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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

  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. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 2468–2481 (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}

    Comments (3)

    Comment #1506 by jojo on June 16, 2015 a 7:03 am UTC

    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 June 25, 2015 a 3:39 pm UTC

    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 June 25, 2015 a 3:41 pm UTC

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

    There are also 2 comments on Section 10.14: Commutative Algebra.

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