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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?

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

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The answer is "yes" (and I believe the argument works with minor modifications if you do not assume that there are infinitely many prime ideals). By definition, either of your two subschemes has at least 2 points. This means that their closures can not be zero-dimensional (a closed subscheme of a Noetherian scheme is Noetherian; a zero-dimensional Noetherian scheme is Artinian; an integral Artinian scheme is the spectrum of a field).

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.

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