Is is true that $\mathbb{Q}(x^2)(x) = \mathbb{Q}(x^2)[x] = \mathbb{Q}(x)$?

Question

I am working on a proof and want to show this equality. It appears to be true, but I'm not sure of how to show it rigorously. I know the first equality hold since $x \in \mathbb{Q}(x)$ is algebraic over $\mathbb{Q}(x^2)$.

Answer 1

We obviously have $\Bbb Q(x^2)(x) = \Bbb Q(x)$, as $x^2$ belongs to $\Bbb Q(x)$.

Thus it suffices to show that $\Bbb Q(x^2, x) = \Bbb Q(x^2)[x]$.

But this is simly because the element $x$ is algebraic (of degree $2$) over $\Bbb Q(x^2)$, as it is root of the polynomial $T^2 - x^2 \in \Bbb Q(x^2)[T]$.