1
$\begingroup$

I'm reading a paper 'D'Andrea, Carlos(RA-UBA), Dickenstein, Alicia(RA-UBA)Explicit formulas for the multivariate resultant. (English summary) Effective methods in algebraic geometry (Bath, 2000). J. Pure Appl. Algebra 164 (2001), no. 1-2, 59–86.'

Let $f_1,\ldots,f_n$ be homogeneous polynomials of degrees $d_1,\ldots,d_n$ and $A=\mathbb Z[a_{\alpha}]$, where $a_{\alpha}$ denotes the all coefficients of $f_1,\ldots,f_n$. Let $S_m$ be a $A$-free module generated by the monomials in $A[x_1,\ldots,x_n]$ with degree $m$.

In p.73, the author define a Koszul complex $K^{\bullet}(t;f_1,\ldots,f_n)$ by

$$ \{ 0 \to K(t)^{-n} \to \cdots \to K(t)^{-1} \to K(t)^0 \}, $$

where

$$ K(t)^{-j}=\bigoplus_{i_1 < \cdots < i_j} S_{t-d_{i_1}-\cdots-d_{i_j}}, $$

and the morphisms from $K(t)^{-j}$ to $K(t)^{-j+1}$ are the standard Koszul morphisms.

Here what does "standard Koszul morphism" mean? I couldn't catch it.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ To $f_i$ corresponds an element, say $F_i$, of $S_{d_i}$. Multiplying by $F_i$ one gets complexes $0\to S_m \to S_{m+d_i}\to 0$. Now try a tensor product of such complexes. $\endgroup$
    – Wilberd van der Kallen
    Aug 8, 2019 at 7:21
  • $\begingroup$ Wilberd van der Kallen // Could you let me know details of 'try a tensor product of such complexes?' $\endgroup$
    – LWW
    Aug 8, 2019 at 7:31
  • 3
    $\begingroup$ en.wikipedia.org/wiki/Koszul_complex $\endgroup$
    – Denis Nardin
    Aug 8, 2019 at 8:21
  • $\begingroup$ Nardin // My question is that, tensor product means tensor product over i=1,...,n? $\endgroup$
    – LWW
    Aug 8, 2019 at 9:25

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.