The kernel of the quotient map $R \to R[x]/(ax-1)$ This is a problem from Algebra by Michael Artin.

Let $a$ be an element of a ring $R$, and let $R_0 = R[x]/(ax − 1)$ be
  the ring obtained by adjoining an inverse of $a$ to $R$. Prove that
  the kernel of the map $R → R_{0}$ is the set of elements $b ∈ R$ such
  that $a^n b = 0$ for some $n > 0$.

http://isites.harvard.edu/fs/docs/icb.topic256346.files/Set%202.pdf
There is a solution in the pdf above (labelled Artin 10.5.14), but I cannot follow it yet. I started to be confused after the third line of the proof. And it seems unclear to express $b$ and the image of $b$ in the quotient ring with the same notation.
Could someone show explicitly what the author is doing after then? Many thanks!
 A: Let $f:R\to R_0$ be the map described, which is obtained by restricting the quotient map $\pi:R[x]\to R_0$ to the constant polynomials $R\subseteq R[x]$. If $\alpha\in R$ is in the kernel of $f$, then it is in the kernel of $\pi$, which is the ideal $(ax-1)R[x]$ by definition.  So $\alpha=(ax-1)Q(x)$, for some polynomial $Q(x)$ which we can write in the form $$Q(x)=b_0+b_1x+\dots+b_nx^n.$$  Now we have $$\alpha=(ax-1)Q(x),$$ and remember that this is an equation of polynomials, where the left-hand side is a constant polynomial.  For two polynomials to be equal, their coefficients must be equal.  Expanding out the right-hand side we get $$(ax-1)Q(x)=-b_0+(ab_0-b_1)x+(ab_1-b_2)x^2+\dots+(ab_{n-1}-b_n)x^n+ab_nx^{n+1}.$$
For this to be equal to $\alpha$ as a polynomial, the constant coefficient $-b_0$ must be $\alpha$, and all the other coefficients must be $0$.  That is, $\alpha=-b_0$, $ab_0=b_1$, $ab_1=b_2$, and so on, ending with $ab_n=0$.  Putting these equations together we find:
\begin{align*}
\alpha&=-b_0 \\
a\alpha&=-ab_0=-b_1 \\
a^2\alpha&=-ab_1=-b_2 \\
&\vdots \\
a^n\alpha&=-ab_{n-1}=-b_n \\
a^{n+1}\alpha&=-ab_n=0.
\end{align*}
That is, some power of $a$ multiplied by $\alpha$ is $0$.
(The solution you linked to is basically the same, but has an error where it writes $\alpha=ab_0$ instead of $\alpha=-b_0$.)
