TITLE: Intersection of two subfields of the Rational Function Field in characteristic $0$
QUESTION [9 upvotes]: 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 $L_1\cap L_2 = F$?

REPLY [7 votes]: This is a variant of Jyrki Lahtonen's solution, which avoids explicit computations with polynomials.  You have $K=F(x)$ and $L_1=F(x^2)$ and $L_2=F(x^2+x)$, where $F$ is a field of characteristic $0$.  Then $K/L_1$ and $K/L_2$ are Galois, with Galois groups $G_1=\langle\sigma_1\rangle$ and $G_2=\langle\sigma_2\rangle$ where $\sigma_1(x)=-x$ and $\sigma_2(x)=-1-x$.  Next, every element of $G:=\langle\sigma_1,\sigma_2\rangle$ fixes $L_1\cap L_2$. But $G$ is infinite, since $\sigma_1\sigma_2$ maps $x\mapsto x-1$ and thus has infinite order.  Hence $[K:L_1\cap L_2]$ is infinite.  It follows that $L_1\cap L_2=F$, since $[F(x):F(f(x))]$ is finite for any nonconstant $f(x)\in F(x)$ (proof: $x$ is a root of the numerator of $f(X)-f(x)$, and this numerator is a nonconstant polynomial in $X$ with coefficients in $F(f(x))$).
Additional remark: if $f(x)$ and $g(x)$ are rational functions in $F(x)$ for which $F(x)/F(f(x))$ and $F(x)/F(g(x))$ aren't Galois, it is generally quite difficult to determine whether $F(f(x))\cap F(g(x))=F$.  The only situation where this is completely understood is when $F$ has characteristic zero and $f,g$ are polynomials.  Then, thanks primarily to work of Ritt from the 1920's, we know that the intersection is $F$ unless there are polynomials $h,\mu,\nu\in F[x]$ with $\text{deg}(\mu)=\text{deg}(\nu)=1$ for which one of these holds:

 $f=\mu\circ D_n(x,a)\circ h \quad\text{and}\quad g=\nu\circ D_m(x,a)\circ h$
 $f=\mu\circ x^n\circ h \quad\text{and}\quad g=\nu\circ x^i p(x^n)\circ h$
 $f=\mu\circ x^i p(x^n)\circ h \quad\text{and}\quad g=\nu\circ x^n\circ h$.

Here $p(x)\in F[x]$, and $D_n(x,a)$ is the degree-$n$ Dickson polynomial with parameter $a$; this means that $a\in F$ and that $D_n(x,a)$ is the unique polynomial in $F[x]$ satisfying $D_n(x+a/x,a)=x^n+(a/x)^n$.  There is no known analogue of this result for either rational functions in characteristic zero, or polynomials in characteristic $p$.  Such analogues would have important applications in number theory, algebraic geometry, logic, and complex analysis.  For more information about such questions, see for instance this paper and pp.31-32 of this paper.
Added later: let me mention one more result.  If you're given a field $K$, and two subfields $L_1,L_2$ of $K$ such that each $K/L_i$ is finite and separable, you can try to use Galois theory to determine whether $[K:L_1\cap L_2]$ is finite.  For, if $[K:L_1\cap L_2]$ is finite then there is a finite extension $M/K$ such that both $M/L_1$ and $M/L_2$ are Galois.  (This isn't too difficult, but isn't trivial.)  Conversely, if such an $M$ exists then $[K:L_1\cap L_2]$ is finite if and only if the Galois groups of $M/L_1$ and $M/L_2$ generate a finite group.  Now, if such an $M$ exists, you can construct the minimal such $M$ by letting $M_1$ be the Galois closure of $K/L_1$, $M_2$ be the Galois closure of $M_1/L_2$, $M_3$ be the Galois closure of $M_2/L_1$, and so on.  This gives a tower $K\subseteq M_1\subseteq M_2\subseteq M_3\subseteq...$, and $M$ exists if and only if the tower eventually becomes an infinite chain of field equalities, in which case the minimal $M$ is this common field.  Now here's an amazing result due to David Goldschmidt (1980): suppose that both $K/L_1$ and $K/L_2$ have degree 3, and you do this procedure to build $M_1,M_2,...$. If you ever see an instance where $[M_i:K]$ does not divide $2^7$, then the tower $K\subset M_1\subset M_2\subset M_3\subset\dots$ will continue increasing forever!!!