Solution: We need to show that the fields $\mathbb{Q} \left( \sqrt{2} \right) $ and $\mathbb{Q} \left( \sqrt{3} \right) $ are not isomorphic. If they are isomorphic, then let $f: \mathbb{Q} \left( \sqrt{2} \right) \to \mathbb{Q} \left( \sqrt{3} \right) $ is an isomorphism. Let \[ f\left( \sqrt{2} \right) = a + b\sqrt{3},~ a,b \in \mathbb{Q} . \] Since $f(1) = 1$, so for any $p \in \mathbb{Z} \setminus \{ 0 \} $, \[ f (p) = p f(1) = p, \text{ and } f\left( \frac{1}{p} \right) = \frac{1}{p}, \] therefore, $f$ fixes the elements of $\mathbb{Q} $. (Note that we only used $f$ being homomorphism). Therefore, \begin{align*} f(2) = 2 & \implies f\left( \sqrt{2} ^2 \right) = 2 \\ & \implies f\left( \sqrt{2} \right) ^2 = 2 \\ & \implies \left( a + b\sqrt{3} \right) ^2 = 2 \\ & \implies a^2 + 3b^2 + 2ab \sqrt{3} = 2. \end{align*} Thus we have \[ a^2 + 3b^2 = 2,~ \text{ and } 2ab = 0 \implies a = 0 \text{ or } b =0. \] If $a = 0$, then \[ b^2 = \frac{2}{3} \implies b = \pm \sqrt{\frac{2}{3}} \notin \mathbb{Q} , \] so $a\neq 0$. Therefore, $b = 0$, which implies \[ a^2 = 2 \implies a = \pm\sqrt{2}, \] which is not a rational number. Thus we arrived at a contradiction and hence the fields $\mathbb{Q} \left( \sqrt{2} \right) $ and $\mathbb{Q} \left( \sqrt{3} \right) $ are not isomorphic.