Division Rings, Herstein's Little Theorem

Finite Projective Space

Let K be a division ring and let V be a left K vector space with dimension > 1. The projective space P is the set of lines passing through the origin of V. We've seen this with fields, but it's valid for division rings too. Just remember that K acts on the left. So all the left multiples of a nonzero point s in V form a line in V, which is a point in P.

If V is finite then the quotient set P is finite. Conversely, assume P is finite. Choose two independent vectors s and t in V. Let a map f take K into P by x → s+x*t. Since s and t are linearly independent, f(x) can never be 0, hence f always produces a well defined value in P.

If x and y produce the same point in P, then s+xt = z*(s+yt). By linear independence, the coefficients on s must agree, hence z = 1. Then the coefficients on t must agree, hence x = y. Therefore the map is injective.

Since K embeds in P, K is finite. The points of V are the points of P cross the nonzero scalars in K, plus the origin. This too is finite.

In summary, V is finite iff P is finite.

Herstein's Little Theorem

Let E be a division ring containing the division subring D. For a given c in E, let the D-conjugates of c be xc/x for x in D. Call this set S. Now D^* acts on S by conjugation. By convention, xy acts on S by y, and then x. This makes perfect sense when the action is conjugation.

If D commutes with c then there is but one D-conjugate, namely c. So assume D and c do not commute.

Let K be the division ring in D that stabilizes, or commutes with, c. The size of S is now the index of K^* in D^*.

View D as a right K vector space. Since K ≠ D, D has dimension greater than 1. Note that each D-conjugate in S is a point in projective space. Apply the theorem at the top of this page, and S is finite iff D is finite.

Herstein's little theorem says a noncentral element in an infinite division ring has infinitely many conjugates. (Remember that a finite division ring is a field, so a noncentral element automatically implies an infinite division ring.) Set D = E in the above, and S is infinite, just like D.