## Tag `00DS`

Chapter 10: Commutative Algebra > Section 10.14: Miscellany

**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. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 2464–2477 (see updates for more information).

```
\begin{lemma}[Prime avoidance]
\label{lemma-silly}
\begin{slogan}
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.
\end{slogan}
Let $R$ be a ring. Let $I_i \subset R$, $i = 1, \ldots, r$,
and $J \subset R$ be ideals. Assume
\begin{enumerate}
\item $J \not\subset I_i$ for $i = 1, \ldots, r$, and
\item all but two of $I_i$ are prime ideals.
\end{enumerate}
Then there exists an $x \in J$, $x\not\in I_i$ for all $i$.
\end{lemma}
\begin{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 $I_2$.
\medskip\noindent
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 \ref{lemma-product-ideals-in-prime}) a contradiction.
Pick $y \in J I_1 \ldots I_{r - 1}$, $y \not \in I_r$. Then $x + y$ works.
\end{proof}
```

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