Intersection of two subfields of the Rational Function Field in characteristic $0$ Let $K=F(x)$ be the rational function field over a field $F$ of characteristic $0$, let $L_1=F(x^2)$, and $L_2=F(x^2+x)$. How to show that $L_1\cap L_2 = F$?

Hint $\:$ Exploit parity: $\rm\:h(x) = f(x^2+x)\in F[x]$ is even $\rm (h(-x) = h(x))$ $\rm\:\Rightarrow\: f\in F\:$  is constant, since otherwise its highest degree term yields an odd term, namely 
$$\rm\:f_n (x^2+x)^n\! + f_{n-1} (x^2+x)^{n-1}+\cdots\:  =\ f_n x^{2n}\! + n\:f_n\:x^{2n-1}\! + g(x),\ \ deg\ g\: \le\: 2n\!-\!2$$
The odd monomial $\rm\:x^{2n-1}$ has nonzero coefficient $\rm\:n, f_n\ne 0\:$ $\Rightarrow$ $\rm\:n\:f_n \ne 0\:$ by $\rm\:char\ F = 0,\:$ hence $\rm\:f(x)\:$ is not even. Ditto for rational functions: if $\rm\: h(x) = f(x^2+x)/g(x^2+x)\:$ is even then $\rm\:h(-x) = h(x)\:$  $\Rightarrow$ $\rm\:f(x^2+x)g(x^2-x) = f(x^2-x)g(x^2+x)\:$ is even, so $\rm\in F,\:$ so $\rm\:f,g\in F.$
