I have to find a necessary condition on $k$ so that $\mathbb{Q}(\sqrt{k})$ produces the same field for different values of $k$, where $k$ is a positive real number such that $\sqrt{k}\notin \mathbb{Q}$.
Attempt: It is easy to see that $\mathbb{Q}(\sqrt{p})$ and $\mathbb{Q}(\sqrt{q})$ are different for distinct primes $p,q$, for if, suppose $\sqrt{p} \in \mathbb{Q}(\sqrt{q})$. Then $\sqrt{p}=a+b\sqrt{q}$ for $a,b \in \mathbb{Q}$ $\qquad \implies p=a^2+b^2q+2ab\sqrt{q} \qquad \implies \sqrt{q} \in \mathbb{Q},\,\,$ a contradiction.
Proceeding similarly in the general case, suppose $\mathbb{Q}(\sqrt{k_1})=\mathbb{Q}(\sqrt{k_2})$. Then $\sqrt{k_1}=a+b\sqrt{k_2}$ for some $a,b \in \mathbb{Q}$. Thus, $k_1=a^2+b^2k_2+2ab\sqrt{k_2}$
Clearly $b\neq 0$. If $a \neq 0$, we get $\sqrt{k_2} \in \mathbb{Q}$, a contradiction. Hence the only possibility for the fields to be same is $a=0$ and $\sqrt{k_1}=b\sqrt{k_2}$ and thus $\sqrt{\frac{k_1}{k_2}} \in \mathbb{Q}$
Is this correct or am I missing some justification/generalization?
The question asks to cover all cases where $k\in\mathbb{R}^+$.
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Had the question imposed $k\in\mathbb{Q}^+$,
then $\mathbb{Q}(\sqrt{k_1})=\mathbb{Q}(\sqrt{k_2})\iff\dfrac{k_2}{k_1}$ is the square of a rational number.
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But generally for $k\in\overline{\mathbb{Q}}\cap\mathbb{R}^+$,
it's much harder to write down something simple that's equivalent to $\mathbb{Q}(\sqrt{k_1})=\mathbb{Q}(\sqrt{k_2})$.
Thinking along these lines, how would you even characterise $\mathbb{Q}(j_1)=\mathbb{Q}(j_2)$?
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As for transcendental $k$, one has to be careful.
For example, $\mathbb{Q}(\sqrt{k_1})=\mathbb{Q}(\sqrt{k_2})$ does not imply that $\mathbb{Q}(k_1)=\mathbb{Q}(k_2)$. Consider the example $k_1=e^2$ and $k_2=(e+1)^2$.
So one may as well forget the square root and ask what condition is equivalent to $\mathbb{Q}(j_1)=\mathbb{Q}(j_2)$, for $j_1$ and $j_2$ transcendental. The answer is $j_1$ is either $aj_2+b$ or $\dfrac{1}{aj_2+b}$, where $a\in\mathbb{Q}-\{0\}$ and $b\in\mathbb{Q} $.