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  "body": "\n## Statement\n\nLet $R$ be a commutative [[Noetherian ring|0670T59HGE]], let $I$ be an [[ideal|096Z9MP2ME]] of $R$, and let $M$ be a [[finitely generated|05QY1VEBYD]] $R$-module.\nThen\n$$\\bigcap_{n \\ge 0} I^n M = \\set{x \\in M : (1 + a) x = 0 \\text{ for some } a \\in I}.$$\nIn particular:\n\n1. if $I$ is contained in the [[Jacobson radical|XA0DCNH7WY]] of $R$, then $\\bigcap_{n} I^n M = 0$; this applies when $R$ is a [[local ring|MZ02BKPHJA]] and $I \\neq R$;\n2. if $R$ is an [[integral domain|0QVD7WBFGN]] and $I \\neq R$, then $\\bigcap_{n} I^n = 0$.\n\n## Remarks\n\nLet $N = \\bigcap_n I^n M$.\nBy the [[Artin–Rees lemma|30H6B0G2FT]], $N = I^n M \\cap N = I^{n-k}(I^k M \\cap N) \\subseteq I N$ for $n \u003e k$, so $N = IN$.\nAs $N$ is finitely generated, the [[determinant trick|A8HRT9H3XA]] gives $a \\in I$ with $(1 + a) N = 0$.\nConversely, if $(1 + a) x = 0$ then $x = -a x = (-a)^n x \\in I^n M$ for every $n$.\nExample: in $R = \\Z/6\\Z$ with $I = (2)$, one has $I^2 = I$, so the intersection is $I = \\set{0, 2, 4}$, and indeed $3 = 1 + 2$ kills $2$ and $4$.\nThe Noetherian hypothesis is needed.\nIn the local ring of germs at $0$ of smooth functions $\\R \\to \\R$, consider the germ of $e^{-1/x^2}$, extended by $0$ at $0$.\nIt is nonzero, but it lies in every power of the maximal ideal.\n\n## Explanations\n\n### L1 human\n\nAn element divisible by $p$, by $p^2$, by $p^3$ and so on forever must be $0$ in the integers.\nThe theorem says that the same holds in any Noetherian local ring and in any Noetherian domain: an element that lies in every power of a proper ideal is $0$.\nIn other words, the $I$-adic topology separates points.\n\n### L2 human\n\nClause 1 shows that a Noetherian local ring embeds into its $\\mathfrak m$-adic [[completion|KBN13NPYJQ]].\nIt also shows that two elements of a Noetherian local ring are equal if they agree modulo every power of $\\mathfrak m$.\nThis is the basis of many arguments by successive approximation.\n",
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