3
$\begingroup$

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.

  • 0
    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 $\mathfrak{m}$.2011-04-01

1 Answers 1

4

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.