Prove that $\operatorname{res}(A(ax),B(ax)) = a^{de}\operatorname{res}(A(x),B(x))$, using the fact that $\sum(i-\sigma(i))=0$ I have been asked to prove the following result:
$$\operatorname{res}(A(ax),B(ax)) = a^{de}\operatorname{res}(A(x),B(x))$$
where $a,e$ are the degrees of the polynomials $A(x)$ and $B(x)$, using the fact that $\sum(i-\sigma(i))=0$. Furthermore, we are given the hint
that the product on the main diagonal $a_{1,1}\cdot . . . \cdot a_{n+m,n+m}$ has degree nm.
I understand it has something to do with the permutation with the entries. However, there is clearly a point I am missing.
Very grateful if someone could thoroughly explain their solution.
 A: Before we begin, a note on my convention for Sylvester matrices: I take the version with $e$ columns of $a_i$s followed by $d$ columns of $b_j$s.
Recall that $\det M = \sum_{\sigma\in S_n} \operatorname{sgn}(\sigma)\prod_{i=1}^n m_{i\sigma(i)}$ for a $n\times n$ matrix $M$. Letting $S$ be the Sylvester matrix of the polynomials $A(x)$ and $B(x)$ and $T$ be the Sylvester matrix of the polynomials $A(ax)$ and $B(ax)$, I claim that $$\prod_{i=1}^n t_{i\sigma(i)} = a^{de}\prod_{i=1}^n s_{i\sigma(i)}.$$
The big observation to make here is that when $s_{ij}\neq 0$, then $t_{ij}=a^{i-j}s_{ij}$ or $t_{ij}=a^{i-j+e}s_{ij}$, depending on whether $j\leq e$ or not. Here's an example to help you see this, where $A$ is of degree $3$ and $B$ is of degree $2$: $$S=
\begin{pmatrix} a_0 & & b_0 & & \\
 a_1 & a_0 & b_1 & b_0 & \\
 a_2 & a_1 & b_2 & b_1 & b_0 \\
 a_3 & a_2 & & b_2 & b_1 \\
 & a_3 & & & b_2 \end{pmatrix}, 
T= 
\begin{pmatrix} a_0 & & b_0 & & \\
 aa_1 & a_0 & ab_1 & b_0 & \\
 a^2a_2 & aa_1 & a^2b_2 & ab_1 & b_0 \\
 a^3a_3 & a^2a_2 & & a^2b_2 & ab_1 \\
 & a^3a_3 & & & a^2b_2 \end{pmatrix}.$$
We therefore obtain $$\prod_{i=1}^n t_{i\sigma(i)} = \left(\prod_{i=1}^{e} a^{i-\sigma(i)}s_{i\sigma(i)}\right)\left(\prod_{i=e+1}^{d+e} a^{i-\sigma(i)+e}s_{i\sigma(i)}\right)=a^{de} \prod_{i=1}^n a^{i-\sigma(i)}s_{i\sigma(i)},$$ which simplifies to $$\prod_{i=1}^n t_{i\sigma(i)} = a^{de} \prod_{i=1}^n s_{i\sigma(i)}$$ from the fact $\sum i-\sigma(i)=0$.
