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Dear all,

I am interested in the following basic and fundamental question in elimination theory: given a variety in some product space $Z\subseteq X\times Y$, how could I explicitly calculate the polynomial equation system cutting out (the closure of) the image of $Z$'s projection to $X$ or $Y$? The cases I am interested in are ususally like $Z=\{(x,y)\in X\times Y: x\subseteq y\}$ (i.e. incidence varieties) or $Z=\{(x,y)\in X\times Y: x\cap y\neq\emptyset\}$ where $X$ and $Y$ are usually affine/projective spaces, grassmanians, or their Veronese embeddings.

Here are two simple examples for which I know the images of projections explicitly: The projection of $\{(A,x)\in\mathbb{A}^{n\times n}\times\mathbb{P}^n:Ax=0\}$ in $\mathbb{A}^{n\times n}$ is cutting out by the determinant $\det(A)=0$. And the projection of $\{((f_0,\dots,f_n),x)\in (S^d_{n+1})^{n+1}\times \mathbb{P}^n:f_0(x)=\dots=f_n(x)=0 \}$ is cutting out by the resultant $\mathrm{Res}(f_0,\dots,f_n)=0$ where $S^d_{n+1}$ denotes the projective space of degree-$d$ homogeneous polynomials in $n+1$ variables. (In both cases the projections have codimension 1, but of course they don't have to in general.)

I am aware that Gröbner bases algorithms can do the job, but these are not what I am looking for, as they do not show conceptually what the images of projections look like. I ask this question because I need to find certain explicit object that I can prove it is not in the image, and I can't hope to prove it if I only know the image can be computed by some algorithms.

On the other hand, it is shown in the book Discriminants, Resultants and Multidimensional Determinants by Gelfand et al. that classical objects like discriminants and resultants can be viewed as the determinants of certain complexes. This point of view is conceptually clear to me, and it gives (at least in principle) a way of writing out the explicit forms of these objects. So such kind of solutions are what I am looking for. More specifically, I hope it will show that the image of a projection is cutting by some determinant. Then for my goal of picking an object not contained in the projection, I just need to make sure its determinant does not vanish (e.g. using Vandermonde matrices).

I also noticed that a procedure using resultant systems is given in Lang's Algebra, IX, §4. However I do not understand it very well. Any explanation will be appreciated.

If the question looks too general, I have a more concrete one in my mind: it is well known that for $d>2n-3$, a generic degree-$d$ hypersurface in $\mathbb{P}^n$ contains no lines. This can be shown by considering the incidence variety $X=\{(\ell,f)\in \mathbb{G}(1,n)\times S^d_{n+1}:\ell\subseteq V(f)\}$ and arguing that its image under the projection to $S^d_{n+1}$ is a proper closed subset, which is simply dimension counting. Again I use $S^d_{n+1}$ to denote the projective space of degree-$d$ homogeneous polynomials in $n+1$ variables. Now is it possible to explicitly calculate the polynomial systems cutting out this closed subset?

Thank you very much!

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  • $\begingroup$ Do you know "On the closed image of a rational map and the implictization problem" by L.Busé and J.P. Jouanolou ? arXiv:math/0210096 (based on the work of Jouanolou in elimination theory). $\endgroup$ – Al-Amrani Oct 10 '18 at 12:31
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I will give the answer for the concrete question. Let $V$ be the vector space of dimension $n+1$ such that $P^n = P(V)$. Let $U$ be the tautological rank 2 bundle on the Grassmannian $G$ and $O(-1)$ the tautological line bundle on $P(S^dV^*)$. Consider the product $G\times P(S^dV^*)$ and the morphism $$ S^dU \boxtimes O \to S^dV \otimes O \boxtimes O \to O \boxtimes O(1) $$ on it (the first map is the $d$-th symmetric power of the tautological embedding $U \to V\otimes O$ and the second map is dual to the tautological embedding $O(-1) \to S^dV^*\otimes O$). This map gives a global section of the vector bundle $S^dU^*\boxtimes O(1)$. Clearly this section vanishes precisely on (the cone over) your incidence variety $X$. Since the rank of the bundle $d+1$ equals the codimension of $X$, the section is regular and so $O_X$ has a Koszul resolution $$ \dots \to \Lambda^2(S^dU\boxtimes O(-1)) \to S^dU\boxtimes O(-1) \to O\boxtimes O \to O_X \to 0. $$ If you want now to compute the image of $X$ in $P(S^dV^*)$ you can just push forward this resolution by using Borel--Bott--Weil theorem to compute the cohomology on $G$.

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