Sylvester resultant of two positive-degree binary forms

Let $R$ be a commutative ring (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution) and let $d,e\ge 1$. For $m\ge 0$ let $R[X,Y{]}_m=\{{\sum }_{i=0}^m{c}_i{X}^{m-i}{Y}^i:c_i\in R\}$ be the $R$-module of homogeneous polynomials of total degree $m$ in $X,Y$, with the ordered basis $X^m,X^{m-1}Y,\dots ,X{Y}^{m-1},Y^m\mathrm{.}$

Fix homogeneous elements $F\in R[X,Y{]}_d$ and $G\in R[X,Y{]}_e$ of the nominated degrees $d$ and $e$; the zero polynomial is allowed, and it is homogeneous of every degree. The Sylvester resultant ${\mathrm{Res}}_{d,e}(F,G)\in R$ is the determinant (For $n\ge 1$, the determinant over a commutative ring by the Leibniz formula, and $\mathrm{\mid }\det A\mathrm{\mid }$ for a real matrix) of the $R$-linear map ${\mathrm{\Phi }}_{F,G}\text{\hspace{0.17em}⁣}:R[X,Y{]}_{e-1}\oplus R[X,Y{]}_{d-1}⟶R[X,Y{]}_{d+e-1},(A,B)⟼AF+BG,$ in the ordered bases above: the $e$ basis vectors of the $F$-block $R[X,Y{]}_{e-1}$ are listed first, then the $d$ basis vectors of the $G$-block $R[X,Y{]}_{d-1}$. The target has $d+e$ basis vectors, so the matrix is square.

Since $F$ and $G$ have no constant term when they are nonzero and positive degree, the map is well defined and $R$-linear, and no hypothesis on the leading coefficients of $F$ or $G$ is imposed. For two linear forms $F=aX+bY$ and $G=cX+dY$ the definition gives ${\mathrm{Res}}_{1,1}(F,G)=ad-bc$, because ${\mathrm{\Phi }}_{F,G}(1,0)=aX+bY$ and ${\mathrm{\Phi }}_{F,G}(0,1)=cX+dY$ are the two matrix columns.

The degrees $d,e$ belong to the data: if the coefficient of $X^d$ in $F$ vanishes, then ${\mathrm{Res}}_{d,e}(F,G)$ is still read off the degree-$d$ Sylvester matrix. In particular dehomogenising to $f(T)=F(T,1)$ and $g(T)=G(T,1)$ never replaces $d$ or $e$ by the actual degrees of $f$ or $g$.