Moreover, A C F(a,, . . . ,a, ) since a11 the cosets a,F, are contained
in F(a,, . . . , a,). Since F(a,, . . . ,a,) is by assumption left fixed by
o, a(a) = a which contradicts a/o(a) f 1. It follows, therefore, that
F(a,,...,a,) = E.
˜ If E is a normal extension of F, of prime order p, and
Corollary.
if F contains a primitive pth root of unity, then E is splitting field of

an irreducible polynomial xn  a in F.
˜˜
E is generated by elements ai, . . . ,a,, where ay 6 F. Let ai be
not in F.. Then xp  a is irreducible, for otherwise F(a 1 ) would be an
intermediate field between F and E of degree less than p, and by the
product theorem for the degrees, p would not be a prime number, con
trary to assumption. E = F ( a1 ) is the splitting field of xp  a.
M. Simple Extensions.
We consider the question of determining under what conditions
an extension field is generated by a single element, called a primitive.
We prove the following
THEOREM 26. A finite extension E of F is primitive over F if

65
and only if there are only a finite number of intermediate fields.
˜
(a) Let E = F(a) and cal1 f(x) = 0 the irreducible equation for
a in F. Let B be an intermediate field and g(x) the irreducible equa
tion for Q in B. The coefficients of g(x) adjoined to F Will generate a
field B ™ between F and B. g ( x ) is irreducible in B, hence also in B ™ .
Since 15 = B™(a) we see (E/B) = (E/B™). This proves B™ = B. SO B
is uniquely determined by the polynomial g(x). But g(x) is a divisor
of f(x), and there are only a finite number of possible divisors of f(x)
in E. Hence there are only a finite number of possible B™s.
(b) Assume there are only a finite number of fields between E
and F. Should F consist only of a finite number of elements, then E is
generated by one element according to the Corollary on page 53. We
may therefore assume F to contain an infinity of elements. We prove:
TO any two elements a, p there is a y in E such that F(a, /? ) = F (y).
Let y = a + ap with a in F but for the moment undetermined. Con
sider a11 the fields F(y) obtained in this way. Since we have an
infinity of a™s at our disposal, we cari find two, say a, and a2, such
that the corresponding y™s, yr = a + a,/3 and y2 = a + a,@, yield
the same field F ( y1 ) = F ( y2 ). Since both y1 and y2 are in F ( y1 ),
their difference (and therefore 6) is in this field. Consequently also
Y1  a,P := a. SO F(a,P) C F(y,). Since F(y,) C F(a,B) our con
tention is proved. Select now I] in E in such a way that ( F ( T] )/F ) is
as large as possible. Every element E of E mustbe in F( ˜7) or else we
could fïnd an element 6 such that F(6) contains both 7 and 6. This
proves E = F(n).
66
THEOREM 27. If E = F(a,, a2, . . . ,a,) is a finite extension of

the field F, and ai, a*, . . . ,a, are separable elements in E, then there
exists a primitive 19 in E such that E = F (0).
 
Proof: Let fi(x) be the irreducible equation of ai in F and let B
be an extension of E that splits fi(x) fz( x) . . . f,(x). Then B is normal
over F and contains, therefore, only a finite number of intermediate
fields (as many as there are subgroups of G). SO the subfield E contains
only a finite number of intermediate fields. Theorem 26 now compietes
the proof.
N. Existence of a Normal Basis.
˜
The following theorem is true for any field though we prove it
only in the case that F contains an infinity of elements.
THEOREM 28. If E is a normal extension of F and hi, oz, . . . , on
 
are the elements of its group G, there is an element 8 in E such that
˜
the n elements o,(t9),a,(ti), . . . , un( 0) are linearly independent with
˜
resuect to F.
According to Theorem 27 there is an a such that E = F(U). Let
f(x) be the equation for a, put o,(a) = ci,
and gi(x) = o,(g(x>> = (xaf,\˜?ca.)
1 1
g i( x y) is a polynomial in E having ak as root for k f i and hence
g,(x)g,(x) = 0 (mod f(x)) for i f k.
(1)
In the equation
g*(x) + g,(x) + . . . + g,(x)  1 = 0
(2)
the left side is of degree at most n  1. If (2) is true for n different
values of x, the left side must be identically 0. Such n values are
67
ai) gk(ai ) = 0
al,a2,. . . ,a,, since gi( = 1 and for k f i.
Multiplying (2) by gi( x) and using (1) shows:
(3) (g,(X))™ = gi(X) (mod f(x))
We next compute the determinant
D ( x ) = loiok(g(x))l i , k = 1,2 ,..., n
(4)
and prove D(x) f 0. If we square it by multiplying column by column
and compute its value (mod f(x)) we get from (l), (2), (3) a determi
nant that has 1 in the diagonal and 0 elsewhere.
SO
(D(x))˜ = 1 ( m o d f ( x ) ) .
D (x 1) cari have only a finite number of roots in F. Avoiding them
we cari find a value a for x such that D(a) f 0. Now set 8 = g(a).
Then the determinant
(5) l”iak(e)l f O™
C:onsider any linear relation
x1a,(8) + xp*(8> + . . . + ˜,.,a,( 0) = 0 where the xi are in F. Apply
ing the automorphism oi to it would lead to n homogeneous equations for
the n unknowns xi. (5) shows that xi = 0 and our theorem is proved.
0. ˜˜ on Natural Irrationalities.
Theorem
L,et F be a field, p(x) a polynomial in F whose irreducible factors
are separable, and let E be a splitting field for p(x). Let B be an arbi
trary extension of F, and let us denote by EB the splitting field of p(x)
al, . . . ,as
when p(x) is taken to lie in B. If are the roots of p(x) in
EB, then F(a,, . . . , as) is a subfield of EB which is readily seen to
..., as )
form a splitting field for p(x) in F. By Theorem 10, E and F(a*,
68
are isomorphic. There is therefore no loss of generality if in the sequel
wetakeE = F(al,..., a,) and assume therefore that E is a subfield
o f E B . A l s o , E B = B ( a *,...> as).
Let us denote by E A B the intersection of E and B. It is readily
seen that E n B is a field and is intermediate to F and E.
THEOREM 29. If G is the group of automorphisms of E over F,
and H the group of EB over B, then H is isomorphic to the subgroup of
 
G having E n B as its fixed field.
Each automorphism of EB over B simply permutes al, . . . , as in
some fashion and leaves B, and hence also F, fixed. Since the ele
ments of EB are quotients of polynomial expressions in al, . . . , as with
coefficients in B, the automorphism is completely determined by the
permutation it effects on a 1, . . . , as. Thus, each automorphism of EB
over B defines an automorphism of E = F ( al, . . . , as ) which leaves F
fixed. Distinct automorphisms, since (x1,. . . , as belong to E, have
different effects on E. Thus, the group H of EB ovet B cari be con
sidered as a subgroup of the group G of E over F. Each element of H
leaves E n B fixed since it leaves even a11 of B fixed. Howevet, any
element of E which is not in E n B is not in B, and hence would be
moved by at least one automorphism of H. It follows that E r\ B is the
fixed field of H.
Corollaty. If, undet the conditions of Theorem 29, the gtoup G is of

prime order, then either H = G or H consists of the unit element alone.

69
III APPLICATIONS
by
A. N. Milgram
A. Solvable Groups.
Before proceeding with the applications we must discuss certain
questions in the theory of groups. We shall assume several simple propo
sitions: (a) If N is a normal subgroup of the group G, then the mapping
f(x) = xN is a homomorphism of G on the factor group G/N. f is called
the natural homomorphism. (b) Th e image and the inverse image of a
normal subgroup under a homomorphism is a normal subgroup. (c) If f
is a homomorphism of the group G on G™ , then setting N™ = f(N), and
defining the mapping g as g( xN ) = f(x) N ™ , we readily see that g is
a homomorphism of the factor group G/N on the factor group G ˜/NI .
Indeed, if N is the inverse image of N™ then g is an isomorphism.
We now prove
˜˜ (Zassenhaus). If U and V are subgroups of G, u and
THEOREM 1.
v normal subgroups of U and V, respectively, then the following three
factor groups are isomorphic: u (U nV) /II (U nv),
v(UnV)/V(UnV), (unv)/(unv)(vnu).
It is obvious that U n v is a normal subgroup of U n V. Let f
be the natural mapping of U on V/u. Cal1 f(UnV) = H and f(Unv) = K.
Then f˜(H) = u(UnV) and f“(K) = u(Unv) from which it follows
that u( UnV)/u( Unv) is isomorphic to H/K. If, however, we view f as
defined only over U n V, then f˜(K) = [un(UnV)](Unv) =
(unV)(Unv) that (UnV)/(unV)(Unv) is also isomorphic to H/K.
SO
70
Thus the first and third of the above factor groups are isomorphic to
each other. Similarly, the second and third factor groups are isomorphic.
Corollary 1. If H is a subgroup and N a normal subgroup of the

group G, then H/HnN is isomorphic to HN/N, a subgroup of G/N.
Proof: Set G = U, N = u, H = V and the identity 1 = v in
Theorem 1.
Corollary 2. Under the conditions of Corollary 1, if G/N is

abelian, SO also is H/HnN.
Let us cal1 a group G solvable if it contains a sequence of sub
groupsG = Go 1 G, I... 1 Gs = 1, each a normal subgroup of the
preceding, and with G,r /Gi abelian.
THEOREM 2. Any subgroup of a solvable group is solvable. For

let H be a subgroup of G, and cal1 Hi = HnG,. Then that Hir/H, is
abelian follows from Corollary 2 above, where Gi_r, Gi and Hi_, play
the role of G, N and H.
THEOREM 3. The homomorph of a solvable group is solvable.

Let f(G) = G™ , and define G 1 = f ( Gi) where Gi belongs to a
a sequence exhibiting the solvability of G. Then by (c) there exists a
homomorphism mappingGi_r/Gi on Gi_,/Gi . But the homomorphic image
of an abelian group is abelian that the groups GI exhibit the
SO
solvability of G™ .
B. Permutation Groups.

Any one to one mapping of a set of n abjects on itself is called
a permutation. The iteration of two such mapping is called their product.
 
71
It may be readily verified that the set of a11 such mappings forms a
group in which the unit is the identity map. The group is called the
˜ group on n letters.
symmetric
Let us for simplicity denote the set of n abjects by the numbers
1,2,..., n. The mapping S such that S(i) = i + 1 mod n Will be de
noted by (12!3. . .n) and more generally (i j . . . m) Will denote the map
pingTsuchthatT(i)=j,...,T(m)=i.If(ij...m)hasknumbers,
then it Will be called a k cycle. It is clear that if T = (i j . . . s) then
Ti = ( s . . .ji).
We now establish the
˜ If a subgroup U of the symmetric group on n letters
Lemma.
(n > 4) contains every 3cycle, and if u is a normal subgroup of U
˜˜
˜˜ is abelian, then u contains
such that U,/u every 3cycle.
Proof: Let f be the natural homomorphism f(U) = U/u and let
x = (ijk), y = (krs) be twoelements of U, where i, j, k, r, s are 5
numbers. Then since V/u is abelian, setting f(x) = x™ , f(y) = y ™
we have f(x˜y˜xy) = ˜˜˜y™rx™y™ = 1, so that x˜y˜xy E u. But
x˜y˜xy = (kji).(srk).(ijk).(krs) = (kjs) and for each k, j, s we
have (kjs) #F u.
THEOREM 4. The symmetric group G on n letters is not solvable
˜˜
for n > ,4.
If there were a sequence exhibiting the solvability, since G con
tains every 3cycle, would each succeeding group, and the sequence
SO
could not end with the unit.
72
C. Solutïon of Equations by Radicals.
 
The extension field E over F is called an extension by radicals
if there exist intermediate fields B, , B, , . . . , Br = E and Bi = Bir( a i)
where each ai is a root of an equation of the form xni  ai = 0,
ai E Bi, . A polynomial f ( x ) in a field F is said to be solvable by
radicals if its splitting field lies in an extension by radicals. We assume
unless otherwise specified that the base field has characteristic 0 and
that F contains as many roots of unity as are needed to make our sub
sequent statements valid.
Let us remark first that any extension of F by radicals cari always
be extended to an extension of F by radicals which is normal over F.
Indeed B, is a normal extension of B, since it contains not only ar,
but cal, where E is any n,root of unity, that B, is the splitting field
SO
of xnl  a,. If fr(x) =TT(xn2  c( a2 )), where 0 takes a11 values in
u
the group of automorphisms of B, over BO, then f, is in BO, and ad
joining successively the roots of xn2  o( a 2) brings us to an exten
sion of B2 which is normal over F. Continuing in this way we arrive at
an extension of E by radicals which Will be normal over F. We now
prove
THEOREM 5. The polynomial f(x) is solvable by radicals if

and only if its group is solvable.
Suppose f(x) is solvable by radicals. Let E be a normal exten
sion of F by radicals containing the splïtting field B of f(x), and cal1
G the group of E over F. Since for each i, Bi is a Kummer extension of
Bi_r, the group of Bi over B,r is abelian. In the sequence of groups
73
3 . . . 3 GB = 1 each is a normal subgroup of the
G = GB 11 GB
1 r
precediig since G is the group of E over Bis1 and Bi is a normal
Bil
extension of B,i. But GB, /GB, is the group of B, over B,i and hence
1l 1
is abelian. Thus G is solvable. However, G, is a normal subgroup of
G, and G/G, is the group of B over F, and is therefore the group of the
polynomial f(x). But G/G, is a homomorph of the solvable group G and
hence is itself solvable.
On the other hand, suppose the group G of f(x) to be solvable
and let E be the splitting field. Let G = Go 1 G, 1 . . . 1 Gr = 1 be a
sequence with abelian factor groups. Cal1 Bi the fixed field for Gi.
Since G,i is the group of E over B,i and Gi is a normal subgroup of
Gil, then Bi is normal over B,i and the group Cii/G, is abelian. Thus
Bi is a Kummer extension of Bi_i, hence is splitting field of a polynomial
that by forming the successive
oftheform(x”a,)(x”a2)...(xnas) SO
splitting fields of the x”  ak we see that Bi is an extension of Biml by
radicals, from which it follows that E is an extension by radicals.
Remark. The assumption that F contains roots of unity is not
necessary in the above theorem. For if f(x) has a solvable group G,
then we may adjoin to F a primitive nth root of unity, where n is, say,
equal to the order of G. The group of f(x) when considered as lying in
F™ is, by the theorem on Natural Irrationalities, a subgroup G™ of G,
and hence is solvable. Thus the splitting field over F™ of f(x) cari be
obtained by radicals. Conversely, if the splitting field E over F of f(x)
cari be obtained by radicals, then by adjoining a suitable root of unity
E is extended to E™ which is still normal over F™ . But E™ could be
74
obtained by adjoining first the root of unity, and then the radicals, to
F; F would first be extended to F ™ and then F ™ would be extended to
E ™ . Calling G the group of E ™ over F and G ™ the group of E ™ over F ™ ,
we see that G ™ is solvable and G/G ™ is the group of F ™ over F and
hence abelian. Thus G is solvable. The factor group G/G, is the
group of f(x) and being a homomorph of a solvable group is
also solvable.
D. The General Equation of Degree n.
If F is a field, the collection of rational expressions in the
variables ui, up, . . . , un with coefficients in F is a field F(ui, u2, . . . , un).
By the general equation of degree n we mean the equation
f ( x ) = x”  ulx”i + l12x”2  + . . . + (l)“u,.
(1)
Let E be the splitting field of f(x) over F(ui, II˜,. . . , un). If
v, are the roots of f(x) in E, then
l™VZ>...>
V
u1 = v1 + v2 + . . . + vn> u2 = v1v2 + v1v3, + . . . + vnel Vn™. .
. . . , un = VI . v2 . . . . . vn .
We shall prove that the group of E over F(ui, u2, . . . , un) is the
symmetric group.
LetF(x,,x,,..., xn) be the field generated from F by the
variables xi, x2,. . . ,x,. Let ai = xi + x2 + . . . + x”,
xn be the ele
= x1x*. ..
= x1x2 + x1x3 +.. . + XnelXn >...> an
a2
mentary symmetric functions, i.e., (xx ,)(xx 2). . . (xx,) =
 alx”™ +  . . .(l)% ” = f*(x). If is a
g(al,a2, . . . , a , )
X”
polynomial in ai, . . . ,a,, then g(a, ,a2, . . . ,a,) = 0 only if g is the
75
zero polynomial. For if g( xx,, cxixk, . . .) = 0, then this relation would
hold also if the xi were replaced by the vi. Thus,
g(cvi,cviv,,...) = Oorg(ul,uz,...,un) = Ofromwhichitfollows
that g is identically zero.
Between the subfield F(a,, . . . , an) of F( xi, . . . , xn) and
F(ul,u2,. . . > un) we set up the following correspondence: Let
f(u,, *. . ,11,)/g(u,, . . . >un) be an element of F(ui,. . . ,un). We make
this correspond to f(a,, . . . ,a,)/g(a,, . . . ,a,). This is clearly a map
. . ,u,) on a11 of F(ai , . . . ,a,). Moreover, if
ping of F(ui,uz,.
f(al,a2,...,an)/g(a1,a˜,...,a˜)
= fi (al,az!, . . . ,a,)/gl(al,a2,. . . ,a*), then fg,  gf, = 0. But this
implies by the above that
f(u,, . ..9U,h$Ul,. . .>U,)  g(u,, . ..,Un)˜f,(U,, . . .>U,) = 0
that f(u,, . . . ,u,)/g(u1,u2,. . . ,u,)
SO
= fi(U],..., un). It follows readily from this that
U”)/6cl(UI>U2>...>
,..., un) on F(a,,a,,. ..,an)is an isomor
themappingof F(u,,u,
phism. But under this correspondence f(x) corresponds to f * ( x).
SinceE:andF(x,,x,,..., xn) are respectively splitting fields of f(x)
and f * I[X), by Theorem 10 the isomorphism cari be extended to an iso
morphism between E and F (xi, x2, . . . , xn). Therefore, the group of E
over F(ur,uz,. . . , un) is isomorphic to the group of F ( x1, x2, . . . , xn)
over F(al,a2,. . . ,a,).
E:ach. permutation of xi, x2,. . . , xn leaves al,az, . . . ,a,, fixed
and, therefore, induces an automorphism of F( xi, x2, . . . , x”) which
leaves F(a,,a*, . . . , a,.,) fixed. Conversely, each automorphism of
F(xl>xZ>...> x,, ) which leaves F(a 1, . . . , a,, ) fixed must permute the
roots xi, x:!, . . . , xn of f*(x) and is completely determined by the
76
permutation it effects on x1, x2, . . . , xn. Thus, the group of F( x1, x2, . . . , xn)
over F(a1,a2,. . .,a,)is th e s y mmetric group on n letters. Because of
the isomorphism between F ( x1, . . . , xn ) and E, the group for E over
W++. . . >u,,) is also the symmetric group. If we remark that the
symmetric group for n > 4 is not solvable, we obtain from the theorem
on solvability of equations the famous theorem of Abel:
THEOREM 6. The group of the general equation of degree n is
the symmetric group on n letters. The general equation of degree n is
˜˜
not solvable by radicals if n > 4.
E. Solvable Equations of Prime Degree.
The group of an equation cari always be considered as a permu
tation group. If f(x) is a polynomial in a field F, let a,, a2, . . . , c, be
the roots of f(x) in the splitting field E = F( ar, . . . , an). Then each
automorphism of E over F maps each root of f(x) into a root of f(x),
that is, permutes the roots. Since E is generated by the roots of f(x),
different automorphisms must effect distinct permutations. Thus, the
group of E over F is a permutation group acting on the roots
of f(x).
al,a2,...,Qn
For an irreducible equation this group is always transitive. For
a a™
let and be any two roots of f(x), where f(x) is assumed irreduci
ble. F(a ) and F(a ™ ) are isomorphic where the isomorphism is the
identity on F, and this isomorphism cari be extended to an automorphism
of E (Theorem 10). Thus, there is an automorphism sending any given
root into any other root, which establishes the “transitivity” of the group.
77
A permutation o of the numbers 1,2, , . . , q is called a linear
˜˜modulo q if there exists a number b b 0 modulo q such
substitution
that o(i) :E bi + c(mod q), i = 1,2,. . . ,q.
THEOREM 7. Let f( x ) be an irreducible equation of prime de

˜ field F. The group G of f( x) (which is a permutation group
gree q in a
of the roots, or the numbers 1,2, . . . , q) is solvable if and only if,
˜˜
after a suitable change in the numbering of the roots, G is a group of
˜˜
linear substitutions modulo q, and in the group G a11 the substitutions
˜˜
withb = l,o(i) = c + l(c = 1,2 ,..., q)occur.
˜
Let G be a transitive substitution group on the numbers
1,2,. . . , q and let G, be a normal subgroup of G. Let 1,2,. . . , k be the
images of 1 under the permutations of G,; we say: 1,2, . . . , k is a
e transitivity of G,. If i <q is a number not belonging to this
domain of
domain of transitivity, there is a o E G which maps 1 on i. Then
0(1,2,..., k) is a domain of transitivity of oGlu˜. Since G, is a
normal subgroup of G, we have G, = oG,o˜. Thus, (T( 1,2,. . . , k) is
again a domain of transitivity of G, which contains the integer i and
has k elements. Since i was arbitrary, the domains of transitivity of
G, a11 contain k elements. Thus, the numbers 1,2, . . . , q are divided
into a collection of mutually exclusive sets, each containing k ele
ments, that k is a divisor of q. Thus, in case q is a prime, either
SO
k = 1 (and then G, consists of the unit alone) or k = q and G, is
also transitive.
TO prove the theorem, we consider the case in which G is
solvable. Let G = G0 7> G, 3 . . . 3 Gs+l = 1 be a sequence exhibiting
the solvability. Since G, is abelian, choosing a cyclic subgroup of it
78
would permit us to assume the term before the last to be cyclic, i.e.,
Gs is cyclic. If ˜7 is a generator of Gs, CJ must consist of a cycle con
taining a11 q of the numbers 1,2, . . . , q since in any other case Gs
would not be transitive [ if <z = ( lij . . . m)( n . . . p) . . . then the powers
of (T would map 1 only into 1, i, j . . ..m, contradicting the transitivity of
Gs 1. By a change in the number of the permutation letters, we
may assume
o(i) = i + 1 (mod q)
oc(i) E i + c (modq)
Now let r be any element of Gsel. Since Gs is a normal subgroup
of Gs.., >7˜7 l isanelementofGs,sayrm1=ob.Let7(i) = jorrl(j) = i,
then rorl( j) = ob( j) = j + b (mod q). Therefore,
Ta(i) E r(i) + b (mod q) or r(i+l) = r(i) + b for each i. Thus,
setting T(O) = c, we have r(l) = c + b, r(2) = r( 1) + b = c + 2b
and in general 7(i) E c + ib (mod q). Thus, each substitution in G sl
is a linear substitution. Moreover, the only elements of Gsml which
leave no element fixed belong to Gs, since for each a f 1, there is an
i such that ai + b = i (mod q) [ take i such that (al) i z  b].
We prove by an induction that the elements of G are a11 linear
substitutions, and that the only cycles of q letters belong to Gs. Sup
pose the assertion true of Gsq. Let r c Gsmnml and let v be a cycle
which belongs to Gs (hence also to G,,). Since the transform of a
cycle is a cycle, r107 is a cycle in Gs, and hence belongs to Gn.
Thus T˜UT = ub for some b. By the argument in the preceding para
graph, r is a linear substitution bi + c and if 7 itself does not belong to
Gs, then 7 leaves one integer fixed and hence is not a cycle of q elements.
79
We now prove the second half of the theorem. Suppose G is a
group of linear substitutions which contains a subgroup N of the form
c(i) 5: i + c. Since the only linear substitutions which do not leave
an integer fixed belong to N, and since the transform of a cycle of q
elements is again a cycle of q elements, N is a normal subgroup of G.
In each coset N . r where r(i) = bi + c the substitution 0l˜ occurs,
where (T E i + c. But oir( i) = (bi + c)  c F bi. Moreover, if
r(i) z biandr™(i) = b™i thenrr™(i) E bb™i. Thus, thefactorgroup
(G/N) is isomorphic to a multiplicative subgroup of the numbers
1,2,. . . , q1 mod q and is therefore abelian. Since (G/N) and N are
both abelian, G is solvable.
˜ If G is a solvable transitive substitution group on q . *
Corollary 1.
letters (q prime), then the only substitution of G which leaves two or
˜
˜˜ fixed is the identity.
more letters
This follows from the fact that each substitution is linear modula
q and bi + c E i (mod q) has either no solution (b z 1, c + 0) or
exactly one solution(b f 1) unless b = 1, c = 0 in which case the sub
stitution is the identity.
˜ A solvable, irreducible equation of prime degree in
Corollary 2.
˜˜ is a subset of the real numbers has either one real root
a field which
or a11 its roots are real.
The group of the equation is a solvable transitive substitution
group on q (prime) letters. In the splitting field (contained in the field
of complex numbers) the automorphism which maps a number into its
complex conjugate would leave fixed a11 the real numbers. By Corollary
80
1, if two roots are left fixed, then a11 the roots are left fixed, that
SO
if the equation has two real roots a11 its roots are real.
F. Ruler and Compass Constructions.
Suppose there is given in the plane a finite number of elementary
geometric figures, that is, points, straight lines and circles. We seek
to construct others which satisfy certain conditions in terms of the
given figures.
Permissible steps in the construction Will entai1 the choice of
an arbitrary point interior to a given region, drawing a line through two
points and a circle with given tenter and radius, and finally intersec
ting pairs of lines, or circles, or a line and circle.
Since a straight line, or a line segment, or a circle is determined
by two points, we cari consider ruler and compass constructions as con
structions of points from given points, subject to certain conditions.
If we are given two points we may join them by a line, erect a
perpendicular to this line at, say, one of the points and, taking the dis
tance between the two points to be the unit, we cari with the compass
lay off any integer n on each of the lines. Moreover, by the usual
method, we cari draw parallels and cari construct m/n. Using the two
lines as axes of a cartesian coordinate system, we cari with ruler and
compass construct a11 points with rational coordinates.
Ifa,b,c,... are numbers involved as coordinates of points which
determine the figures given, then the sum, product, difference and
quotient of any two of these numbers cari be constructed. Thus, each
81
element of the field R( a, b, c, . . .) which they generate out of the
rational numbers cari be constructed.
It is required that an arbitrary point is any point of a given region.
If a construction by ruler and compass is possible, we cari always
choose our arbitrary points as points having rational coordinates. If we
join two points with coefficients in R( a, b, c, . . . ) by a line, its equa
tion Will have coefficients in R( a, b, c, . . .) and the intersection of two
such lines Will be a point with coordinates in R( a, b, c, . . . ). The equa
tion of a circle Will have coefficients in the field if the circle passes
through three points whose coordinates are in the field or if its tenter
and one point have coordinates in the field. However, the coordinates
of the intersection of two such circles, or a straight line and circle, Will
involve square roots.
It follows that if a point cari be constructed with a ruler and com
pass, its coordinates must be obtainable from R( a, b, c, . . . ) by a formula
only involving square roots, that is, its coordinates Will lie in a field
RS 3 Rsi 3 . . . 3 R, = R(a,b,c,... ) where each field Ri is splitting
field over Rir of a quadratic equation x2  a = 0. It follows (Theorem
6, p. 21) since either Ri = Rir or ( Ri/Rir ) = 2, that (RJR, ) is a
power of two. If x is the coordinate of a constructed point, tben
(Rr( x)/R, ) * ( RS/R, (x)) = (RJR, ) = 2” that Rr( x)/R, must also
SO
be a power of two.
Conversely, if the coordinates of a point cari be obtained from
R(a,b,c,... ) by a formula involving square roots only, then the point
cari be constructed by ruler and compass. For, the field operations of
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addition, subtraction, multiplication and division may be performed by
ruler and compass constructions and, also, square roots using 1: r =
r : rl to obtain r = d rI may be performed by means of ruler and
compass instructions.
As an illustration of these considerations, let us show that it is
impossible to trisect an angle of 604 Suppose we have drawn the unit
circle with tenter at the vertex of the angle, and set up our coordinate
system with Xaxis as a side of the angle and origin at the vertex.
Trisection of the angle would be equivalent to the construction
of the point (COS 20”, sin 209 on the unit circle. From the equation
38 = 4 cos3 0  3 8, the abscissa would satisfy
COS COS
4x3 . 3x = 1/2. The reader may readily verify that this equation has
no rational roots, and is therefore irreducible in the field of rational
numbers. But since we may assume only a straight line and unit
length given, and since the 60° angle cari be constructed, we may take
R(a,b,c,. . ..) to be the field R of rational numbers. A root a of the
irreducible equation 8x3  6x  1 = 0 is such that (R(a)/R) = 3,
and not a power of two.