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I'm trying to find a general proof to an exercise given in Garrity et al's book, Algebraic Geometry: A problem-solving approach.

The problem is this: Given two polynomials f and g, show that for each pair of roots, $f(r) = 0, g(s) = 0$, that $(r - s)$ divides the resultant.

There is a book of selected answers, but somewhat disappointingly, the solution is given as a brutal appeal to algebra. Moreover, the result is only given for quadratic polynomials.

It seems cited in a few places that the resultant, defined as the determinant of the Sylvester matrix of two polynomials $f = \lambda_1\prod (x - r_i)$ and $g = \lambda_2 \prod (x - s_i)$, is equal to the product $\prod r_i - s_i$. But so far, I have been unable to find a general proof of this fact.

Would anyone either mind sketching the proof, or else pointing me to a resource which does?

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  • $\begingroup$ This is such a basic property that I imagine it wouldn't be too hard to find it in online lecture notes. You can try algebraic number theory. $\endgroup$ Commented Mar 15, 2015 at 7:52
  • $\begingroup$ The notes by Svante Janson are very detailed and clear: www2.math.uu.se/~svante/papers/sjN5.pdf $\endgroup$
    – Sha Vuklia
    Commented Jan 5, 2023 at 15:10

3 Answers 3

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The following is really only a sketch. Feel free to ask for more details.

The coefficients of a polynomial $f$ are equal to elementary symmetric polynomials in the roots of $f$. Since the resultant is a polynomial function in the coefficients of two polynomials $f$ and $g$, it is a symmetric polynomial function in the roots of $f$ and $g$. The definition of the resultant is made in such a way that $\operatorname{res}(f,g)=0$ if (and only if) $f$ and $g$ share a common root. Now, we use the following:

Lemma. Let $R$ be an integral domain and denote by $K$ the algebraic closure of its quotient field. Let $p\in R[X,Y]$ a polynomial such that $p(a,a)=0$ for all $a\in K$. Then, $(Y-X)$ divides $p$.

Proof. Write $p\in R[X][Y]$ as a polynomial in $Y$, i.e. $p=\sum_{i=0}^n p_i Y^i$ with certain $p_i\in R[X]$. In this integral domain, perform division with remainder of $p$ by $Y-X$ to obtain $p=q(Y-X)+r$ for $q\in R[Y,X]$ and $r\in R[X]\subseteq R[X,Y]$. Since $r(a)=r(a,a)=p(a,a)=0$ for all $a\in K$, we must have $r=0$. Indeed, $K$ is an infinite field because it is algebraically closed and any nonzero polynomial has only finitely many roots. Consequently, $p=(X-Y)\cdot q$ as required.

Applying this lemma to the resultant as a polynomial in the zeros of $f$ and $g$, you get the desired statement.

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  • $\begingroup$ Thank you. This proof seems to be much more palatable than the one I found involving the determinant of the product of the Sylvester matrix with the Vandermonde matrix! $\endgroup$
    – Tac-Tics
    Commented Mar 16, 2015 at 17:45
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We take a monic polynomial

$$g(x):=c_0+c_1x+\dots+c_nx^n$$

with $c_n=1$ and a second polynomial

$$h(x):=d_0+d_1x+\dots+d_mx^m$$

with $d_m\ne0$. Every polynomial

$$f(x):=a_0+a_1x+\dots+a_{m+n-1}x^{m+n-1}$$

of degree less than $m+n$ can uniquely be written to

$\tag{1}f(x)=q(x)g(x)+r(x)$

with $\deg(r)<n$. Therefore there exist two basis

$$B:=K[1,x,\dots,x^{n+m-1}]$$

and

$$\hat B:=K[1,x,\dots,x^{n-1},g(x),xg(x),\dots,x^{m-1}g(x)]$$

in which this polynomial can be expressed. The polynomial

$$f(x):=\hat r_0+\hat r_1x+\dots+\hat r_{n-1}x^{n-1}+\hat q_0g(x)+\hat q_{1}xg(x)+\dots+\hat q_{m-1}x^{m-1}g(x)\in\hat B$$

can be transformed to an element $f(x)\in B$ by the matrix $\hat T$

$$\tag{2}\hat T \begin{pmatrix} \hat q_{m-1} \\ \vdots \\ \hat q_{0} \\ \hat r_{n-1} \\ \vdots \\ \hat r_{0} \end{pmatrix} := \begin{pmatrix} c_{n} \\ c_{n-1} & c_{n} \\ \vdots & c_{n-1} & \ddots \\ c_{0} & \vdots & \ddots & c_{n} \\ & c_{0} & \ddots & c_{n-1} & 1 & 0 & 0 \\ & & \ddots & \vdots & \vdots & \ddots & 0 \\ & & & c_{0} & 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} \hat q_{m-1} \\ \vdots \\ \hat q_{0} \\ \hat r_{n-1} \\ \vdots \\ \hat r_{0} \end{pmatrix} = \begin{pmatrix} a_{m+n-1} \\ \vdots \\ a_{0} \end{pmatrix}$$

with $m$ columns of the coefficients of the polynomial $g(x)$ and with a $n\times n$ identy matrix in the right bottom of the matrix $\hat T$. With $c_n=1$ the matrix $\hat T$ has determinant $1$ and it is invertible.

Now we introduce the homomorphism

$$\tag{3}p(x):=R(f):=g(x)q(x)+h(x)r(x)$$

with the unique polynomials $q(x)$ and $r(x)$ in equation $(1)$. We write the homomorphism $R(f)$ from the basis $\hat B$ of the polynomial $f(x)$ to the basis $B$

$$\hat S \begin{pmatrix} \hat q_{m-1} \\ \vdots \\ \hat q_{0} \\ \hat r_{n-1} \\ \vdots \\ \hat r_{0} \end{pmatrix}= \begin{pmatrix} p_{m+n-1} \\ \vdots \\ p_{0} \end{pmatrix}$$

with the matrix $\hat S$

$$\tag{4}\begin{pmatrix} c_{n} & & & & d_{m} \\ c_{n-1} & c_{n} & & & d_{m-1} & d_{m} \\ \vdots & c_{n-1} & \ddots & & \vdots & d_{m-1} & \ddots \\ c_{0} & \vdots & \ddots & c_{n} & d_{0} & \vdots & \ddots & d_{m} \\ & c_{0} & \ddots & c_{n-1} & & d_{0} & \ddots & d_{m-1} \\ & & \ddots & \vdots & & & \ddots & \vdots \\ & & & c_{0} & & & & d_{0} \end{pmatrix} \cdot \begin{pmatrix} \hat q_{m-1} \\ \vdots \\ \hat q_{0} \\ \hat r_{n-1} \\ \vdots \\ \hat r_{0} \end{pmatrix} = \begin{pmatrix} p_{m+n-1} \\ \vdots \\ p_{0} \end{pmatrix}$$

with $m$ columns with the coefficients of the polynomial $f(x)$ and $n$ columns with the coefficients of the polynomila $h(x)$ and with the polynomial

$$p(x):=p_0+p_1x+\dots+p_{m+n-1}x^{m+n-1}\in B.$$

In multiplying $(4)$ with $\hat T^{-1}$ we get

$$\hat T^{-1}\hat S \begin{pmatrix} \hat q_{m-1} \\ \vdots \\ \hat q_{0} \\ \hat r_{n-1} \\ \vdots \\ \hat r_{0} \end{pmatrix}= \begin{pmatrix} \hat p_{q,m-1} \\ \vdots \\ \hat p_{q,0} \\ \hat p_{r,n-1} \\ \vdots \\ \hat p_{r,0} \end{pmatrix}$$

with

$$p(x):=\hat p_{r,0}+\hat p_{r,1}x+\dots+\hat p_{r,n-1}x^{n-1}+\hat p_{q,0}g(x)+\hat p_{q,1}xg(x)+\dots+\hat p_{q,m-1}x^{m-1}g(x)\in\hat B.$$

Obviously the matrix $\hat S$ is Sylvester's matrix (or the transpose) and we have developed all means in order to deal with the question. We will take two different looks into the question.

I. The domain $K[x]$ is a Euclidean domain. Therefore the gratest common divisor $d(x):=\gcd(g(x),h(x))$ exists and we get the identity of the ideals

$$\langle d(x)\rangle=\langle g(x), h(x)\rangle.$$

This means that any polynomial $f(x)$ divisible by $d(x)$ can be written to

$$u(x)g(x)+v(x)h(x)=f(x).$$

with two polynomials $u(x)$ and $v(x)$. For any such equation the polynomial $f(x)$ must be divisible by the greatest common divisor $d(x)$. But for any polynomial $f(x)$ of degree less than $n+m-1$ the equation $(4)$ just renders a solution to this equation (with $u(x)=\hat q(x)$, $v(x)=\hat r(x)$ and $p(x)=f(x)$ in this equation). Because the polynomial $f(x)$ must be divisible by the polynomila $d(x)$ there exists a solution for polynomials $f(x)=f_1(x)\cdot d(x)$ with a polynomial $f_1(x)$ of degree less than $n+m-\deg(d(x))$ or the image of Sylvester's matrix has rank $n+m-\deg(d(x))$. In particular Sylvester's matrix has determinant unequal to $0$ if the greatest common divisor $d(x)$ has the degree $0$.

II. In the module $M[x]:=K[X]/ g(X)K[X]=K[X]/\langle g(X)\rangle$ there exists a homorphism $R$ that is a multiplication with the polynomial $h(x)$. For any polynomial $w(x)$ of degree less than $n$ we write

$$R(w):=h(x)\cdot w(x).$$

This renders a polynomial $R(w)$ of degree less than $n$ over the module $M[x]$. In equation $(3)$ we obtain

$$\tag{5}R(w):=q(x)g(x)+h(x)w(x)=h(x)w(x)$$

with $q(x)=0$. The homomorphism $R$ over the module $M[x]$ can be written to

$$\begin{pmatrix} a_{11} & \dots & a_{1,n-1} \\ \vdots & \ddots & \vdots \\ a_{n-1,1} & \dots & a_{n-1,n-1} \\ \end{pmatrix} \cdot \begin{pmatrix} w_{n-1} \\ \vdots \\ w_{0} \end{pmatrix} = \begin{pmatrix} p_{n-1} \\ \vdots \\ p_{0} \end{pmatrix}$$

with

$$w(x):=w_0+w_1x+\dots+w_{n-1}x^{n-1}$$

and

$$R(w)=p(x):=p_0+p_1x+\dots+p_{n-1}x^{n-1}.$$

With

$$\tag{6}h(x)x^j={\sum_{i=0}^{n-1}a_{ij}x^î}+g_j(x)g(x),\;\;0\le j\le n-1$$

and an arbitrary polynomial $q(x)$ we write the equation $(5)$ in the basis $\hat B$ by a matrix and obtain

$$(r_{ij})\cdot \begin{pmatrix} q_{m-1} \\ \vdots \\ q_0 \\ w_{n-1} \\ \vdots \\ w0 \end{pmatrix} := \begin{pmatrix} E_{m\times m} & G \\ 0 & (a_{ij}) \end{pmatrix} \cdot \begin{pmatrix} q_{m-1} \\ \vdots \\ q_0 \\ w_{n-1} \\ \vdots \\ w_0 \end{pmatrix} = \begin{pmatrix} \eta_{m-1} \\ \vdots \\ \eta_0 \\ p_{n-1} \\ \vdots \\ p0 \end{pmatrix}$$

with the identity matrix $E_{m\times m}$ and the matrix $G$ that contains the coefficients of the polynomials $g_j(x)$ in equation $(6)$. But the matrix $(r_{ij})$ is just the matrix $\hat T^{-1}\hat S$ and we get

$$det(r_{ij})=det(a_{ij})=det(\hat T^{-1}\hat S)=det(\hat T^{-1})\cdot det(\hat S)=det(\hat S)$$ with $det(\hat T^{-1})=1$. Therefore the determinant of the homophism $R(w)$ is $det(a_{ij})=det(\hat S)$. Moreover this determinant is the norm $N_K^{M[x]}(h(x))$ of the polynomial $h(x)$.

For a linear polynomial $h_1(x)=x-\alpha$ we obtain

$$\det\hat S= \det\begin{pmatrix} c_{n} & 1 \\ c_{n-1} & -\alpha & 1 \\ c_{n-2} & 0 & -\alpha & \ddots \\ \vdots & \vdots & \vdots & \ddots & 1 \\ c_{0} & 0 & 0 & 0 & -\alpha \\ \end{pmatrix}= {c_n(-\alpha)^n- \det\begin{pmatrix} c_{n-1} & 1 \\ c_{n-2} & -\alpha & 1 \\ c_{n-3} & 0 & -\alpha & \ddots \\ \vdots & \vdots & \vdots & \ddots & 1 \\ c_{0} & 0 & 0 & 0 & -\alpha \\ \end{pmatrix}}=\dots= {(-1)^n\{c_n\alpha^n+c_{n-1}\alpha^{n-1}+\dots+c_0\}}=(-1)^ng(\alpha).$$

In the splitting field $L\supseteq K$ of the polynomial $h(x)$ we get $h(x)=d_m\prod_{i=1}^m(x-\xi_i)$ with the zeros $\xi_i$. The norm is multiplicative over the module $M[x]$. For $h(x)=h_1(x)\cdot h_2(x)$ we obtain two matrices $(a_{ij}^{(1)})$ for the product with $h_1(x)$ and $(a_{ij}^{(2)})$ for the product with $h_2(x)$. And the matrix $(a_{ij})$ for the product with $h(x)$ must be the product $(a_{ij}^{(1)})\cdot (a_{ij}^{(2)})$ because it is equal to the product $h_1(x)\cdot h_2(x)$. Therefore we obtain with $M_1[x]:=L[x]/g(x)L[x]$

$$N_K^{M_1[x]}(h(x))=N_K^{M_1[x]}(d_m)\prod_{i=1}^{m}N_K^{M_1[x]}(x-\xi_i)=(-1)^{mn}d_m^n\prod_{i=1}^{m}g(\xi_i)$$

and Sylvester's matrix is $0$ if one of the zeros $\xi_i$ is also a zero of the polynomial $g(x)$. In particular we get with $g(x)=c_n\prod_{j=1}^n(x-\zeta_j)$

$$\det(\hat S)=c_n^md_m^n\prod_{i=1}^{n}\prod_{j=1}^{m}(\zeta_i-\xi_j)=(-1)^{mn}c_n^md_m^n\prod_{i=1}^{m}\prod_{j=1}^{n}(\xi_i-\zeta_j).$$

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I found the same formula in Frederic Butin's book (Polynomials, Galois Theory and application). I am not sure whether those one liners or two liners after that is a proof or not. If they are then definitely I did not get that. So I tried myself.

If there are two polynomials $P = \Sigma_{i=0}^{m} p_iX^i, Q = \Sigma_{j=0}^{n} q_jX^j$, then the following matrix

$ \begin{pmatrix} p_m & p_{m-1} & . & . & . & p_0 & . & . & 0 \\ 0 & p_m & . & . & . & p_0 & . & . & 0 \\ .\\ .\\ q_n & q_{n-1} & . & . & . & q_0 & . & . & 0 \\ 0 & q_n & . & . & . & q_0 & . & . & 0 \\ .\\ .\\ \end{pmatrix} \\ $

is called Sylvester matrix $S_{p,q}$. This is the form there in Wikipedia. And the determinant of this matrix is called resultant of $P,Q$ ($R_x(P,Q)$).

Step 1 - Let the two polynomials are $P$ and, $PQ$, i.e. first polynomial divides second polynomial. Then

$PQ = p_mq_nX^{m+n}+ (p_{m}q_{n-1}+p_{m-1}q_{n})x^{m+n-1}+...+p_0q_0$

The first $m+n$ rows will be occupied by coefficients of $P$, next $m$ rows will be occupied by coeficients of $PQ$. In the $m+n+1$ th row, the coefficients of $PQ$ will be occupying first $m+n+1$ spaces. $i$ th row is denoted as $R_i$. We will count elements in any row, considering 1st element as $0$ th element. So the $0$ th element of $R_{m+n+1}$ is $ p_mq_n$, $1st$ element is $p_{m}q_{n-1}+p_{m-1}q_{n}$ etc.

We call $kth$ element of $R_{m+n+1}$ as $c_k$ (coefficient of $x^{m+n-k}$).

If $n \geq m$ then

$ c_k = \begin{cases} p_{m}q_{n-k}+...+p_{m-k}q_{n} & \text{if } k\leq m \\ P_{m}q_{n-k}+...+p_{0}q_{n-(k-m)} & \text{if } m < k \leq n \\ p_{m-(k-n)}q_{0}+...+p_{0}q_{n-(k-m)} & \text{if } n < k \\ \end{cases} $

If $m > n$ then

$ c_k = \begin{cases} p_{m}q_{n-k}+...+p_{m-k}q_{n} & \text{if } k\leq n \\ p_{m-(k-n)}q_{0}+...+p_{m-k}q_{n} & \text{if } n < k \leq m \\ p_{m-(k-n)}q_{0}+...+p_{0}q_{n-(k-m)} & \text{if } m < k \\ \end{cases} $

So in $R_{m+n+1}, c_0 = p_mq_n$ etc.

And for the first $m$ rows,, $kth$ element in $R_x$ is $p_{m-(k-(x-1))} $, $p_{m-(k-(x-1))} =0 $ if $m-(k-(x-1)) < 0 $ or $m-(k-(x-1)) > m$ (obviously). So in first row (i.e. in $R_1$) $c_k = p_{m-k}$.

We do $R_{m+n+1} - q_{n-1}R_2 - q_{n-2}R_3 -...- q_{0}R_{n+1} $

For $n\geq m$ and $k\leq m$, we get

$p_{m}q_{n-k}+...+p_{m-k+1}q_{n-1}+ p_{m-k}q_{n} - q_{n-1}p_{m-(k-1)} - q_{n-2}p_{m-(k-2)}-...- q_{n-k}p_{m-(k-k)}-...$

Note if $ l > m \text{ or } l < 0, p_{l}= 0$ in above expression. (This information is used in following paragraph also).

So $c_k$ will be after this $c_k = p_{m-k}q_{n}$

For $n\geq m$ and $m < k \leq n$ we have $c_k - q_{n-1}p_{m-(k-1)} - q_{n-2}p_{m-(k-2)}-...- q_{n-k}p_{m-(k-k)}-... =\\ p_{m}q_{n-k}+...+p_{0}q_{n-(k-m)} - q_{n-1}p_{m-(k-1)} - q_{n-2}p_{m-(k-2)}-...- q_{n-k}p_{m-(k-k)}-... $

Note that in $- q_{n-1}p_{m-(k-1)} - q_{n-2}p_{m-(k-2)}-...- q_{n-k}p_{m-(k-k)}-...$ we will have for some $x$, $k-x=m$, so we will have $- q_{n-x}p_{m-(k-x)} = - q_{n-(k-m)}p_{0}$, this term and all the term right side of it will cancell all the terms in $p_{m}q_{n-k}+...+p_{0}q_{n-(k-m)}$. And all the terms left side of it are zeros. So $c_k = 0$

For $m > n$ and $n < k \leq m$ we have $c_k - q_{n-1}p_{m-(k-1)} - q_{n-2}p_{m-(k-2)}-...- q_{n-k}p_{m-(k-k)}-... =\\ p_{m-(k-n)}q_{0}+...+p_{m-k}q_{n} - q_{n-1}p_{m-(k-1)} - q_{n-2}p_{m-(k-2)}-...- q_{n-n}p_{m-(k-n)}-...\\ = p_{m-k}q_{n} $

For $n\geq m$ and $n < k $ we have $ p_{m-(k-n)}q_{0}+...+p_{0}q_{n-(k-m)} - q_{n-1}p_{m-(k-1)} - q_{n-2}p_{m-(k-2)}-...- q_{0}p_{m-(k-n)}$ Again this will be zero.( if $ l > m \text{ or } l < 0, p_{l}= 0$, also $q_i = 0$ for $i<0$ ).

So we see that $c_k = p_{m-k}q_{n}$ if $k \leq m $ (irrespective of whether $n>m$ or $m>n$) or else $c_k = 0 $

So $R_{m+n+1} - q_{n-1}R_2 - q_{n-2}R_3 -...- q_{0}R_{n+1} = q_nR_1$
Similarly we can see that $R_{m+n+2} - q_{n-1}R_3 - q_{n-2}R_4 -...- q_{0}R_{n+2} = q_nR_2$ .
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For row $m+n+m$ similarly we can see that $R_{m+n+m} - q_{n-1}R_{m+1} - q_{n-2}R_{m+2} -...- q_{0}R_{n+m} =q_nR_m$.

So the determinant is zero.
Fact1: So every row from $R_{m+n+1}$ to $R_{m+n+m}$ is linearly dependent on the first $m+n$ rows. This fact will be used shortly.

Step 2: If $Q=PU+H$, then $R_x(P,Q) = p_m^{m+n-r}R_x(P,H)$ where $m,n,r$ are degrees of $P,U,H$ respectively. (This is also mentioned in the book, without elaborate proof or any proof)

Consider $R_x(P,PU+H)$. Note that order of the Sylvester matrix for $P \text{ and } PU +H$ is same as that of $ P \text{ and } PU$ as degree of $ PU$ is greater than $H$. From the rule of sum of determinants we know that we can write the above determinant as sum of multiple determinants where one determinant will consist only the coefficients of $P$ and $H$, in rest of the determinants at least one row will be occupied by coefficients of $PU$, and from Fact1 we know that those determinants will be zero. The determinant which will consist only the coefficients of $P$ and $H$ will look like

$ \begin{vmatrix} p_m & p_{m-1} & . & . & p_0 & . & . & . & . & 0 \\ 0 & p_m & . & . & . & p_0 & . & . & . & 0 \\ .\\ .\\ 0 & . & 0 & p_m & . & . & . & p_0 & . & 0 \\ .\\ .\\ 0 & . & 0 & h_r & . & h_0 & . & . & . & 0 \\ ,\\ .\\ 0 & . & . & 0 & . & . & h_r & . & . & h_0 \\ \end{vmatrix} $ Note that in $R_{m+n+1}$, number of zeros before $h_r$ is $m+n-r$. Also diagonal element of first $m+ n - r$ rows is $p_m$ (and all zeros left side of $p_m$). So we can take out first $m+n-r$ rows and $m+n-r$ columns. And the determinant will look like

$p_m^{m+n-r} \begin{vmatrix} p_m & . & . & . & p_0 & . & 0 \\ .\\ .\\ h_r & . & h_0 & . & . & . & 0 \\ ,\\ .\\ 0 & . & . & h_r & . & . & h_0 \\ \end{vmatrix} = p_m^{m+n-r}R_x(P,H) $

For example, suppose degree of $P$ is two (m=2) and degree of $Q$ is three (m+n = 3) then $Q=PU+H = S + H$ , degree of $S = PU$ is three and degree of $H$ say one (r = 1), then the determinant

$ \begin{vmatrix} p_2 & p_{1} & p_{0} & 0 & 0 \\ 0 & p_2 & p_1 & p_0 & 0 \\ 0 & 0 & p_2 & p_1 & p_0 \\ s_3 & s_2& s_1+h_1 & s_0+h_0 & 0 \\ 0 & s_3 & s_2 & s_1+h_1 & s_0+h_0 \\ \end{vmatrix} = \\ \begin{vmatrix} p_2 & p_{1} & p_{0} & 0 & 0 \\ 0 & p_2 & p_1 & p_0 & 0 \\ 0 & 0 & p_2 & p_1 & p_0 \\ s_3 & s_2& s_1 & s_0 & 0 \\ 0 & s_3 & s_2 & s_1 & s_0\\ \end{vmatrix} + \begin{vmatrix} p_2 & p_{1} & p_{0} & 0 & 0 \\ 0 & p_2 & p_1 & p_0 & 0 \\ 0 & 0 & p_2 & p_1 & p_0 \\ s_3 & s_2& s_1 & s_0 & 0 \\ 0 & 0 & 0 & h_1 & h_0\\ \end{vmatrix} + \begin{vmatrix} p_2 & p_{1} & p_{0} & 0 & 0 \\ 0 & p_2 & p_1 & p_0 & 0 \\ 0 & 0 & p_2 & p_1 & p_0 \\ 0 & 0& h_1 & h_0 & 0 \\ 0 & s_3 & s_2 & s_1 & s_0\\ \end{vmatrix} + \begin{vmatrix} p_2 & p_{1} & p_{0} & 0 & 0 \\ 0 & p_2 & p_1 & p_0 & 0 \\ 0 & 0 & p_2 & p_1 & p_0 \\ 0 & 0& h_1 & h_0 & 0 \\ 0 & 0 & 0 & h_1 & h_0\\ \end{vmatrix} =\\ \begin{vmatrix} p_2 & p_{1} & p_{0} & 0 & 0 \\ 0 & p_2 & p_1 & p_0 & 0 \\ 0 & 0 & p_2 & p_1 & p_0 \\ 0 & 0& h_1 & h_0 & 0 \\ 0 & 0 & 0 & h_1 & h_0\\ \end{vmatrix} =p_2^{3-1} \begin{vmatrix} p_2 & p_{1} & p_{0} \\ h_1 & h_0 & 0 \\ 0 & h_1 & h_0\\ \end{vmatrix} = p_m^{m+n-r}R_x(P,H)$

(The first three determinants are zero by Fact1.)

Suppose two poynomials have a common root $\alpha$, we may write them as $(x-\alpha)P,(x-\alpha)Q$. We may also assume that degree of $Q$ is greater than degree of $P$, so we may write $Q = SP +T$ where degree of $T$ is less than degree of $P$, $m,n,r$ are degrees of $(x-\alpha)P,(x-\alpha)S,(x-\alpha)T$ respectively.

So finally we can write $R_x( (x-\alpha)P,(x-\alpha)Q )$ = $R_x ( (x-\alpha)P,(x-\alpha)SP +(x-\alpha)T) $

Applying the previous result we have $R_x ( (x-\alpha)P,(x-\alpha)SP +(x-\alpha)T) = p_m^{m+n-r}R_x ( (x-\alpha)P,(x-\alpha)T)$. As we did it for $P$ and $Q$, we do it now for $T$ and $P$. That is we do now $R_x ( (x-\alpha)T,(x-\alpha)P) $ ( as sign of the determinant does not matter as we want to know it is zero or not.) Note that as $P$ and $Q$ have finite degrees, this process has to stop at some point. In The final form we will have something like $R_x( (x-\alpha)V,(x-\alpha)VW )$ and this determinant will be zero as was proved earlier.

Step 3: If $\alpha_i=\beta_j$ then the determinant is zero. So $\alpha_i-\beta_j$ must be a factor of the determinant $R_x(P,Q)$. (See the answer by Jesko Hüttenhain above)

Step 4: When we consider monic polynomials then $p_m=q_n = 1$ in the first matrix. Let $\alpha_1,...,\alpha_m$ are roots of $P$ and $\beta_1,...,\beta_n$ are roots of $Q$. So $\Pi_{i=1}^m\Pi_{j=1}^n(\alpha_i-\beta_j)$ is a factor of the determinant. Also if the equations are not monic then we factor out $p_m$ from first $n$ rows and $q_n$ from last $m$ rows to make the coefficients inside the determinant, coefficients of monic polynomials. By this we see that determinant has a factor $p_m^nq_n^m$, let the determinant is $p_m^nq_n^m\Pi_{i=1}^m\Pi_{j=1}^n(\alpha_i-\beta_j)G$ then writing the determinant in terms of roots, it looks like

$p_m^nq_n^m\Pi_{i=1}^m\Pi_{j=1}^n(\alpha_i-\beta_j)G = \\ p_m^nq_n^m \begin{vmatrix} 1 & -(\alpha_1+..+\alpha_m) & . & . & . & (-1)^m(\alpha_1..\alpha_m) & . & . & 0 \\ 0 & 1 & -(\alpha_1+..+\alpha_m) & . & . & . & (-1)^m (\alpha_1..\alpha_m) & . & 0 \\ .\\ .\\ 1 & -(\beta_1+..+\beta_n) & . & . & . &(-1)^n(\beta_1..\beta_n) & . & . & 0 \\ 0 & 1 & . & . & . & . & (-1)^n( \beta_1..\beta_n) & . & 0 \\ .\\ .\\ \end{vmatrix} \\ $
If we expand both side of the equation, we see that $n$ is highest degree of $\alpha_i$ in the left as well as in the right. So $G$ does not contain any $\alpha_i$. Similarly $G$ does not contain any $ \beta_j$.
Also cancelling out $p_m^nq_n^m$ from both side of the equation, we check the coefficient of $(\alpha_1..\alpha_m)^n$; in the left it is $(\alpha_1..\alpha_m)^nG$, in the right it is $(-1)^{mn}(-1)^{m(m+n+1)}(\alpha_1..\alpha_m)^n$. Now $(-1)^{mn}(-1)^{m(m+n+1)} = 1$, always, irrespective of values of $m,n$. From this we conclude that $G=1$

So finally $R_x(P,Q) = p_m^nq_n^m\Pi_{i=1}^m\Pi_{j=1}^n(\alpha_i-\beta_j)$

$\endgroup$

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