Prove that any finite division ring is a field (that is, prove commutativity). Give an example of a (necessarily infinite) division ring which is NOT a field.
Solution
Let \(D\) be a finite division ring with center \(Z\), a finite field of order \(q\ge2\), and let \(n=\dim_ZD\), so \(|D|=q^n\).
1. For \(x\in D^\times\) noncentral, \(|C_D(x)|=q^d\) with \(d\mid n\) and \(d<n\).
\(C_D(x)=\{y:yx=xy\}\) is a division subring containing \(Z\), so \(|C_D(x)|=q^d\) with \(d=\dim_ZC_D(x)\). \(D\) is a vector space over \(C_D(x)\), so \(q^n=(q^d)^m\) for some \(m\), and \(d\mid n\); \(d<n\) because \(x\) is not central.
2. \(q-1=(q^n-1)-\sum_{i=1}^m\frac{q^n-1}{q^{d_i}-1}\) for proper divisors \(d_i\) of \(n\).
This is the class equation of \(D^\times\): \(|D^\times|=q^n-1\), \(|Z(D^\times)|=|Z^\times|=q-1\), and the class of a noncentral \(x_i\) has size \([D^\times:C_D(x_i)^\times]=\frac{q^n-1}{q^{d_i}-1}\) with \(d_i\) as in step 1.
3. \(\Phi_n(q)\mid q-1\), where \(\Phi_n\) is the \(n\)th cyclotomic polynomial.
In \(\ZZ[x]\), \(x^n-1=\prod_{c\mid n}\Phi_c(x)\) and \(x^{d}-1=\prod_{c\mid d}\Phi_c(x)\). For a proper divisor \(d\) of \(n\), \(\Phi_n\) does not occur in the second product, so \(\Phi_n(x)\) divides \(\frac{x^n-1}{x^d-1}\) in \(\ZZ[x]\). Evaluating at \(q\), \(\Phi_n(q)\) divides \(q^n-1\) and every \(\frac{q^n-1}{q^{d_i}-1}\), hence \(q-1\) by step 2.
4. \(n=1\), so \(D=Z\) is a field.
Suppose \(n>1\). For a primitive \(n\)th root of unity \(\zeta\neq1\), \(\operatorname{Re}\zeta<1\), so \[|q-\zeta|^2=q^2-2q\operatorname{Re}\zeta+1>(q-1)^2 .\] Hence \(|\Phi_n(q)|=\prod_\zeta|q-\zeta|>(q-1)^{\varphi(n)}\ge q-1\), contradicting step 3 since \(q-1>0\).
5. The real quaternions \(\mathbb H=\{a+bi+cj+dk:a,b,c,d\in\RR\}\), with \(i^2=j^2=k^2=ijk=-1\), form a division ring that is not a field.
For \(z=a+bi+cj+dk\neq0\), put \(\bar z=a-bi-cj-dk\); then \(z\bar z=\bar zz=a^2+b^2+c^2+d^2>0\), so \(z^{-1}=\bar z/(a^2+b^2+c^2+d^2)\). Since \(ij=k\neq-k=ji\), \(\mathbb H\) is not commutative.