An example of a quotient ring

Posted: December 26, 2009 in Noncommutative Ring Theory Notes, Quotient Rings
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Let $A$ be a ring and $\alpha \in \text{Aut}(A).$ Let $R=A[x,\alpha]$ and $S=\{1,x,x^2, \cdots \}.$ Clearly $S$ is a multiplicatively closed subset of $R.$ Let $Q=A[x,x^{-1}, \alpha].$ The claim is that $Q=S^{-1}R.$ To see this we define $\varphi : R \longrightarrow Q$ to be the inclusion map, i.e. $\varphi (f) = f,$ for all $f \in R.$ We will show that the three conditions in the definition of a left qutient ring, mentioned in the previous section, are satisfied:

1) For any $x^n \in S$ we have $(\varphi (x^n))^{-1}=(x^n)^{-1}=x^{-n} \in Q.$

2) Let $g(x) = \sum_{j=m}^n a_j x^j \in Q,$ where $m,n \in \mathbb{Z}$ with $m \leq n.$ If $m \geq 0,$ then $g(x) \in R$ and $g(x) = (\varphi (1))^{-1} \varphi (g(x)).$ If $m < 0,$  let $b_j = \alpha^{-m}(a_j)$ and $f(x)=b_m + \cdots + b_n x^{n-m}.$ See that $g(x) = (\varphi (x^{-m}))^{-1} \varphi (f(x)).$

3) Since $\varphi$ is the inclusion map, we have $\ker \varphi = (0).$ On the other hand if $f(x)=\sum_{j=0}^n c_jx^j \in R,$ then $x^n f(x)=\sum_{j=0}^n \alpha^n (c_j) x^{n+j}$ and hence, since $\alpha$ is an automorphism, we have $x^n f(x) = 0$ if and only if $f (x)=0.$

It’s even easier to prove that $Q=RS^{-1}.$ So $Q$ is both the left and the right quotient ring of $R$ with respect to $S.$