6
$\begingroup$

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 $c \in (0,1)$ .

$\endgroup$
8
  • 3
    $\begingroup$ Do you want $C((0,1))$, i.e. all the continous functions (possibly unbounded), or $C_b((0,1))$ (only the bounded continuous ones)? $\endgroup$ – Alex M. Apr 27 '18 at 12:11
  • 4
    $\begingroup$ 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. $\endgroup$ – Martin Sleziak Apr 27 '18 at 12:21
  • 2
    $\begingroup$ 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. $\endgroup$ – Alex M. Apr 27 '18 at 12:22
  • 11
    $\begingroup$ 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. $\endgroup$ – Martin Sleziak Apr 27 '18 at 13:36
  • 1
    $\begingroup$ @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. $\endgroup$ – LSpice Dec 8 '20 at 15:49
5
$\begingroup$

Since $\mathbb{R}$ and $(0,1)$ are homeomorphic, you can look at maximal ideal in $C(\mathbb R)$.

There is the idea form Dummit and Foote Abstract Algebra, page 259 Ex 34.

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 $f\notin M_{c}$.

$\endgroup$
5
  • 1
    $\begingroup$ This is one of those occasions when you have the sinking feeling that Zorn's lemma isn't such a great idea. $\endgroup$ – Eugene Z. Xia Dec 8 '20 at 9:47
  • 1
    $\begingroup$ Of course the maximal ideal here is really a maximal ideal, right, or is there only one? \\ Notice that this point of view was already mentioned by @MartinSleziak, who linked to mathoverflow.net/q/3871, although this particular answer does not seem to appear there. $\endgroup$ – LSpice Dec 8 '20 at 15:16
  • $\begingroup$ Specifically, this construction is mentioned by @AntonGeraschenko in a comment. $\endgroup$ – LSpice Dec 8 '20 at 15:42
  • $\begingroup$ @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 $M$. $\endgroup$ – Matthew Daws Dec 8 '20 at 20:14
  • $\begingroup$ @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.) $\endgroup$ – LSpice Dec 8 '20 at 21:11

Your Answer

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