Let $\Bbbk$ be an algebraically closed field, let $R$ denote the graded ring $\Bbbk[x_0, \dotsc, x_N]$, and let $f_1, \dotsc, f_n \in R_m$ be nonconstant homogeneous polynomials. Then the common vanishing locus $V(f_1, \dotsc, f_n) \subset \mathbb{P}^N$ is empty if and only if the saturation of the homogeneous ideal $I = (f_1, \dotsc, f_n)$ is the entire irrelevant ideal $R_+$. This is true iff for some $d > 0$, $I_d = R_d$ (the degree-$d$ parts are equal), in which case $I_e = R_e$ for all $e \geq d$.

It is not hard to compute the vector subspace $I_d \subset R_d$ for successive values of $d$: if $I_d$ is generated as a $\Bbbk$-module by $a_1, \dotsc, a_k \in R_d$, then $I_{d+1}$ is generated by the $x_i a_j$, plus any of the $f_i$ that are of degree $d+1$.

If you want to show that $I$ does, in fact, have saturation equal to the entire ideal $R_+$, you can start computing the vector subspaces $I_d$; if you're right, then sooner or later you'll get $I_d = R_d$ and have your answer. But suppose you go on and on, and $I_d$ remains stubbornly a proper subspace of $R_d$. Is there some point--when $d$ is a thousand, a million, $10^{100}$—at which you can say, "If $I_d$ does not contain all $R_d$ by now, it never will"?

Does there exist $D$, depending only on the degrees of the $f_i$, sufficiently large that if $I_d = R_d$ for any $d$, then $I_D = R_D$?

I'm reasonably confident that the answer to that question is yes, based on the following sketch: Look at the space $V$ of all $n$-tuples of homogeneous polynomials $(f_1, \dotsc, f_n)$ with fixed degrees $d_1, \dotsc, d_n$. Let $S_d \subset V$ be the subset of those for which $I_d = R_d$. Since the condition on $S_d$ comes down to the condition that some linear map of vector spaces is surjective, $S_d$ is Zariski-open. Thus, $S_d \subset S_{d+1} \subset S_{d+2} \subset \dotsb$ is an increasing union of Zariski-open sets, and consequently must stabilize at some $D$.

Unfortunately, this argument is entirely non-effective. We have no idea what the value of $D$ is, and so if we actually want to show that $I^{sat} \neq R_+$, we're out of luck. This motivates the following question:

Assuming an affirmative answer to the previous question, what is a (preferably computable) function $$D = D(d_1, \dotsc, d_n)$$ that works?

The Geometry of SyzygiesOr you might try Dave Bayer's thesis? $\endgroup$ – Alexander Woo Apr 3 '11 at 21:14