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The problem is:

Suppose that you have $f(z)=z^2-10 \in \mathbb{Q}(z)$ and denote by $f^3=f\circ f\circ f$. Define $$\phi(z)=\frac{f^3(z)-z}{f(z)-z}.$$ If $z$ and $w$ are roots of $\phi$, then $$[(f^3)'(z) - (f^3)'(w)]^2,$$ is rational.

I tried to resolve this but I couldn't. You can suppose that $f(z) - z$ divides $f^3(z) - z$ and $f(z)-z$ does not have shared root with $\phi$.

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    Your expression is a symmetric polynomial (with rational coefficients) of the roots of $\phi$, so it can be expressed as a polynomial (with rational coefficients) of the coefficients of $\phi$, which has rational coefficients (see http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial and the fundamental theorem of symmetric polynomials).2011-12-21
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    @Joel Cohen: it's an expression in only two of the six roots of $\phi$, how can it be symmetric (i.e symmetric in all six of them)?2011-12-21
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    For the interested, $\phi(z) = z^6 + z^5 - 29z^4 - 19z^3 + 271z^2 + 81z - 809$ and its Galoisgroup is $C_3\times S_3$. If it brings up any ideas.2011-12-21
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    @Myself : You're right. My mistake :)2011-12-21
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    It seems like somehow you should be able to use that if $u$ is a root of $\phi$ then $f(u)$ and $f(f(u))$ are other (non-equal) roots of $\phi$, as are $\bar u$, $f(\bar u)$, and $f(f(\bar u))$. In particular, if there is a complex root, they are all complex, and $\phi(z)=(z-u)(z-f(u))(z-f(f(u)))(z-\bar u)(z-f(\bar u))(z-f(f(\bar u)))$2011-12-21
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    Specifically, the roots of $\phi$ are the values such that $f^3(u)=u$ and $f(u)\neq u$2011-12-21
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    Okay, I think I have a finished proof.2011-12-22

2 Answers 2