A property of the resultant Let $A$ be a domain. There's a problem in Lorenzini's An Invitation to Arithmetic Geometry, which states that if $f,g\in A[y]$, then we can find $u,v\in A[y]$ with $\deg u<\deg g$ and $\deg v<\deg f$ s.t.
$$uf+vg = \textrm{Res}(f,g).$$
Is there an easy way to see this?
 A: Suppose that $f$ has degree $m$ and $g$ has degree $n$, say $$
f = a_0y^m + a_1y^{m-1} + \cdots + a_m$$ $$g = b_0y^n + b_1y^{n-1}+ \cdots + b_n$$ Then the resultant of $f$ and $g$ is the determinant of the $(n + m)\times (n+m)$ Sylvester matrix: 
$$Res(f,g) = \det \left(\begin{matrix} a_0 & a_1 & \cdots & a_m & 0 &\cdots & 0\\
0 & a_0 & a_1 & \cdots & a_m &\cdots & 0\\
& & & \ddots\\
0 & \cdots & 0 & a_0 & a_1 &\cdots & a_m \\
b_0 & b_1 &\cdots & b_n &0 &\cdots & 0\\
0 &b_0 &b_1&\cdots & b_n &\cdots & 0\\
& & & \ddots\\
0 & \cdots & 0 & b_0 & b_1 & \cdots &b_n
\end{matrix}\right).$$ To make notation easy, let $A$ denote this matrix. Note that $$A\left(\begin{matrix} y^{n+m-1}\\ y^{n+m-2}\\\vdots\\ y\\ 1\end{matrix}\right) = \left(\begin{matrix}y^{n-1}f\\ y^{n-2} f\\\vdots\\ yf\\f\\y^{m-1}g\\y^{m-2}g\\\vdots\\yg\\ g\end{matrix}\right).$$ If $B$ is the adjugate matrix of $A$, that is, $BA = \det(A)\,Id = Res(f,g)\,Id$, then multiplying both sides of this equality on the left by $B$ gives $$ 
Res(f,g)\left(\begin{matrix} y^{n+m-1}\\ y^{n+m-2}\\\vdots\\ y\\ 1\end{matrix}\right) = B \left(\begin{matrix}y^{n-1}f\\ y^{n-2} f\\\vdots\\ yf\\f\\y^{m-1}g\\y^{m-2}g\\\vdots\\yg\\ g\end{matrix}\right).$$ If we then look in particular at the bottom row of this equality, you get an expression of the form $$Res(f,g) = uf + vg,$$ as desired.
A: How easy this is may depend on how you define the resultant.  A key fact is that if $f = q g + r$ for polynomials $q$ and $r$ with $\deg(r) \le \deg(f)$, then $\text{Res}(f,g) = a^{\text{deg}(f)-\text{deg}(r)} \text{Res}(r,g)$ where $a$ is the leading coefficient of $g$.
