Here is the statement of the problem:
Let $(x^2 - 3)$ be the ideal of $\mathbb{Z}[x]$ generated by $x^2 - 3$ and let $\mathbb{Z}[\sqrt{3}] = \{a + b\sqrt{3} : a,b \in > \mathbb{Z}\}$. Prove that $\mathbb{Z}[x]/(x^2 - 3) \cong > \mathbb{Z}[\sqrt{3}]$.
I'm asking this question here because some of these objects are not very familiar to me (and I've having trouble finding similar problems elsewhere). So, I was just wondering if my approach here was correct (and, I suppose, in the "spirit" of the problem—though this is obviously context-dependent).
Here is my approach:
Define a map $\phi: \mathbb{Z}[x] \to \mathbb{Z}[\sqrt{3}]$ by $\phi(f(x)) = f(\sqrt{3})$. It is clear (right?) that this is a well-defined epimorphism. Moreover, $f \in \ker\phi \iff f(\sqrt{3}) = 0 \iff f \in (x^2 - 3)$; and so $\ker\phi = (x^2 - 3)$. Thus, by the First Isomorphism Theorem, $$ \mathbb{Z}[x]/(x^2 - 3) \cong \mathbb{Z}[\sqrt{3}], $$ as desired.