Why is $\operatorname{res}(fg,h)=\operatorname{res}(f,h)\cdot\operatorname{res}(g,h)$, where $\operatorname{res}$ stands for resultant? I'm learning Computer Algebra and met an exercise asking me to prove that
$$
\operatorname{res}(fg,h)=\operatorname{res}(f,h)\cdot\operatorname{res}(g,h)
$$
where $f(x)$, $g(x)$ and $h(x)$ are polynomials and $\operatorname{res}$ stand for resultant.
I know if we use the following fact:
$$\operatorname{res}(f,g)=\operatorname{lc}(f)^{\operatorname{deg}(g)}\operatorname{lc}(g)^{\operatorname{deg}(f)}\prod_{(x,y):f(x)=0,g(y)=0} (x-y)$$
the proof would become obvious.
However, in our book the resultant was defined as the determinant of the Sylvester matrix. So I just want to find a proof using this definition directly. (I don't want to prove the fact above first.)
(Supposing that $\deg f=m,\deg g= n, \deg h = p$)
I first tried the multiplication of matrices but found that the Sylvester matrix of $(f,h)$ is $(m+p)\times(m+p)$, the Sylvester matrix of $(g,h)$ is $(n+p)\times(n+p)$, thus they cannot be multiplied. I even tried to  extend their Sylvester matrix to $(m+n+p)\times(m+n+p)$. But I still can't get any useful result.
Can you please help? Thank you! 
 A: The identity you are asking is an immediate consequence of the following interpretation of the resultant (since $M_{fg}=M_fM_g$ and $\deg(fg)=\deg(f)+\deg(g)$):
Proposition. For nonzero polynomials $h,f$ in $X$ over a field $K$, one has
$$
  \mathrm{res}(h,f)=\mathrm{lc}(h)^{\deg(f)}\det(M_f),
$$
where $M_f$ is the endomorphism of the $K$-vector space $K[X]/(h)$ defined by multiplication by (the image modulo $h$ of) $f$.
To prove this, fix the polynomial $h=h_mX^m+\cdots+h_0$, fix another
$n\in\mathbf{N}$ (later taken to be the degree of $f$), and consider any
$m$-tuple $(P_1,\ldots,P_m)$ of polynomials of degree less than $m+n$, say $P_i=\sum_{j=0}^{m+n-1}c_{i,j}X^j$. Let $(R_1,\ldots,R_m)$ be their images in $K[X]/(h)$; expressed on the basis $(X^{m-1},\ldots,X^0)$ of that space, the coefficients of $R_i$ are those of the remainder of $P_i$ after division by $h$. Now I claim that $h_m^n\det\nolimits_{(X^{m-1},\ldots,X^0)}(R_1,\ldots,R_m)$ equals
$$
  \left| \begin{matrix}
  h_m&h_{m-1}&\ldots&h_0&0&\ldots&0\\
  0&h_m&h_{m-1}&\ldots&h_0&\ddots&\vdots \\
  \vdots&\ddots&\ddots&\ddots&\ddots&\ddots&0\\
  0&\ldots&0&h_m&h_{m-1}&\ldots&h_0\\
  c_{1,m+n-1}&\ldots&&&\ldots&c_{1,1}&c_{1,0}\\
  c_{2,m+n-1}&\ldots&&&\ldots&c_{2,1}&c_{2,0}\\
  \vdots &&&&&&\vdots \\
  c_{m,m+n-1}&\ldots&&&\ldots&c_{m,1}&c_{m,0}\\
 \end{matrix} \right|.
$$
It suffices to check that as a function of $(P_1,\ldots,P_m)$ the latter determinant


*

*does not change when reducing any of its polynomials modulo $h$, so that it defines a function on $(K[X]/(h))^m$ (this is a consequence of the presence of the first $n$ rows);

*is $m$-linear over $K$ and alternating;

*takes the value $h_m^n$ when $(P_1,\ldots,P_m)=(X^{m-1},\ldots,X^0)$.


Now to obtain the proposition, apply this with $n=\deg(f)$ and $P_i=fX^{m-i}$ for $i=1,\ldots,m$; in this case $\det\nolimits_{(X^{m-1},\ldots,X^0)}(R_1,\ldots,R_m)$ computes $\det(M_f)$ on the basis $(X^{m-1},\ldots,X^0)$.
