What are the maximal ideals in $C(0,1)$ apart from $M_c = \{f\in C(0,1) | f(c)=0\}$?

I have studied about maximal ideals in $C[0,1]$ . They are precisely of the form $$\{f\in C[0,1] | f(c)=0\} \text{ for } c\in [0,1].$$

If we replace $[0,1]$ by $(0,1)$ and then look at $C(0,1)$, then obviously $M_c = \{f\in C(0,1) | f(c)=0\}$ are maximal ideals in $C(0,1)$ , $\forall c \in (0,1)$ , but since we do not have the compactness of $[0,1]$ anymore, I guess there are some other maximal ideals as well.

I am trying to solve the problem by first giving an existential argument and then by showing an explicit maximal ideal other that of the form $M_c$ for some $c \in (0,1)$ .

• Do you want $C((0,1))$, i.e. all the continous functions (possibly unbounded), or $C_b((0,1))$ (only the bounded continuous ones)? – Alex M. Apr 27 '18 at 12:11
• For bounded functions this seems to be directly linked to the Stone-Cech compactification. And you could probably find a bit more in the classical text Gillman, Jerison: Rings of Continuous Functions. (I guess somebody who know more about this will post a more detailed answer soon.) However, I am not sure whether this is too helpful, since Stone-Cech compactification of $(0,1)$ seems like a quite complicated object. – Martin Sleziak Apr 27 '18 at 12:21
• In the case of $C_b(X)$ for non-compact Hausdorff $X$, the Gelfand-Naimark theorem tells us that this must be isomorphic to $C_b(Y)$, where $Y$ is the compact space of all maximal ideals (equivalently - continuous characters). It turns out that $Y$ is precisely the Stone-Čech compactification of $X$, which is notoriously difficult to give explicitely even for simple spaces. – Alex M. Apr 27 '18 at 12:22
• Since $(0,1)$ and $\mathbb R$ are homeomorphic, this question might be worth looking at: Maximal ideals in the ring of continuous real-valued functions on R. – Martin Sleziak Apr 27 '18 at 13:36
• @AlexM., re, according to @‍EricWofsey (who cites Johnstone's "Stone spaces" (MSN)), the maximal ideals in $C(X)$ and $C_b(X)$ are the same. – LSpice Dec 8 '20 at 15:49

Since $$\mathbb{R}$$ and $$(0,1)$$ are homeomorphic, you can look at maximal ideal in $$C(\mathbb R)$$.
Let $$I\subset C(\mathbb R)$$ be the set of all continuous functions with compact support. It can be shown that $$I$$ is an ideal. Let $$M$$ be a maximal ideal containing $$I$$. Then $$M\neq M_{c}$$ for any $$c\in \mathbb R$$. For suppose $$M=M_{c}$$, and let $$r=|c|+1$$. There is a function $$f\in C(\mathbb R)$$, such that $$f(x)>0$$ when $$x\in (-|c|,|c|)$$ and $$f(x)=0$$ when $$x>r$$ or $$x<-r$$. Then $$f\in I$$ but $$f\notin M_{c}$$.
• @LSpice But notice that the answer there shows that maximal ideals correspond to points of $\beta\mathbb R$. So if $x\in\beta\mathbb R \setminus\mathbb R$ then $f(x)=0$ for any $f\in I$, and so $M_x$ contains $I$. Thus there are many, many choices for $M$. – Matthew Daws Dec 8 '20 at 20:14
• @MatthewDaws, indeed, thanks. Your 'but' suggests disagreement, but I meant to argue against the terminology "the maximal ideal containing $I$", not for it. (Anyway, @‍AlexM. has now edited out that terminology, so I would delete my comment except that I think that the other half of it might still be worthwhile.) – LSpice Dec 8 '20 at 21:11