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  "body": "\n## Statement\n\nLet $\\F_q$ be a [[finite field|0ZXZ82FTX2]] with $q$ elements and let $n \\ge 1$.\nA *Kakeya set* in $\\F_q^n$ is a set $K \\subseteq \\F_q^n$ such that for every $x \\in \\F_q^n \\setminus \\set{0}$ there is $y \\in \\F_q^n$ with $\\set{y + t x : t \\in \\F_q} \\subseteq K$.\nThen every Kakeya set $K$ in $\\F_q^n$ satisfies\n$$\\abs{K} \\ge \\binom{q + n - 1}{n} \\ge \\frac{q^n}{n!}.$$\n\n## Remarks\n\nThis is Dvir's theorem, which answered the finite field analogue of the [[Kakeya problem|6PEN7AKC7M]] asked by Thomas Wolff.\nThe proof uses the polynomial method.\nIf $\\abs{K} \u003c \\binom{q + n - 1}{n}$, the number of monomials of degree at most $q - 1$ in $n$ variables, then some nonzero [[polynomial|07GNX8GCYP]] $P$ of degree $d \\le q - 1$ vanishes on $K$.\nLet $P_d$ be the top-degree homogeneous part of $P$.\nRestricting $P$ to a line $y + t x$ inside $K$ gives a polynomial in $t$ of degree at most $d \u003c q$ with $q$ roots, so it is zero.\nIts coefficient of $t^d$ is $P_d(x)$, so $P_d(x) = 0$.\nSo $P_d$ vanishes on all of $\\F_q^n$, which is impossible for a nonzero polynomial of degree less than $q$ in each variable.\n\nFor $n = 1$ the bound gives $\\abs{K} \\ge q$, and indeed $K$ must be the whole line $\\F_q$.\nFor $n = 2$ it gives $\\abs{K} \\ge q(q + 1)/2$.\n\n## Explanations\n\n### L1 human\n\nA Kakeya set is a set that contains a full line pointing in every direction.\nOver the real numbers such sets can have measure zero, which is surprising.\nOver a finite field the theorem says the opposite: such a set must contain a fixed positive fraction, about $1/n!$, of all the points of the space.\n\n### L2 human\n\nDvir's proof is a few lines long and made the polynomial method a standard tool in combinatorics.\nThe idea is that a set too small cannot avoid being the zero set of a low-degree polynomial, while lines in all directions force the polynomial to vanish identically.\nThe constant $1/n!$ has since been improved, but the order $q^n$ is optimal.\nThe finite field result does not solve the Euclidean Kakeya conjecture, whose difficulty comes from the multiscale structure of $\\R^n$.\n",
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