Ore condition

Posted: December 26, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
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 In this section, R is a unitary ring, S \subset R is multiplicatively closed with 0 \notin S and 1_R \in S.

Definition. S is called a left Ore set if Rs \cap Sr \neq \emptyset, for all r \in R, \ s \in S. Similarly S is called a right Ore set if sR \cap rS \neq \emptyset, for all r \in R, \ s \in S.

Remark 1. If Q is the left (resp., right) quotient ring of R with respect to S, then  S is a left (resp., right) Ore set. The reason is that if \varphi : R \longrightarrow Q is the corresponding map, then for any r \in R, \ s \in S we must have \varphi (r) (\varphi (s))^{-1}=(\varphi (s'))^{-1} \varphi (r'), for some r' \in R, \ s' \in S. Thus \varphi (s'r)=\varphi (r's) and hence s'r - rs \in \ker \varphi. So s''(s'r - r's)=0, for some s'' \in S. Hence s''s'r=s''r's \in Rs \cap Sr.

Remark 2. If S is a left Ore set, then I=\{r \in R : \ sr = 0, \ \text{for some} \ s \in S \} is an ideal of R. (Recall that if R has the left quotient ring Q, then I is nothing but the kernel of the corresponding map from R to Q). To see this, we note that I is clearly closed under right multiplication by elements of R. also:

i) for every r,r' \in I: \ r+r' \in I. To prove this, we have sr=s'r'=0, for some s,s' \in S. By Remark 1, Rs \cap Ss' \neq \emptyset. Therefore r_1s=s''s', for some r_1 \in R, \ s'' \in S. So s''s'r=r_1sr=0, \ s''s'r'=0 and hence s''s'(r+r')=0, i.e. r+r' \in I.

ii) for every r \in I, \ r' \in R: \ r'r \in I. To prove this one, we have sr=0, for some s \in S. Now, by Remark 1: Rs \cap Sr' \neq \emptyset. So r''s=s'r', for some r'' \in R, \ s' \in S. Thus 0=r''sr=s'r'r, i.e. r'r \in I.

iii) if S is a right Ore set, then J=\{r \in R : \ rs = 0, \ \text{for some} \ s \in S \} \lhd R.

Remark 3. Suppose S is left Ore and s_1, \cdots , s_n \in S. There exist r_1, \cdots , r_n \in R such that r_1s_1 = \cdots = r_ns_n \in S. The proof is by induction over n: if n = 1, then we may choose r_1=1. Suppose n > 1 and the claim is true for n-1. Choose r_1', \cdots , r_{n-1}' so that r_1's_1 = \cdots = r_{n-1}' s_{n-1}=s \in S. Also there exist r_n \in R and t \in S such that r_ns_n = ts, since Rs_n \cap Ss \neq \emptyset. Let r_j = tr_j' for j=1, \cdots , n-1. Then for all 1 \leq j \leq n-1 we have r_j s_j = t r_j' s_j = ts = r_ns_n and ts \in S. Similarly, if S is right Ore and s_1, \cdots , s_n \in S, then there exist r_1, \cdots , r_n \in R such that s_1r_1 = \cdots = s_n r_n \in S.


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