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.