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  "body": "\n## Proof\n\n1. Let $p$ be a [[prime number|0SN0MAEKYX]] and let $R$ be a [[commutative ring|0GYGHKX834]] of [[characteristic|0P7CH1CND5]] $p$. So $p \\cdot 1 = 0$ in $R$. Let $x, y \\in R$.\n2. By the [[binomial theorem|0EHWPWGHB2]], $(x + y)^p = \\sum_{k=0}^{p} \\binom{p}{k} x^k y^{p-k}$.\n3. Fix an integer $k$ with $0 \u003c k \u003c p$. By the [[factorial formula|033KTKKRV6]], $\\binom{p}{k} \\cdot k! \\cdot (p-k)! = p!$.\n4. The prime $p$ divides $p!$. So $p$ divides the product $\\binom{p}{k} \\cdot k! \\cdot (p-k)!$.\n5. The number $k! \\, (p-k)!$ is a product of integers from $1$ to $p - 1$. None of these factors is divisible by $p$.\n6. So $p \\nmid k! \\, (p-k)!$, by [[Euclid's lemma|051VYKA1EX]] applied to one factor at a time.\n7. By Euclid's lemma and steps 4 and 6, $p$ divides the [[binomial coefficient|092GEF38YD]] $\\binom{p}{k}$. Write $\\binom{p}{k} = p t$ with an integer $t \\ge 1$.\n8. Let $z = x^k y^{p-k}$. Then $\\binom{p}{k} z = t \\cdot (p \\cdot z)$, a sum of $t$ copies of $p \\cdot z$.\n9. By the distributive law, $p \\cdot z = (p \\cdot 1) z = 0 \\cdot z = 0$. So $\\binom{p}{k} x^k y^{p-k} = 0$.\n10. Thus only the terms with $k = 0$ and $k = p$ remain in step 2. Their coefficients are $\\binom{p}{0} = \\binom{p}{p} = 1$.\n11. Hence $(x + y)^p = y^p + x^p = x^p + y^p$.\n\n## Idea\n\nExpand with the binomial theorem. Every middle coefficient $\\binom{p}{k}$ is a multiple of $p$, because $p$ appears in $p!$ but not in $k! \\, (p-k)!$. In characteristic $p$, multiples of $p$ act as zero, so only $x^p$ and $y^p$ survive.\n",
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    "name": "Proof of the freshman's dream",
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    "sources": [
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        "locator": "Prime characteristic",
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      "0GYGHKX834",
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  "markdown": "https://math2.tamnd.com/node/MHBEP4R11X.md",
  "msc": [
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  "name": "Proof of the freshman's dream",
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