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If $O_p$ is the ring of germs of smooth functions at point $p$ and $\alpha\in O_p$ represents $(U,f)$, and the map $I:O_p→R,(U,f)\xrightarrow{}f(p)$ is an $R$-algebra homomorphism. It says that if $I(\alpha)$ is not zero, then $\alpha$ is a unit.

I don't really get this statement. How can I prove this? I am thinking about using the continuity of $f$ but I am not getting the picture.

hslo
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If $f(p)\neq 0$ then, since it is continuous, there exists a neighbourhood of $p$, let's say $W$, where $f(q)\neq 0$ for $q\in W$. Then $\alpha=[(U,f)]=[(W,f|_W)]$ has the inverse $\alpha^{-1}=[(W,1/f|_W))]$.

By the way, the existence of the neighbourhood $W$ can be proven by contradiction. If every neighbourhood $V$ of $p$ contains a point $q_V$ such that $f(q_V)=0$ you can create a sequence $(q_n)$ of points convergent to $p$ such that $f(q_n)=$ and then, by continuity, $f(p)=0$.

A. J. Pan-Collantes
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  • If it is about the kernel of I, then the contrapositive will mean that the kernel is all the non-units right? – hslo Oct 12 '22 at 15:52
  • Is this related to the maximal ideal of $O_p$? – hslo Oct 12 '22 at 15:59
  • Yes, the kernel of $I$ is the maximal ideal, since $O_p/\ker(I)\approx \mathbb{R}$, which is a field. See here:https://math.stackexchange.com/questions/97928/if-i-is-a-maximal-ideal-of-r-why-is-r-i-a-field – A. J. Pan-Collantes Oct 12 '22 at 16:29