This question comes up after I read the chapter 7 : Banach algebras and spectrum theory from Conway's book. As we have known that if $X$ is compact space, then all the maximal ideals of $C(X)=\{ f : X\to \mathbb{C}\}$ are of the forms $m_x =\{f\in C(X) : f(x)=0\}$. Now assume that $X$ is just locally compact and let $X^{+}$ and $\overline{X}$ be the onepoint compactification, and StoneČech compactification of $X$ respectively. Let $I_{0}(X)$, $I(X^+), I(\overline{X})$ be the set of all maximal ideals of $C_{0}(X), C(X^+), C(\overline{X})$ respectively. My question here is that: do we have any relation between these sets $I_{0}(X)$, $I(X^+), I(\overline{X})$ ? For example, if we know $I(X^+)$, how can we find $I_0(X)$ (because if we know the answer for this, it seems for me that one proof of the StoneČech theorem could be found from here) ?

1$\begingroup$ No problem :) I should have explained: you need to use \\{ and \\} for getting braces (that's because one backslash gets "eaten" by the parser). In similar situations just try to double the backslashes, or include LaTeX in backticks. You may want to read the hints on FAQ as well: mathoverflow.net/faq#latex $\endgroup$ – Theo Buehler Jul 26 '11 at 11:04
The set $I(X^+)$ is just the set of all points of $X^+$, and similarly the set $I(\overline{X})$ is the set of points of $\overline{X}$.
On the other hand the set $I_0(X)$ is somewhat mysterious. In general the ideals of a commutative but nonunital Banach algebra are harder to deal with.
When $X$ is not compact, the maximal ideals of $C_0(X)$ are more than just the points of $X$. The points of $X$ are in 11 correspondence with the kernels of (nonzero) characters $C_0(X) \to \mathbb{C}$. These are known as modular ideals (see Exercises 6 and 7 in $\S VII.2$ of Conway), or also known as regular ideals, see e.g. Exercise 4.2.1 of the excellent book Analysis Now, by Gert Pedersen.
Too see the sort of thing that goes wrong in the nonunital situation, for instance let $$ M = \{ f \in C_0(X) \mid \mathrm{supp}(f) \text{ is compact} \}. $$ Then certainly $M$ is an ideal. But no evaluation functional vanishes on $M$, and in fact $M$ is dense in $C_0(X)$.
In a unital Banach algebra, proper ideals are always contained in maximal ones, but this shows that the same is not the case in the nonunital situation. I don't know if this $M$ is actually contained in a maximal proper ideal (which would necessarily be nonclosed) but if so it would certainly not correspond to a point of $X$ and I imagine any such ideal would be impossible to describe in a constructive way.
Sorry if this was a bit incoherent. It's been a long time since I've thought about these things. I don't think I actually answered your question but hopefully it's given you something to think about.

1$\begingroup$ +1 for a clear example of a nonmodular ideal. In fact it is certainly contained in a maximal proper ideal by Zorn's lemma, the point is that this maximal ideal is not closed in the algebra. I have a vague recollection of something like this question being discussed elsewhere on MO (ideals in rings of continuous functions) $\endgroup$ – Yemon Choi Jul 26 '11 at 11:00

4$\begingroup$ Well, the Zorn's lemma thing is not so clear to me. I thought about that, but it seems to me that you need the algebra to be unital in order to conclude that the union of a chain of proper ideals is again proper. $\endgroup$ – MTS Jul 26 '11 at 14:07

$\begingroup$ MTS: oh, good point, was being too hasty. $\endgroup$ – Yemon Choi Jul 26 '11 at 15:17