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Is the ring $\mathcal{O}$ of germs of $C^{\infty}$ functions defined on the neighborhoods of $0\in\mathbb{R}$ the localization of the ring of $C^{\infty}$ functions on $\mathbb{R}$ at the maximal ideal $\mathfrak{m}$ of those functions which vanish at the origin? I can construct an injective map from $C^{\infty}_\mathfrak{m}$ to $\mathcal{O}$, but I'm having trouble showing surjectivity. I know that this is true for regular functions (those which are locally quotient of polynomials) over an algebraically closed field. One big difference I see between regular functions and $C^{\infty}$ functions is that $\mathcal{O}$ is not integral domain.

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Can you show your proof of injectivity? Surjectivity seems a weird place to get hung up, as trivially any element of $\mathcal{O}$ can be realized as the equivalence class of an element of $C^\infty$, and equivalence in $\mathcal{O}$ respects equivalence in $C^\infty$ because if the difference of two functions in $C^\infty$ is zero on some neighborhood of $0$ then they must also be zero at $0$. – Alex Becker Apr 1 '11 at 4:09
If a $C^{\infty}$ function vanishes around the origin, one can multiply it by a bump function at the origin to make it zero everywhere. – ashpool Apr 1 '11 at 4:21
I meant $C^\infty_{\mathfrak{m}}$. – Alex Becker Apr 1 '11 at 4:24
When I said $C^{\infty}$ function vanishes around the origin, I'm referring to an element in $\mathcal{O}$, and the bump function refers to an element in $C^{\infty}$ outside $\mathfrak{m}$. – ashpool Apr 1 '11 at 13:32
up vote 3 down vote accepted

Suppose $\xi$ is a germ, and let $\phi:U\to\mathbb R$ be a function defined on some open set $U$ containing $0$ which represents $\xi$. Pick open sets $V$, $W$ such that $0\in V\subseteq\bar V\subseteq W\subseteq \bar W\subseteq U$, and pick a smooth function $\psi:\mathbb R\to\mathbb R$ such that $\psi|_V\equiv1$ and $\psi|_{\mathbb R\setminus W}\equiv0$. Then $\psi\phi$, which in principle is defined only on $U$, can be extended to all of $\mathbb R$ by zero and remain smooth. The germ of $\psi\phi$ is still $\xi$.

This argument can be embellished to prove surjectivity.

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