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  "body": "\n## Proof\n\nWe follow the proof of Witt. Steps 1 to 9 prove facts on [[cyclotomic polynomials|0W0KVBM54S]] $\\Phi_m$. Steps 10 to 26 apply them to the division ring.\n\n1. Let $m \\ge 1$ and $\\zeta_k = e^{2 \\pi i k / m}$ for $0 \\le k \u003c m$. These are all the $m$th roots of unity, [[by their description|057BH3JC9S]]. There are [[exactly $m$ of them|0E021GF2B8]], so they are pairwise distinct.\n2. Claim: $x^m - 1 = \\prod_{k=0}^{m-1} (x - \\zeta_k)$ in $\\C[x]$. By the [[factor theorem|0K3MWA7DMC]], $x^m - 1 = (x - \\zeta_0) h_1$ for some $h_1 \\in \\C[x]$.\n3. Evaluating at $\\zeta_1$ gives $0 = (\\zeta_1 - \\zeta_0) h_1(\\zeta_1)$. Since $\\zeta_1 \\neq \\zeta_0$ and [[$\\C$ is a field|0FRSR10DWF]], $h_1(\\zeta_1) = 0$. Repeating this $m$ times gives $x^m - 1 = c \\prod_{k=0}^{m-1} (x - \\zeta_k)$ with $c \\in \\C$. [[Comparing degrees|0T233AZ1GE]] and leading coefficients gives $c = 1$.\n4. For $0 \\le k \u003c m$, let $g = \\gcd(k, m)$, the [[greatest common divisor|02WF878BH1]], with $\\gcd(0, m) = m$. Put $d = m / g$ and $j = k / g$. Then $d \\mid m$, $\\gcd(j, d) = 1$ and $\\zeta_k = e^{2 \\pi i j / d}$. If $j = 0$, then $d = 1$, and we replace $j$ by $1$; the value is unchanged by [[periodicity|02P2PYH3X3]].\n5. This sends $k$ to a pair $(d, j)$ with $d \\mid m$, $1 \\le j \\le d$ and $\\gcd(j, d) = 1$. It is a bijection onto these pairs. Its inverse sends $(d, j)$ to the remainder of $j m / d$ modulo $m$, since $\\gcd(j m / d, m) = (m / d) \\gcd(j, d) = m / d$.\n6. By steps 3 and 5 and the definition of $\\Phi_d$, $x^m - 1 = \\prod_{d \\mid m} \\Phi_d(x)$.\n7. Each $\\Phi_m$ is a [[monic polynomial|0BDX4MBKT4]] with integer coefficients. We use strong induction on $m$. It is monic as a product of monic factors, and $\\Phi_1 = x - 1$. For $m \u003e 1$, let $F = \\prod_{d \\mid m, d \u003c m} \\Phi_d$, which is monic in $\\Z[x]$ by induction. By step 6, $x^m - 1 = \\Phi_m F$.\n8. By the [[division algorithm for polynomials|0MES3KS0ZC]] in $\\Z[x]$, $x^m - 1 = Q F + r$ with $Q, r \\in \\Z[x]$ and $\\deg r \u003c \\deg F$. The same algorithm in $\\C[x]$ has a unique result, and $x^m - 1 = \\Phi_m F + 0$ is one. So $\\Phi_m = Q \\in \\Z[x]$.\n9. Let $d \\mid m$ with $d \u003c m$. The divisors of $d$ are divisors of $m$ other than $m$. So by step 6, $x^m - 1 = (x^d - 1) \\Phi_m(x) H(x)$, where $H$ is the product of $\\Phi_e$ over the $e \\mid m$ with $e \\nmid d$ and $e \\neq m$. By step 7, $H \\in \\Z[x]$.\n10. Let $D$ be a [[division ring|0WZXK05ME7]] with finitely many elements. Let $Z = \\set{z \\in D : z x = x z \\text{ for all } x \\in D}$.\n11. If $z, w \\in Z$, then $z - w$ and $z w$ commute with every $x \\in D$, and $1 \\in Z$. So $Z$ is a [[subring|0859D58WPD]] of $D$, and it is commutative. If $z \\in Z$ and $z \\neq 0$, then $z^{-1} x = z^{-1} x z z^{-1} = z^{-1} z x z^{-1} = x z^{-1}$ for all $x$. So $z^{-1} \\in Z$, and $Z$ is a [[field|0QA8CHH8MG]]. Let $q = \\abs{Z}$; then $q \\ge 2$, since $0 \\neq 1$.\n12. For $a \\in D$, let $C(a) = \\set{x \\in D : x a = a x}$. It contains $Z$, and it is closed under addition and under left multiplication by elements of $Z$. With these operations, $C(a)$ is a [[vector space|05EGHEQZAW]] over $Z$. The axioms follow from the ring axioms of $D$. In the same way, $D$ is a vector space over $Z$.\n13. Let $V$ be a vector space over $Z$ with finitely many elements. The set $V$ spans itself, so it [[contains a basis|06MRNP4BNT]] $v_1, \\dots, v_r$. Each vector is $c_1 v_1 + \\dots + c_r v_r$ for exactly one choice of $c_1, \\dots, c_r \\in Z$, by the definition of a [[basis|0VTQTPQGAV]]. So $\\abs{V} = q^r$.\n14. So $\\abs{D} = q^n$ for some $n \\ge 1$, and $\\abs{C(a)} = q^{d(a)}$ for some $d(a) \\ge 1$. We assume $n \u003e 1$ and derive a contradiction.\n15. The nonzero elements of $D$ are units, and $0$ is not a unit. So $D^* = D \\setminus \\set{0}$ is the [[group of units|05THZKR9AN]] of $D$, of order $q^n - 1$.\n16. The [[centre|0S7Y9Q990Q]] of the group $D^*$ is $Z \\setminus \\set{0}$, since every element commutes with $0$. It has order $q - 1$. The [[centraliser|0Q0Z1BEXTF]] of $a \\in D^*$ in $D^*$ is $C(a) \\setminus \\set{0}$, of order $q^{d(a)} - 1$.\n17. Fix $a \\in D^*$ and write $d = d(a)$. By [[Lagrange's theorem|3C53TD]], $q^d - 1$ divides $q^n - 1$. Write $n = s d + t$ with integers $s \\ge 0$ and $0 \\le t \u003c d$, by the [[division algorithm|0CHQKZFK5S]].\n18. By the [[geometric sum|05APWSX53S]], $q^{s d} - 1 = (q^d - 1)(1 + q^d + \\dots + q^{(s-1) d})$, so $q^d - 1$ divides $q^{s d} - 1$. Since $q^n - 1 = q^t (q^{s d} - 1) + (q^t - 1)$, it also divides $q^t - 1$. As $0 \\le q^t - 1 \u003c q^d - 1$, we get $q^t = 1$, so $t = 0$ and $d \\mid n$.\n19. By the [[class equation|08BPSC6YTE]] for $D^*$, $q^n - 1 = (q - 1) + \\sum_{i=1}^{l} [D^* : C_{D^*}(a_i)]$. Here $a_1, \\dots, a_l$ represent the conjugacy classes with more than one element.\n20. Such an $a_i$ is not central, since a central element is alone in its class. So $C(a_i) \\neq D$, and $d_i = d(a_i) \u003c n$. By [[the index formula|04DC0SRN2K]] and step 16, $[D^* : C_{D^*}(a_i)] = (q^n - 1)/(q^{d_i} - 1)$. By step 18, $d_i \\mid n$.\n21. Evaluate steps 6 and 9 at $x = q$, with $m = n$. The integer $\\Phi_n(q)$ divides $q^n - 1 = \\Phi_n(q) F(q)$, where $F$ is the polynomial of step 7 for $m = n$, so $F(q)$ is an integer. It also divides $(q^n - 1)/(q^{d_i} - 1) = \\Phi_n(q) H(q)$ for each $i$. By step 19, $\\Phi_n(q)$ divides $q - 1$. As $q - 1 \\ge 1$, $\\abs{\\Phi_n(q)} \\le q - 1$.\n22. Now let $\\zeta = e^{2 \\pi i j / n}$ with $1 \\le j \\le n$ and $\\gcd(j, n) = 1$. Since $n \u003e 1$, $j \\neq n$, so $\\zeta \\neq 1$ by step 1. Write $\\zeta = u + i v$ with $u, v$ real. Its [[modulus|0KW2X0TFBX]] is $1$, so $u^2 + v^2 = 1$, and $\\zeta \\neq 1$ forces $u \u003c 1$.\n23. Then $\\abs{q - \\zeta}^2 = (q - u)^2 + v^2 = q^2 - 2 q u + 1 \u003e q^2 - 2 q + 1 = (q - 1)^2$. So $\\abs{q - \\zeta} \u003e q - 1 \\ge 1$.\n24. The modulus [[is multiplicative|0HYNZE940T]], so $\\abs{\\Phi_n(q)}$ is the product of the numbers $\\abs{q - \\zeta}$ of step 22. There is at least one factor, for $j = 1$. Each factor exceeds $1$, and one exceeds $q - 1$. So $\\abs{\\Phi_n(q)} \u003e q - 1$.\n25. This contradicts step 21. Hence $n = 1$, so $\\abs{D} = q = \\abs{Z}$ and $D = Z$.\n26. So the multiplication of $D$ is commutative. A commutative division ring is a field, by the definitions.\n\n## Idea\n\nThe centre $Z$ is a field with $q$ elements, and $D$ has $q^n$ elements. The class equation of the group $D^*$ writes $q^n - 1$ as $q - 1$ plus terms $(q^n - 1)/(q^d - 1)$ with $d$ a proper divisor of $n$. The integer $\\Phi_n(q)$ divides all of these, so it divides $q - 1$. But each root of $\\Phi_n$ is farther than $q - 1$ from $q$, which is impossible unless $n = 1$.\n",
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