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 $L_1\cap L_2 = F$?
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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. |
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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 $\rm\:f,g\in F.$ |
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