4
$\begingroup$

I asked this question at MSE but I did not received any answer, so I repeat it here at MO:

What is an example of a Hausdorff topological space $X$, not a singleton, such that the ring $C(X)$ of all real (or complex) valued continuous functions on $X$ is a clean ring?

A clean ring is a ring in which every element is a sum of a unit and an idempotent.

$\endgroup$
0

2 Answers 2

5
$\begingroup$

Well, the two point set will do. Also an infinite discrete set will do. To give a less obvious example, let me argue that every compact space with a basis of clopen sets will do too.

Let $X$ be such and consider a real valued function $f$ on $X$. Consider the compact subset $\{f \leq 1/3\}$ and cover it with clopen sets that are contained in $\{f<2/3\}$. Choose a finite subcover and let $p$ be the characteristic function of its union. This is a continuous idempotent. $u=f-p$ is a continuous function that is bounded away from 0, hence a unit. $f=p+u$.

$\endgroup$
12
$\begingroup$

For a Tychonoff space $X$, the rings $C(X)$ and $C(X,\mathbb{C})$ are each clean if and only if the space $X$ is strongly zero-dimensional (s.z.d). This is a result of Azarpanah, me, and most recently Arora and Kundu.

A Tychonoff space $X$ is s.z.d if and only if every cozeroset is the countable union of clopen subsets. In particular, s.z.d implies a base of clopen sets. Examples of s.z.d spaces include the rationals, the Cantor space, and W.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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