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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 and the fundamental theorem of symmetric polynomials). – Joel Cohen Dec 21 '11 at 19:51
@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)? – Myself Dec 21 '11 at 20:00
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. – Myself Dec 21 '11 at 20:03
@Myself : You're right. My mistake :) – Joel Cohen Dec 21 '11 at 20:06
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 $\phi(z)=(z-u)(z-f(u))(z-f(f(u)))(z-\bar u)(z-f(\bar u))(z-f(f(\bar u)))$ – Thomas Andrews Dec 21 '11 at 20:42

The roots of $\phi$ are the values $u$ such that $f^3(u)=u$ and $f(u)\neq u$.

If $u$ is a root of $\phi$, then $f(u)$ and $f^2(u)$ are also different roots of $\phi$, since:

$$f^3(f(u))=f^4(u)=f(u)$$ $$f^3(f^2(u)) = f^5(u)=f^2(u)$$

Applying the chain rule to both sides of $f^3\circ f = f\circ f^3$, we see that:

$$(f^3)'(f(z))f'(z) = f'(f^3(z))(f^3)'(z)$$

So if $f^3(z)=z$, then:

$$(f^3)'(f(z)) = (f^3)'(z)$$

So, in particular, if $u$ is a root of $\phi$ then $(f^3)'(u)=(f^3)'(f(u))=(f^3)'(f^2(u)))$

So given two roots in $\{u,f(u),f(f(u))\}$ your expression is not only rational, but zero.

So, now you can partition of the sets of roots of $\phi$ into two disjoint sets $\{u,f(u),f^2(u)\}$ and $\{v,f(v),f^2(v)\}$.

If $g(z)=(f^3)'(z)$, then we know that $g(u)=g(f(u))=g(f^2(u))$ and $g(v)=g(f(v))=g(f^2(v))$.

The final step is to look at the expression $$\sum_{w_1,w_2} {(g(w_1)-g(w_2))^2}$$ where $w_1$ and $w_2$ are taken over all roots of $\phi$. By the above-noted property, this expression is equal to $18(g(u)-g(v))^2$.

But since this expression is symmetric in the roots of $\phi$, and $\phi$ has rational coefficients, this expression is necessarily rational. Therefore, $(g(u)-g(v))^2$ is rational, which is what we needed to prove.

[There are a few more steps you need to fill in here - specifically, that the $\phi$ does not have $f(z)-z$ as a factor, or, alternatively, that $f^3(z)-z$ has no repeated roots. You can prove that by brute force, but I wonder if you can come up with a more general proof for (most?) quadratic $f$.]

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I think the key observation one needs to make is

$$\frac{d}{dX}f^n(X) = 2^n\prod_{i=0}^{n-1} f^i(X).$$

So any question about evaluating the derivative of an iterative power of $f$ at a point of $\overline{\mathbb{Q}}$ can be reduced to an examination of the orbit of that element under the monoid generated by $f.$ Let's persue such an analysis.

Consider $\mathbb{N}$ acting on $\overline{\mathbb{Q}}$ via the iterative powers of $f.$ Let $\Lambda_n$ be the set of fixed points of $f^n.$ Then the action of $\mathbb{N}$ on $\Lambda_n$ factors through $\mathbb{Z}/n.$ It follows that the orbits of the action of $\mathbb{N}$ on $\Lambda_n$ have size a divisor of $n.$ So in the particular case $n =3,$ the orbits have size $1$ or $3.$ Let $\Lambda'_3 = \Lambda_3 \setminus \Lambda_1.$ The elements of $\Lambda'_3$ are exactly the roots of $\phi,$ so $|\Lambda'_3| = 6.$ It follows $\mathbb{N}$ partitions $\Lambda'_3$ into two disjoint subsets of size 3.

Now consider $G_{\mathbb{Q}} = Gal(\overline{\mathbb{Q}}/\mathbb{Q}).$ Because $\Lambda'_3$ is cut out as the vanishing of a polynomial over $\mathbb{Q},$ the group $G_{\mathbb{Q}}$ acts on $\Lambda'_3.$ Furthermore, as $\mathbb{N}$ acts through the iterative powers of $f,$ and these maps are defined over $\mathbb{Q},$ the $G_{\mathbb{Q}}$ action and $\mathbb{N}$ action on $\Lambda'_3$ commute. It follows the $G_{\mathbb{Q}}$ action on $\Lambda'_3$ induces an action on the set of orbits of $\Lambda'_3$ under $\mathbb{N}.$

Let $z_1,z_2 \in \Lambda'_3$ and $O_{z_1}$ and $O_{z_2}$ be the orbits of $z$ and $w,$ respectively, under $\mathbb{N}.$ Then

$$(\frac{df^3}{dX}(z_1) - \frac{df^3}{dX}(z_2))^2 = 64(\prod_{i=0}^{2} f^i(z_1) - \prod_{i=0}^{2} f^i(z_2))^2 = 64(\prod_{p \in O_{z_1}} p - \prod_{p \in O_{z_2}} p)^2.$$

It follows

$$(\frac{df^3}{dX}(z_1) - \frac{df^3}{dX}(z_2))^2 = 0$$

if $O_{z_1} = O_{z_2}.$ On the other hand, if $O_{z_1} \ne O_{z_2},$ then for any $\sigma \in G_{\mathbb{Q}},$

$$\sigma(\frac{df^3}{dX}(z_1) - \frac{df^3}{dX}(z_2))^2 = 64(\prod_{p \in \sigma(O_{z_1})} p - \prod_{p \in \sigma(O_{z_2})} p)$$

which is equal to $64(\prod_{p \in O_{z_1}} p - \prod_{p \in O_{z_2}} p)^2$ if $\sigma$ acts trivially on $\Lambda'_3/\mathbb{N}$ and $64(\prod_{p \in O_{z_2}} p - \prod_{p \in O_{z_1}} p)^2,$ otherwise. Hence, $(\frac{df^3}{dX}(z_1) - \frac{df^3}{dX}(z_2))^2$ is fixed by $G_\mathbb{Q}$. We conclude $(\frac{df^3}{dX}(z_1) - \frac{df^3}{dX}(z_2))^2$ is rational.

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I would be interested to know if such an expression being rational has any significance in dynamics. – jspecter Dec 22 '11 at 0:54

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