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  "body": "\n## Statement\n\nLet $p$ be a [[prime number|0SN0MAEKYX]].\nThen the ring $\\mathbb{Z}/p\\mathbb{Z}$ of the [[integers modulo p|01N3AYAA6N]] is a field.\nThat is, each nonzero element of $\\mathbb{Z}/p\\mathbb{Z}$ has a multiplicative inverse.\n\n## Remarks\n\nThe proof: if $\\overline{a} \\neq \\overline{0}$, then $p \\nmid a$, so $\\gcd(a, p) = 1$. [[Bézout's identity|0DQVHMYAGA]] gives $a x + p y = 1$, and then $\\overline{a} \\cdot \\overline{x} = \\overline{1}$.\nThe converse holds: if $n \u003e 1$ is composite, with $n = a b$ and $1 \u003c a, b \u003c n$, then $\\overline{a} \\cdot \\overline{b} = \\overline{0}$, so $\\mathbb{Z}/n\\mathbb{Z}$ is not a field.\nThe field $\\mathbb{Z}/p\\mathbb{Z}$ is often written $\\mathbb{F}_p$ or $\\mathrm{GF}(p)$. It is the finite field with $p$ elements.\n\n## Explanations\n\n### L1 human\n\nModulo $7$, you can divide by each nonzero number. For example, $3 \\cdot 5 = 15 \\equiv 1 \\pmod 7$, so $5$ plays the part of $1/3$.\nModulo $6$ this fails: no number $x$ gives $2 x \\equiv 1 \\pmod 6$, because $2 x$ is always even.\n",
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