I was asked to show that given a rational function field $E = F(x)$ over $F$ with $char(F) = 0$, the following intersection of its subfields must be $F$: $F(x^2) \cap F(x^2-x) = F$.
I managed to show (using the fundamental theorem of Galois Theorem) to show that the extension $E/(F(x^2) \cap F(x^2-x))$ must be infinite (thus either not finitely-generated or transcendental), but am not sure how to continue.
$ F(x^2) $ is invariant under the automorphism $ x \to -x $ of $ F(x) $, whereas $ F(x^2 - x) $ is invariant under the automorphism $ x \to 1 - x $. Then, any element $ f(x) $ in their intersection must be invariant under the composition of these automorphisms, which is $ x \to -x \to 1 - (-x) = x + 1 $. You can check that the only elements of $ F(x) $ invariant under this translation are the constant rational functions - indeed, if $ a \in F $ is any element and $ f $ is invariant under this translation, then it follows that $ f(x) = f(a) $ has infinitely many solutions in $ F $ (since $ F $ has characteristic $ 0 $), implying that the equality must hold identically, i.e $ f(x) = f(a) $ as rational functions.