3
$\begingroup$

Assume that $X_1,\ldots,X_r\subseteq\mathbb P^n$ are irreducible, reduced hypersurfaces in complex projective space, each of the same degree $d$. In other words, $X_i=Z_\ast(f_i)$ for certain irreducible, homogeneous polynomials $f_i\in\mathbb C[X_0,\ldots,X_n]_d$. Let $X:=X_1\cap\cdots\cap X_r$ be their intersection. For the moment, let's just say scheme-theoretic intersection. If it helps, assume that the codimension of $X$ is $r$, i.e. it is the complete intersection of the $X_i$. I would like to know what is known, if anything, about the Hilbert function of the closed subscheme $X$, or equivalently, the ideal $I=(f_1,\ldots,f_r)$. Can it be expressed in terms of the $f_i$ somehow?

Not very important, but still interesting: What about passing to $\sqrt I$?

$\endgroup$

1 Answer 1

6
$\begingroup$

It sure does help to assume complete intersection. It's nicer to describe the Hilbert series $H(t) := \sum_k h(k) t^k$ than the Hilbert function $h(k)$. For the polynomial ring itself, it's $H(t) = 1/(1-t)^{n+1}$. For the complete intersection, it's $(1-t^d)^r/(1-t)^{n+1}$ (with an obvious generalization if the degrees are different).

If it's not a complete intersection, there's much less you can say -- you have to look at the syzygies of your generators, and the higher syzygies, and so on inclusion/exclusion.

Going to $\sqrt{I}$, all that's easy to say is that the degree of $h(k)$ stays the same but the leading coefficient can drop.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .