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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!

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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 $Q_i.$ –  only Nov 14 '12 at 14:11
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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. –  only Nov 14 '12 at 14:18
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1 Answer

This is not really an answer - which I leave to the experts - just a (possibly useless) thought. Let $R=k[x_1,\dots,x_n]$ and $I=(Q_1,\dots,Q_s)\subset R$ the ideal defining $V$. Each $V_i$ corresponds to a prime ideal $\mathfrak p_i\subset R$ (minimal above those) containing $I$. Let us set $\mathfrak p=\mathfrak p_1$ and work with $V_1=V(\mathfrak p)$ where, as we said, $\mathfrak p=(f_1,\dots,f_r)\supset I$.

Now, the first question is about primary decomposition of $I$. The number $m$ is the number of minimal primes in Ass$_R(R/I)$. I don't know whether there is a bound on $m$. Of course it is $\leq |\textrm{Ass}_R(R/I)|$, but I do not feel like this number could depend on the degree of the $Q_j$'s. Indeed (but it's just a naive idea), consider the integers $\mathbb Z$ and look at some $n=\prod_{i=1}^Mp_i^{e_i}$ where the $e_i$'s can be very large powers, and $p_i$ are prime numbers. Well, if $l=\prod_{i=1}^Mp_i$, then $(n)$ and $(l)$ have the same associated primes, and the number $M$ of such associated primes doesn't depend on how huge the generator $n$ is (whence my thought that, in our situation, $m$ should not depend on how huge is the degree of the generators).

For your second question, I can only observe that for every $j$ one has $Q_j\in\mathfrak p$ so $Q_j=\sum_{i=1}^ra_if_i^{b_i}$ so that $\deg Q_j=\max\,(b_i\cdot\deg f_i)\geq\deg f_h$ for all $1\leq h\leq r$.

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Thanks for your thoughts! Regarding the your last observation, couldn't there be cancellation of the high degree terms in the sum? That is, there can be $f_h$'s such that $\deg f_h \gg \deg Q_j$, but in the sum they cancel. –  Henry Yuen Nov 14 '12 at 0:40
    
Dear Henry, of course you are right. –  Brenin Nov 14 '12 at 17:15
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