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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.

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    @user10: there is a big difference between polynomials of degree at most $n-1$ and $R[x]/(x^n)$. They are isomorphic as $R$-modules, but the former is a submodule of $R[x]$ and the latter is a quotient, and there is no natural ring structure on the former (it is not closed under multiplication as a submodule of $R[x]$).2012-08-16

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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 $\bar\phi^{-1}\circ q$.

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    This looks a bit contorted, precisely to avoidn the problems I hinted at in the comments to the question..2012-08-16