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.

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

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