show/hide this revision's text 2 added 36 characters in body

The localization $RS^{-1}$ always exists due to abstract nonsense: The subfunctor of $\hom(R,-)$ of homomorphisms mapping $S$ to units is representable since it is continuous and the solution set condition is satisfied, so that we can use Freyd's criterion for representability. Specifically, it consists of elements of the form $r_1 s_1^{-1} r_2 s_2^{-1} \dotsc$. dotsc$, and sums of elements of these form. However, for practical uses, one wants elements of the form $r s^{-1}$ (or $s^{-1} r$, or both options) and an easy condition for equality of such fractions (some people put this into the definition of the localization, but this is artificial). This is contained in the Ore condition. You can find this everywhere (just google for "Ore condition"), for example in "An Introduction to Noncommutative Noetherian Rings" by K. R. Goodearl, Robert B. Warfield, Chapter 6.

show/hide this revision's text 1

The localization $RS^{-1}$ always exists due to abstract nonsense: The subfunctor of $\hom(R,-)$ of homomorphisms mapping $S$ to units is representable since it is continuous and the solution set condition is satisfied, so that we can use Freyd's criterion for representability. Specifically, it consists of elements of the form $r_1 s_1^{-1} r_2 s_2^{-1} \dotsc$. However, for practical uses, one wants elements of the form $r s^{-1}$ (or $s^{-1} r$, or both options) and an easy condition for equality of such fractions (some people put this into the definition of the localization, but this is artificial). This is contained in the Ore condition. You can find this everywhere (just google for "Ore condition"), for example in "An Introduction to Noncommutative Noetherian Rings" by K. R. Goodearl, Robert B. Warfield, Chapter 6.