Let $X$ be either the spectrum, or the punctured spectrum of a Noetherian regular local ring (by punctured spectrum I mean Spec$(R)\setminus \{\mathfrak m\}$ where $\mathfrak m$ is the unique maximal ideal of $R$ ) . Let $d=\dim X$.
If $U,V$ are non-empty subschemes of $X$ with $\dim U + \dim V \ge d$ , then is it necessarily true that $U \cap V$ is non-empty ?
If this is not true in general, then what if we also assume both $U,V$ are open or both $U,V$ are closed ?
When $U,V$ are closed in the spectrum of a local ring , then $U\cap V$ must be non-empty because for any two proper ideals $I,J$ of a local ring, $I+J$ is also proper … also if both $U,V$ are open in Spec$(R)$ then also we're done since Spec$(R)$ is irreducible so any two non-empty open sets have non-empty intersection … I'm not sure what happens in other cases though like if one of them is open and the other is closed …
Motivation: since in many ways, the punctured spectrum of a regular local ring behaves like $\mathbb P^n_k$ and for $\mathbb P^n_k$ it is indeed true that for subschemes $U,V$ with $\dim U +\dim V \ge n$ implies $U \cap V$ is non-empty, hence my motivation for asking this question …
Best Answer
In general, one asks for a result like $U,V$ must intersect if something is true about their dimensions when $U,V$ are both closed inside some irreducible space - for instance, without requiring that they're closed, if they're both contained in lower-dimensional closed subvarieties and these subvarieties have no common components, you can often just delete the intersection points from one of $U$ or $V$ without changing anything about the circumstances of the problem. (The case of $U,V$ both open is trivial: a space is irreducible iff any two open subsets intersect, and any regular noetherian local ring is a domain, hence has irreducible spectrum, and any open subset of an irreducible space is again irreducible, so the punctured spectrum is again irreducible.)
In the case that $U,V$ are both closed, you know that they must intersect inside the punctured spectrum - there's only one closed point and every closed subset contains it, so every pair of closed subsets must intersect. This lets us get at the problem for the punctured spectrum in much the same manner as we treat intersections in projective space via the affine cone. The same proof will apply here - I leave it to you to verify the details.