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**hatsoff** Next show that any nonzero homomorphism $\displaystyle \varphi:\mathbb{Q}[\sqrt{2}]\to\mathbb{Q}[\sqrt{5}]$ has $\displaystyle \varphi(1)=1$. It follows that $\displaystyle \varphi(\sqrt{2})^2=\varphi(2)=\varphi(1)+\varphi( 1)=2$, which means $\displaystyle \varphi(\sqrt{2})=\sqrt{2}$. But if $\displaystyle \sqrt{2}\in\mathbb{Q}[\sqrt{5}]$, then $\displaystyle \sqrt{2}=a+b\sqrt{5}$. Use this to show that $\displaystyle \sqrt{5}$ is rational, a contradiction. So there are no nonzero homomorphism, i.e. no isomorphisms.