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  "body": "\n## Proof\n\nLet $q = \\abs{\\mathbb{F}}$. We use the convention $x^0 = 1$ for all $x \\in \\mathbb{F}$, including $x = 0$.\n\n1. By [[the order of a finite field|0H5WWC8RZX]], $q = p^m$ for some $m \\ge 1$. So $q \\cdot 1 = 0$ in $\\mathbb{F}$, by the definition of the [[characteristic|0P7CH1CND5]] $p$.\n2. Lemma: $\\sum_{x \\in \\mathbb{F}} x^k = 0$ for each integer $k$ with $0 \\le k \u003c q - 1$.\n3. For $k = 0$, the sum is $q \\cdot 1 = 0$ by step 1.\n4. Let $1 \\le k \\le q - 2$. The polynomial $X^k - 1$ is nonzero of degree $k$, so it has at most $k \u003c q - 1$ roots in $\\mathbb{F}$ by [[the root bound|0KJVQV3E0F]].\n5. The set $\\mathbb{F} \\setminus \\set{0}$ has $q - 1$ elements. So there is $c \\ne 0$ in $\\mathbb{F}$ with $c^k \\ne 1$.\n6. Let $S = \\sum_{x \\in \\mathbb{F}} x^k$. Since $x \\mapsto cx$ is a bijection of $\\mathbb{F}$, $S = \\sum_{x} (cx)^k = c^k S$.\n7. So $(c^k - 1) S = 0$ with $c^k - 1 \\ne 0$, and $S = 0$. This proves the lemma.\n8. Let $g = X_1^{e_1} \\cdots X_n^{e_n}$ be a monomial of total degree $e_1 + \\dots + e_n \u003c n(q - 1)$. Then $e_j \u003c q - 1$ for some $j$.\n9. The sum of $g$ over $\\mathbb{F}^n$ factors: $\\sum_{a \\in \\mathbb{F}^n} g(a) = \\prod_{j=1}^{n} \\sum_{x \\in \\mathbb{F}} x^{e_j}$. By the lemma the factor for the $j$ of step 8 is $0$, so the sum is $0$.\n10. A polynomial $P$ of total degree less than $n(q-1)$ is a linear combination of such monomials. By step 9 and linearity, $\\sum_{a \\in \\mathbb{F}^n} P(a) = 0$.\n11. Now put $P = \\prod_{i=1}^{r} \\left(1 - f_i^{\\,q-1}\\right)$. Its total degree is at most $(q - 1)(d_1 + \\dots + d_r) \u003c (q - 1) n$, by the hypothesis $d_1 + \\dots + d_r \u003c n$.\n12. By [[the identity $x^q = x$ in a finite field|0QWR8WTZEW]], $y^{q-1} = 1$ for each $y \\ne 0$ in $\\mathbb{F}$. Also $0^{q-1} = 0$, since $q - 1 \\ge 1$.\n13. So for $a \\in \\mathbb{F}^n$, the factor $1 - f_i(a)^{q-1}$ is $1$ if $f_i(a) = 0$ and $0$ otherwise.\n14. Hence $P(a) = 1$ if $a$ is a common zero of $f_1, \\dots, f_r$, and $P(a) = 0$ otherwise.\n15. Let $Z$ be the number of common zeros. By step 14, $\\sum_{a \\in \\mathbb{F}^n} P(a) = Z \\cdot 1$ in $\\mathbb{F}$.\n16. By steps 10 and 11, this sum is $0$. So $Z \\cdot 1 = 0$ in $\\mathbb{F}$.\n17. Write $Z = sp + t$ with $0 \\le t \u003c p$. Then $t \\cdot 1 = Z \\cdot 1 - s (p \\cdot 1) = 0$. Since $p$ is the least positive integer $m$ with $m \\cdot 1 = 0$, we get $t = 0$.\n18. So $p$ divides $Z$, the number of common zeros.\n\n## Idea\n\nThe polynomial $\\prod_i (1 - f_i^{q-1})$ is the indicator function of the common zero set, because $y^{q-1}$ is $1$ for $y \\ne 0$ and $0$ for $y = 0$.\nSo summing it over $\\mathbb{F}^n$ counts the zeros, as an element of $\\mathbb{F}$.\nBut a polynomial of low degree sums to zero over $\\mathbb{F}^n$, because some variable appears to a power below $q - 1$, and power sums of such exponents vanish.\n",
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