How can I prove that $\mathbb Z_3[i]\cong \mathbb Z_3[x]/\langle x^2+1\rangle$? 
Question:
Let $\mathbb Z_{3}\left [ i \right ]=\left \{ a+bi\mid a,b \in \mathbb{Z}_{3} \right \}.$
Show that the field $\mathbb Z_{3}\left [ i \right ]$ is ring isomorphic to the field $\mathbb Z_{3}\left [ x \right ]/\left \langle x^{2}+1 \right \rangle$

Here's my thought process:

*

*I need to first show that a ring Homomorphism exists from the ring $Z_{3}\left [ i \right ]$ to the Quotient ring.


*Then Check that a bijection exists.


*Would the First Isomorphism theorem for rings work here?
Am I correct?
 A: Start with the map 
$$\begin{align}\Bbb{Z}_3\left[x\right]&\to \Bbb{Z}_3\left[i\right]\\P&\to P(i)\end{align}$$
It is a ring morphism by definition of the substitution that is surjective. Indeed take any $a+bi\in \Bbb{Z}_3\left[i\right]$ it is the image of the polynomial $bx+a$.
The kernel of this morphism is the ideal generated by $x^2+1$  because any polynomial such that $p(i)=0$ has $i$ as a root and therefore $-i$ because it has integer (real) coefficients and is divisible by $x^2+1$ so we can factor and we get the isomorphism required.
A: You are basically correct. If you find a homomorphism between the rings, you only need to show if that particular homomorphism is bijective. (This approach is quite standard and in no way restricted to your particular case.) 
The first isomorphism theorem is now trivial, so it can be left out.
A: Consider the homomorphism $\varphi:\mathbb{Z}_3[x]\to \mathbb{Z}_3[i]$ with 
$$\varphi(x)=i \text{ and } \varphi(n)=n \text{ for all } n\in \mathbb{Z}_3.$$
Then $\varphi(x^2+1)=i^2+1=0.$ Also, if $\varphi(f(x))=0$ for some $f(x)\in \mathbb{Z}_3[X]$, then $f(\varphi(x))=0\Rightarrow f(i)=0\Rightarrow x^2+1|f(x)$.
Hence, $\ker{\varphi}=(x^2+1)$. Also, $\varphi$ surjective, since for any $a+bi$, we have $\varphi(a+bx)=a+bi$. Hence, $\mathbb{Z}_3[i]\cong \mathbb{Z}_3[x]/(x^2+1)$.
In general, let $K$ be a field, if $a$ is a root of some monic irreducible polynomial $f(x)\in K[x]$, then $K[a]\cong K[x]/(f(x))$. We can consider the homomorphism $\varphi:K[x]\to K[a]$ with 
$$\varphi(x)=a \text{ and } \varphi(n)=n \text{ for all } n\in K.$$
Then $\ker{\varphi}=(f(x))$. Here, $f(x)$ is called the minimal polynomial of $a$ over $K$.
