\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

The Stacks project

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

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

  2. 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 (3)

Comment #1506 by jojo on

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

Comment #1544 by jojo on

Suggested slogan: In an affine scheme if a finite number of points is contained in an open subset then they are contained in a smaller principal open subset

Comment #1545 by jojo on

Oops sorry I tried the slogan generator and it did the same thing as my comment.

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  • 2 comment(s) on Section 10.14: Miscellany

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