Stalk and local property of a scheme This has confused me for a long time: say if a scheme $(X,O_X)$ has property P at the stalk $O_{X,x}$, does it imply that there exists a neighborhood $U$ of $x$ such that $P$ satisfies on $(U, O_X|_U)$?
For example, I want to prove that a necessary condition for a scheme to be reduced is that the stalk $O_{X,x}$ at $x$ has no nilpotent element.

For the proof of $\Leftarrow$ direction, assume there is an $f$ such
  that $f_x^m=0$, then there is an $f\in \Gamma(V,O_X)$ such that $f^m =
 0$. Hence $\Gamma(V,O_X)$ has a nilpotent element.

So in the proof above we can just pass from stalk to some neighborhood $V$ of $x$, can we always do the same for other properties?
Edit: another example:Is invertibility a stalk-local property?
 A: To give a simple answer, a lot of these properties say that if $f = g$ at $x$ then $f = g$ in a neighbourhood of $x$. For instance nilpotent: $f^m = 0$ at $x$ or invertible: $fg = 1$ at $x$. The reason for this comes down to the definition of the stalk as a direct limit.

Quick refresher on direct limits if you need it:
$$ \mathcal{O}_{X,x} = \lim_{\substack{\longrightarrow \\ U \ni x}} \mathcal{O}_X(U) = \bigsqcup_{U \ni x} \mathcal{O}_X(U) \Big/\sim.$$
You should think of a directed limit roughly as a disjoint union where if $a \in A \cap B$ then $a$ as an element of $A$ is equivalent to $a$ as an element of $B$. In this specific case, the equivalence is
$$ (f \in \mathcal{O}(U)) \sim (g \in \mathcal{O}(V)) \text{ if } f = g \text{ on } U \cap V. \tag{1} $$
We write an element of $\bigsqcup \mathcal{O}(U)$ as $(f, U)$ where $f \in \mathcal{O}(U)$ (so we remember which open set it comes from) and $f_x \in \mathcal{O}_x$ is the equivalence class of some $(f, U)$.

You can see that $(1)$ captures the idea that if $f_x = g_x$ then $f = g$ in a neighbourhood of $x$. Namely if $(f, U)$ represents the equivalence class $f_x$ and $(g, V)$ represents $g_x$ then $(f, U) \sim (g, V)$ means that $f|_{U \cap V} = g|_{U \cap V}$.
