Internat. J. Math. & Math. Scl.
VOL. 11 NO. 3 (1988) 523-526
523
COMPUTATION OF RELATIVE INTEGRAL BASES
FOR ALGEBRAIC NUMBER FIELDS
MAHMOOO HAGHIGHI
Department of Computer Science
Bradley University
Peoria, IL 61625
(Received March 2, 1987 and in revised form September 21, 1987)
ABSTRACT.
At first we are given conditions for existence of relative integral bases
for extension (K;k)
extensions
Then we will construct relative integral bases for
n.
OK6 (6/)/Ok2(/) OK (6/)/Ok3(3/), OK6(6/) / Z.
6
Integral Bases and Principal Ideal.
KEY WORDS AND PHRASES.
1980 AMS SUBJECT CLASSIFICATION CODE. 115.
EXISTENCE OF A RELATIVE INTEGRAL BASES.
i.
The following criterion has been shown in [I] for existence of a Relative
Integral Bases, for any finite extension K/k.
THEOREM I.I.
Let (K;k)
COROLLARY 1.2.
Ok+-dK/k
P.I.D., then h
If O
K
every finite extension of k where O
Let k
so
Q
OK1, OK2 OK3
extensions
and
k
dK/k
k
K6
Q(6/)
are P.I.D. and then by corollary 1.2,
K6/kl, K6/k2, K6/k
P.I.
Therefore for
P.I.D., a relative integral bases exists.
Q(3)
Q(/)
k2
be an odd integer then O has a
K
k
is a principal ideal. See also [2].
n, and let h
"relative integral bases" over
hkl hk2 hk
relative integral bases for
Since
exists.
3
Now, we will compute the relative discrlmlnant for the extensions.
and for some 8 e K, O
in
Ok(8)
K
and 8 satisfies an equation F(8)
(n)
8(t)), where 8, 8 (I) 8 (2)
8
DK/k
Since extensions K2/K
I, K3/K have dlscrlminant divisible
[3] dlscrimlnants K6/k2, K63, K6/ are also divisible by
(8
(F(8)
Then
ramified in
Let (K;k)= n
0 of degree n.
are conjugates [3].
by 3 [3], by theorem
3 and 3 is completely
kl,klk 3.
For extension
K6/k 2,
DK6/k
2
DK6 /k
2
6/
8
(8-
8(I))(8
(-3)
4/3 for
we therefore have:
8
(2))
dK6/k NK6/k (DK6/k2)
2
2
I,
3
(6-D 6//) (6/ _02 6)
-I +
2
(-3) 4
By the definition in [4],
M. HAGHIGHI
524
For extension
dK6/k
K6/k 3,
By theorem in [4],
II
(-3)
6,
0
DK6
(0-0
DK6 /k
(-3)
DK6/k Uk2/k
/ kl
(I))
(-3)
4/3
I/6
(-3)
/k
dK6
II/6
(-3)
(-3)
then
I/2
then
2
See also [5]
Now we will construct relative integral bases for the extensions.
for associated work.
For
For
2.
K3/k I,
K2/k I,
(1,3/, 3/)2).
OK3
OK2
II
disc(l,e,)
3p) 2
Now 282(3p2
We may take e
they are in
Since
06
(-3)4
and
6/
and
3/
N6/3 (e)
(I,
06
3.
06(6)/O2(/).
(l,,B)O 2 for ,B in
06
(i,)O
06
-
06
By
theorem2 in
Pa
P28
dK6/k
e
for
4
are in
03
e.B
/
and
we have:
02
O6(6/)Io3(3/).
Again by theorem [6]
06
2
4a 2
Jl -=J
because
(-3)
dK6/k2,
2
because they satisfy an
3/(-3)5
6, 6V’(,3)2
[6], disc (l,,B)
/.
here
6{(-3),
8
N6/3(8)
and
disc(l,)
6/
Note
3
06
rom
RELATIVE INTEGRAL BASES FOR
Let
Z, [3].
1+/z )- Z, [3].
(I,
2
RELATIVE INTEGRAL BASES FOR
Let
N6/3()
dK6/k3
3V’,
_6/
_3,/’i-"
2
4
E:
03
Hence
,o.) is
not a relative integral bases.
We define
and
06
for
2
B
03 such that N6/3() is divisible by 2.2
03 it satisfies the conditions, this is
3(-3)2
If we take
4
because
2
2
Also, disc(l,a)
4.
4
dk6 /k 3(-3)2 + 6/
[I
Let 0
6/i
Q(6),
6
e
0
2
RELATIVE INTEGRAL BASES FOR
Since K
3/
03
22
22.
by theorem [6],
so that:
06
I/2
3
06(6/Z)/Z.
at first we start by:
06
(1,0,02,03,04,05
06
Since disc(l 0
Z
02 03 0 W, 05)
22
dK6/k
we can apply
COMPUTATION OF RELATIVE INTEGRAL BASES
22 22 22
theorem [3] in order to cancel out
bases (**
i
We will build a new
(*i
first bases element
(*I
,
{(*i:
(*0/2
(*0"
is going to be changed.
525
and generate a new bases.
0
5}.
By the theorem [3] we check which
I/2
06
Thus there is no change for the
I.
(*0
gi(*O + (*I
gi(*0 +
2
2
8
for
(*i*
I. For any value of gi’
0gi
is not in
06
This is because
N6/3((*I
I+6/:
I-6Z
I-3
2
2
4
is no change for
(*2*
(*3
2
for
e
For any value of gi’
for
06"
I.
Ogi
(*4
4
gl(*O + g2(*l + g3(.2 +
,
6/ + 6/(_3)
g4(.3" + (*4
,
6,/ + 6/
6,/
03
and for other values
In this case for g2
6
2
4
gl(*0 + g2l + g3(.2 + g4u3 *+
*+(*5
2
64(-3)2 + 64(-3) 5
This is because
06
2
gi’(*5* 06"
4.3
2
g4
I,
for
g2
e
03,
e
gi’(*4* 606"
I,
g5
N6/3(
and for other
03
and for other values
This last assertion is since
22
disc((*0’(*l’(*2’(*3*’(*4*’(*5*)
22
22. 2
2
22
dK6/k I,
22
and each
(*i’
ei* are
in
06
by theorem [6].
06
I,
This is because
2
(*5*
of
g3
g2
2
N6/3((’4")
(*5
then there will
This is because:
gi’(*3* 06"
,
(*2* 06
In this case for gl
e
of
so there
03
(*2"
gl(*0 + g2(*l + g3(.2 + (*3
64(-3)
I.
0Sgi
2
(*3*
N6/3(8/2)
and also since
I"
gl(*O + g2(*l + (*2
be no change for
,
03
2
2
2
"Z.
then
526
M. HAGHIGHI
REFERENCES
I.
NARIEWICZ, W. Elementary and Analytic Theory of
1976.
2.
ARTIN, E.
Questions de Base Minimal dans la Theorie des Nombres Algebriques,
National de la Recherche Scientifiques XXIV, (1950), 19-20.
3.
COHN, H.
A Classical Invitation to Alebraic Numbers and Class Field Theory,
Springer-Verlag, New York, 1978.
4.
COHN, H.
5.
6.
Alebraic
Numbers, Pwin, Warsaw,
Introduction to the Construction of Class Fields, Cambridge University
Press, New York, 1985.
HAGHIGHI, M. Relative Integral Bases for Algebraic Number Fields, Internat. J.
Math. and Math. Scl. 9 (1986), 97-104.
RIBENBOIM, P.
Algebraic Number Theory, John Wiley and Sons, New York, 1972.