Finding an ideal such that $\mathbb{Z}[x]/I \cong \mathbb{Z}[i]$. On this released exam, it asks at 2g (slightly modified wording):

Give a brief example or show there does not exist an ideal $I$, $I \subseteq \mathbb{Z}[x]$ such that $\mathbb{Z}[x]/I$ is isomorphic to $\mathbb{Z}[i]$ the Gaussian integers.

I have a lot of trouble with quotient groups. I have a somewhat decent suspicion that:
$$I = \langle x^2+1 \rangle$$
Is the solution, since the zero of this polynomial is $i$. Can someone help clarify why this is the case if it is, and point me to the right direction if it is not?
 A: You might for example consider the mapping
$$f :  \mathbb{Z}[X] \to \mathbb{C}$$
defined by $f(P) = P(i)$, and use some "isomorphism theorem"...
A: It is useful to recall the very definitions of the rings in question.
$\mathbb Z[i]$ is by definition the smallest subring of $\mathbb C$ that contains $\mathbb Z$ and also $i$. Its elements are all obtainable by finitely many ring operations, i.e., adding, subtracting, multiplying of integers and/or $i$. Likewise the ring $\mathbb Z[x]$ has a similar property: All elements are obtainable from integers and $x$ in finitely many steps via adding, subtracting, multiplying. Moreover, $\mathbb Z[x]$ has by definition the univresal property that for any ring $R$, element $r\in R$, ring homomrphism $\phi\colon \mathbb Z\to R$, there exists a unique homomorphism $\Phi\colon\mathbb Z[x]\to R$ such that $\Phi|_{\mathbb Z}=\phi$ and $\Phi(x)=r$.
To construct $\Phi\colon\mathbb Z[x]\to\mathbb Z[i]$ such that $\Phi$ is onto it is therefore quite natural to consider as $\phi\colon\mathbb Z\to\mathbb Z[i]$ the embedding and to pick $i$ as image of $x$. Then let $I$ be the kernel of $\Phi$. Per isomorphism theorem, $\mathbb Z[x]/I$ is isomorphic to the image of $\Phi$. As our choice guaranteed surjectivity, we conclude $\mathbb Z[x]/I\cong \mathbb Z[i]$ as desired.
