Let $E=F(x)$ for a field $F$ of characteristic $0$. Show that $F(x^2) \cap F(x^2-x) = F$ as subfields of $F(x)$.
I could use a hand with this... Thanks
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1This has been asked here and answered here though with a minor change in the sign. You can easily accomodate for that. – Jyrki Lahtonen May 12 '16 at 06:57
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Here’s another argument, assuming that $x$ is transcendental over $F$, i.e. just an indeterminate.
We have two automorphisms of $E$, namely $\sigma:x\mapsto-x$ and $\tau:x\mapsto1-x$. Both are involutions, so the fixed fields are both of degree two beneath $E$, and as you see, they are $F(x^2)$ and $F(x^2-x)$ respectively.
Certainly any element of $F(x^2)\cap F(x^2-x)$ will be fixed under both $\sigma$ and $\tau$, thus fixed under their composition. What about $\sigma\circ\tau$? It sends $x$ to $x+1$, so in a field of characteristic zero, it’s not torsion, i.e. not of finite order.Let $f(x)$ be a polynomial (or rational function) fixed by $\sigma\circ\tau$. If $\xi$ is a root of $f$ in any finite extension of $F$, then $\xi+n$ is also a root, so $f$ has infinitely many roots or none. Thus it’s constant.
I think you’ll find that in characteristic $p\ne2$, $(x^p-x)^2$ is fixed under both your automorphisms.
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Any non-constant polynomial over F in $x^2$ has only terms of even degree in $x$. Any non-constant polynomial over F in $x^2 - x$ necessarily has terms of odd degree in $x$. In particular, a polynomial $f(y)$ of degree N, evaluated at $x^2 - x$, has a term in $x$ of degree 2N-1 generated by the degree N term in $y$, and no lower-degree term in $y$, evaluated at $x^2 - x$, can generate a term in $x$ of equal degree, so the coefficients of $f(y)$ cannot 'conspire' to cancel out all the odd-degree terms in $x$. It follows that the intersection must be F.
[I think this argument even works when the characteristic is non-zero]
PMar
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