Showing truncation is a ring homomorphism Let $R$ be a commutative ring with $1$ and let $n$ be any positive integer. Denote the ring of formal power series by $R[[x]]$ and define a map as follows:
$f: R[[x]] \rightarrow R[x]/(x^{n})$
by sending $g$ to $g+(x^{n})$.
It is clear this map preserves sums, but why preserves products? i.e a ring homomorphism.
 A: Let $I$ be the ideal of $R[[x]]$ generated by $x^n$. One can easily see that $I$ is the set of all series in $R[[x]]$ whose first non-zero coefficient is that of $x^m$ with $m\geq n$.
There is a canonical map $q:R[[x]]\to R[[x]]/I$, mapping each series to its coset module $I$, and it is a ring homomorphism.
Now, the inclusion gives us a ring morphism $\phi:R[x]\to R[[x]]$. If $J\subseteq R[x]$ is the ideal of $R[x]$ generated by $x^n$, then clearly $\phi(J)\subseteq I$, so $\phi$ induces a well-defined ring morphism $\bar\phi:R[x]/J\to R[[x]]/I$.
Let us check that $\bar\phi$ is an isomorphism. Its kernel is $\phi^{-1}(I)/J$, but this is zero because $\phi^{-1}(I)$ is actually equal to $J$. On the other hand, $\bar\phi$ is surjective: if $g\in R[[x]]$ is any series, there exist $a_0$, $\dots$, $a_{n-1}\in R$ and $h\in R[[x]]$ such that $g=a_0+a_1x+\cdots+a_{n-1}x^{n-1}+x^nh$, so that $g\cong a_0+a_1x+\cdots+a_{n-1}x^{n-1}\mod I$ and therefore $$\bar\phi(a_0+a_1x+\cdots+a_{n-1}x^{n-1}+J)=g.$$
This does all you want, because what you are calling truncation is the composition $\bar\phi^{-1}\circ q$.
