## 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 $s_1r_1 = \cdots = s_n r_n \in S.$