Existence of quotient rings

Posted: December 27, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
Tags: , , , ,

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.


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s