Jacobson radical as intersection of maximal ideals

Equivalently, an element $r\in R$ lies in $J(R)$ if and only if its image in every residue field $R/\mathfrak m$ is zero (i.e. $r\in\mathfrak m$ for every maximal ideal $\mathfrak m$).

1. The integers. In $R=\mathbb Z$, maximal ideals are precisely $(p)$ for primes $p$. Their intersection is $(0)$, so $J(\mathbb Z)=0$.

1. A local example from localization. Let $R=\mathbb Z_{(p)}$ be the localization of $\mathbb Z$ at the prime $(p)$. This is a local ring whose unique maximal ideal is $p\mathbb Z_{(p)}$, hence
   $$J(\mathbb Z_{(p)}) = p\mathbb Z_{(p)}.$$

1. Dual numbers. Let $R=k[\varepsilon]/(\varepsilon^2)$ for a field $k$. The ideal $(\bar\varepsilon)$ is maximal (indeed $R/(\bar\varepsilon)\cong k$ is a field), and it is the unique maximal ideal. Therefore $J(R)=(\bar\varepsilon)$. In particular, $\bar\varepsilon$ lies in every maximal ideal, so it lies in the intersection.

A particularly important consequence is: if $R$ is a local ring with maximal ideal $\mathfrak m$ (see the characterization of local rings by a unique maximal ideal), then

This description is compatible with the module-theoretic viewpoint: compare Jacobson radical annihilates simples.

These examples illustrate the philosophy: $J(R)$ consists of the elements that vanish in every residue field (and, in a local ring, precisely the nonunits).