Resultants - if $a,b$ have degrees $m,n$, why are there polynomials $u,v$ of degrees $m-1,n-1$ so that $Res(a,b)=av-bu$? Let $R$ be a commutative ring. Suppose I have polynomials $a\left(x\right),b\left(x\right) \in R\left[x\right]$ of degrees $m,n$, respectively. My reading states that

there then exist polynomials $u\left(x\right),v\left(x\right) \in R\left[x\right]$ of degrees $m-1,n-1$, respectively, such that $a(x)v(x)-b(x)u(x) = \operatorname{Res}_{m,n}(a(x), b(x))$.

My question is why must this be true and how can it be proven? I have tried small examples by hand and it seems to be true, but I have no idea how to approach a proof. I think there might be a way via the division algorithm, but that isn't obvious to me.
 A: 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)  $).
A: I'll assume we're working over a field.
If the resultant is $0$, then $u = v = 0$ works.
If the resultant is nonzero, then $a$ and $b$ are coprime (since they don't have any common roots) so by the euclidean algorithm we see there are polynomials $u'$ and $v'$ so that
$$au' + bv' = 1 = \text{gcd}(a,b)$$
So multiplying both sides by the resultant $R$ we see
$$a Ru' + b Rv' = R$$
and taking $u = Ru'$ and $v = -Rv'$ does the job. I'll let you check that the polynomials we get from the euclidean algorithm really are one less than the degrees of the polynomials we started with. If you like, you can also find some discussion here.

I hope this helps ^_^
