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 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? 1 # An effective way to tell if the saturation of a homogeneous ideal is the irrelevant ideal 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?