You cannot demand $u\left( x\right) $ and $v\left( x\right) $ to have
degrees exactly $m-1$ and $n-1$. Indeed, if $R=\mathbb{Q}$ and $m=2$ and $n=2$
and $a\left( x\right) =x^{2}$ and $b\left( x\right) =x^{2}-1$, then the
only pair $\left( u\left( x\right) ,\ v\left( x\right) \right) $ of two
polynomials $u\left( x\right) $ and $v\left( x\right) $ of degree $\leq1$
satisfying $a\left( x\right) v\left( x\right) -b\left( x\right) u\left(
x\right) =\operatorname{Res}_{2,2}\left( a\left( x\right)
,\ b\left( x\right) \right) =1$ is $\left( 1,\ 1\right) $, so the degrees
are $0$ here.
So the appropriate thing to ask for is to have two polynomials $u\left(
x\right) $ and $v\left( x\right) $ of degrees $\leq m-1$ and $\leq n-1$,
respectively, that satisfy
\begin{align}
a\left( x\right) v\left( x\right) -b\left( x\right) u\left( x\right)
=\operatorname{Res}_{m,n}\left( a\left( x\right) ,\ b\left(
x\right) \right) .
\end{align}
To prove that two such polynomials exist (in full generality, assuming that $m+n > 0$), we recall the
linear-algebraic meaning of the resultant first.
For each nonnegative integer $k$, we let $R\left[ x\right] _{\deg<k}$ denote
the $R$-submodule of $R\left[ x\right] $ spanned by $x^{0},x^{1}
,\ldots,x^{k-1}$. This is a free $R$-module of rank $k$; it consists of those
polynomials that have degree $<k$. Now, consider the map
\begin{align*}
\Phi:R\left[ x\right] _{\deg<n}\times R\left[ x\right] _{\deg<m} &
\rightarrow R\left[ x\right] _{\deg<n+m},\\
\left( r\left( x\right) ,\ s\left( x\right) \right) & \mapsto a\left(
x\right) r\left( x\right) +b\left( x\right) s\left( x\right) .
\end{align*}
This is an $R$-linear map between two free $R$-modules of the same rank.
Moreover, with respect to the "monomial" bases of both $R$-modules $R\left[
x\right] _{\deg<n}\times R\left[ x\right] _{\deg<m}$ and $R\left[
x\right] _{\deg<n+m}$ (that is, the bases formed by the monomials), this
linear map $\Phi$ is represented precisely by the Sylvester matrix
\begin{align}
S:=\left(
\begin{array}
[c]{c}
\begin{array}
[c]{ccccccccc}
a_{0} & 0 & 0 & \cdots & 0 & b_{0} & 0 & \cdots & 0\\
a_{1} & a_{0} & 0 & \cdots & 0 & b_{1} & b_{0} & \cdots & 0\\
\vdots & a_{1} & a_{0} & \cdots & 0 & \vdots & b_{1} & \ddots & \vdots\\
\vdots & \vdots & a_{1} & \ddots & \vdots & \vdots & \vdots & \ddots & b_{0}\\
a_{m} & \vdots & \vdots & \ddots & a_{0} & \vdots & \vdots & \ddots & b_{1}\\
0 & a_{m} & \vdots & \ddots & a_{1} & b_{n} & \vdots & \ddots & \vdots\\
\vdots & \vdots & \ddots & \ddots & \vdots & 0 & b_{n} & \ddots & \vdots\\
0 & 0 & 0 & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & a_{m} & 0 & 0 & \cdots & b_{n}
\end{array}
\\
\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }
_{n\text{ columns}}
\underbrace{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }
_{m\text{ columns}}
\end{array}
\right)
\end{align}
(where $a_{0},a_{1},\ldots,a_{m}$ are the coefficients of $a\left( x\right)
$, and where $b_{0},b_{1},\ldots,b_{n}$ are the coefficients of $b\left(
x\right) $). The determinant $\det S$ of this matrix $S$ is precisely the
resultant $\operatorname{Res}_{m,n}\left( a\left( x\right)
,\ b\left( x\right) \right) $.
Now, we need the following fact:
Lemma 1. Let $M$ and $N$ be two free $R$-modules of the same rank, which
is finite. Let $f:M\rightarrow N$ be an $R$-linear map. Let $A$ be a matrix
that represents this map $f$ with respect to some pair of bases of $M$ and of
$N$. Then, $\left( \det A\right) \cdot v\in f\left( M\right) $ for each
$v\in N$.
Proof of Lemma 1. Let $k$ be the rank of the free modules $M$ and $N$. We
WLOG assume that both $M$ and $N$ are the $R$-module $R^{k}$ of column vectors
of size $k$, and that the $R$-linear map $f$ is actually the left
multiplication by the matrix $A$ -- that is, we have $f\left( w\right) =Aw$
for each vector $w\in R^{k}$. (This is an assumption we can make, because we
can always replace $M$ and $N$ by $R^{k}$ via the two bases we have chosen,
and then the linear map $f$ becomes the left multiplication by the matrix that
represents it with respect to these bases; but this matrix is precisely $A$.)
Let $\operatorname*{adj}A$ be the adjugate of the matrix $A$. It is well-known
that $A\cdot\operatorname*{adj}A=\left( \det A\right) \cdot I_{k}$ (where
$I_{k}$ denotes the $k\times k$ identity matrix). Now, let $v\in N$. Then,
$v\in N=R^{k}$ and $\left( \operatorname*{adj}A\right) \cdot v\in R^{k}=M$.
Furthermore,
\begin{align*}
f\left( \left( \operatorname*{adj}A\right) \cdot v\right) &
=\underbrace{A\cdot\left( \operatorname*{adj}A\right) }_{=\left( \det
A\right) \cdot I_{k}}\cdot v\ \ \ \ \ \ \ \ \ \ \left( \text{since }f\left(
w\right) =Aw\text{ for each }w\in R^{k}\right) \\
& =\left( \det A\right) \cdot I_{k}v=\left( \det A\right) \cdot v,
\end{align*}
so that
\begin{align}
\left( \det A\right) \cdot v=f\left( \left( \operatorname*{adj}A\right)
\cdot v\right) \in f\left( M\right) .
\end{align}
This proves Lemma 1. $\blacksquare$
We can now easily finish our proof: Applying Lemma 1 to $M=R\left[ x\right]
_{\deg<n}\times R\left[ x\right] _{\deg<m}$ and $N=R\left[ x\right]
_{\deg<n+m}$ and $f=\Phi$ and $A=S$ and $v=1$, we conclude that
\begin{align}
\left( \det S\right) \cdot1\in\Phi\left( R\left[ x\right] _{\deg<n}\times
R\left[ x\right] _{\deg<m}\right)
\end{align}
(since $R\left[ x\right] _{\deg<n}\times R\left[ x\right] _{\deg<m}$ and
$R\left[ x\right] _{\deg<n+m}$ are two free $R$-modules of the same rank,
which is finite). Since $\left( \det S\right) \cdot1=\det
S=\operatorname{Res}_{m,n}\left( a\left( x\right) ,\ b\left(
x\right) \right) $, we can rewrite this as
\begin{align}
\operatorname{Res}_{m,n}\left( a\left( x\right) ,\ b\left(
x\right) \right) =\Phi\left( R\left[ x\right] _{\deg<n}\times R\left[
x\right] _{\deg<m}\right) .
\end{align}
In other words, there exists a pair $\left( r\left( x\right) ,\ s\left(
x\right) \right) \in R\left[ x\right] _{\deg<n}\times R\left[ x\right]
_{\deg<m}$ such that $\operatorname{Res}_{m,n}\left( a\left(
x\right) ,\ b\left( x\right) \right) =\Phi\left( r\left( x\right)
,\ s\left( x\right) \right) $. Consider this pair. Then,
\begin{align*}
\operatorname{Res}_{m,n}\left( a\left( x\right) ,\ b\left(
x\right) \right) & =\Phi\left( r\left( x\right) ,\ s\left( x\right)
\right) \\
& =a\left( x\right) r\left( x\right) +b\left( x\right) s\left(
x\right) \ \ \ \ \ \ \ \ \ \ \left( \text{by the definition of }\Phi\right)
\\
& =a\left( x\right) r\left( x\right) -b\left( x\right) \left( -s\left(
x\right) \right) .
\end{align*}
Moreover, from $\left( r\left( x\right) ,\ s\left( x\right) \right) \in
R\left[ x\right] _{\deg<n}\times R\left[ x\right] _{\deg<m}$, we obtain
$r\left( x\right) \in R\left[ x\right] _{\deg<n}$, so that $\deg\left(
r\left( x\right) \right) <n$ and thus $\deg\left( r\left( x\right)
\right) \leq n-1$. Similarly, $\deg\left( s\left( x\right) \right) \leq
m-1$, so that $\deg\left( -s\left( x\right) \right) \leq m-1$ as well.
Thus, we have found two polynomials $u\left( x\right) $ and $v\left(
x\right) $ of degrees $\leq m-1$ and $\leq n-1$, respectively, that satisfy
\begin{align}
a\left( x\right) v\left( x\right) -b\left( x\right) u\left( x\right)
=\operatorname{Res}_{m,n}\left( a\left( x\right) ,\ b\left(
x\right) \right)
\end{align}
(namely, $u\left( x\right) =-s\left( x\right) $ and $v\left( x\right)
=r\left( x\right) $).
