Division Rings, Algebraic over the Reals

Algebraic over the Reals

If the reals, denoted R, are in the center of a division ring K, such that K is algebraic over R, what can we say about K? As it turns out there are only three possibilities: K must be the reals, the complex numbers, or the quaternions.

If K = R then we are done, so pick any z in K-R. This satisfies a monic irreducible polynomial in R, which must be quadratic. Thus R[z] = C, the complex numbers.

Let i² = -1 as usual.

If K = C we are done. Otherwise build two sets as follows. K⁺ is the set of elements that commute with i, and K^- is the set of elements that anticommute with i, so that i*x = -x*i.

Since C commutes with i, both sets are C vector spaces. Show that their intersection is 0.

For any x in K, x*i+i*x commutes with i, and x*i-i*x anticommutes with i. Divide these by 2i, and x becomes the sum of two elements, one from K⁺ and one from K^-.

If a+b equals c+d, terms from K⁺+K^-, subtract them to find x+y = 0. This means x = -y; yet K⁺ and K^- intersect in 0. Thus a = c and b = d. Elements in the cross product are distinct, and K is isomorphic to the direct sum of the two vector spaces.

Since C is algebraically closed, K⁺ = C.

Take any z in K^-, and right multiplication by z becomes a C linear map from K^- into K⁺. This because the product of two anticommuting elements commutes with i. Since there are no zero divisors, this map is 1-1. Thus the dimension of K^- is 1, and K^- is isomorphic to C.

We know that z, as selected above, is algebraic over R. Since there are no zero divisors, z satisfies an irreducible polynomial, namely a quadratic. Hence z² = az+b. Now z² lies in K⁺, hence az+b lies in K⁺, and that can happen only if a = 0. In other words, z² is a real number.

If z² > 0, write z² = s², where s is real. This means z = ±s, and z is real, which is impossible. Therefore z² = -s². Note that this does not lead to the same contradiction. We cannot write (z-si)*(z+si) = 0, whence z = ±si, because z and i do not commute.

Set j = z/s. thus j² = -1.

Now put K⁺ and K^- together. The former is C, which is a real vector space spanned by 1 and i. The latter is isomorphic to C, a real vector space spanned by j and ij. Therefore K has dimension 4, as a real vector space, with basis 1, i, j, and ij. This is the quaternions.

As a corollary, let K be an algebraic integral domain over R. An algebraic integral domain over a field is a field, hence K equals R or C.