Jacobson radical as intersection of maximal ideals

An element $r\in R$ lies in $J(R)$ if and only if its image in every residue field $R/\mathfrak m$ is zero; equivalently, $r\in\mathfrak m$ 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 the unique maximal ideal because $R/(\bar\varepsilon)\cong k$. Therefore $J(R)=(\bar\varepsilon)$.

If $R$ is a local ring with maximal ideal $\mathfrak m$, then

Thus, in a local ring, the Jacobson radical is precisely the set of nonunits. Compare the module-theoretic characterization that the Jacobson radical annihilates simple modules.