[This post is part of a series on the long-standing conjecture that all finite projective planes have prime power order. See here for a list of the other posts. ]

Recall that a “division ring” is a field except that the multiplication is not necessarily commutative. The theorem is that any finite division ring is a field.

The theorem was first proved in 1905. When I was young, MacLagan Wedderburn got the credit because his proof was published first. But there have since been claims (which I have not checked) that his proof was wrong, so that the credit should go to Leonard Dickson, a colleague at Princeton, who published another proof soon afterwards.

So let \(D\) be a finite division ring. For any element \(g\in D\) let \(C(g)\), the centraliser be \(\{d\in D:dg=gd\}\). The centre \(Z\) is the set of elements which commute with all elements of \(D\). Evidently \(\{0,1\}\subseteq C(G)\subseteq D\), and \(C(g)\) is a division ring, whilst \(Z\) is a field.

Let \(k=|Z|\), so \(k\) is an integer \(\ge 2\). Let \(S=\{a_1,\dots,a_n\}\) be (one of ) the smallest sets of elements such that every element of \(D\) can be written as a linear combination \(\sum z_ia_i\) with all \(z_i\in Z\). Such sets clearly exist, for example, we could take \(S=D\). All of the \(k^n\) linear combinations must be distinct, for otherwise we would have \(\sum z_ia_i=\sum z’_ia_i\) and \(z_j\ne z’_j\) for some \(j\) but then, subtracting \((z_j-z’_j)a_j\) is a linear combination of other \(a_i\), and multiplying through by \((z_j-z’_j)^{-1}\), we have that \(a_j\) is a linear combination of the other \(a_i\) so we could get a set smaller than \(S\) by deleting \(a_j\). Contradiction. Hence \(D\) has \(k^n\) elements. Similarly, \(C(g)\) has \(k^m\) elements for some \(m\le n\). If \(D\) is not a field, then for some \(g\) we have \(C(g)\ne D\) and hence \(m\lt n\).

Recall that for any group \(G\) we can define a conjugacy relation: \(g\)~\(h\) if and only if \(g=a^{-1}ha\) for some \(a\) in the group. It is easy to check that this is an equivalence relation, so that it divides the group into disjoint “conjugacy classes”. The elements in the centre \(Z\) are precisely those elements that belong to a conjugacy class with just one element. Hence we have: \(|G|=|Z|+\sum |K_i|\), where the sum is taken over all conjugacy classes with at least two elements. That is known as the class equation. Moreover, it is easy to check that \(|K_i|\) is \(|G|/|C(g)|\) where \(g\in K_i\).

Thus in this case we have $$|k^n-1|=(k-1)+\sum{\frac {k^n-1}{k^m_i-1}}$$ where the sum is taken over the conjugacy classes with at least two elements and hence \(m_i\lt n\). If we write \(n=qm+r\) with \(0\le r\lt m\) then \(k^n-1=(k^m-1)(k^{m(q-1)}+\dots+1)k^r+k^r-1\), so if \(k^m-1\) divides \(k^n-1\) then it also divides \(k^r-1\) which is impossible unless \(r=0\). So each \(m_i\) must divide \(n\).

Now we know from elementary results on cyclotomic polynomials \(\Phi_n(x)\in\mathbb{Z}[x]\) that \(\Phi_n(x)\) is a factor of \(x^n-1\) and of \((x^n-1)/(x^m-1)\) for \(m\lt n\) and \(m|n\) (see Exercise 7 for the latter). We may set \(x\) to be the integer \(k\) and we now have as relations between integers that \(\Phi_n(k)\) is a factor of \(k^n-1\) and of \((k^n-1)/(k^{m_i}-1)\). Hence the class equation implies that it is also a factor of \(k-1\).

Suppose that \(z=a+ib\) is any (complex) root of unity of order \(n\). Then \(|z|=1\) so \(|a|\lt 1\). Hence \(|k-z|^2=k^2-2ak+1\gt k^2-2k+1=(k-1)^2\gt 1\). Hence \(|\Phi_n(k)|=\prod_z|k-z|\gt k-1\). Contradiction. \(\Box\)

## Post a Comment