# Intersection of subschemes of the (punctured) spectrum of regular local rings

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 ...

• -1 for a lack of research effort on the first question: have you tried any examples? You should be able to find counterexamples very quickly. – KReiser Oct 9 at 8:50
• @KReiser: yes, only with regular schemes, one can find counterexamples, even in the affine case, since two proper ideals can add up to give the whole ring ... however, that can't happen in affine local case with closed subschemes , so I've suitably modified the question ... – user Oct 9 at 17:48
• Why not just take $R = k[x_1,\ldots,x_d]_{(x_1,\ldots,x_d)}$ whose punctured spectrum is the prime ideals other than the unique maximal ideal $(x_1,\ldots,x_d)$ ? With $d=2$ then $V(x_1)\cap V(x_2) = V(x_1,x_2)$. – reuns Oct 9 at 19:45
• @reuns: I think that doesn't work since $\dim V(x_i)=0$ in the punctured spectrum, so the dimension condition is not satisfied ... – user Oct 9 at 22:24
• Why $\dim V(x_i) =0$ ? – reuns Oct 9 at 23:09

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.
• For closed subschemes in the punctured spectrum case, I guess the only thing I'm thinking is why can't it happen that we have two ideals $I,J$ of $R$ such that $\mathfrak m$ is the only prime ideal above $I+J$ ? This will happen iff $V(I)\cap V(J)=\{\mathfrak m \}$ iff $\dim R/(I+J)=0$ ... I'm not sure whether this would contradict the dimension condition or not ... also this answer of yours doesn't address the question of what happens when one of $U,V$ is open and the other is closed ... – user Oct 10 at 1:41
• Now that I think about it more closely, for any ideal $I$ of a Noetherian ring $R$ of finite dimension , there is a prime ideal $P$ lying above $I$ such that $\dim R/I=\dim R/P$. So by Serre's theorem, in a regular local ring, for any ideals $I,J$ , we should always have $\dim R/I + \dim R/J \le \dim R$ ... now my dimension condition for the punctured spectrum states that $\dim (R/I) -1 + \dim (R/J) -1 \ge \dim R -1$, so $\dim (R/I) + \dim (R/J) \ge \dim R +1$ , which is never true ... Please let me know if I'm saying this right or not ... – user Oct 10 at 2:10
• Again, asking about the case when $U$ is open and $V$ is closed is a bit silly: just remove $V$ from $U$ and then they don't intersect. You want to ask about $U,V$ closed. – KReiser Oct 10 at 3:36
• Your second comment is very puzzling to me. Your claim is not correct: consider $R=k[x,y,z]$ with $I=(x)$ and $J=(y)$. Then $\dim R/I+\dim R/J = 2+2=4 \not\leq 3= \dim R$ and all of this is preserved upon going to the localization at the origin. I don't understand what you're up to here. – KReiser Oct 10 at 3:43