How to schat that if $\mathbb{Q}_p(\sqrt{a}) \cong \mathbb{Q}_p(\sqrt{b})$ ($a,b$ non-squares), then $a/b$ is a square?
Approach: I'm able to prove that $\mathbb{Q}_p(\sqrt{a})\neq \mathbb{Q}_p(\sqrt{b})$ as follows : We know
\begin{equation}
\sqrt{a} = x + y \sqrt{b}
\end{equation}
By squaring both sides : $a = x^2 + 2xy\sqrt{b} + y^2 b$. Then $x$ or $y$ is zero. If $x\neq 0$, then $y=0$ and hence $a=x^2$. Contradiction. So $a=by^2$. But how to prove that they are not isomorphic?
Answer
Of course you are correct to be careful about the difference between two field extensions being equal or only isomorphic. But in the case of a quadratic extension, this is the same notion after we fixed an algebraic closure (If we do not fix such, then it does not make sense at all to say what equal means).
Let $K$ be a field and $K \subset L \subset \overline K$ a quadratic extension. Of course $L=K(a)$ for some $a \in \overline K$.
Another quadratic extension $K(b)$ for $b \in \overline K$ is isomorphic to $L$ if and only if the minimal polynomials of $a$ and $b$ coincide. But the minimal polynomial of $a$ is quadratic, thus then $a$ and $b$ are either equal or $-(a+b)$ is equal to the linear coefficient. In both cases $K(a)=K(b)$ follows.
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