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  "body": "\n## Definition\n\nLet $R$ be an [[integral domain|0QVD7WBFGN]] with [[field of fractions|0F2PTGX1A2]] $K$.\nA *fractional ideal* of $R$ is a nonzero $R$-[[submodule|05WDSN597C]] $I$ of $K$ such that $dI \\subseteq R$ for some nonzero $d \\in R$.\nThe product $IJ$ of two fractional ideals is the set of finite sums $\\sum a_k b_k$ with $a_k \\in I$ and $b_k \\in J$.\nA fractional ideal $I$ is *invertible* if there is a fractional ideal $J$ with $IJ = R$.\nAn [[ideal|096Z9MP2ME]] of $R$ is invertible if it is nonzero and invertible as a fractional ideal.\n\n## Remarks\n\nIf $I$ is invertible, its inverse is unique and equals $I^{-1} = (R : I) = \\set{x \\in K : xI \\subseteq R}$.\nEquivalently, a fractional ideal is invertible exactly when it is a [[projective|BFFA03GE3V]] $R$-module.\nFor a Noetherian domain, a nonzero ideal is invertible exactly when its localization at every maximal ideal is principal.\n\nExample: let $R = \\Z[\\sqrt{-5}]$ and $P = (2, 1 + \\sqrt{-5})$.\nThen $P^2 = (4, 2 + 2\\sqrt{-5}, -4 + 2\\sqrt{-5})$ contains $6$ and $4$, hence $2$, and all its generators are multiples of $2$, so $P^2 = (2)$.\nThus $J = \\frac{1}{2} P$ is a fractional ideal with $PJ = \\frac{1}{2}(2) = R$, and $P$ is invertible.\n\nNon-example: let $R = k[x, y]$ for a field $k$ and $\\mathfrak{m} = (x, y)$.\nIf $f \\in K$ satisfies $fx \\in R$ and $fy \\in R$, write $f = a/x = b/y$ with $a, b \\in R$; then $ay = bx$, so $x$ divides $a$ and $f \\in R$.\nHence $(R : \\mathfrak{m}) = R$ and $\\mathfrak{m} (R : \\mathfrak{m}) = \\mathfrak{m} \\ne R$, so $\\mathfrak{m}$ is not invertible.\n",
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