Germ of a smooth function

Remarks

With these operations $\mathcal{E}_p$ is a commutative $\R$-algebra. It is a local ring: its unique maximal ideal $\mathfrak{m}_p$ consists of the germs with $f(p) = 0$. Indeed, if $f(p) \ne 0$, then $f \ne 0$ on a neighbourhood of $p$ and $[1/f]_p$ is an inverse of $[f]_p$. The value $f(p)$ and all partial derivatives of $f$ at $p$ in a chart depend only on the germ.

Example: on $M = \R$, let $f(x) = x^2$ on $\R$ and $g(x) = x^2$ on $(-1, 1)$. They agree on $W = (-1, 1)$, so $[f]_0 = [g]_0$.

Non-example: let $f = 0$ on $\R$, and let $h(x) = e^{-1/x^2}$ for $x \ne 0$ with $h(0) = 0$. All derivatives of $f$ and $h$ at $0$ agree, but $h(x) \ne 0$ for every $x \ne 0$. So $f$ and $h$ agree on no neighbourhood of $0$, and $[f]_0 \ne [h]_0$. A germ is thus finer information than the Taylor series at $p$.