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 $n>1$?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
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 $R$.
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.