## 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. $RS^{-1})$.