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We know that the resultant of two polynomials can be computed as the determinant of their Sylvester matrix ( ). How do we compute the resultant of more than two polynomials? Can we generalise the use of the Sylvester matrix for this case?

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It all depends on why you want this -- there are a number of quite different generalizations. – Igor Rivin Jan 9 '11 at 20:49
I need to use it to check whether a system of polynomial equations has solutions. – Andrew Jan 9 '11 at 23:32
up vote 7 down vote accepted

To speak of a (single) resultant of several polynomials one must allow for several variables as well and the situation becomes trickier that in the one variable case.

Let $f=(f_0,\ldots, f_k)$ be homogeneous polynomials (say over an algebraically closed field) of degrees $d_0,\ldots, d_k$ in $k+1$ variables. One can view $f$ as a section of $E=\mathcal{O(d_0)}\oplus\cdots\oplus\mathcal{O(d_k)}$, a vector bundle on $\mathbf{P}^k$. Starting from this bundle form the positive Koszul complex $$K=[0\to \mathcal{O}_{\mathbf{P}^k}\to E\to \wedge^2 E\to\cdots\to\wedge^{k+1}E\to 0]$$

where $\mathcal{O}_{\mathbf{P}^k}$ is in degree 0 and the differential $\wedge^iE\to\wedge^{i+1}E$ is given by the wedge with $f$. This complex is (stalkwise) acyclic iff $f$ is everywhere non-zero see e.g. Proposition 1.4 of Gelfand, Kapranov, Zelevinsky, Discriminants, resultants and multidimensional determinants, chapter 2. If $K$ is acyclic, then its hypercohomology vanishes; on the other hand the higher cohomology of each of the sheaves in $K$ vanishes (see e.g. Hartshorne, chapter 3). So if $K$ is acyclic, then the complex $\Gamma K$of the global sections of $K$ is acyclic. On the other hand, since all sheaves in $K$ are globally generated, the converse is also true: if $\Gamma K$ is acyclic, then $K$ is acyclic.

The complex $\Gamma K$ can be described explicitly as follows: the degree $i$ part is $\bigoplus_{|A|=i}C_{A}$ where $A\subset{ \{ } 0,\ldots, k\}$, $C_{A}$ is the space of all homogeneous polynomials of degree $\sum_{j\in A} d_{j}$ and the differential restricted to $ C_A$ is the sum for all $j\not\in A$ of the maps $ C_A\to C_{A\cup\{j\}}$ given by multiplication by $f_j$ times $(-1)^{b}$ where $b$ is the number of the elements of $A$ that are greater than $j$.

The value of the resultant on $f$ is the hyperdeterminant of $\Gamma K$. (Hyperdeterminants are a generalization of the usual determinants to complexes; they are described e.g. in Gelfand, Kapranov, Zelevinsky, Appendix A.)

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A good entry in the subject of Sylvester like formulae for the multivariate resultant is the paper Explicit formulas for the multivariate resultant by D'Andrea and Dickenstein.

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If your goal is to check if a set of polynomial equations has solutions, resultants are not what you want. There is a number of extremely sophisticated algorithms to do this (from the variations on subgresultant algorithms (due initially to Collins) to Grobner bases. Grobner bases work particularly well when the variety you are trying to determine is zero-dimensional (I have no way of knowing if this is your situation). All the computer algebra systems (Mathematica, Maple, Singular) have these implemented. In the case of mathematica the function Solve[] is what you want.

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Thanks. I know I can use Groebner bases but I have been asked to do it with resultants. – Andrew Jan 10 '11 at 11:49
Hi Igor, why do you say resultants are not good? – Abdelmalek Abdesselam Jan 10 '11 at 12:14

An other (down-to-earth ?) way to construct The Resultant (n homogeneous plynomials ,n variables) can be found in : HAL : hal-00912907, version 1 , An Introduction to Trägheitsformen (PDF). Its main computational properties are proven . Many examples are given.

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A direct link is (but it loads very slowly, at least on my computer). – Gerry Myerson Feb 3 '14 at 23:42

Many years ago Macauley (sp?) gave a way to find a resultant of $n+1$ polynomials in $n$ variables by computing a bunch of Sylvesterish determinants. It looked very complicated to me, and I never succeeded in working through a non-trivial example, but maybe others have better luck. Sorry I don't have the reference to hand.

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