Definition of a quotient ring

Posted: December 26, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
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Throughout this section R is a ring, unitary as usual, and S \neq \emptyset a multiplicatively closed subset of R such that 0 \notin S, \ 1_R \in S.

Definition 1.  A ring Q is said to be a left quotient ring of R with respect to S if there exists a ring homomorphism \varphi : R \longrightarrow Q such that the following conditions are satisfied:

1) \varphi (s) is a unit in Q for all s \in S;

2) every element of Q is in the form (\varphi (s))^{-1} \varphi (r), for some r \in R, \ s \in S;

3) \ker \varphi = \{r \in R: \ sr = 0, \ \text{for some s} \in S \}.

Remark. If rs = 0, for some r \in R, \ s \in S, then s'r =0, for some s' \in S. This is because

0 = \varphi (rs) = \varphi (r) \varphi (s)

and thus \varphi (r)=0, since, by the condition 1), \varphi (s) is a unit of Q.

Definition 2. A ring Q is said to be a right quotient ring of R (with respect to S) if there exists a ring homomorphism \varphi : R \longrightarrow Q such that the following conditions are satisfied:

1) \varphi (s) is a unit in Q for all s \in S;

2) every element of Q is in the form \varphi (r)(\varphi (s))^{-1}, for some r \in R, \ s \in S;

3) \ker \varphi = \{r \in R: \ rs = 0, \ \text{for some s} \in S \}.

Lemma. Suppose f : R \longrightarrow R_1 is a ring homomorphism and Q is a left (resp. right) quotient ring of R with respect to S. If f(s) is a unit in R_1, for every s \in S, then there exists a (unique) ring homomorphism g : Q \longrightarrow R_1 which extends f.

Poof. If Q is a left quotient ring of R and \varphi is the map in Definition 1, then for all r \in R, \ s \in S we  define

g((\varphi (s))^{-1} \varphi (r))=(f(s))^{-1}f(r).

Similarly, if Q is a right quotient ring, then we define  g(\varphi (r) (\varphi (s))^{-1}) = f(r) (f(s))^{-1}, for all r \in R and s \in S. Proving that g is a well-defined ring homomorphism is lengthy. I will only prove that g is well-defined:

Suppose (\varphi (s))^{-1} \varphi (r) = (\varphi (s'))^{-1} \varphi (r'), for some r,r' \in R and s,s' \in S. Then

\varphi (s') (\varphi (s))^{-1} \varphi (r)=\varphi (r'),

which gives us \varphi (r''r)=\varphi (tr'), for some t \in S and r'' \in R which satisfy

(\varphi (t))^{-1} \varphi (r'')= \varphi (s') (\varphi (s))^{-1}.

So r''r - tr' \in \ker \varphi and ts' - r'' s \in \ker \varphi. Hence ur''r=utr', \ vts' = vr''s, for some u, v \in S. Therefore

f(r'')f(r)=f(t)f(r') and f(t)f(s')=f(r'')f(s),

which will give us (f(s))^{-1}f(r)=(f(s'))^{-1}f(r'). \Box

Theorem. A left (resp. right) quotient ring R with respect to S, if it exists, is unique up to isomorphism. If R has a left quotient ring Q and a right quotient ring Q' (with respect to S), then Q \cong Q'.

Proof. An easy consequence of the lemma. \Box

Notation. “The” left (resp. right) quotient ring of R w.r.t. S, if it exists, is called the left (resp. right) localization of R at S and it is denoted by S^{-1}R (resp. RS^{-1}).

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