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  "body": "\n## Definition\n\nLet $(R, \\mathfrak m)$ be a [[local ring|MZ02BKPHJA]] with [[residue field|NJ3APQ8BHH]] $k = R/\\mathfrak m$.\nThe quotient $\\mathfrak m / \\mathfrak m^2$ is an $R$-module killed by $\\mathfrak m$, hence a [[vector space|000AYPA9AJ]] over $k$ with $(a + \\mathfrak m)(x + \\mathfrak m^2) = ax + \\mathfrak m^2$.\nThe *cotangent space* of $R$ is this $k$-vector space $\\mathfrak m / \\mathfrak m^2$.\nIts dual $\\Hom_k(\\mathfrak m / \\mathfrak m^2, k)$ is the *Zariski tangent space* of $R$.\n\n## Remarks\n\nThe scalar multiplication is well defined because $\\mathfrak m \\cdot \\mathfrak m \\subseteq \\mathfrak m^2$.\n\nExample: let $k$ be a field and $R = k[x, y]_{(x, y)}$, the [[localization|X6JD4AG973]] of the polynomial ring at $(x, y)$.\nThen $\\mathfrak m = (x, y)R$ and $\\mathfrak m / \\mathfrak m^2$ has the basis given by the classes of $x$ and $y$, so it has dimension $2$.\nFor the cusp $R = k[x, y]_{(x, y)} / (y^2 - x^3)$, the classes of $x$ and $y$ still form a basis, because $y^2 - x^3 \\in \\mathfrak m^2$; the dimension is $2$ although the cusp is a curve.\n\nNon-example: the quotient $\\mathfrak m / \\mathfrak m^3$ is in general not a vector space over $k$.\nIn $k[\\![t]\\!]$ with $\\mathfrak m = (t)$, the element $t$ acts on $\\mathfrak m / \\mathfrak m^3$ by sending the class of $t$ to the nonzero class of $t^2$, so $\\mathfrak m$ does not kill it.\n",
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