Witt rings of quadratically presentable fields

TL;DR

This paper develops a new axiomatic framework for quadratic forms using presentable partial orders, enabling a uniform construction of Witt rings across all field characteristics.

Contribution

It introduces the concept of quadratically presentable fields and demonstrates their use in constructing Witt rings uniformly for all characteristics.

Findings

Witt rings over fields are isomorphic to those over quadratically presentable fields.

The framework applies to fields of any characteristic, including characteristic 2.

A new categorical approach to quadratic forms is established.

Abstract

This paper introduces a novel approach to the axiomatic theory of quadratic forms. We work internally in a category of certain partially ordered sets, subject to additional conditions which amount to a strong form of local presentability. We call such partial orders presentable. It turns out that the classical notion of the Witt ring of symmetric bilinear forms over a field makes sense in the context of quadratically presentable fields, that is fields equipped with a presentable partial order inequationaly compatible with the algebraic operations. As an application, we show that Witt rings of symmetric bilinear forms over fields, of both characteristic 2 and not 2, are isomorphic to Witt rings of suitably built quadratically presentable fields, which therefore provide a uniform construction of Witt rings for all characteristics.

Equations

$$(a\le b)\wedge (c\le d)\Rightarrow (a+c\le b+d)$$

$$s$$

$$({\mathcal{P}}^{\ast }(M),\subseteq ,\{0\})$$

$$\begin{array}[]{lll}\boldsymbol{+}:\mathcal{P}^{\ast}(M)\times\mathcal{P}^{\ast}(M)&\longrightarrow&\mathcal{P}^{\ast}(M)\\ (A,B)&\mapsto&\{a+b\mid a\in A,b\in B\}\end{array}$$

$$\left(\bigcup _{i\in I}{A}_i\right)+\left(\bigcup _{j\in J}{B}_j\right)$$

$$+:M\times M\to {\mathcal{P}}^{\ast }(M)$$

$$\begin{array}[]{lll}\boldsymbol{+}:\mathcal{P}^{\ast}(M)\times\mathcal{P}^{\ast}(M)&\longrightarrow&\mathcal{P}^{\ast}(M)\\ (A,B)&\mapsto&\bigcup\{a+b|a\in A,b\in B\}\end{array}$$

$$(s\le t+u)\Rightarrow (t\le s+(-u))$$

$$(a\le b)\Rightarrow (-a\le -b)$$

$$0⩽a+(-a)$$

$$0$$

$$\{1,3\}\subseteq \{0,1\}+\{0,2\}=\{0,1,2,3\}$$

$$\{0,1\}⊈\{1,3\}-\{0,2\}=\{1,-1,3\}$$

$$\begin{array}[]{rcl}-:\mathcal{P}^{\ast}(G)&\longrightarrow&\mathcal{P}^{\ast}(G)\\ A&\mapsto&\{-a|a\in A\}\end{array}$$

$$\begin{array}[]{lll}-:\mathcal{P}^{\ast}(G)&\longrightarrow&\mathcal{P}^{\ast}(G)\\ A&\mapsto&\{-a|a\in A\}\end{array}$$

$${\mathcal{S}}_{ab}$$

$$A\cdot B\stackrel{def.}{=}\{a\cdot b|a\in A,b\in B\}$$

$$A\cdot B=\{a\cdot b|a\in A,b\in B\}$$

$$x\sim y\stackrel{\mathrm{def}.}{⟺}x\cdot s=y\cdot t\text{for some }s,t\in T$$

$$\bar{y}+^{\prime}\bar{z}=\left\{\begin{array}[]{lll}\bar{y}+\bar{z}&&\text{if }\bar{y}=\bar{0}\text{or }\bar{z}=\bar{0}\\ \bar{y}+\bar{z}\cup\{\bar{y},\bar{z}\}&&\text{if }\bar{y}\neq\bar{0},\bar{z}\neq\bar{0},\bar{y}\neq-\bar{z}\\ k/_{m}k^{\ast 2}&&\text{if }\bar{y}\neq\bar{0},\bar{z}\neq\bar{0},\bar{y}=-\bar{z}\end{array}\right.$$

$$\begin{array}[]{|c|c|c|c|}\hline\cr+&\bar{0}&\bar{1}&\overline{-1}\\ \hline\cr\bar{0}&\bar{0}&\bar{1}&\overline{-1}\\ \hline\cr\bar{1}&\bar{1}&\bar{1}&\{\bar{0},\bar{1},\overline{-1}\}\\ \hline\cr\overline{-1}&\overline{-1}&\{\bar{0},\bar{1},\overline{-1}\}&\overline{-1}\\ \hline\cr\end{array}$$

$$\begin{array}[]{llll}\theta=\{\bar{0}\}&I=\{\bar{1}\}&\kappa=\{-1\}&\beta=\{\bar{0},\bar{1},\overline{-1}\}\\ \alpha_{1}=\{\bar{0},\bar{1}\}&\alpha_{2}=\{\bar{0},\overline{-1}\}&\alpha_{3}=\{\bar{1},\overline{-1}\}&\end{array}$$

$$\begin{array}[]{|c|c|c|c|c|c|c|c|}\hline\cr+&\theta&I&\kappa&\alpha_{1}&\alpha_{2}&\alpha_{3}&\beta\\ \hline\cr\theta&\theta&I&\kappa&\alpha_{1}&\alpha_{2}&\alpha_{3}&\beta\\ \hline\cr I&I&I&\beta&I&\beta&\beta&\beta\\ \hline\cr\kappa&\kappa&\beta&\kappa&\beta&\kappa&\beta&\beta\\ \hline\cr\alpha_{1}&\alpha_{1}&I&\beta&\alpha_{1}&\beta&\beta&\beta\\ \hline\cr\alpha_{2}&\alpha_{2}&\beta&\kappa&\beta&\alpha_{2}&\beta&\beta\\ \hline\cr\alpha_{3}&\alpha_{3}&\beta&\beta&\beta&\beta&\beta&\beta\\ \hline\cr\beta&\beta&\beta&\beta&\beta&\beta&\beta&\beta\\ \hline\cr\end{array}\text{\quad}\begin{array}[]{|c|c|c|c|c|c|c|c|}\hline\cr&\theta&I&\kappa&\alpha_{1}&\alpha_{2}&\alpha_{3}&\beta\\ \hline\cr\theta&\theta&\theta&\theta&\theta&\theta&\theta&\theta\\ \hline\cr I&\theta&I&\kappa&\alpha_{1}&\alpha_{2}&\alpha_{3}&\beta\\ \hline\cr\kappa&\theta&\kappa&I&\alpha_{2}&\alpha_{1}&\alpha_{3}&\beta\\ \hline\cr\alpha_{1}&\theta&\alpha_{1}&\alpha_{2}&\alpha_{1}&\alpha_{2}&\beta&\beta\\ \hline\cr\alpha_{2}&\theta&\alpha_{2}&\alpha_{1}&\alpha_{2}&\alpha_{1}&\beta&\beta\\ \hline\cr\alpha_{3}&\theta&\alpha_{3}&\alpha_{3}&\beta&\beta&\alpha_{3}&\beta\\ \hline\cr\beta&\theta&\beta&\beta&\beta&\beta&\beta&\beta\\ \hline\cr\end{array}$$

$$⟨a_1,\dots ,a_n,b_1,\dots ,b_m⟩$$

$$⟨a_1{b}_1,\dots ,a_1{b}_m,a_2{b}_1,\dots ,a_2{b}_m,\dots ,a_n{b}_1,\dots ,a_n{b}_m⟩$$

$$\varphi \oplus m\times ⟨1,-1⟩\equiv \psi \oplus n\times ⟨1,-1⟩$$

$$\bar{\varphi }+\bar{\psi }=\overline{\varphi \oplus \psi }\bar{\varphi }\cdot \bar{\psi }=\overline{\varphi \otimes \psi }$$

$$\begin{array}[]{lll}\omega:W(k)&\longrightarrow&W(\mathcal{P}^{\ast}(Q(k)))\\ \overline{\langle a_{1},\ldots,a_{n}\rangle}&\mapsto&\overline{\langle\{a_{1}\},\ldots,\{a_{n}\}\rangle}\end{array}$$

$$(a_1,\dots ,a_n){\cong }_n(b_1,\dots ,b_n)$$

Full text

1. Introduction

In this work we approach the axiomatic theory of quadratic forms by generalising the underlying principles of hyperrings [16] to certain partial orders we call presentable. Roughly speaking, presentable posets generalise the behaviour of pierced powersets, that is powersets excluding the empty set with order given by inclusion. The most salient order-theoretic feature of pierced powersets is that they exhibit a generating set of minimal elements, since a non-empty set is a union of singletons. It is precisely this feature which is captured in the definition of presentable posets. The objective here is to build an axiomatic theory of quadratic forms by describing the behaviour of their value sets. Although we do not address this question explicitly in the present paper, the main motivation underlying this construction is concerned with the concrete problem of assigning a $K$-theory spectrum to the category of hypermodules over a hyperring and, in turn, introducing $K$-theoretic methods to the axiomatic theory of quadratic forms.

In Section 2 we formally introduce presentable posets and provide some  examples including the set of integers greater or equal $1$ augmented with a point at infinity and ordered by division, as well as the set of proper ideals of a Noetherian ring reversely ordered by inclusion.

In Sections 3 and 4 we introduce presentable monoids, groups, rings and fields. In particular, we exhibit presentable groups, rings and fields arising in a natural way from hypergroups, hyperrings and hyperfields, respectively. This provides the main link between our theory and already existing axiomatic theories of quadratic forms. A word of caution might be in order here as far as the terminology is concerned: a presentable group is not a group, not even a cancellative monoid. We just choose to stick to established terminology. A similar comment can be made about the notion of hypergroup underlying the notion of hyperring [16]. On the other hand, the Witt rings we construct are rings without further ado.

In Section 5 we define pre-quadratically and quadratically presentable fields which share certain similarities with groups of square classes of fields, endowed with partial order and addition. We then exhibit a Witt ring structure naturally occuring in quadratically presentable fields. As an application, for every field one can form a hyperfield by defining on the multiplicative group of its square classes multivalued addition that corresponds to value sets of binary forms. The presentable field induced by this hyperfield is quadratically presentable, and its Witt ring in our sense is isomorphic to its standard Witt ring. What makes this construction of interest is the fact that it uniformely works for fields of both characteristic $2$ and $\neq 2$. It is technically reminiscent of the one used by Dickmann and Miraglia to build Witt rings of special groups [6].

In Sections 6 and 7 we explain how a pre-quadratically presentable field can be obtained from any presentable field. For that purpose we introduce in Section 6 quotients of presentable fields, the quotienting being performed with respect to the multiplicative structure. In Section 7 we use these quotients in order to build pre-quadratically presentable fields from presentable fields. The techniques here heavily rely on the connection between presentable algebras and hyperalgebras.

2. Presentable posets

Recall that a partially-ordered set or poset is a set equipped with a reflexive, transitive and anti-symmetric relation. Let $A$ be a poset. We shall write $\bigvee X$ for the supremum of $X\subseteq A$ and $x\vee y$ for $\bigvee\{x,y\}$. We shall denote by $\mathcal{S}_{A}$ the set of $A$’s minimal elements, by $\mathcal{S}_{a}\overset{}{}$ the set of all minimal elements below $a\in A$, and by $\mathcal{S}_{X}\overset{\operatorname{def}.}{=}\bigcup_{x\in X}\mathcal{S}_{x}$ the set of minimal elements below $X\subseteq A$.

Definition 1**.**

A poset $A$ is weakly presentable if

1. i.

every nonempty subset $S\subseteq\mathcal{S}_{A}$ has a supremum; 2. ii.

$x=\bigvee\mathcal{S}_{x}$* for every $x\in A$.*

Remark 1*.*

Let $A$ be a weakly presentable poset.

Any non-empty subset $X\subseteq A$ has a supremum. To be more specific, $\bigvee X=\bigvee\bigcup_{x\in X}\mathcal{S}_{x}$. 2. 2.

$x\leqslant y\Longleftrightarrow\mathcal{S}_{x}\subseteq\mathcal{S}_{y}$ for all $x,y\in A$.

Definition 2**.**

Let $A$ be a poset and $Y\subseteq A$ a non-empty subset. An element $x\in A$ is compact if $x\leqslant\bigvee Y$ entails that there is an element $y\in Y$ such that $x\leqslant y$.

Remark 2*.*

This definition of (order-theoretic) compactness is more general than the standard one, which only takes suprema of directed subsets into account.

Definition 3**.**

A poset $X$ equipped with a distinguished element $\wp_{X}$ is called pointed. $\wp_{X}$ is called basepoint.

Definition 4**.**

A pointed poset $A$ is presentable if

1. i.

$A$* is weakly presentable;* 2. ii.

$\wp_{A}$* is minimal;* 3. iii.

every $a\in\mathcal{S}_{A}$ is compact.

We shall call the minimal elements of a presentable poset supercompacts.

Proposition 1**.**

Let $A$ be a weakly presentable poset. The following are equivalent.

1. i.

Every minimal element is compact; 2. ii.

given $x\in A$, if $x=\bigvee S$ for some $S\subseteq\mathcal{S}_{A}$ then $S=\mathcal{S}_{x}$.

Proof.

(i.$\Rightarrow$ii.): Let $x\in A$ with $x=\bigvee S$, for some $S\subseteq\mathcal{S}_{A}$. Clearly $S\subseteq\mathcal{S}_{x}$. Suppose that, for some $a\in\mathcal{S}_{x}$, $a\notin S$. But $a\leq x=\bigvee S$, so, by compactness of $a$, $a\leq b$, for some $b\in S$. As $b$ is a minimal element, this yields $a=b$, which leads to a contradiction.

(ii.$\Rightarrow$i.): Let $a\in\mathcal{S}_{A}$ and assume that $a\leq\bigvee Y$, for some $Y\subseteq A$. Thus $a\leq\bigvee\bigcup_{y\in Y}\mathcal{S}_{y}$ and hence $a\in\bigcup_{y\in Y}\mathcal{S}_{y}$, say $a\leq y$ for some $y\in Y$. ∎

Example 1* (Walking supremum).*

Perhaps the smallest example of a non-discrete presentable poset is the poset $\mathcal{W}(x,p)$ given by

with $p=\wp_{\mathcal{W}(x,p)}$. We call a presentable poset of this type walking supremum.

Definition 5**.**

Let $X$ be a set. The set $\mathcal{P}^{\ast}(X)\overset{\operatorname{def}.}{=}2^{X}\backslash\{\varnothing\}$ is called pierced powerset of $X$.

Example 2*.*

A pierced powerset $(\mathcal{P}^{\ast}(X)$ together with a distinguished point is a presentable poset when ordered by inclusion, with supercompacts the singletons.

Example 3*.*

The set $\mathcal{Z}_{\geq 2}^{\infty}$ of greater and equal than $2$ and square-free integers with a point at infinity added is a presentable poset with respect to the ordering by division. Prime numbers are the supercompacts. Clearly $\mathcal{Z}_{\geq 2}^{\infty}$ here can be replaced with conjugation classes of square-free non-units of any unique factorization domain with classes of nonzero irreducibles as supercompacts.

Example 4*.*

The set $(\mathcal{I}^{\ast}(R),\supseteq)$ of all ideals of an absolutely flat (i.e. von Neumann regular) Noetherian ring $R$ is a presentable poset with respect to the ordering by reverse inclusion. The primary ideals in absolutely flat rings are maximal, and so is the trivial ideal $R$, hence they constitute the supercompacts. Indeed, every element of $\mathcal{I}^{\ast}(R)$ is either contained in a maximal ideal (which, in particular, is primary), or is equal to $R$. Every proper ideal is an intersetion of some primary ideals due to the Noether-Lasker theorem, and, clearly, an intersection of any family of proper ideals is a proper ideal. Clearly, this example can be also phrased in the language of affine algebraic sets that also satisfy some extra conditions.

3. Presentable groups

Definition 6**.**

A presentable monoid $(M,\leq,0,+)$ is a presentable poset $(M,\leq,0)$ with a distinguished supercompact [math] and a suprema-preserving binary addition $+:M\times M\rightarrow M$ such that

1. i.

$a+(b+c)=(a+b)+c$* for all $a,b,c\in M$;* 2. ii.

$a+0=0+a=a$* for all $a\in M$;* 3. iii.

$a+b=b+a$* for all $a,b\in M$.*

Remark 3*.*

The addition is in particular monotone, that is

[TABLE]

for all $a,b,c,d\in M$. 2. 2.

Suppose $a\leq b+c$. Let $s\in\mathcal{S}_{a}$. We have

[TABLE]

Hence by Proposition 1 there are $t\in\mathcal{S}_{b}$ and $u\in\mathcal{S}_{c}$ such that

[TABLE]

Example 5*.*

Let $(M,0,+)$ be a commutative monoid. The pointed pierced powerset

[TABLE]

(c.f. Example 2) is a presentable monoid with addition given by

[TABLE]

The singleton $\{0\}$ is the neutral element. The addition preserves suprema:

[TABLE]

We shall abuse the notation by using the same symbol $+$ for addition $+$ in $M$ and $\boldsymbol{+}$in $\mathcal{P}^{\ast}(M)$

Definition 7**.**

A hypermonoid is a pointed set $(M,0,+)$ equipped with a multivalued addition

[TABLE]

such that

1. i.

$a+0=a=0+a$* for all $a\in M$;* 2. ii.

$a+b=b+a$* for all $a,b\in M$;* 3. iii.

$(a+b)+c=\bigcup\{a+x|x\in b+c\}=a+(b+c)$* for all $a,b,c\in M$.*

Remark 4*.*

Let $(M,0,+)$ be a hypermonoid. The pointed pierced powerset ($\mathcal{P}^{\ast}(M),\subseteq,\{0\}$) is a presentable monoid with addition given by

[TABLE]

Again, in what follows we shall use the same symbol $+$ for addition $+$ in $M$ and $\boldsymbol{+}$ in $\mathcal{P}^{\ast}(M)$.

Definition 8**.**

A presentable group $G$ is a presentable monoid equipped with a suprema preserving involution $-:G\rightarrow G$ called inversion, verifying

[TABLE]

for all $s,t,u\in\mathcal{S}_{G}$.

Remark 5*.*

Assume a presentable group $(G,\leqslant,0,+,-)$.

Notice that the inversion is, in particular, monotone, so we have quite counterintuitively

[TABLE]

for all $a,b\in G$. 2. 2.

We have $0\leqslant s+(-s)$, for all $s\in\mathcal{S}_{G}$, since $s\leqslant 0+s$ implies that $0\leqslant s+(-s)$. This entails that, in fact

[TABLE]

for any $a\in G$. Since $\mathcal{S}_{a}\neq\varnothing$ there is a supercompact $s\in\mathcal{S}_{a}$ such that $s\leqslant a$, hence

[TABLE] 3. 3.

It is in general not true that $a\leq b+c$ implies $b\leq a-c$ for arbitrary $a,b,c\in G$. Take the presentable group $(\mathcal{P}^{\ast}(\mathbbm{Z}),\subseteq,\{0\},+)$, where $\mathbbm{Z}$ is endowed with the usual addition. Then

[TABLE]

but

[TABLE]

Example 6*.*

Let $(G,0,+)$ be an abelian group and denote by $-a$ the opposite element of $a$ with respect to $+$. The presentable monoid $\mathcal{P}^{\ast}(G)$ as defined in Example 5 is a presentable group with inversion given by

[TABLE]

Definition 9**.**

A hypergroup $G$ is a hypermonoid together with a map $-:G\rightarrow G$ such that

1. i.

$0\in a+(-a)$* for all $a\in G$;* 2. ii.

$(a\in b+c)\Rightarrow(c\in a+(-b))$* for all $a,b,c\in G$.*

Example 7*.*

Let $(G,0,+,-)$ be a hypergroup. The presentable monoid $(\mathcal{P}^{\ast}(G),\subseteq,\{0\},+)$ is a presentable group with inversion given by

[TABLE]

4. Presentable rings and fields

Definition 10**.**

A presentable ring $R$ is a presentable group $(R,\leq,0,+,-)$ as well as a commutative monoid $(R,\cdot,1)$, such that $\cdot$ is compatible with $\leq$ and $-$, distributative with respect to $+$, and verifies

[TABLE]

for all $a,b\in R$. A presentable ring $R$ such that $\mathcal{S}_{R}^{\ast}=\mathcal{S}_{R}\setminus\{0\}$ is a multiplicative group will be called a presentable field.

Example 8*.*

Let $(R,0,+,\cdot,1)$ be a ring. The presentable group $\mathcal{P}^{\ast}(R)$ (c.f. Example 6) is  a presentable ring with identity $\{1\}$ and with multiplication given by

[TABLE]

If $R$ is a field, then $\mathcal{P}^{\ast}(R)$ becomes a presentable field.

Remark 6*.*

Assume a presentable ring $(R,\leq,0,+,-,\cdot,1)$.

The element $1\in R$ is uniquely defined. 2. 2.

$1\neq 0$. 3. 3.

$-1\in\mathcal{S}_{R}$.

Items $1$ and $2$ are immediate. For item $3$ fix $s\in S_{-1}$. Then $s\leq-1$ and, consequently, $-s\leq 1$. But $1$ is a supercompact, so $-s=1$, hence $-1=s\in\mathcal{S}_{R}$.

Definition 11**.**

A hyperring $(R,0,+,-,1,\cdot)$ is a hypergroup $(R,0,+,-)$ such that $(R,1,\cdot)$ is a commutative monoid and

1. i.

$0\cdot a=0$* for all $a\in R$;* 2. ii.

$a\cdot(b+c)=(a\cdot b)+(a\cdot c)$* for all $a,b,c\in R$;* 3. iii.

$0\neq 1$.

If, in addition, every non-zero element has a multiplicative inverse, then $R$ is called a hyperfield.

Example 9*.*

Let $(R,0,+,-,1,\cdot)$ be a hyperring (or a hyperfield). The presentable group $\mathcal{P}^{\ast}(R)$ is a presentable ring (or a presentable field, respectively) with multiplication given by

[TABLE]

for $A,B\in\mathcal{P}^{\ast}(R)$. The identity is $\{1\}$.

Remark 7*.*

Let $(F,0,+,-,1,\cdot)$ be a hyperfield and $T$ be a subgroup of the multiplicative group $(F,1,\cdot)$. The relation $\sim$  on $F$ given by

[TABLE]

is an equivalence. Let $F/_{m}T$ be its set of equivalence classes and $\bar{x}$ the class of $x\in F$. The induced operations

• $-$

$\bar{x}\in\bar{y}\bar{+}\bar{z}\overset{\operatorname{def}.}{\Longleftrightarrow}x\cdot s\in(y\cdot t)+(z\cdot u)\quad\text{for some }s,t,u\in T$;

• $-$

$\bar{x}\bar{\cdot}\bar{y}=\overline{\operatorname{xy}}$;

• $-$

$\bar{-}\bar{x}=\overline{-x}.$

are well-defined and $(F/_{m}T,\bar{0},\bar{+},\bar{-},\bar{1},\bar{\cdot})$ is  a hyperfield that we shall call the quotient hyperfield of $F$ modulo $T$ [16]. Whenever clear from the context, we shall use the same symbols for $+$, $-$ and $\cdot$ both in $F$ and $F/_{m}T$.

Example 10*.*

Let $k$ be a field with $\operatorname{char}(k)\neq 2$ and $k\neq\mathbbm{F}_{3},\mathbbm{F}_{5}$. This yields an example of a hyperfield with $a+b=\{a+b\}$. Let $T=k^{\ast 2}$ be $k$’s multiplicative group of squares and $x,y,z\in k$. It is easy to see that the following are equivalent

1. i.

$x=s^{2}y+t^{2}z\text{ for some }s,t\in k$; 2. ii.

$\bar{x}\in\bar{y}+\bar{z}\text{ in }k/_{m}k^{\ast 2}$.

We thus have $\bar{y}+\bar{z}=D(y,z)\cup\{\bar{0}\},$ where $D(y,z)$ is the value set of the binary quadratic form $(y,z)$.

Assume now $\operatorname{char}(k)=2$ or $k\in\{\mathbbm{F}_{3},\mathbbm{F}_{5}\}$. In this case the above assertions are in general not equivalent anymore. However, $k/_{m}k^{\ast 2}$ with a modified addition given by

[TABLE]

is again a hyperfield. Observe that this addition is well-defined for any hyperfield [9, Proposition 2.1], we just have $+^{\prime}=+$ whenever $\operatorname{char}(k)\neq 2$ and $k\neq\mathbbm{F}_{3},\mathbbm{F}_{5}$.

Definition 12**.**

Let $k$ be a field. $Q(k)\overset{\operatorname{def}.}{=}(k/_{m}k^{\ast 2},+^{\prime})$is called $k$’s quadratic hyperfield.

Example 11*.*

Let $k$ be a field with two square classes, for instance when $k$ is formally real. The two square classes are represented by $1,-1$, so that $k$ is Euclidean (for example, $k=\mathbb{R}$, or the field of real algebraic numbers, or the field of real constructible numbers etc), and $Q(k)=\{\bar{0},\bar{1},\overline{-1}\}$ with multivalued addition given by

[TABLE]

along with the obvious multiplication. The presentable ring $\mathcal{P}^{\ast}(Q(k))$ with identity $I$ consists of 7 elements

[TABLE]

with arithmetic given by

[TABLE]

The partial order in $\mathcal{P}^{\ast}(Q(k))$ is generated by

5. Witt rings of quadratically presentable fields

Definition 13**.**

Let $(R,\leq,0,+,-,\cdot,1)$ be a presentable field. We shall call $R$ pre-quadratically presentable, if the following conditions hold

1. i.

$a\leq a+b$* for all $a\in\mathcal{S}_{R}^{\ast},b\in\mathcal{S}_{R}$;* 2. ii.

$(a\leq 1-b)\wedge(a\leq 1-c)\Rightarrow(a\leq 1-bc)$* for all $a,b,c\in\mathcal{S}_{R}$;* 3. iii.

$a^{2}=1$* for all $a\in\mathcal{S}_{R}\setminus\{0\}$.*

Remark 8*.*

Note that in the axiom $i$. the assumption that $a\in\mathcal{S}_{R}^{\ast}$ is cruicial: if $a=0$ then $a\leq a+b$ is just $0\leq 0+b=b$, which means $b=0$ for all $b\in\mathcal{S}_{R}$.

Example 12*.*

Let $k$ be a field, $Q(k)$ be its quadratic hyperfield and $\mathcal{P}^{\ast}(Q(k))$ be the induced presentable field (c.f. Examples 10 and 11). It is easy to see that this presentable field is pre-quadratically presentable.

Example 13*.*

The presentable field $\mathcal{P}^{\ast}(R)$ constructed from a field $(R,0,+,\cdot,1)$ (c.f. Example 8) is usually not pre-quadratically presentable, since it is, in general, not true that $\{a\}\subset\{a\}+\{b\}=\{a+b\}$.

Definition 14**.**

A form on a pre-quadratically presentable field $R$ is an $n$-tuple $\langle a_{1},\ldots,a_{n}\rangle$ of elements of $\mathcal{S}_{R}^{\ast}$. The relation $\cong$ of isometry of forms of the same dimension is given by induction:

• $-$

$\langle a\rangle\cong\langle b\rangle$*  $\overset{def.}{{\Longleftrightarrow}}\ a=b$;*

• $-$

$\langle a_{1},a_{2}\rangle\cong\langle b_{1},b_{2}\rangle$*  $\overset{def.}{{\Longleftrightarrow}}\ a_{1}a_{2}=b_{1}b_{2}$ and $b_{1}\leq a_{1}+a_{2}$;*

• $-$

$\langle a_{1},\ldots,a_{n}\rangle\cong\langle b_{1},\ldots,b_{n}\rangle$*  $\overset{def.}{{\Longleftrightarrow}}$  there exist $x,y,c_{3},\ldots,c_{n}\in\mathcal{S}_{R}^{\ast}$ such that*

1. i.

$\langle a_{1},x\rangle\cong\langle b_{1},y\rangle$; 2. ii.

$\langle a_{2},\ldots,a_{n}\rangle\cong\langle x,c_{3},\ldots,c_{n}\rangle$; 3. iii.

$\langle b_{2},\ldots,b_{n}\rangle\cong\langle y,c_{3}\ldots,c_{n}\rangle$.

Proposition 2**.**

The relation $\cong$ is an equivalence on the sets of all unary and binary forms of a pre-quadratically presentable field $R$.

Proof.

The statement is clear for unary forms. For binary forms, reflexivity follows from the axiom $(i.)$. For symmetry assume that $\langle a,b\rangle\cong\langle c,d\rangle$, for $a,b,c,d\in\mathcal{S}_{R}^{\ast}$. Thus $ab=cd$ and $a\leq c+d$. But then $a=bcd$, so that $bcd\leq c+d$. Thus $b\leq cd(c+d)=c+d$. For transitivity assume $\langle a,b\rangle\cong\langle c,d\rangle$ and $\langle c,d\rangle\cong\langle e,f\rangle$, for $a,b,c,d,e,f\in\mathcal{S}_{R}^{\ast}$. This means $ab=cd$, $cd=ef$, $a\leq c+d$ and, by symmetry, $e\leq c+d$. Therefore, $c\leq a-d$ and $c\leq e-d$, which gives $-cd\leq 1-ad$ and $-cd\leq 1-ed$. By (2) this implies $-cd\leq 1-ae$. Since $cd=ef$, this is just $-ef\leq 1-ae$, or, equivalently, $ef\leq ae-1$. But this is the same as $ae\leq 1+ef$, so $a\leq e+f$. ∎

Definition 15**.**

A pre-quadratically presentable field $(R,\leq,0,+,-,\cdot,1)$ will be called quadratically presentable if the isometry relation is an equivalence on the set of all forms of the same dimension.

Example 14*.*

The pre-quadratically presentable field $\mathcal{P}^{\ast}(Q(k))$, for a field $k$, is quadratically presentable. That $\cong$ is an equivalence relation on the set of all forms of the same dimension follows from the well-known inductive description of the isometry relation of quadratic forms.

Definition 16**.**

Let $R$ be a pre-quadratically presentable field, let $\phi=\langle a_{1},\ldots,a_{n}\rangle$, $\psi=\langle b_{1},\ldots,b_{m}\rangle$ be two forms. The orthogonal sum $\phi\oplus\psi$ is defined as the form

[TABLE]

and the tensor product $\phi\otimes\psi$ as

[TABLE]

We will write $k\times\phi$ for the form $\underbrace{\phi\oplus\ldots\oplus\phi}_{k\leavevmode\nobreak\ times}$.

Proposition 3**.**

**

Let $R$ be a pre-quadratically presentable field. The direct sum and the tensor product of isometric forms are isometric. 2. 2.

(Witt cancellation) Let $R$ be a quadratically presentable field. If $\phi_{1}\oplus\psi\cong\phi_{2}\oplus\psi$, then $\phi\cong\psi$.

Proof.

Induction on the dimension of forms. ∎

Definition 17**.**

Let $R$ be a quadratically presentable field. Two forms $\phi$ and $\psi$ will be called Witt equivalent, denoted $\phi\sim\psi$, if, for some integers $m,n\geq 0$:

[TABLE]

Remark 9*.*

It is easily verified that $\sim$ is an equivalence relation on forms over $R$, compatible with (and, clearly, coarser than) the isometry. One also easily checks that Witt equivalence is a congruence with respect to orthogonal sum and tensor product of forms. Denote by $W(R)$ the set of equivalence classes of forms over $R$ under Witt equivalence, and by $\bar{\phi}$ the equivalence class of $\phi$. With the operations

[TABLE]

$W(R)$ is a commutative ring, having as zero the class $\overline{\langle 1,-1\rangle}$, and $\overline{\langle 1\rangle}$ as multiplicative identity.

Definition 18**.**

Let $R$ be a quadratically presentable field. $W(R)$ with binary operations as defined above is called the Witt ring of $R$.

As one might expect, the main example of a Witt ring of a quadratically presentable field, is the Witt ring of the quadratically presentable field induced by the quadratic hyperfield of a field.

Theorem 1**.**

For a field $k$, $W(\mathcal{P}^{\ast}(Q(k)))$ is just the usual Witt ring $W(k)$ of non-degenerate symmetric bilinear forms of $k$.

Proof.

The map

[TABLE]

is well-defined and an isomorphism of rings. ∎

Remark 10*.*

Notice that Theorem 1 provides a uniform construction of the Witt ring for all charateristics as well as for $\mathbb{F}_{3}$ and $\mathbb{F}_{5}$.

Definition 19** (Dickmann-Miraglia).**

A pre-special group [6, Definition 1.2] is a group $G$ of exponent $2$ together with a distinguished element $-1$ and a binary operation $\cong$ on $G\times G$ such that, for all $a,b,c,d\in G$:

1. i.

$\cong$* is an equivalence relation;* 2. ii.

$(a,b)\cong(b,a)$; 3. iii.

$(a,-a)\cong(1,-1)$; 4. iv.

$[(a,b)\cong(c,d)]\Rightarrow[ab=cd]$; 5. v.

$[(a,b)\cong(c,d)]\Rightarrow[(a,-c)\cong(-b,d)]$; 6. vi.

$[(a,b)\cong(c,d)]\Rightarrow\forall x\in G[(xa,xb)\cong(xc,xd)]$.

Remark 11*.*

Let $(G,\cong,-1)$ be a pre-special group. The relation $\cong$ can be extended to the set $\underbrace{G\times\ldots\times G}_{n}$ as follows:

[TABLE]

provided that there exist $x,y,c_{3},\ldots,c_{n}\in G$ such that

1. i.

$(a_{1},x)\cong(b_{1},y)$; 2. ii.

$(a_{2},\ldots,a_{n})\cong_{n-1}(x,c_{3},\ldots,c_{n})$; 3. iii.

$(b_{2},\ldots,b_{n})\cong_{n-1}(y,c_{3}\ldots,c_{n})$.

A special group [6, Definition 1.2] is a pre-special group $(G,\cong,-1)$ such that $\cong_{n}$ is an equivalence relation for all $n\in\mathbb{N}$.

Remark 12*.*

Let $(R,\leq,0,+,-,\cdot,1)$ be a quadratically presentable field. Then $(\mathcal{S}_{R}^{\ast},\cong,-1)$ is a special group. The only non-trivial parts to check are that $\langle a,-a\rangle\cong\langle 1,-1\rangle$ and that $\langle a,b\rangle\cong\langle c,d\rangle$ implies $\langle a,-c\rangle\cong\langle-b,d\rangle$, for $a,b,c,d\in\mathcal{S}_{R}^{\ast}$. The first statement follows from the fact that $a\leq a+1$ implies $1\leq a-a$. For the second assume $ab=cd$ and $a\leq c+d$. Thus $d\leq a-c$ by the exchange law, so that $\langle d,-b\rangle\cong\langle a,-c\rangle$.

Now, given suitable notions of morphisms, it can be shown that the assignment

[TABLE]

is functorial and that $S$ is, in fact, an equivalence of categories. Hence the construction of Witt rings we provide (c.f. Theorem 1 and Remark 10) carries over to special groups. Conversely, the relevant constructions could be carried out directly in the category of special groups. However, the formalism of presentable algebras is of independent interest, this since the category of presentable groups as well as the category of presentable modules over a presentable ring (not formally introduced here) exhibit quite good properties. This circle of ideas will be addressed in a forthcoming paper.

6. Quotients in presentable fields

In order to investigate Witt rings of presentable fields, one needs to know how to pass from presentable fields to quadratically presentable fields. We are “almost” able to do that, and will show how one can build a pre-quadratically presentable field from arbitrary presentable field – it is, however, an open question when the resulting presentable field is quadratically presentable. The main tool to be used are quotients of presentable fields. Before we proceed to general quotients, we focus on a rather special case of quotients “modulo” multiplicative subsets of supercompacts. These are, in fact, the only quotients that we need in the sequel, which explains why we choose to present our exposition in this particular manner.

Theorem 2**.**

Let $(R,\leq,0,+,-,\cdot,1)$ be a presentable field. Let $T\subseteq\mathcal{S}_{R}^{\ast}$ be a multiplicative set i.e. for all $s,t\in T$, $st\in T$. Define the relation $\sim$ on $\mathcal{S}_{R}$ by

[TABLE]

This is an equivalence relation, whose equivalence classes will be denoted by $\bar{a}$, $a\in\mathcal{S}_{R}$. Let

[TABLE]

and let

[TABLE]

Then $(\mathcal{S}_{R}/\sim,\bar{0},\overline{+},\overline{-},\bar{1},\overline{\cdot})$ is a hyperfield.

Proof.

The relation $\sim$ is clearly reflexive and symmetric, and for transitivity assume $as=bt$ and $bu=cv$, for some $a,b,c\in\mathcal{S}_{R}$, $s,t,u,v\in T$. Then $asu=btu$ and $btu=cvt$ with $su,tu,vt\in T$ thanks to the multiplicativity of $T$.

Next, the operation $\overline{\cdot}$ is clearly well-defined, and to see that so is $\overline{+}$, assume $\bar{b}=\overline{b^{\prime}}$ and $\bar{c}=\overline{c^{\prime}}$, say, $vb=v^{\prime}b^{\prime}$ and $wc=w^{\prime}c^{\prime}$, for some $v,v^{\prime},w,w^{\prime}\in T$. Then

[TABLE]

In order to show that $\mathcal{S}_{R}/\sim$ with operations defined as above is, indeed, a hyperring, we note that both the commutativity of $\overline{+}$ and the fact that $(\mathcal{S}_{R}/\sim,\bar{1},\overline{\cdot})$ forms a commutative group are obvious, that $\bar{0}\in\bar{a}\overline{-}\bar{a}$, for all $\bar{a}\in\mathcal{S}_{R}/\sim$, follows immediately from $0\leq a-a$ for all $a\in\mathcal{S}_{R}$, that $\bar{0}\overline{\cdot}\bar{a}=\bar{0}$ is clear in view of $0\cdot 1=0$, and that $\bar{0}\neq\bar{1}$ is apparent, as $1\cdot t=0$, for some $t\in T$, leads to $0=1$. It remains to show the neutrality of $\bar{0}$, associativity of $\overline{+}$, cancellation and distributativity of $\overline{\cdot}$ and $\overline{+}$.

Assume $\bar{b}\in\bar{a}\overline{+}\bar{0}$, so $bs\leq at+0=at$, for some $s,t\in T$. But then $bst^{-1}\leq a$, and, since $a$ is a supercompact, this yields $bst^{-1}=a$ and, consequently, $\bar{b}=\bar{a}$.

Assume $\bar{d}\in\bar{a}\overline{+}(\bar{b}\overline{+}\bar{c})$, so that $\bar{d}\in\bar{a}\overline{+}\bar{e}$ with $\bar{e}\in\bar{b}\overline{+}\bar{c}$. Hence $ds\leq at+eu$ and $es^{\prime}\leq bt^{\prime}+cu^{\prime}$, for some $s,t,u,s^{\prime},t^{\prime},u^{\prime}\in T$. Thus $dss^{\prime}\leq ats^{\prime}+eus^{\prime}$ and $eus^{\prime}\leq but^{\prime}+cuu^{\prime}$, so that $dss^{\prime}\leq ats^{\prime}+(but^{\prime}+cuu^{\prime})=(ats^{\prime}+but^{\prime})+cuu^{\prime}$. It follows that there exist supercompacts $d^{\prime},f,c^{\prime}\in\mathcal{S}_{R}^{\ast}$ with $d^{\prime}\leq dss^{\prime}$, $f\leq ats^{\prime}+but^{\prime}$ and $c^{\prime}\leq cuu^{\prime}$ with $d^{\prime}\leq f+c^{\prime}$. Using the same argument as in the proof of neutrality of $\bar{0}$, we easily check that $d^{\prime}=d$ and $c^{\prime}=c$. Therefore $d\leq f+c$ and $f\leq ats^{\prime}+but^{\prime}$. This yields $\bar{d}\in\bar{f}\overline{+}\bar{c}$ with $\bar{f}\in\bar{a}\overline{+}\bar{b}$, so that $\bar{d}\in(\bar{a}\overline{+}\bar{b})\overline{+}\bar{c}$.

Assume $\bar{a}\in\bar{b}\overline{+}\bar{c}$, so that $at\leq bs+cu$, for some $s,t,u\in T$. Then there are supercompacts $a^{\prime}\leq at$, $b^{\prime}\leq bs$ and $c^{\prime}\leq cu$ such that $a^{\prime}\leq b^{\prime}+c^{\prime}$. Using the same trick as before we conclude $a=a^{\prime}$, $b=b^{\prime}$, $c=c^{\prime}$, so that, in fact, $a\leq b+c$, and thus $b\leq a-c$, which implies $\bar{b}\in\bar{a}\overline{+}\bar{c}$.

Finally, if $\bar{d}\in\bar{a}\overline{\cdot}(\bar{b}\overline{+}\bar{c})$, then $\bar{d}=\overline{ae}$ with $\bar{e}\in\bar{b}\overline{+}\bar{c}$, and thus $es\leq bt+cu$, for some $s,t,u\in T$. But then $aes\leq abt+acu$, so $\overline{ae}\in\overline{ab}\overline{+}\overline{ac}$. ∎

Remark 13*.*

Observe that the above works, in fact, for any presentable ring $(R,\leq,0,+,-,\cdot,1)$ and a subgroup $T\subseteq\mathcal{S}_{R}^{\ast}$ of the multiplicative monoid $\mathcal{S}_{R}^{\ast}$. That is, we only need to be able to invert the elements of $T$ for the argument to go through.

Definition 20**.**

The quotient of $(R,\leq,0,+,-,\cdot,1)$ modulo the multiplicative set $T$ is the presentable field $(\mathcal{P}^{\ast}(\mathcal{S}_{R}/_{\sim}),\subseteq,\{\bar{0}\})$ with the hyperfield $(\mathcal{S}_{R}/_{\sim},\bar{0},\overline{+},\overline{-},\bar{1},\overline{\cdot})$ defined in Theorem 2 and will be denoted by $R/_{m}T$.

Theorem 2, as remarked before, is a special case of the following, more general result:

Theorem 3**.**

Let $(R,\leq,0,+,-,\cdot,1)$ be a presentable field. Let $\sim$ be a nontrivial congruence on the set $\mathcal{S}_{R}^{\ast}$ of supercompacts of $R$, i.e. an equivalence relation such that $0\nsim 1$, and, for all $a,a^{\prime},b,b^{\prime}\in\mathcal{S}_{R}^{\ast}$, if $a\sim a^{\prime}$, and $b\sim b^{\prime}$ then

[TABLE]

Denote by $\bar{a}$ the equivalence class of $a\in\mathcal{S}_{R}$. Let

[TABLE]

and let

[TABLE]

Then $(\mathcal{S}_{R}/\sim,\bar{0},\overline{+},\overline{-},\bar{1},\overline{\cdot})$ is a hyperfield.

The proof mimics the one of Theorem 2. That $\bar{0}\neq\bar{1}$ follows from the fact that $0\nsim 1$.

7. From presentable fields to pre-quadratically presentable fields

Remark 14*.*

Let $(R,\leq,0,+,-,\cdot,1)$ be a presentable field and define the following multivalued addition on the set $\mathcal{S}_{R}$ of supercompacts of $R$:

[TABLE]

Then $(\mathcal{S}_{R},0,\boldsymbol{+},-,1,\cdot)$ is a hyperfield. Further, define the prime addition on $\mathcal{S}_{R}$ as follows:

[TABLE]

Then $(\mathcal{S}_{R},0,\boldsymbol{+}^{\prime},-,1,\cdot)$ is again a hyperfield [9, Proposition 2.1], called the prime hyperfield of $(R,\leq,0,+,-,\cdot,1)$. The induced presentable field $(\mathcal{P}^{\ast}(\mathcal{S}_{R}),\subseteq,\{0\},+^{\prime},-,\cdot,\{1\})$, that will be called the prime presentable field, satisfies the condition:

[TABLE]

Theorem 4**.**

Let $(R,\leq,0,+,-,\cdot,1)$ be a presentable field such that

[TABLE]

$T\overset{\operatorname{def}.}{=}\{s\in\mathcal{S}_{R}^{\ast}\mid s\leq a^{2}\text{for some }a\in R\}$* is a multiplicative set and the quotient $R/_{m}T$ of $R$ modulo $T$ is a pre-quadratically presentable field.*

Proof.

$T$ is multiplicative, for if $s\leq a^{2}$ and $t\leq b^{2}$, for some $s,t\in\mathcal{S}_{R}^{\ast}$, $a,b\in R$, then $st\leq a^{2}b^{2}=(ab)^{2}$ and $st\neq 0$, since $\mathcal{S}^{\ast}_{R}$ is a group. The condition

[TABLE]

carries over to $R/_{m}T$, non-zero supercompacts of $R/_{m}T$ form a group, since in the process of taking a quotient modulo multiplicative set we end up with a presentable field, and, finally, squares of all non-zero supercompacts of $R/_{m}T$ are equal to identity, as they are just classes of squares of non-zero supercompacts in $R$, which are, by definition, equivalent to $1$.

It remains to show that for all supercompacts $\{\bar{a}\}$, $\{\bar{b}\}$ and $\{\bar{c}\}$ in $R/_{m}T$, if $\{\bar{a}\}\subseteq\{\bar{1}\}-\{\bar{b}\}$ and $\{\bar{a}\}\subseteq\{\bar{1}\}-\{\bar{c}\}$, then $\{\bar{a}\}\subseteq\{\bar{1}\}-\{\overline{bc}\}$. Fix three supercompacts as above and assume the antedecent. This is equivalent to $\bar{a}\in\bar{1}\overline{-}\bar{b}$ and $\bar{a}\in\bar{1}\overline{-}\bar{c}$ in the hyperfield $\mathcal{S}_{R}/\sim$, which, in turn, is equivalent to

[TABLE]

for some non-zero supercompacts $s,s^{\prime},t,t^{\prime},u,u^{\prime}\in R$ such that $s\leq x^{2}$, $s^{\prime}\leq x^{\prime 2}$, $t\leq y^{2}$, $t^{\prime}\leq y^{\prime 2}$, $u\leq z^{2}$, $u^{\prime}\leq z^{\prime 2}$, for some $x,x^{\prime},y,y^{\prime},z,z^{\prime}\in R$. Since $\mathcal{S}^{\ast}_{R}$ is a group, the elements $sa,s^{\prime}a,ub,u^{\prime}c$ are also supercompacts, which allows switching terms between both sides of the above inequalities, and gives

[TABLE]

and, in turn

[TABLE]

Hence

[TABLE]

Let $v$ and $w$ be supercompacts with $uu^{\prime}bc\leq v-w$ and $v\leq(yy^{\prime}+axx^{\prime})^{2}$ and $w\leq a(x^{\prime}y+xy^{\prime})^{2}$. If both $v$ and $w$ are equal to zero, then one of $b$ or $c$ is zero, so $\bar{a}\in\bar{1}\overline{-}\overline{bc}$ is just $\bar{a}\in\bar{1}\overline{-}\bar{b}$ or $\bar{a}\in\bar{1}\overline{-}\bar{c}$. If $v=0$ and $w\neq 0$, then $w=w^{\prime}w^{\prime\prime}$ for $w^{\prime},w^{\prime\prime}\in\mathcal{S}_{R}^{\ast}$ with $w^{\prime}\leq a$ and $w^{\prime\prime}\leq(x^{\prime}y+xy^{\prime})^{2}$. Thus $w^{\prime}=a$, since $a$ is a supercompact itself, and hence a minimal element, and $w^{\prime\prime}\in T$, so that $uu^{\prime}bc\leq-aw^{\prime\prime}$, $w^{\prime\prime}\in T$. But $-aw^{\prime\prime}$ is again a supercompact, as $\mathcal{S}_{R}^{\ast}$ is a group, so $uu^{\prime}bc=-aw^{\prime\prime}$. But $-aw^{\prime\prime}\leq-aw^{\prime\prime}+1$, so $uu^{\prime}bc=-aw^{\prime\prime}\leq 1-aw^{\prime\prime}$, yielding $\bar{a}\in\bar{1}\overline{-}\overline{bc}$. Similarly, if $v\neq 0$ and $w=0$, then $\overline{bc}=\bar{1}\in\bar{1}-\bar{a}$.

This leaves us with the case $v\neq 0$ and $w\neq 0$. Then $v\in T$ and $w=w^{\prime}w^{\prime\prime}$, for some $w^{\prime},w^{\prime\prime}\in\mathcal{S}_{R}^{\ast}$ with $w^{\prime}\leq a$ and $w^{\prime\prime}\leq(x^{\prime}y+xy^{\prime})^{2}$. But then $w^{\prime}=a$, and $w^{\prime\prime}\in T$. So, at the end we obtain

[TABLE]

with $uu^{\prime},v,w^{\prime\prime}\in T$, or, equivalently

[TABLE]

which is the same as $\{\bar{a}\}\subseteq\{\bar{1}\}-\{\overline{bc}\}$. ∎

Example 15*.*

Let $k$ be a field, let $(\mathcal{P}^{\ast}(k),\subseteq,\{0\})$ be the induced presentable field. It follows from the construction that applying Remark 14 and Theorem 4 we obtain the pre-quadratically presentable field $\mathcal{P}^{\ast}(\mathcal{S}_{\mathcal{P}^{\ast}(k)})/_{m}T$, where

[TABLE]

$\mathcal{P}^{\ast}(\mathcal{S}_{\mathcal{P}^{\ast}(k)})/_{m}T$ is isomorphic to $\mathcal{P}^{\ast}(Q(k))$, hence quadratically presentable. We have in particular

[TABLE]

Remark 15*.*

It is an open question when the resulting pre-quadratically presentable field is quadatically presentable.