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  "body": "\n## Definition\n\nLet $R$ be a Noetherian [[integral domain|0QVD7WBFGN]] with [[field of fractions|0F2PTGX1A2]] $K$, and let $A$ be a finite-dimensional [[associative algebra|3CTW1P4HBP]] over $K$.\nAn *$R$-order* in $A$ is a [[subring|0GNCRPYXYV]] $\\Lambda$ of $A$, with the same identity, such that:\n\n1. $\\Lambda$ is a [[finitely generated|05QY1VEBYD]] $R$-module.\n2. $\\Lambda$ contains a $K$-basis of $A$, that is $K \\Lambda = A$.\n\n## Remarks\n\nExample: $\\Lambda = \\Z[i]$ in $A = \\Q(i)$, with $R = \\Z$.\nIt is a subring containing $1$, it is generated as a $\\Z$-module by $1$ and $i$, and $1, i$ is a $\\Q$-basis of $\\Q(i)$.\nOther examples are $M_n(\\Z)$ in $M_n(\\Q)$ and the group ring $\\Z G$ of a finite group $G$ in $\\Q G$.\nNon-example: $\\Z[\\tfrac{1}{2}]$ is a subring of $\\Q$ that spans $\\Q$, but it is not a finitely generated $\\Z$-module, so clause 1 fails.\nNon-example: $\\Z$ inside $\\Q(i)$ is finitely generated but fails clause 2, since $\\Q \\cdot \\Z = \\Q \\ne \\Q(i)$.\nWhen $R$ is a principal ideal domain, clause 1 makes $\\Lambda$ a free $R$-module, and clause 2 then says its rank is $\\dim_K A$.\nEvery element of an order is integral over $R$.\n",
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