TITLE: An effective way to tell if the saturation of a homogeneous ideal is the irrelevant ideal
QUESTION [7 upvotes]: 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?

REPLY [4 votes]: Yes, there exists an effective bound on $D$. I am not sure who first found such a bound, but here is a nice reference :
MR2198324 : Jelonek, Z. On the effective Nullstellensatz. Invent. Math.  162  (2005),  no. 1, 1--17.
The minimal number $e=e(I)$ such that $I \supset (\sqrt{I})^e$ is called the Noether exponent of $I$. The above article gives an effective bound for $e(I)$.
More precisely, in the situation at hand, one may assume $n>N$ and also $d_1 \geq d_2 \geq \cdots \geq d_n$. Then Jelonek proves that $e(I) \leq (d_1 \cdots d_N) \cdot d_n$ (see Corollary 1.4 with $X=\mathbf{P}^N$).
Thus the function $D(d_1,\ldots,d_n)=(d_1 \cdots d_N) \cdot d_n$ (with $d_1 \geq \cdots \geq d_n$) works.