basic question on integral schemes If $X$ is a (reduced) scheme and $P$ is a point of $X$ (not necessarily closed) such that the local ring $\mathcal{O}_{X,P}$ is a regular domain, then must there exist an open affine neighborhood $U = \text{Spec }A$ of $P$ such that $A$ is an integral domain?
I'm almost certain this is true, since the local ring being regular means that it doesn't sit in the intersection of irreducible components, and hence it must be "locally" irreducible...but I can't think how to prove it.
If needed, we can assume $X$ is also Noetherian.
 A: The theory of irreducible components can be tricky when $X$ is not Noetherian,
so I will suppose that $X$ is Noetherian, and so is the union of finitely many
irreducible components.  
Now the components on which $P$ lies are in bijection with the minimal primes 
of $\mathcal O_{X,P}$, and hence $P$ lies on a single component.  The complement of all the other components is an integral open subscheme of $X$ containing $P$,
and (like an scheme) it contains an open affine n.h. containing $P$, which will also be integral.
If we don't assume that $X$ is Noetherian, there can be topological complications
in the theory of irreducible compoments.  E.g. if $A$ is any Boolean ring (i.e. a ring of characteristic $2$ in which every element is idempontent), then each local ring of Spec $A$ is a domain, in fact a copy of $\mathbb F_2$, but Spec $A$ can contain points that are not contained in any integral open subscheme; one concrete example is given by $A = \prod_{n = 1}^{\infty} \mathbb F_2$ (and there are many others).  
