8
$\begingroup$

Let $Q_1,\ldots,Q_s \in k[x_1,\ldots,x_n]$, where $k$ is not necessarily algebraically closed (I'm thinking of $k$ as some field with positive characteristic $p$). I'm somewhat new to the world of classical algebraic geometry, so the following may be trivial questions:

Let $V = V(Q_1,\ldots,Q_s)$. Decompose $V = V_1 \cup \cdots \cup V_m$, where each of the $V_i$ are irreducible.

  1. Is there a bound on the number of irreducible components, $m$, that depends on the degrees of the $Q_j$'s?
  2. Let each of the $V_i$'s be defined by polynomials $f_{i1},f_{i2},\ldots$. If the the original $Q_j$'s are of low degree, can we bound the degrees of the $f_{ik}$'s?

Thanks!

  • 1
    If you work with projective space instead of affine space, I believe the answer to both questions is yes. (But both require a fair bit of machinery). The first utilizes the fact that Chow ring of projective space is just $Z[t]$, so the degree of the intersection is the product of the degrees of the polynomials, and each $V_i$ has positive degree, so $m \le $ the product of the degrees of the $Q_i.$2012-11-14
  • 2
    As to the second question, I am much less sure if this works, but I think that it can be done by some fiddling with Hilbert polynomials and invoking some standard lemmas (which use Castelnuovo-Mumford regularity) used in the construction of the Hilbert scheme. You can probably reduce the affine case by taking closures to the projective case, and I see no issues, but I have no access to paper at the moment, and can't check anything. So, neither of these questions are trivial at all; the second one is strongly related to a crucial step in the construction of the Hilbert scheme.2012-11-14

1 Answers 1