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  "body": "\n## Statement\n\nLet $R$ be a [[principal ideal domain|0CMD0BCH8T]] and let $M$ be a [[finitely generated|05QY1VEBYD]] [[$R$-module|0NZ8JA25WK]].\n\n1. (Invariant factors) There are an integer $r \\ge 0$ and nonzero non-unit elements $d_1, \\ldots, d_k \\in R$ with $d_1 \\mid d_2 \\mid \\cdots \\mid d_k$ such that\n$$M \\cong R^r \\oplus R/(d_1) \\oplus \\cdots \\oplus R/(d_k).$$\nThe integer $r$ and the ideals $(d_1), \\ldots, (d_k)$ are uniquely determined by $M$.\n2. (Elementary divisors) There are an integer $r \\ge 0$, [[prime elements|01VH21EV6F]] $p_1, \\ldots, p_s \\in R$, not necessarily distinct, and integers $e_1, \\ldots, e_s \\ge 1$ such that\n$$M \\cong R^r \\oplus R/(p_1^{e_1}) \\oplus \\cdots \\oplus R/(p_s^{e_s}).$$\nThe integer $r$ and the ideals $(p_j^{e_j})$, counted with multiplicity, are uniquely determined by $M$ up to order.\n\n## Remarks\n\nSince $R$ is Noetherian, $M$ has a finite presentation $R^m \\xrightarrow{A} R^n \\to M \\to 0$.\nRow and column operations bring the matrix $A$ to Smith normal form, a diagonal matrix with entries $d_1 \\mid d_2 \\mid \\cdots$; this gives part 1.\nPart 2 follows from part 1 by the Chinese remainder theorem, which splits $R/(d)$ into $\\bigoplus_p R/(p^{e})$ over the prime powers $p^e$ exactly dividing $d$.\nThe part $R/(d_1) \\oplus \\cdots \\oplus R/(d_k)$ is the [[torsion submodule|6PBS6KW0NC]] of $M$, and $r$ is the rank of the [[free|0DZ9DKAKZR]] quotient $M$ modulo torsion.\n\nExample: over $\\Z$, the module $\\Z/2\\Z \\oplus \\Z/6\\Z$ has invariant factors $2 \\mid 6$ and elementary divisors $2, 2, 3$, since $\\Z/6\\Z \\cong \\Z/2\\Z \\oplus \\Z/3\\Z$.\nFor $R = \\Z$ the theorem is the [[structure theorem for finitely generated abelian groups|0RY0B72XAS]].\nFor $R = k[x]$ and $M = k^n$ with $x$ acting by a matrix, it gives the rational canonical form and the [[Jordan normal form|085HQ5G5TE]].\n\nThe principal ideal hypothesis is necessary.\nOver $R = \\Z[x]$, the ideal $I = (2, x)$ is a finitely generated torsion-free module.\nAny two elements of $I$ are linearly dependent, so a basis would have one element, and $I$ would be principal.\nIt is not principal, so $I$ is not free and not of the form above.\n\n## Explanations\n\n### L1 human\n\nOver a principal ideal domain, every finitely generated module breaks into simple pieces.\nThere is a free part, a copy of $R^r$, and a torsion part made of cyclic modules $R/(d)$.\nThe list of pieces is unique, so it is a complete invariant: two modules are isomorphic exactly when their lists agree.\nFor abelian groups, this says each finitely generated abelian group is $\\Z^r$ times a product of finite cyclic groups.\n\n### L2 human\n\nThe theorem unifies two classical results: the classification of finitely generated abelian groups, for $R = \\Z$, and the normal forms of matrices, for $R = k[x]$.\nIn the second case, a matrix $A$ makes $k^n$ a $k[x]$-module, the invariant factors are the invariant polynomials of $A$, and $d_k$ is the minimal polynomial.\nThe proof by Smith normal form is algorithmic when $R$ is Euclidean.\nOver a Dedekind domain that is not principal, the torsion part still decomposes, but the torsion-free part is a sum of ideals rather than a free module.\n",
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