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  "body": "\n## Statement\n\nLet $n \\ge 2$, let $D \\subseteq \\C^n$ be an open set, and let $f$ be a [[holomorphic function|507C34W0ZT]] on $D$.\nIf $p \\in D$ and $f(p) = 0$, then every neighbourhood of $p$ contains a point $q \\ne p$ with $f(q) = 0$.\n\n## Remarks\n\nSuppose instead that $p$ is the only zero of $f$ in a small open ball $B$ around $p$.\nThen $1/f$ is holomorphic on $B \\setminus \\set{p}$, which is connected.\nBy [[Hartogs's extension theorem|FSGES670AZ]] with $K = \\set{p}$, it extends to a holomorphic function $g$ on $B$.\nThen $g f = 1$ on $B \\setminus \\set{p}$, hence at $p$ by continuity, which contradicts $f(p) = 0$.\n\nExample: $f(z_1, z_2) = z_1^2 + z_2^2$ vanishes at the origin, and its zeros near the origin form the two complex lines $z_1 = i z_2$ and $z_1 = -i z_2$.\n\nThe hypothesis $n \\ge 2$ is necessary: $f(z) = z$ on $\\C$ has the isolated zero $0$.\n\n## Explanations\n\n### L1 human\n\nIn one complex variable, a nonzero holomorphic function has isolated zeros, like the roots of a polynomial.\nIn two or more variables this is never the case: a zero always lies on a whole curve, surface or larger set of zeros.\nSo one cannot hope to describe zero sets by listing points.\n\n### L2 human\n\nThe zero set of a holomorphic function that is not identically zero is an analytic hypersurface, of complex dimension $n - 1$ at each of its points.\nThis proposition is the first step toward that fact.\nDivisors in several variables are made of hypersurfaces instead.\n",
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