2
$\begingroup$

Let $\mathscr{A}_p=\{f\in C[0,1]:f(p)=0\}$. Then we all know that every maximal ideal of $C[0,1]$ is of the form $\mathscr{A}_p$ for some $p\in [0,1]$.

Instead of considering compact set $[0,1]$, if we take an open set $(0,1)$, then $\mathscr{M}_p=\{f\in C(0,1):f(p)=0\}$ is also maximal ideal in $C(0,1)$ for $p\in (0,1)$.

But can we have a maximal ideal of $C(0,1)$ which is not of the form $\mathscr{M}_p$ for all $p\in (0,1)$ ? The answer is YES. I have constructed such ideal.

Let $\mathscr{M}=\{f\in C(0,1):f\text{ is of compact support}\}$. Clearly this is an ideal. Now take any $p\in (0,1)$. Then we can clearly construct a continuous function which is non-zero at point $p$ and has a compact support. Hence we have showed that $\mathscr{M}\not= \mathscr{M}_p$ for any $p\in (0,1)$. Thus we have a maximal ideal which is not of the form $\mathscr{M}_p$ for all $p$.

Now the $\textbf{first question}$ is : Can we have an infinite number of maximal ideals which are not of the form $\mathscr{M}_p$ ?

The $\textbf{second question}$ is : Consider the map $\mathbb{R}\to C(0,1)$ which sends $a\in\mathbb{R}$ to a constant function with value $a$. If $\mathscr{N}$ is a maximal ideal of $C(0,1)$ such that $$\mathbb{R}\longrightarrow C(0,1) \longrightarrow C(0,1)/\mathscr{N}$$ is an isomorphism, then show that $\mathscr{N}$ has to be $\mathscr{M}_p$ for some $p\in (0,1)$.

I don't know how to even start. Any help or hint will be appreciated. Thank you in advance.

$\endgroup$
  • $\begingroup$ You probably meant $\mathscr{M}$ is not included in $\mathscr{M}_p$, not $\neq$ . $\endgroup$ – Max Feb 11 at 9:01

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.