Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
2  
The degree you're looking for is related to the Castelnuevo--Mumford regularity of the ideal. I seem to recall there is a doubly exponential bound for the C-M regularity in terms of the degrees of the generators. As a first reference, let me point you towards Eisenbud's book The Geometry of Syzygies Or you might try Dave Bayer's thesis? –  Alexander Woo Apr 3 '11 at 21:14

2 Answers 2

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.

share|improve this answer

As Alexander Woo correctly points out, this is the Castelnuovo-Mumford regularity, and my answer is basically an elaboration of this fact. First of all, the saturation of $I$ equals the irrelevant ideal if and only if $R/I$ is Artinian.

In other words, $I_d=R_d$ if and only if $(R/I)_d=0$, and hence you are looking for the minimal degree $d$ such that $(R/I)_d=0$. This is precisely the Castelnuovo-Mumford regularity of $R/I$, (which is just the CM regularity of $I$ plus 1). See Corollary 4.4 of Eisenbud's "The Geometry of Syzygies" for a reference.

So you'd like to bound $reg(I)$ in terms of $d_1, \dots, d_n$. There is a fair amount of literature on this subject, but I'm not an expert. I know that Marc Chardin has lots of results in this direction, as well as some expository papers on this subject, so I would look at his work. I believe that the bounds are something horrible (perhaps doubly exponentional in $\max {d_1, \dots, d_n\}$??), but I would look at Chardin's work for more precise results.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.