which homogeneous polynomials split into linear factors? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:16:35Z http://mathoverflow.net/feeds/question/109334 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109334/which-homogeneous-polynomials-split-into-linear-factors which homogeneous polynomials split into linear factors? Mark C. Wilson 2012-10-10T22:20:53Z 2012-10-11T14:15:56Z <p>Let $R$ be the set of homogeneous polynomials of degree $n$ in $d$ variables over $\mathbb{C}$. When $n>2$, the set of elements of $R$ that split into a product of linear factors forms a proper subset $S$ of $R$. </p> <ol> <li>Is $S$ an algebraic variety, or something almost as nice?</li> <li>If so, how can $S$ be described implicitly, in terms of the original coefficients, without using a factorization algorithm? In other words, is there a finite set of polynomials in the coefficients which vanish if and only if $p\in S$ (or something almost as nice)?</li> <li>If so, how do I compute these polynomials for each fixed $d$ and $n$?</li> <li>Are there any other shortcuts for checking whether a homogeneous polynomial splits into linear factors?</li> </ol> http://mathoverflow.net/questions/109334/which-homogeneous-polynomials-split-into-linear-factors/109337#109337 Answer by Jürgen Böhm for which homogeneous polynomials split into linear factors? Jürgen Böhm 2012-10-10T22:38:54Z 2012-10-10T22:38:54Z <p>The questions 1. to 3. can be answered by using the theory of Gröbner bases. Let</p> <p>(1) $f(x_1,\ldots,x_d) = \sum a_{i_1\ldots i_d}\, x^{i_1} \cdots x^{i_d}$</p> <p>with indeterminate coefficients $a_{i_1\ldots i_d}$.</p> <p>Now assume there is a factorization</p> <p>(2) $f(x_1,\ldots,x_d) = \prod_{i=1}^n (b_{i1} x_1 + \cdots + b_{id} x_d)$</p> <p>again with indeterminate coefficients $b_{ij}$</p> <p>Multipliying (2) out and equating coefficients of like monomials $x_1^{i_1} \cdots x_d^{i_d}$ in (1) and (2) gives a set of polynomials</p> <p>(3) $G_\nu(\ldots,a_{i_1\ldots i_d},\ldots, b_{ij}, \ldots) = 0$</p> <p>Eliminating from the $G_\nu$ all the $b_{ij}$ by Gröbner basis methods gives a set of equations $F_\mu(\cdots,a_{i_1\ldots i_d},\cdots) = 0$. These describe $S$ as an algebraic variety.</p> http://mathoverflow.net/questions/109334/which-homogeneous-polynomials-split-into-linear-factors/109340#109340 Answer by Abdelmalek Abdesselam for which homogeneous polynomials split into linear factors? Abdelmalek Abdesselam 2012-10-10T23:29:36Z 2012-10-11T14:15:56Z <p>1) This is the Chow variety of degree $n$ zero cycles in $\mathbb{P}^{d-1}$.</p> <p>2) Yes, this collection of polynomials can be bundled together into the Brill form or covariant.</p> <p>3) Rather explicit descriptions of the Brill equations can be found in the book by Gelfand, Kapranov and Zelevinsky on resultants. There is also a paper by Rota and Stein. But first check out <a href="http://emmanuel.jean.briand.free.fr/publications/" rel="nofollow">Emmanuel Briand's page</a> and in particular the articles "Covariants decomposing on totally decomposable forms" and "Brill's equations for the subvariety of factorizable forms" and if you read French (or German) the translation of the original article by Gordan (respectively the article itself).</p> <p>As an aside, analogues of the Brill equations for the variety of forms which are powers of forms of degree dividing $n$ have been given recently in my paper with Chipalkatti <a href="http://dx.doi.org/10.4153/CJM-2012-046-1" rel="nofollow">"On Hilbert covariants"</a>.</p> http://mathoverflow.net/questions/109334/which-homogeneous-polynomials-split-into-linear-factors/109341#109341 Answer by jvp for which homogeneous polynomials split into linear factors? jvp 2012-10-10T23:31:38Z 2012-10-10T23:31:38Z <p>There are Brill's equations. Look for them at the book by Gelfand, Kapranov, and Zelevinski. In general Brill's equations do not generate the ideal of totally decomposable polynomials, see <a href="http://emmanuel.jean.briand.free.fr/publications/eaca2004/eaca2004.pdf" rel="nofollow">this paper</a>.</p>