Semi-local ring

Remarks

If $\mathfrak m_1, \ldots, \mathfrak m_n$ are the maximal ideals, then the Chinese remainder theorem gives $$R / J(R) \cong R/\mathfrak m_1 \times \cdots \times R/\mathfrak m_n,$$ where $J(R) = \mathfrak m_1 \cap \cdots \cap \mathfrak m_n$ is the Jacobson radical. For a commutative ring $R$, being semi-local is equivalent to $R/J(R)$ being a finite product of fields. For noncommutative rings, the usual definition asks that $R/J(R)$ be semisimple.

Example: let $S = \Z \setminus (2\Z \cup 3\Z)$, the set of integers prime to $6$. It is a multiplicative set, and by prime ideals of a localization the prime ideals of $S^{-1}\Z$ correspond to the primes of $\Z$ disjoint from $S$. These are $(0)$, $2\Z$ and $3\Z$, so $S^{-1}\Z$ has exactly the two maximal ideals generated by $2$ and $3$. Every local ring is semi-local, and so is $\Z/n\Z$ for $n \ge 2$, with one maximal ideal for each prime divisor of $n$.

Non-example: $\Z$ is not semi-local, because $p\Z$ is a maximal ideal for each of the infinitely many primes $p$.