Let $R$ be a (non-commutative) domain satisfying the (right) Ore condition; i.e. for all $a,b\in R$ one can find $\beta_1,\beta_2\in R$ such that $a\beta_1=b\beta_2$. In the well known construction of Ore, one considers pairs of elements $(a,b)$ (with $a,b\in R$, $b\neq 0$) together with the equivalence relation $(a,b)\sim(c,d)$ iff there exist $\beta_1,\beta_2\in R$ such that $a\beta_1=c\beta_2$ and $b\beta_1=d\beta_2$. The pair $(a,b)$ corresponds to an element of the type $ab^{-1}$. By introducing addition and multiplication on equivalence classes of pairs, one finds a division ring $F$ that contains $R$. An element $a\in R$ is embedded as $(a,1)$, and its inverse is simply $(1,a)$.

Now, assume that $R$ is also a $\ast$-algebra (over $\mathbb{C}$). Is there a canonical way of introducing a $\ast$-operation on $F$, such that $F$ becomes a $\ast$-algebra (over $\mathbb{C}$)?

I tried the following: Since $(ab^{-1})^\ast$ should equal ${b^\ast}^{-1} a^\ast$ one defines $(a,b)^\ast=(1,b^\ast)(a^\ast,1)$. It is easy to see that $((a,b)^\ast)^\ast=(a,b)$ and $(\lambda(a,b)+\mu(c,d))^\ast=\bar{\lambda}(a,b)^\ast+\bar{\mu}(c,d)^\ast$.

The problematic property is to show that $((a,b)(c,d))^\ast=(c,d)^\ast(a,b)^\ast$ (and right now I'm not even sure it is true). Does anyone have experience with (or can point me to a reference for) skew fraction fields of $\ast$-algebras?