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 $R$.

  1. Is $S$ an algebraic variety, or something almost as nice?
  2. 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)?
  3. If so, how do I compute these polynomials for each fixed $d$ and $n$?
  4. Are there any other shortcuts for checking whether a homogeneous polynomial splits into linear factors?
  • $\begingroup$ I think (though I may be wrong!) that S is the Hilbert scheme of n points in $X=\mathbb{P}^{d−1}$: a degree n homogeneous form F that is decomposable into linear forms corresponds to a union of n hyperplanes in X (counted with multiplicity). Considering the dual projective space this corresponds to n points. I am not an expert in Hilbert schemes so perhaps someone who knows more (=anything!) about these objects will verify this/show me the error of my ways. A quick search in Google reveals a substantial amount of literature on these objects. $\endgroup$ Oct 10, 2012 at 23:01
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    $\begingroup$ I think it's Sym^n P^{d-1}, not the Hilbert scheme. $\endgroup$
    – JSE
    Oct 11, 2012 at 0:57
  • $\begingroup$ Thanks for the responses, all of which were helpful. The first paper of Briand mentioned by Abdelmalek was the most helpful single reference. It seems that Brill's covariant is not the last word, and other covariants exist. I don't know whether explicit Groebner computations would help to discover new ones. $\endgroup$ Oct 11, 2012 at 2:28
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    $\begingroup$ @Mark: Finding all polynomials which vanish on S is an open problem related to the Foulkes-Howe conjecture which says the minimal degree of such polynomials is n+1. Also, I think Briand has checked that the Brill equations typically do not generate the degree n+1 part of the ideal. Finally a recent paper by Mueller and Neunhoffer seems to disprove the conjecture for d=n=5. $\endgroup$ Oct 11, 2012 at 14:11

3 Answers 3


1) This is the Chow variety of degree $n$ zero cycles in $\mathbb{P}^{d-1}$.

2) Yes, this collection of polynomials can be bundled together into the Brill form or covariant.

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 Emmanuel Briand's page 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).

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 "On Hilbert covariants".

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    $\begingroup$ Is the degree of Chow variety known in this case? $\endgroup$ Mar 8, 2018 at 6:23
  • $\begingroup$ @GoravJindal: Good question! I don't know, although I would be surprized if no one computed this degree. This Chow variety is very similar to that of binary forms which are powers. For the latter the degree was determined by Hilbert in eudml.org/doc/157325 $\endgroup$ Mar 8, 2018 at 15:15

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 this paper.


The questions 1. to 3. can be answered by using the theory of Gröbner bases. Let

(1) $f(x_1,\ldots,x_d) = \sum a_{i_1\ldots i_d}\, x^{i_1} \cdots x^{i_d}$

with indeterminate coefficients $a_{i_1\ldots i_d}$.

Now assume there is a factorization

(2) $f(x_1,\ldots,x_d) = \prod_{i=1}^n (b_{i1} x_1 + \cdots + b_{id} x_d)$

again with indeterminate coefficients $b_{ij}$

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

(3) $G_\nu(\ldots,a_{i_1\ldots i_d},\ldots, b_{ij}, \ldots) = 0$

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.

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    $\begingroup$ This answer may be incorrect. The equations $F_\mu=0$ describe the Zariski closure of the sets of polynomials which factor as a product of linear factors, but nothing guarantees that this set is Zariski closed. It is of course constructible for the Zariski topology, since it is the image of a map between affine schemes (this is essentially what you show in your answer), but I don't see why it should be Zariski closed. BTW, Gröbner bases are a tool to compute the elimination ideal, which is simply an intersection, but conceptually they are unnecessary. $\endgroup$ Oct 10, 2012 at 23:01
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    $\begingroup$ @Fernando: Can't we view the set of completely reducible homogeneous polynomials of degree n in d variables as the image of the multiplication map sending (P(S1))n to P(Sn), where S=C[x0,…,xd−1]? The image of this morphism should be a projective subvariety of P(Sn), in particular it is a Zariski-closed set. The affine cone over this subvariety gives the affine variety that Jeurgen described. $\endgroup$ Oct 11, 2012 at 15:26

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