/
I
GROUPS OF SPECIAL GALOIS
DOMAINS OF RATIONALITY
BY
HARRY ALBERT BENDER
A. B. Ohio University, 1918
THESIS
Submitted in Partial Fulfillment of the Requirements for the
Degree of
MASTER OF ARTS
IN MATHEMATICS
IN
THE GRADUATE SCHOOL
OF THE
UNIVERSITY OF ILLINOIS
1921
Digitized by the Internet Archive
in 2015
https://archive.org/details/groupsofspecialgOObend
UNIVERSITY OF ILLINOIS
THE GRADUATE SCHOOL
I HEREBY RECOMMEND THAT THE THESIS PREPARED UNDER MY
BE ACCEPTED AS FULFILLING THIS PART OF THE REQUIREMENTS FOR
THE DEGREE OF
/3 \CH&f/&L.
In Charge of Thesis
'/ Head of Department
Recommendation concurred in* *
Committee
on
Final Examination*
*Required for doctor’s degree but not for master’s
Let f(x)=0 be an irreducible equation of degree n and let
x,,x^ ,Xj , ,x^ be its roots. Let p* be sin arbitrary rational
prime and let us suppose that in k(p)
(l) f (x)=f, (x) .f 2 (x) .fj (x) fs-(x) (p)
where f c -(x) is of degree n^ ,and irreducible in k(p) .
In k(p),the domain of the rational p-adic numbers,the
number x, cannot satisfy an equation of degree less than n (x, is
taken in this sense as any one of the roots of f(x)). For suppose
that
0 ( x ) =a 0 x*'4a ; x**
is a polynomial of degree less than n.with p-adic coefficients,
such that 0( x ; )=0 (p) ,and let
X/'-a^w, -h a^w.,* + a^w* ( i=1 ,2,3, ,n) .
We then have
<P(*f =A, w, -f -+A^w^
A i= a fl ^'+ a / + + \. f a /c (i=1 >2,3, ,n) .
Since an integer of k(x,) is divisible by p when and only when
each coefficient in its representation by a fundamental system
is a multiple of p,we conclude that <^(x y )=0 (p) when and only
when A^=0 (p) (i=1,2,3, --,n),and from this system of
equations it follows that J a t - J.a^=0 (p) .
But the a ( ^are ordinary rational integers and hence | a iy.'|
is a number of k ( 1 ) and since it is not zero there exists a
*By p without subscripts we shall mean an arbitrary rational
prime, and p with subscripts we shall mean the prime divisors
of p in the algebraic domain.
.
/ 4 yM~ a 4 =,a -4-° (p) -
Hence ^(x) must vanish Identically , and x, cannot in k(p) satisfy
an equation of degree less than n.
From ( 1 ) we have
f ( x,) =f,(x,) •f JL (x l ) .f 3 [x,) . fj( Xj) =0 ( p )
while no one of the factors is zero. We therefore conclude that
in this case B(p,x,) is not a domain.
Since f(x) is irreducible in the ordinary sense its
discriminant cannot vanish and hence it must also be different
from zero for the domain of p. Consequently no two of the £
factors can be equal.
We shall now introduce , corresponding to each of the s
factors of f(x) ,s new systems of values for the numbers of R(p,X/)
as follows. If/?=B(x / ) is any number of R(p,x / ) and if
B(x)=Qt(x) .f c - (x) + R< (x) (p),
we shall call R i(x f ) the value of (3 for the domain of p- cor-
responding to the factor f^(x) and shall write /3=Rc (x, ) (p- ) .
Two numbers f3,- B / (x / ) and/^=B^(x^ ) are said to be equal
for the domain p^ when and only when B/(x)-B^(x) is divisible
by U (x) .
We thus have s new rings R(p^ ,x, ) (i= 1 , 2 , 3 * -,s) such
that each number of R(p,x,) is for the domain of p^- equal to
some number of R(jj£ ,x / ) and the sum , difference , and product of
two numbers of R(p,x^ ) is for the domain of equal to the sum,
difference , and product respectively of the corresponding numbers
of R(p^ ,x / ) . Evidently % (x, )=.0 (p^ )
. . « £JH
■
5
We shall next see that these p -adlc values of the numbere
of R(p,Xy) constitute a domain. We need only show that every
number/3 ^0 (p^ ) has a uniquely determined reciprocal in R(p x -,x ; ).
Let us therefore suppose that ft =B(xy ) rjfcO Since/? £0 (p^-)
it follows that B(x) is not divisible by f/ (x) and hence since
fy (x) is irreducible we know that they are relative prime. Hence
there are two polynomials \f^(x) and such that
(p^x) .fy (x) -fl£(x) .B(x) = 1 (p)
and since rational numbers are equal for the domain of p - ,we
can write
(p/x) .fy (x)-/-(^(x) .B(x) = 1 (py )
the coefficients of the polynomials being rational numbers. But
f/ (x f )=0 (p^ )
and hence (x, ) ). Therefore/? has a reciprocal which is
unique, for if q and /?, are two numbers such that/? ./> -n,A- 1 (p- ) ,
j / y A.
then /0 / - ^=0 (py ) and hence /?,/?( - /^) -fi, -fy= 0 (Py ) •
Therefore /? = /? (p.) and the p -adic values of the numbers of
R(p,x / ) form a domain which we shall denote by k(py ,x ( ) .
In k(x^) p is divisible by s distinct prime divisors
%'/ 'tyx » ,v:here the notation is so chosen that
xy ) =C^-j. Since in k(p) xy cannot satisfy an equation of degree
less than n and since t: (x) (j=1,2,3, -,s) are all diBtinct,
) (^rj) • But fj (xy ) .f^ (xy ) . -f 5 (x^)=0 (p) and
hence f, (x^ ) .f^Uy ) -- .f^ (xy )=° (jc-^ ) (i=1,2,— ,n) (j=1»2,-s)
( k= 1 , 2 , — , s ) •
Let us consider the array J constructed with the various
distinct prime divisors of p in each of the n domains k(xy)
'
.
4
(i=1,2,3» ,n) in which each row contains the distinct prime
divisors of p in a certain one of the domains.
p //
■P/JL ’
p ,J •
1
1
1
1
m
L
,v xx *
p « •
™*P«
p *,
P *J •
p yy *"
* 9 h s
and associated with this the array of ns numbers from the domains
fy ( X/ ) > f 2 ( Xy ) , f J ( X/ ) > , f^( X, )
f / ( ) > fz ( ^2. ^ ) i » f j- ( )
n
f / ^ ) > f ) j fj ( X^ ; f — t f j- ( X* ) •
We observe that this arrangement has been so made that any
element in II is zero for the domain of the corresponding element
of I.
Let us next suppose that F(V)=0 is the Galoisian resolvent
of f(x)=0 for the domain R(l),and that t is any one of its roots,
It's degree we shall suppose to be g and when we need to dis-
tinguish between the roots we shall denote them by V ; , V 3L ,----,V^.
The substitutions on the n roots by which these V's are derived
from V / form a transitive group G of order g. The distinct prime
divisors of p in k(V) we shall denote by P , , ? 3 , , .
Since k(x, ) is included in k(v) we see that t^s .
Let us suppose that in k(p)
F(V)=F, (V) -F a (v) -Fj (V) F* (V) (pj
where F,; (V) is of degree^'and F^ (V)=0 (P^- ) ( i=1 ,2,3, ,t) •
And since the factors are all irreducible, they are all distinct
and only one of them is zero for the domain of (P^ ) according to
.
5
the same discussion as in the case of f(x)=f / (x) .f a (x) . .f^ (x)(pj
Hence each is a root of some F^(v)=0 (P^) and since this
equation cannot have more roots than its degree we know that at
most of the numbers V # , , , are roots of F^(V)=0
where is the degree of f^(V). Since this is true for all
(i=1,2, ,g) and for all F^{V) (k=1,2, ,t) and since H ( H---+flpg
we know that each of the equations F^(v)=0 (P A ) (k=1,2, ,t)
has in the domain k(P A ,V) exactly roots.
Let S be a substitution of G which transforms V. into
and let us suppose that F^(V, )=0 (P A ). The substitution S then
transforms F^CV, ) into ^(V^) and this is again zero for the
domain of some P^. If F^V^ )=0 (P^) th en/^=\. In this case we
shall say that S transforms the prime divisor P^ into the prime
divisor P A ,or the substitution S leaves p* invariant. If however
F^(V t )£0 (P^‘ but F^(V^- )=0 (i^J^^^we shall say that S trans-
forms the prime divisor P A into the prime divisor
Hence every substitution of G either leaves P^ invariant
or transforms it into another prime divisor of p in k(V) •
If. we now consider that
F / (V) -F ^(V) F a (V)=(V-fy )(V-^) (V-.^)(^ )
and factorization of a polynomial in a domain is unique we know
that
F; ( V) = ( V- V„ ) ( V-V /JL ) ( V- V /3 ) ( V-V^) ( P A )
F A ( V) = ( V- V x , ) ( V-V xx ) ( V- V^) - (V-^) ( P )
in *
E, (V) = { V-V„ ) ( V-T^) (V-V^) (V-\r„) {P* ) .
The substitutions of G which leave P-, invariant form a
'
•
•
6
subgroup of G of order say g^ . Let us call this subgroup G / . The
substitutions of G / form a group, since G contains all the sub-
stitutions which when operated on gives all the other V's,it
then contains all the substitutions which when operated on V^y
gives all the roots of F^(V) in and by hypothesis all these
substitutions are in G / , thus from the theorem in group theory
these substitutions of G form a group*.
Every substitution of G transforms each Vy into some i£j
unless the substitution is the identical substitution because
each is a primitive number of k(V) and hence under any sub-
stitution S goes over into one of its conjugates since k(V) is a
galois field or domain.
Let us suppose that H ; ^H^ (1=1,2, ,t) and 1, S a , ,
are substitutioms such that operating on V ; gives V </( * . Let us
represent this operation as IV^V,, , , SjV /f =V /Jt , S^V fl =V ; i S .
Thus there are at least Hy substitutions that leave P^ invariant.
Hence the order of Gy is <£Hy . Moreover P^ cannot be left invariant
by any other substitution of G because if such were the case then
this substitution S transforms V / into and since and S both
transform V /y into V,^ and s/s leaves V 7/ invariant. But this is as
we have said above, only true when s/s is the identical substitu-
tion or when S~.'s= 1 or =S . Hence the order of Gy is H, .
G contains at least t- 1 substitutions which do not leave
invariant , namely the substitutions which carry over into V^y
(1=1,2,-— ,t) or t x =T£ ( , =V J( , , Then t,G,
consists of Hy substitutions which do not leave invariant , for
*Miller ,Blichf eldt ,and Dickson 'Finite Groups' page 286.
»•
.
-
7
since Greaves P^ invariant it will give a root of Fj L (V)=0 (P^)
when operating on V^and all of these roots are distinct from
those of Fy (V) , therefore t^Gy transforms P^ into another prime
divisor. The same argument holds with t 3 Gy , t G^ , , t r G / . Thus
G contains at least Hy substitutions that leave P^ invariant , and
(t-i)H substitutions that will transform P^ into some other prime
divisor. Therefore G contains at least tHy substitutions , that is
S^tH / . Since g=H,-f H 3 + -fti; and Hyis^H* (i=1,2,3, ,t) we
see that g^tHy . We therefore conclude that g=tHy and Hy =H 3L =H J( ==-=:H^
Hence Fy (V) , F^(V) , F^V), ,F r (V) are all of the same degree,
and the substitutions of Gy permutes the V^y in TXT with the same
first subscript among themselves.
Since G(V) of degree nj ,and whose roots are derived from
the nj valued function of Vy is reducible in k(p) ,and Fy(V) is
that irreducible factor in k(p) for which 0 (P^),the sub-
stitutions on Xy , x^, Xj , — , x^ by which the roots of
Fy (V)=0 (P^) are derived from V f is called the group of the
given equation in k(p) .
Since k(Xy ) is included in k(V)
fy (x) = (x-x // ) (x— x /JL ) (x-x yJ )-- (x-x^)=0 (P^)
f^ (x) = (x-x J _ / ) (x-x^Kx-Xjj) ( x-x^J) =0 (P^ )
fj (x) - (x-x^) (x-Xjj,) (x-x^) (x-Xj^=0 (PA)
where xj^ ( j=1 ,2,3, ,s) (i=1 ,2,3, ,n^ ) is some one of the
roots of f ( x ) =0 • The coefficients of f (x)=fy (x) *fj_(x) «r-- .fj (x)
being symmetric functions of x^ (k=1,2, ,s) (i=1,2, ,n )
are unaltered numerically by the substitutions of G, , because Gy
8
leaves invariant , and hence equals numbers in R. It has been
shown that Cr / contains all the substitutions that leave
invariant .
Since the roots of f^ (x)=0 (P^ ) are x </? x^ , x^ , , x^
and the roots of f^(x) in k(p) are, say y^ , , , y^,
and from the simple isomorphism that exists between the roots in
k(p) and those in P^ ,we see that the element x-- in the substitu-
<7
tions of G, can be replaced by y.-,and we have the group of f(x)
in k(p) in terms of the roots of f(x)=0 in k(p) .
Thus the Theorem: If f(x)=0 is irreducible in R( 1 )
and its group in R is G,then in k(p) the group of f(x)=f (x) .fj_(x)
fj (x) ‘fj (x) (p) is simple isomorphic to a subgroup G / of G.
Since this group is simple isomorphic to a subgroup of G,
we shall discuss some of the properties of this subgroup instead
of the group in k(p) •
Since f(x) is irreducible in R( 1 ) its group is transitive
in the n roots. If f(x) is irreducible in k(p) its group will be
transitive also. Thus in this case is G or a transitive sub-
group of G. If f ( x ) is reducible in k(p) its group is intran-
sitive in the n roots. Since (x) , f^(x), f^ (x) , •, f s (x)
have no common roots, and are irreducible in k(p) ,the groups of
f/ (x) , f^(x) , fj(x), ,and fj (x) will not have any common
element and will be transitive in their roots. G, being the group
of f(x)=0 in k(p) it will be composed of these transitive con-
stituents .
If G is a regular group then in k(p) all of the factors of
f(x) will be of the same degree , because a subgroup of a regular
group will have cycles all of the same degree.
If the group G is not regular, then all of the substitutions
do not contain all of the letters, and from a theorem* in group
theory, that the order of the sobgroup of a transitive group formed
by all of the substitutions which omit a given letter is equal to
the order of the group divided by its degree. Hence in this case
the order of the group of fj (x) , f^(x) , fj(x), ,and f 5 (x)
separately can not exceed g/n, that is, the order of the transitive
constituents can not exceed g/n. Likewise if In k(p) f(x) has
one linear factor, this root will not be an element of G ; ,and
therefore G, will be of order g/n. If there are several linear
factors in k(p) the order of Gj may be less.
Since Ft (V) (i=1,2,3, ,or t) is irreducible in k(p) , if
the discriminant of F^(V) is not divisible by p,from the theory
of p-adic numbers, we have that the roots of F^(V)=0 mod p are
4> 3L jb , / a
M X, - Xi/ > \i . . % • If S V£,=V Xl then s\, =SV^={SV // ) =V/J ,
eto. Thus the group G / is cyclic and thus transitive. Therefore
f(x) has all linear factors but one of degree ,and since Q f
does not contain all the roots its order is g/n, or a subgroup
of this group of order g/n.
If the group of f(x)=0,f(x) being irreducible in R( 1 ) , is
known, then by studying the subgroups of G one can form some idea
of the factors of f(x)=0 in k(p),and in many cases determine
exactly the form of the factors, as in the case when G is regular.
By extending this work it may be possible to determine the
exact subgroup for certain classes of p and thus determine exact-
ly the factors of f(x) in k(p) from the group standpoint of view.
*Miller ,Blichf eldt ,and Dickson 'Finite Groups’ page 32.