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. 


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