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My strategy is something like this: suppose $\phi: \mathbb{Q}(\sqrt{p}) \to \mathbb{Q}(\sqrt{q})$ is a isomophism such that $\phi(x) = x$ for all $x \in \mathbb{Q}$ and let $\phi(\sqrt{p}) = a + b \sqrt{q}$. Then $p = \phi(p) = \phi(\sqrt{p}^2) = \phi(\sqrt{p})^2 = (a + b \sqrt{q})^2 = a^2 + 2ab \sqrt{q} + b^2 p$. I feel like '$\phi(x) = x$ for all $x \in \mathbb{Q}$' can be proved from the supposition that $\phi: \mathbb{Q}(\sqrt{p}) \to \mathbb{Q}(\sqrt{q})$ is an isomorphism, but I am not sure how to prove it. Can anyone help me?

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Hint $\phi(1)=1$ implies $\phi(n)=n$.. Now use $\phi(mx)=m \phi(x)$ to deduce that $\phi(x)=x$ for all $x \in \mathbb Q$.

Also note that $\phi(\sqrt{p})$ has to be a root of $X^2-p=0$, and the roots of this are...

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