A property regarding $\mathbb{Q}(\sqrt{x})$

Question

Given any non-zero $x,y \in \mathbb{Q}$, I would like to show that $\mathbb{Q}(\sqrt{x}) = \mathbb{Q}(\sqrt{y})$ if and only if there exists a rational number z such that $z^2 = \frac{x}{y}$.

My take for one way is that since $\mathbb{Q}(\sqrt{x}) = \{a + b\sqrt{x} \:{:}\: a,b \in \mathbb{Q}\}$ and $\mathbb{Q}(\sqrt{y}) = \{c + d\sqrt{y} \:{:}\: c,d \in \mathbb{Q}\}$, let us choose $b$ and $d$ such that $b\sqrt{x} = d\sqrt{y}$, then $z = \frac{d}{b}$, but I am not sure about the other way.

Any help will be greatly appreciated.

Answer 1

If $\mathbb Q[\sqrt x ] = \mathbb Q[\sqrt y]$ then there exist $a + b\sqrt {y} = \sqrt {x}$ so $\frac a{b\sqrt x} + \frac{\sqrt y}{\sqrt x}= 1$. $\frac {\sqrt x}{\sqrt y} = 1 - \frac a{b\sqrt x}$ so $\frac xy = 1 + \frac {a^2}{b^2 x} - \frac {2a}{b\sqrt x} \in \mathbb Q$.

So either $\sqrt x$ is rational (and thus the result is trivial as $x$ and $y$ are square rations and these aren't extensions of the rationals at all), or $a = 0$. So $b \sqrt{y} =\sqrt{x}$ so $b = \frac {\sqrt x}{\sqrt y}$.

That's one way.

If $b = \frac {\sqrt{x}}{\sqrt{y}}$ then $b\sqrt y = \sqrt x$ and $\mathbb Q[\sqrt x]= \mathbb Q[\sqrt y]$.

The other way

Answer 2

Perhaps it would be worthwhile to argue Galois-wise : let $K=\mathbf Q (\sqrt a)=\mathbf Q (\sqrt b)$ be a quadratic field, with Galois group generated by a single element $s$ of order 2. Then $s(\sqrt a)=-\sqrt a$ and $s(\sqrt b)=-\sqrt b$, so that $s(\sqrt a/\sqrt b)=\sqrt a/\sqrt b$, hence $\sqrt a/\sqrt b \in \mathbf Q $, or equivalently, $a/b \in {{\mathbf Q}^*}^2$. Note that this proof is valid over any field of characteristic $\neq 2$. The advantage is that only the multiplicative structure of the field is used, which allows to avoid the computational mix with the additive structure when using e.g. linear independence arguments. This approach can be extended to multiquadratic fields by considering subgroups (instead of elements) of ${\mathbf Q}^*/{{\mathbf Q}^*}^2$ : this is just a particular case of Kummer theory.