Lemma 10.15.2 (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.
1. 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. 2. Affine opens are cofinal among the neighborhoods of a given finite set of an affine scheme
Lemma 10.15.2 (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 10.15.1) a contradiction. Pick $y \in J I_1 \ldots I_{r - 1}$, $y \not\in I_ r$. Then $x + y$ works. $\square$
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$
). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (9)
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
Comment #5495 by comment_bot on
Comment #5499 by Johan on
Comment #5506 by comment_bot on
Comment #5699 by Johan on
There are also: