Let a be an element of a ring R. Let R' be the ring R[x]/(ax-1). Show that every element B in R' can be written as $B=α^kb$ with b in R Let a be an element of a ring R, and let $R' = R[x]/(ax − 1)$ be the ring obtained by adjoining an inverse of a to R. Let α denote the residue of x (the inverse of a in R').
Show that every element of B in R' can be written as $B=α^kb$ with b in R
I have a few questions about this just to make sure I understand the problem. Firstly, can someone briefly explain how ax-1 is the inverse of a in R[x]/(ax − 1)?
Secondly, by residue of x, don't we mean the result of R[x] modulo (ax-1)? or am I misunderstanding? 
so basically α = ax-1? 
And I need to show that every element in this quotient ring can be written as $(ax-1)^kb$? If someone can give a very brief pointer on how to start the proof or how I want to think about this problem, that would be great. 
 A: In congruence language we have $\,x\equiv 1/a\,$ and the idea is simply to put a sum of fractions over a common denominator, e.g. in the degree $\,k=2\,$ case we have
$$ bx^2+cx+d\ \equiv\, \dfrac{b}{a^2}+\dfrac{c}{a}+d\  \equiv\, \dfrac{\overbrace{b + c a + d a^2}^{\large :=\ r\ \in\ R}}{a^2}\, \equiv\, \dfrac{r}{a^2}\, \equiv\ r\, x^2$$
Obviously the same method works for any degree, e.g. as below
$$\begin{align} 
&\qquad\ \ \ r_k x^k + \cdots + r_1 x + r_0\\
\equiv &\ \ x^k a^k(r_k x^k + \cdots + r_1 x + r_0)\\ 
\equiv  &\ \ x^k (\underbrace{r_k + \cdots + r_1 a^{k-1}\!+r_0 a^k}_{\large :=\ r\ \in\ R})\\ 
\end{align}$$
A: Let $\alpha$ denote the residue class of $x$ in $R'$. That is, $\alpha = x + (ax-1)$ where $(ax-1)$ is the ideal generated by $ax-1$. Notice that 
\begin{align*}
[a + (ax-1)]\alpha - [1 + (ax-1)] &= [a + (ax-1)][x + (ax-1)] - [1 + (ax-1)] \\
&= [ax-1] + (ax-1) \\
&= 0 + (ax-1).
\end{align*}
Now, all that coset notation is cumbersome, which is why we defined $\alpha$ in the first place. Let me rewrite that last computation, dropping the coset notation:
\begin{align*}
a\alpha - 1 &= ax-1 \\
&= 0.
\end{align*}
All we are saying is that in the ring $R'$, $x + (ax-1)$ acts as the inverse of $a + (ax-1)$. But to write things more succinctly we like to just think of this thing $\alpha$ that acts as the inverse of $a$.
A: If $\alpha$ denotes the image of $x$ under the homomorphism $R[x]\to R[x]/(ax-1)$, we have that $a\alpha=1$ (we “identify” elements of $R$ with their image under the obvious homomorphism $R\to R[x]/(ax-1)$).
The image of $f(x)=b_0+b_1x+\dots+b_kx^k\in R[x]$ can be written
$$
b_0+b_1\alpha+\dots+b_k\alpha^k=
\alpha^k(a^kb_0+a^{k-1}b_1+\dots+b_k)
$$
If $R$ is an integral domain, then the homomorphism $R\to R[x]/(ax-1)$ is injective; however it need not be. Precisely, $b\in R$ is mapped to $0$ if there exist $k$ with $a^kb=0$.
