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?

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    $\begingroup$ 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? $\endgroup$ Apr 3, 2011 at 21:14

2 Answers 2


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.


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.


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