X is a prevariety, then prove local ring of a point x ∈ X is integral if and only if x lies on a unique irreducible component of X? I actually do not even understand this problem? How can the product of two regular function be zero? I mean in what case local ring at x is not an integral domain?
I have no idea about how to prove this.
 A: If $x \in X$ lies on a unique irreducible component $V \subset X$, one easily sees $\mathcal O_{X,x} = \mathcal O_{V,x}$. The latter is an integral domain, since $V$ is an irreducible and reduced variety.
The reason for the equality of the two local rings is the following: Let $X = V \cup Y$ with $Y$ closed and $x \notin Y$ ($Y$ might possibly chosen to be empty, but then there is nothing to show). For any open set $U \ni x$ in $X$, we can consider the open set $U \cap Y^c$. This is an open set contained in $V$ and still containing $x$. Hence it does not matter whether computing the direct limit over all open sets containing $x$ or only over open sets containing $x$ and contained in $V$.
For the converse, you should first reduce to the affine case and then it is quite easy: Let $X_1, \dotsc, X_s$ be the irreducible components, that $x$ lies in. By definition of Zariski topology, we have $X_i=V(f_i)$. If $i > 1$, the stalks of the $f_i$ at $x$ are non-zero, but the stalk of the product $f_1 \dotsb f_s$ is zero.
