Let $X$ be a scheme, $a$ a closed point of $X$ considered as a reduced closed subscheme $\{a\} := Z$ of $X$ and let $\alpha \subset \mathcal{O}_X$ the corresponding ideal sheaf.
So we have the exact sequence
$0 \to \alpha \to \mathcal{O}_X \to \mathcal{O}_Z$
Obviously $\mathcal{O}_Z$ is a skyscraper sheaf. I read that
\begin{align} \mathcal{O}_{Z,x} = \left\{ \begin{array}{cc} \kappa(a) = \mathcal{O}_{X,x}/ m_a, & \hspace{5mm} \text{if }x=a \\ 0 & \hspace{5mm} \text{else} \\ \end{array} \right. \end{align}
My quetion is why $\mathcal{O}_{Z,x}$ is a field at $x=a$.
Or more generally: If $Z = \{z_1, ... z_n\}$ is a discrete set of closed points why for their skyscraper structure sheaf $\mathcal{O}_Z$ the stalk $\mathcal{O}_{Z,x}$ for each $x = z_i$ is a field?
The question only depends on an open neighborhood of $a$, so we may assume $X=\operatorname{Spec} R$ is affine, and our sheaves are determined by their global sections as $R$-modules. Then since $a$ is a closed point, it is a maximal ideal of $R$, and that ideal is exactly $\alpha(X)$. So $\mathcal{O}_Z(X)=R/\alpha(X)$ is the quotient of $R$ by a maximal ideal, which is a field. Since $\mathcal{O}_Z$ is a skyscraper sheaf at $a$, its global sections are the same as its stalk at $a$, so $\mathcal{O}_{Z,x}$ is a field. (Or, the stalk $\mathcal{O}_{Z,x}$ is just the localization of $R/\alpha(X)$ with respect to the maximal ideal $\alpha(X)$, but $R\setminus\alpha(X)$ already acts on $R/\alpha(X)$ by units so this does not change it.)
The more general question is exactly the same, since to compute $\mathcal{O}_{Z,z_i}$ you can choose an affine open neighborhood that contains $z_i$ and no other points of $Z$.