## Quotient rings; some facts (1)

Posted: January 1, 2010 in Noncommutative Ring Theory Notes, Quotient Rings
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Throughout $R$ is a ring with unity, $S \subset R$ is a regular submonoid which is left Ore. So $Q,$ the left quotient ring of $R$ with respect to $S,$ exists and $R$ is a subring of $Q.$ The following facts also holds for the “right” version:

1) If $I \lhd_{\ell} Q$, then $I \cap R \lhd_{\ell} R$ and if $J \lhd_{\ell} Q,$ then $I \cap R = J \cap R \Longleftrightarrow I=J.$ Also $I=Q(I \cap R).$

Proof. The first part is trivial. Now suppose that $s^{-1}a=x \in I.$ Then $a = sx \in I \cap R = J \cap R$ and so $x=s^{-1}a \in J.$ Hence $I \subseteq J.$ Similarly $J \subseteq I.$ For the last part, if $x = s^{-1}a \in I,$ then $a=sx \in I \cap R$ and so $x=s^{-1}a \in Q(I \cap R).$

2) If $I \lhd_{\ell} R,$ then $QI \lhd_{\ell} Q$ and $QI=\{s^{-1}a : \ s \in S, \ a \in I \}.$ It is also true that $QI \cap R = \{a \in R : \ sa \in I \ \text{for some} \ s \in S \}.$

Proof.  Again, the first part is trivial. To prove the non-trivial side of the second part, let $x = \sum_{i=1}^n x_i a_i \in QI,$ where $x_i \in Q$ and $a_i \in I.$ As we’ve showed before, there exist some $s \in S$ and $b_i \in R$ such that $x_i = s^{-1}b_i$ and thus $x = s^{-1}a,$ where $a=\sum_{i=1}^n b_ia_i \in I.$ For the last part, suppose that $s^{-1}b \in QI \cap R,$ where $s \in S, \ b \in I,$ by the second part. So we have $s^{-1}b=a \in R$ and thus $sa = b \in I.$

3) For any $I, J \lhd_{\ell} R$ we have $I \cap J = (0) \Longrightarrow QI \cap QJ = (0).$

Proof. Let $x=s^{-1}a = t^{-1}b \in QI \cap QJ,$ where $a \in I, \ b \in J.$ We have $ts^{-1} \in Q$ and thus $ts^{-1}=u^{-1}c,$ for some $u \in S, \ c \in R.$ Then, since $ts^{-1}a=b,$ we get $ca=ub \in I \cap J = (0),$ which gives us $b=0$ because $u$ is a unit in $Q$ (or because $u$ is regular in $R$). Thus $x = 0.$

4) If $R$ is left Noetherian, then so is $Q.$ If $I \lhd R$ and $Q$ is left Noetherian, then $QI \lhd Q.$

Proof. For the first part, first note that, by 1) and 2), left ideals of $Q$ are exactly in the form $QI,$ where $I$ is a left ideal of $R.$ Let $I \lhd_{\ell} R$ and $I=\sum_{i=1}^n Ra_i.$ Then $QI=\sum_{i=1}^n Qa_i.$ For the second part, let $I$ be a two-sided ideal of $R$ and $s \in S.$ Then $I \subseteq Is^{-1} \subseteq Is^{-2} \subseteq \cdots$ and thus $QI \subseteq QIs^{-1} \subseteq QIs^{-2} \subseteq \cdots.$ So, since $Q$ is left Noetherian, there exists some $n \geq 0$ such that $QIs^{-n} = QIs^{-n-1},$ and thus $QI=QIs^{-1}.$ Therefore $QIs^{-1}a = QIa \subseteq QI,$ for all $s \in S, \ a \in R.$ Hence $QIQ \subseteq QI$ and so $QI$ is a two-sided ideal of $Q.$ (In fact $QIQ=QI.$)

5) If $R$ is prime, then so is $Q.$

Proof. Suppose that $x_1Qx_2=(0),$ for some $x_1=s_1^{-1}a_1, x_2=s_2^{-1}a_2 \in Q.$ Then $s_1^{-1}a_1(as_2)s_2^{-1}a_2 = 0,$ for all $a \in R.$ Thus $s_1^{-1}a_1aa_2 = 0,$ which gives us $a_1aa_2=0,$ i.e. $a_1Ra_2=(0).$ Therefore either $a_1=0$ or $a_2=0,$ which means either $x_1=0$ or $x_2=0.$ This also proves that if $R$ is semiprime, then so is $Q.$