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?

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    $\begingroup$ What if your ring is a product of a Jacobson ring, such as $\mathbb{C}[x,y]$ and a non-Jacobson ring, such as $\mathbb{C}[t]_{\langle t \rangle}$? Then what do you get for both sides of your inequality? $\endgroup$ – Jason Starr Dec 12 '15 at 18:22
  • $\begingroup$ $R $ is not any special ring in general, such as a product of Jacobson rings. $\endgroup$ – Anderia silva Dec 12 '15 at 18:38
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    $\begingroup$ I think that you did not understand the point of my comment. For a product of two rings, the prime spectrum (with the Zariski topology) is the disjoint union of the prime spectra of the factors. Thus, even if one factor ring fails to be Jacobson, if the factor ring with larger dimension is Jacobson, forming the quotient by the Jacobson radical will not change the dimension. $\endgroup$ – Jason Starr Dec 12 '15 at 19:29

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.


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