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  "body": "\n## Statement\n\nLet $(R, \\mathfrak m)$ be a [[regular local ring|63XXTF6B5M]].\nThen $R$ is an [[integral domain|0QVD7WBFGN]].\n\n## Remarks\n\nProof idea: let $d = \\dim R$ and $k = R/\\mathfrak m$.\nA regular system of parameters $x_1, \\ldots, x_d$ gives a surjection $k[X_1, \\ldots, X_d] \\to \\operatorname{gr}_{\\mathfrak m} R = \\bigoplus_n \\mathfrak m^n / \\mathfrak m^{n+1}$ onto the [[associated graded ring|8HFAEGFQT1]].\nComparing dimensions shows that this surjection is an isomorphism, so $\\operatorname{gr}_{\\mathfrak m} R$ is a [[polynomial ring|0SDC934R52]], hence a domain.\nBy the [[Krull intersection theorem|1Q1AH2V41S]], $\\bigcap_n \\mathfrak m^n = 0$, so each nonzero $a \\in R$ has an order $n$ with $a \\in \\mathfrak m^n \\setminus \\mathfrak m^{n+1}$.\nThe leading forms of two nonzero elements multiply to a nonzero form, so their product is nonzero.\n\nExample: $k[\\![x, y]\\!]$ and $\\Z_{(p)}[x]_{(p, x)}$ are regular local rings, hence domains.\n\nThe hypothesis $\\dim R = \\operatorname{edim} R$ cannot be dropped.\nThe Noetherian local ring $R = k[x, y]_{(x, y)}/(xy)$ has dimension $1$ and [[embedding dimension|DP3KSCRKFP]] $2$, and it is not a domain, since $x \\cdot y = 0$ with $x, y \\ne 0$.\n\n## Explanations\n\n### L1 human\n\nA regular local ring describes a smooth point.\nNear a smooth point a space cannot be made of two pieces that cross, so the ring has no zero divisors.\nThe example $xy = 0$, two lines crossing at the origin, is not smooth at the crossing point, and there $x \\cdot y = 0$.\n\n### L2 human\n\nThe statement is the first step in the theory of regular local rings.\nThe stronger theorem that [[regular local rings are UFDs|1HD3EK5392]] uses it together with homological tools such as [[Serre's theorem on regular local rings|93RQRG9R1P]].\nIt also implies that a regular local ring of dimension $1$ is a [[discrete valuation ring|MGM3CNJW9V]].\n",
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