Let $R$ be a principal ideal domain which is not a field, and let $M$ be a maximal ideal of the polynomial ring $R[X_1,\dots,X_n]$. If $n=1$ it is very easy to see that $M \cap R \neq 0$. Is this also true for $n>1$?

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    $\begingroup$ Dear Jose, Your "very easy to see" claim is wrong. E.g. if $R = \mathbb Z_p$ and $M$ is the ideal $(p x - 1)$ in $\mathbb Z_p[x]$, then $M$ is maximal, but has trivial intersection with $\mathbb Z_p$. Regards, $\endgroup$ – Matt E Aug 1 '11 at 13:18

The premise of the question is incorrect (and the "very easy to see" claim for $n =1$ is false).

Suppose e.g. that $R$ is a DVR, with uniformizer $\pi$. (E.g. $R = \mathbb Z_p$ and $\pi = p$.) If we consider the principal ideal $(\pi x - 1)$ in $R[x]$, then the quotient of $R[x]$ by this ideal is isomorphic to $R[1/\pi]$, the fraction field of $R$. Thus this principal ideal is maximal, but has trivial intersection with $R$.

These two posts are relevant.

If the PID $R$ has infinitely many distinct prime ideals, then the claim of the question is true for any $n$. The proof uses the fact that such a ring $R$ is Jacobson, together with the general form of the Nullstellensatz for Jacobson rings.

  • $\begingroup$ This was an interesting problem. It seems a general fact is that an morphism of Jacobson rings of finite type, $\phi: R \to R'$ is such that the map $\mathrm{Spec} R' \to \mathrm{Spec} R $ sends closed points to closed points (which I assume is what you had in mind). $\endgroup$ – Akhil Mathew Aug 1 '11 at 14:54
  • $\begingroup$ (of which a quick argument is as follows: let $\mathfrak{m}' \subset R'$ be a maximal ideal, and quotient by it. So wlog, $R'$ is a field. By quotienting by the kernel of $\phi$, we can assume $\phi$ is injective. We thus want to prove that if $R$ is a Jacobson ring and $R'$ a field which is a f.g. $R$-algebra, then $R$ is a field. But $R'$ must be a finite extension of the quotient field of $R$ (as a field extension which is f.g. as a map of algebras is finite). In particular the field of fractions of $R$ is itself a finitely generated $R$-module, which means (unless $R$ is a field) $\endgroup$ – Akhil Mathew Aug 1 '11 at 15:08
  • $\begingroup$ that said field of fractions is $R_f$ for some $f$. This means that $R$ can't be Jacobson (topologically, Jacobson is that the maximal ideals are dense, but if $R_f$ is a field for some $f$, then the generic point is an isolated point).) I "cheated" by checking that the relevant statement was in EGA, but perhaps this will be helpful to the OP! $\endgroup$ – Akhil Mathew Aug 1 '11 at 15:10
  • $\begingroup$ Well, the implication that $K(R)$ (field of fractions) was a f.g. $R$-algebra because $R'$ was is perhaps tenuous if $R$ is not noetherian. $\endgroup$ – Akhil Mathew Aug 1 '11 at 15:12

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