I'm reading this and on page 443, Example 2, they said that $\mathbb{Z}\oplus \mathbb{Z}_n\oplus \mathbb{Z}_n\oplus...\cong \mathbb{Z}[x]/(nx) $.
I have three questions:
(1) Is it true that $\mathbb{Z}\oplus \mathbb{Z}_n\oplus \mathbb{Z}_n\oplus ...\cong \mathbb{Z}\oplus (\mathbb{Z}_n[x])$?
(2) Why is $\mathbb{Z}\oplus (\mathbb{Z}_n[x])\cong \mathbb{Z}[x]/(nx)$?
(3) What is the class of $p(x)=x^2+x+1$ in $\mathbb{Z}[x]/(nx)$?
For (1), I think it's true since for a ring $R$, $R\oplus R\oplus ....\cong R[x]$. So, $\mathbb{Z}\oplus \mathbb{Z}_n\oplus \mathbb{Z}_n\oplus ...=\mathbb{Z}\oplus (\mathbb{Z}_n\oplus \mathbb{Z}_n\oplus ...)\cong \mathbb{Z}\oplus (\mathbb{Z}_n[x])$.
I don't know how to do (2) and (3).
For (2), define $f : \mathbb{Z}[x] \to \mathbb{Z} \oplus \mathbb{Z}_n[x]$ by $$f \left( \sum_{k=0}^r a_k x^k \right) = \left( a_0, ~ \sum_{k=1}^r (a_k \bmod n) x^k \right)$$ Then $f$ is surjective and $\mathrm{ker}(f) = nx\mathbb{Z}[x]$, so you can apply the first isomorphism theorem.
It's not clear what kind of answer is expected for (3). The class consists of those polynomials of the form $1+x+x^2+nxq(x)$, where $q(x) \in \mathbb{Z}[x]$. You can describe these as those polynomials $\sum_{k=0}^r a_kx^k$ for which ($r \ge 2$ and) $a_0=1$, $a_1,a_2 \equiv 1 \bmod n$ and $a_k \equiv 0 \bmod n$ for all $k>2$.