Prove that $\mathbb{Q}[\sqrt{2}]=a+b\sqrt{2}$ is isomorphic to $\mathbb{Q}[x]/(x^2-2)$.

You can prove that they are isomorphic, using the first isomorphism theorem, for rings. Consider the ring homomorphism $\phi:\mathbb{Q}[x] \to \mathbb{Q}[\sqrt{2}]$ such that $\phi\left(\displaystyle\sum_{i=1}^na_ix^i\right)=\displaystyle\sum_{i=1}^na_i{\sqrt{2}}^i$ i.e the substitution map. Clearly, $\phi(a+bx)=a+b\sqrt{2}$ so, $\phi$ is surjective. The kernel is the set of rational polynomials which vanish at $\sqrt{2}$. Since $\mathbb{Q}[x]$ is a PID, and $x^2-2$ is a polynomial of lowest degree vanishing at $\sqrt{2}$, $\ker{\phi}=\langle x^2-2 \rangle$. Thus, $\mathbb{Q}[x]/\langle x^2-2 \rangle \cong \mathbb{Q}[\sqrt{2}]$.