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 $n>1$?
Contraction of maximal ideals in polynomial rings over PIDs
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ring-theory
commutative-algebra
principal-ideal-domains
maximal-and-prime-ideals
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4Dear 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, – 2011-08-01
1 Answers
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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.
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0Well, 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. – 2011-08-01