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.


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