Every $x \in K$ can be expressed in the form $x = ab^{-1}$, with $a$ and $b$ in $A$ and $b \neq 0$. If $x$ is positive, then $a$ and $b$ have the same sign, and conversely. Thus we see that, if there exists an ordering on $K$ satisfying the prescribed conditions, then it is unique, and the set $P$ of positive elements is identical to the set of $ab^{-1}$, where $a$ and $b$ are elements of $A$ of the same sign, and $b \neq 0$. It remains to show that $P$ satisfies conditions (AP$_I$), (AP$_{II}$), (AP$_{III}$) and (AP$_{IV}$). This is obvious for (AP$_{II}$) and (AP$_{IV}$). For (AP$_I$), consider $ab^{-1} + cd^{-1}$, where we may assume that $a$, $b$, $c$ and $d$ are positive; this sum is $(ad + bc)(bd)^{-1}$, and $ad + bc$ and $bd$ are positive.

To show $(\mathrm{AP}_{III})$, consider an identity of the form $ab^{-1} = -cd^{-1}$, so that $ad + bc = 0$. If we assume that $a$ and $b$ have the same sign and that $c$ and $d$ have the same sign, then the sign rules show that $ad$ and $bc$ have the same sign; whence $ad = bc = 0$, so $a = c = 0$; hence $P$ does indeed satisfy $(\mathrm{AP}_{III})$.

Condition (OE) is clearly equivalent to: (OE') The element – 1 is not a sum of elements of the form $px^2$ ($x \in E, p \in K, p \geq 0$).

Condition (OE) is necessary: if E is an ordered extension of K then the elements $p_i x_i^2$ are positive in E, so zero if their sum is zero. On the other hand $p_i x_i^2 = 0$ is equivalent to $p_i x_i = 0$.

Conversely, suppose condition (OE) is satisfied, then we will define an ordering on E by constructing a subset P of E which satisfies conditions (AP_I), (AP_{II}), (AP_{III}) and (AP_{IV}), and which contains the set $K_+$ of positive elements of K. Such a subset P will certainly make E an ordered extension of K, for we will have $K \cap P = K_+$; indeed, if P were to contain an element $-a < 0$ of K, then a would belong to $P \cap (-P)$, contradicting (APIII).

To define P, let us consider the set $\mathscr{M}$ of subsets of E which satisfy (AP_I), (AP_{II}) and (AP_{III}), and which contain the union of $K_+$ and the set C of squares of elements of E. This set $\mathscr{M}$ is nonempty, for it contains the set $P_0$ of elements of the form $\sum p_i x_i^2$ (that $P_0$ satisfies (AP_{III}) follows immediately from (OE)).

Moreover $\mathscr{M}$ is inductive (Set Theory, III, p. 154, Def. 3). Thus there exists, by Th. 2 of Set Theory, III, p. 154, a maximal element in $\mathscr{M}$, which it remains for us to prove satisfies (AP_{IV}); now this follows from the following lemma:

Take $P' = P - xP$, and check that $P'$ has the required properties. Since $0 \in C \subset P$, we have $P \subset P'$. Whence $C \subset P'$ and $K_+ \subset P'$. Since $1 \in C \subset P$ we have $-x \in P'$. We have

whence (AP_I). We have whence (AP_{II}). Finally, let us check (AP_{III}): suppose given an identity of the form $p - xq = -(r - xs)$ where $p, q, r, s$ belong to P; we deduce from this the relation $x(s + q) = p + r$; if $s + q \neq 0$ we have contrary to the hypothesis; hence $s + q = 0$, whence $p + r = 0$; since P satisfies (AP_{III}) we deduce that $s = q = r = p = 0$, which completes the proof.

The necessity is obvious. Conversely, the stated condition implies that E is of characteristic zero, hence an extension of Q; then condition $(OE)$ is satisfied, and Th. 1 shows that there exists on E the structure of an ordered extension of Q, that is an ordered field structure.

There does not exist any ordered field structure on a field E in which $-1$ is a square, in particular on an algebraically closed field.

The condition is obviously sufficient; it is also necessary, for (in the notation of the proof of Th. 1), if $x \notin P_0$ there exists a maximal element P of $\mathscr{M}$ such that $x \notin P$; then $-x \in P$ by the Lemma, and x is not positive under the ordering defined by P, since $x \neq 0$.

We will argue by induction on the degree n off. For $n = 1$ the proof is trivial. Suppose the result holds for degrees $\leq n - 1$, and let us prove it for degree n by contradiction; by Th. 1 we are thus assuming a relation of the form

Without loss of generality we may suppose that the $f_i$ have degrees $\leq n - 1$ (IV, p. 11, Cor). Then

where $h \neq 0$ has degree at most $n - 2$. Replacing X by a and b in the above inequality, we see that $h(a) f(a) > 0$ and $h(b) f(b) > 0$. Since f changes sign between a and b by hypothesis, we conclude that $h(a) h(b) < 0$. Then we have a similar inequality for one of the irreducible factors $g(X)$ of $h(X)$: that is $g(a)\ g(b) < 0$. But $1 + \sum_i P_i f_i^2(X) \equiv 0 \pmod{g(X)}$, which shows that the field $K[X]/(g)$ cannot be an ordered extension of $K$ (Th. 1), contrary to the induction hypothesis.

We can immediately reduce ourselves to the case of a monic polynomial; then one can write $f(x) = x^n(1 + a_1 x^{-1} + \ldots + a_n x^{-n})$ for $x \neq 0$. Let

For $|x| > M$ we have $1 + a_1 x^{-1} + \ldots + a_n x^{-n} > 0$, which completes the proof of the proposition.

Such an extension, being monogenous (V, p. 40, Th. 1), is isomorphic to $K[X]/(f)$, where $f$ is an irreducible polynomial of odd degree. Then it is enough to show that $f$ changes sign in $K$ (Prop. 3), which follows immediately from Prop. 4.

The result is trivial if $a$ is a square in $K$. Otherwise the polynomial $f(X) = X^2 - a$ is irreducible and changes sign, since $f(0) < 0$ and $f(x)$ has the same sign as $x^2$, so positive, for $x$ in the complement of some interval of $K$. We can now complete the proof by applying Prop. 3.

One can show that this ordered extension is unique up to K-isomorphism (VI, p. 40, Ex. 15).

Let $\Omega$ be an algebraic closure of K, and let $\mathfrak{N}$ be the set of pairs $(A, w)$, where $A$ is a sub-K-extension of $\Omega$, and $w$ is an ordering on $A$ making $A$ an ordered extension of $K$. Order $\mathfrak{N}$ by the relation « L is an ordered extension of M » between M and $L$. Equipped with this ordering, $\mathfrak{N}$ is an inductive ordered set: indeed if $(L_i)$ is a totally ordered family of elements of $\mathfrak{N}$, then the field $L = \bigcup L_i$, ordered by taking $L_1 = \bigcup (L_i)_1$, is an upper bound for the $L_i$. Then $\mathfrak{N}$ has a maximal element, by Set Theory, III, p. 154, Th. 2, which completes the proof.

At least one of the irreducible factors off, say $h$, changes sign between $a$ and $b$. Then the field $K[X]/(h)$ admits the structure of an ordered extension of $K$ (VI, p. 23, Prop. 3), and $h$ has degree 1 (Def. 4). Since $h(a) h(b) < 0$, the unique root $x$ of $h$ is such that $a < x < b$, since a polynomial function of degree 1 is monotonic.

This follows immediately from Cor. 2 and 1 to Prop. 4 of VI, p. 24.

Indeed the positive elements of K are determined by its algebraic structure: they are the squares.

It is clear that a) implies b): indeed K has only two algebraic extensions up to isomorphism, the field K itself and $K(i)$, which cannot be ordered since -1 is a square.

The fact that b) implies c) is nothing other than Prop. 6 of VI, p. 25.

It remains for us to prove that c) implies a). That will follow from the next two propositions.

Let us show first that the second assertion reduces to the first. One can put the second degree polynomial $aX^2 + bX + c$ ($a \neq 0$) in the following form, often called the canonical trinomial form:

If d is a square root of $(b^2 - 4ac)/4a^2$, then $d - (b/2a)$ is a root of the quadratic polynomial under consideration.

Now we show that every element $a + bi$ ($a \in K, b \in K$) is a square; we are looking for an element $x + yi$ such that

this translates into $x^2 - y^2 = a$ and $2xy = b$. From this we deduce that

Let c denote the positive square root of $a^2 + b^2$; then $c \geq |a|, c \geq |b|$ and $x^2 + y^2 = c$. Whence $x^2 = (c + a)/2$ and $y^2 = (c - a)/2$. Since $c \geq |a|$ these equations are soluble in K, and if $x_0$ and $y_0$ are two solutions then $x_0^2 - y_0^2 = a$ and $2x_0y_0 = \pm b$. We obtain the desired square root by taking $x = x_0$ and $y = b/2x_0$.

b) every polynomial in $K'[X]$ of degree 2 has a root in $K'$. Then $K'$ is algebraically closed.

Note first that it is enough to prove that every non-constant polynomial in $K[X]$ has a root in $K'$: this is indeed clear if $K' = K$; if $K' \neq K$ then $[K':K] = 2$; let $a \mapsto a$ denote the unique $K$-automorphism of $K'$ distinct from the identity map; if $f \in K'[X]$ and if $\bar{f}$ denotes the polynomial obtained by applying $a \mapsto \bar{a}$ to the coefficients of $f$, then $f \bar{f} \in K[X]$; if $a \in K'$ is a root of $f \bar{f}$ then $a$ is either a root of $f$ or of $\bar{f}$; thus either $a$ or $a$ is a root of $f$.

Thus let $f$ be a polynomial over $K$ of degree $2^n p$, $p$ odd. We will proceed by induction on $n$, the property being true for $n = 0$ by hypothesis a). Let E be an extension of $K$ in which $f$ splits into linear factors:

Let $b \in K$; put $y_{ij} = a_i + a_j + ba_i a_j \in E$ and

The coefficients of this polynomial are symmetric functions in the $a_i$, with coefficients in $K$; it therefore belongs to $K[X]$ (IV, p. 62, Th. 1); since it has degree $2^n p (2^n p - 1)/2 = 2^{n-1} p'$ ($p'$ odd), it has a root $y_{}$, in $K'$ by inductive hypothesis. If we note that this holds for all $b \in K$, and that $K$ is an infinite field (indeed a finite field, which has monogenous extensions of arbitrarily large odd degree (V, p. 94, Prop. 3), cannot satisfy a)), then we can deduce the existence of at least one pair $(i, j)$ such that

with $b \neq b'$. Then $a_i + a_j$ and $a_i a_j$ are elements of $K'$, hence so are $a_i$ and $a_j$, since they are the roots of the quadratic equation

Q.E.D.

For a generalisation and an alternative proof of Prop. 8, based on Galois theory, see VI, p. 46, Ex. 33.

Let $K$ be an ordered field and let $K' = K(i)$; for every element $z = a + bi$ of $K'$, the norm $z \overline{z} = a^2 + b'$ of $z$ relative to $K$ (III, p. 544, example 1) is a positive element of $K$, which vanishes only for $z = 0$. If every positive element in $K$ is a square (in particular if $K$ is a maximal ordered field), then the positive square root of the norm $z \overline{z}$ is called the absolute value of $z$, and is written $|z|$. Since $|zz'|^2 = |z|^2 |z'|^2$ we have $|zz'| = |z| \cdot |z'|$.

Moreover, the triangle inequality

holds for every pair of elements $z, z'$ of $K'$. Indeed, if $z = a + bi$ and $z' = a' + b'i$, then this inequality is equivalent to

and hence also to

which can be written $(ab' - ba')^2 \geq 0$.

Th. 3 enables us to determine all the irreducible polynomials over a maximal ordered field:

Since $K(i)$ is algebraically closed, every algebraic extension of K, and hence also every irreducible polynomial over K, has degree 1 or 2. To see which second degree polynomials are irreducible, it is enough to consider the canonical form $a((X + (b/2a))^2 - (b^2 - 4ac)/4a^2)$ (cf. VI, p. 26, Prop. 7).