5
$\begingroup$

A minimal ideal of a commutative ring $R$ is a nonzero  ideal which contains no other nonzero ideal. 

Let $X $ be a completely regular topological space and $C (X) $ the ring of all real valued continuous functions over $X $. Is there any characterization for minimal ideals of $C (X) $?

$\endgroup$
1
  • 6
    $\begingroup$ Let $I \subseteq R = C(X)$ be such a minimal ideal and let $f \in I \setminus\{0\}$. Then $\{0\} \neq Rf^2 \subseteq Rf \subseteq I$ so that $I = Rf = Rf^2$. Thus $I$ is generated by function $f$ such that $1/f$ extends continuously from the set-theoretic support of $f$ to $X$. $\endgroup$ – Luc Guyot Mar 28 '18 at 15:59
7
$\begingroup$

Let us first observe that an ideal $I$ of a commutative ring $R$ with identity is minimal in OP's sense if and only if $I$ is a non-zero simple module over $R$. From now on, we will favour this terminology.

Note also that a topological space $X$ with trivial topology, i.e., the open sets of $X$ are $X$ and $\emptyset$, is completely regular for a trivial reason. Let $C(X)$ be the ring of continuous real-valued functions on $X$. If $X$ is a topological space with trivial topology, then the ring $C(X)$ is isomorphic to $\mathbb{R}$ and hence is a simple ring.

This leads to

Claim. Let $X$ be a completely regular topological space. Then $I \subseteq C(X)$ is a non-zero simple ideal if and only if $I$ is the set of continuous functions $f$ such that either $f = 0$ or $\{ x \in X \,|\, f(x) \neq 0 \} = Y$ where $Y = Y(I) \subseteq X$ is an open subspace depending only on $I$ and whose induced topology is trivial. In particular, any non-zero simple ideal of $C(X)$, if it exists, is isomorphic to $\mathbb{R}$.

If $X$ is a completely regular topological space and $Y$ is an open subspace of $X$ with trivial induced topology, then $Y$ is also closed. In particular, if $x \in X$ is an isolated point, then $\{x\}$ is closed and the ideal consisting of the continuous functions $f$ such that $f(y) = 0$ for every $y \neq x$ is simple. I am indebted to Zach Teitler who mentioned this fact first.

Proof. Let $R = C(X)$ and let $Y \subseteq X$ be a topological subspace whose induced topology is trivial. Then any $f \in R$ is constant on $Y$ and since $Y$ is both closed and open, any continuous function $f$ defined on $Y$ can be continuously extended to $X$ by setting $f(x) = 0$ for $x \notin Y$. Thus the ideal consisting of all functions which are zero on $X \setminus Y$ is a simple ideal. Consider now a simple ideal $I = Rf$ with $f \neq 0$ and let $Y = \{ x \in X \,|\, f(x) \neq 0 \}$. Since $Rf = Rf^2$, the function $1/f$ defined over $Y$ can be continuously extended to $X$. From this we infer that $Y$ is closed. Indeed, if there is $x \in \overline{Y} \setminus Y$, then the continuous extension of $1/f$ cannot be bounded on an open subset containing $x$, a contradiction. Now let $g \in R$ be such that its restriction to $Y$ is not zero. Since $Rfg = Rf = I$, we deduce that $g$ has no zero in $Y$. Let us prove that $g$ is constant. Reasonning by contradiction, we consider $x,y \in Y$ such that $g(x) \neq g(y)$. Let $F$ be the pre-image under $g$ of a closed segment in $\mathbb{R}$ that contains $g(x)$ but not $g(y)$. Then $F$ is a closed set that contains $x$ but not $y$. Since $X$ is completely regular, we can find a function $h \in R$ such that $h$ vanishes on $F$ and $h(y) = 1$. This is impossible since we have established before that $h$ cannot have any zero in $Y$. Hence every $g \in R$ is constant over $Y$. Since $Y$ is both closed and open, every continuous function on $Y$ extends continuously to $X$. Therefore any continuous function on $Y$ is constant. As $Y$ is also a completely regular topological space, its topology is trivial.

$\endgroup$
2
  • 1
    $\begingroup$ So e. g. $C(\mathbb R)$ does not have any minimal ideals? $\endgroup$ – მამუკა ჯიბლაძე Mar 29 '18 at 19:48
  • $\begingroup$ @მამუკაჯიბლაძე Indeed, $C(\mathbb{R})$ doesn't have any minimal ideal. If $X$ is any connected completely regular space with a non-trivial topology, the ring $C(X)$ has no minimal ideals. $\endgroup$ – Luc Guyot Mar 29 '18 at 20:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.