Compute $[F(x):F (\frac{x^2}{x-1})]$ in a simpler way.

Question

I have to compute $[F(x):F (\frac{x^2}{x-1})]$. I proved it by setting a isomorphism between them. I think setting homomorphism between two fields cannot be an answer if degree is not $1$. Is there any other simple way to compute it?

Answer 1

Set $y = \frac{x^2}{x-1}$. Then we have $x = \frac12(y\pm\sqrt{y(y-4)})$. This means that $x$ is (one of) the root(s) of the following polynomial in $F(y)[t]$: $$ (2t - y)^2 -y(y-4) = 4t^2 - 4ty + 4y = 4(t^2 - ty + y) $$ so the degree is (at most) $2$.