Minimal prime ideals of $\mathcal O_{X,x}$ correspond to irreducible components of $X$ containing $x$ Let $X$ be an algebraic variety over an algebraically closed field $K$.  By definition, $X$ is a separated prevariety, and $x \in X$.  I'm trying to show 

(i): The minimal primes of $\mathcal O_{X,x}$ are in bijection with the irreducible components of $X$ containing $x$.  

Since $\mathcal O_{X,x}$ is reduced, this shows in particular that

(ii): $\mathcal O_{X,x}$ is an integral domain if and only if $x$ lies in only one irreducible component.

The affine case is pretty easy.  Here $X = \textrm{Max } A$ for some reduced finitely generated $K$-algebra $A$.  If $m \in X$, the stalk $\mathcal O_{X,m}$ is canonically isomorphic to $A_m$, and prime ideals here are in bijection with primes $P$ of $A$ contained in $m$, i.e. primes $P$ for which $m \in V(P) = \{ \mathfrak m \in X : P \subseteq \mathfrak m\}$.  Primes $P$ correspond to irreducible closed sets $V(P)$, with minimal primes corresponding to irreducible components, done.  
This is apparently supposed to generalize to the nonaffine case, but I'm just not seeing it.  If $X_1, ... , X_s$ are all the irreducible components of $X$, and $x$ lies only in $X_1, ... , X_t$, and if I pick an affine open set $U$ containing $x$, then a subset of $X_1 \cap U, ... , X_s \cap U$ are the irreducible components of $U$, and a subset of $X_1 \cap U, ... , X_t \cap U$ are the components containing $x$.  On account of the fact that intersecting the $X_i$ with $U$ may produce some repetition or loss of information, I do not think that (i) is true in general. What do you think?
 A: Let $(X,\mathcal{O}_X)$ be an algebraic variety over a field $\mathbb{K}$ and let $x\in X$; by your definition, there exists an open neighbourhood $U$ of $x$ in $X$ such that $U$ is an affine algebraic variety.
Let $\mathscr{X}=\{X_i\}_{i\in I}$ be the family of irreducible components of $X$, and let $\mathscr{U}=\{U_j\}_{j\in J}$ be the family of irreducible components of $U$; as affirmed by Ayman Hourieh: the function $U_j\in\mathscr{U}\to\overline{U_j}^X\in\mathscr{X}$ is a bijection, with inverse $X_i\in\mathscr{X}\to X_i\cap U\in\mathscr{U}$; in particular, there is a bijection between the irreducible components of $X$ cointaining $x$ and the irreducible components of $U$ cointaining $x$, and furthermore: this bijection is independent from the particular $U$!
By your definition: $U$ is (isomorphic as locally ringed space to) the maximal spectrum of a reduced finitely generated $\mathbb{K}$-algebra $A$; in particular: $A$ is a Noetherian ring, therefore by Noether-Lösker theorem, $A$ admits minimal prime ideals; by reducedness of $A$, these are all and only associated prime ideal of $(0)$, therefore $A$ has finitely many minimal prime ideals!
By a topological reasoning, $U$ has finitely many irreducible components too, and these are in bijection with the (finitely many) minimal prime ideals of $A$; in particular, the irreducible components of $U$ containing $x$ are in bijection with the minimal prime ideal of $A$ contained in $\mathfrak{m}_x$, the maximal ideal of $A$ corresponding to $x$; that is they are in bijection with the minimal prime ideals of $A_{\mathfrak{m}_x}$.
Because $\mathcal{O}_X(U)=A$ and $\mathcal{O}_{X,x}$ is indepedent from $U$, then $\mathcal{O}_{X,x}\cong A_{\mathfrak{m}_x}$; that is the irreducible components of $X$ containing $x$ are in bijection with the minimal prime ideals of $\mathcal{O}_{X,x}$.
P.S.1: The irreducibility of $X$ is wrong, otherwise the answer to your first question is obvious!
P.S.2: We do not need to work over an algebraically closed field!, otherwise, by separatedness of $X$, one can prove that $X$ has finitely many points.
