The binary Sylvester resultant detects a common geometric projective root

The inclusion $k\to K$ is a unital ring homomorphism, and an element of the field $k$ is zero exactly when its image in $K$ is zero, so ${\mathrm{Res}}_{d,e}(F,G)=0$ in $k$ if and only if its image in $K$ is zero; by the specialization clause of [L2] that image is ${\mathrm{Res}}_{d,e}(F_K,G_K)$, where $F_K,G_K\in K[X,Y]$ are the coefficientwise images, that is, the same forms viewed over $K$, homogeneous of the same nominated degrees. Evaluating $F$ and $G$ at a point of ${\mathbf{P}}^1(K)$ gives the same field elements as evaluating $F_K$ and $G_K$, so also the common-zero condition is unchanged. Hence it suffices to prove the equivalence for forms over the algebraically closed field $K$, and from here on we work over $K$.

For $m\ge 0$ let ${\delta }_m\text{\hspace{0.17em}⁣}:K[X,Y{]}_m\to K[T]$ be the substitution $H\mapsto H(T,1)$, a $K$-linear map by [L4]. It carries the ordered monomial basis $X^m,X^{m-1}Y,\dots ,Y^m$ to the ordered list $T^m,T^{m-1},\dots ,1$, and it is bijective onto the polynomials of degree $\le m$, with two-sided inverse $p(T)\mapsto Y^{m}p(X\mathrm{/}Y)$; it is also multiplicative in the sense ${\delta }_{m+n}(H{H}^{\mathrm{\prime }})={\delta }_m(H){\delta }_n(H^{\mathrm{\prime }})$. Writing $f:={\delta }_d(F)$ and $g:={\delta }_e(G)$, of degrees $\le d$ and $\le e$, the map $\mathrm{\Psi }\text{\hspace{0.17em}⁣}:(u,v)\mapsto uf+vg$ from $P_{\le e-1}\oplus P_{\le d-1}$ to $P_{\le d+e-1}$ (polynomials in $K[T]$ of the indicated degree bounds) has, in the bases transported by $\delta$, exactly the matrix $M$ of the resultant map $(A,B)\mapsto AF+BG$; hence ${\mathrm{Res}}_{d,e}(F,G)=\det M$ by [L1].

By [L6] the coordinates of $\mathrm{\Psi }(u,v)$ are $M$ times the coordinates of $(u,v)$; coordinates are unique, so $\mathrm{\Psi }$ is bijective if and only if the multiplication map $x\mapsto Mx$ on column vectors is bijective, which by [L6] holds if and only if $M$ is invertible. By [L5] and the fact that $K$ is a field, $M$ is invertible if and only if $\det M\mathrm{\ne }0$. Since domain and target of $\mathrm{\Psi }$ both have dimension $d+e$ by [L1], [L7] gives that $\mathrm{\Psi }$ is bijective if and only if $\mathrm{\Psi }$ is injective. Chaining these equivalences, ${\mathrm{Res}}_{d,e}(F,G)=0$ if and only if $\mathrm{\Psi }$ is not injective.

Suppose that $f$ and $g$ have a common root $\alpha \in K$; if $f=g=0$ every $\alpha$ qualifies, and then $\mathrm{\Psi }$ is the zero map on a nonzero space by [L1], so it is not injective. Otherwise, by [L8] there are $f_1,g_1\in K[T]$ with $f=(T-\alpha )f_1$ and $g=(T-\alpha )g_1$, where $f_1=0$ if $f=0$ and $g_1=0$ if $g=0$; by [L9] and $\mathrm{deg}f\le d$, $\mathrm{deg}g\le e$ we get $\mathrm{deg}{f}_1\le d-1$ and $\mathrm{deg}{g}_1\le e-1$. Then $(g_1,-f_1)$ lies in $P_{\le e-1}\oplus P_{\le d-1}$ and is nonzero, because if $f\mathrm{\ne }0$ then $f_1\mathrm{\ne }0$ by [L9], and if $f=0$ then $g\mathrm{\ne }0$ and $g_1\mathrm{\ne }0$ by [L9]; and $\mathrm{\Psi }(g_1,-f_1)=g_{1}f-f_{1}g=g_1(T-\alpha )f_1-f_1(T-\alpha )g_1=0$. So $\mathrm{\Psi }$ is not injective.

Suppose that $f$ and $g$ have no common root in $K$, that they are not both zero, and that $\mathrm{deg}f=d$ or $\mathrm{deg}g=e$. By [L8] take $A,B$ with $Af+Bg=\gcd (f,g)$; the gcd is monic and divides both $f$ and $g$. If the gcd were not $1$, it would be nonconstant, hence by [L3] would have a root $\alpha \in K$, and by [L8] applied to the divisibility that $\alpha$ would be a common root of $f$ and $g$; so the gcd is $1$ and $Af+Bg=1$. Now let $\mathrm{\Psi }(u,v)=uf+vg=0$. Multiplying $1=Af+Bg$ by $v$ gives $v=A(vf)+B(vg)=A(vf)-B(uf)=f(Av-Bu)$, so $f$ divides $v$; since $\mathrm{deg}v\le d-1$ and, when $\mathrm{deg}f=d$, a nonzero multiple $v=fw$ would have $\mathrm{deg}v=d+\mathrm{deg}w\ge d$ by [L9], we get $v=0$ in that case. Symmetrically, multiplying $1=Af+Bg$ by $u$ gives $u=g(Bu-Av)$, so $g$ divides $u$; since $\mathrm{deg}u\le e-1$ and, when $\mathrm{deg}g=e$, a nonzero multiple $u=gw$ would have $\mathrm{deg}u=e+\mathrm{deg}w\ge e$ by [L9], we get $u=0$ in that case. At least one of the two cases holds. If both cases hold, the two conclusions just displayed give $u=v=0$. If only $\mathrm{deg}f=d$ holds, then $v=0$ and therefore $\mathrm{\Psi }(u,0)=uf=0$, which forces $u=0$ because $K[T]$ is a domain by [L9] and $f\mathrm{\ne }0$; if only $\mathrm{deg}g=e$ holds, the same argument with the roles of $f$ and $g$ interchanged gives $u=0$, and then $vg=0$ forces $v=0$ because $g\mathrm{\ne }0$. Hence $(u,v)=(0,0)$ and $\mathrm{\Psi }$ is injective.

Suppose the $X^d$-coefficient of $F$ and the $X^e$-coefficient of $G$ both vanish. In the homogeneous expansion only the pure terms $X^d$ and $X^e$ avoid a factor $Y$, so $Y$ divides $F$ and $Y$ divides $G$; write $F=Y{F}^{\mathrm{\prime }}$ and $G=Y{G}^{\mathrm{\prime }}$ with $F^{\mathrm{\prime }},G^{\mathrm{\prime }}$ homogeneous of degrees $d-1$ and $e-1$ (the zero polynomials allowed). Then $f=F^{\mathrm{\prime }}(T,1)$ and $g=G^{\mathrm{\prime }}(T,1)$, so $\mathrm{\Psi }(G^{\mathrm{\prime }}(T,1),-F^{\mathrm{\prime }}(T,1))=G^{\mathrm{\prime }}(T,1)F^{\mathrm{\prime }}(T,1)-F^{\mathrm{\prime }}(T,1)G^{\mathrm{\prime }}(T,1)=0$: if $(F,G)\mathrm{\ne }(0,0)$ this displays a nonzero kernel vector because $F^{\mathrm{\prime }}$ or $G^{\mathrm{\prime }}$ is nonzero, and if $F=G=0$ then $\mathrm{\Psi }$ is the zero map on a nonzero space by [L1]. Either way $\mathrm{\Psi }$ is not injective, so ${\mathrm{Res}}_{d,e}(F,G)=0$ by 1.3; and conversely $[1:0]$ is a common zero of $F$ and $G$ in ${\mathbf{P}}^1(K)$ by the chart clause of [L2], because the two top coefficients vanish. Thus in this case the resultant vanishes and a common projective zero exists.

Suppose now that the $X^d$-coefficient of $F$ or the $X^e$-coefficient of $G$ is nonzero, so that $\mathrm{deg}f=d$ or $\mathrm{deg}g=e$ and $f,g$ are not both zero. If $f$ and $g$ have a common root in $K$, then ${\mathrm{Res}}_{d,e}(F,G)=0$ by 1.4 and 1.3; if they do not, then $\mathrm{\Psi }$ is injective by 1.5 and ${\mathrm{Res}}_{d,e}(F,G)\mathrm{\ne }0$ by 1.3. Hence ${\mathrm{Res}}_{d,e}(F,G)=0$ if and only if $f,g$ have a common root in $K$, and by the chart clause of [L2] the common roots of $f$ and $g$ are exactly the affine common zeros $[a:1]$, the point $[1:0]$ being excluded because it is a common zero only when both top coefficients vanish, which is not the case here. So ${\mathrm{Res}}_{d,e}(F,G)=0$ if and only if $F$ and $G$ vanish together at a point of ${\mathbf{P}}^1(K)$.

Every pair $(F,G)$ satisfies the hypothesis of 2.2 or the hypothesis of 2.1, and in both cases ${\mathrm{Res}}_{d,e}(F,G)=0$ is equivalent to the existence of a common zero in ${\mathbf{P}}^1(K)$, which proves the equivalence over the algebraically closed field $K$; undoing the coefficient extension of 1.1 gives the equivalence over the original field $k$. No $k$-rational point was produced anywhere: the points obtained are points of ${\mathbf{P}}^1(K)$ over the supplied algebraic closure.