Free ordered field? 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?
 A: 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 $K\mapsto \{\text{algebraic elements}\}$.
A: 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 $K \longrightarrow K$.
