$k$th power of ideal of of germs Well,We denote set of germs at $m$ by $\bar{F_m}$ A germ $f$ has a well defined value $f(m)$ at m namely the value at $m$ of any representative of the germ. Let $F_m\subseteq \bar{F_m}$ be the set of germs which vanish at $m$. Then $F_m$ is an ideal of $\bar{F_m}$ and let $F_m^k$ denotes its $k$th power. Could any one tell me how the elements look like in this Ideal $F_m^k$? and they said all finite linear combination of $k-fold$ products of elements of $F_m$ But I dont get this. and they also said These forms $\bar{F_m}\supsetneq F_m\supsetneq F_m^2\supsetneq\dots \supsetneq$
 A: If $x_1,\ldots,x_n$ are local coordinates at the point $m$, then any smooth germ at $m$ has an associated Taylor series in the coordiantes $x_i$.  The power $F_m^k$ is precisely the set of germs whose degree $k$ Taylor polynomial vanishes, i.e. whose Taylor series has no non-zero terms of degree $\leq k$.
A: Let $R$ be a commutative ring and $I\subseteq R$ an ideal of $R$. The $k$th power $I^k$ of that ideal is defined to be the set of all elements of $R$ that can be written as finite sums of elements of the form $a_1\cdot a_2\cdot\ldots\cdot a_k$ with $a_i\in I$. One can easily check, that with $I^k$ is itself an ideal of $R$. It is then also clear that for all $k\in\mathbb{N}$ one has $I^{k+1}\subseteq I^k$, while in general one does not have $I^{k+1}\neq I^k$. For the latter one needs more asumptions concerning the ring $R$ or the ideal $I$.
A simple guess in the case you are considering is this one: on a smooth manifold there exist functions $f$ that are smooth in a neighborhood of $m$ possessing a simple zero at $m$. The germ defined by such a function then lies in $F_m$ but not in any $F_m^k$, $k>1$. Similar for the germ defined by $f^2$ etc.
