The natural map $R\mapsto R[t]/(ft-1)$ is injective Let $R$ be a commutative ring and $f\in R$ such that $f$ is not a zero divisor (thanks Darij for correction). How can I rigorously show that 

The canonical map $R\mapsto R[t]/(ft-1)$ is injective.

Context: This step shows up in Rabinowitch trick (when proving Nullstellensatz) where $R=k[x_1, x_2, …, x_n]$. 

An element $a$ belongs to the kernel of the canonical morphism if and only if $a=(ft-1)P(t)$, where $P(t)\in R[t]$. Suppose $a\ne0$ and $P(t)=r_0+r_1t+\dots+r_nt^n$ exists, with $r_n\ne0$. Then
$$
(ft-1)P(t)=
-r_0+(fr_0-r_1)t+(fr_1-r_2)t^2+\dots+(fr_{n-1}-r_n)t^n+fr_nt^{n+1}
$$
so
\begin{align}
r_0&=-a\\
r_1&=-fa\\
r_2&=-f^2a\\
&\;\;\vdots\\
r_n&=-f^na\\
0&=f^{n+1}a
\end{align}
If $f$ is not a zero divisor, we have a contradiction.
Note that if $f$ is a zero divisor, the statement is false: if $fb=0$, then taking $P(t)=-b$, we have
$$
(ft-1)(-b)=fbt+b=b
$$
so $b$ belongs to the ideal generated by $ft-1$.
