** In an affine scheme if a finite number of points are contained in an open subset then they are contained in a smaller principal open subset. **

Lemma 10.14.2 (Prime avoidance). Let $R$ be a ring. Let $I_ i \subset R$, $i = 1, \ldots , r$, and $J \subset R$ be ideals. Assume

$J \not\subset I_ i$ for $i = 1, \ldots , r$, and

all but two of $I_ i$ are prime ideals.

Then there exists an $x \in J$, $x\not\in I_ i$ for all $i$.

**Proof.**
The result is true for $r = 1$. If $r = 2$, then let $x, y \in J$ with $x \not\in I_1$ and $y \not\in I_2$. We are done unless $x \in I_2$ and $y \in I_1$. Then the element $x + y$ cannot be in $I_1$ (since that would mean $x + y - y \in I_1$) and it also cannot be in $I_2$.

For $r \geq 3$, assume the result holds for $r - 1$. After renumbering we may assume that $I_ r$ is prime. We may also assume there are no inclusions among the $I_ i$. Pick $x \in J$, $x \not\in I_ i$ for all $i = 1, \ldots , r - 1$. If $x \not\in I_ r$ we are done. So assume $x \in I_ r$. If $J I_1 \ldots I_{r - 1} \subset I_ r$ then $J \subset I_ r$ (by Lemma 10.14.1) a contradiction. Pick $y \in J I_1 \ldots I_{r - 1}$, $y \not\in I_ r$. Then $x + y$ works.
$\square$

## Comments (5)

Comment #1506 by jojo on

Comment #1544 by jojo on

Comment #1545 by jojo on

Comment #4333 by comment_bot on

Comment #4334 by Pieter Belmans on

There are also: