# Why is any maximal minor of the Bezoutian matrix divisible by the resultant?

I'm referring to Emiris and Mourrain's paper "Matrices in Elimination Theory," Theorem 3.13. Toward the end of the proof, it says that, just because $(f_1,\ldots,f_{n+1})$ is dense in ${\cal Z}({\rm Res}_X(f_1,\ldots,f_{n+1})=0)$, all maximal minors of the Bezoutian matrix vanish on the resultant variety. Why is this true? What does "dense" mean? And what's relationship between being dense and having maximal minors vanish on the resultant variety?

In general, it's very hard to see a maximal minor can be divisible by the resultant, if you look at the actual matrices. It's very messy and any maximal minor will do. It's just unbelievable.

My work follows the above result. It's about picking up the extraneous factors in the determinant of a maximal minor, so you get precisely the resultant.

In fact, it's difficult to see the extraneous factors are inside maximal minors, let alone the resultant.

• [PDF]Représentations matricielles en théorie de l'élimination ... tel.archives-ouvertes.fr/docs/00/59/36/03/PDF/hdr-buse.pdf‎ de L Busé - ‎2011 - ‎Autres articles 16 mai 2011 - M. Jean-Pierre JOUANOLOU Professeur émérite à l'Université de Strasbourg ... – Al-Amrani Aug 12 '13 at 10:00
• I think the preceding Habilitation Thesis should be helpful for you. If not ask Jouanolou; I am sure he has an answer to your question. – Al-Amrani Aug 12 '13 at 10:04
• @Al-Amrani: What's his email address? – Zirui Wang Aug 12 '13 at 12:26
• I suggest that you read the proof, and even better, the whole paper, once again more carefully. It explains the word dense fairly clearly a few times. If it is still troubling you, maybe you should ask at MSE, not MO. – Vladimir Dotsenko Aug 26 '13 at 6:24
• @Vladimir Dotsenko: You got the focus wrong. It's not about "dense", which I have a vague sense about, but about what that has to do with maximal minors. Imagine there is a set of dependent columns. Removing any of them gives a maximal minor. And yet they all give the resultant (+ extraneous factors). That's amazing and I'm not convinced by the proof that this happens. Any maximal minor will do, but why? That's my question. – Zirui Wang Aug 26 '13 at 15:05

A proof of your question is given by Mohamed Elkadi and Bernard Mourrain : Theorem 7.3, page 24 in : "HAL - INRIA :: [inria-00073109, version 1] Some Applications of Bezoutians in Effective Algebraic Geometry... hal.inria.fr/inria-00073109/‎ de M Elkadi - ‎1998 - ‎Cité 28 fois - ‎Autres articles We also show how Bezoutian matrices, enable us to compute a non-trivial multiple of the resultant on any irreducible algebraic variety and decompose an ..."

• That paper doesn't appear in Math Reviews. The review of Busé, Laurent; Elkadi, Mohamed; Mourrain, Bernard; Generalized resultants over unirational algebraic varieties, J. Symbolic Comput. 29 (2000), no. 4-5, 515–526, MR1769653 (2001m:13046) makes explicit reference to resultants as factors of maximal minors of Bezoutian matrices. – Gerry Myerson Aug 26 '13 at 23:32
• @Al-Amrani: This paper is a duplicate to what I linked, so long as this theorem is concerned. Both are written by Mourrain. I need another presentation, so as to understand why the theorem is true. – Zirui Wang Aug 27 '13 at 6:21
• Stop giving (almost) identical proofs, including Gerry's. – Zirui Wang Aug 27 '13 at 6:29
• Or is it false? – Zirui Wang Aug 27 '13 at 6:41
• The reference I mentioned above is written by Mohamed Elkadi and Bernard Mourrain. It is a publication by INRIA, number 3572,Dec. 1998 : "Some applications of Bezoutians in Effective Algebraic Geometry". The answer to your question is given by the proof of Theorem 7.3 (§7 Bezoutians and resultants): pages 24-25. – Al-Amrani Aug 27 '13 at 15:43