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  "body": "\n## Proof\n\n1. Let $p$ be a [[prime number|0SN0MAEKYX]]. For an integer $a$, write $\\overline{a}$ for its [[residue class|0QRV8FPK5M]] modulo $p$.\n2. By the definition of a residue class, $\\overline{a} = \\overline{b}$ if and only if $a \\equiv b \\pmod p$, that is, $p$ [[divides|096PW7Q014]] $a - b$.\n3. The ring $\\mathbb{Z}/p\\mathbb{Z}$ of [[integers modulo p|01N3AYAA6N]] is commutative: $\\overline{a} \\cdot \\overline{b} = \\overline{a b} = \\overline{b a} = \\overline{b} \\cdot \\overline{a}$.\n4. We have $\\overline{1} \\ne \\overline{0}$. Indeed, $p \u003e 1$, so $p$ does not divide $1 - 0 = 1$, and step 2 applies.\n5. Let $\\overline{a}$ be a nonzero element of $\\mathbb{Z}/p\\mathbb{Z}$. By step 2, $p$ does not divide $a$.\n6. Let $d = \\gcd(a, p)$ be the [[greatest common divisor|02WF878BH1]]. It is a positive divisor of $p$, so $d = 1$ or $d = p$, since $p$ is prime.\n7. The integer $d$ divides $a$. Since $p$ does not divide $a$ by step 5, $d \\ne p$. So $d = 1$.\n8. By [[Bézout's identity|0DQVHMYAGA]], there are integers $x$ and $y$ with $a x + p y = 1$.\n9. Then $a x - 1 = - p y$, so $p$ divides $a x - 1$. By step 2, $\\overline{a x} = \\overline{1}$.\n10. Hence $\\overline{a} \\cdot \\overline{x} = \\overline{a x} = \\overline{1}$, and $\\overline{x}$ is a multiplicative inverse of $\\overline{a}$.\n11. By steps 3, 4 and 10, $\\mathbb{Z}/p\\mathbb{Z}$ is a commutative ring with $1 \\ne 0$ in which each nonzero element has an inverse. So it is a [[field|0QA8CHH8MG]].\n\n## Idea\n\nA class $\\overline{a}$ is nonzero exactly when $p \\nmid a$. For a prime $p$, this forces $\\gcd(a, p) = 1$, and Bézout's identity then writes $1$ as $a x$ plus a multiple of $p$, which is an inverse of $\\overline{a}$.\n",
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