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  "body": "\n## Statement\n\nLet $p$ be a [[prime number|01826KMG2H]], and let $A$ and $B$ be nonempty subsets of the [[integers modulo $p$|03VFZ2WETT]], $\\Z/p\\Z$.\nThen the [[sumset|XNBN31HQWB]] $A + B$ satisfies\n$$\\abs{A + B} \\ge \\min(p, \\ \\abs{A} + \\abs{B} - 1).$$\n\n## Remarks\n\nOne proof is by induction on $\\abs{B}$.\nIt replaces $(A, B)$ by $(A \\cup (B + e), \\ B \\cap (A - e))$ for a suitable $e$, which keeps $\\abs{A} + \\abs{B}$ and does not enlarge the sumset.\nAnother proof uses the [[combinatorial Nullstellensatz|9ZW7BCJMDT]] with a polynomial that vanishes on a too small set $C \\supseteq A + B$.\nThe bound is sharp: for $A = \\set{0, 1, \\dots, k-1}$ and $B = \\set{0, 1, \\dots, l-1}$ with $k + l - 1 \\le p$, the sumset is $\\set{0, \\dots, k + l - 2}$.\nThe hypothesis that $p$ is prime is needed.\nIn $\\Z/4\\Z$, take $A = B = \\set{0, 2}$; then $A + B = \\set{0, 2}$ has $2$ elements, while $\\min(4, 3) = 3$.\n\n## Explanations\n\n### L1 human\n\nWork with numbers modulo a prime $p$, on a clock with $p$ hours.\nAdd every element of a set $A$ to every element of a set $B$.\nYou get at least $\\abs{A} + \\abs{B} - 1$ different results, unless you already get all $p$ of them.\nSo sums modulo a prime cannot bunch up too much.\n\n### L2 human\n\nThe Cauchy–Davenport theorem is the analogue for $\\Z/p\\Z$ of the easy bound $\\abs{A + B} \\ge \\abs{A} + \\abs{B} - 1$ for sets of integers.\nOver composite moduli, subgroups give smaller sumsets, and Kneser's theorem describes this.\nThe Dias da Silva–Hamidoune theorem bounds the set of sums of two distinct elements of $A$ by $\\min(p, 2\\abs{A} - 3)$, as conjectured by Erdős and Heilbronn.\n",
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