# 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.

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This is not a ring homomorphism (unless $n=0$). – Olivier Bégassat Aug 16 '12 at 22:39
It's not a ring homomorphism. The correct target is $k[x]/x^{n+1}$. – Qiaochu Yuan Aug 16 '12 at 22:43
It is neither a ring homomorphism if $n=0$ nor if the domain is what Qiaochu proposes :-) The problem is that it is not even well defined, unless $I$ contains some power of $x$. – Mariano Suárez-Alvarez Aug 16 '12 at 22:49
After the edit: you still need to check that rule you used to define the map is well-defined... Notice there are no series in the codomain of your map, so in principle what you wrote does not make any sense: if $g$ is a series in $R[[x]]$, then $g+(x^n)$ does not make sense in $R[x]/(x^n)$. – Mariano Suárez-Alvarez Aug 16 '12 at 22:54
@Mariano Suárez-Alvarez: I'm not sure what you mean, basically you take an infinite series and forget the terms that have powers of $n$ or higher, no? you'll get a polynomial with degree at most $n-1$ which is essentially $R[x]/(x^{n})$ – user10 Aug 16 '12 at 22:57

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