Theorem B on the first page of this paper states the following : If $R$ is a Noetherian ring and $M$ is a maximal ideal in $R[X_1,X_2,...,X_n]$, then $M \cap R$ is a prime ideal of $R$.
With this, as every prime ideal of a ring is not maximal, I am wondering if there is a counterexample to the following : If $R$ is a Noetherian ring and $M$ is a maximal ideal of $R[X]$, then $M \cap R$ is a maximal ideal of $R$. This claim has been speculated to be true by the instructor in my algebraic geometry course.
Thank you !
For a counterexample, take $R=\mathbb{Z}_{(p)}$ and $M=(pX-1)$. Then $R[X]/M\cong R[1/p]=\mathbb{Q}$ is a field so $M$ is maximal, but $M\cap R=0$ is not maximal in $R$.
A commutative ring $R$ with the property that every maximal ideal in $R[X]$ has maximal intersection with $R$ is called a Jacobson ring. This property has many equivalent formulations; for instance, it is equivalent to saying that every prime ideal in $R$ is an intersection of maximal ideals. A general version of the Nullstellensatz says that any finitely generated algebra over a Jacobson ring is Jacobson. It follows that if $R$ is Jacobson, then any maximal ideal in any finitely generated $R$-algebra pulls back to a maximal ideal in $R$. (In the case that $R=k[X]$ for an algebraically closed field $k$, this implies the classical Nullstellensatz over $k$.)