On the definition of free algebra and localization of a non-commutative ring Below is a  construction (p15) of the localization of a non-commutative ring $A$ by a subset $S$. 

Construction 3.1 Form the free algebra on a set which is in bijection with $S$ $$A\langle i_s \colon s \in S\rangle$$ and impose the relation that $i_s$ is a two-sided inverse of $s \in S$ for each $s \in S$: $$S^{-1}A := \frac{A\langle i_s \colon s \in S\rangle}{\langle si_s-1, i_s s-1 \colon s \in S \rangle}\,.$$ Then define $\phi A \to S^{-1}A$ by letting $\phi(a)$ be the image of $a$ in $S^{-1}A$. By definition $\phi(S)$ consists of the units of $S^{-1}A$.

What I am confused about is what is precisely meant here by 

Form the free algebra on a set which is in bijection with $S$, $A \langle i_s \colon s \in S \rangle$. 

What is the definition of free algebra over a non-commutative ring $A$? 
 A: There's no standard definition of a free $A$-algebra when $A$ is non-commutative, but my guess is that it can only really mean one thing in this context: the smallest ring that contains $A$ and all of the $i_s$, with no extra relations imposed upon them (in particular, $i_s a \neq a i_s$ for $s\in S, a\in A$). Elements of $A$ multiply with each other in the usual way, and that's all.
This ring is obviously horrendously large. Even after you quotient out by the relation $``i_s = s^{-1}"$, elements of the resulting ring $S^{-1}A$ might still look like $a_1 (s_1 s_2 s_3)^{-1} a_2 (s_4 s_5)^{-1} a_3 \dots a_m s_n^{-1}$, with no further simplification possible. This is a horrible outcome if what we were hoping for (like the commutative case) was a ring consisting of simple fractions like $``a/s"$.
Luckily (see Definition 3.4 and the remark after it), in many 'nice' cases you can in fact simplify everything to a simple left fraction $s^{-1} a$, or a simple right fraction $a's'^{-1}$, or either.
