the q™s are written, p,(x) = ciqi(x), i = 1,2,. . ,r, and ci 6 F.
˜
Let F(a) be an extension of F in which p i(a) = 0. We may
suppose the leading coefficients of the pi( x ) and the qi( x ) to be 1, for,
by factoring out a11 leading coefficients and combining, the constant
multiplier on each side of the equation must be the leading coefficient
of p (x ) and hence cari be divided out of both sides of the equation.
Since 0 = ˜˜(a).˜˜(a). . . . .˜,(a) = p(a) = si(a).. . . .4,(a) and
since a product of elements of F(a) cari be 0 only if one of these is 0,
it follows that one of the qi( a), say qi( a), is 0. This gives (see page
25)p,(x) = si(x). Thus˜,(x).˜,(x).....˜,(x)
= Pdx).q,(x). . . .q,(x) or
34
pi(x).[p,(x).. . . .p,(x)  q*(x).. . . .qs(x)] = 0. Since the product
of two polynomials is 0 only if one of the two is the 0 polynomial, it
follows that the polynomial within the brackets is 0 that
SO
p,(x) . . . .p,(x) = q*(x). . .: q.(x). If we repeat the above argument
r times we obtain p,(x) = si(x), i = 1,2,. . , r. Since the remaining
q™s must have a product 1, it follows that r = s.
F. Group Characters.
˜ 
If G is a multiplicative group, F a field and o a homomorphism
mapping G into F, then o is called a character of G in F. By homomor
phism is meant a mapping u such that for a, fi any two elements of G,
o(a).a(B) = a(a.@)ando(a) f Oforanya.
(If o(a) = 0 for one element a, then o(x) = 0 for each x t G, since
o( ay) = o(a). o(y) = 0 and ay takes a11 values in G when y assumes
a11 values in G).
The characters or, 02,. . . , onare called dependent if there exist
elements a r, a,, . . . , a,, not a11 zero in F such that
a,o,(x) + a202(x) + . . . + anon = 0 for each x t G. Such a de
pendence relation is called nontrivial. If the characters are not
dependent they are called independent.
THEOREM 12. If G is a group and or, u2,. . . , on are n mutu
ally distinct characters of G in a field F, then oi, 02,. . . , on
are independent.
One character cannot be dependent, since a rcr( x) = 0 implies
a1 = 0 due to the assumption that or(x) f 0. Suppose n > 1.
35
We make the inductive assumption that no set of less than n distinct
characters is dependent. Suppose now that
. + angn( x) = 0 is a nontrivial dependence
aru, i a,o,(x> + .
between the u™s. None of the elements ai is zero, else we should have
a dependence between less than n characters contrary to our induc
tive assumption. Since or and un are distinct, there exists an element
a in G; such that or (a) f o”(a). Multiply the relation between the
u™s b y n ar e obtain a relation
W
(*) bru,(x) + . . . + b,.r onr(x) + o,(x) = 0, bi = air ai f 0.
Replace in this relation x by ax. We have
b,o,(a)ol(x) + . . + b,, un., (a>un.,(x> + un(a (x> = 0,
o r a, ( a j™b,u,(a)u, ( x ) + . + = 0.
U,(X)
Subtracting the latter from (*) we have
(**> [b,  un (a)˜blul (a>la,(x) t  . + cn.lun.l (x) = 0 .
The c™oefficient of ur (x ) in this relation is not 0, otherwise we should
h a v e b, = u, (a)˜b,al (a), that
SO
q, (a)b, = blo,(a) = u,(a)b,
and since b, f 0, we get a,( a) = ur (a) contrary to the choice of a.
Thus, (* * ) is a nontrivial dependence between u r, g2, . ,v” 1 which
is contrary to our inductive assumption.
Corollary. If E and E™ are two fields, and q , u2, . . , un are n
mutually distinct isomorphisms mapping E into E ™ , then u, , . , u,
are independent. (Where “independent” again means there exists no
nontrivial dependence a ru r (x ) + . + anun (x ) = 0 which holds for
every x 6 E).
This follows from Theorem 12, since E without the 0 is a group
36
and the u™s defined in this group are mutually distinct characters.
If oi >a2 > . . . , u, are isomorphisms of a field E into a field E™ ,
then each element a of E such that o*(a) = o,(a) = . . . = on(a)
is called a fixed point of E under oi , 02, . . . , o,., . This name is
chosen because in the case where the u™s are automorphisms and ui
is the identity, i.e., u1 (x) = x, we have ui (x) = x for a
fixed point.
Lemma. The set of fixed points of E is a subfield of E. We
shall cal1 this subfield the fixed field.
For if a and b are fixed points, then
ui(a 1 b) = u,(a) + u,(b) = uj (a) + oj (b) = uj (a + b) and
uj(a.b) = ui(a).ui(b) = uj (a).uj(b) = uj (a.b).
Finally from u,(a) = aj (a) we have (uj(a))™ = (u,(a))™
= ˜,(a˜) = uj ( a  ˜ ) .
Thus, the sum and product of two fixed points is a fixed point, and
the inverse of a fixed point is a fixed point. Clearly, the negative of a
fixed point is a fixed point.
THEOREM 13. If cri,. . . >un are n mutually distinct isomorphisms
of a field E into a field E™ , and if F is the fixed field of E, then
(E/F:) L
> n.
Suppose to the contrary that (E/F) = r < n. We shall show that
we are led to a contradiction. Let w i, o 2, . . . , o, be a generating sys
tem of E over F. In the homogeneous linear equations
37
ul(ol)xl + u*(w1)x2 + . . . + u,(wl)x, = 0
+ u2(o*)x2 + . . . + u,(o*)xn = 0
OI(OZ)XI
UI(W,)XI + u*(or)x2 + . . . + un(o,)xn =0
there are more unknowns than equations that there exists a non
SO
trivial solution which, we may suppose, x r, x 2, . . . ,x, denotes. For
any element a in E we cari find ar, a2,. . . , a, in F such that
a = aIci>, f . . . + a,o,. We multiply the first equation by o 1 ( a 1 ),
the second by o1 (a ,J, and on. Using that ai 6 F, hence that
SO
ol(ar) = oj (ai)and also t h a t oj(ai) oj(ai) = oj(aiwi),
we obtain
ol(a,wI)x, + . . . + on(a,o,)x = 0
n
oI(ar˜,)xl + . . . + on(arWr)x n = 0 .
Adding these last equations and using
oi(a,o,) + oi(a2W2) + . . . + oi(aro,) = oi(a,o, + . . . + aru,) = ai(Q)
we obtain
o,(a)x, + o,(a>x, + . . . + un(a)xn = 0.
This, however, is a nontrivial dependence relation between or, 02, . . . , on
which cannot exist according to the corollary of Theorem 12.
Ciorollary. If or, u2, . . . , cr,, are automorphisms of the field E, and

F is the fixed field, then (E/F) > n. 
If F is a subfield of the field E, and 0 an automorphism of E, we
shall say that u leaves F fixed if for each element a of F, o(a) = a.
38
If o and r are two automorphisms of E, then the mapping o( r( x)) written
briefly ur is an automorphism, as the reader may readily verify.
[E.g., u˜(x.Y) =o(r(˜.˜)) = o(r(˜>.˜(y)> = a(˜(x))˜o(s(y))l.
We shall cal1 UT the product of o and r. If o is an automorphism
(o(x) = y), then we shall cal1 0l the mapping of y into x, i.e.,o˜(y) =x
the inverse of o. The reader may readily verify that or is an automor
phism. The automorphism 1 (x ) = x shall be called the
unit automorphism.
 
˜ If E is an extension field of F, the set G of automorphisms
Lemma.
which leave F fixed is a group.
The product of two automorphisms which leave F fixed clearly
leaves F fixed. Also, the inverse of any automorphism in G is in G.
The reader Will observe that G, the set of automorphisms which
leave F fixed, does not necessarily have F as its fixed field. It may be
that certain elements in E which do not belong to F are left fixed by
every automorphism which leaves F fixed. Thus, the fixed field of G
may be larger than F.
G. Applications and Examples to Theorem 13.
Theorem 13 is very powerful as the following examples show:
1) Let k be a field and consider the field E = k (x ) of a11
rational functions of the variable x. If we map each of the functions
f (x ) of E onto f(L) we obviously obtain an automorphism of E. Let us
X
consider the following six automorphisms where f (x ) is mapped onto
f (x ) (identity), f ( lx), f ($), f (li), f (&) and f (5) and cal1 F the
39
fixed point field. F consists of a11 rational functions satisfying
f ( x ) = f ( l  x ) = f(i) = f(l$) = f(A) = f($$.
(1)
It suffices to check the first two equalities, the others being conse
quences. The function
1 = I(x) = (x2  x+1j3
(2)
x*(xl)*
belongs to F as is readily seen. Hence, the field S = k (1) of a11
rational functions of 1 Will belong to F.
We contend: F = S and (E/F) = 6.
Indeed, from Theorem 13 we obtain (E/F) 1 6. Since S C F it
suffices to prove (E/S) < 6. Now E = S(x). It is thus sufficient to

find some 6th degree equation with coefficients in S satisfied by x.
The following one is obviously satisfied;
(x2  x+1)3 1.x2(x1)2 = 0.
The reader Will find the study of these fields a profitable exer
cise. At a later occasion he Will be able to derive a11 intermediate fields.
2 ) LetkbeafieldandE = k(x,,x2,...,x,)thefieldofall
rational functions of n variables x1, x2, . . . , xn. If (vi , v2, . . . , Vu ) is a
permutation of (1,2, . . . , n) we replace in each function f (x 1, x *, . . . , xn)
of E the variable x, by x xn by x1/ . The mapping of E
x2 by x 1/2™.
Vl ™ .™ n
onto itself obtained in this way is obviously an automorphism and we
may construct n ! automorphisms in this fashion (including the identity).
Let F be the fixed point field, that is, the set of a11 socalled
“symmetric functions.” Theorem 13 shows that (E/F) > n ! . Let us in
troduce the polynomial:
(3) f(t) = (txr)(tx2). . . (tx,) = t” + a itni + . + an
40
wherear =  (x1 + x2 + . . . + x,); a2 .: + (x1x2 + x1x3 + . . . + xnrxn)
and more generally ai is (  1)™ times the sum of a11 products of i differ
erent variables of the set x1, x2, . . . , xn. The functions a,, a*, . . . , an
are called the elementary symmetric functions and the field
S = k(a,,a,,..., an ) of a11 rational functions of a,, a2, . . . , ati is
obviously a part of F. Should we suceed in proving ( E/S ) <n ! we
would have shown S = F and (E/F) = n ! .
We construct to this effect the following tower of fields:
S = S, c SnT1 c Snm2 . . . c S, c S, = E
by the definition
( 4 ) sn = s; si = s(xi+l ˜xi+*˜...˜xn) = si+l (xi+l 1.
It would be sufficient to prove ( Sir/S, ) < i or that the generator xi

for Sir out of S satisfies an equation of degree i with coefficients
in S, .
Such an equation is easily constructed. Put
Fi+, (t)
f(t)
(5) Fi (t> =
(txi+l )(tXi+* ). . . (tx,) = (txi+l )
and Fn ( t ) = f ( t ). Performing the division we see that Fi (t ) is a
polynomial in t of degree i whose highest coefficient is 1 and whose
coefficients are polynomials in the variables
a1 , azF . . . , a,, and xi+r , xi+2 , . . . , x,. Only integers enter as coefficients
in these expressions. Now xi is obviously a root of Fi (t ) = 0.
xn) be a polynomial in x1,x2,. . . , xn.
Nowletg(x,,x,,...,
Since Fr ( xr ) = 0 is of first degree in xr , we cari express x, as a
polynomial of the a, and of x2, x3, . . . , xn . We introduce this expression
ing(x,,x,,..., xn). Since F, (x2 ) = 0 we cari express x2 or higher
41
powers as polynomials in x3, . . . , xn and the ai. Since F, ( xg) = 0
we cari express xi and higher powers as polynomials of x4, x5,. . . , x,,
and the ai. Introducing these expressions in g( xi, x2,. . . , xn) we see
that we cari express it as a polynomial in the xr, and the ar, such that
the degree in xi is below i. SO g( xi, x2, . . . , xn) is a linear combination
of the following n ! terms:
Un
Vl V2
where each vi 5 i  1.
x2 . . . xn
(6) x1
The coefficients of these terms are polynomials in the ai . Since the
expressions (6) are linearly independent in S (this is our previous
result), the expression is unique.
This is a generalization of the theorem of symmetric functions in
its usual form. The latter says that a symmetric polynomial cari be
written. as a polynomial in ai, a,, . . . , a,,. Indeed, if g(x,, . . . ,x,) is
symmetric we have already an expression as linear combination of the
terms (,6) where only the term 1 corresponding to vi = v2 = . . . = v,, = 0
has a coefficient # 0 in S, namely, g( xi,. . . , x,.,). SO g( xi, x2,. . . , xn)
is a polynomial in a,, a2,. . . , a,.
Hut our theorem gives an expression of any polynomial, symmetric
or not.
H. Normal Extensions.
An extension field E of a field F is called a normal extension if
the group G of automorphisms of E which leave F fixed has F for its
fixed field, and (E/F) is finite.
Mthough the result in Theorem 13 cannot be sharpened in general,
42
there is one case in which the equality sign Will always occur, namely,
in the case in which CT˜, 02, . . . , an is a set of automorphisms which
form a group. We prove
THEOREM 14. If ol> g2>. . . , o,, is a group of automorphisms of a
field E and if F is the fixed field of g1 ,u*,. . . , on, then (E/F) = n.
 
Ifa,,o,,..., is a group, then the identity occurs, say, u1 = 1.
un
The fixed field consists of those elements x which are not moved by
any of the U™S, i.e., ai(x) = x, i = 1,2, . . . n. Suppose that (E/F ) > n.
Then there exist n + 1 elements al, a*, . . . , a,,, of E which are
linearly independent with respect to F. By Theorem 1, there exists a
nontrivial solution in E to the system of equations
x1 q (a1 > + x2 q (a, > + . . . + xn+, q (a,,, > = 0
x1 u2 (a1 > + x2 o2 (a2 > + . . . + xn+l a2 ( a,+l > = 0
(™ >
..................................
xlun(al) + x2un(a2) + . . . + xn+lun(an+l> = 0
We not.e that the solution cannot lie in F, otherwise, since u1 is the
identity, the first equation would be a dependence between a,, . . . , a n+l .
Among a11 nontrivial solutions x1, x2, . . . , x,+˜ we choose one
which has the least number of elements different from 0. We may sup
pose this solution to be al, a*, . . . , a,, 0, . . . , 0, where the first r
terms are different from 0. Moreover, r # 1 because a, u1 (a1 ) = 0
implies a1 = 0 since u1 (a, ) = a1 f 0. Also, we may suppose ar = 1,
since if we multiply the given solution by a;™ we obtain a new solution
in which the rth term is 1. Thus, we have
43
(*> alUi(al > + a2Ui(a2 > + a. + ar.,ui(a,_l > + oi(ar> = 0
for i = 1,2, . . . , n. Since ai, . . . , ar_i cannot a11 belong to F, one of
these, say a 1, is in E but not in F. There is an automorphism ok for
which o,,( a, ) # a1 . If we use the fact that oi, 02, . . . , o,, form a group,
we see ok. gi, ok* 02,. . . ,ak. an is a permutation of Oi,oz,. .  tu,.
Applying ok to the expressions in (*) we obtain
> + *f. + ˜˜(arl)˜˜Qj(a,l) + =0
Ukk(a,).UkQj(al @kUj(ar)
for j = ˜l,2,. . . , n, that from okaj = (T.1
SO
(**> Uk(al >Ui(al> + . ..+ Uk(arl)Ui(a,l > + Ui(a, > = 0
and if we subtract ( * * ) from ( * ) we have
[a,  uk(al >l. Qi(Q1 > + . .. + [arl  uk(arel >l˜i(Qrl) = 0
which is a nontrivial solution to the system (™ ) having fewer than r
elements different from 0, contrary to the choice of r.
Corollary 1. If F is the fixed field for the finite group G, then

each automorphism u that leaves F fixed must belong to G.
(E/F) = order of G = n. Assume there is a u not in G. Then F
would remain fixed under the n + 1 elements consisting of u and the
elements of G, thus contradicting the corollary to Theorem 13.
Corollary 2. There are no two finite groups G, and G, with the
same fixed field.
This follows immediately from Corollary 1.
If f(x) is a polynomial in F, then f(x) is called separable if its
irreducible factors do not have repeated roots. If E is an extension of
44
the field F, the element a of E is called separable if it is root of a
separable polynomial f(x) in F, and E is called a separable extension
if each element of E is separable.
THEOREM 15. E is a normal extension of F if and only if E is
the splitting field of a separable polynomial p(x) in F.
 
Sufficiency. Under the assumption that E splits p (x) we prove
that E is a normal extension of F.
If a11 roots of p(x) are in F, then our proposition is trivial, since
then E = F and only the unit automorphism leaves F fixed.
Let us suppose p(x) has n > 1 roots in E but not in F. We make
the inductive assumption that for a11 pairs of fields with fewer than n
roots of p(x) outside of F our proposition holds.
Let p(x) = pr(x).p,(x).. . . . pr( x) be a factorization of p(x)
into irreducible factors. We may suppose one of these to have a degree
greater than one, for otherwise p(x) would split in F. Suppose deg
p,(x) = s > 1. Let ai be a root of p,(x). Then (F(a, )/F) = deg p,(x) = s.
If we consider F (ai ) as the new ground field, fewer roots of p( x) than
n are outside. From the fact that p(x) lies in F(a, ) and E is a split
ting field of p(x) over F( a1 ), it follows by our inductive assumption
that E is a normal extension of F(a, ). Thus, each element in E which
is not in F(a, ) is moved by at least one automorphism which leaves
F(a, ) fixed.
p (x) being separable, the roots ai ,aZ, . . . , as of pi (x) are a
distinct elements of E. By Theorem 8 there exist isomorphisms
45
Ul>ff2>. f *, os mappingF(a,)onF(a,), F(a,),...,F(a,),
respect.ively, which are each the identity on F and map or on
respectively. We now apply Theorem 10. E is a splitting
a1,a2,...,as
field of p(x) in F(a, ) and is also a splitting field of p(x) in F(ai ).
Hence, the isomorphism oi, which makes p( x ) in F ( a1 ) correspond to
the same p(x) in F( ai ), cari be extended to an isomorphic mapping of
E onto :E, that is, to an automorphism of E that we denote again by oi.
Hence, u1,02,. . . , os are automorphisms of E that leave F fixed and
map a1 ont0 a1,a2,. . . a,.
Now let 8 be an element that remains fixed under a11 automor
phisms of E that leave F fixed. We know already that it is in F (a 1 )
and hence has the form
8 = CO + CIaI + C2 a: + . . . + csml asl
where the ci are in F. If we apply ui to this equation we get, since
Ui(0) = 8:
8 = C, + CIai + C2af + . . . + Csml a:™
The polynomial esr xs1 + csd2x sz + . . . + crx + (c Cl  0)
has therefore the s distinct roots or, 02, . . . , as. These are more than
its degree. SO a11 coefficients of it must vanish, among them c 0  8.
This shows 8 in F.
Necessity. If E is a normal extension of F, then E is spiitting
field of a separable polynomial p(x). We first prove the
˜ If E is a normal extension of F, then E is a separable
Lemma.
extension of F. Moreover any element of E is a root ofan equation over
F which splits completely in E.
46
Let (5, , u2, . . . , o, be the group G of automorphisms of E whose
fixed field is F. Let a be an element of E, and let Q, a2, a3, . . . ,cr be
the set of distinct elements in the sequence oi( a), O˜(U), . . . , os(c).
Since G is a group,
Uj(Qi) = Uj(c7,(U)) = ajo&) = a,(u) = a,.
Therefore, the elements u,u2, . . . , cr are permuted by the automorphisms
of G. The coefficients of the polynomial f(x) = (˜a)( *). . . (xa,)
XU
are left fixed by each automorphism of G, since in its factored form the
factors of f(x) are only permuted. Since the only elements of E which
are left fixed by a11 the automorphisms of G belong to F, f(x) is a
polynomial in F. If g(x) is a polynomial in F which also has a as root,
then applying the automorphisms of G to the expression g (u ) = 0 we
obtain g(a,) = 0, that the degree of g(x) > s. Hence f(x) is irre
SO
ducible, and the lemma is established.
TO complete the proof of the theorem, let or, 02,. . . , ut be a gen
erating system for the vector space E over F. Let fi(x) be the separable
polynomial having oi as a root. Then E is the splitting field of
p ( x ) = f,(x).f,(x).....f,(x).
If f(x) is a polynomial in a field F, and E the splitting field of
f (x ), then we shall cal1 the group of automorphisms of E over F the
˜ the equation f(x) = 0. We corne now to a theorem known in
group of
algebra as the Fundamental Theorem of Galois Theory which gives the
relation between the structure of a splitting field and its group
of automorphisms.
THEOREM 16. (Fundamental Theorem). If p(x) is a separable
polynomial in a field F, and G the group of the equation p(x) = 0
where E is the
  
47
splitting field of p(x), then: (1) Each intermediate field, B,is the
fixed field for a subgroup G, of G, and distinct subgroups have dis
tinct fixed fields. We say B and G, “belong” to each other. (2) The
intermediate field B is a normal extension of F if a,nd only if the sub
group G, is a normal subgroup of G. In this case the group of automor
phisms of B which leaves F fixed is isomorphic to the factor group
(G/G ). (3) For each intermediate field B, we have (B/F) = index of

G, and (E/B) = order of G,.
The first part of the theorem cornes from the observation that E
is splitting field for p(x) when p(x) is taken to be in any intermediate
field. Hence, E is a normal extension of each intermediate field B, SO
that B is the fixed field of the subgroup of G consisting of the automor
phisms which leave B fixed. That distinct subgroups have distinct fixed
fields is stated in Corollary 2 to Theorem 14.
Let B be any intermediate field. Since B is the fixed field for
the subgroup G, of G, by Theorem 14 we have (E/B) = order of G,.
Let us cal1 o(G) the order of a group G and i(G) its index. Then
o(G) =: o(GaBut (E/F) = o(G), and (E/F) = (E/B)*(B/F)
from which (B/F) = i (G, ), which proves the third part of the theorem.
The number i( G, ) is equal to the number of left cosets of G,.
The elements of G, being automorphisms of E, are isomorphisms of B;
that is, they map B isomorphically into some other subfield of E and
are the identity on F. The elements of G in any one coset of G, map B
in the same way. For let U. a1 and o. o2 be two elements of the coset
uG,. Since or and o2 leave B fixed, for each a in B
48
we have (ror( a ) = a( cz ) = oo2( a ). Elements of different cosets give
different isomorphisms, for if o and r give the same isomorphism,
o(a) = r(c) for each a in B, then o˜r(a) = a for each a in B. Hence,
dT == al> where or is an element of G,. But then T = oo, and
rG, := oorGs= aG, SO that o and r belong to the same coset.
Each isomorphism of B which is the identity on F is given by an
automorphism belonging to G. For let Q be an isomorphism mapping B
on B™ and the identity on F. Then under c, p(x) corresponds to p(x),
and E is the splitting field of p(x) in B and of p(x) in B ™ . By
Theorem 10, a cari be extended to an automorphism o™ of E, and since
0™ leaves F fixed it belongs to G. Therefore, the number of distinct
isomorphisms of B is equal to the number of cosets of G, and is there
fore equal to (B/F).
The field aB onto which g maps B has obviously oGso1 as cor
responding group, since the elements of aB are left invariant by
precisely this group.
If 13 is a normal extension of F, the number of distinct automor
phisms of B which leave F fixed is (B/F) by Theorem 14. Conversely,
if the number of automorphisms is (B/F) then B is a normal extension,
because if F™ is the fixed field of a11 these automorphisms, then
F C F™ (1 B, and by Theorem 14, (B/F ˜) is equal to the number of
automorphisms in the group, hence (B/F ˜) = (B/F). From ( B/F) =
(B/F™)(F™/F) we have (F™/F) = 1 or F = F™. Thus, B is a normal
extension of F if and only if the number of automorphisms of B is (B/F).
B is a normal extension of F if and only if each isomorphism of
B into E is an automorphism of B. This follows from the fact that each
of the above conditions are equivalent to the assertion that there are
49
the same numberof isomorphisms and automorphisms. Since, for each
u, B = aB is equivalent to oG,o * C G,, we cari finally say that B is
a normal extension of F and only if G, is a normal subgroup of G.
As we have shown, each isomorphism of B is described by the
effect of the elements of some left coset of G,. If B is a normal exten
sion these isomorphisms are a11 automorphisms, but in this case the
cosets are elements of the factor group (G/G, ). Thus, each automor
phism of B corresponds uniquely to an element of ( G/G, ) and con
versely. Since multiplication in ( G/G, ) is obtained by iterating the
mappings, the correspondence is an isomorphism between (G/G, ) and
the group of automorphisms of B which leave F fixed. This completes
the proof of Theorem 16.
1. Finite F i e l d s .
It is frequently necessary to know the nature of a finite subset
of a field which under multiplication in the field is a group. The
answer to this question is particularly simple.
THEOREM 17. If S is a finite subset (f 0 ) of a field F which

is a group under multiplication in F, then S is a cyclic group.
˜˜
The proof is based on the following lemmas for abelian groups.
L,emma 1. If in an abelian group A and B are two elements of

˜ b, and if c is the least common multiple of a and b, then
orders a and
˜˜element C of order c in the group.
there is an
50
Proof: (a) If a and b are relatively prime, C = AB has the re
quired order ab. The order of C a = B” is b and therefore c is divisible
by b. Similarly it is divisible by a. Since Ceb = 1 it follows c = ab.
(b) If d is a divisor of a, we cari find in the group an element of
order d. Indeed Aa/d is this element.
(c) Now let us consider the general case. Let pr, pz, . . . , p, be
the prime numbers dividing either a or b and let
a = p, “1 pz n2
. . . prnr
b = pimlpzm;, . ..prm™.
Cal1 ti the larger of the two numbers ni and 1. Then
t
t1 t2
c = p* p2 . . . prr*.
According to (b) we cari find in the group an element of order pni and
one of order pi?. Thus there is one of order pi ti. Part (a) shows that
the product of these elements Will have the desired order c.
Lemma 2. If there is an element C in an abelian group whose

order c is maximal (as is always the case if the group is finite) then c
 
is divisible by the order a of every element A in the group; hence
 
xc = 1 is satisfied by each element in the group.

Proof: If a does not divide c, the greatest common multiple of a
and c would be larger than c and we could find an element of that order,
thus contradicting the choice of c.
We now prove Theorem 17. Let n be the order of S and r the
largest order occuring in S. Then x™  1 = 0 is satisfied for a11 ele
51
ments of S. Since this polynomial of degree r in the field cannot have
more than r roots, it follows that r  n. On the other hand r  n be
> <
cause the order of each element divides n. S is therefore a cyclic
group consisting of l,˜, ˜2,. . . , c”I where 6” = 1.
Theorem 17 could also have been based on the decomposition
theorem for abelian groups having a finite number of generators. Since
this theorem Will be needed later, we interpolate a proof of it here.
Let G be an abelian group, with group operation written as +.
The element g,, . . . , g, Will be said to generate G if each element g of
G cari be written as sum of multiples of g,, . . . , g,, g = n,g, + . . . + nkgk.
If no set of fewer than k elements generate G, then g,, . . . , G Will be
called a minimal generating system. Any group having a finite genera
ting system admits a minimal generating system. In particular, a finite
group al.ways admits a minimal generating system.
F r o m t h e i d e n t i t y nl( g, + mg,) + (n2  n,m)g2 = n,g, + n2g,
it follows that if g,, g,, . . . , g, generate G, also g, + mg,,
g,, *. . >g, generate G.
An equation m,g, + m,g, + . . . + m,g, = 0 Will be called a re
lation between the generators, and ml, . . . , mk Will be called the co
efficients in the relation.
We shall say that the abelian group G is the direct product of its
subgroups G,, G,, . . . , G, if each g E G is uniquely representable as a
s u m g = x1 + x2 + . . . + x,,wherexi E Gi,i = l,..., k..
52
Decomposition Theorem. Each abelian group having a finite num

ber of generators is the direct product of cyclic subgroups G,, . . . , G,
˜˜
where the order of Gi divides the order of Gi+i, i = 1, . . . , nl and n is
˜
the number of elements in a minimal generating system. ( Gr, Gr+i , . . . , Gn
may each be infinite, in which case, to be precise,
= 1,2,...,r2).
O(Gi)lO(Gi+,)fori
We assume the theorem true for a11 groups having minimal genera
ting systems of kl elements. If n = 1 the group is cyclic and the
theorem trivial. Now suppose G is an abelian group having a minimal
generating system of k elements. If no minimal generating system satis
fies a nontrivial relation, then let g,, g,, . . . , g, be a minimal generating
system and G,,G,, . . . , G, be the cyclic groups generated by them.
For each g 6 G, g = n,g, + . . . + nkgk where the expression is
uniqu.e; otherwise we should obtain a relation. Thus the theorem would
be true. Assume now that some nontrivial relations hold for some mini
mal generating systems. Among a11 relations between minimal genera
ting systems, let
m,g, + . . . + mkg, = 0
(1)
be a relation in which the smallest positive coefficient occurs. After
an eventual reordering of the generators we cari suppose this coefficient
to be mi. In any other relation between g,, . . . , g,.
ni g, + . . . + nkgk = 0
(2)
we must have mi/ni. Otherwise n 1 = qmi + r, 0 < r < mi and q times
relation (1) subtracted from relation (2) would yield a relation with a
coefficient r < mi. Also in relation (1) we must have m,/m,, i = 2,. . . , k.
53
For suppose mi does not divide one coefficient, say m, . Then
= qm, + r, 0 < r < mr. In the generating system
m2
g, + g,, k$˜...> g, we should have a relation
mi( g, + qg,) + rg, + m,g, + . . . + mkq, = 0 where the coefficient
r contradicts the choice of mi. Hence m2 = q2m1, m3 = q,m,, . . . , mk = q,m,.
The system & = g, + q,g, + . . . + qkgk, g,, . . . , g, is minimal gen
erating, and m,gr = 0. In any relation 0 = n,Fi + n2g2 + . . . + nkgk
since mr is a coefficient in a relation between gi, g,, . . . , g, our pre
vious argument yields mr / nr , and hence nr gr = 0.
Let G™ be the subgroup of G generated by g,, . . . , g, and G, the
cyclic group of order m, generated by gr . Then G is the direct product
of G, and G™ . Each element g of G cari be written
g = nigi + n2g2 + . . . + nkg, = nrgr + g™.
The representation is unique, since n,g, + g™ = nr™gl + g” implies
the relation (nr  nr™)g, + (g™  g ” ) = 0 , h e n c e
(nl  ni )E, = 0, that nrgr = n;gi and also g™ = g”.
SO
E3y our inductive hypothesis, G ™ is the direct product of kl
cyclic groups generated by elements g2, ES, . . . , gk whose respective
orders t,, . . . , t, satisfy ti / ti+r , i = 2, . . . , kl. The preceding argu
ment applied to the generators gr, g2, . . . , g, yields m, j t,, from which
the theorem follows.
I3y a finite field is meant one having only a finite number
of elements.
Corollary. The nonzero elements of a finite field form a cyclic
˜˜
group.
If a is an element of a field F, let us denote the nfold of a, i.e.,
54
the element of F obtained by adding a to itself n times, by na. It is ob
vious that n.(m.a) = (nm).a and(n.a)(m.b) = nmeab. If for one
element a f 0, there is an integer n such that na a = 0 then n. b = 0
foreachbinF,sincen.b=(n.a)(a˜b)=O.a™b=O..Ifthereisa
positive integer p such that p. a = 0 for each a in F, and if p is the
smallest integer with this property, then F is said to have the charac
teristic  p. If no such positive integer exists then we say F has charac
teristic 0. The characteristic of a field is always a prime number. for if
p = r.s t h e n p a = rs.a = r.(s.a). H o w e v e r , s.a = b # Oif a f 0
and r b + 0 since both r and s are less than p, that pa f 0 contrary
SO
to the definition of the characteristic. If na = 0 for a f 0, then p divides
n, for n = qp + r where 0  r < p and na = (qp + r)a = qpa + ra.
<
Hence na = 0 implies ra = 0 and from the definition of the characteristic
since r < p, we must have r = 0.
If F is a finite field having q elements and E an extension of F
such that (E/F) = n, then E has q” elements. For if or, 02,. . . ,W n is
a basis of E over F, each element of E cari be uniquely represented as
a linear combination xiwr + xZoZ + . . . + x,w, where the xi belong to
F. Since each xi cari assume q values in F, there are qn distinct possi
ble choices of x1, . . . , xn and hence qn distinct elements of E. E is
finite,, hence, there is an element a of E that E = F(a). (The non
SO
zero elements of E form a cyclic group generated by a).
5 [0,1,2 ,..., pl] the set of multiples of the
IfwedenotebyP
unit element in a field F of characteristic p, then P is a subfield of F
having p distinct elements. In fact, P is isomorphic to the field of
integers reduced mod p. If F is a finite field, then the degree of F over
55
P is finite, say (F/P) = n, and F contains p* elements. In other
words, the order of any finite field is a power of its characteristic.
If F and F™ are two finite fields having the same order q, then
by the preceding, they have the same characteristic since q is a power
of the characteristic. The multiples of the unit in F and FI form two
fields P and P™ which are isomorphic.
The nonzero elements of F and F™ form a group of order ql
and, therefore, satisfy the equation xql  1 = 0. The fields F and FI
are splitting fields of the equation x ql = 1 considered as lying in P
and P™ respectively. By Theorem 10, the isomorphism between P and
P ™ cari be extended to an isomorphism between F and F ™ . We have thus proved
THEOREM 18. Two finite fields having the same number of ele
ments are isomorphic.
˜
gifferentiation. If f(x) = ao + six + . . . + anxn i s a poly
nomial in a field F, then we define f™ = a, + 2a,x + . . . + nanx”i .
The reader may readily verify that for each pair of polynomials f and
g we have
(f + g)™ = f™ + g™
= fg™ + gf™
(f g>™
(f^)™ = nf”˜. f™
THEOREM 19. The polynomial f has repeated roots if and only

if in the splitting field E the polynomials f and f™ have a common
root. This condition is eauivalent to the assertion that f and f™ have a
56
common factor of degree greater than 0 in F.
˜
If a is a root of multiplicity k of f(x) then f = (xa )kQ (x) where
Q(a) f 0. This gives
f ™=( )kQ ™ ( ) + k ( xa ) k1 Q ( ) = ( xa ) k1 [ ( ) Q™ ( ) + kQ ( x ) 1.
xa x x xa x
If k > 1, then a is a root of f™ of multiplicity at least kl. If
k = 1, then f™(x) = Q(x) + (xa)Q™(x) and f™(a) = Q(a) f; 0. Thus,
f and f™ have a root a in common if and only if a is a root of f of
multiplicity greater than 1.
If f and f™ have a root a in common then the irreducible polynomial
in F having a as root divides both f and f™ . Conversely, any root of a
factor common to both f and f™ is a root of f and f ™ .
˜ If F is a field of characteristic 0 then each irreducible
Corollary.
˜˜ in F is separable.
polynomial
Suppose to the contrary that the irreducible polynomial f(x) has
a root a of multiplicity greater than 1. Then, f ™ (x) is a polynomial
which is not identically zero (its leading coefficient is a multiple of
the leading coefficient of f(x) and is not zero since the characteristic
is 0) and of degree 1 less than the degree of f(x). But a is also a root
of f™ (x) which contradicts the irreducibility of f(x).
J. Roots  of Unity.

If F is a field having any characteristic p, and E the splitting
field of the polynomial x D  1 where p does not divide n, then we
shall refer to E as the field generated out of F by the adjunction of a
˜˜ nth root of unity.
primitive
The polynomial x” ˜ 1 does not have repeated roots in E, since
its derivative, nx”™ , has only the root 0 and has, therefore, no roots
57
in common with x n  1. Thus, E is a normal extension cf F.
IfE,,t* ,..., Cn are the roots of x n  1 in E, they form a group under
multiplication and by Theorem 17 this group Will be cyclic. If
1 ,C,C2,a. .,c n1 are the elements of the group, we shall cal1 c a primi
tive n th root of unity. The smallest power of E which is 1 is the n th.
THEOREM 20. If E is the field generated from F by a primitive
˜
nth root of unity, then the group G of E over F is abelian for any n and
˜˜
˜ a prime number.
cyclic if n is
We have E = F(É), since the roots of x”  1 are powers of 6.
Thus, if o and r are distinct elements of G, o(c) f r(e). But ˜(6) i s a
root of xn  1 and, hence, a power of 6. Thus, o(c) = c ““where no
is an integer 1 5 no < n. Moreover, ru(˜) = r( r”o) = (r(t)) “ o =
cnr “u = or(˜). T h u s , nor = snr mod n. Thus, the mapping of o on no
is a homomorphism of G into a multiplicative subgroup of the integers
mod n. Since r f o implies T(E) f o(c), it follows that r f o implies
“a f nr mod n. Hence, the homomorphism is an isomorphism. If n is a
prime number, the multiplicative group of numbers forms a cyclic group.
K. Noether Equations.
If E is a field, and G = ( CJ, r, . . .) a group of automorphisms of E,
any set of elements x,, xr , . . . in E Will be said to provide a solution to
Noether™s equations if xo . D (xr ) = xor for each o and r in G. If one
˜˜
element x0 = 0 then xr = 0 for each r 6 G. As T traces G, or assumes
a11 values in G, and in the above equation xor = 0 when xo = 0. Thus,
in any solution of the Noether equations no element xo = 0 unless the
solution is completely trivial. We shall assume in the sequel that the
58
trivial solution has been excluded.
THEOREM 21. The system xo, xr , . . . is a solution to Noether™s

equations if and only if there exists an element a in E, such that
CL/(˜( a ) for each o.
=fJ
X
For any a, it is clear that xo = a/o( a) is a solution to the
equations, since
a/o(a)˜o(a/r(a>) = a/a(a>*o(a)/ui(a) = a/o7(a).
Conversely, let x0, xr , . . . be a nontrivial solution. Since the
automorphisms o, T , . . . are distinct they are linearly independent, and
the equation x0 ˜U(Z) + XrT(Z) + . . . = 0 does not hold identically.
Hence, there is an element a in E such that
xo.cr(a) + x,r(a) + . . . = a f 0. Applying u to a gives
o(a) = T,CG a(x,).m(a>.
Multiplying by ˜0 gives
xu.u(u) = C xoo(x,).or(a>.
7CG
Replacing xc. c( xr ) by b and noting that or assumes a11 values in
G when r does, we have
Xrlr . o(a) = C x,r(a) = a
TEG
that
SO
xu = a/o(a).
A solution to the Noether equations defines a mapping C of G
into E, namely, C(a) = x,,. If F is the fixed field of G, and the ele
ments xc lie in F, then C is a character of G. For
C(OT)= xm = x,.a(x,) = x0x7= C(a).C(r)sinceo(x,) = x,if
xr c F. Conversely, each character C of G in F provides a solution
59
to the Noether equations. Cal1 C(o) = xo. Then, since xr 6 F,
we have a( xr) = xr . Thus,
xu .a(s:,) == xoexr = C(˜)*C(T) = C ( o r ) = xor. W e t h e r e f o r e h a v e ,
by combining this with Theorem 21,
THEOREM 22. If G is the group of the normal field E over F,
then for each character C of G into F there exists an element Q in E
such that C(a) = a/o(a) and, conversely, if a/o( a) is in F for each
˜˜
(T, then C ((7) = a/o( a ) is a character of G. If r is the least common
a™ c
multiple of the orders of elements of G, then F.
˜˜
We have already shown a11 but the last sentence of Theorem 22.
o(ar)
TO prove this we need only show = a™ for each o E G. But
ar/a(ar) = (a/a(a))= = (C(o))r = C(or) = C(1) = 1.
L. Kummer™s Fields.
If F contains a primitive nth root of unity, any splitting field E
of a polynomial (x”  a,)(x”  a*). . .(x”  ar) where ai 6 F for
i = 1,2,. . . , r Will be called a Kummer extension of F, or more briefly,
a .z
Kummer field
If a field F contains a primitive nth root of unity, the number n
is not divisible by the characteristic of F. Suppose, to the contrary, F
has characteristic p and n = qp. Then yn  1 = (y  1 )n since in the
expansion of (y  1 )P each coefficient other than the first and last is
divisible by p and therefore is a multiple of the pfold of the unit of F
and thus is equal to 0. Therefore x”  1 = (x q)P  1 = (x4  1 )n
and x”  1 cannot have more than q distinct roots. But we assumed
that F has a primitive n th root of unity and 1, 6,˜ 2,. . . , E n1 would be
60
n distinct roots of x”  1. It follows that n is not divisible by the
characteristic of F. For a Kummer field E, none of the factors
 ai, ai f 0 has repeated roots since the derivative, nx”l , has
X”
only the root 0 and has therefore no roots in common with xn  ai.
Therefore, the irreducible factors of x”  a, are separable, that 
E
SO
is a normal extension of F.
Let ai be a root of x”  ai in E. If ci, c2, . . . , tn are the n dis
tinct nth roots of unity in F, then ai+,, aic2, . . . , a,˜,, Will be n distinct
roots of x”  ai, and hence Will be the roots of x”  a,, that
SO
E = F(a,,a,,..., ar). Let o and r be two automorphisms in the group
G of E over F. For each ai, both o and r map ai on some other root of
x”  ai. Thus r(ai) = ciTai and o(ai) = cio ai where Q, and tir are nth
roots of unity in the basic field F. It follows that
= T(tiaai) = ciuT = tiocirai = o(T(ai). Since oand r
r(o(ai))
are commutative over the generators of E, they commute over each ele
ment of E:. Hence, G is commutative. If ˜7 c G, then o(ai) =
a'(a,) = eio 2ai, etc. Thus, oni = ai for ni such that
ciGai,
“i
= 1. Since the order of an n th root of unity is a divisor of n, we
%J
have ni a divisor of n and the least common multiple m of nr, n2,. . . , nr
is a divisor of n. Since o ya,) = ai for i = 1,2,. . . , r it follows that m
is the order of CJ. Hence, the order of each element of G is a divisor of
n and, therefore, the least common multiple r of the orders of the ele
ments of G is a divisor of n. If c is a primitive nth root of unity, then
en/™ is a primitive r th root of unity. These remarks cari be summarized
in the following.
61
THEOREM 23. If E is a Kummer field, i.e., a splitting field of

p(x) = (x”  a,)(x”  az). . .(x”  ar) whete a, lie in F, and F
˜
contains a primitive nth root of unity, then: (a) E is a normal extension
˜˜
˜ group G of E over F is abelian, (c) the least common multi
of F; (b) the
ple of the orders of the elements of G is a divisor of n.
˜˜
Corollary. If E is the splitting field of xp  a, and F contains a
 
primitive pth root of unity where p is a prime number, then either E = F
˜˜
and xp  a is split in F, or xp  a is irreducible and the group of E
over F is cyclic of order p.
The order of each element of G is, by Theorem 23, a divisor of p
and, hence, if the element is not the unit its order must be p. If Q is a
root of xp  a, then a,ey, . . . ,cplc are a11 the roots of xP  a that
SO
F(a ) = E and ( E/F) <  . Hence, the order of G does not exceed p
p SO
that if G has one element different from the unit, it and its powers must
constitute a11 of G. Since G has p distinct elements and their behavior
is determined by their effect on a, then a must have p distinct images.
Hence, the irreducible equation in F for a must be of degree p and is
therefore xp  a = 0.
The properties (a), (b) and (c) in Theorem 23 actually characterize
Kummer fields.
Let us suppose that E is a normal extension of a field F, whose
group G over F is abelian. Let us further assume that F contains a
primitive r th root of unity where r is the least common multiple of the
orders of elements of G.
The group of characters X of G into the group of r th roots of
62
unity is isomorphic to G. Moreover, to each cr 6 G, if 0 f 1, there exists
a character C 6 X such that C (0) f 1. Write G as the direct product of
the cyclic groups G,, G,, . . . , G, of orders mi / m2 / . . . / m,. Each u E G
may be w r i t t e n D = 0r 0z v2 . . .0r Vt . Cal1 Ci the character sending cri
Vl
into ci, a primitive m.Ith root of unity and 5 into 1 for j f i. Let C be
any character. C(oi) = cri, then we have C = CFi . C 2 i. . . C,™ i.
Conversely, C P1 . . . Ct Pr defines a character. Since the order of Ci is
1
mi, the character group X of G is isomorphic to G. If 0 f 1, then in
u=o 1 9 u2 1.˜2 . ..u t Vt at least one say vr , is not divisible by mi.
vi,
T h u s Cr( u) = civl f 1.
Let A denote the set of those nonzero elements cz of E for which
a™ 6 F and let Fi denote the nonzero elements of F. It is obvious that
A is a multiplicative group and that Fi is a subgroup of A. Let A™ de
note the set of rth powers of elements in A and Fi the set of r th powers
of elements of Fi. The following theorem provides in most applications
a convenient method for computing the group G.
THEOREM 24. The factor groups (A/F, ) and (A™/F ;) are iso
morphic to each other and to the groups G and X.
a™ 6
We map A on A™ by making a 6 A correspond to A™. If a™ t Fi,
where a 6 F, then b c A is mapped on ar if and only if br = a=, that is,
if b is a solution to the equation x™  a™ = 0. But a, ca, c2a,. . . , cr™ a
are distinct solutions to this equation and since 6 and a belong to Fi,
it follows that b must be one of these elements and must belong to Fi.
Thus, the inverse set in A of the subgroup Fi of A™ is Fi , that the
SO
factor groups (A/F, ) and (A™/F; ) are isomorphic.
63
If a is an element of A, then (a/o(˜))˜ = ar/o(a*) = 1. Hence,
a/o(a) is an rth root of unity and lies in F 1. By Theorem 22, ˜/a (a )
defines a character C (0) of G in F. We map a on the corresponding
character C. Each character C is by Theorem 22, image of some a.
Moreover, u . a ™ is mapped on the character C * (0) =
a.a™/o(a.a™) = a.a™/o(a>a(a™> = C(o>.C™(a> = C.C™(u), SO
that the mapping is homomorphism. The kernel of this homomorphism
is the set of those elements a for which a/a(a) = 1 for each o, hence
is Fi. It follows, therefore, that (A/F, ) is isomorphic to X and hence
also to G. In particular, (A/F, ) is a finite group.
We now prove the equivalence between Kummer fields and fields
satisfying (a), (b) and (c) of Theorem 23.
THEOREM 25. If E is an extension field over F, then E is a
Kummet field if and only if E is normal, its group G is abelian and F
contains a primitive rth root 6 of unity where r is the least common
multinle of the orders of the elements of G.
The necessity is already contained in Theorem 23.. We prove the
sufficiency. Out of the group A, let a,F,,a,F,, . . . ,a,F, be the cosets
of Fi. Since ai E A, we have af = ai 6 F. Thus, ai is a root of the
equation x™  ai = 0 and since coi, c2ai, . . . , c'*ai are also roots,
 ai must split in E. We prove that E is the splitting field of
X™
(x™  a1 )(x™  a2 ). . . (x™  a,) which Will complete the proof of the
theorem. TO this end it suffices to show that F ( ai, a2, . . . , ut) = E.
64
Suppose that F(a,,a*, . . . ,a,) f E. Then F(a,, . . . ,a,) is an
intermediate field between F and E, and since E is normal over
F ( a , , . . . , ut) there exists an automorphism o 6 G, D f 1, which
leavesF(a,,..., a,) fixed. There exists a character C of G for which
C(a) f 1. Finally, there exists an element a in E such that