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 $L_1\cap L_2 = F$?

Solutions Collecting From Web of "Intersection of two subfields of the Rational Function Field in characteristic $0$"

The mapping $\sigma_1:x\mapsto -x$ extends uniquely to an $F$-automorphism of $K$. Clearly $L_1$ is contained in the fixed field $Inv(G_1)$ of $G_1=\langle \sigma_1\rangle\le Aut(K/F)$.
Because $|G_1|=2$, we have $[K:Inv(G_1)]=2$. As $K=L_1(x)$ is also a quadratic extension of $L_1$, we can conclude that $L_1=Inv(G_1)$. Similarly, if $\sigma_2$ is the $F$-automorphism of $K$ determined by $x\mapsto -x-1$ (warm thanks to Georges Elencwajg for the correct automorphism), then $G_2=\langle\sigma_2\rangle$ is also cyclic of order two, and $L_2$ is its fixed field.

Assume that $z=p(x)/q(x)\in L_1\cap L_2$, where $p(x),q(x)\in F[x]$ are coprime polynomials. Then $z$ must be fixed by both $\sigma_1$ and $\sigma_2$. But
$$
(\sigma_2\circ\sigma_1)(x)=\sigma_2(\sigma_1(x))=\sigma_2(-x)=-(-x-1)=x+1,
$$
so we must have
$$
\frac{p(x)}{q(x)}=z=(\sigma_2\circ\sigma_1)(z)=\frac{p(x+1)}{q(x+1)}.
$$
By induction we also have, for all $n\in\mathbf{Z}$, $p(x)/q(x)=p(x+n)/q(x+n)$.
Converting that to a polynomial identity, we have for all $n\in\mathbf{Z}$ that
$$
p(x) q(x+n)=p(x+n)q(x).
$$
Assume that $p(x)$ is not a constant polynomial. Then it has a zero $\alpha$ in some finite extension $E$ of $F$, and by our assumption $q(\alpha)\neq0$. From the above identities we get that $p(\alpha+n)=0$ for all $n\in\mathbf{Z}$. As we assumed that $char F=0$, there are infinitely many distinct elements $\alpha+n$ in the field $F(\alpha)$, and we arrive at the absurd conclusion that $p(x)$ has infinitely many zeros. Therefore $p(x)$ must be a constant. Similarly we see that the denominator $q(x)$ must also be a constant.
This proves the claim.

Note that the assumption $char F=0$ was absolutely essential. Indeed, the claim is false without that assumption, as the group generated by $\sigma_1$ and $\sigma_2$ is finite in that case. As a concrete example I offer the following. Assume $p=char F=2$. Then
$$
x^4+x^2=(x^2+x)^2\in L_1\cap L_2.
$$
Note: $\sigma_1$ is the identity mapping when $p=2$, and the extension $K/L_1$ is then purely inseparable. If $p>2$ we easily see that the group $G$ generated by $\sigma_1$ and $\sigma_2$ is the dihedral group of order $2p$, and the fixed field $L=L_1\cap L_2$ of that group satisfies $[K:L]=2p$, and is a transcendental extension of $F$. Georges kindly calculated that in that case we have $L=F((x^p-x)^2)$. This follows from the observation that
$$
\prod_{t=0}^{p-1}(\sigma_2\circ\sigma_1)^t(x)=\prod_{t=0}^{p-1}(x+t)=x^p-x.
$$
As $G=H\cup \sigma_1 H$, where $H=\langle \sigma_2\circ\sigma_1\rangle$, we then get that the element
$$
(x^p-x)\sigma_1(x^p-x)=-(x^p-x)^2
$$
is invariant under all of $G$. Clearly $[K:F((x^p-x)^2)]=2p$, so the claim follows.

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:

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

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!!!

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 $\rm\:f,g\in F.$