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  "body": "\n## Proof\n\n1. Let $n \\ge 2$ be an integer. Write $\\overline{a}$ for the class of an integer $a$ in the ring $\\Z/n\\Z$ of [[integers modulo n|01N3AYAA6N]].\n2. The ring $\\Z/n\\Z$ is a [[commutative ring|0GYGHKX834]], because the operations on classes come from the commutative operations on $\\Z$.\n3. We have $\\overline{a} = \\overline{0}$ if and only if $n \\mid a$. In particular, $\\overline{1} \\neq \\overline{0}$, since $n \\ge 2$ does not divide $1$.\n4. First suppose that $n$ is not a [[prime number|0SN0MAEKYX]]. Since $n \\ge 2$, $n$ is [[composite|0N3ADXHNTQ]]. So $n$ has a positive divisor $a$ with $a \\neq 1$ and $a \\neq n$. Put $b = n / a$. Then $1 \u003c a \u003c n$, and so $1 \u003c b \u003c n$.\n5. By step 3, $\\overline{a} \\neq \\overline{0}$ and $\\overline{b} \\neq \\overline{0}$, since $0 \u003c a, b \u003c n$. But $\\overline{a} \\cdot \\overline{b} = \\overline{n} = \\overline{0}$.\n6. Suppose that $\\overline{a}$ had an inverse $\\overline{c}$, so that $\\overline{c} \\cdot \\overline{a} = \\overline{1}$. Then $\\overline{b} = \\overline{c} \\cdot \\overline{a} \\cdot \\overline{b} = \\overline{c} \\cdot \\overline{0} = \\overline{0}$.\n7. This contradicts step 5. So $\\overline{a}$ is a nonzero element without inverse, and $\\Z/n\\Z$ is not a [[field|0QA8CHH8MG]].\n8. Now suppose that $n = p$ is a prime number. By steps 2 and 3, it remains to show that every nonzero class has an inverse.\n9. Let $\\overline{a} \\neq \\overline{0}$, so $p \\nmid a$ by step 3. Let $d = \\gcd(a, p)$ be the [[greatest common divisor|02WF878BH1]].\n10. Since $d$ is a positive divisor of the prime $p$, $d = 1$ or $d = p$. Since $d \\mid a$ and $p \\nmid a$, $d \\neq p$. So $d = 1$.\n11. By [[Bézout's identity|0DQVHMYAGA]], there are integers $x, y$ with $a x + p y = 1$.\n12. Passing to classes, $\\overline{a} \\cdot \\overline{x} + \\overline{p} \\cdot \\overline{y} = \\overline{1}$. Since $\\overline{p} = \\overline{0}$, this gives $\\overline{a} \\cdot \\overline{x} = \\overline{1}$.\n13. So $\\overline{x}$ is an inverse of $\\overline{a}$, and $\\Z/p\\Z$ is a field.\n14. By steps 7 and 13, $\\Z/n\\Z$ is a field if and only if $n$ is prime.\n\n## Idea\n\nA factorization $n = a b$ gives two nonzero classes with product zero, and such a class cannot be invertible. For a prime $p$, any $a$ not divisible by $p$ is coprime to $p$. Bézout's identity then writes $1$ as $a x$ plus a multiple of $p$, so $x$ inverts $a$ modulo $p$.\n",
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