4

I haven't been able to find a proof for this theorem in the literature:

Let $f,g\in k[x_0,\dots,x_k]$ be homogeneous polynomials, of degree $m$ and $n$ respectively. Then $R_{x_0}(f,g)$ is homogeneous of degree $mn$, where $R_{x_0}(f,g)$ stands for the resultant of $f$ and $g$ viewed as elements of $k[x_1,\dots,x_k][x_0]$.

Fernando Martin
  • 5,847
  • Maybe you should change this sorry circumstance and write a proof yourself :) The resultant of two polynomials is defined as a certain determinant, and a determinant is defined as a huge sum of certain products over permutations. Show that, in your case, the products all have the same degree. – darij grinberg Dec 14 '14 at 23:53
  • Yes, I've thought about that, but it's not obvious that the products have the right degree. – Fernando Martin Dec 14 '14 at 23:55
  • Notice that the $\left(i,j\right)$-th entry of the Sylvester matrix is homogeneous of degree $m-j+i$ if $i \leq n$, and $-j+i$ otherwise. Or did I make a mistake? – darij grinberg Dec 14 '14 at 23:58
  • That's correct. Some of the products in the sum are zero and others are not; the ones that are not should be of degree $mn$, but I don't find that obvious at all. – Fernando Martin Dec 15 '14 at 00:02
  • 1
    The product for permutation $\sigma$ has degree $\sum\limits_{i=1}^{n} \left(m-\sigma\left(i\right)+i\right) + \sum\limits_{i=n+1}^{m+n} \left(-\sigma\left(i\right)+i\right) = mn + \sum\limits_{i=1}^{m+n} \left(-\sigma\left(i\right)+i\right)$. Remains to show that $\sum\limits_{i=1}^{m+n} \left(-\sigma\left(i\right)+i\right) = 0$. But $\sum\limits_{i=1}^{m+n} \left(-\sigma\left(i\right)+i\right) = - \sum\limits_{i=1}^{m+n} \sigma\left(i\right) + \sum\limits_{i=1}^{m+n} i = - \sum\limits_{i=1}^{m+n} i + \sum\limits_{i=1}^{m+n} i = 0$. – darij grinberg Dec 15 '14 at 00:04
  • 1
    :) Thanks a lot @darijgrinberg. Feel free to post the comment as an answer so I can accept it. – Fernando Martin Dec 15 '14 at 00:08
  • Or you could do it -- I am lacking time these days. – darij grinberg Dec 15 '14 at 00:15

1 Answers1

4

This theorem is not true. Take $$f=x_0x_1x_2 - x_1x_2x_3 ~\text{ and }~ g=x_0x_3x_4-x_0x_4x_5.$$ These polynomials are both of homogeneous degree 3, so according to the theorem the resultant $Res_{x_0}(f,g)$ should be of degree $9$, however $$Res_{x_0}(f,g)=-x_1x_2x_3^2x_4+x_1x_2x_3x_4x_5$$ is of homogeneous degree $5$.

The mistake in the proof in the comments under the question is in determining the homogeneous degree of the $(i,j)$-the entry of the Sylvester matrix, specifically in this part

the $(i,j)$-th entry of the Sylvester matrix is homogeneous of degree $m−j+i$ if $i\leq n$, and $−j+i$ otherwise

as it doesn't take into account that certain entries of the matrix are $0$, e.g. if $m=2$ and $n=3$ then at position $(0,3)$ the entry is $0$, but $m+i-j=2+0-3=-1$.

Edit: the correct homogeneous degree is $$m\cdot\deg_{x_0}g+n\cdot \deg_{x_0}f - \deg_{x_0}g\cdot\deg_{x_0}f$$ because if $f=a_m+a_{m-1}x+\cdots+a_{m-k}x^k$ and $g=b_n+b_{n-1}x+\cdots+b_{n-l}x^l$ (the coefficients are indexed like so because then the homogeneous degree of $a_i$ and $b_i$ is exactly $i$) then the term $a_m^lb_{n-l}^k$ generically appears in the resultant, up to sign. This generic term may not appear for specific choices of $f$ and $g$ (e.g. as in my example above), but all other terms that do appear have the same degree anyway.

Randy Marsh
  • 2,857
  • The theorem is however true if you assume the polynomials do not vanish at (0,0,1), then you have enough non-zero coefficients in the matrix to show that the resultant's degree is the product of the degrees – Azur Nov 29 '20 at 23:41
  • 1
    @Azur Yes, this follows directly from the formula $m\deg_z g+n\deg_z f-\deg_zf\deg_zg$ in my answer. If neither $f$ nor $g$ vanish at $(0,0,1)$, then there is a term in $f$ and $g$ that contains only $z$ to power $m$ and $n$, respectfully. Therefore $m\deg_z g+n\deg_z f-\deg_zf\deg_zg = m n$. – Randy Marsh Nov 30 '20 at 00:41