Scaling, specialization, and the affine and infinite charts of a binary resultant

The map ${\mathrm{\Phi }}_{uF,vG}$ sends $(A,B)$ to $uAF+vBG$. In the ordered bases of [L1] its matrix is obtained from the matrix of ${\mathrm{\Phi }}_{F,G}$ by multiplying the $e$ columns of the $F$-block by $u$ and the $d$ columns of the $G$-block by $v$. Column multilinearity [L2] therefore gives $\det {\mathrm{\Phi }}_{uF,vG}=u^e{v}^d\det {\mathrm{\Phi }}_{F,G},$ which is the scaling formula.

Let $\phi \text{\hspace{0.17em}⁣}:R\to R^{\mathrm{\prime }}$ be a unital ring homomorphism to a commutative ring $R^{\mathrm{\prime }}$. By [L3] applied to the polynomial rings $R[X]$, $R[X,Y]$ and to $R^{\mathrm{\prime }}$, there is a unique ring homomorphism $\tilde{\phi }\text{\hspace{0.17em}⁣}:R[X,Y]\to R^{\mathrm{\prime }}[X,Y]$ extending $\phi$ with $X\mapsto X$ and $Y\mapsto Y$; it sends every basis monomial $X^{m-i}{Y}^i$ to the corresponding basis monomial and every entry of the matrix of ${\mathrm{\Phi }}_{F,G}$ to the corresponding entry of the matrix of ${\mathrm{\Phi }}_{\phi (F),\phi (G)}$. The Leibniz formula exhibits the determinant as a polynomial with integer coefficients in the matrix entries, so it commutes with $\tilde{\phi }$, and $\phi ({\mathrm{Res}}_{d,e}(F,G))={\mathrm{Res}}_{d,e}(\phi (F),\phi (G))$.

By [L4] every point of ${\mathbf{P}}^1(K)$ is a class $[a_0:a_1]$ with $(a_0,a_1)\mathrm{\ne }(0,0)$. If $a_1\mathrm{\ne }0$ then $[a_0:a_1]=[a:1]$ with $a=a_0\mathrm{/}{a}_1\in K$, and if $a_1=0$ then $a_0\mathrm{\ne }0$ and $[a_0:0]=[1:0]$. Thus every point is $[a:1]$ for some $a\in K$ or is $[1:0]$, and the two kinds are disjoint.

Fix $a\in K$. The point $[a:1]$ is a common zero of $F$ and $G$ exactly when $F(a,1)=G(a,1)=0$. Indeed a general representative of the class is $(\lambda a,\lambda )$ with $\lambda \in K^{\times }$, and by [L5] $F(\lambda a,\lambda )={\lambda }^{d}F(a,1),G(\lambda a,\lambda )={\lambda }^{e}G(a,1);$ as ${\lambda }^d,{\lambda }^e\mathrm{\ne }0$, both vanish exactly when $F(a,1)=G(a,1)=0$. By definition $F(a,1)=f(a)$ and $G(a,1)=g(a)$.

Evaluating the defining linear combinations at $(X,Y)=(1,0)$ shows that $F(1,0)$ is the coefficient of $X^d$ in $F$ (all remaining monomials have a factor $Y$) and $G(1,0)$ is the coefficient of $X^e$ in $G$. Hence the point $[1:0]$ is a common zero of $F$ and $G$ exactly when both of these coefficients vanish.

Steps 1.3, 1.4 and 1.5 describe every point of ${\mathbf{P}}^1(K)$ and decide when it is a common zero, so the common zeros are exactly the affine zeros $[a:1]$ with $f(a)=g(a)=0$ together with the possible point $[1:0]$ detected by the two vanishing top coefficients. Nothing in the argument replaces $d$ or $e$ by the actual degrees of $f$ or $g$: the zero forms have all coefficients zero, so they contribute the point $[1:0]$ as claimed, and a form whose top coefficient vanishes contributes no condition beyond the vanishing already recorded.