Resultant of homogeneous polynomials and composition I've been reading through Silverman's Arithmetic of Dynamical Systems and have encountered the following exercise about resultants: Let $K$ be a field, $F(X,Y)$ and $G(X,Y)$ two degree $D$ homogeneous polynomials with coefficients in $K$, and $f(X,Y)$ and $g(X,Y)$ homogeneous polynomials with coefficients in $K$ each of degree $d$.  Define $A(X,Y) := F(f(X,Y),g(X,Y))$ and $B(X,Y) := G(f(X,Y), g(X,Y))$.  Show that 
$$
\mathrm{Res} (A,B) = \mathrm{Res} (F,G) ^d \cdot \mathrm{Res} (f,g) ^{D^2}
$$

I've just finished giving an argument for the case $D=2$ and arbitrary $d$ and it was a nightmare: it took two entire whiteboards, half a sheet of paper, another exercise, and a fact which I don't believe appears anywhere in the text.  If anyone knows of a simpler argument, I would be extremely grateful.  

Definitions and basic properties of resultants included for your convenience: if $P(X,Y) = a_0 X^n + a_1 X^{n-1} Y + \ldots + a_{n} Y^n$ and $Q(X,Y) = b_0 X^m + b_1 X^{m-1} Y + \ldots + b_m Y^m$ are two polynomials, then their resultant is the matrix determinant 
$$
\left| 
\begin{array}{ccccccccc}
a_0 & a_1 & a_2 & \ldots & a_n &  &  &  &  \\
 & a_0 & a_1 & a_2 & \ldots & a_n &  &  &  \\ 
 & & a_0 & a_1 & a_2 & \ldots & a_n &  &  \\
 & & & \ddots & & & & \ddots & \\ 
 & & & & a_0 & a_1 & a_2 & \ldots & a_n \\ 
\hline{} b_0 & b_1 & b_2 & \ldots & b_m &  &  &  &  \\
 & b_0 & b_1 & b_2 & \ldots & b_m &  &  &  \\ 
 & & b_0 & b_1 & b_2 & \ldots & b_m &  &  \\
 & & & \ddots & & & & \ddots & \\ 
 & & & & b_0 & b_1 & b_2 & \ldots & b_m \\  
\end{array}
\right|
$$
where the top "half" has $m$ rows and the bottom "half" has $n$ rows.
If we can factor $P(X,Y) = a_0 \prod_{i = 1} ^n (X - \alpha_i Y)$ and $Q(X,Y) = b_0 \prod_{j = 1} ^m (X - \beta_j Y)$, then (assuming $a_0 b_0 \ne 0$) we have the formula
$$
\mathrm{Res} (P , Q) = a_0 ^{m} b_0 ^n \prod_{i = 1} ^n \prod_{j = 1} ^m (\alpha_i - \beta_j)
$$
Another useful property which appears as an exercise in the text of the same section says that if $m = n$, then
$$
\mathrm{Res} (\alpha P(X,Y) + \beta Q(X,Y) , \gamma P(X,Y) + \delta Q(X,Y)) = (\alpha \delta - \beta \gamma )^{n} \mathrm{Res} (P(X,Y), Q(X,Y))
$$
for all $\alpha , \beta , \gamma , \delta \in K$.
Finally my argument made use of the fact (which is not in the text, near as I can tell) that 
$$
\mathrm{Res} (pq , r) = \mathrm{Res} (p,r) \mathrm{Res} (q , r)
$$
for any homogeneous $p,q,r$.
 A: We have a solution at long last.  
First a very general comment: we can freely move back and forth between the worlds of "homogeneous degree $d$ polynomials in $X,Y$" and "polynomials in $z$ of degree $d$" by homogenizing and de-homogenizing.  In fact what we have is a one-to-one correspondence which preserves all the nice resultant statements we work with.  Strictly speaking, the question only defines the resultant for homogeneous polynomials in $X$ and $Y$, but I bet you can easily guess the correct definition for the "inhomogeneous resultant."  
Here are the facts about resultants I used to solve the problem: let $f(z), g(z) \in K[z]$ be polynomials of degree $d$ and $e$ respectively, and suppose that we can factor 
$$
f(z) = a_0 \prod_{i=1}^d (z - \alpha_i), \ \ g(z) = b_0 \prod_{j=1} ^e (z - \beta_j)
$$
Then we have:
(1) $\mathrm{Res}(f , g) = a_0^e b_0 ^d \prod_{i=1}^d \prod_{j=1}^e (\alpha_i - \beta_j)$; 
(2) $\mathrm{Res}(f_1 f_2 , g) = \mathrm{Res}(f_1 , g) \cdot \mathrm{Res}(f_2 , g)$ for all $f_1 , f_2 \in K[z]-\{0\}$;
(3) if $e=d$, then $\mathrm{Res}(f-\alpha g , f-\beta g) = (\alpha - \beta)^d \cdot \mathrm{Res}(f , g)$, and also $\mathrm{Res}(\lambda f , g) = \lambda^d \mathrm{Res}(f , g)$ and $\mathrm{Res}(f , \mu g) = \mu ^d \mathrm{Res}(f , g)$ for all $\lambda$ and $\mu$. 
These statements have equally valid counterparts for homogeneous resultants according to our "homogenize/de-homogenize" machinery.

Suppose that $F$ and $G$ factor as 
$$
  F(X,Y) = a_0 \prod_{i=1}^D (X-\alpha_i Y), \ \ G(X,Y) = b_0 \prod_{j=1}^D (X- \beta_j Y)
$$
Then by definition 
$$
  A(X,Y) = a_0 \prod_{i=1}^D (f-\alpha_i g), \ \ B(X,Y) = b_0 \prod_{j=1}^D (f- \beta_j g)
$$
Using the above and the properties (1),(2), and (3), we see that 
\begin{align*}
  \mathrm{Res}(A , B) &= \mathrm{Res}\left(a_0 \prod_{i=1}^D (f - \alpha_i g) , b_0 \prod_{j=1}^D (f - \beta_j g)\right) \\[0.5em]
   &= (a_0 b_0)^{dD} \mathrm{Res}\left( \prod_{i=1}^D (f - \alpha_i g) ,  \prod_{j=1}^D (f - \beta_j g)\right) \\[0.5em]
   &= (a_0 b_0)^{dD}  \prod_{i=1}^D \prod_{j=1}^D \left( (\alpha_i - \beta_j)^d \mathrm{Res}(f , g) \right) \\[0.5em]
   &= (a_0 b_0)^{dD} \mathrm{Res}(f , g)^{D^2} \prod_{i=1}^D \prod_{j=1}^D ((\alpha_i - \beta_j)^d) \\[0.5em]
   &=  \mathrm{Res}(f , g)^{D^2} \left((a_0 b_0)^D \prod_{i=1}^D  \prod_{j=1}^D (\alpha_i - \beta_j)\right)^d \\[0.5em]
   &= \mathrm{Res}(f , g)^{D^2} \cdot \mathrm{Res}(F , G)^d  
\end{align*}
as desired.
