How can I show that $\mathbb{Q}(t,\sqrt{t^3-t})$ is not purely transcendental over $\mathbb{Q}$ I tried assuming that it is purely transcendental and since its transcendence degree is $1$, then it is generated by a single element. Further we have $E=\mathbb{Q}(x)$ for some $x \in E$, but I am unable to reach to a contradiction. Can anyone help me with this?
I read a solution on MSE, but couldn't understand as it was fast paced.
Edit: I am able to do this, see if anyone can help me complete the argument.
Suppose $E=\mathbb{Q}(x)$ where $x\in E$, then $t,\sqrt{t^3-t} \in E$ implies the existence of rational functions $f,g$ such that $t=f(x)$ and $\sqrt{t^3-t}=g(x)$. Hence $$g^2(x)=f^3(x)-f(x)$$ Further note that we also have $x \in E=\mathbb{Q}(t,\sqrt{t^3-t})=\mathbb{Q}(f(x),g(x))$ and hence $x=h(f(x),g(x))$ where $h$ is a rational function in two variables. But I am unable to proceed from here.