Remark in Atiyah and Macdonald proving $C^\infty(\mathbb{R})$ is not Noetherian. On page 110 of Atiyah and Macdonald's Commutative Algebra, they remark that the ring $C^\infty(\mathbb{R})$ of smooth functions on $\mathbb{R}$ is not Noetherian by observing that if $\mathfrak{a}$ is the ideal of functions vanishing at $0$, then the kernels of the canonical map into the $\mathfrak{a}$-adic completion and the kernel of the canonical map into the localization at $1+\mathfrak{a}$ do not coincide.
Here is the setup: $A=C^\infty(\mathbb{R})$, $\mathfrak{a}$ the ideal of functions vanishing at $0$, which is generated by the identity function $x$, so $\mathfrak{a}=(x)$.
I'm confused where they state $f\in A$ is annihilated by some element $1+\alpha$ ($\alpha\in\mathfrak{a})$ iff $f$ vanishes identically in some neighborhood of $0$.
Writing $\alpha=g\cdot x$ for some smooth $g$, $(1+gx)f=0$ implies $-f=fgx$. So if $f$ vanishes on a neighborhood $U$ of $0$, then probably using bump functions/partition of unity argument or something related, one can construct a smooth function $g$ such that $g$ coincides with $-1/x$ off of $U$, but is say constant on $U$, so that $-f=fgx$ everywhere, and thus $1+gx$ annihilates $f$.
I think this is okay, but I can't see why $(1+\alpha)f=0$ for some $\alpha\in\mathfrak{a}$ implies $f$ vanishes on a neighborhood of $0$.
 A: For one direction, assuime $f$ vanishes on $0\in U$. Your idea is correct, but generally cannot guarantee that $g(x)=-1/x$ on $\mathbb{R}\setminus U$ and that $g$ is constant on $U$. Thankfully, what $g$ does on $U$ doesn't matter, so pick a smaller neighborhood $0\in V\subseteq U$, such that $\overline{V}\subseteq U$. Then, $\mathbb{R}\setminus\overline{V}$ and $U$ is an open cover of $\mathbb{R}$, so let $\rho_1,\rho_2$ be a subordinate partition of unity. The function $g(x)=\rho_2(x)-\frac{\rho_1(x)}{x}$ is well-defined, since $\rho_1$ vanishes on the neighborhood $0\in V$ and smooth. For $x\in\mathbb{R}\setminus U$, $\rho_2(x)=0$, so $\rho_1(x)=1$ and $1+g(x)x=0$. For $x\in U$, $f(x)=0$. So $(1+g(x)x)f(x)=0$ for all $x\in\mathbb{R}$.
For the other direction, note that $\alpha\in\mathfrak{a}$ implies $\alpha(0)=0$, by definition. By continuity, there exists an open neighborhood $0\in U$, such that $\alpha(x)\in\mathbb{R}\setminus\{-1\}$ for $x\in U$. Then, $1+\alpha(x)\neq0$ for $x\in U$, but $(1+\alpha)f=0$; so $f(x)=0$ for all $x\in U$, i.e. $f$ vanishes identically on $U$.
