By Way of Ignorance: Ore Localization

Prologue~
" In order to arrive there,
To arrive where you are, to get from where you are not,
You must go by a way wherein there is no ecstasy.
In order to arrive at what you do not know 
You must go by a way which is the way of ignorance.
In order to possess what you do not possess
You must go by the way of dispossession.
In order to arrive at what you are not
You must go through the way in which you are not."
— T. S. Eliot, Four Quartets (Part II: East Coker)

In the last post we dealt with the concept of localization, categorically, this post, as promised earlier, delves into the Ore condition which helps us extend the concept of localization to non-commutative rings. 

Localization of commutative rings relies heavily on commutativity, but for non-commutative rings multiplication does to commute, and the familiar construction breaks down. The Ore condition provides precisely the compatibility needed to recover a workable theory of fractions in this case. 

Verse I: We begin, again~
To begin with our discussion of the concept of localization for non-commutative rings (NCRs), we rely on the Ore condition.
Let us begin our construction by considering a non-commutative ring $R$ and a multiplicative subset $S$. We seek a ring $S^{-1}R$ and a homomorphism $\iota :R \to S^{-1}R$ such that:

1. $\iota(s)$ is a unit for all $s \in S$,
2. every element of $S^{-1}R$ is of the form $a s^{-1}$ (right fractions) or $s^{-1} a$ (left fractions), and
3. the universal property holds.

Verse II: Some terminologies~

Now, a multiplicative subset $S \subseteq R$ satisfies the right Ore condition if for all $r \in R$ and $s \in S$, the intersection $rS \cap sR$ is nonempty: $$\forall r \in R,\, \forall s \in S,\, \exists\, r' \in R,\, s' \in S \text{ such that } rs' = sr' $$ This condition guarantees the existence of common right multiples, allowing right fractions to be represented over a common denominator. 

Given $a s^{-1}$ and $b t^{-1}$, we can find $s' \in S$, $t' \in R$ such that $s t' = t s'$, allowing us to write: $$a s^{-1} + b t^{-1} = (a t' + b s') (s t')^{-1}$$ $S$ is right reversible if for all $r \in R$, $s \in S$:$$rs = 0 \implies \exists\, t \in S \text{ such that } tr = 0 $$

This ensures the equivalence relation on fractions is well-defined and that $\iota:R \to S^{-1}R$ is injective when $S$ consists of regular elements (non-left and non-right zero divisors).

Verse III: Certain assumptions~
Let us assume that $S$ satisfies the right Ore condition. We define the set of right fractions:$$S^{-1}R = \left\{ (a,s) \mid a \in R,\, s \in S \right\} / \sim $$ where $(a,s) \sim (b,t)$ iff $\exists\, u,v \in R$ with $su = tv \in S$ such that $au = bv$. 

Equivalence relation is then expressed as $\frac{a}{s} = \frac{b}{t} \iff \exists\, u \in R,\, v \in R$ such that $su = tv $ and $au = bv$
Thus, $$S^{-1} R = \{ (a,s) \mid a \in R, s \in S\} / \sim.$$

The Ore condition here, substitues for the lack of commutativity in our ring and defines localization on common, 'agreeable' grounds.

Indeed, given two fractions $\frac{a}{s}, \frac{b}{t},$ the Ore condition provides us with elements $u,v \in R$ such that $su=tv \in S.$ 
Both fractions may therefore be expressed over the common denominator $su=tv,$ and we define their sum by $$\frac{a}{s} + \frac{b}{t} = \frac{a u + b v}{su}.$$

Multiplication gets a bit tricky under these circumstances, since the denominator $s$ simply cannot be moved past the numerator $b$. But (again!) using the Ore condition, we have the existence of a $b' \in R$ and $s' \in S$ such that $sb'=bs'$, and this enables us to define the product as $$\frac{a}{s} \cdot \frac{b}{t} = \frac{ab'}{ts'}$$Symmetrically, $S$ satisfies the left Ore condition if: $$\forall r \in R,\, \forall s \in S,\, \exists\, r' \in R,\, s' \in S \text{ such that } s'r = r's $$

This yields a left ring of fractions $RS^{-1}$ with elements of the form $s^{-1}a$.

If $S$ satisfies both left and right Ore conditions, together with the corresponding reversibility conditions, then the left and right constructions agree canonically, $S^{-1}R \cong RS^{-1},$ giving rise to a two-sided localization.

Verse IV: We have arrived~
The Ore localization is the universal solution to inverting $S$ in the category of (not necessarily commutative) rings:

1. For any $f \colon R \to T$ with $f(S)$ invertible, there exists a unique $\tilde{f} \colon S^{-1}R \to T$ with $\tilde{f} \circ \iota = f$.
2. More generally, for a category $\mathcal{C}$ and class of morphisms $\mathcal{Mor}$, the localized category $\mathcal{C}[\mathcal{Mor}^{-1}]$ exists when $\mathcal{Mor}$ satisfies the Ore conditions (calculus of fractions).
   That is to say: the ring-theoretic construction is a special case of a much broader categorical phenomenon. 

If $R$ is commutative, the Ore condition is guaranteed: $rs = sr$ for all $r \in R$, $s \in S$. Thus every multiplicative subset admits localization.

We are essentially saying that given $R$ to be a ring and $S \subseteq R$ to be a multiplicative subset. The following are equivalent:

1. $S$ satisfies the right Ore condition and right reversible condition.
2. There exists a ring $S^{-1}R$ (the right ring of fractions) and homomorphism $\iota :R \to S^{-1}R$ such that: $\iota(s)$ is a unit for all $s \in S$, every element of $S^{-1}R$ is of the form $a s^{-1}$ with $a \in R$, $s \in S$, $\iota$ satisfies the universal property for inverting $S$.

Epilogue~
" And what you do not know is the only thing you know
And what you own is what you do not own
And where you are is where you are not."
— T. S. Eliot, Four Quartets (Part II: East Coker)