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  "body": "\n## Proof\n\n1. Let $F$ be a [[finite field|08RTDHX1SK]], with zero $0$ and identity $1$. For an integer $k$, write $k \\cdot 1$ for the $k$-fold sum of $1$, and its negative when $k \u003c 0$.\n2. The map $\\chi \\colon \\Z \\to F$, $k \\mapsto k \\cdot 1$, is a [[ring homomorphism|0GGE0Z0JQ1]].\n3. The set $\\Z$ is infinite and $F$ is finite, so $\\chi$ is not injective. So there are integers $j \u003c k$ with $j \\cdot 1 = k \\cdot 1$, and then $(k - j) \\cdot 1 = 0$ with $k - j \u003e 0$.\n4. So the [[characteristic|0P7CH1CND5]] of $F$ is a positive integer $p$. A field is an integral domain, so [[$p$ is a prime number|00758QM6JJ]].\n5. The kernel of $\\chi$ is the ideal of the integers $k$ with $k \\cdot 1 = 0$. It contains $p$. If $k = sp + r$ with $0 \\le r \u003c p$ is in it, then $r \\cdot 1 = 0$, so $r = 0$ by the minimality of $p$. So the kernel is $p\\Z$.\n6. By the [[first isomorphism theorem for rings|0ZMYX7GMH0]], the image $P = \\chi(\\Z)$ is a subring of $F$ isomorphic to $\\Z/p\\Z = \\F_p$. So $P$ is a field with $p$ elements.\n7. Every subfield of $F$ contains $1$, hence all the elements $k \\cdot 1$, hence $P$. So $P$ is the [[prime subfield|0VPFBAY7MW]] of $F$, and we identify it with $\\F_p$.\n8. The field $F$ is a vector space over $P$. It is spanned by the finite set $F$, so it has a finite [[basis|0VTQTPQGAV]] $b_1, \\dots, b_n$. Here $n = [F : \\F_p]$ is the [[degree|0GS74FGFEK]].\n9. Each element of $F$ is $c_1 b_1 + \\dots + c_n b_n$ for exactly one tuple $(c_1, \\dots, c_n) \\in P^n$, because $b_1, \\dots, b_n$ is a basis.\n10. There are $p^n$ such tuples. So $\\abs{F} = p^n$ with $n = [F : \\F_p]$.\n\n## Idea\n\nA finite field cannot contain an infinite copy of $\\Z$, so some multiple of $1$ is zero and the characteristic is a prime $p$. The multiples of $1$ form a copy of $\\F_p$, and the whole field is a finite-dimensional vector space over it. Counting coordinates gives $p^n$ elements.\n",
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