What is the correct way of adjoining an "infinite" point to a totally-ordered field, such that the end result is still totally-ordered?

Question

Let $Q$ denote a totally-ordered field. Now adjoin an infinite element $\infty$ to $Q$, by equipping the field of rational functions $Q(\infty)$ in the indeterminate $\infty$ with the least preorder $\lesssim$ such that the obvious axioms hold:

• If $x \in Q$, then $x \lesssim \infty$.
• If $x,y \in Q$ and $x \leq y$, then $x \lesssim y$.
• If $x \lesssim y$, then $x+z \lesssim y+z$.
• If $0 \lesssim x$ and $0 \lesssim y$, then $0 \lesssim xy$.

Question. Is $\lesssim$ necessarily a total ordering of $Q(\infty)$? (i.e. antisymmetric and linear). And if not, what is the correct way of adjoining an infinite point to a totally-ordered field, such that the end result is still totally-ordered?

Answer 1

For a more general construction : let $\Gamma$ be a totally ordered abelian group, and $\chi: \Gamma\to \{ \pm 1\}$ be a group morphism.

Then we can form the Hahn series field $Q((\Gamma))$ as the set of formal series $\sum_{\gamma\in \Gamma} a_\gamma X^\gamma$ with $a_\gamma\in Q$ such that the support (ie $\{\gamma\in \Gamma\,|\, a_\gamma\neq 0\}$) is well-ordered. The product is then given by usual convolution, which is well-defined precisely thanks to the well-ordering condition on the support( this guarantees that the sums involved in the convolution product will be finite).

For instance, $\Gamma = \mathbb{Z}$ gives $Q((\Gamma)) = Q((X))$ the usual Laurent series field (the well-ordering condition just says that the series have bounded below degree terms).

Then for $f = \sum_{\gamma\in \Gamma} a_\gamma X^\gamma\in Q((\Gamma))$, we define $f\geqslant 0$ iff $\chi(\gamma)a_\gamma\geqslant 0$ in $Q$, where $\gamma$ is the least element of the support of $f$.

This gives a total ordering on $Q((\Gamma))$, and if you specialize to $\Gamma = \mathbb{Z}$ and $\chi:\mathbb{Z}\to \{\pm 1\}$ the reduction mod $2$ morphism, this gives an ordering on $Q((X))$ wich restricts to the ordering you want on $Q(X)$.

(I took this construction from Efrat's book "Valuations, orgerings, and Milnor K-Theory.)

Answer 2

We give a model-theoretic version of the construction. Let $Q$ be an ordered field. Construct the language $L$ as follows. Among the non-logical symbols of $L$ are two binary function symbols (for addition and multiplication) and a binary relation symbol $\lt$. In addition, $L$ has a constant symbol $k_a$ for every element $a$ of $Q$, and an additional constant symbol $B$ (for "big").

We construct a theory $T$ over the language $L$ as follows:

(i) Among the axioms of $T$ are the usual axioms of the theory of ordered fields.

(ii) In addition, $T$ contains as axioms $k_a+k_b=k_{c}$ for every triple $(a,b,c)$ of elements of $Q$ such that $a+b=c$, and $k_a+k_b\ne k_{c}$ for every triple such that $a+b\ne c$. We also add similar axioms for multiplication, and for $\lt$. We also add $k_a\ne k_b$ for all pairs $(a,b)$ such that $a\ne b$. (We are adding the complete diagram of $Q$.)

(iii) Finally, we add $k_a\lt B$ for every $a\in Q$.

It is not hard to see that the above theory $T$ is consistent, since every finite subset of the axioms has $Q$ as a model if the $k_a$ and $B$ are interpreted suitably.

Thus by the Compactness Theorem, the theory $T$ has a model $M$. Any model of $T$ is an ordered field that has a subfield naturally isomorphic to $Q$. The interpretation of $B$ in $M$ is larger than every element of $Q$.

Remark: The above may not look elementary if one has had no exposure to model theory. But if one has, the argument writes itself.

One can make a somewhat more explicit-looking construction by using a suitable ultrapower of $Q$.

Answer 3

Here is another construction.

If $(M,+,0)$ is a commutative monoïd and $A$ is a set, let $M^{(A)}$ denote the set of maps $f: A \rightarrow M$ whose support $s(f) := A - f^{-1}(\{0\})$ is finite. For $f \in M^{(A)}, f \neq 0$, define $d^{\circ}(f) = \max(s(f))$.

If $(G,+,0,<)$ is an ordered commutative group and $(E,<)$ is a non empty linear order, then $(G^{(E)},+,0,\prec)$ is an ordered commutative group, where $+$ is pointwise addition, and $u \prec v \longleftrightarrow (u \neq v \wedge 0 <(v-u)(d^{\circ}(v-u))$. You get a similar construction for an ordered commuative monoïd $(M,+,0,<)$ using its Grothendieck group and then looking at the substructure $(M^{(A)},+,0,\prec)$ of maps whose range is in $M$. The map $d^{\circ}: G^{(A)} - \{0\} \rightarrow E$ behaves like the degree: $d^{\circ}(u+v) \preceq \max(d^{\circ}(u),d^{\circ}(v))$ and there's equality when the degrees differ.

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Let $F$ be an ordered field, let $E$ be a non empty linear order. $(\mathbb{N}^{(E)},\prec)$ is a linear order and $F$ is a commutative ordered group so one can consider the commutative ordered group $F[E]:= F^{(\mathbb{N}^{(E)})}$. Now define a product and an external law $. : F \times F[E] \rightarrow F[E]$ by $\forall f,g \in F[E], \forall x \in F, \forall u \in \mathbb{N}^{(E)}, (fg)(u) = \sum \limits_{u = v+w} f(v)g(w)$, and $(x.f)(u) = xf(u)$.

Then $F[E]$ is an ordered $F$-algebra, $d^{\circ}$ is an epimorphism $(F[E] - \{0\}, \times) \rightarrow (\mathbb{N}^{(E)},+)$ and every element $f \in F[E]$ is equal to $\sum \limits_{u \in s(f)} f(u).\prod \limits_{e \in s(u)} {X_e}^{u(e)}$ where for $e\in E$, $X_e$ is the map $\chi_{\{\chi_{\{e\}}\}}: \mathbb{N}^{(E)} \rightarrow F$.

$F[E]$ is also factorial, and every element in its fraction field denoted $F(E)$ is the quotient of two unique relatively prime polynomials $f,g \in F[E]$ with $g(d^{\circ}(g)) = 1$.

In $F(E)$, every element of the form $X_e, e \in E$ is a strict upper bound of the ($F$-algebra) span of $\left\langle X_{e'} \, e' < e\right\rangle_F$. So when $E$ has only one element, you basically get the field of fractions in one positive infinite inderterminate described by André Nicolas (which is the tiniest satisfying your condition). In general, $F(E)$ is just the ordered field of fractions with indeterminates in $E$ ordered following the order on $E$.

$F(E)$ embeds in an extension $F'$ of $F$ (which is naturally an $F$-algebra) iff $E$ embeds in the poset $(F',<_{al})$ where $x <_{al} y \longleftrightarrow \left\langle x \right\rangle_F < y$.

You can also replace $\mathbb{N}$ with another ordered commutative monoïd $M$ and get those properties except the last one, but you then have to define ${X_e}^m$ for $(e,m) \in E \times M$ as $\chi_{\{m\chi_{\{e\}}\}}$.