A division ring is a ring where every nonzero element is invertible. In other words, every nonzero element is a unit.

A commutative division ring is a field. You can explore field theory if you like, but for now we are going to explore noncommutative division rings. The classic example of a division ring is the quaternions. A basic understanding of the quaternions will help, since I will refer to them from time to time.

It is logical to start with something simple, like a finite division ring, but there's nothing simple about this topic. Every finite division ring is a field, yet the proof is far from obvious. Read on and see.

The center of a division ring K is the set of elements that commute with all of K. If x and y are two such elements then c*(x+y) = cx+cy = xc+yc = (x+y)*c. In other words, the center is closed under addition. The same holds for multiplication. Finally, start with cx = xc and multiply by x inverse on the left and the right to show the inverse of x lies in the center. Thus the center of K is a field. It may not be the largest field however, as shown by the complex numbers in the quaternions.

Let K be a finite division ring and let F be the center, a field of characteristic p. Suppose K is larger than F. Thus K is an F vector space of dimension n > 1.

The multiplicative group K* is a finite nonabelian group. Consider its class equation. In the following equation, a ranges over the nontrivial orbits, the nontrivial conjugacy classes of K. Here c(a) is the size of that conjugacy class.

|K*| = |F*| + ∑c(a)

The size of an orbit is the index of its stabilizer, hence c(a) = |K*| divided by the size of the centralizer of a.

Let q = |F|, the size of the central field. Obviously |F*| = q-1. Since K is an F vector space, |K*| = qn-1.

For a fixed element a, consider all the elements that commute with a. Certainly F commutes with a. If x and y commute with a then so does x+y. Fill in the details, and the centralizer of a is an F vector space. Since a is not in F, and a commutes with a, the vector space has dimension > 1. Say its dimension is r. Since a does not commute with all of K (a not being in F), r < n. Setting zero aside, the centralizer has size qr-1, and c(a) = qn-1 over qr-1.

If x and y commute with a then so does xy. Fill in the details, and the centralizer of a is a division ring E inside K. Now K is an E vector space, and E is an F vector space, and K is an F vector space. Combine dimensions, and r divides n. We already showed qr-1 goes into qn-1; and now we know that r goes into n.

In 1903 Wedderburn developed a complicated theorem to finish the proof, but the cyclotomic polynomials provide a simpler conclusion. We are interested in the quotient qn-1 over qr-1. I'll insert the value of q later. For now, step back and write xn-1 as a product of cyclotomic polynomials. The primitive nth roots produce a polynomial of degree φ(n). Roots of lesser order combine to form their own cyclotomic polynomials. The root 1, with order 1, forms its own polynomial, x-1. Multiply these together to get xn-1.

Do the same thing for xr-1. Since r divides n, every cyclotomic polynomial in xr-1 appears in xn-1. Divide these out, and we are left with a subset of cyclotomic polynomials drawn from xn-1. Multiply them together and replace x with q to get c(a).

Since n is strictly larger than r, there are no primitive nth roots in xr-1. This means ζn(x) remains. Other cyclotomics may remain as well, but we know ζn(q) is part of c(a).

Let t = ζn(q). Now t divides each term c(a), and it also divides qn-1, which is the left side of the class equation. Therefore t divides |F*|, or q-1.

Move to the complex plane, and write ζn(q) as a product of factors q-y, where y ranges over the primitive nth roots of 1. This is an equivalent formula for t.

Take the absolute value of the product, i.e. its distance from the origin. This is the product of the absolute values of the individual factors q-y. We know that q is at least 2, and each y lies on the unit circle. Thus q-y has absolute value at least 1. Each factor makes |t| a little larger. Since 1 is not a primitive nth root, we don't have to worry about q-1. All the factors q-y are larger, in absolute value, than q-1. Multiply them together and |t| exceeds q-1. Therefore t cannot divide q-1. This is a contradiction, thus K is a field.

Earlier in ring theory we showed that a finite domain is a division ring. Combine that result with this theorem, and every finite domain is a field.

Apart from fields, and the quaternions, division rings come up very rarely, and the rest of this chapter is only partly successful at characterizing them, so if you want to move on, I will understand.

Let G be a finite, multiplicative subgroup in the division ring R. We will show that G is cyclic, but only if R has characteristic p. In the quaternions, a ring with characteristic 0, G might be ±(1,i,j,k). This multiplicative subgroup isn't even abelian, much less cyclic.

This is a generalization of the same theorem for fields, assuming characteristic p.

Let W be the set of linear combinations of group elements with integer coefficients. A typical element of W might look like 7g1 + 11g5 + 2g9. Of course 2g9 is shorthand for g9+g9. Show that W forms a ring inside R.

Since the coefficients are bounded by p, and since G is a finite group, W is a finite subring of R. Since R is a division ring, W is a finite domain. By the previous theorem, W is a field. The nonzero elements of a finite field form a cyclic multiplicative group, and G is a subgroup of F*, hence G is cyclic.

Note, we only need R to be a domain for this to work.

Let K be a domain that is algebraic over a field F. This means each w in K is algebraic over F, satisfying a certain polynomial; and w commutes with F, so we don't have to worry about whether the coefficients are on the left or the right. In other words, F is in the center of K. Furthermore, each polynomial is irreducible, else K has zero divisors.

Select w from K and restrict attention to the subring F[w]. Since w is algebraic, this is a simple field extension, and w is invertible. Do this for every w, and K becomes a division ring. An example is the quaternions over the rationals, where every element is quadratic over Q.

If F is a finite field then F[w] is also a finite field. Everything is torsion, i.e. it a root of 1. I will show, later on in this chapter, that K becomes a field.

If F is infinite, like the reals, then K could be the quaternions, which is not a field.

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. If x is in both sets, then ix is both xi and -xi, hence 2xi = 0, and x = 0. The sets intersect in 0.

Since C commutes with i, both sets are C vector spaces.

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-. As a C vector space, K is the direct product of K+ and K-.

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 injective. Thus the dimension of K- is 1, and K- is isomorphic to C.

We know that z 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 our quadratic polynomial was not irreducible. Therefore z2 = -s2. This does not lead to the same contradiction. You 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.

Let K be a division ring with characteristic p. Let u be an element not in the center of K, such that un = 1. We shall exhibit an element y such that yu/y = ui, where ui ≠ u. Also, y has the form xu-ux for some x. This is herstein's lemma.

Let f be the function f(x) = xu-ux, which maps K into itself.

Since u is a root of 1, it satisfies an irreducible polynomial with integer coefficients. Let C be (Z/p)[u], a finite field inside K. Multiply by C on the left, and K becomes a left C vector space.

Since C commutes with u, f(C) = 0. Also, f is linear, and f(C*x) = C*f(x). Thus f is a linear map on K, when viewed as a left C vector space.

Let R be the ring of endomorphisms of K, when viewed as a left C vector space. Note that R, like K, has characteristic p.

Think of f as the difference between two endomorphisms, namely x*u and u*x. Within R, these two endomorphisms commute; either way we get u*x*u. Since they commute, one can expand fp by the binomial theorem. The result is (xu)p-(ux)p. Remember that fp does not mean f times f times f … p times, but rather, p applications of f in R. Similarly, (xu)p is the application of the function xu, p times, which is xup. Therefore fp = xup-upx.

As a finite field, let C have order pm. Call this order t. Invoke f t times, and since u raised to the t = u, the result is xu-ux, or f. Within R, f to the t = f.

Write ft-f = 0, the 0 endomorphism, and factor ft-f in R. Let b be an element of C and consider the endomorphism bx. Since b and C commute, bx is a valid function on a left vector space. Verify that bx and f commute; you can apply them in either order. Expand the product of f-bx for every b in C. Set f-0 to the side. That leaves all the nonzero elements of C. These are the distinct roots of 1, and the solutions to the polynomial xt-1 - 1. Thus the expansion of this product is ft-1-1, and when we bring f-0 back in, the result is ft-f. Therefore the composition of endomorphisms f-bx yields 0.

Since these functions commute, apply f first, then the other functions. Suppose all the other functions f-bx are injective. Their composition is injective, and the result cannot be 0 unless f is 0. Since u is not central, f cannot be 0. Therefore f-bx is not injective for some b.

Let x be a nonzero element in the kernel. For this particular x, write the following:

f(x) = bx

xu-ux = bx

xu = (u+b)x

xu/x = u+b

Conjugation by x keeps C within C, and is reversible, hence it is a permutation on C. In fact it performs an automorphism on the multiplicative group of C.

Since b is nonzero, u and u+b are distinct elements, having the same order in C, and in K. They both generate the same multiplicative cyclic subgroup of C*. Thus u+b is some power of u beyond u1. In other words, xu/x = ui.

Finally, set y = xu-ux. Since u and x don't commute, y is nonzero. Multiply xu = uix on the right by u, and on the left by u, and subtract the resulting equations. The result is yu = uiy. Divide by y on the right and yu/y = ui. That completes the proof.

Although a commutator in a group is ab/a/b, a commutator in a ring is ab-ba. Like its group theory counterpart, a commutator is equal to the identity iff a and b commute.

An element y in a division ring K is central iff it commutes with all the commutators.

One direction is obvious, so let's prove the converse. Suppose y is not central, so that xy-yx is nonzero. Let c = xy-yx. Further suppose y commutes with all the commutators, including c, and also with x(xy)-(xy)x. This latter commutator is xc. Write yxc = xcy = xyc, and cancel c, whence xy = yx. This is a contradiction, and that completes the proof.

If every ab-ba is central, then every y commutes with every ab-ba, every y is central, and K is a field. But if K is a field then every ab-ba is 0. In that case (ab-ba)2 = ab-ba. Conversely, assume each commutator has some exponent n > 1 such that (ab-ba)n = ab-ba. Thus each comutator is 0, or a torsion element in the group K*. Suppose K is not a field, hence there is some w = ab-ba that is not central. Now w is nonzero, hence w is a torsion element. For any nonzero c in the center of K, cw is a nonzero commutator, and is torsion. Write cnwn = 1. If wj = 1, raise this to the j, whence cnj = 1. Thus the center is a torsion group. All integers lie in the center, and are torsion, hence K has finite characteristic p.

Apply herstein's lemma, and there is a commutator y such that yw/y = wi, for i > 1. The subgroup generated by w is a finite cyclic group, and conjugating by y maps this group into itself. Thus y and w together span a subgroup, which, as a set, is the cross product of the powers of w times the powers of y. Since y is another commutator it is torsion, and the subgroup spanned by w and y is finite. A finite group of a division ring is cyclic. However, this group is not even abelian, since w and y do not commute. This is a contradiction, hence K is a field.

In summary, K is a field iff every commutator w has some n > 1 such that wn = w.

Assume each nonzero w has an n such that wn = 1. Equivalently, there is some n > 1 such that wn = w. Set w to any commutator and apply the previous result. Therefore K is a field. I referenced this result earlier, when K was algebraic over a field.

Let F be a division subring of the division ring K. We say F is normal in K if the multiplicative group F* is normal in K*. When this happens either F = K, or F is a field in the center of K.

Choose c in F and d in K-F, and suppose c and d do not commute. This means c and d are not 0, and not 1.

Set b = d-1, thus c and b do not commute either. Note that b is nonzero.

By normality, (1/d)cd - (1/b)cb lies in F. Left multiply by d and get cd - (d/b)cb. Replace d with b+1 and get c - (1/b)cb. This too lies in F. Since c and b do not commute, this expression is nonzero. Recall that it is the product of two elements, namely d and (1/d)cd-(1/b)cb. Divide by the latter and find that d lies in F, which is a contradiction. Therefore F commutes with everything outside of F.

We still need to prove F is a field. Let d lie outside of F, as it did before. Note that d is nonzero. Let c and e be any two elements of F. Make e nonzero (since 0 commutes with everything). Since F is a division ring, e*d lies outside of F. Now c commutes with d and with e*d.

ced = edc = ecd

Cancel d, and c and e commute. That completes the proof. Of course F could equal K, whence F need not be a field at all. We need F to be smaller, and normal, for this to work.

A multiplicative commutator has the form xy/x/y. If the ring is commutative each commutator equals 1. This is the case when K is a field.

Conversely, assume each commutator is central. Fix an element c, and it commutes with every commutator. In particular, c commutes with 1/x*c*x/c, hence c/x*c*x/c = 1/x*c*x. In other words, c commutes with every conjugate of c.

Suppose c and a do not commute, and let b = a-1. Note that c commutes with the following expression e, which is the difference between two conjugates of c.

e = 1/a*c*a - 1/b*c*b

Multiply by a on the left, and see if c commutes with ae.

ae = c*a - a/b*c*b =
c*a - (b+1)/b*c*b =
c*a - c*b - 1/b*c*b =
c - 1/b*c*b

Yes indeed, c commutes with ae. Furthermore, since c and b do not commute, ae is nonzero, and e is nonzero. Thus c commutes with two elements whose quotient is a, and c commutes with a after all. This shows c is central.

If every commutator is central then every c commutes with every commutator, every c is central, and K is a field.

Herstein's conjecture, still open, states that K is a field if every multiplicative commutator can be raised to some power to produce a central element.

Within a division ring K, that is not a field, let w be noncentral, and let F be the division ring generated by the conjugates of w. By generated, we mean as a division ring, closed under +-*/.

Look at the structure of F. It includes w, in the form of 1*w/1. By closure it includes w/w, or 1. This brings in all the integers, and the rationals if K has characteristic 0.

Show that F is normal in K. When an atom, i.e. a conjugate of w, is conjugated, the result is another conjugate of w, which is in F, so we're off to a good start. Expressions are built recursively, by closure. Show that the conjugate of u+v is the conjugate of u plus the conjugate of v. Since u and v remain in F after conjugation, so does u+v. The same holds for u-v and u*v. How about an arbitrary conjugate of 1/u? Write x*(1/u)/x = 1/y and invert both sides, whence y = xu/x. Since y is in F (by assumption), 1/y is in F, and the conjugate of u inverse is in F. Therefore conjugation maps F into F.

Since F is normal in K, an earlier theorem says F is central in K, or F is all of K. Since F contains w, the former is impossible, hence F is all of K. Therefore K is generated by the conjugates of any noncentral element w.

Again, let K be a division ring that is not a field. Let the multiplicative commutators generate F. If all the commutators are central then the previous theorem tells us K is a field. Therefore some commutator w = xy/x/y is not central. For any z, F includes zw/z/w, and w, hence it includes zw/z. F includes all the conjugates of a noncentral element. These conjugates generate all of K, hence F = K. The multiplicative commutators generate K.

Let G be the multiplicative group K*, and let H be the center of G. Let J be the preimage of the center of G/H in G. J includes H, but in this case J = H.

Suppose some c lies in J-H. This means there is some a that does not commute with c. Yet c commutes with everything when mapped into G/H. Every commutator, such as 1/a*c*a/c, lies in H.

Set b = a-1. Thus (1/a*c*a - 1/b*c*b)/c is the difference between two commutators, and is central in K. Multiply by a on the left and get 1-1/b*c*b/c, which is another element in the center of K. Since c and b do not commute, this element is nonzero. Now a is the quotient of two elements in H, hence a lies in H, which is a contradiction. Therefore J = H.

Mod out by the center of G, and the quotient group has no center beyond 1.

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. There are finitely many points in projective space. 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.

Therefore V is finite iff P is finite.

Let D be a division subring of the division ring E, such as the quaternion rationals in the quaternion reals. For a fixed 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 is convenient 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. For w in D, all of wK moves c to the same point in its orbit. For x not in wK, x moves c to some other point in its orbit. The lines through D, i.e. the points of D as a projective space, are the elements of S. As per the previous result, S is finite iff D is finite.

Herstein's little theorem says a noncentral element in an infinite division ring has infinitely many distinct 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.

Let D be an infinite division ring (the finite ones aren't too interesting), and let w be an element in D. The centralizer of w is the set of elements that commute with w. Verify that this is a division subring of D. We will prove this centralizer is infinite.

This is automatically true if w is central, or if w has infinite order in D*, so assume these conditions do not hold. Let w have order q, and let K be the (allegedly finite) centralizer of w.

Since w commutes with all the integers, let D have finite characteristic p. Of course K inherits the same characteristic.

By Herstein's lemma, there is some y in D such that yw/y = wi, where wi is different from w. Suppose y is torsion. Let H be the subgroup generated by w and y. Represent an element of H as a power of w times a power of y. When two such elements are concatenated, replace each yw with wiy. Pull y's and w's past each other until the result is in standard form. Since the exponents on w and y are bounded, |H| is finite.

Let R be the set of linear combinations of elements of H, using coefficients in Z/p. (This is called a group ring.) Some of these combinations may represent the same element in D; that's ok. In any case, R is a finite ring that embeds in D, hence it has no zero divisors. A finite domain is a field. This means w and y commute, which is impossible. Therefore the powers of y go on forever.

Since w and wi are conjugate, they have the same order mod q, hence i and q are coprime. Let m = φ(q), so that im = 1 mod q. Conjugate w by ym, which applies the inner automorphism m times, and w maps to w. Therefore ym and w commute. In the same way, y raised to any multiple of m commutes with w. Since y is torsion free, the powers of y are distinct, and the centralizer of w is infinite.

Let the field F be the center of a division ring D. If D is finite dimensional over F then D is centrally finite, else D is centrally infinite. I'll illustrate with some twisted rings.

Let K be a field, and let D be the twisted laurent series K[[x]], such that xc = σ(c)x for some field automorphism σ of K. Let K′ be the fixed field of σ. Each series is invertible, hence D is a division ring.

First, let the order of σ be infinite. Let the series ∑ aixi lie in the center, hence it commutes with every scalar b in K. Therefore b*ai = ai*σi(b). For every i other than 0, σi moves some b in K. For this b, the equation cannot be satisfied, unless ai = 0. Therefore the series is merely a constant term a0. To commute with x, a0 lies in K′. Therefore the center of D is the fixed field K′, and D is centrally infinite.

Next assume σ has order m. Commuting a series with every scalar b in K shows only powers of xm are allowed. If any coefficient lies outside of K′, the series does not commute with x. Therefore F, the center, consists of series whose exponents are multiples of m, with coefficients in K′. By galois theory, the index of K′ in K is m. Build a basis for D, as an F vector space, by crossing the basis of K over K′ with the powers of x from 0 to m-1. the dimension of D over F is m2, and D is centrally finite.