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  "body": "\n## Proof\n\n1. Let $R$ be an [[integral domain|09S3PFBEPM]] with finitely many elements. Then $R$ is commutative and $1 \\neq 0$ in $R$.\n2. By the definition of a [[field|0QA8CHH8MG]], it remains to show that each $a \\in R$ with $a \\neq 0$ has some $b \\in R$ with $a b = 1$.\n3. Fix $a \\in R$ with $a \\neq 0$. Define $m \\colon R \\to R$ by $m(x) = a x$.\n4. The map $m$ is [[injective|0PVG1CRAJN]]. Indeed, if $a x = a y$, then $x = y$ by [[cancellation in an integral domain|0GJM24EXP9]], since $a \\neq 0$.\n5. We claim that $m$ is [[surjective|0XKH881BZ7]]. Suppose not, and let $B = m(R)$ be its image. Then $B$ is a proper subset of the finite set $R$, so $\\abs{B} \u003c \\abs{R}$.\n6. Consider $m$ as a function from $R$ to $B$. Since $\\abs{R} \u003e \\abs{B}$, the [[pigeonhole principle|0HGVCWQGY3]] gives $x \\neq y$ in $R$ with $m(x) = m(y)$.\n7. This contradicts step 4. So $m$ is surjective.\n8. In particular, $1 = m(b) = a b$ for some $b \\in R$.\n9. So every nonzero element of $R$ has an inverse, and $R$ is a field.\n\n## Idea\n\nMultiplication by a fixed nonzero $a$ is injective, because one can cancel $a$. On a finite set, an injective map to itself is also surjective. So $1$ is a multiple $a b$ of $a$, and $b$ is the inverse of $a$.\n",
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