Is the function ring $C^{\infty}(M)$ noetherian?

Question

Let $M$ be a smooth manifold and $C^{\infty}(M)$ be its function ring. Is this a noetherian ring?

Answer 1

No, the ring $C^{\infty}(M)$ is never noetherian if $\dim M\gt 0$.
Indeed consider a strictly decreasing sequence of closed subsets $$M=C_0\supsetneq C_1 \supsetneq C_2 \supsetneq C_3\supsetneq \cdots$$ (Such a sequence is easy to construct: take the inverse image in a chart of closed concentric balls in $\mathbb R^n$.)
A theorem of Whitney ensures that there exist a smooth function $f_j\in C^{\infty}(M)$ whose zero locus is precisely $C_j$.
Now if you consider the ideal $I_j=Z(C_j)\subset C^{\infty}(M)$ of functions zero on $C_j$ you get a strictly increasing sequence of ideals of $C^{\infty}(M)$ destroying all hopes of noetherianness for that ring: $$I_0 \subsetneq I_1 \subsetneq I_2 \subsetneq I_3 \subsetneq \cdots \subsetneq C^{\infty}(M)\quad (f_{j+1}\in I_{j+1}\setminus I_j)$$

Answer 2

I think that this is not true in general.

Consider the ring $C^{\infty}(\mathbb{R})$ and the ideal $I$ consisting of smooth functions that vanish on a neighborhood of $0$. I claim that $I$ is not finitely generated.

Assume by contradiction that $I=(f_{1},\ldots,f_{n})$, where each $f_{i}$ vanishes on a neighborhood $V_{i}$ of $0$. Then each element of $(f_{1},\ldots,f_{n})$ vanishes on $V:=\cap_{i=1}^{n}V_{i}$. Though we can construct smooth functions in $C^{\infty}(\mathbb{R})$ that vanish on an arbitrarily small neighborhood of $0$, in particular strictly smaller than $V$. So $I\neq (f_{1},\ldots,f_{n})$. Hence $I$ is not finitely generated.

Answer 3

Consider the sequene of ideals of $C^{\infty}(\mathbb {R}) $, for instance, given by: $$m \mapsto J_m =\langle \{ e, e^x, e^{x^2}, \cdots e^{x^m}\} \rangle $$ This is an ascending chain of ideals with no upper bound. The ring is not Noetherian (neither Artinian).