ideal in the ring of smooth functions

Question

What is an ideal $I$ of the ring of smooth functions $C^{\infty}(\mathbb R)$ which is not finitely generated and for all $x\in\mathbb R$ there exist $f\in I$ such as $ f(x)\neq 0$.

Answer 1

Consider the function

$$h(x)=\begin{cases}e^{-1/x^2},&\text{ for }x>0\\0,&\text{ for }x\leq0 \end{cases}$$

then $g(x)=h(x)h(2-x)$ and finally $g_n(x):=g(x+n)$.

Put

$$I=(\{g_n\}_{n\in\mathbb{Z}}).$$

The functions $g_n$ are smooth, have a bump in an interval and are zero everywhere else. For each $x\in\mathbb{R}$ we can choose a $g_n$ with a bump over that $x$.

If $I$ were generated by $f_1,f_2,...,f_m$ then every function in $I$ must vanish on $Z(I):=\bigcap_{i=1}^{m}Z(f_i)$, where $Z(f)$ is the set of zeros of $f$. But no $x$ is in the zero set of all $a\in I$. This means that $Z(I)=\emptyset$. But this can't be the case since all functions in $I$ must vanish outside some big compact set. In fact all functions of $I$ are of the form $p:=a_1g_{n_1}+a_2g_{n_2}+...+a_kg_{n_k}$, for some $a_1,a_2,...,a_k\in C^{\infty}(\mathbb{R})$. Such a $p$ vanishes outside, say $\{|x|< n_1+n_2+...+n_k\}$.

Answer 2

The ideal

$$I = \{ f \ | \text{ there exists } M \ \text{so that } f(x) = 0 \ \forall\, x \ge M\}$$

has no zeroes since for every $M$ we have $f_M \in I$ where

$$f_{M}(x)=\begin{cases}e^{\frac{1}{x-M}},&\text{ for }x<M\\0,&\text{ for }x \ge M \end{cases}$$

If $f_1$, $\ldots$, $f_n \in I$ then there exists $M_0$ so that for all $f \in (f_1, \ldots f_n)$ and $x\ge M_0$ we have $f(x)= 0$, that is, $[M_0, \infty)$ is in zero set of the ideal $(f_1, \ldots f_n)$. Since $I$ has no zeroes we conclude that $(f_1, \ldots, f_n) \ne I$. Thus $I$ is not finitely generated.