The ring $C^\infty(M)$ of smooth functions on a smooth manifold $M$ is a topological ring with respect to the Whitney topology and the usual ring operations. Is it possible to describe, maybe under some conditions on $M$, the ideals and the closed ideals of $C^\infty(M)$?

4$\begingroup$ J'imagine il sera utile d'ajouter quelque detail supplementaire. En plus, j'ai vu que avant vous avez ecrit votre questions en Anglais. Pour la plupart des particpants dans ce siteoueb (je crois) il est plus agreable de lire en Anglais. Alors, si ce n'est pas tres penible pour vous, il sera gentil d'ecrire en Anglais. Merci beaucoup par avance. $\endgroup$– user9072Oct 5, 2011 at 18:44

2$\begingroup$ Cette question m'ennerve moins que les questions affreuses deja posees en Anglais par l'auteur... mais enfin, elle n'est pas une bonne question $\endgroup$– Yemon ChoiOct 5, 2011 at 19:53

4$\begingroup$ Je crois que cette question a ete fermee plutot a cause des questions precedentes posees par son auteur que parce qu'elle est mauvaise en soi. C'est vrai qu'elle n'est pas tres bien formulee, mais je pense qu'on peut la sauver. C'est ce que je vais essayer de faire ; cependant, je la reecrirai en anglais pour la rendre plus accessible aux participants de ce site. $\endgroup$– algoriOct 5, 2011 at 20:19

2$\begingroup$ unknowngoogle: now that it has been edited to include some precise definitions, the question looks much better. The original version was in itself inoffensive, but the lack of detail was not encouraging $\endgroup$– Yemon ChoiOct 5, 2011 at 20:50

3$\begingroup$ META: tea.mathoverflow.net/discussion/1163/usersuspension $\endgroup$– Will JagyOct 6, 2011 at 0:47
2 Answers
In 1948 Whitney proved his ideal (spectral) theorem [1] describing the closed ideals
Let $M$ be an $n$dimensional manifold. for each point $p \in M$ and each natural $k$ we define $N(k)$ to be the number of (up to $n$) tuples $m$ such that $m \leq k$. Define the map $J_p^k: C^\infty(M) \rightarrow \mathbb{R}^{N(k)}$ by assigning to $f$ the $m$jets of $f$ at $p$ up to $m=k$.
If $I$ is an ideal of $C^\infty(M)$ then its closure is the ideal of functions $f$ such that for each $p$ in $M$ and $k \geq 0$ then $J^k_p f \in J^k_p(I)$.
So in some sense the closed ideals are like $I_\infty$ in Neil's answer.
[1] H.Whitney. On ideals of differentiable functions. American Journal of Mathematics. Vol. 70, No. 3, pp. 635658 (1948)

3$\begingroup$ I was about to upvote this question when I saw the edit history.. I guess I should upvote algori :) EDIT: well, couldn't resist, it's the question we vote, not who asks it. $\endgroup$ Oct 6, 2011 at 1:21
Here are some examples, for the case $M=\mathbb{R}$.
For each $n\geq 0$ we have a closed ideal $$ I_n=\{f: f^{(i)}(0)=0 \text{ for } 0\leq i < n\} $$ We can write $I_\infty$ for the intersection of these, which is again closed. We can also put $$ J = \{ f : f(x)=0 \text{ for all } x \leq 0\} $$ and note that this is closed and contained in $I_\infty$.
Next, for $n,a>0$ with $n\in\mathbb{Z}$ we can let $K_{n,a}$ be the principal ideal generated by the function $\exp(a/x^{2n})$. These are all different and contained in $I_\infty$. I am not sure whether they are closed.
For another kind of example, let $\mathcal{U}$ be a free ultrafilter on $\mathbb{R}$ and put $$ L = \{f : f^{1}\{0\} \in \mathcal{U} \}. $$ This is a nonclosed maximal ideal.
UPDATE:
Now let $A$ be an arbitrary closed ideal in $C^\infty(\mathbb{R})$. Put $$ X_n = \{ x\in\mathbb{R} : f^{(i)}(x)=0 \text{ for all } i \leq n \text{ and } f\in A\}. $$ Specialising Reimundo's answer to the case $M=\mathbb{R}$, we see that $$ A = \{ f : f^{(i)}=0 \text{ on } X_n \text{ for all } i\leq n \}. $$ The sets $X_n$ are closed, with $X_n\supseteq X_{n+1}$. Moreover, if $x$ is a nonisolated point of $X_n$ (so it is in the closure of $X_n\setminus\{x\}$) then it is easy to see that $x\in X_{n+1}$. I would guess that the closed ideals biject with chains of subsets with these properties, but I have not tried to prove that.

$\begingroup$ Am I missing something here? Take $A:=\{ f\in C^{\infty}(\mathbb{R}):f(0)=0\}$. Then, $X_0=\{ 0\}$ and $X_n=\emptyset$ for $n\geq 1$. Hence, while we do have $A=\{ f\in C^{\infty}(\mathbb{R}):f^{(i)}=0\text{ on }X_n\text{ for all }i\leq n\}$ if $n=0$, do we not have $\{ f\in C^{\infty}(\mathbb{R}):f^{(i)}=0\text{ on }X_n\text{ for all }i\leq n\} =C^{\infty}(\mathbb{R})\neq A$ for $n\geq 1$? $\endgroup$ Dec 1, 2015 at 22:48