I'm trying to obtain this $R[\sqrt d] \cong R[x]/(x^2-d)$ isomorphism.
I was thinking about using the Chinese remainder theorem to obtain this isomorphism (maybe I have a mistake there): $R[\sqrt d] \cong R[x]/(x^2-d) \cong R[x]/(x-\sqrt d) \oplus R[x]/(x+ \sqrt d)$.
But I can't get it from here.
Any help will be greatly appreciated.
It's easy to see that $R[\sqrt{d}] = \{a + b \sqrt{d}: a, b \in R\}$. Now, if $p(\sqrt{d}) + \langle x^2 - d \rangle \in R[x]/\langle x^2 - d \rangle$ then we always have an expression $p(\sqrt{d}) + \langle x^2 - d \rangle = r(\sqrt{d}) + \langle x^2 - d \rangle$ where $r(x)$ is in reduced form (divide $p(x)$ by $x^2 - d$ and then $r(x)$ is just the remainder). The isomorphism is given by the correspondence $r(\sqrt{d}) \leftrightarrow r(\sqrt{d}) + \langle x^2 - d \rangle$
Just out of interest, what is $R$ in your example? I think it needs to be (at least) an integral domain for this to work.
AND: as the commenter pointed out, $x^2 - d$ must be irreducible over $R$.