7 Further Remarks 125a square or−1, then it is a primitive root for infinitely many primesp. (A quantitative
form of the conjecture is considered in Chapter IX.) If the conjecture is not true, then
it is almost true, since it has been shown by Heath-Brown [16] that there are at most
3 square-free positive integersafor which it fails.
A finite subgroup of the multiplicative group of a division ring need not be cyclic.
For example, ifHis the division ring of Hamilton’s quaternions,H×contains the
non-cyclic subgroup{± 1 ,±i,±j,±k}of order 8. All possible finite subgroups of the
multiplicative group of a division ring have been determined (with the aid ofclass field
theory) by Amitsur [2].
For Carmichael numbers, see Alfordet al.[1].
Galois (1830) showed that there were other finite fields besidesFpand indeed, as
Moore (1893) later proved, he found them all. Finite fields have the following basic
properties:
(i) The number of elements in a finite field is a prime powerpn,wheren∈Nand
the primepis the characteristic of the field.
(ii) For any prime powerq = pn, there is a finite fieldFqcontaining exactlyq
elements. Moreover the fieldFq is unique, up to isomorphism, and is the
splitting field of the polynomialtq−toverFp.
(iii) For any finite fieldFq, the multiplicative groupF×qof nonzero elements is cyclic.
(iv) Ifq=pn,themapσ:a→apis an automorphism ofFqand the distinct auto-
morphisms ofFqare the powersσk(k= 0 , 1 ,...,n− 1 }.
The theorem of Chevalley and Warning (Proposition 34) extends immediately to
arbitrary finite fields. Proofs and more detailed information on finite fields may be
found in Lidl and Niederreiter [30] and in Joly [22].
A celebrated theorem of Wedderburn (1905) states that any finite division ring is
a field, i.e. the commutative law of multiplication is a consequence of the other field
axioms if the number of elements is finite. Here is a purely algebraic proof.
Assume there exists a finite division ring which is not a field and letDbe one
of minimum cardinality. LetCbe the centre ofDanda∈ D\C.ThesetMof all
elements ofDwhich commute withais a field, since it is a division ring but not the
whole ofD. EvidentlyMis a maximal subfield ofDwhich containsa.If[D:C]=n
and [M:C]=mthen, by Proposition I.32, [D:M]=mandn=m^2. Thusmis
independent ofa.
IfChas cardinalityq,thenDhas cardinalityqn,Mhas cardinalityqmand the
number of conjugates ofainDis(qn− 1 )/(qm− 1 ). Since this holds for everya∈
D\C, the partition of the multiplicative group ofDinto conjugacy classes shows that
qn− 1 =q− 1 +r(qn− 1 )/(qm− 1 )for some positive integerr. Henceq−1 is divisible by
(qn− 1 )/(qm− 1 )= 1 +qm+···+qm(m−^1 ).Sincen>m>1, this is a contradiction.
For the history of the Chinese remainder theorem (not only in China), see
Libbrecht [29].