The ring of germs of functions $C^\infty (M)$

Define $C^\infty (M)_x := \{ (U,f) | x \in U $ open $ , f \in C^\infty (U) \} / \sim $ where $M$ is a manifold and $(U,f) \sim (V,g)$ if $\exists W$ open, $x \in W$ such that $W \subset V \cap U$ and $f|_W = g|_W$.

This is a ring with the following operations: $[(U,f)] + [(V,g)] := [(U \cap V, f + g)]$ and $[(U,f)]\cdot [(V,g)]:= [(U \cap V, fg)]$.

I’m trying to understand what multiplicative inverses look like, i.e. if I have $[(U, f)]$ then I want to show that there exists an open set $V$ containing $x$ such that $\frac{1}{f}$ is smooth on $V$.

Is this right? And can someone explain to me how I can show this? Many thanks for your help.

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There are two rings involved in your question:

1) The ring you described, which should actually be denoted $\mathcal C^\infty_{M,x}$
2) The ring $\mathcal C^\infty(M)_{\frak m}$, described as follows. Call $\frak m$ the ideal in $\mathcal C^\infty(M)$ consisting of global functions zero at $x$, that is ${\frak m}=\lbrace f\in \mathcal C^\infty(M): f(x)=0\rbrace$. This is a maximal ideal in $\mathcal C^\infty(M)$, and you can localize $\mathcal C^\infty(M)$ at this ideal to get $\mathcal C^\infty(M)_{\frak m}$. The elements of this localized ring are formal fractions
$f/g$ with $f,g\in C^\infty(M)$
and $g(x)\neq 0$ . Beware that these formal fractions are by no stretch of the imagination interpretable as functions on $M$
, since $g$ might very well vanish outside, say, the interval $(x-1,x+1)$ !

There is a canonical ring morphism $\mathcal C^\infty(M)_{\frak m} \to \mathcal C^\infty_{M,x}$ and the strange, perhaps underappreciated, result is that it is an isomorphism.

To answer your question:
1) In the first incarnation an invertible element is a germ $[(U,h)]$ with $h(x)\neq o$ and its inverse is $[(U’,1/h)]$, where $U’\subset U$ is a neighbourhood of $x$ on which $h$ has no zero.
2) In the second incarnation an invertible element of $\mathcal C^\infty(M)_{\frak m}$ is a fraction of the form $f/g$ with $f(x)\neq 0$ (in addition to $g(x)\neq 0$, of course), and its inverse is $g/f$.