When does such isomorphism hold? $$\mathbb{Z}[x]/(x^2+5)\cong \mathbb{Z}[\sqrt{-5}]$$
Edit: Let $R$ be a ring. Consider $p(x)\in R[x]$. Is $R[x]/(p(x)) \cong R[\alpha]$ (where $p(\alpha)=0$) always true? Saw this isomorphism being used many places but never encountered a proper theorem/result proving this isomorphism.
Please give a reference. Thanks.
(There must be a post on this site which answers this query but can't find it)
Hints : take the homomorphism $F :\mathbb{Z}[x] \to \mathbb{Z}[\sqrt{-5}] $ defined by $ F(p(x))=p(\sqrt{-5}) $ with kernel $\langle(x^2+5)\rangle$ .
It's not always true. As a counterexample, take $R=\mathbb Q$, $p=(X^2+5)(X^2-5)$. A root of $p$ is $\sqrt5$, and $\mathbb Q[\sqrt5]$ is known to be a field. But $(p)$ is not a maximal ideal (it's contained in $(X^2+5)$, but $X^2+5$ is not in $(p)$). And $R/I$ is a field if and only if $I$ is maximal. So $\mathbb Q/(p)$ is not a field, while $\mathbb Q[\sqrt5]$ is. So they can't be isomorphic.
In general, for a commutative ring with unity $R$, $R[X]/(p)$ is isomorphic to $R[\alpha]$ iff there exists a surjection $R[X]\longrightarrow R[\alpha]$ with kernel $(p)$. That's the fundamental theorem on homomorphisms. For $R=\mathbb Z$ or if $R$ is a field, this is true if $p$ is irreducible, since then $(p)$ is the kernel of the surjective evaluation homomorphism $f\mapsto f(\alpha)$. For arbitrary rings, it's a bit more complicated, since $R[X]$ might not be a PID, so the kernel might not be a principle ideal of the form $(p)$.
