The Prime Avoidance Theorem is very clean to state in algebraic terms:

Let $I \subset R$ be an ideal (with $R$ noetherian) and $I \subseteq \bigcup_{i=1}^r P_i$, where each $P_i$ is prime. Then $I \subseteq P_i$ for some $i$.

Is there a nice way to rephrase this in geometric terms? As it stands I have little (geometric) intuition for when prime avoidance might be useful, in part because I don't have a good visual picture of this statement. The weird part is $\bigcup P_i$, since this doesn't correspond to a simple geometric object.

The best I can do is:

Let $Y \subset X$ be a subscheme, and suppose there are integral subschemes $\{Z_1, \ldots, Z_r\}$ such that every hypersurface $H \supseteq Y$ has $H \supseteq Z_i$ for some $i$. Then $Y$ contains some $Z_i$.

This is basically stating the containment $I \subseteq \bigcup_{i=1}^r P_i$ element-by-element. Maybe there is something simpler (directly in terms of $V(I)$ and $V(P_1), \ldots, V(P_r)$)?

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    $\begingroup$ More along the line of avoidance: say you have a bunch of subvarieties $Y_1, \dots, Y_n$ and then some functions $f_1, \dots, f_m$ on the ambient space; and for each $f_i$ there is some $Y_j$ such that $f_i$ doesn't vanish identically on $Y_j$. Then you can find some linear combination of the $f_i$ which doesn't vanish identically on any of the $Y_j$. $\endgroup$
    – Hoot
    Jun 11 '15 at 16:04
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    $\begingroup$ [I think I got this out of Eisenbud; it's a clever translation. I don't think you need to assume any sort of noetherianity for the algebraic statement, by the way. Liked your talk at MAGGC!] $\endgroup$
    – Hoot
    Jun 11 '15 at 16:05
  • $\begingroup$ Hm, that's a nice way of phrasing it. Thanks for the comment. $\endgroup$ Jun 11 '15 at 18:48
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    $\begingroup$ Glad you liked the talk -- have I met you in person? $\endgroup$ Jun 11 '15 at 18:53

The contrapositive has a nice geometric interpretation. The algebraic statement is:

If $I\nsubseteq \mathfrak{p}_{i}$ for any $i$, then $I\nsubseteq \cup_{i} \mathfrak{p}_{i}$, i.e. there is some $x\in I$ which is not in any of the $\mathfrak{p}_{i}$.

Now consider the affine scheme $X=\operatorname{Spec}R$. Then $I\nsubseteq \mathfrak{p}_{i}$ for any $i$ translates into $\mathfrak{p}_{i}\in X\setminus V(I)$ for all $i$. That is, the points $\mathfrak{p}_{i}$ are all contained in the open set $X\setminus V(I)$. What the prime avoidance tells us is that we can now find an element $x\in I$ which is not in any of the prime ideals $\mathfrak{p}_{i}$, which translates into saying that $$ \mathfrak{p}_{i} \in D(x)\subseteq X\setminus V(I)$$ So if we have finitely many points contained in an open set of an affine scheme, we can always find a smaller principal open set containing them.

Reference: I found this on the Stacks Project.


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