Non-noetherian scheme with infinitely many irreducible components passing through a point? Point still a domain? Is there a non-noetherian scheme with infinitely many irreducible components passing through a point? (I expect the answer to be yes, but I do not know of an example.)
For extra internet points, I would love it if somehow the local ring of that point was still a domain - that the usual procedure for building zero divisors at a point meeting a finite number of irreducible components fails somehow. For instance, this would require that there was no way to subdivide these infinitely many irreducible components into a finite collection of closed sets.
 A: Let $R$ be the quotient of a polynomial ring $k[x_1,x_2,\dots]$ in infinitely many variables over a field by the ideal generated by all products $x_ix_j$ for $i\neq j$.   Note that if $P\subset R$ is a prime ideal, then there can be at most one $i$ such that $x_i\not\in P$.  It follows that if $P_i$ is the ideal generated by all the $x_j$ for $j\neq i$, every minimal prime of $R$ is of the form $P_i$ (each $P_i$ is prime since $R/P_i\cong k[x_i]$).  All of these minimal primes are contained in the maximal ideal $M$ generated by all the $x_i$.  So $\operatorname{Spec} R$ has a point $M$ which is in all of the infinitely many irreducible components of $\operatorname{Spec} R$.  (Geometrically, you should think of $\operatorname{Spec} R$ as infinitely many lines meeting at the point $M$.  Indeed, there is a natural bijection between $\operatorname{Spec} R$ and infinitely many copies of $\mathbb{A}^1_k$ with the origin in all of them identified.)
On the other hand, it is impossible for the local ring at an intersection point of infinitely many irreducible components to be a domain.  To see this, we may assume our scheme is affine; say it is $\operatorname{Spec} R$ for some ring $R$, and the intersection point is some prime $P\subset R$.  The irreducible components of $\operatorname{Spec} R$ correspond to the minimal prime ideals of $R$, so there are infinitely many minimal primes of $R$ contained in $P$.  But there is an inclusion-preserving bijection between the prime ideals of the localization $R_P$ and the prime ideals of $R$ that are contained in $P$.  So every minimal prime of $R$ corresponds to a minimal prime of $R_P$, and $R_P$ must therefore have infinitely many minimal primes.  In particular, it cannot be a domain.
