The question is as follows:
Consider the group $(\Bbb{Z}[x], +)$. Is the group $\Bbb Z[x]/\Bbb Z$ isomorphic to $\Bbb Z[x]$, where $\Bbb Z[x]$ is a polynomial ring?
I am not able to develop any particular isomorphism for this.
I would appreciate any help.
Consider the principal ideal $(x) \subseteq \mathbb{Z}[x]$ as an abelian group. Then, the homomorphism $\phi: (x) \to \mathbb{Z}[x]/\mathbb{Z}$ given by $\phi(f) = [f]$ is an isomorphism. If $\phi(f) = 0$, then given some expansion $f = \sum_{k = 0}^n c_k x^k $ we have that all $c_k$ are zero except $c_0$. Hence, $f \in \mathbb{Z}$. Since $\mathbb{Z} \cap (x) =\{0\}$, it follows that $f = 0$. Thus $\phi$ is injective.
Conversely, if some $[g] \in \mathbb{Z}[x]/\mathbb{Z}$ we can expand $g = \sum_{k = 0}^n b_kx^k$. Then, let $h = \sum_{k = 1}^n b_kx^k \in (x)$. It follows from their definitions that $g - h = b_0 \in \mathbb{Z}$. Hence $\phi(h) = [g]$ proving that $\phi$ is bijective.