---
tag: Q256RTTNTN
kind: definition
name: Semi-local ring
msc: [13H99]
defines: 6DYAGW68AV
uses: [0310REJJY8, 0PV74TR5FP, XA0DCNH7WY]
sources:
  - url: https://en.wikipedia.org/wiki/Semi-local_ring
---

## Definition

A [[commutative ring|0310REJJY8]] $R$ is *semi-local* if it has only finitely many [[maximal ideals|0PV74TR5FP]].

## 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|XA0DCNH7WY]].
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|W8QNZZ4QJS]] 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|MZ02BKPHJA]] 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 $p$.