Formally Real Division Rings

Q: Why does Albert's theorem need Wedderburn's factorisation theorem?

Because the contradiction is produced by summing conjugates. (16.9) says that the minimal polynomial of an element algebraic over the centre splits in D[t] into linear factors whose roots are all conjugates of that element; comparing the coefficient of t^n-1 then shows those n conjugates sum to zero. Without a factorisation inside D itself there is nothing to compare.

Q: Does Albert's theorem say a formally real division ring has no proper algebraic extensions?

No — it is a statement about elements of D, not about extensions of D. It says D contains no element algebraic over its centre except the central ones. In particular D contains no square root of -1, no primitive cube root of unity, and no noncentral element satisfying a polynomial over F.

Q: Is the ring A of (18.7) ever formally real?

Yes, whenever the twist is order-preserving. If c is positive in some ordering of k that preserves, (18.5) orders A and it is formally real. The construction becomes interesting exactly when c is chosen negative, which destroys the hypothesis of (18.5) and, by (18.7)(2), formal reality itself.

Q: Why is the level of a field always a power of 2 but not that of a division ring?

Pfister's proof uses the multiplicativity of sums of 2^n squares — a quadratic form identity that holds only in the commutative setting. In a division ring the relevant set is the group of square-products, whose behaviour is governed by commutators rather than by quadratic forms, and no multiplicativity constraint survives.

Q: What is the significance of the element x^2(x^-1y)^2(y^-1)^2?

It is the commutator xyx^-1y^-1 rewritten as a product of three squares, using the identity from (18.1). That rewriting is what converts a purely multiplicative fact — the twist introduces the scalar c as a commutator — into the statement that c is a square-product, which is what (18.2) tests against.

Q: Are there formally real division rings that are noncommutative and finitely generated?

Yes: Hilbert's example is generated over Q by x and y as a division ring. What (18.11) forbids is finite dimension over the centre, not finite generation. Being centrally infinite is compatible with being a very concrete, two-generator object.

Executive Summary

Two questions are settled here. First, is the passage from squares to square-products in the Szele–Pickert criterion a real phenomenon or a technicality? $(18.7)$ answers with an explicit family: in $A = k ((y)) ((x; σ))$ with $σ (y) = c y$ , a sum of squares vanishes only trivially, yet for suitable $c$ the element $- 1$ is a sum of $r + 1$ square-products. So $- 1$ can be a sum of square-products while failing to be a sum of squares.

Second, what do formally real division rings look like? Albert's theorem $(18.10)$ answers sharply: the centre $F$ is **algebraically closed in $D$ **. Every element of $D$ algebraic over $F$ is central. Consequently a formally real division ring that is algebraic over a central subfield — in particular any centrally finite one — is a field $(18.11)$ . Noncommutative formally real division rings are necessarily infinite-dimensional over their centres.

$r + 1$ Level realised by

(18.9)

$2^{n}$ Only levels of fields (Pfister)

CentralEvery algebraic element

(18.10)

InfiniteDimension over the centre

Overview

Recall from Ordered Division Rings that $D$ is formally real when $0$ is not a sum of square-products, equivalently when $- 1$ is not, equivalently — by $(18.2)$ — when $D$ admits an ordering. Two directions of study open up.

How badly can formal reality fail? Measured by the level $s (D)$ : the least $s$ with $- 1$ a sum of $s$ square-products. For fields Pfister proved in 1965 that $s$ is always a power of $2$ , and every power of $2$ occurs. For division rings the answer is completely different: Scharlau and Tschimmel showed in 1983 that every positive integer occurs.
What does formal reality force? Albert's 1940 theorem: the centre is algebraically closed in $D$ . This is a genuine restriction with no commutative content — for a field it says nothing at all.

Both halves rest on the ordering machinery. The level computations use the twisted series construction of Constructing Ordered Division Rings with a deliberately order-reversing twist; Albert's theorem uses the fact that an ordering of a division ring is a normal subgroup of $D^{*}$ , so conjugates of a positive element are positive.

Learning Objectives

Set up $A = k ((y)) ((x; σ))$ with $σ |_{k} = id$ , $σ (y) = c y$ , and compute leading terms of squares.
Prove $(18.7) (1)$ : $\sum_{i} H_{i}^{2} = 0$ in $A$ forces all $H_{i} = 0$ .
Prove $(18.7) (2)$ : if $c = - (1 + c_{1}^{2} + \dots + c_{r}^{2})$ then $- 1$ is a sum of $r + 1$ square-products.
State $(18.9)$ and explain which extra input makes $r + 1$ optimal.
Prove Albert's theorem $(18.10)$ using $(16.9)$ and the normality of the cone.
Deduce $(18.11)$ and explain why Hilbert's example must be centrally infinite.

Definitions

Definition—Level

For a division ring $D$ that is not formally real, the level $s (D)$ is the least integer $s \geq 1$ such that $- 1$ is a sum of $s$ square-products in $D$ . For formally real $D$ one sets $s (D) = \infty$ .

$k ((y))$: Formal Laurent series in $y$ over a field $k$ ; here $k$ is always formally real.
$k ((y)) ((x; σ))$: Twisted Laurent series in $x$ over $k ((y))$ , with $x r = σ (r) x$ .
$σ$: In this section, the automorphism of $k ((y))$ fixing $k$ pointwise with $σ (y) = c y$ , for a fixed $c \in k^{*}$ .
Algebraically closed in $D$: $F \subseteq D$ is algebraically closed in $D$ if every $d \in D$ satisfying a nonzero polynomial over $F$ lies in $F$ .
Centrally finite: $\dim_{Z (D)} D < \infty$ . Contrast centrally infinite, the situation forced on formally real noncommutative examples.

A division ring of characteristic $0$ that is algebraic over its centre need not be centrally finite in general; the point of $(18.11)$ is that formal reality collapses both notions to commutativity.

Core Concepts

The twist that breaks formal reality

Fix a formally real field $k$ and $c \in k^{*}$ . Let $R = k ((y))$ and let $σ \in Aut (R)$ fix $k$ pointwise with $σ (y) = c y$ . Form $A = k ((y)) ((x; σ))$ , so that

x y = σ (y) x = c y x, hence x y x^{- 1} y^{- 1} = c .

(18.7a)

The scalar $c$ is realised as a multiplicative commutator inside $A$ , therefore as a square-product by the identity $a b a^{- 1} b^{- 1} = a^{2} (a^{- 1} b)^{2} (b^{- 1})^{2}$ .

This is the whole mechanism. Any element of $k^{*}$ that we can arrange to be a commutator becomes a square-product in $A$ , no matter how negative it is in $k$ . Choosing $c$ negative therefore injects negativity into the group $Σ (A)$ of square-products — while, as the next paragraph shows, the additive structure of squares is untouched.

Why sums of squares still cannot vanish

Leading-term arithmetic controls squares in $A$ at two levels. If $H = h x^{m} + \dots$ with $h \in k ((y))$ nonzero, then $H^{2} = h σ^{m} (h) x^{2 m} + \dots$ . Summing over $i$ with a common lowest exponent $m$ gives a leading coefficient $\sum_{i} h_{i} σ^{m} (h_{i})$ in $k ((y))$ ; repeating the computation one level down with $h_{i} = a_{i} y^{n} + \dots$ produces $c^{m n} (\sum_{i} a_{i}^{2}) y^{2 n}$ , and $\sum_{i} a_{i}^{2} \neq 0$ because $k$ is formally real.

\sum_{i} H_{i}^{2}

A

(\sum_{i} h_{i} σ^{m} (h_{i})) x^{2 m}

c^{m n} (\sum_{i} a_{i}^{2}) y^{2 n}

\neq 0

since

k

is formally real

Conjugates of a positive element are positive

By $(18.1)$ an ordering $P$ is a normal subgroup of $D^{*}$ . So if $a \in P$ and $u \in D^{*}$ then $u a u^{- 1} \in P$ . Since an ordering is also closed under addition and misses $0$ , a sum of conjugates of a positive element can never be $0$ — the exact contradiction that drives Albert's theorem.

Key Results

Proposition(18.7)Squares versus square-products in a twisted series ring

Let $k$ be a formally real field, $c \in k^{*}$ , and let $A = k ((y)) ((x; σ))$ where $σ \in Aut (k ((y)))$ fixes $k$ pointwise and satisfies $σ (y) = c y$ . Then:

if $\sum_{i} H_{i}^{2} = 0$ in $A$ (a finite sum), then $H_{1} = H_{2} = \dots = 0$ ; in particular $- 1$ is not a sum of squares in $A$ ;
if $c = - (1 + c_{1}^{2} + \dots + c_{r}^{2})$ for some nonzero $c_{1}, \dots, c_{r} \in k$ , then $- 1$ is a sum of $r + 1$ square-products in $A$ , so $A$ is not formally real.

Proof

(2). From $x y = σ (y) x = c y x$ we get $c = x y x^{- 1} y^{- 1}$ , and the commutator identity of $(18.1)$ rewrites this as $c = x^{2} (x^{- 1} y)^{2} (y^{- 1})^{2}$ , a product of three squares, hence a square-product. Substituting $c = - (1 + c_{1}^{2} + \dots + c_{r}^{2})$ and rearranging,

- 1 = c + c_{1}^{2} + \dots + c_{r}^{2} = x^{2} (x^{- 1} y)^{2} (y^{- 1})^{2} + c_{1}^{2} + \dots + c_{r}^{2},

(18.7b)

A sum of $r + 1$ square-products: one genuine three-fold product and $r$ ordinary squares.

(1). Suppose $\sum_{i} H_{i}^{2} = 0$ with the $H_{i}$ not all zero. Let $m$ be the least $x$ -exponent occurring in any $H_{i}$ and write $H_{i} = h_{i} x^{m} + (higher powers of x)$ , so that $h_{i} \in k ((y))$ and some $h_{i} \neq 0$ . Multiplying out and using $x^{m} h = σ^{m} (h) x^{m}$ ,

\sum_{i} H_{i}^{2} = (\sum_{i} h_{i} σ^{m} (h_{i})) x^{2 m} + (higher powers of x) .

(18.8)

Now repeat the same manoeuvre inside $k ((y))$ . Let $n$ be the least $y$ -exponent occurring in any $h_{i}$ and write $h_{i} = a_{i} y^{n} + \dots$ with $a_{i} \in k$ and some $a_{i} \neq 0$ . Since $σ$ fixes $k$ and sends $y$ to $c y$ , we have $σ^{m} (h_{i}) = a_{i} c^{m n} y^{n} + \dots$ , so

\sum_{i} h_{i} σ^{m} (h_{i}) = c^{m n} (\sum_{i} a_{i}^{2}) y^{2 n} + (higher powers of y) .

(18.8a)

Because $k$ is formally real and the $a_{i}$ are not all zero, $\sum_{i} a_{i}^{2} \neq 0$ ; and $c^{m n} \neq 0$ . Hence $\sum_{i} h_{i} σ^{m} (h_{i})$ is a nonzero element of $k ((y))$ , so by $(18.8)$ the sum $\sum_{i} H_{i}^{2}$ is a nonzero element of $A$ — contradiction. Finally, if $- 1$ were a sum of squares, say $- 1 = \sum_{i} H_{i}^{2}$ , then $1^{2} + \sum_{i} H_{i}^{2} = 0$ with a nonzero summand, contradicting what has just been proved.

Theorem(18.9)Scharlau–Tschimmel level theorem

In the situation of $(18.7) (2)$ , assume in addition that $- c = 1 + c_{1}^{2} + \dots + c_{r}^{2}$ is not a sum of squares of $r$ elements of $k$ . Then the representation $- 1 = x^{2} (x^{- 1} y)^{2} (y^{- 1})^{2} + c_{1}^{2} + \dots + c_{r}^{2}$ is a shortest representation of $- 1$ as a sum of square-products in $A$ ; that is, $s (A) = r + 1$ .

The proof, which Lam omits, combines leading-term calculations like those in $(18.7)$ with a nontrivial fact about fields: over any field, the product of a sum of $p$ squares and a sum of $q$ squares is a sum of $p + q - 1$ squares. To realise a given level $r + 1$ one needs a formally real $k$ with $1 + c_{1}^{2} + \dots + c_{r}^{2}$ not a sum of $r$ squares; by a theorem of Cassels, $k = ℝ (c_{1}, \dots, c_{r})$ with independent indeterminates $c_{i}$ works.

Theorem(18.10)Albert's theorem

Let $D$ be a formally real division ring with centre $F$ . Then $F$ is algebraically closed in $D$ : every $d \in D$ that is algebraic over $F$ lies in $F$ .

Proof

Since $D$ is formally real it has an ordering by $(18.2)$ , hence $char D = 0$ by $(17.4)$ and $ℚ \subseteq F$ . Let $d \in D$ be algebraic over $F$ and suppose, for a contradiction, that $d \notin F$ . Its minimal polynomial over $F$ , say $t^{n} + c_{1} t^{n - 1} + \dots + c_{n}$ , then has degree $n \geq 2$ .

Put $a = d + c_{1} / n$ , legitimate because $char F = 0$ and $c_{1} \in F$ is central. Substituting $d = a - c_{1} / n$ into the minimal equation and expanding, the coefficient of $a^{n - 1}$ becomes $- n \cdot (c_{1} / n) + c_{1} = 0$ , so the minimal polynomial of $a$ over $F$ has the form

f (t) = t^{n} + e_{2} t^{n - 2} + \dots + e_{n} \in F [t],

(18.10a)

Degree $n \geq 2$ , with vanishing coefficient in degree $n - 1$ . Also $a \neq 0$ : otherwise $d = - c_{1} / n \in F$ .

Let $Δ$ be the conjugacy class of $a$ in $D$ . It is algebraic over $F$ with minimal polynomial $f$ , so Wedderburn's factorisation theorem $(16.9)$ applies and gives elements $a_{1} = a, a_{2}, \dots, a_{n} \in Δ$ with

f (t) = (t - a_{n}) \dots (t - a_{1}) in D [t] .

(18.10b)

Comparing coefficients of $t^{n - 1}$ on both sides and using $(18.10 a)$ gives $a_{1} + a_{2} + \dots + a_{n} = 0$ .

Now use the ordering. Fix an ordering $P \subseteq D^{*}$ . Since $a \neq 0$ , either $a \in P$ or $a \in - P$ . Suppose $a \in P$ . Each $a_{i}$ is a conjugate $u_{i} a u_{i}^{- 1}$ , and $P$ is a normal subgroup of $D^{*}$ by $(18.1)$ , so every $a_{i} \in P$ ; then $a_{1} + \dots + a_{n} \in P$ by additive closure, contradicting the fact that this sum is $0 \notin P$ . If instead $a \in - P$ , apply the same argument to $- a$ , whose conjugates are the $- a_{i}$ and which again sum to $0$ . Either way we have a contradiction, so $d \in F$ .

Remark—The ordering is scaffolding

Albert originally stated the result for ordered division rings. The ordering appears only inside the proof; the conclusion mentions none. Stating the hypothesis as formally real — legitimate once $(18.2)$ is available — is the sharper formulation, because formal reality is an intrinsic arithmetic condition on $D$ .

Corollary(18.11)Formally real algebraic division algebras are fields

Let $D$ be a formally real division ring that is an algebraic division algebra over a field $F \subseteq Z (D)$ — that is, every element of $D$ is algebraic over $F$ . Then $D$ is a field. In particular, any formally real centrally finite division ring is a field.

Proof

Every $d \in D$ is algebraic over $F$ , hence over the larger field $Z (D) \supseteq F$ . By $(18.10)$ , $d \in Z (D)$ . So $D = Z (D)$ is commutative. For the second statement, a centrally finite $D$ is finite-dimensional over $Z (D)$ and therefore algebraic over it.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Two-level leading-term induction

In an iterated series ring, prove a statement about the lowest $x$ -coefficient, then rerun the same argument for the lowest $y$ -coefficient. The twist only contributes a nonzero scalar $c^{m n}$ , which cannot cause cancellation.

Move 2

Manufacture scalars as commutators

A twist $σ (y) = c y$ makes $c$ a multiplicative commutator, and commutators are square-products. This is the standard way to force a prescribed element into $Σ (D)$ without touching the additive theory.

Move 3

Kill trace-zero conjugate sums

Normalise so the degree $n - 1$ coefficient vanishes, factor over $D$ by $(16.9)$ , and read off that the conjugates sum to zero. An ordering then forbids that, because the cone is normal and additively closed.

Move 3 is the reusable core of Albert's theorem, and it explains why the result has no commutative analogue: over a field the conjugates of $a$ live in an extension, not in $D$ itself, and $(16.9)$ has no content.

It is worth noticing what is not used: no chain condition, no finiteness over the centre, and no assumption about the number of orderings. A single ordering suffices, and $(18.2)$ supplies it from arithmetic alone.

Worked Example

The smallest case: $c = - 1$

Take $k = ℝ$ and $r = 0$ , so $c = - 1$ . Then $σ$ is the automorphism of $ℝ ((y))$ fixing $ℝ$ with $σ (y) = - y$ , and $A = ℝ ((y)) ((x; σ))$ satisfies $x y = - y x$ .

- 1 = c = x y x^{- 1} y^{- 1} = x^{2} (x^{- 1} y)^{2} (y^{- 1})^{2} .

(E.1)

$- 1$ is a single square-product, so $s (A) = 1$ — the smallest possible level.

Check the identity directly: $x^{2} (x^{- 1} y)^{2} (y^{- 1})^{2} = x x x^{- 1} y x^{- 1} y y^{- 1} y^{- 1} = x y x^{- 1} y^{- 1}$ , and $x y x^{- 1} = σ (y) = - y$ , so the product is $- y y^{- 1} = - 1$ .

By $(18.7) (1)$ , however, $- 1$ is not a sum of squares in $A$ : any relation $\sum_{i} H_{i}^{2} = 0$ forces all $H_{i} = 0$ , and $ℝ$ is formally real. This is the promised example of a division ring in which $- 1$ is a square-product but not a sum of squares, and it shows the wording of $(18.2)$ cannot be weakened.

What sort of division ring is it?

Here $σ^{2} = id$ , so $σ$ has order $2$ , and $(14.2)$ gives $Z (A) = ℝ ((y^{2})) ((x^{2}))$ — the fixed field of $σ$ inside $ℝ ((y))$ , extended by $x^{2}$ . So $A$ is centrally finite of dimension $4$ , spanned by $1, x, y, x y$ over its centre: a quaternion algebra with $x^{2}, y^{2}$ central and $x y = - y x$ . There is no conflict with $(18.11)$ , because $A$ is not formally real.

A higher level

Take $r = 1$ , $k = ℝ (c_{1})$ with $c_{1}$ an indeterminate, and $c = - (1 + c_{1}^{2})$ . Then $k$ is formally real, and by Cassels' theorem $1 + c_{1}^{2}$ is not a square in $k$ , so $(18.9)$ gives $s (A) = 2$ for $A = k ((y)) ((x; σ))$ with $σ (y) = c y$ . For a field, level $2$ is also attainable — $ℚ (i)$ has level $1$ , $ℚ$ has level $\infty$ , $𝔽_{5}$ has level $1$ and $𝔽_{3}$ has level $2$ — but level $3$ is attainable only for division rings.

Comparison and Classification

Levels: fields versus division rings
Question	Fields	Division rings
Possible finite values of $s$	powers of $2$ only (Pfister, 1965)	every positive integer (Scharlau–Tschimmel, 1983)
What is being summed	squares	square-products
Are the two the same?	yes, $a^{2} b^{2} = (a b)^{2}$	no — $(18.7)$
Standard witnesses	$𝔽_{p}$ , $ℚ_{p}$ , function fields	$k ((y)) ((x; σ))$ with $σ (y) = c y$
$s = \infty$ means	formally real, orderable	formally real, orderable

Which properties can coexist with formal reality
	Formally real	Noncommutative	Centrally finite	Possible?
An ordered field such as $ℚ$	yes	no	yes	yes
Hilbert's $ℚ ((y)) ((x; σ))$ , $σ (y) = 2 y$	yes	yes	no	yes
A formally real quaternion-like algebra	yes	yes	yes	no — forbidden by (18.11)
$ℍ$	no	yes	yes	yes
$ℝ ((y)) ((x; σ))$ , $σ (y) = - y$	no	yes	yes	yes

Which properties can coexist with formal reality

The third row is the content of Albert's theorem: formal reality plus finite dimension over the centre forces commutativity, so that combination simply does not occur.

Relationship Map

Albert's theorem is a hinge: everything above it is arithmetic in a constructed example, everything below it is structural.

D

formally real

(18.2)

: an ordering existscone normal in

D^{*}

(18.10)

: algebraic over

F

\Rightarrow

central

(18.11)

: centrally finite

\Rightarrow

field

Formally real division ring $D$ , centre $F$
- must have
  - characteristic $0$ , so $ℚ \subseteq F$
  - at least one ordering
  - $F$ algebraically closed in $D$
  - no roots of unity other than $\pm 1$
- cannot have
  - a solution of $t^{2} + 1 = 0$
  - any noncentral algebraic element
  - finite dimension over $F$ unless $D = F$
  - an archimedean ordering unless $D = F$ embeds in $ℝ$
- typical example
  - $ℚ ((y)) ((x; σ))$ with $σ$ order-preserving and $σ \neq id$

The final entry of the second branch is $(17.21)$ ; the third is exactly the family from Constructing Ordered Division Rings, and $(18.11)$ explains why it had to be centrally infinite. Exercise 17.13 confirms this independently: an order-preserving automorphism of finite order is the identity, so $σ$ has infinite order and $(14.2)$ makes the centre a subfield of $k$ .

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

Very little here is computable in the usual sense, and it is worth being precise about why.

Deciding formal reality of an abstractly presented division ring is not effective; the question is whether $- 1$ lies in a subgroup generated by infinitely many squares, and the word problem for finitely presented rings is already undecidable.
Deciding the level of a field is effective for many concrete fields: for a number field $k$ , $s (k) \in {1, 2, 4, \infty}$ and is determined by local conditions. For general function fields it is bounded via Pfister theory but not always computed.
**Computing in $k ((y)) ((x; σ))$ ** is done with truncated series: each element is a finite $x$ -expansion whose coefficients are truncated $y$ -series. Multiplication costs $O (N M \log)$ coefficient operations for truncation orders $N$ , $M$ , plus the cost of applying $σ^{m}$ , which for $σ (y) = c y$ is a single scalar power.
**Verifying an identity such as $(E . 1)$ ** requires only leading-term arithmetic and is exact; no truncation error arises because the identity involves finitely many terms.

Failure Modes and Common Mistakes

Do not apply $(16.9)$ without checking its hypotheses: the conjugacy class must be algebraic over the centre with minimal polynomial of the stated degree.
Do not forget the normalisation $a = d + c_{1} / n$ in Albert's proof; without a vanishing $t^{n - 1}$ coefficient the conjugates sum to $- c_{1}$ , not $0$ , and no contradiction follows.
Do not conclude from $(18.11)$ that formally real division rings are rare — they are abundant, but all noncommutative ones are infinite-dimensional over their centres.
Do not read $s (D) = \infty$ as a defect: it is precisely formal reality, hence orderability.

Historical Notes and Lessons Learned

1940AlbertProves that the centre of an ordered division ring is algebraically closed in it — the first structural theorem about ordered division rings, and still the sharpest.
1952Szele and PickertThe orderability criterion for division rings, phrased with square-products, makes it possible to restate Albert's hypothesis intrinsically as formal reality.
1965PfisterThe level of a non-formally-real field is a power of $2$ , and every power of $2$ occurs. A landmark of quadratic form theory, resting on the theory of multiplicative forms.
1964–1970sCassels and the sums-of-squares machineryCassels' theorem on $1 + c_{1}^{2} + \dots + c_{r}^{2}$ over rational function fields supplies the fields needed to realise prescribed levels.
1983Scharlau and TschimmelEvery positive integer is the level of some division ring, via twisted series with a prescribed commutator. The contrast with Pfister's theorem is total.

The lesson is that the two halves of the classical theory come apart in the noncommutative setting. Additive facts about squares are robust and transfer almost unchanged, as $(18.7) (1)$ shows; multiplicative facts — Pfister's multiplicativity, the power-of-two levels — depend on commutativity and fail completely. Knowing which half a classical theorem belongs to is the practical skill.

Quick Reference

The family

A = k ((y)) ((x; σ))

k

formally real,

σ |_{k} = id

σ (y) = c y

Key commutator

x y x^{- 1} y^{- 1} = c = x^{2} (x^{- 1} y)^{2} (y^{- 1})^{2}

$(18.7) (1)$

\sum H_{i}^{2} = 0 \Rightarrow

all

H_{i} = 0

;

- 1

is never a sum of squares in

A

$(18.7) (2)$

c = - (1 + c_{1}^{2} + \dots + c_{r}^{2}) \Rightarrow - 1

is a sum of

r + 1

square-products

$(18.9)$ if

- c

is not a sum of

r

squares in

k

, then

s (A) = r + 1

Levelsfields: powers of

2

; division rings: every positive integer

$(18.10)$ Albert

D

formally real,

F = Z (D)

: algebraic over

F

\Rightarrow

F

$(18.11)$ formally real

+

algebraic over a central subfield

\Rightarrow

field

Hypotheses at a glance
Result	Needs	Concludes
(18.7)(1)	$k$ formally real, any $c \in k^{*}$	no nontrivial vanishing sum of squares in $A$
(18.7)(2)	$c = - (1 + c_{1}^{2} + \dots + c_{r}^{2})$	$- 1$ is a sum of $r + 1$ square-products
(18.9)	additionally $- c$ not a sum of $r$ squares in $k$	$s (A) = r + 1$ exactly
(18.10)	$D$ formally real, $F = Z (D)$	$F$ algebraically closed in $D$
(18.11)	$D$ formally real, algebraic over a central subfield	$D$ is a field

Frequently Asked Questions

Why does Albert's theorem need Wedderburn's factorisation theorem?

Because the contradiction is produced by summing conjugates. $(16.9)$ says that the minimal polynomial of an element algebraic over the centre splits in $D [t]$ into linear factors whose roots are all conjugates of that element; comparing the coefficient of $t^{n - 1}$ then shows those $n$ conjugates sum to zero. Without a factorisation inside $D$ itself there is nothing to compare.

Does Albert's theorem say a formally real division ring has no proper algebraic extensions?

No — it is a statement about elements of $D$ , not about extensions of $D$ . It says $D$ contains no element algebraic over its centre except the central ones. In particular $D$ contains no square root of $- 1$ , no primitive cube root of unity, and no noncentral element satisfying a polynomial over $F$ .

Is the ring

A

(18.7)

ever formally real?

Yes, whenever the twist is order-preserving. If $c$ is positive in some ordering of $k$ that $σ$ preserves, $(18.5)$ orders $A$ and it is formally real. The construction becomes interesting exactly when $c$ is chosen negative, which destroys the hypothesis of $(18.5)$ and, by $(18.7) (2)$ , formal reality itself.

Why is the level of a field always a power of 2 but not that of a division ring?

Pfister's proof uses the multiplicativity of sums of $2^{n}$ squares — a quadratic form identity that holds only in the commutative setting. In a division ring the relevant set is the group of square-products, whose behaviour is governed by commutators rather than by quadratic forms, and no multiplicativity constraint survives.

What is the significance of the element

x^{2} (x^{- 1} y)^{2} (y^{- 1})^{2}

It is the commutator $x y x^{- 1} y^{- 1}$ rewritten as a product of three squares, using the identity from $(18.1)$ . That rewriting is what converts a purely multiplicative fact — the twist introduces the scalar $c$ as a commutator — into the statement that $c$ is a square-product, which is what $(18.2)$ tests against.

Are there formally real division rings that are noncommutative and finitely generated?

Yes: Hilbert's example is generated over $ℚ$ by $x$ and $y$ as a division ring. What $(18.11)$ forbids is finite dimension over the centre, not finite generation. Being centrally infinite is compatible with being a very concrete, two-generator object.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §18, pp. 288–291.
A. A. Albert, “On ordered algebras”, Bulletin of the American Mathematical Society 46 (1940).
A. Pfister, “Zur Darstellung von −1 als Summe von Quadraten in einem Körper”, Journal of the London Mathematical Society 40 (1965).
W. Scharlau and A. Tschimmel, “On the level of skew fields”, Archiv der Mathematik 40 (1983).
T. Y. Lam, The Algebraic Theory of Quadratic Forms, W. A. Benjamin, Reading, Massachusetts, 1973, Chapters 10 and 11.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.

AI Suggested Questions

Reconstruct the omitted proof of $(18.9)$ , including the lemma that a product of a sum of $p$ squares and a sum of $q$ squares is a sum of $p + q - 1$ squares.
Prove Cassels' theorem that $1 + c_{1}^{2} + \dots + c_{r}^{2}$ is not a sum of $r$ squares in $ℝ (c_{1}, \dots, c_{r})$ .
Compute the centre of $k ((y)) ((x; σ))$ for $σ (y) = c y$ with $c$ not a root of unity in $k^{*}$ .
Give a self-contained proof that a formally real division ring contains no roots of unity other than $\pm 1$ .
What is the analogue of Albert's theorem for ordered rings that are not division rings?
How do levels behave under field extension, and is there a division-ring analogue of the Pfister multiplicativity theorem?
Which quaternion algebras over a formally real field are themselves formally real, and why does $(18.11)$ apply?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

The twist that breaks formal reality

Why sums of squares still cannot vanish

Conjugates of a positive element are positive

06Key Results

07Proof Techniques and Method

Two-level leading-term induction

Manufacture scalars as commutators

Kill trace-zero conjugate sums

08Worked Example

The smallest case: c=−1

What sort of division ring is it?

A higher level

09Comparison and Classification

10Relationship Map

11Computational Notes

12Failure Modes and Common Mistakes

13Historical Notes and Lessons Learned

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

The smallest case: $c = - 1$

Comparison and Classification

Relationship Map

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions