Number Theory: An Introduction to Mathematics

9 Vector Spaces and Associative Algebras 69

the ring structure, we can regardAas a vector space overF. The associative algebra

is said to befinite-dimensionalif it is finite-dimensional as a vector space overF.

For example, the setMn(F)of alln×nmatrices with entries from the fieldF

is a finite-dimensional associative algebra, with the usual definitions of addition and

multiplication, and withα∈Fidentified with the matrixαI.

More generally, ifDis a division ring containingFin its centre, then the set

Mn(D)of alln×nmatrices with entries fromDis an associative algebra overF.It

is finite-dimensional ifDitself is finite dimensional overF.

By the definition for rings, an associative algebraAissimpleifA={ 0 }andA

has no ideals except{ 0 }andA. It is not difficult to show that, for any division ringD

containingFin its centre, the associative algebraMn(D)is simple. It was proved by

Wedderburn (1908) that any finite-dimensional simple associative algebra has the form

Mn(D),whereDis a division ring containingFin its centre and of finite dimension

overF.

IfF=C, the fundamental theorem of algebra implies thatCis the only suchD.If

F=R, there are three choices forD, by the following theorem of Frobenius (1878):

Proposition 31If a division ring D contains the real fieldRin its centre and is of

finite dimension as a vector space overR, then D is isomorphic toR,CorH.

Proof Suppose first thatDis a field andD=R.Ifa∈D\Rthen, sinceDis finite-

dimensional overR,ais a root of a monic polynomial with real coefficients, which

we may assume to be of minimal degree. Sincea∈/R, the degree is not 1 and the

fundamental theorem of algebra implies that it must be 2. Thus

a^2 − 2 λa+μ= 0

for someλ,μ∈Rwithλ^2 <μ.Thenμ−λ^2 =ρ^2 for some nonzeroρ ∈Rand
i=(a−λ)/ρsatisfiesi^2 =−1. ThusDcontains the fieldR(i)=R+iR.But,since
Dis a field, the onlyx ∈Dsuch thatx^2 =−1areiand−i. Hence the preceding
argument shows that actuallyD = R(i). ThusDis isomorphic to the fieldCof
complex numbers.
Suppose now thatDis not commutative. Letabe an element ofDwhich is not
in the centre ofD,andletMbe anR-subspace ofDof maximal dimension which is
commutative and which contains bothaand the centre ofD.Ifx∈Dcommutes with
every element ofM,thenx∈M. HenceMis a maximal commutative subset ofD.It
follows that ifx∈Mandx=0thenalsox−^1 ∈M,sincexy=yxfor ally∈Mim-
pliesyx−^1 =x−^1 yfor ally∈M. Similarlyx,x′∈Mimpliesxx′∈M. ThusMis a
field which properly containsR. Hence, by the first part of the proof,Mis isomorphic
toC. ThusM=R(i),wherei^2 =−1, [M:R]=2andRis the centre ofD.
Ifx∈D\M,thenb=(x+ixi)/2 satisfies

bi=(xi−ix)/ 2 =−ib= 0.

Henceb∈D\Mandb^2 i=ib^2. But, in the same way as before,N =R+Rbis

a maximal subfield ofDcontainingbandR,andN=R(j),wherej^2 =−1. Thus

b^2 =α+βb,whereα,β∈R. In fact, sinceb^2 i=ib^2 ,wemusthaveβ=0. Similarly