Dimension of a commutative ring

Question

Let $R$ be a commutative ring with identity having finite Krull dimension (denoted $\dim$). Let $Nil(R)$ be the set of all nilpotent elements in $R$, and let $J(R)$ the intersection of all maximal ideal of $R$.

If $Nil(R)\not= J (R)$, can we deduce than $\dim (R/J (R))<\dim (R)$? Or if the inequality is not true, under which conditions it can be true?

Answer 1

Picking up a simplified version of Jason Starr's idea:

$R = \mathbb{Z}_p \times \mathbb{Z}$ has dimension $1\,\,( \mathbb{Z}_p$ denotes the p-adic integes) and Jacobson radical $(p) \times 0 \not= 0 = Nil(R)$. Hence $R/J(R)=\mathbb{F}_p \times \mathbb{Z}$ has dimension $1$ as well. $$$$

Background: The prime ideals of the product $R=R_1 \times R_2$ of commutative rings are $p_1 \times R_2$ and $R_1 \times p_2$. These are maximal iff $p_i$ are maximal in $R_i$. This shows:

• $\dim R = \max(\dim R_1, \dim R_2)$
• $J(R) = J(R_1) \times J(R_2)$
• $\dim R/J(R) = \max( \dim R_1/J(R_1), \dim R_2/J(R_2))$

In particular any domains $R_1, R_2$ with $J(R_1) \neq 0 = J(R_2)$ and $\dim R_2 \ge \dim R_1$ will give a counterexample.