Does there exist any maximal ideal $M^p$ in $C(X)$ (the ring of continuous functions on a topological space $X$) such that each element of $M^p$ is a divisor of zero but $M^p≠O^p$?
-
3$\begingroup$ what is $O^p$? Does the question start with "does there exists $X$ such that"? $\endgroup$– YCorCommented Apr 26, 2017 at 8:01
-
$\begingroup$ If $O^p$ means the zero ideal, then yes, such a space and an ideal exist: take $X$ to be the set $\{1,2\}$ with the discrete topology and let $M^p$ be the ideal of functions $f$ with $f(1)=0$. $\endgroup$– user1688Commented Apr 26, 2017 at 10:51
1 Answer
Assuming that $M^p$ is the set of elements of $C(X)$ which vanish at $p$, and $O^p$ is the set of elements of $C(X)$ which vanish on a neighborhood of $p$, the answer depends on the space $X$. For example, if $X = [0,1]$, since every singleton is a zero-set, there is no such maximal ideal. On the other hand if $X = D \cup \{\infty\}$ is the one-point compactification of the discrete space of size $\aleph_1$ and $p = \infty$, then $M^p \ne O^p$ but each element of $M^p$ is a zero divisor. The reason is that if $\{x_n : n = 1, 2, \cdots\}$ is a countable subset of $D$, the function $f \in C(X)$ given by $f(x_n) = \frac{1}{n}$ and $f(x) = 0$ otherwise is not $0$ on any neighborhood of $p$ (because neighborhoods of $p$ are cofinite). On the other hand, since zero-sets are G$_\delta$ sets and $\{\infty \}$ is not a G$_\delta$ set, every element of $C(X)$ which vanishes at $\infty$ also vanishes at an isolated point and, therefore, is a zero-divisor.
-
-
$\begingroup$ No, {\infty} is not a G_\delta set, because open sets containing \infty are cofinite, so G_\delta sets containing \infty are cocountable. $\endgroup$ Commented May 3, 2017 at 15:27
-
$\begingroup$ consider the intersection of all such sets ({infinity} union (N-{n})) $\endgroup$– S.BCommented May 4, 2017 at 6:05
-
$\begingroup$ It is not clear what N is supposed to be, but in the answer above D is an uncountable discrete space. The intersection of all sets of the form {\infty} \cup (D - {x}) is not a G_\delta set. $\endgroup$ Commented May 4, 2017 at 17:13