[proofplan] We interpret the resultant as the determinant of the Sylvester multiplication map. Its vanishing is equivalent to the existence of a nonzero bounded-degree relation $u f+v g=0$. Such a relation is equivalent, by the Euclidean gcd theory in $k[T]$, to $f$ and $g$ having a nonconstant common divisor. Since $k$ is algebraically closed, a nonconstant common divisor has a linear factor $T-\alpha$, which is exactly the same as a common root. [/proofplan] [step:Translate vanishing of the resultant into a bounded polynomial relation] Let \begin{align*} m:=\deg f \end{align*} and let \begin{align*} n:=\deg g. \end{align*} For an integer $r\geq 1$, define \begin{align*} k[T]_{<r}:=\{a\in k[T]: \deg a<r\}\cup\{0\}. \end{align*} Thus $k[T]_{<r}$ is a $k$-[vector space](/page/Vector%20Space) of dimension $r$. By the Sylvester determinant definition of the resultant, $\operatorname{Res}_T(f,g)$ is the determinant of the $k$-[linear map](/page/Linear%20Map) \begin{align*} \Phi:k[T]_{<n}\times k[T]_{<m}\to k[T]_{<m+n} \end{align*} defined by \begin{align*} \Phi(u,v)=u f+v g. \end{align*} Both the domain and codomain have dimension $m+n$ over $k$. Therefore \begin{align*} \operatorname{Res}_T(f,g)=0 \end{align*} if and only if $\Phi$ is not injective, equivalently if and only if there exist polynomials $u\in k[T]_{<n}$ and $v\in k[T]_{<m}$, not both zero, such that \begin{align*} u f+v g=0. \end{align*} [/step] [step:Show that a bounded relation is equivalent to a nonconstant common divisor] We prove that such a nonzero pair $(u,v)$ exists if and only if $f$ and $g$ have a nonconstant common divisor in $k[T]$. Assume first that $u\in k[T]_{<n}$ and $v\in k[T]_{<m}$ are not both zero and satisfy \begin{align*} u f+v g=0. \end{align*} Let $d\in k[T]$ be a greatest common divisor of $f$ and $g$, chosen monic. Write \begin{align*} f=d f_1 \end{align*} and \begin{align*} g=d g_1, \end{align*} where $f_1,g_1\in k[T]$ are coprime. Since $k[T]$ is a [Euclidean domain](/page/Euclidean%20Domain), Euclid's lemma applies: from \begin{align*} u f_1=-v g_1 \end{align*} and $\gcd(f_1,g_1)=1$, it follows that $f_1$ divides $v$. If $d$ were constant, then $\deg f_1=\deg f=m$. Since $f_1\mid v$, either $v=0$ or $\deg v\geq m$. The degree bound $v\in k[T]_{<m}$ rules out $\deg v\geq m$, so $v=0$. Then the relation gives $u f=0$, and since $k[T]$ is an [integral domain](/page/Integral%20Domain) and $f\neq 0$, we get $u=0$, contradicting that $(u,v)$ is nonzero. Hence $d$ is nonconstant. Conversely, suppose $f$ and $g$ have a nonconstant common divisor $d\in k[T]$. Choose $d$ monic and write \begin{align*} f=d f_1 \end{align*} and \begin{align*} g=d g_1. \end{align*} Let $r:=\deg d$. Since $r\geq 1$, we have \begin{align*} \deg g_1=n-r<n \end{align*} and \begin{align*} \deg f_1=m-r<m. \end{align*} Define $u:=g_1$ and $v:=-f_1$. Then $u\in k[T]_{<n}$, $v\in k[T]_{<m}$, and \begin{align*} u f+v g=g_1 d f_1-f_1 d g_1=0. \end{align*} The pair $(u,v)$ is nonzero because $f$ and $g$ have positive degree and $d$ is a proper common factor of their displayed quotients only in the harmless case where one quotient is constant; in all cases at least one of $f_1,g_1$ is nonzero. Thus a bounded relation exists. [guided] The point of the degree bounds is that they turn a common divisor into an actual kernel vector for the Sylvester map, and conversely prevent a kernel vector from existing when $f$ and $g$ are coprime. First assume there are $u\in k[T]_{<n}$ and $v\in k[T]_{<m}$, not both zero, such that \begin{align*} u f+v g=0. \end{align*} Let $d\in k[T]$ be the monic greatest common divisor of $f$ and $g$. Define $f_1,g_1\in k[T]$ by \begin{align*} f=d f_1 \end{align*} and \begin{align*} g=d g_1. \end{align*} By the definition of a greatest common divisor, $f_1$ and $g_1$ are coprime. Substituting the factorizations into the relation and cancelling the nonzero polynomial $d$ in the integral domain $k[T]$ gives \begin{align*} u f_1=-v g_1. \end{align*} Since $k[T]$ is a Euclidean domain, the usual Bezout identity holds for coprime polynomials: there exist $a,b\in k[T]$ such that \begin{align*} a f_1+b g_1=1. \end{align*} Multiplying this identity by $v$ gives \begin{align*} v=a v f_1+b v g_1. \end{align*} From $u f_1=-v g_1$, the term $b v g_1$ is a multiple of $f_1$, and the term $a v f_1$ is also a multiple of $f_1$. Therefore $f_1$ divides $v$. Now suppose, toward a contradiction, that $d$ is constant. Then $\deg f_1=\deg f=m$. Since $f_1\mid v$, either $v=0$ or $\deg v\geq m$. But $v\in k[T]_{<m}$ means $\deg v<m$ unless $v=0$, so we must have $v=0$. The relation then becomes \begin{align*} u f=0. \end{align*} Because $k[T]$ is an integral domain and $f$ is nonzero, this forces $u=0$. This contradicts the assumption that $(u,v)$ is not the zero pair. Hence $d$ cannot be constant, so $f$ and $g$ have a nonconstant common divisor. Conversely, assume $f$ and $g$ have a nonconstant common divisor $d$. Write \begin{align*} f=d f_1 \end{align*} and \begin{align*} g=d g_1. \end{align*} Let $r:=\deg d$. Since $d$ is nonconstant, $r\geq 1$. Therefore \begin{align*} \deg g_1=\deg g-\deg d=n-r<n \end{align*} and \begin{align*} \deg f_1=\deg f-\deg d=m-r<m. \end{align*} Thus $g_1\in k[T]_{<n}$ and $-f_1\in k[T]_{<m}$. Define \begin{align*} u:=g_1 \end{align*} and \begin{align*} v:=-f_1. \end{align*} Then \begin{align*} u f+v g=g_1 d f_1-f_1 d g_1=0. \end{align*} At least one of $f_1$ and $g_1$ is nonzero because $f$ and $g$ are nonzero polynomials. Hence $(u,v)$ is a nonzero bounded-degree relation. [/guided] [/step] [step:Convert nonconstant common divisors into common roots] We now use the hypothesis that $k$ is algebraically closed. If $f$ and $g$ have a nonconstant common divisor $d\in k[T]$, then $d$ has a root $\alpha\in k$. Hence $T-\alpha$ divides $d$ by the [factor theorem](/theorems/3235). Since $d$ divides both $f$ and $g$, the polynomial $T-\alpha$ divides both $f$ and $g$, so \begin{align*} f(\alpha)=g(\alpha)=0. \end{align*} Conversely, if there exists $\alpha\in k$ such that \begin{align*} f(\alpha)=g(\alpha)=0, \end{align*} then the factor theorem gives that $T-\alpha$ divides both $f$ and $g$. Therefore $f$ and $g$ have the nonconstant common divisor $T-\alpha$. [/step] [step:Combine the equivalences] From the Sylvester determinant interpretation, \begin{align*} \operatorname{Res}_T(f,g)=0 \end{align*} is equivalent to the existence of a nonzero bounded-degree relation $u f+v g=0$. By the gcd argument, this is equivalent to $f$ and $g$ having a nonconstant common divisor. Since $k$ is algebraically closed, this is equivalent to the existence of $\alpha\in k$ such that \begin{align*} f(\alpha)=g(\alpha)=0. \end{align*} This proves the claimed equivalence. [/step]