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  "body": "\n## Statement\n\nLet $p$ be a prime number and let $R$ be a [[commutative ring|0GYGHKX834]] of [[characteristic|0P7CH1CND5]] $p$.\nThen $(x + y)^p = x^p + y^p$ for all $x, y \\in R$.\n\n## Remarks\n\nThe proof uses the binomial theorem. For $0 \u003c k \u003c p$, the prime $p$ divides the binomial coefficient $\\binom{p}{k}$, so these terms vanish in $R$.\nBy induction, $(x + y)^{p^n} = x^{p^n} + y^{p^n}$ for every $n \\ge 0$.\nThe identity gives the [[Frobenius endomorphism|0GHDDVAH8P]].\nIn $\\mathbb{R}$ the identity is false: $(1 + 1)^2 = 4$ and $1^2 + 1^2 = 2$.\n\n## Explanations\n\n### L1 human\n\nThe name refers to a common error: $(x + y)^2 = x^2 + y^2$ is false for real numbers.\nIn characteristic $2$, the middle term $2 x y$ is $0$, and the identity becomes true.\nThe same thing happens for every prime $p$ in characteristic $p$.\n",
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  "name": "Freshman's dream",
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