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(All my rings are unital, but not-necessarily commutative.)

According to wikipedia:

Localizing non-commutative rings is more difficult; the localization does not exist for every set S of prospective units. One condition which ensures that the localization exists is the Ore condition.

I don't get this. Suppose $R$ is a ring and $S \subseteq R$ is a subset. Write $S' = \{s' : s \in S\}$ for a set assumed disjoint from $R$, canonically in bijection with $S$. Okay, can't we just write:

$$S^{-1}R = R[S']/\{ss'=1,s's=1 : s \in S\}?$$

(By $R[S']$, I mean the (not-necessarily-commutative) $R$-algebra freely generated by the set $S'$.)

In more ring-theoretic notation:

$$S^{-1}R = R[S']/\left[\left(\bigvee_{s \in S}(ss'-1)\right) \vee \left(\bigvee_{s \in S}(s's-1)\right)\right]$$

(Everything in the above line is dealing with two-sided ideals; so $(k)$ denotes the two-sided ideal generated by $k$, and $\vee$ denotes the join in the poset of two-sided ideals, and similarly with $\bigvee$.)

Question. What goes wrong with this construction? In other words, what does "the localization of $R$ at $S \subseteq R$ doesn't exist" actually mean?

goblin GONE
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1 Answers1

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Wikipedia is being imprecise. The localization always exists in the sense that there is always a ring $S^{-1} R$ equipped with a map $R \to S^{-1} R$ which is universal with respect to the property that the image of every element of $S$ is invertible, and it is constructed in the obvious way by formally adjoining inverses to every element of $S$. The general element of $S^{-1} R$ is a linear combination of elements of the form

$$s_1^{-1} r_1 s_2^{-1} r_2 \dots $$

and what is not true in general is that every element of $S^{-1} R$ is equal to an element of the form $s^{-1} r$ or $r s^{-1}$. The Ore condition is a condition which ensures something like this.

In the more general context of localization of categories, where the same issue arises but even more seriously (in general the localization of a locally small category can a priori fail to even be locally small) see calculus of fractions.

Qiaochu Yuan
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  • How do we write an expression like $s_1^{-1}+s_2^{-1}$ in the form prescribed? (Working on it as we speak.) – goblin GONE Apr 01 '15 at 08:21
  • @goblin: if $s_1$ and $s_2$ commute, that would just be $\frac{s_1 + s_2}{s_1 s_2}$. In general you need some hypothesis, e.g. the Ore condition. – Qiaochu Yuan Apr 01 '15 at 08:27
  • Okay, that particular expression can be written $s_1^{-1}(s_2+s_1)s_2^{-1}.$ But what about the case of $s_1^{-1}+s_2^{-1}+s_3^{-1}$? The way your answer is written, it says that the normal form $s_1^{-1}r_1s_2^{-1}r_2\ldots$ should work without any further assumptions on $R$. – goblin GONE Apr 01 '15 at 08:29
  • @goblin: oh, sorry, I misspoke above. I meant "the general element is a linear combination of elements of the form...." I had in mind the localization of a monoid, or more generally a category. – Qiaochu Yuan Apr 01 '15 at 08:30
  • Sweet, that makes more sense. So essentially that's just an explicit description of the ring that I denote $S^{-1} R$ in my question, right? – goblin GONE Apr 01 '15 at 08:32
  • @goblin: well, in fact it's not at all explicit, because I haven't given you a method for deciding when two elements are equal. In fact this is quite hard in general and one reason you want a nice calculus of fractions is that in that case it can be much easier (e.g. in the commutative case it's easy). – Qiaochu Yuan Apr 01 '15 at 08:36