Number Theory: An Introduction to Mathematics

70 I The Expanding Universe of Numbers

j=γ+δb,whereγ,δ∈Randδ=0. Sincej^2 =γ^2 + 2 γδb+δ^2 α=−1, we must
haveγ=0. Thusj=δbandji=−ij.
If we putk=ij, it now follows that

k^2 =− 1 , jk=i=−kj, ki=j=−ik.

Since noR-linear combination of 1,i,jhas these properties, the elements 1,i,j,k
areR-linearly independent. But, by Proposition 32 below, [D:M]=[M:R]=2.
Hence [D:R]=4and1,i,j,kare a basis forDoverR. ThusDis isomorphic to the
division ringHof quaternions.

To complete the proof of Proposition 31 we now prove

Proposition 32Let D be a division ring which, as a vector space over its centre C , has
finite dimension[D:C]. If M is a maximal subfield of D, then[D:M]=[M:C].

Proof Putn=[D:C]andlete 1 ,...,enbe a basis forDas a vector space overC.
Obviously we may supposen>1. We show first that ifa 1 ,...,anare elements ofD
such that

a 1 xe 1 +···+anxen=0foreveryx∈D,

thena 1 =···=an=0. Assume that there exists such a set{a 1 ,...,an}with not all
elements zero and choose one with the minimal number of nonzero elements. We may
suppose the notation chosen so thatai=0fori≤randai=0fori>rand, by
multiplying on the left bya 1 −^1 , we may further suppose thata 1 =1. For anyy∈D
we have

a 1 yxe 1 +···+anyxen= 0 =y(a 1 xe 1 +···+anxen)

and hence

(a 1 y−ya 1 )xe 1 +···+(any−yan)xen= 0.

Sinceaiy=yaifori=1andfori>r, our choice of{a 1 ,...,an}implies thataiy=
yaifor alli. Since this holds for everyy∈D, it follows thatai∈Cfor alli. But this
is a contradiction, sincee 1 ,...,enis a basis forDoverCanda 1 e 1 +···+anen=0.
The mapTjk:D→Ddefined byTjkx=ejxekis a linear transformation of
Das a vector space overC. By what we have just proved, then^2 linear mapsTjk
(j,k= 1 ,...,n) are linearly independent overC. Consequently every linear transfor-
mation ofDas a vector space overCis aC-linear combination of the mapsTjk.
Suppose now thatT:D→Dis a linear transformation ofDas a vector space
overM.SinceC⊆M,Tis also a linear transformation ofDas a vector space overC
and hence has the form

Tx=a 1 xe 1 +···+anxen

for somea 1 ,...,an∈D.ButT(bx)=b(Tx)for allb∈Mandx∈D. Hence

(a 1 b−ba 1 )xe 1 +···+(anb−ban)xen=0foreveryx∈D,