## Existence of quotient rings

Posted: December 27, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
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We will assume throughout that $R$ is a ring with unity and $S \neq \emptyset$ a multiplicatively closed subset of $R$ with $0 \notin S$ and $1_R \in S.$

Definition 1. $S$ is called a left (resp. right) denominator set if:

1) $S$ is left (resp. right) Ore;

2) for $r \in R$ and $s \in S$ with $rs = 0,$ there exists $s' \in S$ such that $s'r=0$  (resp. for $r \in R$ and $s \in S$ with $sr = 0,$ there exists $s' \in S$ such that $rs'=0$).

Theorem. The left (resp. right) quotient ring of $R$ with respect to $S$ exists if and only if $S$ is a left (resp. right) denominator set.

Proof. (Sketch) We’ve already proved that if the left (resp. right) quotient ring of $R$ with respect to $S$ exists, then $S$ is a left (resp. right) denominator set. For the converse, assuming that $S$ is a left denominator (the “right” version is the same), we’ll construct the left quotient ring of $R$ by first defining a relation on $S \times R$: we say $(s_1,r_1) \sim (s_2,r_2)$ if and only if there exist $r_1',r_2' \in R$ such that $r_1'r_1=r_2'r_2$ and $r_1's_1=r_2' s_2 \in S.$

Claim. The relation $\sim$ is an equivalence relation.

Proof. The reflexive property is proved by choosing $r_1'=r_2'=1_R.$ Obviously $\sim$ is symmetric. In order to prove that $\sim$ is transitive suppose that $(s_1,r_1) \sim (s_2,r_2) \sim (s_3,r_3).$ So there exist $r_1',r_2'$  such that $r_1'r_1=r_2'r_2$ and $r_1's_1 = r_2's_2 \in S.$ There also exist $r_2'',r_3'' \in R$ such that $r_2''r_2=r_3''r_3$ and $r_2''s_2=r_3''s_3 \in S.$ On the other hand, since $S$ is left Ore, we have $Rr_2's_2 \cap S r_2''s_2 \neq\emptyset$ and thus there exist $r \in R$ and $s \in S$ such that $rr_2's_2 = sr_2''s_2.$ So $(rr_2' - sr_2'')s_2=0$ and hence there exists $t \in S$ such that $t(rr_2' - sr_2'')=0,$ i.e. $trr_2' = tsr_2''.$ Finally if we put $x=trr_1', \ y=tsr_3'',$ then $xr_1=yr_3, \ xs_1=ys_3 \in S,$ i.e. $(r_1,s_1) \sim (r_3,s_3). \ \Box$

Next, we will show the equivalence class of the element $(s,r) \in S \times R$ by $s^{-1}r$ and we let $Q$ be the set of all $s^{-1}r.$ We are going to put a ring structure on $Q.$ Let $\alpha = s_1^{-1}r_1, \ \beta = s_2^{-1}r_2$ be two elements of $Q.$ By left Ore condition, $Rs_1 \cap Ss_2 \neq \emptyset$ and thus there exist $r \in R$ and $s \in S$ such that $rs_1=ss_2=t \in S.$ Now define $\alpha + \beta = t^{-1}(rr_1 + sr_2).$ Also, since $Rs_2 \cap Sr_1 \neq \emptyset,$ there exist $r' \in R$ and $s' \in S$ such that $r's_2=s'r_1.$ Let $s's_1=t' \in S$ and define $\alpha \beta = t'^{-1}r'r_2.$ It is proved that the addition and multiplication that we’ve defined are well-defined and satisfy all the conditions needed to make $Q$ a ring. Let $1=1_R.$ It’s easy to see that $1^{-1}0 = 0_Q$ and $1^{-1}1 = 1_Q.$ Finally define $\varphi : R \longrightarrow Q$ by $\varphi (r)=1^{-1}r.$ See that $\varphi$ is a ring homomorphism. Now $r \in \ker \varphi$ if and only if $(1,r) \sim (1,0)$ if and only if there exist $r_1,r_2 \in R$ such that $r_1r=0$ and $r_1=r_2 = s \in S.$ Thus $\ker \varphi = \{r \in R : \ sr = 0, \text{for some} \ s \in S \}.$ Therefore $Q=S^{-1}R$ is the left quotient ring of $R$ with respect to $S. \ \Box$

Remark. Every finite subset of $S^{-1}R$ can be written as $\{s^{-1}x_1, \cdots , s^{-1}x_n \}.$ This is easy to see: let $\{s_1^{-1}y_1, \cdots s_n^{-1}y_n \} \subset S^{-1}R.$ By Remark 3 in this post, there exist $r_1, \cdots , r_n \in R$ such that $r_1s_1 = \cdots = r_ns_n = s \in S.$ Let $x_j = r_jy_j,$ for every $1 \leq j \leq n$, and see that $(s_j , y_j) \sim (s , x_j).$  So $s_j^{-1}y_j = s^{-1}x_j,$ for all $j$.