# Subschemes of the affine line over PID

Let $$R$$ be a PID with infinitely many prime ideals. Suppose we have two integral locally closed subschemes of $$\mathrm{Spec}\,R[x]$$ such that

• both have non-empty intersection with the affine open $$\mathrm{Spec}\, R[x, \frac{1}{x}]$$; moreover, both intersections have the same set of points;
• both have non-empty intersection with the subscheme $$\mathrm{Spec}\,R$$ defined by $$x=0$$.

EDIT: Is the triple intersection of these two subschemes and $$\mathrm{Spec}\,R$$ necessarily non-empty?

• Let $p$ be a closed point of $\operatorname{Spec }R\subseteq\operatorname{Spec }R[x]$ and consider the schemes $\operatorname{Spec }R[x]$ and $\operatorname{Spec }R[x]\setminus\{p\}$. Apr 9, 2019 at 14:43
• FWIW, the user appears to have been part of a group of accounts controlled by the same individual Jun 24, 2019 at 3:12

Your hypotheses imply that $$R$$ has Krull dimension 1 and $$R[x]$$ has Krull dimension 2. So the dimensions of the closures of your subschemes can be either $$(2, 2)$$, $$(2, 1)$$, or $$(1, 1)$$.
Case $$(2, 2)$$. The closure of an irreducible set is irreducible; a top-dimensional closed irreducible subset of an irreducible space (which $$\mathrm{Spec}\, R[x]$$ is) has to be the whole space. So both of your subschemes are actually open subschemes. The intersection of either of them with $$\mathrm{Spec}\,R$$ is thus open in $$\mathrm{Spec}\,R$$ (and the intersection of non-empty opens in $$\mathrm{Spec}\,R$$ is non-empty because it contains the generic point).
Case $$(2, 1)$$. This case can not happen. As we remarked above, one of your subschemes has to be open in $$\mathrm{Spec}\,R[x]$$. Your hypotheses imply that this subscheme lies in the union of $$\mathrm{Spec}\,R$$ and the second subscheme. Recall that closure commutes with finite union; so the closure of the open subscheme (which is $$\mathrm{Spec}\,R[x]$$) would have to lie in the union of closures of $$\mathrm{Spec}\,R$$ and the second subscheme. So $$\mathrm{Spec}\,R[x]$$ would be covered by two irreducible closed subsets of Krull dimension 1, which contradicts irreducibility of $$\mathrm{Spec}\,R[x]$$.
Case $$(1, 1)$$. Here it is proved (Thm. 1.1) that a prime ideal in $$R[x]$$ is either generated by an irreducible polynomial or a prime element $$p\in R$$ and a polynomial irreducible $$\mathrm{mod}\,p$$ (here, the polynomial is allowed to be zero). An irreducible closed subscheme of dimension 1 corresponds to a prime ideal of height 1 which is either $$(p)$$ or $$(f(x))$$ for a non-zero polynomial $$f$$. Since your subschemes have to meet both $$x=0$$ and $$x\neq 0$$, it has to be the former. If the closures of your subschemes were defined by $$(p)$$ and $$(p')$$ for $$p\neq p'$$, their intersection would be empty; so your subschemes actually have the same closure; the closure intersects $$\mathrm{Spec}\,R$$ in a single point (by the classification of primes). Now note that two non-empty subsets of a single-point space have non-empty intersection.