Ideals of the ring of smooth functions

Question

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 $C^\infty(M)$?

Answer 1

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. 635-658 (1948)

Answer 2

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 non-closed 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 non-isolated 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.