TITLE: Is $\mathbb{Z}[x]/(x^2+2x+1)$ isomorphic to a product of non-trivial rings?
QUESTION [6 upvotes]: As in the title: is there an isomorphism from $R=\mathbb{Z}[x]/(x^2+2x+1)$ to a non-trivial product of rings? I know already that there will be such an isomorphism if and only if there exist non-trivial idempotents in $R$. My thoughts so far have been:

try to rearrange the generator of the ideal $x^2+2x+1=0$ to something of the form $a^2=a$, thus finding a non-trivial idempotent. I can't seem to do this.
show somehow that no such element can exist. I'm not at all sure how I'd go about this.

REPLY [2 votes]: Here's a more general result:

If $R$ is a commutative ring, and $\mathfrak{p}\subset R$ is a prime ideal, then for all $n$, $R/\mathfrak{p}^n$ cannot be written as a non-trivial product of rings.

To prove it, first note the $n=1$ case works because then $R/\mathfrak{p}$ is an integral domain. [Although the proof below works for this case, I think it's nice to call it out separately.]
For the $n>1$ case, suppose that $R/\mathfrak{p}^n\cong S\times T$.  Now every prime ideal of $S\times T$ looks like either $\mathfrak{s}\times T$ or $S\times\mathfrak{t}$. 
  Since $\mathfrak{p}/\mathfrak{p}^n$ is prime in $R/\mathfrak{p}^n$, it looks like, say, $\mathfrak{s}\times T$. We have
\begin{align}
0 &= \left(\mathfrak{p}/\mathfrak{p}^n\right)^n\\
&= (\mathfrak{s}\times T)^n\\
&= \mathfrak{s}^n\times T
\end{align}
which implies $T$ is trivial, and hence the product decomposition was trivial.