There is no such thing as a free field, because there are no morphisms between fields of different characteristics. However, ordered fields seem to be much better behaved: There is an initial object ($\mathbb{Q}$, the rational numbers) and a terminal object (No, the surreal numbers). Does this mean a free ordered field exists, or would the need to be able to answer x > y, where x and y are independent objects in the field, make it non-free?
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4$\begingroup$ I think you are right with your last sentence. If F is the free ordered field on two elements x,y, the map of sets which swaps x and y should induce a morphism of ordered fields $F\rightarrow F$ extending that map of sets. Thus it cannot be order preserving. However it might make sense to talk about the free ordered field on an ordered set, i.e. to consider the left adjoint to the forgetful functor from ordered fields to ordered sets (that assigns to an ordered field F the set $F\setminus \{0\}$). $\endgroup$– HenrikRüpingCommented Oct 28, 2019 at 20:52
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$\begingroup$ So if the functor forgets the + and * functions and just keeps the < relation, there may still be a free object construction? $\endgroup$– ZemylaCommented Oct 28, 2019 at 20:57
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$\begingroup$ If you care only about the field structure of the ordered fields: fields that can be ordered are exactly the formally real fields (which implies characteristic $0$), and if I’m not missing something, the field of rational functions $\mathbb Q(X)$ over a set of indeterminates $X$ is a free formally real field over $X$. $\endgroup$– Emil JeřábekCommented Oct 29, 2019 at 9:10
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2 Answers
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I interpret a "free ordered field" to mean the existence of a left adjoint functor to the$^1$ forgetful functor from ordered fields either to sets or to totally ordered sets. In both cases, the answer is "no". You don't even need two points x,y and compare them, a singleton $X=\lbrace\ast\rbrace$ is enough.
I claim that supposing a "free ordered field over $X$" - let's denote it by $\mathbb{Q}_X$ - cannot exist.
Proof: Let's have a look at where the element $\ast\in\mathbb{Q}_X$ is located in the order. There are two standard orderings on the function field $\mathbb{Q}(T)$; one making $T$ positive, but infinitesimal (i.e. monic polynomials of higher degree get progressively smaller), one making it positive and infinite (i.e. monic polynomials of higher degree get progressively larger). Now consider the (order preserving) map $\lbrace\ast\rbrace \to \mathbb{Q}(T), \ast\mapsto T$. If $\mathbb{Q}_X$ really had the universal property of a free ordered field, both maps would extend to maps of ordered fields $\mathbb{Q}_X\to\mathbb{Q}(T)$. This gives the contradiction that both $\ast < 1$ and $\ast > 1$ would have to hold inside $\mathbb{Q}_X$.
$^1$ In fact this proof shows that there is no left adjoint to a number of functors originating on the category of ordered fields. The functors $K\mapsto K$, $K\mapsto K\setminus\{0\}$, $K\mapsto K\setminus\mathbb{Q}$, $K\mapsto \{\text{transcendental elements}\}$, $K\mapsto \{\text{positive elements}\}$, and many more do not possess left adjoints. And slight variations of the proof also exclude other functors one could think of like $K\mapsto \{\text{algebraic elements}\}$.
answered Oct 28, 2019 at 21:02
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There is no terminal object in the category of ordered fields with non-decreasing field morphisms. Indeed morphisms are injective and there are ordered fields of arbitrary cardinality (strictly speaking $\mathbf{No}$ is not a set, hence not an object in this category). Moreover $\mathbb{R}$ embeds into $\mathbf{No}$ in more than one way so $\mathbf{No}$ is in no way terminal.
No ordered field enjoys a non-empty free object, because for any such field $K$ and $\varnothing\neq X \subseteq K$, the map $X \longrightarrow \{0\}$ if $X\neq \{0\}$ or $X\longrightarrow \{1\}$ if $X=\{0\}$ cannot be extended into an embedding $K \longrightarrow K$.
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1$\begingroup$ What are the two morphisms from $\mathbb{R}$ to $\mathbf{No}$? I'm curious now. $\endgroup$– ZemylaCommented Oct 28, 2019 at 23:45
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2$\begingroup$ @Zemyla : Comments to this answer discuss embedding nonstandard models of the reals into the surreals. $\endgroup$ Commented Oct 29, 2019 at 6:41
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6$\begingroup$ It is much easier to embed $\mathbb Q(X)$, the field of rational functions, into No in many ways: for any infinite surreal $\alpha$ there is a map taking $X$ to $\alpha$. $\endgroup$– WojowuCommented Oct 29, 2019 at 7:41
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2$\begingroup$ @Zemyla : Given an embedding of ordered fields $f:K \supset F \rightarrow K'$ and element $\alpha \in K \setminus F$ which is transcendental over $F$, there are many ways to extend $f$ on $F(\alpha)$. In fact $\alpha$ can be sent to any element $\beta$ of $K'$ with $\{f(x):x<\alpha\}<\beta<\{f(x):x>\alpha\}$. (continued) $\endgroup$– nombreCommented Oct 29, 2019 at 12:52
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2$\begingroup$ (cont.) Embed $\mathbb{Q}$ into $\mathbf{No}$, then send $\pi$ onto $\pi+\varepsilon$ where $\varepsilon \in \mathbf{No}$ is any infinitesimal, then go on using the properties of real-closure or this method for transcendental elements, until $\mathbb{R}$ (or indeed any ordered field) is embedded . This yields a copy of $\mathbb{R}$ which contains $\pi + \varepsilon$. $\endgroup$– nombreCommented Oct 29, 2019 at 12:52