Is the ring of germs of $C^\infty$ functions at $0$ Noetherian?

Question

I'm considering the property of the ring $R:=C^\infty(\mathbb R)/I$, where $I$ is the ideal of all smooth functions that vanish at a neighborhood of $0$. I find that $R$ is a local ring of which the maximal ideal is exactly $(x)$. I also want to determine if $R$ is a Noetherian ring, but I have no idea about how to tackle with it. Can anyone help me? Thanks in advance.

Answer 1

I think this should work:

Consider the function $f_n$ defined by $$f_n(x) = e^{-1/x^2}\sin(2^n/x).$$ Intuitively, the exponential vanishes so incredibly quickly that the craziness of the sine term causes no problems. More formally, the $n$th derivative (away from zero) of $f_1$ is, by induction, of the form $$e^{1/x^2}[P(1/x)\sin(1/x)+Q(1/x)\cos(1/x)]$$ which is easily seen to be continuous everywhere by the fact that $e^{1/x^2}/x^n$ goes to zero as $x$ goes to zero, for any given $n$, and the trig terms are bound between $-1$ and $1$. $f_n$ should be similar. But the vanishing locus of $f_{n}$ strictly contains that of $f_{n+1}$ (and intersects $(-\varepsilon, \varepsilon)$ infinitely many times), so $$(0) \leq (f_1) \leq (f_1, f_2) \leq \cdots$$ is an infinite ascending chain.

Answer 2

I think $\langle e^{-\frac 1{x^{2n}}}\rangle_{n\in\mathbb N}$ is an infinitely generated ideal of $R$.