[proofplan] Write $F$ and $G$ as polynomials in $Y$ of degrees $m$ and $n$ and express their resultant as the determinant of the fixed Sylvester matrix over $k[X]$. The forward implication is proved directly at the specialized Sylvester matrix: a common root $b$ gives a nonzero vector annihilated by all Sylvester rows, so the determinant vanishes even if the leading coefficients specialize to zero. For the converse, the degree-preservation hypothesis identifies the evaluated fixed resultant with the ordinary one-variable resultant of $F(a,Y)$ and $G(a,Y)$, and the one-variable resultant criterion gives the required common root. [/proofplan]

[step:Write the Sylvester determinant with fixed $Y$-degrees]Set $m:=\deg_Y F$ and $n:=\deg_Y G$. Since $F$ and $G$ have positive degree in $Y$, we have $m,n\ge 1$. Write $F(X,Y)=\sum_{i=0}^{m} f_i(X)Y^i$ and $G(X,Y)=\sum_{j=0}^{n} g_j(X)Y^j$, where $f_i,g_j\in k[X]$, $f_m\ne 0$, and $g_n\ne 0$. Let $S(X)$ denote the $(m+n)\times(m+n)$ Sylvester matrix whose first $n$ rows are the coefficient rows of \begin{align*} Y^{n-1}F,\ Y^{n-2}F,\ \dots,\ F \end{align*} in the ordered basis \begin{align*} Y^{m+n-1},Y^{m+n-2},\dots,Y,1, \end{align*} and whose last $m$ rows are the coefficient rows of \begin{align*} Y^{m-1}G,\ Y^{m-2}G,\ \dots,\ G \end{align*} in the same ordered basis. By definition of the $Y$-resultant via the Sylvester matrix, \begin{align*} \operatorname{Res}_Y(F,G)=\det S(X)\in k[X]. \end{align*} For $a\in k$, write $S(a)$ for the matrix obtained by evaluating every entry of $S(X)$ at $X=a$. Then \begin{align*} \operatorname{Res}_Y(F,G)(a)=\det S(a), \end{align*} because determinant is a polynomial expression in the entries of the matrix.[/step]

[guided]We first fix the exact matrix whose determinant is the resultant. The important point is that the matrix size is fixed by the original $Y$-degrees $m=\deg_Y F$ and $n=\deg_Y G$. Thus the Sylvester matrix has size $m+n$, even if the leading coefficient of one of the specialized polynomials later vanishes at a particular value $X=a$. Write $F(X,Y)=\sum_{i=0}^{m} f_i(X)Y^i$ and $G(X,Y)=\sum_{j=0}^{n} g_j(X)Y^j$, with $f_i,g_j\in k[X]$, $f_m\ne 0$, and $g_n\ne 0$. The Sylvester matrix $S(X)$ is formed from the coefficient rows of the shifted polynomials \begin{align*} Y^{n-1}F,\ Y^{n-2}F,\ \dots,\ F \end{align*} and \begin{align*} Y^{m-1}G,\ Y^{m-2}G,\ \dots,\ G \end{align*} with respect to the ordered monomial basis \begin{align*} Y^{m+n-1},Y^{m+n-2},\dots,Y,1. \end{align*} This convention gives \begin{align*} \operatorname{Res}_Y(F,G)=\det S(X). \end{align*} Now evaluate at $X=a$. Since each entry of $S(X)$ is a polynomial in $X$, evaluation at $a$ gives a matrix $S(a)$ over $k$. Since the determinant is built from addition and multiplication of entries, evaluation commutes with taking the determinant: \begin{align*} \operatorname{Res}_Y(F,G)(a)=\det S(a). \end{align*} This observation is what lets us handle the forward implication without assuming that the specialized degrees remain $m$ and $n$.[/guided]

[step:Use a common point of the plane curves to force a kernel vector] Assume $(a,b)\in\mathbb A_k^2$ satisfies $F(a,b)=G(a,b)=0$. Define a vector $v_b\in k^{m+n}$ by its coordinates in the ordered basis dual to \begin{align*} Y^{m+n-1},Y^{m+n-2},\dots,Y,1 \end{align*} as \begin{align*} v_b:=(b^{m+n-1},b^{m+n-2},\dots,b,1)^\top. \end{align*} This vector is nonzero because its last coordinate is $1$. Each row of $S(a)$ is the coefficient row of either $Y^rF(a,Y)$ for some integer $r$ with $0\le r\le n-1$, or $Y^sG(a,Y)$ for some integer $s$ with $0\le s\le m-1$. If the row is the coefficient row of $Y^rF(a,Y)$, then its dot product with $v_b$ is \begin{align*} b^rF(a,b)=0. \end{align*} If the row is the coefficient row of $Y^sG(a,Y)$, then its dot product with $v_b$ is \begin{align*} b^sG(a,b)=0. \end{align*} Therefore \begin{align*} S(a)v_b=0. \end{align*} Since $v_b\ne 0$, the matrix $S(a)$ is singular. Hence \begin{align*} \operatorname{Res}_Y(F,G)(a)=\det S(a)=0. \end{align*} This proves the forward implication. [/step]

[step:Identify the evaluated resultant with the specialized one-variable resultant under degree preservation] Now assume $a\in k$ satisfies \begin{align*} \operatorname{Res}_Y(F,G)(a)=0 \end{align*} and \begin{align*} \deg_Y F(a,Y)=m, \qquad \deg_Y G(a,Y)=n. \end{align*} Define the specialized one-variable polynomials $P\in k[Y]$ and $Q\in k[Y]$ by $P(Y):=F(a,Y)$ and $Q(Y):=G(a,Y)$. By the degree-preservation hypothesis, $\deg P=m$ and $\deg Q=n$. Thus the Sylvester matrix defining $\operatorname{Res}_Y(P,Q)$ has the same size and row pattern as $S(a)$, with exactly the specialized coefficient entries. Hence \begin{align*} \operatorname{Res}_Y(P,Q)=\det S(a)=\operatorname{Res}_Y(F,G)(a)=0. \end{align*} [/step]

[step:Apply the one-variable resultant criterion to obtain a common root]The polynomials $P,Q\in k[Y]$ have positive degrees because $\deg P=m\ge 1$ and $\deg Q=n\ge 1$. Since $k$ is algebraically closed and \begin{align*} \operatorname{Res}_Y(P,Q)=0, \end{align*} the [[Resultant Criterion for Common Roots](/theorems/9433)][citetheorem:9433] applied to the one-variable polynomials $P$ and $Q$ gives an element $b\in k$ such that \begin{align*} P(b)=0 \qquad\text{and}\qquad Q(b)=0. \end{align*} By the definitions of $P$ and $Q$, this means \begin{align*} F(a,b)=0\qquad\text{and}\qquad G(a,b)=0. \end{align*} Thus there exists $b\in k$ with $(a,b)$ a common zero of the two plane curves. This proves the converse and completes the proof.[/step]

[guided]We now use the one-variable criterion for resultants. The specialized polynomials $P,Q\in k[Y]$ have positive degrees because $\deg P=m\ge 1$ and $\deg Q=n\ge 1$. The field $k$ is algebraically closed by hypothesis, and the previous step proved \begin{align*} \operatorname{Res}_Y(P,Q)=0. \end{align*} Therefore the [Resultant Criterion for Common Roots][citetheorem:9433] applies to $P$ and $Q$ and gives an element $b\in k$ such that \begin{align*} P(b)=0\qquad\text{and}\qquad Q(b)=0. \end{align*} Unpacking the definitions $P(Y):=F(a,Y)$ and $Q(Y):=G(a,Y)$ gives \begin{align*} F(a,b)=0\qquad\text{and}\qquad G(a,b)=0. \end{align*} This is exactly the required common zero above the chosen value $a$, so the converse follows.[/guided]