TITLE: Geometric statement of Prime Avoidance?
QUESTION [19 upvotes]: 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 $V(P_1), \ldots, V(P_r)$)?

REPLY [14 votes]: 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.