I'm currently learning algebraic geometry from Shafarevich's Basic Algebraic Geometry, and a couple of the exercises ask the reader whether or not certain mappings are birational isomorphisms. I know that a birational isomorphism is simply a rational function whose inverse is a rational function, and I can usually work the algebra out if it is a birational isomorphism, but I have no clue where to go if that doesn't work.
For example, one exercise goes "given $2n$ numbers $\alpha_i(i=1,...,2n) \: \alpha_i \neq 0,$ we set $\alpha_i+\alpha_i^{-1} = a_i.$ Show that the functions $u=x+x^{-1},\:v=y/x^n$ determine a mapping of the curve $$y^2=\prod_{i=1}^{2n}(x-\alpha_i)(x-\alpha_i^{-1})$$ into the curve $$v^2=\prod_{i=1}^{2n}(u-a_i).$$ Is this mapping a birational isomorphism?"
I was easily to verify fairly quickly that it was, in fact, a mapping, but I could not figure out how to show that it isn't a birational isomorphism. In the end, I went to Wolfram Alpha, found the inverse function of $u=x+x^{-1}$, which is $\frac{1}{2}(u\pm\sqrt{u^2-4}),$ and claimed that if it were a birational isomorphism, it would have to be a rational function of $u,$ in which case we could take $u=4$ to imply $\sqrt{12}$ is rational. I do not believe this is a rigorous proof, so I've been trying to figure out how to actually prove it, and I've made basically no headway. Do I just need to algebra harder, or is there some tool I can use to prove the claim?