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  "body": "\n## Definition\n\nA [[commutative ring|0310REJJY8]] $R$ is *semi-local* if it has only finitely many [[maximal ideals|0PV74TR5FP]].\n\n## Remarks\n\nIf $\\mathfrak m_1, \\ldots, \\mathfrak m_n$ are the maximal ideals, then the Chinese remainder theorem gives\n$$R / J(R) \\cong R/\\mathfrak m_1 \\times \\cdots \\times R/\\mathfrak m_n,$$\nwhere $J(R) = \\mathfrak m_1 \\cap \\cdots \\cap \\mathfrak m_n$ is the [[Jacobson radical|XA0DCNH7WY]].\nFor a commutative ring $R$, being semi-local is equivalent to $R/J(R)$ being a finite product of fields.\nFor noncommutative rings, the usual definition asks that $R/J(R)$ be semisimple.\n\nExample: let $S = \\Z \\setminus (2\\Z \\cup 3\\Z)$, the set of integers prime to $6$.\nIt 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 $S$.\nThese are $(0)$, $2\\Z$ and $3\\Z$, so $S^{-1}\\Z$ has exactly the two maximal ideals generated by $2$ and $3$.\nEvery [[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 $n$.\n\nNon-example: $\\Z$ is not semi-local, because $p\\Z$ is a maximal ideal for each of the infinitely many primes $p$.\n",
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