---
abstract: 'In 1853 J. Sylvester introduced a [family]{} of double sum expressions for two finite sets of indeterminates and showed that some members of the family are essentially the polynomial subresultants of the monic polynomials associated with these sets. In 2009, in a joint work with C. D’Andrea and H. Hong we gave the complete description of all the members of the family as expressions in the coefficients of these polynomials. In 2010, M.-F. Roy and A. Szpirglas presented a new and natural inductive proof for the cases considered by Sylvester. Here we show how induction also allows to obtain the full description of Sylvester’s double-sums.'
address:
- 'Departamento de Matemática, Facultad de Ciencias Exactas y Naturales, Universidad de Buenos Aires and IMAS, CONICET, 1428 Buenos Aires, Argentina.'
- 'Department of Mathematics, North Carolina State University, Raleigh NC 27695, USA.'
author:
- Teresa Krick
- Agnes Szanto
title: |
    Sylvester’s double sums:\
    an inductive proof of the general case 
---

[^1]

Sylvester’s double sums, Subresultants.

Introduction
============

Let $A$ and $B$ be non-empty finite lists (ordered sets) of distinct indeterminates. In [@sylv], J. Sylvester introduced for each $0\le
p\le |A|$ and $0\le q \le |B|$ the following univariate polynomial in the variable $x$, of degree $\le p+q$, called the [*double-sum expression*]{} in $A$ and $B$:

$$\operatorname{Sylv}^{p,q}(A,B)  :=
\sum_{\scriptsize{\begin{array}{c} A^{\prime }\subset
A,\,B^{\prime}\subset B
\\[-1.8mm]
|A^{\prime}|=p,\,|B^{\prime}|=q
\end{array}}}R(x,A^{\prime
})\,R(x,B^{\prime})\,\frac{R(A^{\prime},B^{\prime})\,R(A-
A^{\prime},B- B^{\prime})}{R(A^{\prime},A- A^{\prime
})\,R(B^{\prime},B- B^{\prime})},$$ where for sets $Y$, $Z$ of indeterminates, $$R(Y,Z):=\prod_{y\in Y,z\in Z}(y-z), \ \quad R(y,Z):=\prod_{z\in
Z}(y-Z).$$ and by convention $R(Y,\emptyset)=1$.

Let now $f, g$ be monic univariate polynomials such that $$f=\prod_{\alpha\in A}(x-\alpha)=x^m+a_{m-1}x^{m-1}+\ldots+a_0 \ \mbox { and } \
g =\prod_{\beta\in B}(x-\beta)=x^n+b_{n-1}x^{n-1}+\ldots+b_0,$$ where $m:=|A|\geq1$ and $n:=|B|\geq1$. The [*$k$-th subresultant*]{} of the polynomials $f$ and $g$ is defined, for $0\le k <  \min\{m,n\}$ or $k=\min\{m,n\}$ when $m\neq n$, as [ $$\label{defsub}
\Sres_k(f,g)
:=\det\begin{array}{|cccccc|c}
\multicolumn{6}{c}{\scriptstyle{m+n-2k}}\\
\cline{1-6}
a_{m} & \cdots & & \cdots & a_{k+1-\left(n-k-1\right)}& x^{n-k-1}f(x)&\\
& \ddots & && \vdots  & \vdots &\scriptstyle{n-k}\\
&  &a_{m}&\cdots &a_{k+1}& x^0f(x)& \\
\cline{1-6}
b_{n} &\cdots & & \cdots & b_{k+1-(m-k-1)}&x^{m-k-1}g(x)&\\
&\ddots &&&\vdots & \vdots &\scriptstyle{m-k}\\
&& b_{n} &\cdots & b_{k+1} & x^0g(x)&\\
\cline{1-6} \multicolumn{2}{c}{}
\end{array}$$]{} with $a_{\ell}=b_{\ell}=0$ for $\ell<0$. For $k=0$, $ \Sres_0(f,g)$ coincides with the resultant: $$\label{result}\Res(f,g)= \prod_{\alpha\in A}
g(\alpha) = (-1)^{mn}  \prod_{\beta\in B} f(\beta).$$

Also, for instance, $$\label{sresm0} \Sres_m(f,g)=f \mbox{ for } m<n \ \mbox{ and } \ \Sres_n(f,g)=g \mbox{ for } n<m.$$

Relating Sylsvester’s double sums with the polynomials $f$ and $g$, it is immediate that $$\label{00} \operatorname{Sylv}^{0,0}(A,B)=R(A,B)=\Res(f,g),$$ $$\label{m0} \operatorname{Sylv}^{m,0}(A,B)=R(x,A)=f \ \mbox{ and } \  \operatorname{Sylv}^{0,n}(A,B)=R(x,B)=g,\quad$$ $$\label{mn}\operatorname{Sylv}^{m,n}(A,B)=R(x,A)\,R(x,B)\,R(A,B)=\Res(f,g)\,f\,g.$$

More generally, every value of the polynomial $\operatorname{Sylv}^{p,q}(A,B)$, which is symmetric in the $\alpha$’s and in the $\beta$’s, can be expressed as a polynomial in $x$ whose coefficients are rational functions in the $a_i$’s and the $b_j$’s. Sylvester in [@sylv] gave this rational expression for the following values of $(p,q)$:

1.  If $0\le k:=p+q<\min\{m,n\}$ or if $k=m<n$, then [@sylv Art. 21]: $$\operatorname*{Sylv}\nolimits^{p,q}(A,B)=(-1)^{p(m-k)}{k\choose p}\,
    \Sres_k(f,g).$$

2.  If $p+q=m=n$, then [@sylv Art. 22]: $$\operatorname*{Sylv}\nolimits^{p,q}(A,B)={m-1\choose q}\,f+{m-1\choose p}\,g.$$

3.  If $m<p+q<n-1,$ then [@sylv Arts. 23 & 24]): $$\operatorname*{Sylv}\nolimits^{p,q}(A,B)=0.$$

4.  If $m<p+q=n-1$, then [@sylv Art. 25]: $\operatorname*{Sylv}\nolimits^{p,q}(A,B)$ is a “numerical multiplier” of $f$, but the ratio is not established.

In [@LP Th.0.1 and Prop. 2.9], A. Lascoux and P. Pragacz presented new proofs for the cases covered by Items (1) and (2). More recently, in a joint work with C. D’Andrea and H. Hong in [@DHKS09 Th.2.10] we introduced a unified matrix formulation that allowed us to give an explicit formula for all possible values of $(p,q)$, i.e. for $0\le p\le m, 0\le q\le n$. The proofs there were elementary though cumbersome. In 2010, M.-F. Roy and A. Szpirglas, were able to produce in [@RoSp Main theorem] a new and natural inductive proof also for the cases covered by Item (1) and (2). The aim of this note is to give, inspired by [@RoSp], an elementary inductive proof for all the cases. We furthermore show how the cases $p+q>\min\{m,n\}$, which seem somehow less natural since there is no “natural" expression associated to them (and were therefore not previously considered by Lascoux and Pragasz and Roy and Szpirglas) immediately yield simple proofs for other known interesting cases, as for instance for the cases $p+q=m<n $ and $p+q=m=n$, which didn’t have simple proofs yet.

Let us now introduce the necessary notation to formulate our main result.\
As in [@DHKS09], we split the last column of the matrix in to write $\Sres_k(f,g)$ as the sum of two determinants, obtaining an expression $$\label{subres}
\Sres_k(f,g)=F_k (f,g)\, f+G_k(f,g)\,g$$ where the polynomials $F_k(f,g)$ and $G_k(f,g)$ are defined for $0\le k <  \min\{m,n\}$ or $k=\min\{m,n\}$ when $m\ne n$ as the determinants of the $(m+n-2k)$-matrices:

[$$F_k(f,g)
:= \det
\begin{array}{|cccccc|c}
\cline{1-6}
a_{m} & \cdots & & \cdots & a_{k+1-\left(n-k-1\right)}& x^{n-k-1}&\\
& \ddots & && \vdots  & \vdots&\scriptstyle n-k \\
&  &a_{m}&\cdots &a_{k+1}& x^0 & \\
\cline{1-6}
b_{n} &\cdots & & \cdots & b_{k+1-(m-k-1)}&0 &\\
&\ddots &&&\vdots & \vdots & \scriptstyle m-k \\
&& b_{n} &\cdots & b_{k+1} & 0 &\\
\cline{1-6}
\end{array}\ , \
G_{k}(f.g)
:=\det\begin{array}{|cccccc|c}
\cline{1-6}
a_{m} & \cdots & & \cdots & a_{k+1-\left(n-k-1\right)}& 0&\\
& \ddots & && \vdots  & \vdots &\scriptstyle n-k \\
&  &a_{m}&\cdots &a_{k+1}& 0 & \\
\cline{1-6}
b_{n} &\cdots & & \cdots & b_{k+1-(m-k-1)}&x^{m-k-1}&\\
&\ddots &&&\vdots & \vdots &\scriptstyle m-k \\
&& b_{n} &\cdots & b_{k+1} & x^0&\\
\cline{1-6}
\end{array}.$$]{} We observe that when $k<\min\{m,n\}$, $\deg
F_k(f,g)\le   n-k-1$ and $\deg G_k(f,g)\le  m-k-1$. Also $$\label{Fm}F_m(f,g)= 1 ,\  \ G_m(f,g)=0 \ \mbox{ for } m<n \ \mbox{ and } \
F_n(f,g)= 0 , \ G_n(f,g)=1 \ \mbox{ for } n<m$$ $$\label{Gm-1}G_{m-1}(f,g)=1 \mbox{ for } m\le n \ \mbox{ and } \ F_{n-1}(f,g)=(-1)^{m-n+1} \mbox{ for } n\le m.$$

We finally introduce the following notation that we will keep all along in this text. Given $m,n\in \N$, $p,q\in \Z$ such that $0\le p\le m$, $0\le q\le
 n$ and $k=p+q$, we set $$\ov p:=m-p,\ \ \ov q:= n-q \ \mbox{ and } \ \ov k:=\ov p + \ov
  q-1
= m+n- k-1.$$ Sylvester’s double sums, for $k$ “too big" w.r.t. $m$ and $n$, will be expressed in our result in terms of the polynomials $F_{\ov k}(f,g)$ and $G_{\ov k}(f,g)$, well-defined since the condition $n-1\le k\le m+n-1$ for $m<n$ is equivalent to $0\le \ov k\le m$, and the condition $m\le k\le 2m-1$ for $m=n$ is equivalent to $0\le \ov k\le m-1$.

\[theorem\] (See also [@DHKS09 Th.2.10])\
Set $1\le m\le n$, and let $0\le p\le m$, $0\le q\le
 n$ and $k=p+q$.\
Then, for $(p,q)\ne (m,n)$,\
– when $m<n$: [$$\operatorname{Sylv}^{p,q}(A,B)
=\left\{\begin{array}{lll}(-1)^{p(m-k)}{k\choose p}\Sres_k(f,g)&\mbox{for}& 0\le k\le m\\
0&\mbox{for}& m+1\le k\le n-2\ \mbox{when }  m\le n-3\\
(-1)^{c}\Big({\ov k \choose \ov p}F_{\ov k}(f,g)\,f - {\ov k\choose
\ov q}G_{\ov k}(f,g)\,g\Big)&\mbox{for}& n-1\le k\le m+n-1
\end{array}\right.$$]{} – when $m=n$: [$$\operatorname{Sylv}^{p,q}(A,B)
=\left\{\begin{array}{lll}(-1)^{p(m-k)}{k\choose p}\Sres_k(f,g)&\mbox{for}& 0\le k\le m-1\\
(-1)^{c}\Big({\ov k\choose \ov p}F_{\ov k}(f,g)\,f - {\ov k\choose
\ov q}G_{\ov k}(f,g)\,g\Big)&\mbox{for}& m\le k\le 2m-1
\end{array}\right. ,$$]{} where $c:=\ov p \,\ov q+ n-p-1+nq$;\
and for $(p,q)=(m,n)$: $$\operatorname{Sylv}^{m,n}(A,B)=\Res(f,g)\,f\,g.$$

Theorem \[theorem\] can be written in a more uniform manner instead of being split in cases: by Identity , for $0\le k\le m$ when $m<n$ and for $0\le k<m$ when $m=n$, $$\operatorname{Sylv}^{p,q}(A,B)
=(-1)^{p(m-k)} \Big({k\choose p}F_k(f,g)f + {k\choose
q}G_k(f,g)g\Big),$$ or for $0\le \ov k\le m$ when $m<n$ and for $0\le \ov k<m$, when $m=n$, $$\begin{gathered}
\label{eqex}\operatorname{Sylv}^{p,q}(A,B)
= (-1)^c\Big( {\ov k\choose \ov p}\Sres_{\ov k}(f,g)-{\ov k+1\choose
\ov q}G_{\ov k }(f,g)\,g\Big)\\  = (-1)^c\Big( {\ov k+1\choose \ov
p}F_{\ov k}(f,g)\,f-{\ov k \choose \ov q}\Sres_{\ov k}(f,g)\Big).\end{gathered}$$

The cases “in between", for $m+1\le k\le n-2$ when $m\le n-3$, are the cases when neither $0\le k\le m$ nor $0\le \ov k\le m$, i.e. the cases when the corresponding matrices $F_k$, $G_k$ and $F_{\ov k}$, $G_{\ov k}$ are not defined (or could be defined as $0$ for uniformity).

We also note that the case $k=m=n-1$ is covered twice: $\Sres_m(f,g)=f= F_m(f,g)f-G_m(f,g)g$ since $\ov k =m$, $F_m=1$ and $G_m=0$. Finally the case $p=m$, $q=n$ is Identity .

The proof of Theorem \[theorem\] is based, as the proof in [@RoSp], on specialization properties.

Specialization properties
=========================

The following specialization property of Sylvester’s double sums is well-known and proved in [@LP Lemma 2.8]. It is also reproved in [@RoSp Prop.3.1], where it is used as one of the key ingredients of their inductive proof for the cases $k\le m<n$ and $k<m=n$. We repeat it here for sake of completeness.

\[sylvester\] For any $\alpha \in A$ and $\beta\in
B$, $$\begin{aligned}
\bullet \ & \operatorname{Sylv}^{p,q}(A,B) (\alpha) =
(-1)^{p}\,{\mbox{coeff}}_{p+q}\big(\operatorname{Sylv}^{p,q}(A-\alpha,B)\big)\,R(\alpha,B)
\ \mbox{ for } \ 0\le p<m \mbox{ and } 0\le q\le n,\\ \bullet \ &
\operatorname{Sylv}^{p,q}(A,B) (\beta) = (-1)^{q+\ov
p}\,{\mbox{coeff}}_{p+q}\big(\operatorname{Sylv}^{p,q}(A,B-\beta)\big)\,R(\beta,A)
\ \mbox{ for }  \ 0\le p\le m  \mbox{ and } 0\le q<n.\end{aligned}$$ Here ${\mbox{coeff}}_{p+q}$ denotes the coefficient of order $p+q$ of $\operatorname{Sylv}^{p,q}(A,B-\beta)$.

$$\begin{aligned}
 \operatorname{Sylv}^{p,q}(A,B)  (\alpha)&=
\sum_{\scriptsize{\begin{array}{c} A^{\prime }\subset
A-\alpha,\,B^{\prime}\subset B
\\[-1.8mm]
|A^{\prime}|=p,\,|B^{\prime}|=q
\end{array}}}R(\alpha,A^{\prime
})\,R(\alpha,B^{\prime})\,\frac{R(A^{\prime},B^{\prime})\,R(A-
A^{\prime},B- B^{\prime})}{R(A^{\prime},A- A^{\prime
})\,R(B^{\prime},B- B^{\prime})}\\
&= (-1)^{p}\,R(\alpha, B)\sum_{\scriptsize{\begin{array}{c}
A^{\prime }\subset A-\alpha,\,B^{\prime}\subset B
\\[-1.8mm]
|A^{\prime}|=p,\,|B^{\prime}|=q
\end{array}}}\frac{R(A^{\prime},B^{\prime})\,R((A-\alpha)-
A^{\prime},B- B^{\prime})}{R(A^{\prime},(A-\alpha)- A^{\prime
})\,R(B^{\prime},B- B^{\prime})}\\
&= (-1)^{p}\,{\mbox{coeff}}_{p+q}
\big(\operatorname{Sylv}^{p,q}(A-\alpha,B) \big)\,R(\alpha,B).\end{aligned}$$

The second identity is a consequence of the fact that $$\operatorname{Sylv}^{p,q}(A,B)=(-1)^{pq}\,(-1)^{\ov p\, \ov q}\, \operatorname{Sylv}^{q,p}(B,A).$$

In the following we replace the specialization property of subresultants proved in [@RoSp Prop. 4.1] by the specialization property of the polynomials $F_k$ and $G_k$. This will allow a more uniform and simpler proof of our main theorem, covering all cases of $p$ and $q$.

\[dos\] For any root $\alpha $ of $f$ and any root $\beta$ of $g$, we have $$\begin{aligned}
\bullet \ &F_k(f,g)(\beta)= - \,{\mbox{coeff}}_{n-k-1}\big(
F_{k-1}(f,\frac{g}{x-\beta})\big)\  \mbox{ for } \ 1\le k\le
\min\{m,n\}-1,\\ \bullet \ &G_k(f,g)(\alpha) = (-1)^{m-k-1}
\,{\mbox{coeff}}_{m-k-1}\big( G_{k-1}(\frac{f}{x-\alpha},g)\big) \ \mbox{ for }
\ 1\le k\le \min\{m,n\}-1.\end{aligned}$$ Here ${\mbox{coeff}}_{n-k-1}$ (resp. ${\mbox{coeff}}_{m-k-1}$) denotes the coefficient of order $n-k-1$ (resp. $m-k-1$) of the corresponding polynomial.

Given a root $\beta$ of $g$, we set $$\frac{g}{x-\beta}:=x^{n-1}+b'_{n-2}x^{n-2}+\cdots + b'_0.$$ The following relationship between the coefficients of $g$ and of $\frac{g}{x-\beta}$ is straightforward: $$\begin{aligned}
\label{bi}
 b_i=b'_{i-1}-\beta b'_i \  \mbox{for } \ 1\le i\le n-1
 \ \mbox{ and } \ b_0=-\beta b'_0.
 \end{aligned}$$ (Here $b_n=b'_{n-1}=1$.)\
First consider [$$\begin{aligned}
& {\mbox{coeff}}_{n-k-1}\big(F_{k-1}(f,\frac{g}{x-\beta})\big)={\mbox{coeff}}_{n-k-1}\big(\det
\begin{array}{|cccccc|c}
\cline{1-6}
a_{m} & \cdots & & \cdots & a_{k-(n-k-1)}& x^{n-k-1}&\\
& \ddots & && \vdots  & \vdots &\scriptstyle (n-1)-(k-1) \\
&  &a_{m}&\cdots &a_k& x^0 & \\
\cline{1-6}
b'_{n-1} &\cdots & & \cdots &b'_{k-(m-k)}&0&\\
&\ddots &&&\vdots & \vdots &\scriptstyle m-(k-1) \\
&& b'_{n-1} &\cdots &b'_k & 0&\\
\cline{1-6}
\end{array}\big)\\
& \ =(-1)^{m+n}\det \begin{array}{|cccccc|c} \cline{1-6}
0 & a_{m}&\cdots & & \cdots & a_{k-(n-k-2)}&\\
& &\ddots  &&& \vdots   &\scriptstyle n-1-k \\
&  &&a_{m}&\cdots &a_k & \\
\cline{1-6}
b'_{n-1}&\cdots&\cdots & &\cdots  &b'_{k-(m-k)}&\\
&\ddots &&& & \vdots&\scriptstyle m-k+1 \\
&& b'_{n-1} &\cdots & &b'_k&\\
\cline{1-6}
\end{array} \
=\ (-1)^{m-k+1}\det \begin{array}{|cccccc|c} \cline{1-6}
a_{m} &\cdots & & \cdots & a_{k-(n-k-2)}& &\\
& \ddots & &\vdots& &  &\scriptstyle n-1-k \\
 &  &a_{m}&\cdots &a_k && \\
\cline{1-6}
b'_{n-1}&\cdots & & \cdots &b'_{k-(m-k-1)}& &\\
&\ddots &&\vdots & &&\scriptstyle m-k \\
&& b'_{n-1} &\cdots &b'_k&&\\
\cline{1-6}
\end{array}.\\\end{aligned}$$]{} We apply elementary column operations on the matrix above, replacing the $j$-th column $C_j$ by $C_j-\beta C_{j-1}$ starting from the last column $C_{n+m-2k-1}$ up to the second column $C_2$, and using the relations in (\[bi\]): [$$\begin{aligned}
\label{coeff}
 {\mbox{coeff}}_{n-k-1}\big(F_{k-1}(f,\frac{g}{x-\beta})\big)=(-1)^{m-k+1}\det
\begin{array}{|cllcccc|c} \cline{1-7}
a_{m} &a_{m-1}-\beta a_m&\cdots & & \cdots & a_{k-(n-k-2)}-\beta a_{k+1-(n-k-2)}& &\\
 &\ddots && &\vdots& &  &\scriptstyle n-1-k \\
   &&a_{m}&a_{m-1}-\beta a_m&\cdots &a_k-\beta a_{k+1} && \\
\cline{1-7}
b_{n}&b_{n-1}&\cdots & & \cdots &b_{k+1-(m-k-1)}& &\\
&\ddots &&&\vdots & &&\scriptstyle m-k \\
&& b_{n} &b_{n-1}& \cdots &b_{k+1}&&\\
\cline{1-7}
\end{array}.\end{aligned}$$]{} Next consider [$$\begin{aligned}
&
F_k(f,g)(\beta)= \det
\begin{array}{|cccccc|c}
\cline{1-6}
a_{m} & \cdots & & \cdots & a_{k+1-(n-k-1)}& \beta^{n-k-1}&\\
& \ddots & && \vdots  & \vdots &\scriptstyle n-k\\
&  &a_{m}&\cdots &a_{k+1}& \beta^0 & \\
\cline{1-6}
b_{n} &\cdots & & \cdots &b_{k+1-(m-k-1)}&0&\\
&\ddots &&&\vdots & \vdots &\scriptstyle m-k \\
&& b_{n} &\cdots &b_{k+1}k & 0&\\
\cline{1-6}
\end{array}.\end{aligned}$$]{} We apply elementary row operations on the matrix above, replacing the $i$-th row $R_i$ by $R_i-\beta R_{i+1}$, starting from the first row $R_1$ up to row $R_{n-k-1}$: [$$\begin{aligned}
\label{Fk} \nonumber
F_k(f,g)(\beta)=\det
\begin{array}{|ccccccc|c}
\cline{1-7}
a_{m} & a_{m-1}-\beta a_m & && \cdots & a_{k+1-(n-k-1)}-\beta a_{k+2-(n-k-1)}& 0&\\
& \ddots && && \vdots  & \vdots &\scriptstyle n-k\\
&  &a_{m}\;\;a_{m-1}-\beta a_m&&\cdots &a_{k+2}-\beta a_{k+1}& 0 & \\
&  &a_{m}&a_{m-1}&\cdots &a_{k+1}& 1& \\
\cline{1-7}
b_{n} &\cdots & && \cdots &b_{k+1-(m-k-1)}&0&\\
&\ddots &&&&\vdots & \vdots &\scriptstyle m-k \\
&& b_{n} &\cdots &&b_{k+1} & 0&\\
\cline{1-7}
\end{array}\\
=(-1)^{m-k}\det\begin{array}{|ccccccc|c} \cline{1-7}
a_{m} & a_{m-1}-\beta a_m & && \cdots & a_{k+1-(n-k-1)}-\beta a_{k+2-(n-k-1)}& &\\
& \ddots && && \vdots  & &\scriptstyle n-k-1\\
&  &a_{m}\;\;a_{m-1}-\beta a_m&&\cdots &a_{k+2}-\beta a_{k+1}& & \\
 \cline{1-7}
b_{n} &\cdots & && \cdots &b_{k+1-(m-k-1)}&&\\
&\ddots &&&&\vdots &  &\scriptstyle m-k \\
&& b_{n} &\cdots &&b_{k+1} & &\\
\cline{1-7}
\end{array} .\end{aligned}$$]{} We obtain the first identity of the statement by comparing (\[coeff\]) and (\[Fk\]).

For the second identity, we have $$\begin{aligned}
G_k(f,g)(\alpha)&=(-1)^{(n-k)(m-k)}F_k(g,f)(\alpha)\\ & =(-1)^{(n-k)(m-k)+1}
{\mbox{coeff}}_{m-k-1}\big(F_{k-1}(g,\frac{f}{x-\alpha})\big)\\ &
=(-1)^{(n-k)(m-k)+1}(-1)^{(m-k)(n-k+1)}{\mbox{coeff}}_{m-k-1}\big(G_{k-1}(\frac{f}{x-\alpha},g)\big)\\
& =(-1)^{m-k-1}{\mbox{coeff}}_{m-k-1}\big(G_{k-1} (\frac{f}{x-\alpha},g)\big).\end{aligned}$$

As an immediate consequence we obtain the specialization property of subresultants which seemed to have been stated and proved for the first time in [@RoSp Prop. 4.1].

\[corrs\] For any root $\alpha $ of $f$, any root $\beta$ of $g$ and any $0\le k < \min\{m,n\}$, we have $$\begin{aligned}
\bullet \ &\Sres_k(f,g)(\beta)  =(-1)^{m-k}\, {\mbox{coeff}}_k \big(\Sres_k(f\,
,\,\frac{g}{x-\beta})\big)\,f(\beta), \\ \bullet \ &
\Sres_k(f,g)(\alpha) = {\mbox{coeff}}_k \big(\Sres_k(\frac{f}{x-\alpha}\,
,\,g)\big)\,g(\alpha).\end{aligned}$$ Here ${\mbox{coeff}}_k$ denotes the coefficient of order $k$ of the corresponding polynomial.

It is sufficient to prove the first identity, since the second identity is a consequence of $$\Sres_k(g,f)=(-1)^{(m-k)(n-k)}\Sres_k(f,g).$$ By and the previous lemma, $$\Sres_k(f,g)(\beta)=F_k(f,g)(\beta)\,f(\beta)=- \,{\mbox{coeff}}_{n-k-1}\big(
F_{k-1}(f,\frac{g}{x-\beta})\big)\,f(\beta).$$ Now it is immediate to verify by the definition of the principal scalar subresultant of order $k$ that $${\mbox{coeff}}_{n-k-1}\big(F_{k-1}(f,\frac{g}{x-\beta})\big)=(-1)^{m-k-1}{\mbox{coeff}}_k\big(\Sres_k(f,\frac{g}{x-\beta})\big).$$

Proof of Theorem \[theorem\]
============================

It turns out that the cases of Theorem 1 where $k$ is “big" are easy to prove by induction and will be used later in the other cases. That is why we start with this case first in the following proposition. The proof will use a lemma for the extremal cases $(p,n)$ and $(m,q)$, which is given after the proposition. We recall that $\ov p=m-p$, $ \ov q= n-q$, and $\ov k=m+n-k-1$.

\[casogrande\] Set $1\le m\le n$ and let $0\le p\le m,\ 0\le q\le n$ and $k=p+q$ be such that $n-1\le k\le m+n-1$, i.e. $0\le \ov k\le m$, when $m<n$ or $m\le k\le 2m-1$, i.e. $0\le \ov k\le m-1$, when $m=n$. Then $$\operatorname{Sylv}^{p,q}(A,B)
= (-1)^{\ov p\,\ov q + n-p-1+nq}\Big({\ov k\choose \ov p}F_{\ov
k}(f,g)\,f - {\ov k\choose \ov q}G_{\ov k}(f,g)\,g\Big).$$

By induction on $\ov k\ge 0$:\
The case $\ov k=0$ implies $(p,q)=(m-1,n)$ or $(p,q)=(m,n-1)$ and will follow from Lemma \[teorema1\].

Now set $\ov k>0$.\
– For $p=m$ and $q<n$ or $p<m$ and $q=n$, also by Lemma \[teorema1\], $$\operatorname{Sylv}^{m,q}(A,B)=(-1)^{n-m-1+nq}F_{\ov q-1}(f,g)f \ \mbox{ and } \ \operatorname{Sylv}^{p,n}(A,B)=(-1)^{p}G_{\ov p-1}(f,g)g$$ accordingly, which matches the statement since in these cases ${\ov
k\choose \ov q}$ or ${\ov k \choose \ov p}$ equals $0$.

– For $p<m$ and $q<n$, we specialize $\operatorname{Sylv}^{p,q}(A,B)$ of degree $k\le m+n-2$ in the $m+n$ elements of $A\cup B$ by means of Lemma \[sylvester\] and the inductive hypothesis: $$\begin{aligned}
\operatorname{Sylv}^{p,q}(A,B) (\alpha)& =
(-1)^{p}\,{\mbox{coeff}}_{k}\big(\operatorname{Sylv}^{p,q}(A-\alpha,B)\big)\,g(\alpha)\\
&= (-1)^{c'+p}\,{\mbox{coeff}}_k\Big({\ov k-1\choose \ov p -1}\Sres_{\ov
k-1}(\frac{f}{x-\alpha},g) -{\ov k\choose \ov q}G_{\ov
k-1}(\frac{f}{x-\alpha},g)g\Big)\,g(\alpha),\end{aligned}$$ by Identity . Here $c'= (\ov p-1)\ov q+n-p-1+nq$.\
Note that we are looking for the coefficient of degree $k$ of the expression between brackets; the condition $\ov k -1\le m-1<n-1\le
k$ in case $m<n$ and $\ov k -1\le m-2 <k$ in case $m=n$ imply in both cases that $\deg(\Sres_{\ov k-1}(\frac{f}{x-\alpha},g))\le\ov
k-1<k$. Then $$\operatorname{Sylv}^{p,q}(A,B) (\alpha)
= (-1)^{c'+p}\,{\mbox{coeff}}_k\Big(-{\ov k\choose \ov q}G_{\ov
k-1}(\frac{f}{x-\alpha},g)g\Big)\,g(\alpha).$$ When $\ov k-1<m-1 $, i.e $k\ge n$, we apply Lemma \[dos\] and get $$\begin{aligned}
\operatorname{Sylv}^{p,q}(A,B) (\alpha)
&= (-1)^{c'+p}\,\Big( -{\ov k\choose \ov q}\,{\mbox{coeff}}_{k-n}\big(G_{\ov k-1}(\frac{f}{x-\alpha},g)\big)\,g(\alpha)\Big)\\
&= (-1)^{c'+p+k-n}\,\Big( -{\ov k\choose \ov q}G_{\ov
k}(f,g)(\alpha)g(\alpha)\Big)\\&= (-1)^{\ov p\,\ov q + n-p-1+nq}
\,\Big( -{\ov k\choose \ov q}G_{\ov
k}(f,g)(\alpha)g(\alpha)\Big).\end{aligned}$$ When $\ov k-1=m-1$, $G_{\ov k-1}(\frac{f}{x-\alpha},g)=0=G_{\ov k}(f,g)$ and therefore we also get $$\operatorname{Sylv}^{p,q}(A,B) (\alpha)=(-1)^{\ov p \,\ov q+ n-p-1+nq} \,
\Big( -{\ov k\choose \ov q}G_{\ov k}(f,g)(\alpha)g(\alpha)\Big).$$

Analogously, $$\begin{aligned}
\operatorname{Sylv}^{p,q}(A,B) (\beta)& = (-1)^{q+\ov
p+c''}\,{\mbox{coeff}}_k\Big({\ov k \choose \ov p}F_{\ov
k-1}(f,\frac{g}{x-\beta})f -{\ov
k-1\choose \ov q-1}\Sres_{\ov k-1}(f,\frac{g}{x-\beta}))\Big)\,f(\beta)\\
&= (-1)^{q+\ov p+c''}\,{\mbox{coeff}}_k\Big({\ov k \choose \ov p}F_{\ov k-1}(f,\frac{g}{x-\beta})f\Big)\,f(\beta)\\
&= (-1)^{q+\ov p+c''}\,{\ov k \choose \ov p}\,{\mbox{coeff}}_{k-m}\big(F_{\ov
k-1}(f,\frac{g}{x-\beta})\big)\,f(\beta)\\ &= (-1)^{q+\ov
p+c''+1}\,{\ov k \choose \ov p}\,F_{\ov k}(f,g)(\beta) \,f(\beta),\end{aligned}$$ where $c''=\ov p\,(\ov q-1) +n-1-p -1+(n-1)q $. Therefore, $$\operatorname{Sylv}^{p,q}(A,B) (\beta)=(-1)^{\ov p \,\ov q+ n-p-1 +nq}\,{\ov k \choose
\ov p}\,F_{\ov k}(f,g)(\beta) \,f(\beta).$$ This concludes the proof.

The next lemma covers the cases $(p,n)$ and $(m,q)$ needed in the proof of the previous result. Observe that $$\begin{aligned}
\operatorname{Sylv}^{p,n}(A,B)  &=
 g\,\sum_{ A^{\prime }\subset
A,
|A^{\prime}|=p}R(x,A^{\prime
})\,\frac{R(A^{\prime},B)}{R(A^{\prime},A- A^{\prime
})}\ \mbox{ for } \ p\le m,\\
\operatorname{Sylv}^{m,q}(A,B)  &=  f\, \sum_{B'\subset B, |B'|=q} R(x,B')\frac{R(A,B')}{R(B',B-B')} \ \mbox{ for } \ q\le n.\end{aligned}$$

\[teorema1\] Set $1\le m\le n$. Then

1.  \[item1\] $\operatorname{Sylv}^{p,n}(A,B) =  (-1)^{p}G_{\ov p-1}(f,g)\,g$ for $0\le p\le m-1$, i.e. $1\le \ov p \le m$.

2.  \[item4\] $\operatorname{Sylv}^{m,q}(A,B) =
    (-1)^{n-m-1+nq}F_{\ov q-1}(f,g)\,f$   for $n-m-1\le q\le n-1$, i.e. $1\le \ov q \le m+1$, when $m<n$ and for $0\le q\le m-1$, i.e. $1\le  \ov q \le m$, when $m=n$.

(\[item1\]) By induction on $m\ge 1$.\
The case $m=1$ is clear from Identities \[m0\] and \[Gm-1\], since in this case $p=0$ and $\ov{p}=1$.

Now set $m>1$ and let $0\le p\le  m-1$. Both $\operatorname{Sylv}^{p,n}(A,B)$ and $G_{\ov p-1 }(f,g)\,g$ are polynomials of degree bounded by $p+n<m+n$ and we compare them by specializing them into the $m+n$ elements $\alpha \in A$ and $\beta\in B$. Clearly both expressions vanish at every $\beta\in B$ and so we only need to compare them at $\alpha\in A$.\
– For $p<m-1$, we apply Lemma \[sylvester\], the inductive hypothesis and Lemma \[dos\] (and the fact that $g$ is monic): $$\begin{aligned}
\operatorname{Sylv}^{p,n}(A,B)(\alpha)&=(-1)^p\,{\mbox{coeff}}_{p+n}\big(\operatorname{Sylv}^{p,n}(A-\alpha,B)\big)\,g(\alpha)\\
&=(-1)^{2p}\,{\mbox{coeff}}_{p+n}\big(G_{(m-1)-p-1}(\frac{f}{x-\alpha},g)\,g\big)\,g(\alpha)\\
&= {\mbox{coeff}}_{p}\big(G_{(m-1)-p-1}(\frac{f}{x-\alpha},g)\big)\,g(\alpha)\
= \ (-1)^{p}\, G_{\ov p-1}(f,g)(\alpha)\,g(\alpha).\end{aligned}$$ – For $p=m-1$: $$\begin{aligned}
\operatorname{Sylv}^{p,n}(A,B)(\alpha)&=R(\alpha, A-\alpha)\frac{R(A-\alpha,B)}{R(A-\alpha,\alpha)}\,g(\alpha)\\
 &=(-1)^{m-1}\prod_{\alpha'\in A} g(\alpha')\ = \ (-1)^{m-1}\Res(f,g)\ = \ (-1)^{m-1}G_0(f,g)(\alpha)g(\alpha)
 ,\end{aligned}$$ by Identity  and the fact that $\Res(f,g)=F_0(f,g)f+G_0(f,g)g$ has degree $0$ in $x$. Therefore $\operatorname{Sylv}^{p,n}(A,B)=(-1)^{p}G_{\ov p-1}(f,g)\,g$.

(\[item4\]) By induction on $n\ge m$.\
For $n=m$, by Item (\[item1\]) we have that for $0\le q\le m-1$, $$\begin{aligned}
\operatorname{Sylv}^{m,q}(A,B)&=(-1)^{mq}\operatorname{Sylv}^{q,m}(B,A)\ = \ (-1)^{mq+q}G_{\ov q-1}(g,f)\,f\\&=(-1)^{mq+q}
 (-1)^{(m-(\ov q-1))(n-(\ov q-1))}F_{\ov q-1}(f,g)\,f \ = \
 (-1)^{nq-1}F_{\ov q-1}(f,g)\,f.\end{aligned}$$ Now set $n\ge m+1$ and let $n-m-1\le q\le n-1$. Both $\operatorname{Sylv}^{m,q}(A,B)$ and $F_{\ov q-1}(f,g)\,f$ are polynomials of degree bounded by $m+q<m+n$ and we compare them by specializing them in the $m+n$ elements $\alpha \in A$ and $\beta\in B$. Clearly both expressions vanish at every $\alpha\in A$ and so we only need to compare them at $\beta\in B$.\
– For $q<n-1$, we apply Lemma \[sylvester\], the inductive hypothesis and Lemma \[dos\]: $$\begin{aligned}
\operatorname{Sylv}^{m,q}(A,B)(\beta)&=(-1)^{q}\,{\mbox{coeff}}_{m+q}\big(\operatorname{Sylv}^{m,q}(A,B-\beta)\big)\,f(\beta)\\
&=(-1)^{q+(n-1-m-1)+(n-1)q}
\,{\mbox{coeff}}_{m+q}\big(F_{(n-1)-q-1}(f,\frac{g}{x-\beta})\,f\big)\,f(\beta)\\
&= (-1)^{(n+m-2+nq)+1}\, F_{\ov q-1}(f,g)(\beta)\,f(\beta).\end{aligned}$$ – For $q= n-1$, $$\begin{aligned}
\operatorname{Sylv}^{m,n-1}(A,B)(\beta)&=f(\beta)\,R(\beta, B-\beta)\frac{R(A,B-\beta)}{R(B-\beta,\beta)}\\
 &=(-1)^{n-1+m(n-1)}\prod_{\beta'\in B} f(\beta')\ = \ (-1)^{(mn+n-m-1)+mn}\Res(f,g)\\ &= \ (-1)^{n-m-1}F_0(f,g)(\beta)f(\beta).\end{aligned}$$ Therefore $\operatorname{Sylv}^{m,q}(A,B)=(-1)^{n-m-1+nq}F_{\ov q-1}(f,g)\,f$ as wanted.\

As a particular case of Proposition \[casogrande\], using Identities  and , we obtain Case (2) and a particular case of Case (4) of the introduction:

\[particular\][ ]{}\

1.  \[k=m=n\] Set $1\le m=n$ and let $ 0\le p, \ 0\le q$ be such that $p+q=m$. Then $$\operatorname{Sylv}^{p,q}(A,B)  =  {m
    -1\choose q} \,f+{m-1\choose p} \,g.$$

2.  \[mm+2\] Set $1\le m=n-2$ and let $0\le p\le m$, $0\le q$ be such that $p+q=n-1$. Then $$\operatorname{Sylv}^{p,q}(A,B)=(-1)^{p+1}{m \choose p}\,f.$$

This allows us to simplify the rather long proofs for the cases when $p+q=m<n$, which appeared previously in [@LP], [@DHKS07] and [@RoSp].

\[k=m&lt;n\] Set $1\le m\le n-1$ and let $p\ge 0, \ q\ge 0$ be such that $1\le p+q=m$. Then $$\operatorname{Sylv}^{p,q}(A,B)  =  {m \choose p} \,f.$$

By induction on $n\ge m+1$, comparing the two expressions at the $n>m$ elements of $B$.\
For $n=m+1$, by Lemma \[sylvester\] and Corollary \[particular\], $$\begin{aligned}
\operatorname{Sylv}^{p,q}(A,B) (\beta)&= {\mbox{coeff}}_{m}\big(\operatorname{Sylv}^{p,q}(A,B-\beta)\big)\,f(\beta)\\
&={\mbox{coeff}}_{m}\big({m -1\choose q} f+{m-1\choose p}
\,\frac{g}{x-\beta}\big) \, f(\beta)\\& =  \Big({m -1\choose q}
+{m-1\choose p}\Big)f(\beta)\ = \ {m \choose p}f(\beta) .\end{aligned}$$ Now set $n>m+1$, $$\operatorname{Sylv}^{p,q}(A,B) (\beta)\ =\ {\mbox{coeff}}_{m}\big(\operatorname{Sylv}^{p,q}(A,B-\beta)\big)\,f(\beta)
\ = \ {\mbox{coeff}}_{m}\big({m\choose p} f\big)\,f(\beta) \ =   \ {m \choose p}\,f(\beta) .$$

We finish the proof of Theorem \[theorem\] by splitting it into the two remaining cases to be proven. The first case is the inductive proof of [@RoSp] that we repeat here for the sake of completeness.

\[kchico\] Set $1\le m\le n$ and let $p\ge 0$, $q\ge 0$ and $k=p+q$ be such that $k\le m$ when $m<n$ and $k<m$ when $m=n$. Then $$\operatorname{Sylv}^{p,q}(A,B)
=(-1)^{p(m-k)}{k\choose p}\Sres_k(f,g).$$

By induction on $m\ge 1$:\
The case $ m=1$ is completely covered by Identities , , and Proposition \[k=m&lt;n\].\
Now set $m>1$ and let $0\le k=p+q\le m$ if $m<n$ and $0\le k=p+q<m$ if $m=n$. We have\
– For $0\le k\le m-1$, we compare $\operatorname{Sylv}^{p,q}(A,B)$ and $\Sres_k(f,g) $, which are both of degree $k<m$, by specializing them into the $m$ elements $\alpha \in A$ by means of Lemma \[sylvester\], the inductive hypothesis and Corollary \[corrs\]: $$\begin{aligned}
\operatorname{Sylv}^{p,q}(A,B)(\alpha)&=
(-1)^p\,{\mbox{coeff}}_k\big(\operatorname{Sylv}^{p,q}(A-\alpha,B)\big)\,g(\alpha)\\
&=(-1)^{p} (-1)^{p(m-1-k)}\,{k\choose
p}\,{\mbox{coeff}}_k\big(\Sres_{k}(\frac{f}{x-\alpha},g)\big) \,g(\alpha)\\ &=
(-1)^{p(m-k)}\,{k\choose p}\,\Sres_k(f,g)(\alpha).\end{aligned}$$ – For $k= m<n$, it is Proposition \[k=m&lt;n\].

Set $1\le m\le n-3$ and let $0\le p\le m$, $0\le q\le n$ be such that $m+1\le p+q\le n-2$. Then $$\operatorname{Sylv}^{p,q}(A,B)=0.$$

By induction on $n\ge m+3$, specializing the expression in the $n>m+1=k$ elements of $B$ by Lemma \[sylvester\].\
For $n=m+3$, by Corollary \[particular\]: $$\operatorname{Sylv}^{p,q}(A,B)(\beta)  =  -\,f(\beta){\mbox{coeff}}_{m+1}\big(\operatorname{Sylv}^{p,q}(A,B-\beta) \big)=-\,f(\beta){\mbox{coeff}}_{m+1} \big((-1)^{p+1}{m \choose p}\,f\big)=0
,$$ since $\deg(f)=m<m+1$.\
The case $n>m+3$ follows immediately.

T. Krick would like to thank the Mittag-Leffler Institute for hosting her in May 2011, during the preparation of this note.

D’Andrea, Carlos; Hong, Hoon; Krick, Teresa; Szanto, Agnes. J. Symb. Comput. Vol. [**42**]{} (2007) 290–297.

D’Andrea, Carlos; Hong, Hoon; Krick, Teresa; Szanto, Agnes. J. Symbolic Comput. Vol. [**44**]{} (2009) 1164–1175.

Lascoux, Alain; Pragacz, Piotr. J. Symbolic Comput. 35 (2003), no. 6, 689–710.

Roy, Marie-Françoise; Szpirglas, Aviva. J. Symbolic Comput. (to appear) (2010).

Sylvester, James Joseph. Philosophical Transactions of the Royal Society of London, Part III (1853), 407–548. Appears also in Collected Mathematical Papers of James Joseph Sylvester, Vol 1, Chelsea Publishing Co. (1973), 429-586.

[^1]: Krick was partially supported by the research projects UBACyT X-113 2008-2010 and CONICET PIP 2010-2012 (Argentina); Szanto’s was partially supported by NSF grant CCR-0347506 (USA).
