I want to proof that show that $\mathbb{Q}[x] \backslash \langle x^2-3 \rangle$ is isomorphic to $\mathbb{Q}(\sqrt{3})$ here is my attempt using the fundamental theorem of homomorphisms.
Let's define $\phi:\mathbb{Q}[x] \rightarrow \mathbb{Q}(\sqrt{3})$ by $\phi(f(x))=f(\sqrt{3})$ note that this represents a homomorphism of rings because, given $f(x)$, $g(x)\in \mathbb{Q}[x]$
$$\phi(f(x)+g(x))=(f+g)(\sqrt{3})=f(\sqrt{3})+g(\sqrt{3})=\phi(f(x))+\phi(g(x))$$
And $$\phi(f(x)g(x))=(fg)(\sqrt{3})=f(\sqrt{3})g(\sqrt{3})=\phi(f(x))\phi(g(x))$$
Thus it is a ring homomorphism. We have to
ker($\phi$)=$\{f(x)\in \mathbb{Q}[x]: f(\sqrt{3})=0 \}=\langle x^2-3 \rangle$
Thus, the map $\mu:\mathbb{Q}[x] \backslash \langle x^2-3 \rangle \rightarrow \mathbb{Q}(\sqrt{3})$ given by $\phi(x)=\mu(f(x)+ \langle x^2-3 \rangle)$ is an isomorphism of rings, and therefore $\mathbb{Q}[x] \backslash \langle x^2-3 \rangle$ is isomorphic to $\mathbb{Q}(\sqrt{3})$.
Is this correct? what can I do better? I appreciate any help!