Ordered Fields
Introduction
An ordered field is a field equipped with a total order that is compatible with addition and multiplication. The compatibility has strong algebraic consequences: the field has characteristic $0$, it contains a canonical copy of $\mathbb{Q}$, every nonzero square is positive, and the positive elements form a cone closed under addition and multiplication. The extra structure that distinguishes one ordered field from another is whether the natural numbers are bounded above, and this Archimedean condition turns out to be equivalent to the density of $\mathbb{Q}$ and to embeddability into $\mathbb{R}$.
This article develops ordered rings and fields, the positive cone that encodes the order, and the Archimedean and non-Archimedean cases, with $\mathbb{Q}$ and $F(t)$ as the two standard examples. The completion theory is the subject of Absolute Values, Valuations and Completions, the theory of real-closed fields that of Real-Closed and Complete Ordered Fields, and the general theory of topological rings that of Topological Rings and Fields; the order-theoretic facts are stated by the order alone, so no distance, no topology and no continuity is used here.
Throughout, an ordered field is written $F$ and its order $\leq$. The axioms of an ordered field are stated from scratch, so no prior theory of order is assumed; the theory of fields, characteristic, and prime fields is from Fields, and divisibility is from Integral Domains.
Ordered Rings and Fields
Definitions
Definition. An ordered ring is a commutative ring $R$ with $1 \neq 0$ together with a total order $\leq$ such that for all $a, b, c \in R$:
(O1) if $a \leq b$ then $a + c \leq b + c$;
(O2) if $a \geq 0$ and $b \geq 0$ then $ab \geq 0$.
An ordered field is an ordered ring that is a field.
The order is translation invariant by (O1) and multiplication by nonnegative elements is monotone by (O2), in the following precise sense.
Proposition. Let $R$ be an ordered ring and $a, b, c, d \in R$.
(a) If $a \leq b$ and $c \leq d$ then $a + c \leq b + d$.
(b) If $a \leq b$ and $c \geq 0$ then $ac \leq bc$. If $a \leq b$ and $c \leq 0$ then $ac \geq bc$.
(c) $a \leq b$ if and only if $b - a \geq 0$.
(d) $0 \leq a \leq b$ implies $a^2 \leq b^2$ and $a^2 \leq ab$.
Proof. (a) Apply (O1) twice. (b) If $c \geq 0$ then $bc - ac = (b-a)c \geq 0$ by (O2). If $c \leq 0$ then $-c \geq 0$ and $(b-a)(-c) \geq 0$, so $ac \geq bc$. (c) Translate by $-a$. (d) Multiply $a \leq b$ by $a \geq 0$ to get $a^2 \leq ab$, and by $b \geq 0$ to get $ab \leq b^2$.
Basic Rules
Theorem. Let $F$ be an ordered field.
(a) $1 > 0$, and $\operatorname{char} F = 0$.
(b) $a^2 > 0$ for every $a \neq 0$, so every nonzero square is positive; in particular negative elements have no square root.
(c) $x > 0$ implies $x^{-1} > 0$, and $0 < x < y$ implies $0 < y^{-1} < x^{-1}$.
(d) If $a < b$ then there are elements strictly between them, for instance $a < \tfrac{a+b}{2} < b$.
(e) $-1$ is not a sum of squares; equivalently $F$ is formally real.
Proof. (a) $1 = 1^2 \geq 0$ by (O2), and $1 \neq 0$; if $1 \leq 0$ then $1 \geq 0$ and $1 \leq 0$ force $1 = 0$, a contradiction, so $1 > 0$. Hence $n \cdot 1 = 1 + \cdots + 1 > 0$ for every $n \geq 1$ by (a) and (a) of the proposition, so no positive integer is $0$ and the characteristic is $0$.
(b) If $a > 0$ then $a^2 = a \cdot a > 0$ by (O2) and $a \neq 0$; if $a < 0$ then $-a > 0$ and $a^2 = (-a)^2 > 0$. If $a^2 = 0$ then $a = 0$ as $F$ is a field.
(c) Since $x x^{-1} = 1 > 0$ and $x > 0$, the inverse cannot be $\leq 0$, because then $x x^{-1} \leq 0$ by (b) of the proposition. Hence $x^{-1} > 0$. If $0 < x < y$ then multiplying by $x^{-1}y^{-1} > 0$ gives $y^{-1} < x^{-1}$.
(d) $2 = 1 + 1 > 0$, so $2^{-1} > 0$ by (c), and $a = \tfrac{a+a}{2} < \tfrac{a+b}{2} < \tfrac{b+b}{2} = b$.
(e) If $-1 = \sum_i a_i^2$ then the right side is a sum of nonnegative elements, hence $\geq 0$; so $-1 \geq 0$, whence $1 \leq 0$, contradicting (a).
Remark (the boundary of the theory). Property (e) is what makes the orderable fields a special class among the fields of characteristic $0$: a field that is not formally real admits no ordering at all, since a sum of squares can never be negative in an ordered field. Conversely, every formally real field is orderable, so the two classes coincide.
Theorem (Artin–Schreier). A field $F$ admits an ordering if and only if it is formally real.
Proof sketch. An ordered field is formally real by (e). Conversely, suppose $-1$ is not a sum of squares, and let $\mathcal{T}$ be the collection of subsets $T \subseteq F$ that contain every square, are closed under addition and under multiplication, and satisfy $-1 \notin T$. It is nonempty, because the set of sums of squares lies in it, and it is closed under unions of chains, so Zorn's lemma gives a maximal $T$. Maximality forces $F = T \cup (-T)$: if $a \notin T \cup (-T)$, the set $T + aT = \{x + ay : x, y \in T\}$ contains $T$, contains every square because $T$ does, and is closed under addition and under multiplication because $a^2 \in T$; since it contains $a = 0 + a\cdot 1$, it contains $T$ properly, so by maximality $-1 \in T + aT$, and the same argument with $-a$ in place of $a$ gives $-1 \in T - aT$. Writing $-1 = x_1 + ay_1 = x_2 - ay_2$ with $x_i, y_i \in T$ and $y_i \neq 0$, multiply the first equation by $y_2^2$ and the second by $y_1^2$: the first becomes $a y_1 y_2^2 = -(y_2^2 + x_1 y_2^2) \in -T$, and the second becomes $a y_1^2 y_2 = y_1^2 + x_2 y_1^2 \in T$. Multiplying the first of these by $y_1^2 y_2 \in T$ and the second by $y_1 y_2^2 \in T$ gives the same element $t = a\, y_1^3 y_2^3$, which is therefore nonzero and lies in both $T$ and $-T$. Then $-1 = (-t^2)(t^{-1})^2 \in T$, because $T$ contains every square and is closed under multiplication, a contradiction. Hence $F = T \cup (-T)$, and $T \cap (-T) = \{0\}$ by the same computation applied to an element of the intersection. Therefore $P = T \setminus \{0\}$ satisfies (C1), (C2) and (C3) of the next section, and the theorem proved there produces an ordering of $F$ with positive cone $P$.
Cones and Positivity
The Positive Cone
Definition. The positive cone of an ordered field $F$ is
$$ P = \{x \in F : x > 0\}. $$
Theorem. The positive cone $P$ of an ordered field satisfies
(C1) $P + P \subseteq P$;
(C2) $P \cdot P \subseteq P$;
(C3) $F$ is the disjoint union $F = (-P) \cup \{0\} \cup P$.
Conversely, if a subset $P \subseteq F$ of a field $F$ satisfies (C1), (C2), (C3), then the relation
$$ a \leq_P b \iff b - a \in P \cup \{0\} $$
is a total order making $F$ an ordered field with positive cone $P$.
Proof. (C1): if $a > 0$ and $b > 0$ then $a + b > a > 0$ by (O1). (C2) is (O2). (C3) is trichotomy for a total order. Conversely, $\leq_P$ is total and antisymmetric by (C3), and transitive because $P \cup \{0\}$ is closed under addition by (C1). Translation invariance: $b - a = (b+c)-(a+c)$. Multiplicativity (O2): if $b - a \in P \cup \{0\}$ and $c \in P \cup \{0\}$ then $(b-a)c \in P \cup \{0\}$ by (C1) and (C2). Hence $\leq_P$ satisfies (O1) and (O2), with positive cone exactly $P$ by (C3).
Corollary. An ordering of a field is equivalent to the choice of a subset $P$ satisfying (C1), (C2), (C3). The notions "$F$ is orderable", "there is a total order compatible with the field structure", and "there is a positive cone in $F$" coincide.
Proposition. Let $F$ be an ordered field.
(a) Every sum of squares is $\geq 0$, and a sum of squares is $0$ only if every term is $0$.
(b) Every element of $F$ is a difference of two squares.
(c) The set of nonzero squares is contained in $P$, and it equals $P$ if and only if every positive element of $F$ has a square root in $F$; every real-closed field satisfies this, but the condition is strictly weaker than real closedness.
Proof. (a) Each square is $\geq 0$ by the basic-rules theorem, a sum of nonnegative elements is $\geq 0$, and a sum of nonnegative elements is $0$ only if every term is $0$, since a positive term would make the sum positive. (b) In characteristic $\neq 2$, and here $\operatorname{char} F = 0$,
$$ x = \left(\frac{x+1}{2}\right)^2 - \left(\frac{x-1}{2}\right)^2 . $$
(c) If $x = a^2 \neq 0$ then $x > 0$, and the converse holds by hypothesis; for the rationals, $2 = 1^2 + 1^2$ is positive and is not a square, so the inclusion of nonzero squares in $P$ is strict there.
Remark. Positivity is a cone condition, not a square condition: in $\mathbb{Q}$ the positive element $2$ is not a square, although it is a sum of squares. The equality $P = \{$nonzero squares$\}$ is a genuine strengthening of the ordered-field axioms; a real-closed field satisfies it.
Order and the Field Operations
The order is compatible with the field operations, and the facts needed later are the ones already proved in §Basic Rules: negation reverses the order, translation preserves it, and multiplication by a positive element preserves it. The interval notation used below is order-theoretic throughout.
Definition. For $a < b$ in an ordered field $F$ the open interval is the set
$$ (a, b) = \{x \in F : a < x < b\}, $$
and the rays are $(a,\infty)=\{x : x>a\}$ and $(-\infty,a)=\{x:x Proposition. Let $F$ be an ordered field and $a, b, c \in F$. (a) If $a < b$ then $-a > -b$ and $a + c < b + c$. (b) If $a < b$ and $c > 0$ then $ac < bc$. (c) $x \in (a,b)$ if and only if $a < x < b$; an interval is order-convex, and it is nonempty exactly when $a < b$, since it then contains $(a+b)/2$. Proof. (a) and (b) restate (O1) and the proposition of §Ordered Rings and Fields; (c) is the definition together with (d) of the basic-rules theorem. No distance, no metric and no topology is used here: the topology generated by these intervals, its interaction with the field operations, and the metric and uniform structure it carries belong to Topological Rings and Fields, and the completeness of an ordered field belongs to Real-Closed and Complete Ordered Fields. The statements above are read off the order alone. Definition. An ordered field $F$ is Archimedean if for every $x \in F$ there is a positive integer $n$ with $$
n \cdot 1 > x.
$$ Equivalently, the set $\{n \cdot 1 : n \geq 1\}$ is unbounded above in $F$. Theorem. For an ordered field $F$ the following are equivalent. (a) $F$ is Archimedean. (b) For every $\epsilon > 0$ there is $n \geq 1$ with $1/n < \epsilon$. (c) $\mathbb{Q}$ is order-dense in $F$: for all $a < b$ in $F$ there is $q \in \mathbb{Q}$ with $a < q < b$. (d) There is no element $x \in F$ with $x > n$ for all $n \geq 1$, and no element $x > 0$ with $x < 1/n$ for all $n \geq 1$. Proof. (a) $\Rightarrow$ (b): apply (a) to $x = 1/\epsilon$. (b) $\Rightarrow$ (a): given $x > 0$, take $n$ with $1/n < 1/x$, i.e. $n > x$. So (a) and (b) are equivalent, and they are negated exactly by the existence of an element as in (d). (b) $\Rightarrow$ (c): given $a < b$, first choose $n$ with $1/n < b - a$; then the multiples $k/n$ form a chain of step $1/n < b-a$, so some $k$ has $k/n \leq a < (k+1)/n \leq a + 1/n < b$, and $(k+1)/n \in \mathbb{Q}$ lies in $(a,b)$. (c) $\Rightarrow$ (b): if $\mathbb{Q}$ is order-dense and $\epsilon > 0$, the interval $(0,\epsilon)$ contains a rational $q$ with $q > 0$; writing $q = m/n$ with $m \geq 1$ gives $1/n \leq m/n = q < \epsilon$. Corollary. Every Archimedean ordered field contains $\mathbb{Q}$ as an ordered subfield, and in an Archimedean ordered field every element is the supremum of the rationals below it and the infimum of the rationals above it. Proof. Containment of $\mathbb{Q}$ is the basic-rules theorem; order-density is (c); the last statement is the definition of order-density together with the order. Example ($F(t)$ with $t$ infinite). Let $F$ be an ordered field, let $F(t)$ be the rational function field, and define for a nonzero $r \in F(t)$ $$
r > 0 \iff \text{the leading coefficient of } r \text{ is positive},
$$ where $r$ is written as a quotient of polynomials and the leading coefficient is that of the quotient of the leading coefficients. This makes $F(t)$ an ordered field in which $t > q$ for every $q \in F$; in particular $t$ exceeds every integer, so $F(t)$ is non-Archimedean, and $1/t$ is a positive infinitesimal: $0 < 1/t < 1/n$ for every $n \geq 1$. Example (formal Laurent series). The field of formal Laurent series $$
F((t)) = \left\{\sum_{k \geq k_0} a_k t^k : k_0 \in \mathbb{Z},\ a_k \in F\right\}
$$ is ordered lexicographically by the lowest-degree nonzero coefficient: $\sum a_k t^k > 0$ if the least $k$ with $a_k \neq 0$ has $a_k > 0$. Then $t$ is a positive infinitesimal and $t^{-1}$ is infinite, so $F((t))$ is non-Archimedean. This ordering is the one induced by the $t$-adic valuation: it orders the subfield $F(t)$ by making $t$ infinitesimal, and $F((t))$ is the completion of $F(t)$ for that valuation in the sense. It is therefore a different ordering from the one of the previous example, in which $t$ is infinite, and the two fields are different as well — for $F = \mathbb{Q}$ the series field is uncountable while the rational function field is countable. The field $F((t))$ is complete as a valued field, but it is not order-complete: an order-complete ordered field is Archimedean, and $t$ is an infinitesimal here. Example (non-Archimedean vs not-formally-real). The ordering of $F(t)$ above is one of many orderings; the field $\mathbb{C}$ has none, and $\mathbb{Q}(\sqrt2)$ has exactly two, one with $\sqrt2 > 0$ and one with $\sqrt2 < 0$. The number of orderings of a field is the subject of the theory of formally real fields and is not needed here. Theorem. Every ordered field $F$ contains a unique subfield isomorphic to $\mathbb{Q}$, and the order induced on it is the usual order of $\mathbb{Q}$. Consequently $\mathbb{Q}$ is, up to isomorphism of ordered fields, the smallest ordered field. Proof. By the basic-rules theorem, $\operatorname{char} F = 0$, so the map $\mathbb{Z} \to F$, $n \mapsto n \cdot 1$, is injective and extends to an embedding $\mathbb{Q} \to F$ by the universal property of the fraction field. Since $n \cdot 1 > 0$ for $n \geq 1$ and inverses of positive elements are positive, the embedding carries positive rationals to positive elements and negatives to negatives; hence it is order-preserving and its image is the prime field. Uniqueness: any subfield isomorphic to $\mathbb{Q}$ contains the prime field, which is exactly the image of this embedding. Theorem. $\mathbb{Q}$ has exactly one ordering, and relative to it $\mathbb{Q}$ is Archimedean. Proof. In any ordering of $\mathbb{Q}$, the element $1 > 0$ by the basic-rules theorem, so $n > 0$ for all positive integers $n$, and the order on $\mathbb{Q}$ is determined by the positive cone, which is then forced: a rational $m/n$ is positive exactly when $mn > 0$ in the usual sense. Hence the ordering is unique. It is Archimedean: for a rational $x$, choose an integer $k > \lvert x \rvert$, which exists by the well-ordering of $\mathbb{N}$, and then $k \cdot 1 = k > x$. Remark. The uniqueness of the ordering of $\mathbb{Q}$ contrasts with the general case: $\mathbb{Q}(\sqrt2)$ has two orderings and $\mathbb{Q}(t)$ has many, one for each way of specifying the sign of $t$ together with a location for $t$ relative to the rationals. The orderings of a field are the points of a compact space, the real spectrum, whose theory belongs to real algebraic geometry. Definition. An order-preserving map of ordered fields (or of ordered sets) $\varphi : F \to K$ satisfies $x \leq y \implies \varphi(x) \leq \varphi(y)$. An ordered-field embedding is an injective field homomorphism that is order-preserving, and its image is then an ordered subfield. Proposition. Let $F$ and $K$ be ordered fields and let $\sigma : F \to K$ be a field homomorphism such that $x > 0$ in $F$ implies $\sigma(x) > 0$ in $K$. Then $\sigma$ is strictly order-preserving and injective. In particular, the embedding of the prime field $\mathbb{Q}$ into any ordered field is order-preserving. Proof. If $x < y$ then $y - x > 0$, so $\sigma(y) - \sigma(x) = \sigma(y-x) > 0$ and $\sigma(x) < \sigma(y)$; hence $\sigma$ is strictly order-preserving. If $\sigma(x) = \sigma(y)$ with $x \neq y$, then either $x < y$, giving $\sigma(x) < \sigma(y)$, or $x > y$, giving $\sigma(x) > \sigma(y)$, a contradiction; so $\sigma$ is injective. On the prime field, $\sigma(n \cdot 1) = n \cdot 1 > 0$ for $n \geq 1$ and $\sigma(1/n) = \sigma(n)^{-1} > 0$, so positive rationals go to positive elements. Remark. A field homomorphism between ordered fields need not be order-preserving. The nontrivial automorphism of $\mathbb{Q}(\sqrt2)$, namely $\sqrt2 \mapsto -\sqrt2$, with the ordering $\sqrt2 > 0$, sends the positive element $\sqrt2$ to the negative element $-\sqrt2$; it is an automorphism of the field but not of the ordered structure. The automorphisms of an ordered field, meaning order-preserving automorphisms, form a subgroup of $\operatorname{Aut}(F)$; for $\mathbb{Q}$ and for $\mathbb{R}$ this subgroup is trivial. Theorem. Let $F$ be an Archimedean ordered field. Then the map $$
\Phi : F \to \mathbb{R}, \qquad \Phi(x) = \sup\{q \in \mathbb{Q} : q < x\},
$$ is an injective order-preserving field homomorphism; hence $F$ is isomorphic, as an ordered field, to a subfield of $\mathbb{R}$. Proof sketch. The set on the right is a nonempty bounded-above subset of $\mathbb{Q}$ because $F$ is Archimedean, and it defines a real number; the map is order-preserving by construction, additive and multiplicative by the arithmetic of suprema of bounded sets of rationals, and its kernel is $0$ because $\Phi(x) = 0$ forces $\{q : q < x\}$ to be the nonpositive rationals, so $x = 0$. Corollary. Up to ordering-preserving isomorphism, the Archimedean ordered fields are exactly the subfields of $\mathbb{R}$, and $\mathbb{Q}$ is the smallest and $\mathbb{R}$ the largest: every Archimedean ordered field embeds as an ordered subfield of $\mathbb{R}$, and every ordered subfield of $\mathbb{R}$ is Archimedean. Proof. The second statement is that $\mathbb{R}$ is Archimedean, which follows from its construction and is proved in The Real Numbers; the first is the theorem. Remark. The theorem separates the two questions of this article and the next. Whether a field embeds in $\mathbb{R}$ is the Archimedean condition; whether it is $\mathbb{R}$ is a completeness condition, and completeness is treated. The rationals are Archimedean but not complete; the real numbers are Archimedean and complete; and $F(t)$ is neither. An ordered ring is a commutative ring with a total order compatible with addition and multiplication, and an ordered field is an ordered ring that is a field. In an ordered field $1 > 0$, the characteristic is $0$, every nonzero square is positive, inverses of positive elements are positive, the order is order-dense, and $-1$ is not a sum of squares; the last condition is formal reality, and a field is orderable exactly when it is formally real (Artin–Schreier). An ordering is equivalent to a choice of positive cone $P$, a subset closed under addition and multiplication and making $F = (-P) \cup \{0\} \cup P$ a disjoint union; equivalently the order is determined by the positive cone, and every element of a field of characteristic different from $2$ is a difference of squares. The order is compatible with the field operations — negation reverses it, translation preserves it and multiplication by a positive element preserves it — and it is order-dense in the sense that between any two elements there is a third; the topology built on the order, and the compatibility of that topology with the field operations, belong to Topological Rings and Fields. An ordered field is Archimedean when the natural numbers are unbounded, equivalently when $\mathbb{Q}$ is order-dense, equivalently when every positive element exceeds some $1/n$, equivalently when there is no infinite element and no infinitesimal; $F(t)$ ordered by leading coefficients and $F((t))$ ordered lexicographically are the standard non-Archimedean examples, with $t$ infinite in the former and infinitesimal in the latter. Every ordered field contains a unique copy of $\mathbb{Q}$ as its prime field, with the unique ordering of $\mathbb{Q}$, and every Archimedean ordered field embeds as an ordered subfield of $\mathbb{R}$, so the Archimedean ordered fields are exactly the subfields of $\mathbb{R}$.
Archimedean Ordered Fields
Definition and Equivalent Conditions
Non-Archimedean Examples
The Rational Numbers as an Ordered Field
The Prime Field
Embeddings and Order
Order-Preserving Maps
Archimedean Fields Embed in $\mathbb{R}$
Summary
Ordered field
Archimedean
$\mathbb{Q}$ order-dense
Contains infinitesimals
$\mathbb{Q}$
yes
yes
no
$\mathbb{R}$
yes
yes
no
$\mathbb{Q}(\sqrt2)$
yes
yes
no
$F(t)$, $t$ infinite
no
no
yes
$F((t))$, $t$ infinitesimal
no
no
yes
Summary of Notation
Symbol
Meaning
$F$, $K$
Ordered fields
$R$
Ordered ring
$\leq$
Total order compatible with the field operations
$P = \{x : x > 0\}$
Positive cone
$(a,b)$
Open interval in the order
$n \cdot 1$
Integer multiple of the identity, $n \in \mathbb{Z}$
$\mathbb{Q}$
Prime field of every ordered field; uniquely ordered
$\mathbb{R}$
Largest Archimedean ordered field (complete)
$F(t)$
Rational function field, ordered by leading coefficients
$F((t))$
Formal Laurent series field, ordered lexicographically
$\operatorname{char} F$
Characteristic of $F$; equals $0$ if $F$ is ordered
$\operatorname{Frac}(R)$
Fraction field
$\sup$, $\inf$
Supremum, infimum in an ordered set
Further Reading