Exercise 17, § 1

a) Let A be an integral domain and $f \in A[T]$, $f \neq 0$. Now prove that there exists a maximal ideal m of $A[T]$ such that $f \notin m$ (take a maximal ideal containing $T.f - 1$). b) Let A be a ring. Now prove that $\dim(\mathrm{Specmax}(A[T])) \geq \dim(A) + 1$. (Let $p_0 \subset p_1 \cdots \subset p_n$ be a chain of prime ideals of A. Consider the chain

of prime ideals of $A[T]$ and put Now prove using a) that $(U_i)_{0 \leq i \leq n+1}$ is a strictly increasing sequence of irreducible closed subsets of $\mathrm{Specmax}(A[T])$. c) Let $k$ be a field, S the subset of $k[X, Y]$ formed by the elements of the form $1 + Xg(X, Y)$, and $A = S^{-1}k[X, Y]$. Now prove that $\dim(\mathrm{Specmax}(A)) = 1$ and $\dim(\mathrm{Spec}(A)) = 2$. Generalize to the case of several indeterminates. d) Now prove that for every pair of integers $0 \leq n < m$, there exists a noetherian ring A such that $\dim(\mathrm{Specmax}(A)) = n$ and $\dim(\mathrm{Specmax}(A[T])) = m$. e) Let A be the noetherian ring defined in exerc. 13. Now prove that $\dim(\mathrm{Specmax}(A)) = 1$ and $\dim(\mathrm{Specmax}(A[T])) = \infty$.

1 This example is due to M. Nagata, Local Rings, Interscience Publishers, 1960.