On the m-torsion Subgroup of the Brauer Group of a Global Field

0.13 MB·English

by  Wen-Chen Chi

ON THE m-TORSION SUBGROUP OF THE BRAUER GROUP OF
A GLOBAL FIELD
WEN-CHEN CHI∗, HUNG-MIN LIAO, AND KI-SENG TAN†
7
0
0 Abstract. In this note, we give a short proof of the existence of certain abelian
2
extension over a given global field K. This result implies that for every positive

n integer m, there exists an abelian extension L/K of exponent m such that the
a m-torsion subgroup of Br(K) equals Br(L/K).
J

2

]
T Assume that a global field K, a prime number p and a natural number n are given.
N
We say that a pro-p abelian extension M/K is granted if the exponent of M/K, by
h. which we mean that of Gal(M/K), divides pn and for every finite place w of M the
at degree [Mw : Kv], w v, also divides pn. A finite place v is said to be supported under
|
m M/K if for each w v, [M : K ] = pn; an archimedean place v is supported under
w v
|
[ M/K if M is complex for every w v. The purpose of this note is to give a simple
w |
1 proof of the following.
v
2 Main Theorem. Suppose K, p and n are given as above. Then there exists a
5 granted extension that supports every place of K.
0
1
We should remind the reader that [KS03, KS06, Po05] put together already gave
0
7 a complete proof of the theorem. However, our proof is much simpler. Furthermore,
0 as these papers have pointed out, using the structure theorem of Brauer groups
/
h over global fields, one can deduce from the main theorem the following corollary
t
a which gives an affirmative answer to the question raised in [AS02]. Recall that the
m
relative Brauer group Br(L/K) is the kernel of the restriction mapBr(K) Br(L)
: ([AS02]). −→
v
i
X Corollary. For a positive integer m, there always exists an abelian extension L/K
r of exponent m such that the m-torsion subgroup of Br(K) equals the relative Brauer
a
group Br(L/K).
We shall prove the theorem by constructing a sequence
M M M
0 1 k
⊂ ⊂ ··· ⊂ ⊂ ···
∞
of granted extensions over K and show that L = M enjoys the desired prop-
Sk=0 k
erty. Wewill frequently usethefact thatifanabelianextension K′/K isofexponent
pn and is unramified at v then the degree [K′ : K ], w v, divides pn. Thus, in
w v |
∗ The author was supported in part by the National Science Council of Taiwan, NSC95-2115-
M-003-005. † The author was supported in part by the National Science Council of Taiwan,
NSC95-2115-M-002-017-MY2.
1
2 CHI, LIAO,ANDTAN
order to check if K′/K is granted, it is sufficient to check the local degrees at the
ramified places.
Consider Pic := K∗ A∗ / ∗, where A∗ is the ideles group, v runs through
all places of KKand for\anK AQrcvhOimvedean plaKce v we let ∗ = K∗. For a finite
place p, let [p] be the image of any prime element π OvK∗ undver the natural
map K∗ A∗ Pic . The abelian group Pic isp fi∈nitelpy generated and by
p → K → K K
the Chabotarev density theorem we can find p ,p ,...,p outside any given finite
1 2 ℓ
set of places such that [p ],[p ],...,[p ] form a set of generators of Pic . Denote
1 2 ℓ K
S = p ,p ,...,p . For later usage, we choose for every idele x A∗ a global
eleme{nt1f 2 K∗, wℓ}hich is unique up to ∗, such that ∈ K
x ∈ OS
x = y f u (0.1)
x
· ·
for some y ℓ K∗ and u ∗.
∈ Qi=1 pi ∈ QvOv
Put Γ = Pic /(pnPic ) = K∗ A∗ /( ∗ (A∗ )pn). Class Field Theory
−1 K K \ K QvOv · K
([AT90]) identifies Γ as the Galois group of an abelian extension, denoted as
−1
M /K, which is everywhere unramified. This together with the obvious fact that
−1
the exponent of Γ divides pn implies that the extension M /K is granted.
−1 −1
Suppose T is a finite set of places with T S = and ∗ is an open
subgroup which contains ∗ via the natural∩embed∅ding KN∗ ֒⊂ Qv∈T OKv∗. Set
OS → Qv∈T v
Γ( ) := K∗ ∗ /((Y ∗ ) (A∗ )pn).
N \AK Ov ×N · K
v∈/T
This group is also of exponent dividing pn. Again, Class Field Theory identifies
Γ( ) as the Galois group of an abelian extension which we denote as M( )/K.
N N
Lemma 0.1. Let be as above. Then we have the exact sequence
N
0 → YOv∗/(N · Y(Ov∗)pn) →i Γ(N) →q Γ−1 → 0, (0.2)
v∈T v∈T
where i is induced from the natural map K∗ A∗ and q is the natural quotient
Qv∈T v → K
map.
Proof. It is enough to show the injectivity of i. Suppose z¯ ker(i) is obtained from
an element z ∗. Then z = f t xpn for some f K∈∗, t ∗ and
∈ Qv∈T Ov · · ∈ ∈ Qv∈/T Ov×N
x A∗ . Writex = y f uas in (0.1). Then we see thatfpn f ∗. Therefore if t
an∈duKaretheT-com·poxn·entsoftandu, thenz = (fpn f)xt· u∈pnOS ( ∗)pnT.
HenceTz¯= 0. x · · T· T ∈ N· Qv∈T Ov (cid:3)
Suppose p = 2 and ,..., , s > 0, are all the real places of K. Put R = K
∞1 ∞s j ∞j
and let sgn♭ be the sign map R∗ R∗/R Z/2Z. As M /K is unramified
j j → j j,+ ≃ −1
everywhere, for each j, we can choose a real place of M sitting over and use
−1 j
∞
it to define the sign map
sgn♭
sgn : M∗ R∗ j Z/2Z.
j −1 −→ j −→
Consider U = f ∗ K∗ s sgn (f) 0 (mod 2) . Then either ∗ = U
{ ∈ OS ⊂ |Pj=1 j ≡ } OS
or ∗ = U gU for some g. Let M′ be the composite of all M (√f), f U.
OS ` −1 −1 ∈
By the Chabotarev density theorem we can find a finite place p splitting completely
ON THE m-TORSION SUBGROUP OF THE BRAUER GROUP OF A GLOBAL FIELD 3
under M′ /K. Since on (M∗ )2 all the values of the map sgn := s sgn equal
−1 −1 Pj=1 j
0 Z/2Z, if g exists then U (M∗ )2 ∗ (M∗ )2. In this case, Kummer’s Theory
sa∈ys that M′ (√g)/M′ is ·a qu−a1draticOeSx·tens−io1n and we choose p so that it does
−1 −1
not completely split under M′ (√g)/K.
−1
Let sgn♭ be the map ∗ ∗/( ∗)2 Z/2Z, for (x ,x ,...,x ) ∗ s R∗
p Op → Op Op ≃ p 1 s ∈ Op×Qi=1 i
define sgn♭(x ,x ,...,x ) = sgn♭(x )+ s sgn♭(x ) and put
p 1 s p p Pi=1 i i
s
N0 = {(xp,x1,...,xs) ∈ Op∗ ×YR∗i | sgn♭(xp,x1,...,xs) ≡ 0 (mod 2)}.
i=1
The extension M( )/K is unramified outside p, ,..., . Since each R∗
N0 { ∞1 ∞s} i 6⊂
, the extension is ramified at each . The map sgn♭ induces an isomorphism
0 i
N∗ s R∗/ Z/2ZandfromLem∞ma 0.1we see that (since psplits completely
Op×Qi=1 i N0 ≃
underM′ /K)thelocalextensionofM( )/K atpisaquadraticextension. There-
−1 N0
fore M( )/K is granted and it supports every real place. We put M = M( ) if
0 0 0
N N
p = 2 and K has a real place. Otherwise, put M = M .
0 −1
We set all finite places of K into a sequence q ,q ,...,q ,.... And we shall
1 2 k
construct M so that it supports q . Assume that M is already constructed. If
k k k−1
it supports q , then we set M = M . Otherwise, assume that the decomposition
k k k−1
subgroup of Gal(M /K) at q is of order pm with m < n. We have m m +m
k−1 k 1 2
≥
where pm1 is the order of the inertia subgroup of Gal(M /K) and pm2 is the degree
k−1
of the residue field extension of M /K at q .
−1 k
Let S = v ,...,v be a finite set of places of K such that q / S and M /K
1 1 r k 1 k−1
{ } ∈
is unramified outside S q . We can assume that S is chosen so that S (S
1 k 1
∪{ } ∩ ∪
q ) = . And denote S = S S , S = S q . Choose a prime element π at
k 2 1 3 2 k
q{ a}nd le∅t f K∗ be the glob∪al element cho∪se{n b}efore. Then the S -units group
k π 3
∗ is the dire∈ct product of ∗ and the infinite cyclic subgroup generated by f .
OS3 OS2 π
To construct M , we will need to find a finite place w outside S , which splits
k 3
completely under M /K, and an open subgroup ∗, which contains ∗
via the natural embekd−d1ing K∗ ֒ K∗, such that theNquwo⊂tieOntw ∗/ is isomorpOhSic2
to Z/pnZ and contains a subg→roupwof order pn−m1 generateOdwbyNtwhe element f
π
(mod ).
w
N
Lemma 0.2. If is as above, then the following hold:
w
N
(1) The extension M( )/K is unramified outside w .
w
N { }
(2) Every finite place q S splits completely under the extension M( )/M .
2 w −1
(3) If q q , then the de∈gree [M( ) : K ] = pn−m1. N
| k Nw q qk
(4) The extension is totally ramified at w with [M( ) : K ] = pn.
w w w
N
Proof. Statements (1) and (4) are from the exact sequence (0.2), since the place w
splits completely under M /K. To prove (2), let π be a prime element at q and
−1 q
assume that in Γ the Frobenius at q is annihilated by pµ. Then πpµ = f t xpn
−1 q · ·
for some f K∗, t ∗ and x A∗ . This and the equation (0.1) imply that
f fpn ∗∈. They∈alQsoviOmvply that∈t ,Kthe w-component of t, is contained in ,
· x ∈ OS2 w Nw
since ∗ ( ∗)pn . Therefore in Γ( ) the Frobenius at q is also annihilated
OS2 · Ow ⊂ Nw Nw
by pµ. And (2) is proved.
4 CHI, LIAO,ANDTAN
Let F Γ( ) be the Frobenius element at q . Using the exact sequence (0.2),
r w k
we deduce∈thaNt pm2F ∗/ and hence πm2 = f t xpn for some f K∗,
t ∗ and x r∗∈. TOhwisNmweans pm2F equals the ·res·idue class t (mo∈d )
an∈d,QtovgOetvher with∈thAeKequation (0.1), this arlso shows that t f−1 (mwod ( ∗)Npnw).
We then write π = z f u′ with z s K∗ and u′ w ≡ ∗ and uOsewit to
· π · ∈ Qi=1 pi ∈ QvOv
deduce that f−1 f−pn fpm2 ∗ . Again, since ∗ ( ∗)pn , we have
· x · π ∈ OS2 OS2 · Ow ⊂ Nw
t fpm2 (mod ). Therefore in the quotient / the order of the residue
cwlas≡s t π(mod N) wequals pn−m1−m2. And this provOeswthNawt F is of order pn−m1. (cid:3)
w w r
N
We then set M = M M( ). It is clear that the exponent of Gal(M /K) is
k k−1 w k
N
pn. The extension M /K is unramified outside S q ,w . Lemma 0.2 implies
k 1 k
∪ { }
that [M : K ] = pn for p q , [M : K ] = pn and if q S , then [M : K ] =
k,p qk | k w w ∈ 1 k,q q
[M : K ]. Therefore M /K is granted and it supports q .
k−1,q q k k
To complete the proof, we need to find w and . Let us first consider the
w
N
case where char.(K) = p. Let q be a place of M sitting over q . Then we
k−1 k
have ord (f ) = pm1 6 ord (f ) = pm1. Since ord (f) = 0 for all f ∗ , the
q π · qk π q ∈ OS2
index ∗ (M∗ )pn : ∗ (M∗ )pn is a multiple of pn−m1. For a global element
|OS2 · k−1 OS3 · k−1 |
f M∗ let M ( p√n f) be the Kummer extension generated by all pnth root
∈ k−1 k−1
of f. Let K′ be the composite of all M ( p√n f), f ∗ and let K′ be the
2 k−1 ∈ OS2 3
composite of all M ( p√n f), f ∗ . Kummer’s theory tells us that the degree
[K′ : K′] is a multipk−le1of pn−m1. ∈ChOooS3se an element σ Gal(K′/K′) Gal(K′/K)
of 3order2 pn−m−1 and apply the Chabotarev density t∈heorem t3o ch2oo⊂se w to3be a
unramified place under K′/K such that the Frobenius at w equals σ. We choose
3
w outside S and put = ( ∗)pn. It is obvious that w splits completely under
M , ∗ 3 andNinw theOqwuotient ∗/ the order of f (mod ) equals
pn−k−m11. OSiSn2ce⊂KN′ wcontains all the primitivOewroNotws of 1 and w splitπs complNetwely under
2
K′/K, the local ring also contains all the primitive roots of 1. This implies that
∗2/ Z/pnZ. ThOusww and satisfy all the required conditions.
Ov Nw ≃ Nw
Finally, we consider the case where char.(K) = p. Apply the Chabotarev density
theorem and choose w to be a place outside S , splitting completely under M /K.
2 k−1
Recall that the p-part = 1+π of ∗ is the direct product of countable many
O1 wOw Ow
copies of Z ([We67]) and the local Leopoldt Conjecture holds ([Kis93]) in the sense
p
that Zp ⊗Z OS∗3 form a direct summand of O1. In other words, we have
∗
= Z Z C W
O1 p ⊗OS2 × p ⊗ ×
where C is the infinite cyclic group generated by f and W is a direct product of
π
countable many copies of Z . Using these, we can easily find an open subgroup
p
∗ so that ∗/ Z/pnZ, f f ∗ and f generates a subgroup
Nofwor⊂deOrwpn−m1 in O∗w/Nw.≃ ∈ Nw ∀ ∈ OS′ π
Ow Nw
References
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[AT90] E.ArtinandJ.Tate,Class Field Theory,AdvancedBookClassics,Vol.47,Addison-Wesley
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ON THE m-TORSION SUBGROUP OF THE BRAUER GROUP OF A GLOBAL FIELD 5
[Kis93] Kisilevsky, H.: Multiplicative independence in function fields. J. Number Theory 44, 352–
355 (1993)
[KS03] H. Kisilevsky and J. Sonn, On the n-torsion subgroup of the Brauer group of a number
field, J. Theor. Nombres Bordeaux 15(2003),199-204.
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[We67] A. Weil, Basic Number theory, Springer-Verlag,Berlin-New York, 1967.
DepartmentofMathematics,NationalTaiwanNormalUniversity,88,Sec.4,Ting-
Chou Road, Taipei, Taiwan 116, E-mail: [email protected]
DepartmentofMathematics,NationalTaiwanNormalUniversity,88,Sec.4,Ting-
Chou Road, Taipei, Taiwan 116, E-mail: [email protected]
Department of Mathematics, National Taiwan University, 1, Sec.4, Roosevelt
Road, Taipei, Taiwan 106, E-mail: [email protected]