TITLE: Is $\mathbb{Z}[x]/(x^2+1)$ isomorphic to $\mathbb{Z}[i]$?
QUESTION [8 upvotes]: Is $\mathbb{Z}[x]/(x^2+1)$  isomorphic to  $\mathbb{Z}[i]$?

My attempt is that try to define a mapping $g$ from $\mathbb{Z}[x]$ to $\mathbb{Z}[i]$ by $g(f(x))= f(i)$, for $f(x)\in\mathbb{Z}[x]$. If it is possible then $\ker g$ is $(x^2+1)$? Am I on the right track? Please Help.

REPLY [11 votes]: You are certainly on the right track.  Let's define a homomorphism $\phi:\mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$ as follows:
$$\phi(f(x)) = f(i)$$
Is $\phi$ a surjective homomorphism?  Certainly: consider any $a+bi \in \mathbb{Z}[i]$.  Note that $(bx + a) \mapsto (a+bi)$.  Next, you should convince yourself that $\ker(\phi) = \langle x^2+1 \rangle$.  At this point, we simply apply the isomorphism theorem:
$$\mathbb{Z}[x]/\langle x^2+1 \rangle \cong \mathbb{Z}[i]$$