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Let $\mathfrak{p}_1, \dotsc, \mathfrak{p}_k$ be relevant homogeneous primes ideals in the graded ring $R := \Bbbk[x_0, \dotsc, x_n]$, where $\Bbbk$ is a field. Prime avoidance (in Eisenbud's terminology) tells us that there exists a nonconstant homogenous polynomial $f \not\in \cup_i \mathfrak{p}_i$. Is there an easy way to see that for some $d \geq 1$, there exist homogeneous $f \in R_d$, $g \in R_{d+1}$, neither of which is contained in $\cup_i \mathfrak{p}_i$?

Note: If $\Bbbk$ is infinite, this is easy: we can use the fact that no vector space is a finite union of proper subspaces to find $\alpha \in R_1 \smallsetminus \cup_i \mathfrak{p}_i$, and then take $f=\alpha$, $g = \alpha^2$.

Motivation: If the $\mathfrak{p}_i$ are the associated primes of a closed subscheme $X$ of $\mathbb{P}^n_{\Bbbk}$ then we can pull back $g/f$ to a nice rational section of $\mathcal{O}_X(1)$. Consequently, we know that $\mathcal{O}_X(1)$ comes from a Cartier divisor. Using the fact that any line bundle can be twisted to a very ample line bundle, this shows that the Cartier divisor group surjects onto $Pic(X)$.

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The problem can be reduced to finding such $f \in R_d$ and $g \in R_m$ with gcd $(m,d)=1$, which seems much easier. If this would not be the case, it would imply that $R \backslash \cup_i p_i \subset \sum_{n\geq 1} R_{dn}$ –  Nick S Feb 10 '11 at 23:08
    
As for the surjection to $Pic(X)$, it is known that it holds for any locally noetherian scheme such that the associted points all live in a same affine open subset (EGA IV, 21.3.5). This contains the case of quasi-projective schemes over a noetherian ring. –  Qing Liu Feb 11 '11 at 12:35
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2 Answers

up vote 6 down vote accepted

The answer is yes. The following is extracted from a preprint of Gabber-Liu-Lorenzini.

Let $B=\oplus_{n\ge 0}B(n)$ be a graded ring. Let $I=\oplus_{n\ge 0}I(n)$ be a homogeneous ideal of $B$. Let $\mathfrak p_1,\dots,\mathfrak p_r$ be homogeneous prime ideals of $B$ not containing $B(1)$ and not containing $I$. Then there exists an integer $n_0 \geq 1$ such that for all $n\ge n_0$, $I(n)\not\subseteq \cup_{1\le i\le r} \mathfrak p_i$.

Proof. We proceed by induction on $r$. If $r=1$, choose $t\in B(1)\setminus \mathfrak p_1$ and a homogeneous element $\alpha\in I\setminus \mathfrak p_1$, say of degree $n_0$. Then $t^{n-n_0}\alpha\in I(n) \setminus \mathfrak p_1$ for all $n\ge n_0$, as desired.

Let $r\ge 2$ and suppose that the lemma is true for $r-1$. We can suppose that $\mathfrak p_i$ is not contained in $\mathfrak p_r$ for all $i\ne r$, so that $I\mathfrak p_1\cdots\mathfrak p_{r-1} \not\subseteq \mathfrak p_r$. Similarly, we can suppose that $\mathfrak p_r$ is not contained in $\mathfrak p_i$ for all $i\ne r$, so that $I\mathfrak p_r\not\subseteq (\mathfrak p_1\cup \ldots\cup\mathfrak p_{r-1})$. Hence, we can apply the case $r=1$ and the induction hypothesis to obtain that there exists $n_0$ such that for all $n\ge n_0$, there are homogeneous elements $f_n\in I\mathfrak p_1\cdots\mathfrak p_{r-1} \setminus \mathfrak p_r$ and $g_n\in I\mathfrak p_r\setminus (\mathfrak p_1\cup \ldots\cup\mathfrak p_{r-1})$ of degree $n$. It is easy to check that $f_n+g_n\in I(n)\setminus \cup_{1\le i\le r}\mathfrak p_i$, as desired.

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@Qing: Very nice! May I ask what preprint? –  Hailong Dao Feb 11 '11 at 12:55
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@Hailong: it is Lemma 5.11 in this very fine paper: math.uga.edu/~lorenz/PaperIndex.pdf. –  Pete L. Clark Feb 11 '11 at 13:28
    
Thanks, Pete! This paper looks very appealing to me. –  Hailong Dao Feb 11 '11 at 13:51
    
Thanks Hailong and Pete. The notation here is different because I extracted the above statement from a second (not yet finished) preprint where we need a more technical form of prime avoidance. –  Qing Liu Feb 11 '11 at 15:41
    
This is a really nice paper. –  Angelo Feb 11 '11 at 16:46
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EDIT: the argument below only works if each $R/\mathfrak{p}_i$ has dimension at least $2$, as pointed out by Angelo.

Suppose $\Bbbk$ is a finite field of size $q$. Let $V_{i,n}= \mathfrak{p}_i \cap R_n$, $d_n= \dim_{\Bbbk} R_n$ and $d_{i,n}=\dim_{\Bbbk}V_{i,n}$. It is enough to show that for $n$ big enough:

$$|V_n| > \sum_1^k |V_{i,n}| $$

Note that for each $i$, $d_n - d_{i,n}$ gives the Hilbert function of $R/\mathfrak{p}_i$, so it eventually becomes a polynomial in $n$ of dimension $={\dim R/\mathfrak{p}_i}-1 \geq 1$ (EDIT: I originally wrote $\dim R/\mathfrak{p}_i$). So for $n\gg 0$, $q^{d_n-d_{i,n}}>k$ for each $i$. Thus

$|V_n|/|V_{i,n}|= q^{d_n-d_{i,n}}> k$

and we are done.

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It seems to me that the degree of the Hilbert polynomial of $R/\mathfrak p_i$ is its dimension minus 1. –  Angelo Feb 11 '11 at 8:31
    
Opps, you are right! Silly me. –  Hailong Dao Feb 11 '11 at 8:56
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