Let $p$ be a prime number. Show that $\Bbb Z_p[i]$ is isomorphic to $\Bbb Z_p[x]/\langle x^2+1\rangle$.
My attempt:
Define $$\phi : \Bbb Z_p[x] \to \Bbb Z_p[i]$$ by $\phi\big(f(x)\big)=f(i)$. Then, it is trivial to show $\phi$ is well defined, homomorphism and onto.
So, I try to show $\ker\phi=\langle x^2+1\rangle$. For given $f(x) \in \ker\phi$ , $f(i)=0$. Then $f(x)$ has quadratic factor $x^2+1$ such that $f(x)=(x^2+1)h(x)$ for some $h(x) \in \Bbb Z_p[x]$. It means that $f$ is in $\langle x^2+1\rangle $.
The other way is trivial to show by definition of principal ideal. Thus, I can conclude $\Bbb Z_p[i]$ is isomorphic to $\Bbb Z_p[x]/\langle x^2+1\rangle $ by the First Isomorphism Theorem.
First, I know $\Bbb Z_p[i]$ is not even an integral domain for $p=5$. So, I want to know which part of above proof is wrong. I'll wait for your comment. Thank you!
Second, by similar way, I can approach to prove that $\Bbb Z_p[\sqrt k]$ is isomorphic to $\Bbb Z_p[x]/\langle x^2-k\rangle$, where $p$ is prime here and $k$ is not square number. Is this right?