Mar 18, 1987 - results of Section 2 and Section 3 for the theory of endoscopic groups are also discussed. I wish to thank Noriko Yui for helpful conversations ...
Internat. J. Math. & Math. Sci. (1988) I-4 VOL. II NO.
QUADRATIC SUBFIELDS OF QUARTIC
EXTENSIONS OF LOCAL FIELDS JOE REPKA Mathematics Department University of Toronto Toronto, Ontario
CANADA M5S IA1
(Received March 18, 1987)
ABSTRACT.
We show that any quartic extension of a local field of odd residue
A consequence of this is that
characteristic must contain an intermediate field.
local fields of odd residue characteristic do not have extensions with Galois
Ah
group
or
Sh
Counterexamples are given for even residue characteristic.
KEY WORDS AND PHRASES.
1980
Local field, quartic extension, endoscopic group.
AMS SUBJECT CLASSIFICATION CODES.
12B25
12B27.
Research supported by the Natural Sciences and Engineering Research Council of Canada.
INTRODUCTION.
i.
In Section 2, a simple application of local class field theory proves the existence of intermediate fields for quartic extensions of local fields with This immediately implies the non-existence of
odd residue characteristic. Galois extensions of type
A
S
h or
h over
such fields.
In Section 3, examples are given of A h
and
S
extensions of fields with
h
even residue characteristic, and of a quartic extension with no intermediate field.
In Section h, the results of Section 2 are used to show that the splitting field of an irreducible quartic polynomial over a local field must have degree
or
8, provided
the residue characteristic is odd.
The implications of the
results of Section 2 and Section 3 for the theory of endoscopic groups are also
discussed.
I wish to thank Noriko Yui for helpful conversations about this work. 2.
EXISTENCE OF INTERMEDIATE EXTENSIONS.
Let
F be a non-archimedean local field.
respectively, be the ring of integers of
F
Let
o
o
F
and
P
PF
and its prime ideal.
Suppose the residue characteristic of F is odd, and [E:F] 4). Then there must be an interis a quartic extension (i.e. THEOREM 2.1.
mediate field
K
i.e.
E
o
K
F
[E:K]
[K:F]
2
E/F
J. REPKA
2 PROOF:
4 to
E/F is unramified, the result is obvious. If the ramifica2 2 and, by Corollary then we must have f E/F is e h of Section of Weil is an unramified [1], there I, chapter 7
If
tion index of Theorem
quadratic intermediate field.
N6w suppose u e o
with
P’
e
of
PE NE/F
F
F
0
h
e
and
so
PE
P
E
Any unit in
1
f
is of the form
+ p’
u
The norm of such an element is
u+p with
So by Hensel’s Lemma the only units contained in the image
PF
In particular,
are fourth powers.
NE/F
is not surjective, so
[1] proves
Corollary 1 to Theorem h of chapter XII, Section 3 of Well
the
theorem.
Translating this into the corresponding result on Galois groups, we obtain the following equivalent formulation
THEOREM 2.2.
PROOF:
A
whose Galois group is isomorphic to
h or S h
h (the cyclic group generated
contains subgroups of index
3-cycle),
by any
(such
Ah
has odd residue characteristic, there cannot be a
F
If
E/F
Galois extension
none of which is properly contained in any proper subgroup
a proper subgroup, if it existed, would be of order
6 and
index 2,
8).
hence normal, hence would contain all 3-cycles, of which there are
An
Sh-extension
F
of
would be an
Ah-extension
of a quadratic exten-
F
sion of
COUNTEREXAMPLE FOR RESIDUE CHARACTERISTIC 2.
3.
F
Let Let
E
and
Gal(E/K)
A
quartic extension
Let
e
(X)
PROOF:
h
L/F
(X)
The norm
S
h
In the process we shall find a
19
is surjective.
N(+l) =.(-1)
3 N
If
F(a)
L
and let
NL/F
Notice that
3(-1)
h-2x-2 eF[X].
with no intermediate field.
characteristic polynomial of
N(a3+l)
X
Gal(E/F)
we shall show that
2 ()
K
where
be a root of
LEMMA 3.1.
(X)
and consider the Eisenstein polynomial
2
be the splitting field of
I
3(X)
is
X
h
$(i) N(e-1) 6X 3 + 12X2
-3
Also the
8X- 8
so
were not surJective, its image would
NL/F
be contained in the image of the norm map from some ramified quadratic
extension of
F
oX
2
modulo
taining
(oX)
Such an image contains exactly two of the four cosets of
We have
Just
contains the three cosets con-
NL/F
shown
l, -3, and 19.
In particular (by Corollary 1 to Theorem h of chapter XII, Section 3 of Well
[1]), L/F
is a quartic extension with no intermediate field.
Factoring the polynomial where
(X)
X 3 + aX
PROPOSITION 3.2.
PROOF:
2
+
(X)
L
over
(X)
we see that
(X-G)(X)
2X + (3-2)
(X)
If all roots of
is irreducible over
(X)
were in
Galois, in contradiction of Lemma 3.1. reducible would be for exactly one root,
L
L then
L
E
would be
The only other way for
’
say, to be in
L
(X)
to be
In this case,
QUADRATIC SUBFIELDS OF QUARTIC EXTENSIONS OF LOCAL FIELDS
F((’ F()
GaI(E/F)
e
F(a)
(c) F(’) F(c)
be such that
g(a’)
implies that
g
’
must equal the only other conjugate of
L
the fixed field 2
X
So
E
So
A
either
Lg[x]
S
or
T(X)
X’
3 2 X3 + X +
The element
PROOF:
3
0
or
’
g(a’) g(’)
and
g(’) Hence
(X-a)(X-(’)
so
L
is quadratic over
e
L
over
X ’3 +
2
W(X)
is
+ (20/27)c 3
(2/3)2X
2
is
(368/27)c6-80 3
2
GaI(E/L)
and
4.9.3 (X)
the discriminant of
mod*(l+4PL)
is a square in
L
is not a square in
3
were a square, truncation of its square root would give
’an element of the form equal to
i.e.
F(’)
is.
LEMMA 3.3. If
Since
L
’
T(X)
2X + 3
(LX)
3
if and only if
L
Hence the discriminant of 3 4.27 +
4.9(i+4PL
Since
g(F())
Then
which shows that
27((20/27)C3-2)2-4((2/3)(2) L
F(a’
hence
This also contradicts Lemma 3.1.
2
X + 2/3
’
in
and
is the splitting field of
3
Now
’
+
contains
(a+’)X + a’ [L:F] [L’Lg]
where
L
contained in
L
Let
a’
F
would be a quartic extension of
3
x
+as+b(2+ca 3
i
3
and so that
1
2
x
a
with
4pL
and
b
each
c
A trivial computation shows
that this is impossible.
GaI(E/L)
Accordingly
4.
F(/)
K
where
S
GaI(E/F)
3
S
and
4
GaI(E/K)
A
4
APPLICATIONS. The splitting field of a quartic polynomial over a local field is
i.
severely constrained by the results of Section 2.
THEOREM 4.1.
f(X)
Let
e
FIx]
F
Let
be a local field with odd residue characteristic.
be an irreducible polynomial with
f(X)
be the splitting field of
GaI(E/F)
PROOF: be
S
4
A
or
4
[E:F]
4
@(X)
F
henc
is a local field, let
elliptic torus in
G
To
T
T
in
that the centralizer of self is isomorphic to
G
and
E: 1
T
{g
GL(2,K)
or
8
2
Theorem
to Theorem
G
4.1
is clearly
2.1).
SL(h,F)
and let
T
be an
{
,
NE/F (x)
}
(cf. Langlands [2], Shelstad [3])
some other groups, among which the most interesting
are constructed as follows:
G’
4
is associated a quartic extension E/F so and T itGL(4,F) is isomorphic to E x
The theory of endoscopic groups
associates to
E
Let
But by Theorem 2.2 it cannot
4
the only possibilities are
result when the residue characteristic is
If
4 f(X) 4 or 8
of Section 3 gives a counterexample to this
equivalent to Theorem 2.2 (and
2.
deg
[E:F]
Then
S
is a subgroup of
Since
The polynomial
F
over
let
NK/F(detg)
E
l}
K m F In
and let
G’
it is possible to find an
J. REPKA
4 elliptic torus
T’
E/K
associated to the quadratic extension
an isomorphism between
T
and
T’
and there is
The hope is to simplify calculations
with orbital integrals over the G-conjugacy class of
t
T
by comparing them
with orbital integrals over the G’-conjugacy class of the corresponding
t’
T’ The example of Section
B
shows that this approach will not apply for
certain tori when the residue characteristic is
2
happily, for these tori
the ordinary orbital integrals are invariant under stable conjugacy, so the
problem does not arise.
The results of Sections 2 encourage optimism in the
case of odd residue characteristic.
REFERENCES
(New York,
I.
WELL, A.
2.
LANGLANDS, R.P. Les D@buts d’une Formule des Traces Stable, Publ. Math. de l’Universit@ Paris VII (1983) 188p.
3.
SHELSTAD, D.
Basic Number Theory, 3rd Edition, Springer-Verlag Heidelberg, Berlin) 197h.
Orbital integrals and a family of groups attached to a real reductive group, Ann. Scient. Ec. Norm. Sup. 1_2 (1979), 1-31.
