Free division rings?

If you had a "free division ring" $F$ on a set $X$, and any division ring $R$, then any set theoretic map $X\to R$ would correspond to a unique division ring homomorphism $F\to R$. If $X$ has at least two elements $x\neq y$, let $R=\mathbb{Q}$. Then you cannot extend both a set theoretic map $f\colon X\to R$ that sends $x$ to $0$ and $y$ to $1$, and a set-theoretic map $g\colon X\to R$ that sends $x$ to $1$ and $y$ to $0$: division rings are simple, so any homomorphism must be either one-to-one or the zero map. (I put both maps, in case one wonders whether you can have $x$ correspond to the zero element of $F$). So, no, you cannot have "free division rings", much like you cannot have "free fields".

There is a notion of free division ring, due to Paul Cohn. I don't know a good online reference, but this AMS bulletin article talks about the construction and gives references. They go by the name of "free skew field" rather than "free division ring".

In the usual category of division rings and ring homomorphisms, there is no free division ring in the sense of category theory, but Cohn's free division rings are pretty close. If you consider the category of division rings with specializations as morphisms, then Cohn's construction gives you precisely the free objects.

Let $D, D'$ be division rings. Then a specialization is a homomorphism $R \to D'$, where $R$ is a subring which generates $D$ as a division ring. (In other words, the smallest division ring within $D$ containing $R$ is $D$ itself.) Specializations are a reasonably natural idea. For example, let $C$ be the complex numbers, and $C(x)$ be the field of rational functions over $C$. Then all specializations between $C(x)$ and $C$ are given by sending $p(x)$ in $C(x)$ to the value $p(a)$ for some $a$ in $C$. This is not defined for all rational functions, but only for the rational functions that don't have a pole at $a$.

If you restrict to the category of fields over a fixed base field $k$ and specializations, the free fields are just of the form $k(x_1, \ldots, x_n)$, where $x_1, \ldots, x_n$ are indeterminates. In the noncommutative case, you get something much more complicated, but free division rings can be realized as subrings of the noncommutative analogue of Laurent series.