I am trying to prove the following:
Two field extensions $\mathbb{Q}[\sqrt{a}]$ and $\mathbb{Q}[\sqrt{b}],$ where $a,b \in \mathbb{Q}^\times,$ are isomorphic if and only if $\sqrt{a/b} \in \mathbb{Q}.$
I started backwards, that is I first assumed the fields were isomorphic. So there exists $\phi$ such that \begin{gather} \phi(\sqrt{a}) = s + t\sqrt{b}. \end{gather} This means that this must satisfy the following \begin{gather} a = s^2 + 2st\sqrt{b} + t^2b. \end{gather} But why is it true that one of $s,t$ must be 0. However, I know that if I can show that then it will be clear that $\sqrt{a/b} \in \mathbb{Q}.$ However, the other direction is not at all obvious, does anyone have any hints?
EDIT: So I started with an element of $u + v\sqrt{a}$ and I tried mapping this to $u + v\sqrt{b/a}.$ But this doesn't satisfy the multiplicative property of the isomorphism.