mathcal{X}^{1}(F/v)$ satisfies
$|\mathcal{X}^{1}(F/v)|\geqslant n(g+1),$ which then immediately yields the
equalities $|G(F)|=2^{n(g+1)}$ and $|\mathcal{X}^{1}(F/v)|=n(g+1).$ In
particular the upper bound in [5, Theorem 3.10] is optimal.

######  4.1 Remark.

Let $n\in\mathbb{N}$ and let $K$ be a field with a henselian
$\mathbb{Z}^{n}$-valuation $v$ and hereditarily euclidean residue field. Then
$K$ is a hereditarily pythagorean field, by [7, Proposition 3.5]. Let $f\in
K[X]$ be a square-free polynomial and let $F=K(X)(\sqrt{f}).$ Let
$K_{1},\ldots,K_{r}$ be the root fields of the distinct nonreal irreducible
factors of $f.$ By [5, Theorem 3.10] we have that

$|G(F)|\leqslant\displaystyle\prod_{i=1}^{r}|K_{i}^{\times}/K_{i}^{\times
2}|.$

Thus $-1\in K_{i}^{\times 2}$ for all $1\leqslant i\leqslant r,$ by [6, III,
Theorem 1]. Since $K_{i}/K$ is a finite extension, there exist a henselian
$\mathbb{Z}^{n}$-valuation $v^{\prime}$ on $K_{i}$ such that
$\mathcal{O}_{v^{\prime}}\cap K=\mathcal{O}_{v}$ and $-1\in
k_{v^{\prime}}^{\times 2}.$ It is well-known that
$|K_{i}^{\times}/K_{i}^{\times
2}|=2^{n}|\kappa_{v^{\prime}}^{\times}/\kappa_{v^{\prime}}^{\times 2}|.$ By
[13, Chap. VII, (7.15)] we have that $\kappa_{v^{\prime}}$ is quadratically
closed, whence $|K_{i}^{\times}/K_{i}^{\times 2}|=2^{n},$ for all $1\leqslant
i\leqslant r.$ Let $g\in\mathbb{N}$ be such that $\deg f=2g+1$ or $2g+2.$ Then
$r\leqslant g+1,$ hence

$|G(F)|\leqslant 2^{n(g+1)}.$

It follows that the unique situation where the bound can be optimal is when
$f\in K[X]$ is a square-free polynomial of degree $2(g+1)$ and has $g+1$
nonreal irreducible factors.

######  4.2 Lemma.

Let $K$ be a real field such that $p(K)=1.$ Let $v$ be a valuation on $K.$ Let
$f\in\mathcal{O}_{v}[X]$ be a monic quadratic irreducible polynomial such that
$K[X]/(f)$ is nonreal. Then there exist $a,b\in\mathcal{O}_{v}$ such that
$f=(X-a)^{2}+b^{2}.$

###### Proof.

Let $\alpha,\beta\in\mathcal{O}_{v}$ be such that $f=X^{2}+\alpha X+\beta.$
Thus, we can write $f=(X-\alpha/2)^{2}+(\beta-\alpha^{2}/4).$ Since
$K[X]/(f)=K(\sqrt{\alpha^{2}-4\beta})$ is nonreal, then
$-(\alpha^{2}-4\beta)\in\mathsf{S}(K)=K^{\times 2},$ by [13, Chap. VIII, Lemma
1.4]. The statement follows considering $a=\alpha/2$ and
$b=\sqrt{(4\beta-\alpha^{2})/4}.$

∎

######  4.3 Lemma.

Assume that $K$ carries a $\mathbb{Z}$-valuation $v.$ Let $f\in K[X]$ be a
non-constant square-free polynomial of degree $d,$ and let $F=K(X)(\sqrt{f}).$
Then $F$ is $K$-isomorphic to

$K(X)(\sqrt{\alpha\cdot q_{1}\cdots q_{r}}),$

for some $r\in\mathbb{N},q_{1},\ldots,q_{r}\in\mathcal{O}_{v}[X]$ monic
irreducible such that $d=\operatorname{\mathsf{\Sigma}}_{i=1}^{r}\deg q_{i},$
and where $\alpha=1$ if $d$ is odd and otherwise $\alpha$ is the leading
coefficient of $f.$

###### Proof.

Since $\textsf{Frac}(\mathcal{O}_{v})=K,$ we can assume that
$f\in\mathcal{O}_{v}[X].$ Let $d=\deg f.$ Let
$a_{0},\ldots,a_{d}\in\mathcal{O}_{v}$ be such that
$f=\displaystyle\operatorname{\mathsf{\Sigma}}_{i=0}^{d}a_{i}X^{i}.$ Since $F$
is the function field of $Y^{2}=f(X),$ multiplying by $a_{d}^{2(d-1)},$ we
have that $(a_{d}^{d-1}Y)^{2}=a_{d}^{d-1}g(X),$ where
$g=\operatorname{\mathsf{\Sigma}}_{i=0}^{d}b_{i}X^{i},$ where
$b_{d}=a_{d}^{d}$ and $b_{i}=a_{i}a_{d}^{d-1}$ for all $0\leqslant i\leqslant
d-1.$ Replacing $X^{\prime}=a_{d}X,$ and $Y^{\prime}=a_{d}^{d-1}Y,$ we have
that $F$ is $K$-isomorphic to
$K(X^{\prime})\left(\sqrt{a_{d}^{d-1}g(X^{\prime})}\right).$ Write
$g(X^{\prime})=q_{1}(X^{\prime})\ldots q_{r}(X^{\prime}),$ where
$q_{i}(X^{\prime})\in K[X^{\prime}]$ are monic irreducible polynomials, for
some $r\in\mathbb{N}.$ Since $\mathcal{O}_{v}$ is a Unique Factorization
Domain, by Gauss’ Lemma, we may assume that
$q_{1}(X^{\prime}),\ldots,q_{r}(X^{\prime})\in\mathcal{O}_{v}[X^{\prime}],$
which concludes the proof. ∎

Let $v$ be a valuation on a field $K.$ By [8, Corollary 2.2.2], there exists a
unique extension $w$ of $v$ to the rational function field $K(X)$ such that
$w(X)=0$ and the residue $\overline{X}\in\kappa_{w}$ is transcendental over
$\kappa_{v}.$ A valuation $w$ with these properties is called the Gauss
extension of $v$ to $F$ with respect to $X.$

######  4.4 Lemma.

Let $n\in\mathbb{N}$ Assume that $K$ is a field carrying a henselian
$\mathbb{Z}^{n}$-valuation with hereditarily euclidean residue field. Let
$F/K$ be a regular function field of genus zero. Then

$|G(F)|=\left\\{\begin{array}[]{ll}2^{n}&\mbox{ if $F$ is nonreal,}\\\
1&\mbox{ if $F$ is real.}\end{array}\right.$

###### Proof.

We assume that $F$ is nonreal. Then $F=K(X)(\sqrt{\alpha\cdot q}),$ where $q$
is a monic irreducible quadratic polynomial over $K,$ by [12, Theorem 5.7.2,
Theorem 5.7.3]. Then $q(X)=(X-a)^{2}+b^{2},$ for some $a,b\in K,$ by Lemma
4.2. Moreover $\alpha\in-K^{\times 2},$ because $F$ is nonreal. Replacing
$X^{\prime}=X-a,$ we have that $F$ is isomorphic to $K(X)(\sqrt{-(X^{2}+1)}).$
We prove that statement by induction on $n.$ For $n=0,$ it follows from [5,
Corollary 4.6] that $G(F)$ is trivial. Let $n\geqslant 1.$ Let
$v_{1}=\pi_{1}\circ v.$ Let $w^{\prime}$ be the Gauss extension of $v_{1}$ to
$K(X)$ with respect to $X$ and let $w$ be an extension of $w^{\prime}$ to $F.$
Since $\kappa_{v_{1}}$ is real, the polynomial $X^{2}+1$ is irreducible over
$\kappa_{v_{1}}.$ Then we have
$\kappa_{w}=\kappa_{v_{1}}(\overline{X})\left(\sqrt{-(\overline{X}^{2}+1)}\right),$
and $\mathcal{O}_{w}\in\mathcal{X}^{1}(F/v)\cap\Omega_{1}(F).$ It follows from
[4, Corollary 3.6] that $\mathcal{O}_{w}$ is the unique valuation ring in
$\mathcal{X}^{1}(F/v)\cap\Omega_{1}(F),$ and hence every
$\mathcal{O}\in\mathcal{X}^{1}(F/v)$ is a refinement of $\mathcal{O}_{w},$ by
Remark 2.3. Let $\overline{v}$ be the residual valuation of $v$ modulo
$v_{1}.$ Thus
$|\mathcal{X}^{1}(F/v)|=1+|\\{\mathcal{O}\in\Omega(\kappa_{w})\mid\mathcal{O}\in\mathcal{X}^{1}(\kappa_{w}/\overline{v})\\}|,$
whereby $|G(F)|=2|G(\kappa_{w})|,$ by 3.6. Note that $\overline{v}$ is a
henselian $\mathbb{Z}^{n-1}$-valuation on $\kappa_{v_{1}}$ with hereditarily
euclidean residue field. Thus, by the induction hypothesis
$|G(\kappa_{w})|=2^{n-1}.$ Therefore $|G(F)|=2^{n}.$

If $F$ is real, then $p(F)=2$ by [18, Theorem 3], whereby $|G(F)|=1.$ ∎

######  4.5 Example.

Let $g,n\in\mathbb{N}.$ Let $F$ be the function field of the curve

$Y^{2}=-\displaystyle\prod_{i=0}^{g}(X^{2}+t_{n}^{2i})$

over $K_{n}:=\mathbb{R}(\\!(t_{1})\\!)\ldots(\\!(t_{n})\\!).$ We will show
that $|G(F)|=2^{n(g+1)}.$ Let $v_{t_{n}}$ be the $t_{n}$-adic valuation on
$K_{n}.$ Note that $\mathcal{O}_{v_{t_{n}}}=K_{n-1}[\\![t_{n}]\\!].$ For
$0\leqslant i\leqslant g,$ let $v_{i}$ be the Gauss extension of $v_{t_{n}}$
to $K_{n}(X)$ with respect to $Xt_{n}^{-i}$ and let $w_{i}$ be an extension of
$v_{i}$ to $F.$

We claim that $\kappa_{w_{i}}\simeq K_{n-1}(X)(\sqrt{-(X^{2}+1)}),$ for all
$0\leqslant i\leqslant g.$ Let $0\leqslant i\leqslant g,$ and let
$Z:=Xt_{n}^{-i}.$ Hence $F=K_{n}(Z)(\sqrt{-g(Z)})$ since $K_{n}(X)=K_{n}(Z),$
where
$g=(t_{n}^{2i}Z^{2}+1)(t_{n}^{2i-2}Z^{2}+1)\cdots(Z^{2}+1)\cdots(Z^{2}+t_{n}^{2(g-i)})$
is a square-free polynomial over $K_{n}.$ Then
$g\in\mathcal{O}_{v_{i}}^{\times}$ and
$\kappa_{w_{i}}=K_{n-1}(\overline{Z})(\sqrt{-\overline{g}})=K_{n-1}(\overline{Z})\left(\sqrt{-(\overline{Z}^{2}+1)}\right).$

Let $v$ be a henselian $\mathbb{Z}^{n}$-valuation on $K_{n}.$ Let
$\overline{v}$ be the residual valuation of $v$ modulo $v_{t_{n}}.$ Since
$|\mathcal{X}^{1}(\kappa_{w_{i}}/\overline{v})|=|\\{\mathcal{O}\in\Omega(F)\mid\mathcal{O}\subsetneq\mathcal{O}_{w_{i}}\>\textit{and}\>\mathcal{O}\in\mathcal{X}^{1}(F/v)\\}|$
by 2.3 and $|\mathcal{X}^{1}(\kappa_{w_{i}}/\overline{v})|=n-1,$ by 4.4, for
every $0\leqslant i\leqslant g,$ we have that
$|\mathcal{X}^{1}(F/v)|\geqslant(g+1)+(n-1)(g+1)=n(g+1).$ Therefore
$|G(F)|\geqslant 2^{n(g+1)}$ by 3.6, whereby

$|G(F)|=2^{n(g+1)},$

by 4.1.

Let $F/K$ be a function field in one variable of genus zero, where $K$ is any
hereditarily pythagorean field, that is $F=K(X)(\sqrt{aX^{2}+b)}),$ for some
$a,b\in K.$ S. Tikhonov and V.I. Yanchevskiĭ in [18, Theorem 3] showed that if
$F$ is real, then $p(F)=2.$ Hence, if we assume that $K$ carries a henselian
$\mathbb{Z}^{n}$-valuation with hereditarily euclidean residue field, for some
$n>0,$ then $|G(F)|=2^{n},$ implies that $F$ is a nonreal field of level two,
by 4.4. This motivates the following question.

######  4.6 Question.

Let $n,g\in\mathbb{N}.$ Let $K$ be a field carrying a henselian
$\mathbb{Z}^{n}$-valuation with hereditarily euclidean residue field. Let
$F/K$ be a hyperelliptic function field of genus $g.$ Does the equality
$|G(F)|=2^{n(g+1)}$ imply that $F$ is a nonreal field ?

We show in 4.9 that this question has an affirmative answer in the case where
$n=1.$

######  4.7 Proposition.

Let $g\in\mathbb{N}.$ Assume that $K$ carries a henselian
$\mathbb{Z}$-valuation $v$ with $\kappa_{v}$ hereditarily pythagorean. Let
$f\in K[X]$ be a square-free polynomial of degree $2g+2$ with all roots in
$K(\sqrt{-1})\smallsetminus K.$ Set $F=K(X)(\sqrt{f}).$ Let $w$ be a
residually transcendental extension of $v$ to $F.$ Assume that $F$ is real.
Then one of the following conditions holds:

  * (a)

$\kappa_{w}$ is nonreal with $s(\kappa_{w})=1.$

  * (b)

$\kappa_{w}/\kappa_{v}$ is ruled.

  * (c)

$\kappa_{w}=\kappa_{v}(X)(\sqrt{h})$ is a real field with $h\in\kappa_{v}[X]$
a square-free polynomial with all roots in
$\kappa_{v}(\sqrt{-1})\smallsetminus\kappa_{v}.$

###### Proof.

By Lemma 4.2 and by Lemma 4.3 we may choose irreducible polynomials
$q_{i}=(X-a_{i})^{2}+b_{i}^{2}$ with $a_{i},b_{i}\in\mathcal{O}_{v},$ and
$\alpha\in K^{\times}$ such that $f=\alpha\cdot q_{0}\cdots q_{g}.$ Since $F$
is real, $\alpha\notin-K^{\times 2}.$ We set $Y_{i}=(X-a_{i})b_{i}^{-1}.$

Let $w$ be a residually transcendental extension of $v$ to $F$ and let
$w_{0}=w|_{K(X)}.$ As $[F:K(X)]=2$ we have that
$[\kappa_{w}:\kappa_{w_{0}}]\leqslant 2.$ Let $\ell$ be the relative algebraic
closure of $\kappa_{v}$ in $\kappa_{w}.$ If $\ell$ is nonreal, since
$\kappa_{v}$ is a hereditarily pythagorean, we have that
$-1\in\kappa_{w}^{\times 2}$ (case $(a)$).

Thus we assume now that $\ell$ is real. If $\kappa_{w}=\kappa_{w_{0}},$ since
$w$ is a residually transcendental extension of $v,$ by Proposition 2.9 we
have that $\kappa_{w}=\kappa_{w_{0}}=\ell(\overline{T}),$ for some
$T\in\mathcal{O}_{w_{0}}^{\times}$ with $\overline{T}$ transcendental over
$\kappa_{v}$ (case $(b)$). Now assume that $[\kappa_{w}:\kappa_{w_{0}}]=2$ and
$\ell$ is real. Then $\Gamma_{w}=\Gamma_{w_{0}}.$ Let $f^{\prime}=q_{0}\cdots
q_{g}.$ Since $f\in F^{\times 2},$ we have that $w_{0}(f)\in 2\Gamma_{w_{0}}$
and since $f^{\prime}\in\mathsf{S}_{2}(K(X))$ we have $w_{0}(f^{\prime})\in
2\Gamma_{w_{0}},$ by [3, Lemma 4.1]. Hence $w_{0}(\alpha)\in 2\Gamma_{w_{0}}.$
Consider $i\in\\{0,\ldots,g\\}.$ Set

$Z_{i}=\left\\{\begin{array}[]{ll}1+Y_{i}^{2}&\mbox{if $w(Y_{i})\geqslant
0$,}\\\ 1+Y_{i}^{-2}&\mbox{ if $w(Y_{i})<0$.}\end{array}\right.$

Note that, since $\ell$ is real, $Z_{i}\in\mathcal{O}_{w}^{\times}$ for all
$0\leqslant i\leqslant g.$ Thus

$\overline{Z_{i}}=\left\\{\begin{array}[]{ll}1+\overline{Y_{i}}^{2}&\mbox{if
$w(Y_{i})=0$,}\\\ \overline{1}&\mbox{ otherwise .}\end{array}\right.$

Let us first assume that $v(\alpha)\notin 2\mathbb{Z}.$ This implies that
$w_{0}$ is a ramified extension of $v.$ Hence $\overline{Y}_{i}$ is algebraic
over $\kappa_{v}$ for all $i\in\\{0,\ldots,g\\},$ because otherwise $w_{0}$
must be a Gauss extension of $v$ with respect to some $Y_{i},$ which would
contradict the ramification. Since $\ell$ is pythagorean, we have
$\overline{Z_{i}}\in\ell^{\times 2}\subseteq\kappa_{w}^{\times 2}$ because
$\ell$ is the relative algebraic closure of $\kappa_{v}$ in $\kappa_{w}.$ By
Proposition [4, Lemma 2.2] we have that
$\kappa_{w}=\kappa_{w_{0}}(\sqrt{\overline{u}}),$ for any $u\in\alpha
f^{\prime}K(X)^{\times 2}$ with $w_{0}(u)=0.$ Since
$\prod_{i=0}^{g}\overline{Z_{i}}\in\kappa_{w_{0}}^{\times 2},$ we have that
$\kappa_{w}=\kappa_{w_{0}}\left(\sqrt{\overline{\alpha h^{2}}}\right)$ for any
$h\in K(X)^{\times}$ such that $w_{0}(\alpha h^{2})=0.$ Let
$K^{\prime}=K(\sqrt{\alpha}).$ Let $w_{0}^{\prime}$ be an extension of $w_{0}$
to $K^{\prime}(X)$ and let $v^{\prime}=w_{0}^{\prime}|_{K^{\prime}}.$ We have
that $\kappa_{w_{0}^{\prime}}=\kappa_{w_{0}}\left(\sqrt{\overline{\alpha
h^{2}}}\right),$ because $K^{\prime}(X)=K(X)(\sqrt{\alpha})$ and $w_{0}(\alpha
h^{2}).$ Furthermore, $w_{0}^{\prime}$ is a residually transcendental
extension of $v^{\prime}$ to $K^{\prime}(X).$ Therefore
$\kappa_{w}/\kappa_{v}$ is ruled, by Proposition 2.9 (case $(b)$).

Let us assume now that $v(\alpha)\in 2\mathbb{Z}.$ Let $J$ be the set of
indices $i\in\\{0,\ldots,g\\}$ such that $w(Y_{i})=0$ and $\overline{Y}_{i}$
is transcendental over $\kappa_{v}.$ Assume first that $J=\emptyset.$ Then
$\overline{Z_{i}}\in\kappa_{w}$ is algebraic over $\ell$ for all $0\leqslant
i\leqslant g,$ and since $\ell$ is pythagorean, we have that
$\kappa_{w}=\kappa_{w_{0}}(\sqrt{\alpha}).$ Therefore $\kappa_{w}/\kappa_{v}$
is ruled (case ($b$)). Now we assume that $J\neq\emptyset.$ Without loss of
generality we put $J=\\{0,\ldots,s\\},$ for some $s\leqslant g.$ For $i\in J,$
let $v_{i}$ be the Gauss extension of $v$ to $K(X)$ with respect to $Y_{i}.$
By [8, Corollary 2.2.2] we have that
$w_{0}=v_{0}=\cdots=v_{s},\;\kappa_{w_{0}}=\kappa_{v}(\overline{Y_{0}})$ and
$\Gamma_{w_{0}}=\mathbb{Z}.$ Thus $w_{0}(\alpha)\in 2\mathbb{Z},$ and we may
consider some $\beta\in\mathcal{O}_{v}^{\times}$ such that $\alpha\in\beta
K^{\times 2}.$ Let $j\in J.$ We have
$Y_{j}^{2}+1\in\mathcal{O}_{w_{0}}^{\times}.$ We set
$c_{j}=b_{0}b_{j}^{-1},d_{j}=(a_{0}-a_{j})b_{j}^{-1}.$ Since
$Y_{j}=c_{i}Y_{0}+d_{j},$ we have
$Y_{j}^{2}+1=c_{j}^{2}Y_{0}^{2}+2c_{j}d_{j}Y_{0}+d_{j}^{2}+1,$ and hence $f\in
h(Y_{0})\cdot K^{\times 2},$ where

$h(Y_{0})=\beta(Y_{0}^{2}+1)\cdots(c_{s}^{2}Y_{0}^{2}+2d_{s}c_{s}Y_{0}+d_{s}^{2}+1)\displaystyle\prod_{i=s+1}^{g}Z_{i}.$

Therefore $F=K(Y_{0})(\sqrt{h(Y_{0})}).$ Since $v$ is henselian and
$\beta\notin-K^{\times 2},$ we have $\overline{\beta}\notin-\kappa_{v}^{\times
2}.$ Finally, since $h\in\mathcal{O}_{w_{0}}^{\times},$ we have that $h\in
fK(Y_{0})^{\times 2}\cap\mathcal{O}_{w_{0}}^{\times},$ and we can conclude
that
$\kappa_{w}=\kappa_{v}(\overline{Y_{0}})\left(\sqrt{\overline{h}(\overline{Y_{0}})}\right)$
is a real field, where $\overline{h}\in\kappa_{v}[\overline{Y_{0}}]$ is a
polynomial with all roots in $\kappa_{v}(\sqrt{-1})\smallsetminus\kappa_{v}$
by Proposition [4, Lemma 2.2] (case $(c)$). ∎

######  4.8 Corollary.

Let $n\in\mathbb{N}^{+}.$ Assume that $K$ carries a henselian
$\mathbb{Z}^{n}$-valuation $v$ such that $\kappa_{v}$ is hereditarily
euclidean. Let $f\in K[X]$ be a nonconstant square-free polynomial with all
roots in $K(\sqrt{-1})\smallsetminus K.$ We set $F=K(X)(\sqrt{f}).$ Assume
that $F$ is real. Then $p(F)=2.$

###### Proof.

We prove the statement by induction on $n.$ Note that by Corollary 3.7 we have
that $p(F)=2$ if and only if $\mathcal{X}^{1}(F/v)$ is empty. Assume $n=1.$
Let $w$ be a residually transcendental extension of $v.$ It follows by
Proposition 4.7 that either $s(\kappa_{w})=1$ or $\kappa_{w}$ is real. Hence
$\mathcal{X}^{1}(F/v)=\emptyset$ and $p(F)=2.$ Assume now that $n>1.$ We show
that $\mathcal{X}^{1}(F/v)$ is empty. Let $v_{1}=\pi_{1}\circ v.$ Let
$\overline{v}$ be the residual valuation of $v$ modulo $v_{1}.$ Then
$\overline{v}$ is a henselian $\mathbb{Z}^{n-1}$-valuation on $\kappa_{v_{1}}$
such that $\kappa_{v}=\kappa_{\overline{v}}.$ It follows by induction
hypothesis and by 4.7 that all the residually transcendental extensions $w$ of
$v_{1}$ to $F$ satisfy $p^{\prime}(\kappa_{w})=2.$ If we had some
$\mathcal{O}\in\mathcal{X}^{1}(F/v),$ then we would obtain that the residue
field of the rank-one coarsening $\mathcal{O}^{\prime}$ of $\mathcal{O}$ would
have $p^{\prime}(\kappa_{\mathcal{O}^{\prime}})>2$ by Proposition 3.2, which
is a contradiction. Therefore $\mathcal{X}^{1}(F/v)=\emptyset,$ whereby
$p(F)=2$ by 3.6. ∎

Note that, if $f\in K[X]$ is assumed to be monic in 4.8, then $F/K(X)$ is a
totally positive quadratic extension and hence the above result follows by [5,
Corollary 4.10].

######  4.9 Corollary.

Let $g\in\mathbb{N}.$ Assume that $K$ carries a henselian
$\mathbb{Z}$-valuation $v$ with hereditarily euclidean residue field. Let
$F/K$ be a hyperelliptic function field of genus $g.$ Let $f\in K[X]$ be a
square-free polynomial such that $F=K(X)(\sqrt{f}).$ If $|G(F)|=2^{g+1},$ then
$F$ is nonreal.

###### Proof.

It follows by Proposition 4.1 and by Lemma 4.2 that we may choose irreducible
polynomials $q_{i}=(X-a_{i})^{2}+b_{i}^{2},$ with
$a_{i},b_{i}\in\mathcal{O}_{v},$ and $\alpha\in K^{\times}$ such that
$f=\alpha\cdot q_{0}\cdots q_{g}.$ For $i\in\\{0,\ldots,g\\}$ let
$Y_{i}:=b_{i}^{-1}(X-a_{i}).$ If $\alpha\notin-K^{\times 2},$ then $p(F)=2,$
by Corollary 4.8. Therefore, under the assumption that $|G(F)|=2^{g+1},$ we
have that $\alpha\in-K^{\times 2}.$ Since $\mathsf{S}_{2}(F)$ is a group,
there exists $\eta_{1},\eta_{2}\in K(X)$ such that
$F=K(X)(\sqrt{-(\eta_{1}^{2}+\eta_{2}^{2})}),$ that is, $F$ is nonreal. ∎

######  4.10 Question.

Let $n,g\in\mathbb{N}.$ Assume that $K$ carries a henselian
$\mathbb{Z}^{n}$-valuation with hereditarily euclidean residue field. Let
$F/K$ be a function field in one variable of genus $g\in\mathbb{N}.$ Is
$|G(F)|\leqslant 2^{n(g+1)}$ for every function field in one variable $F/K$ of
genus $g$?

### Acknowledgments

I thank Marco Zaninelli, Parul Gupta and Nicolas Daans for useful conversation
on these topics. I thank my thesis advisors David Grimm and Karim Johannes
Becher.

This work was supported by CONICYT-PFCHA/Doctorado Nacional/2017-folio
21170477, and by Universidad de Santiago de Chile (Proyecto DICYT, código
041933G), by the FWO Odysseus Programme (project G0E6114N), and by the
Bijzonder Onderzoeksfonds (BOF), Universiteit Antwerpen, (project BOF-DOCPRO4,
2533).

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