The Local-Global Principle in Number Theory
Arnélie Schinck
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OAméiie Schinck, 200 1
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ABSTRACT
The Local-Global Principle in Number Theory
Amélie Schinck
"p-adic fields prmOZrlderernark~ble,easy and natural sol~utions to problems
,which apparently have no relation to p-udic fields and which othenrrise can be
resolved, i f at d l , only b t ~deep and arduous methods".
J .\\-.S. Cassels
The first Local-Global Principle. as formulated by Rasse in 1921, relates the behaviour
of rational quadratic forms in Q (global field) to their behaviour in the p a d i c fields Q,
(local fields).
The notion of using local information as a stepping Stone towards
understanding more dificult global properties has been generalized and applied to many
problemsl making Local-Global met hods a powerful number t heoret ic tool. Even when
the principle fails, m-e can sometimes saIvage some connection between the local and the
global. This thesis aims to give a survey of the basic theory.
Acknowledgements
1 wish to thank my thesis supervisor, Dr. Hershy Kisilevsky.
His guidance and
encouragement helped me stay on course. 1 am also grateful to Dr. Sebastien
Pauli for his help in translating some of Hemel's German text to English, giving
me a better understanding of the history. Finally, 1 would iike to thank Craig T.
Achen for his loving support throughout the writing of this t hesis.
Contents
Preface
1 Introduction
2
1.1
Hensel's Analogy . - - . .
1.2
The Transcendence of e . . . . . . . . - . .
1.3
ThepadicAbsoluteValue. . . . . .
1.4
Hensel.s Lemma .
2.2
2.3
. .. .
. .
.
1
... . .
3
.
3
. .........- - - .
6
9
Helmut Hûsse . . . . . . - . . . - . . . - . . . - . . . . - - - - - The Hasse-Minkowski Theorem . . . . . . . . . . . . . . . . . .
Hilbert's Eleventh Problem . . . . . . . . . . . . . . . . - . - -
Equations modula p
4.2
-
.
. . . ...... . . . .
9
11
13
15
1-5
Irrationality of f i a n d the Three Square Theorem . - - - - .
18
.
. . . . . . . . . .
... ....
. . - . .
.
.
The Hilbert Symbol
4.1
5
... .....
.....
- . ...
Diophantine Equations
3.1
3.2
4
. . . . . . . . . . . . . .
The Hasse Principle
2.1
3
1
The Hilbert Çymbol . . . . . . . . . . . . . . - . . . . - . . . . Hilbert's Tenth ProbIem . . . . . . . . . . . . . . . . . . . . . .
Proof of the Hasse-Minkowski Theorem
20
20
23
24
5.1
Proof for Ternary Quadratic Forms ( n = 3 ) . . - . . . - - . - -
24
5.2
Proof for n >3 . . . . . . . . . . . . . . . . . . . . . . . . - . . . .
Artin's Conjecture . . . , . . . . . . . . . . . . . . . . . . - . . .
Projecting a Conic ont0 a Line . . . . . - - - . - . . . - - . . .
26
5.3
5.4
30
31
6
Equivalence of Quadratic Forms
6.1
6.2
33
Preliminary Results . . . . . . . . . . . . . . . . . . . . . . . . .
A LocaCGlobal Pt-inciple . . . . . . . . . . . . . . . . . . . . . .
7 Failure of the Hasse Principle
7.1
7.2
7.3
7.4
8
9
37
A Simple Counterexample . . . . . . . . . . . . . . . . . . . . .
Seimer's Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Reichard's Equation . . . . . . . . . . . . . . . . . . . . . . . . .
Representation of Integers by Quadratic Forms . . . . . . . .
Finite Extensions of Qp
8.1
8.2
9.2
9.3
9.1
The p-adic Fields . . . . . . . . . . . . . . . . . . . . . . . . . .
Hasse-Minkowski for Algebraic Number FieIds . . . . . . . .
Euler's Criterion . . . . . . . . . . . . . . . . . . . . . . . . . . .
Proof of the Product F o n n d a . . . . . . . . . . . . . . . . . . .
Quadratic Reciprocity . . . . . . . . . . . . . . . . . . . . . . . .
Hasse's Product Formula . . . . . . . . . . . . . . . . . . . . . .
The Hilbert Norm Tlieorern . . . . . . . . . . . . .
The Hasse Norm Theorem for Cyclic Extensions
Abelian Extensions . . . . . . . . . . . . . . . . . .
Some-4pplications . . . . . . . . . . . . . . . . . .
11.2
11.3
11.4
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
First Cohornology Group . . . . . . . . . . . . . . . . . . . . .
Hasse Principle . . . . . . . . . . . . . . . . . . . . . . . . . . .
Principal Homogeneous Spaces . . . . . . . . . . . . . . . . . .
The Tate-Shafarevich Group . . . . . . . . . . . . . . . . . . .
Bibliography
47
49
53
56
57
58
60
11 Measuring the Failure
1 1.1
45
52
10 T h e Hasse Norm Theorem
10.1
10.2
10.3
10.4
37
38
43
47
The Product Formula and Quadratic Reciprocity
9.1
33
34
60
60
62
63
64
64
66
67
68
69
Preface
Chapter 1 is an overview of padic fields in their historical context.
It contains
important results on padic numbers, such as Hemel's Lemma, which will often be
referred to throughout the course of this thesis. Chapter 2 contains a statement of the
Local-Global Principle (Hasse Principle) and traces its historical development. For t his
part of rny study, I a m especially indebted to Dr. Giinther Frei, who kindly sent me a
copy of the article he wrote on the subject for the Proceedings of the Class Field Theory
Conference in Tokyo.
We apply the Local-Global Principle to the study of Diophanthe equations in Chapter
3, easily obtaining seemingly difficult results. In Chapter 4, the Hilbert symbol and its
properties are reviewed.
The proofs of the Hasse-Minkowski Theorem for the
repcesentation of a rational number by a quadratic form and the equivalence of
quadratic forms are given in Chapter 5 and 6. Both chapters were heavily influenced by
Borevich's and Shafarevich's approach in their collaborative t ext Number Theory.
Famous examples of the failure of the Hasse Principle are discussed at length in
Chapter 7, among them, the very h s t counterexample discovered by Reichard in 1942.
The Hasse-Minkowski Theorem is extended to finite extensions of the padic field in
Chapter 8, fully solving Hilbert's eleventh problem. Chapter 9 deduces quadratic
reciprocity from a local-global result; The product formula. The Hasse Norm Theorem,
of which the Hasse-Minkowski theorem is a particdm case, is the focus of Chapter 10.
The failure of the Hasse Principle is again the topic of Chapter 11, but this time, put in
the language of cohomology, allowing us to discuss the group which measures the
obstruction caUed the Tate-Shafarevich group.
vii
Chapter 1
Introduction
In the course of extending Kronecker's work on the factorization of prime ideals in
number fields, one of his studeats was led to the creation of the padic numbers.
This astute student was Kurt Hensel.
Hensel camed over the method of cornplex
analysis of expanding functions locaiiy in order to get information on their global
properties t o the land of Number Theory. H e keenly obsenred that the linear factors
(x - a), the prime ideals of the ring @[XI,play an andogous role in @(x) to the prime
numbers p in Q. Hensel translateci the Laurent series expansion of a h c t i o n f l x )
E
@(x) about a point a E @
flx)
=x
ai(x - a)'
into a closely analogous Laurent series expansion of r E Q in progressive powers of a
prime p
Hensel called the latter the padic expmion of r. Any rational number r can be
written padically with respect to every prime element p of
Z. The padic expansion
gives us local information about r near p, just as each Laurent expansion gives us
/
local information about f l x ) near a. Hensel showed that the collection of al1 such
Laurent series in powers of p forms a field, which he dubbed the field of padic
numbers and labeled K@).
In modem shorthand this field is denoted Qp. An
interesthg historicd side note is that, at the tirne, the dehition of a field
forrnulated by Dedekind required that it be a subfield of the field of complex
numbers, a requirement the padic fields fail to satisfy. The fields of p a d i c numbers
were the motivation behind the fundamental treatise on abstract Field Theory
undertaken by Steinitz. The padic fields were the earliest examples of what are
now known as Local Fields, with the exception of the fields of r d and complex
numbers. Local Fields will be discussed further in Chapter 8.
Hensel introduced p a d i c numbers in 1897 in a short p a p e entitled &r
eine neue
BegrWldung der Theorie der algebraischen Zahlen (About a ~ i e wfmndation of the
themg of algehaic nmbers). Hensel's treatment was extremely formal and failed
to dazzle the mathematicd community. The padics were dismisseci at the time as
imaginative play things, devoid of any real use because no substantial result for
which they were indispensable had been found.
Hilbert mentioried them in his
famous Paris lecture in 1900 yet they went virtually unnoticed for decades. This
attitude was in large part due to an error by Hensel.
1.2 The Transcendence of e
From the beginning, Hensel was intent on applying padic methods to transcendency
questions, a fashionable topic at the end of the 19th century. He gave a flawed
proof of the transcendence of e in 1905 during an invited lecture given in Merano,
Italy, to the Versnmmlung deutscher Nuturforscher und h z e (Meeting of German
Natural Scientists and Physicians). A discussion of Hensel's erroneous argument can
be found in [29]. Hensel later had to admit that ''The proof of the transcendency
of e given in ma( Meran lecture needs an essential campletion." This blunder did
not help the perceiveci legitimacy of p a d i c numbers in mathematical circles.
Jean-Paul
Bézivin
and Philippe Robba
obtained
a padic
proof
of
the
Lindemann-Weierstrass Theorem in a 1987 article [4]. This theorem states that the
values taken by the exponential function at different algebraic points are linearly
independent over the field of algebraic numbers.
theorem impiies that e is transcendental.
The Lindemann- Weierstrass
Referring to their result the authors
wrote:
"We are glad to vindicate Hensel in hzs idea that p-adic nwrnbers could be
used to proue transcendental results althuugh the proof devised here has
nothing to do urith the rnethod that Hensel intended to use."
1.3 The padic abdute value
Nevertheless, Hensel continued to develop his theory more fully.
He introduced
topological notions to the world of padics by defining the p-adic absolute value of a
rational number x
where x = p " ~ ( ~
) a ,+'
a a n d p 1 b. The usual absolute value is written 1. , 1 By
b withp
convention, vp(0) = a, hence 101, = O. It is easy to show that the padic absolute
value is non-archimedean, t hat is it satisfies the property
After laying this topological foundation, Hensel mostly concentrated on investigating
series which converge with respect to 1. ,1
This change of outlook was the h s t step
towards padic numbers becoming an essential tool in Number Theory. Kürschak
generalized Hemel's work to derive the Theory of Vduations in 1913. The padic
valuation endows Q with a metric space structure. From the definition of 1
see that two rational numbers are deemed close
1, we
if their difference (expressed in
lowest terms) has a numerator divisible by a high power of p.
It has been shown that Q is not complete with respect to 1 Ip, that is, not aii Cauchy
sequences converge with respect to 1
lp. Recall that a sequence
{xi),
is Cauchy if
for every
E
> O in Q, there exists an N E PI such that if i, j > N then
bi- xjlp < E.
Following the method of completing Q with respect to the usual absolute value to
obtain
W, Qp can be regarded as the topological field obtained by completing Q with
In this language, R is denoted Q
.,
respect to the padic absolute value.
This
absolute value takes on the same value on Q,, as it does on Q since the padic
absolute value of
a.pn = llpmo is a rational number.
nlmo
The padic nurnbers can then be seen, like the reals, as quivalence classes of Cauchy
padic analysis is quite simple compared to real analysis since a
sequences.
necessary and sufficient condition for a series to converge in Qp is that its terms
approach zero. Two absolute values are equivalent if they induce the same topology
on Q. The trivid absolute value is the absolute value 1 1 such that 101 = O and
for x
+ O.
= 1
A theorem due to Ostrowski states that every non-trivial absolute value
on Q is equivalent to 1 lp where p is a prime or p =
(see [l6], pp.44 for a proof).
The modern presentation of the subject normally follows these lines.
The ring of padic integers Zp is defined to be the set
Zi =
{X E
Qp : & 5 l}
The collection of all invertible elements of Zp forms the group of padic units Zi.
Since we need x and x-' to be elements of
4, the
group of padic units is then
describecl to be the set
25; = {x
Concretely, these are the elements of
where
a0
+ O.
E Qp :
& = l}
Qpthat have a padic expansion a0 + o ip + ...
Every nonzero padic number x can be represented uniquely in the
form
x = piou
where u is a padic unit.
The field Q can be identifid with the subfield of
& consisting of equivalence classes
containhg a constant Cauchy sequence (Le. all of whose terms are equal). This
shows that Q is properly containeci in Qp for any p
Q LI Q*
This is mdogous to the inclusion
@W
-
@((x - 4)
The field C ( x ) is not equal to @((x - a)) since for example the series for
tf =
5
is not an expansion of a rational function. Herein lies a paradox; The bigger field
Q (respectively C ( ( x - a ) ) ) gives us local information about the smaller global field
Q (respectively @ ( x ) ).
In general, given a valuation vp on a number field K, we can extend K to its
completion Kp. Piecing together information about the distinct completions Kp helps
us study properties of K. The following proposition is a simple example of solutions
to a global problem being deduced fiom local information.
Theorem 1.1 (Global Square Theorem)
A rational number x is a square in Q if and only if x is a square in Qp for all p 5 m.
B o of [a] Clear, since Q sits inside 9,.
[e] Suppose x is a square in
We can write x
=npVp(*). If
Qp
for all p 5
m.
x is a square in
If x
=O
W, then x
we are done. Suppose x
+ 0.
ie positive. If x is a square
P<='
at each prime p < a, then vp(x) is even for al1 p, which implies that x is a positive
number and is an even power. Thus x is a square in Q. O
To discuss squares, we will sometimes need the notational ease of the Legendre
symbol introduced in 1798.
Definition 1.1 (Legenàre Symbol)
Let p be an odd prime number and a an integer such that @,a) = 1. The Legendre
symbol of a and p is then:
The Legendre symbol has the multiplicative property
(% ) ($) = (*) .
1.4 Hensel's Lemma
Hensel also produced a theorem which is now a standard tool in padic analysis. It
is amicably known as Hensel's .Lemma.
This lemmo has its mots in Newton's
Method for finding the zeroes of a polynomial equation by successive refinements of
an initial approximate solution.
Hensel's Lemma allows us to lift non-trivial
solutions of congruences modulo p to non-trivial padic solutions of polynomials.
Any field complete under a non-archimedean valuation, also satisfies Hensel's
Lemma. As J.W.S. Cassels states in [Il] pp.83:
"In the literature there i s a varietg of results that go under this name.
lheir comma feature is that the existence o f an approximate solution of
an equatia or system of equatiims in a complete valved field implies the
existence of an exad sohtion to which it i s a n approximation, subjed to
conditions t o the general e f f e d that the approximate solution i s cgood
enougiv
"
.
We will use the following versions of Hensel's Lernma;
Theorem 1.2 (Hensel's Lemma)
Let Xx) = a#
+ ...+ a2x2+ a l x + <IO
derivative of Ax).
Suppose there
-
&,[XI.Let f '(x) = na,,xn-l + ...+al be the
exists a /3 E Zp such that flp) O(modp) and
E
f '(b)+O(modp). Then there exists a unique a
E
Zpsuch that
i) A a ) = O
ii) a
= /3(modp).
Hensel's Lemma, among other things, allows us to d e t e k e the padic squares.
CorollaFy 1.1 p
Let u
E
4'
#
2
be a padic unit. If there exists an a
E
Zp such that
then u is a square in Zp.
fioof
Let u
E
Zi. Suppose there exists a a
E
u2
-
u(modpZ,),
Zp such that a2 = u(modpZ,).
Hensel's Lemma, this solution lifts to a padic solution since p
+ 2 and a E
By
Zpxthus
2a+O(modp).
We d l sometimes need to use the stronger version of Hensel's Lemma.
Theorem 1.3 (Hensel's Lemma)
Let
-
B E Zp satisfy the condition fl/3)lp< I~'(#?)I;.Then there exists an a E 5 such
thatfla) = O and a
p(modp).
The equation x2 - 17 = O is a case in point for the need cf Theorem 1.3 because it
does indeed have mots in
Q2 (try B = 1). An immediate consequence of Theorern
1.3 is the foilowing Corollary.
Corollary 1.2 A 2-adic unit u is a square in Z2 if aad only if u = l(mod8Z2).
Roof Let u
E
Z; andflx) = x 2 - U . Assume u
-
l(mod8Z2). Then R I ) ( 12" is
strictly smaiier than V'(1)I2 = 2-2. By Theorem 1.3 there exists an a E Zt such that
a2 = u. The other direction follows frorn the fact that u2 = (1 +
= l(mod8Z2).
O
Now r e c d that every nonzero padic number x c m be represented uniquely in the
form x = pnu, where u is a padic unit and n = v,(x)
E
2. So, putting Coroilary 1.1
and 1.2 together, we obtain
Theorem 1.4
Let x = upn, u E Zi, n
E
Z , x is a padic square if
= O(mod2)
2. a2 = u(modpZp) for p + 2, u
1. n
Note that squaring u
=
-
l(mod8Zt) for p = 2.
1 + 8k where k
E
Z2 results in the fourth powers in Z;.
For
any k E Z2, we find that (1 + 8kl2 is congruent to 1 modulo 16. It turns out that
the fourth powers in Z; exactly correspond to the elements which are congruent to 1
modulo 16.
The following is an n-variable version of Hemel's Lemma, which will prove quite
useful later on.
Theorem 1.5 (Hensel's Lernma; n-variable version)
A x l , ...,x,,) E Z&, ...,x.]. Suppose there exists BI,PZ,...,B. E 2& such that
ABi, pz,...,pn) = O(modp) and for some i the partial derivative off with respect to xi
satisfies waXi(Pr,
pz,...,&)+O(m~dp), then there exists a 1 , ...,an E Zp S U C that:
~
Let
i) f(a1, ...,an) = O
ii)
ai
= B i ( d p ) , f o r 1 5 i 5 n.
Chapter 2
The Hasse Principle
The theme of this chapter is the link between the behaviour of quadratic forms (with
rational coefficients) over the rational field Q and over the various p a d i c fields Qp
(Q, = IR).
The main result in this area is the Hasse-Minkowski Theorem which
reduces questions over Q about quadratic forms with rational coefficients to the
corresponding questions over Qp.
2.1 Helxnut Hasse
As previously mentioned, the events of 1905 left the impression that p a d i c numbers
were an unfruchtbarer Seitenweg (fkuitless sideway;
Quote of Richard Courant
found in [21],Vol. 1, pp-viii). Fortunately for the legwy of p a d i c numbers, a
mathematicdy gifted young man named Helmut Hasse stumbled upon a copy of
Hensel's 1913 text Zahlentheorie in a Gottingen second-hand bookstore in March of
1920. In the 1st chapter of this book, Hensel applies padic methods to binary and
ternary quadrat ic forms wit h rational coefficients, obtaining necessary conditions for
the representability of a rational number.
Hasse was so fascinated by the theory of p-adic numbers that he decided to study
them a t the source. Less than two months after that fatefd day in March, Hasse
matriculated at Marburg University to study under Hensel. Hensel urged Hasse to
continue the work done in the last chapter of Zahlentheorie. In particular, Hensel
gave his student the task of determinhg whether the necessary conditions he had
found were also sufficient. In addition, Hasse was to examine quaternary quadratic
forms for analogous conditions.
Hasse quickiy delivered. One short year after entering Marburg, Hasse graduatecl as
Dodor F%dzsqphziae. In his doctoral dissertation (1921), Hasse compietely solves
the problem Hensel had given him not only for ternary forms, but for n-ary forms.
A theorem dating back to Legendre was key in Hasse's solution to the problem.
Legendre had reduced the problem of deciding if a homogeneous quadratic form with
rational coefficients has a rational zero to
Theorem 2.1 (Legendre's Theorem)
Let a,b,c be integers other than O and not a l l of the same sign, where abc is square
hee. Then the equation ax2 + b 9
+ a2= O
has a non-trivial solution x, y, z
E Z
if
and only if -bc,-ac,-ab are quadratic residues modulo lai, Ibl, Ici respectively.
Hasse needed a way to translate Legendre's result to padic numbers, which he did
not readily see. He wrote to Hensel for guidance, receiving it in the form of a
postcard (reproduced in vol 1 pp.ix of [21]). Hensel's reply helped Hasse see that the
conditions given by Legendre are equivalent to saying that the ternary form admits
a non-trivial padic representation of zero for each prime p.
Then Legendre's
Theorem itself states that it also admits a non-trivial rational representation of zero.
To see how to interpret Legendre's Theorem padically @ # 2), observe that a, b and
c not all being of the same sign (Le. the form is indefinite) guarantees that
ax2 + b#
+ cz2 = O
has a nomtrivial solution over W.
Furthermore, since the
equation concerned is homogeneous, the distinction between rational and integral
solutions vanishes. So we can assume x, y and z are relatively prime integers.
that if p
1 abc, then
t d
+ by + d = O
Note
has a non-trivial p a d i c solution. To show
this, let t = 1 and consider the congruence t
d s -c
P+L
2
values modulo p and so does -c- b?,
which is the solution modulo p.
- b9(modp).
Since m? takes on
they thus have a value in cornmon
This solution, Say (x~,yo,zo), lifts to a p a d i c
solution since all partial derivatives evaluated at (xo,yo,zo) are not congruent to O
modp.
Suppose ar2+ Z$
+ a2= O has a non-trivial p a d i c solution for ail p. To show -ab
a quadratic residue modc it is necessary and sufficient to show
is
(P)
= 1
for aJl
pic (by the Chinese Remainder Theorem). Assume plc. If p b and ply then pk which
Thus x and y are prime to p.
is a contradiction.
ux2 + by'
The congruence
= O(modp) is solvable in Z which implies that -ab is a quadratic residue
modp for all primes p dividing c. So, -ab is a quadratic residue modc.
NOW, assume -bc
-
-
There exists an integer d such that cd = ~ ( m o d p ) .We then have -bc
thus O = cdt
-
u2(modp) where p ,f bcu and let f =x x ,y,z) = ax2 + by2 + cz2.
+ b(modp).
This results in AO, 1, d ) = b + cd2
c2d2(modp)
O(modp). Since p
#
2
and p 1 b then û$Vy(O,1,d) = 2b+O(modp). B y Hensel's Lemma applied to
f(x,y,z) = ar2+ b y
+ a2there exists ai,az,a3
flai,a2,a3)
E
Zpsuch that
= = a l 2+ b a 2
2
+ca32
=O
where (at,az,a3) = (O,l,d)(modp).
2.2 The NasseMinkowski Theorem
Once Hasse had understood the theorem padically, he proved that it also holds for
any n-ary quadratic form with rational coefficients. Hasse was now in a position to
formulate his first Local-Global Principle fur the representation of numbers by
quadratic forms. Hasse called this the Fundamental Theorem for the representation
of numbers by quadratic forms. His result first appeared in the Crelle Journal of
1923 under the title Cfber die Darstellbarkeit von Zahlen d m h quadratische
F m e n i m Kôrper der rationalen Zahlen (About the representability of a
quadrutic fm in the field of rational numbers) [17]:
Theorem 2.2 (Hasse-Minkowski)
A quadratic form f with rational coefficients represents zero non-trivialiy in the field
of rational numbers Q if and only if f represents zero non-trivially in the field of real
numbers W and in all fields of p a d i c numbers Qp.
The Hasse-Minkowski Theorem is the simplest, purest form of the Local-Global
Principle.
We will see the proof in Chapter 5.
The Hasse-Minkowski Theorem
completely solves the problem of deciding whether a quadratic form with rational
coefficients represents a given rationd number because of the following:
Theorem 2.3
If a nonsingular quadratic form represents zero non-triviaily in some field K of
characteristic not equal t o 2 then it represent all elements of K non-trivially.
R mf
First, note that if K is a field of characteristic different from 2, any
nonsingular quadratic form C a g ~ i( a~gjE K, 1 5 i, j 5 n) can be put into diagonal
form
by a linear change of variables ([6]pp.392). Let f
GI
,...,x,,]
(BI,...,Bn) E K"
and let
l i ,~
. . ~ n=
) a ix 1 2 + ... + anxn2 be in
be a non-trivial solution to f = O (Le.
aiBi2 + . . . + t ~ , , / =
?~
O )~. Without loss of generality, we may assume Bi c O. Let y be
any element of K.
xi = p i ( l + t),
xk
Setting t = yl(4a 1
We want to show that f can be made to represent y.
= Pk(l
2),
- t ) for k = 2, ..A
Substituting in the form f we obtain
we see that f represents y.
O
Write
The Local-Global Principle, also known as the Hasse Rinciple, was the vindication
of Hemel's creation.
It was the h t intimation of the padic numbers' true
potentid. After a reasonably long period of time, their study subsequently regained
importance. "[ ...] the metits of this approach t w k a long tirne to percolate into
As late as 1930 C.E. Dz'cksm m l d
the collective mathematical conscimsness.
write a mmograph a quadratic f m s [...]
and Morde11 m l d review it [...]
withmt either o f t h e m betraying the least awareness of the p d i c viezopaint"
(Ref: [9], pp.vi). Minkowski's name appears in the title cf the theorem because he
had found an equivalent result in 1890.
Theorem 2.4 (Minkowski's Theorem)
Given a quadratic form A x I , ...,x,) with rational coefficients, if A x 1, ...,x,) = O(modpr)
bas a non-trivial solution for ad primes p and positive integer r and i f f l x l , ...,x,) = O
has a non-trivial solution in the reals, then j(x i ,...,xn) = O has a non-trivial rational
solution.
Minkowski's Theorem is equivalent to the Hasse-Minkowski Theorem.
given a quadratic form with rational coefficients f i
,,...,x ~ ) , the
If we are
congruence
f l x i , ...,xn) = O(modpk) is solvable for all k 2 1 if and only if the equation
-
f l x 1, ...,xn) = O is solvable in padic integers. For instance to solve x2 = 2 in
must solve the system of congruences x2
calculation, we
and
2(mod7'),
obtain the square mots of 2:
fi = xi
for all r 2 1.
=3
Q7
we
Doing the
+ 1 x 7 + 2 x 7*+ 6 x 73
= x z = 4 + 5 x 7 + 4 ~ 7 ~ + 0=-XI
~ 7 ~
As J.W.S. Cassels states in his article Diophantine Equations &th Special
Re ference to Elliptic Curves:
'me use of p-ad.ic nurnbers 2s ut first sight rather artificial compared with
the older lanpage of m g m e n c e s but of fers great technical aduantages
because Qp i s a field whe~eas2 modpn i s not even an integral d o m a ~ n . ~
2.3 Hilbert's Eleventh Problem
Hilbert, a close fkiend of Minkowski, had studied at length the quadratic extensions
Q(&)
of Q, which is tantamount to studying binary quadratic forms.
He was
determined to generalize Minkowski's Theorem to quadratic forms with algebraic
numbers as coefficients. This search ied him to prove the existence of certain class
fields. He stressed the importance he placed on extending Minkowski's Theorem by
proposing this as his eleventh problem in his address delivered before the
International Congress of Mathematicians in Paris in 1900. Hilbert believed at the
time that:
11. Quaciratic forms wit h any algebraic numerical coefficients
' ' ~ present
r
lnauledge of the themg of quadratic fields puts us in a
position to attack successfully the theory of quadratic f m s &th any
numbe~of variables and &th any aigebraZc numetiml coef fidents." (341
The Hasse-Minkowski Theorem as stated above does not completely solve Hilbert's
eleventh problem, since we are just over the rationals. We wiil see in Chapter 8
that Hasse s w n generalized the Hasse-Minkowski Theorem to arbitrary algebraic
fields.
Chapter 3
Diophantine Equations
Although padic numbers have a function theoretic origin, they are at their best
when applied to the theory of Diophantine equations. The study of Diophantine
equations is the branch of Number Theory dealing with integer or rational solutions
to polynomial equations.
A curve of genus O d e h e d over Q is birationaily
= 0,
equivalent over Q either to the line or to a conic section ar2 + b f +a2
a,b,c E Q (See [9] pp.255, for a sketch of the proof).
This reduces the study of
Diophantine equations of genus O c w e s to the study of conics.
As one might expect, the Local-Global Principle is often used to answer questions
about rational solutions of Diophantine equations.
An equation f = O is said to
satisfy the Hasse Principle if the existence of non-trivial local solutions of f = O
implies the existence of a non-trivial global solution. it is often extremely difficult
to determine the solvability of a Diophantine equation in the integers or the
ration&, such as xn +yn + zn = O, n > 2. On the other hand, the solvability rnodulo
a prime p, can be determined in a finite number of steps. Also, the existence of real
solutions is easily determined by sign consideratiom. Thus for each prime p 5
00,
Hensel's Lemma then shows that the question of existence of solutions of any given
Diophantine equation over
IR or
Q,, is decidable in the sense of mathematical logic.
So in general approaching a problem locdly is easier than tackling the same problem
globally.
Think of the problem of finding a local minimum of a function in
cornparison to the problem of finding its global minimum. This implies a strategy;
The old idea of divide and conquer prevails.
Theorem 3.1 (Necessary Conditions for Solvability)
A necessaxy condition for the existence of rational solutions to a polynomial
equation f = O with rational coefficients is that it has padic solutions for all primes
p.
If f = O has integer coefficients then a necessary condition for the existence of
integer solutions is that f = O(modp) has solutions for al1 primes p.
Hence we can approach Q through its local fields
Qp and also approach each Qp
through the finite fields IFp by reduction modulo p. Consider the equation
After clearing the denominators, we may look for x, y,z
gcd(x,y,z)
=
1.
If S b and SC then s21(3x2+ 4 9 )
assurnption that gcd(x,y , z ) = 1 . If 5 j' x or 5
E
L and assume that
SR which contradicts o u .
y, the congruence
is impossible since 3 is not a quadratic residue modulo 5. So 3x2 + 4y2 - 5z2 = O has
no non-trivial solution in Q. The quadratic equation
is also clearly insoluble rationally since no real solution exists. It is easy to see by
examples that if a general polynomial equation has a solution modulo p, no
conclusions can be drawn.
The congruence conditions are not sufficient.
For
instance 3x2 + 4y2 - 5z2 = O has a non-trivial solution modulo 7, but no non-trivial
solution in Q.
Many interesting results may be obtained through congruence considerations. For
instance, the equation
x,'
with xi E
+h3
+
4
= hC4
~ 3
~
@(xi
~ 2 ~ 3 ~=x 1 4 )
Pr, only has the trivial solution. This foilows easily from the fact that
o3 = 0,f l ( m o d 9 ) for any integer a, so x , 3
solution xi
~
+ 2x2' + 4 q 3 = O(mod9)
has the sole
= x2 = x3 = O(mod9) which results in x4 = O(mod9).
3.2 Irrationality of f i and the Three Square Theorem
The following is essenti*
the classical proof of the irrationality of 2 stated in the
language of padics.
Theorem 3.2
A-oo f
a is irrational.
Consider the equation x2 = 2. Taking the 2-adic valuat ion on both sides; If
x2 = 2 has a solution a in Qz then a2 = 2. So 2v(a)
Thus x2 = 2 has no solution in
Q2.
= v(2) =
1 . No such a exists.
Since Q is a subfield of Q I , we certainiy do not
have any solutions in Q. Similar reasoning yields that @ is irrational for every
positive prime p.
O
In 1770, Lagrange proved that every positive integer is a sum of squares of four
integers. In 1798, Legendre answered the deeper question: Exadly w hat positive
integers need al1 four squares?
Of course, Legendre's proof did not use the
foliowing Local-Global argument.
Theorem 3.3 (Legendre's Three Square Theorem)
A positive integer n is a sum of squares of three integers if and only if it is not of the
form 4'(8k
+ 7) where r, k E Z,r 2 0.
Roof By the Hasse-Minkowski Theorem, we must show that n is represented by the
form f =Ax,y,z) = x2 +$
2.3, if x2 +?+
+
in IR and all Qp. The real case is clear. By Theorem
9 represents O over Qp, then it represents ail elements of Qp, so it
certainly represents the positive integer n.
We will assume that the condition of
Davenport-Cassels Theorem holds (proof (24) pp.46).
This theorern states that,
under a certain condition, if an integer n is represented by f in Q then it is also
represented by f in Z.
The form f represents O at all completions of Q except perhaps at Qz.
Cl-:
Roof of Claim. Since the equation concerned is homogeneous, we may look for
x,y,z
E
9. Consider the congruence x2 +y + z2 = O(modp) for all p z 2. Suppose
-
If there exists a
there is no solution of the given congruence.
B2 -1 (modp) the congruence
x2 + y2 + z2
would have the solution (1,/3,O), so
some integer a s 0 , -1 (modp), then
find x,y,z
-
E
xZ+y2 + z2
Z such that a
-
-
Z such that
O(modp)
($ ) = -1.
(
E
)=-1
y2(modp),and -(a
+1)
Also, if we assume
(F)
= 1 for
Otherwise we would be able to
-
z2(modp) and the congruence
O(modp) would have the solution ( 1 , a,-(a
+ 1))
= (x,y, z).
We find
that
which implies that every nonzero residue class modulo p is a quadratic residue
moddo p. This is a contradiction since p > 2. The mod p solutions ail lift to padic
solutions by Hemel's Lemma. Therefore f represents O in Qpfor all odd primes p.
= 0 , l or 4(mod8). Hence x2 +y2 +z2
Furthermore, x2 = O,l(mod4) so if
Now, observe that for any integer x we have x2
cannot be congruent to 7 modulo 8.
x2 + y 2 + z 2 = O(mod4), x,y and z are even, Say x = hi,y = Zyi, z = î z i . Hence if
4'(8k+7)=x2+y2+&
then 4 C 1 ( 8 k + 7 ) = x , Z + y 1 22+ z l .So we see that if
(x,y,z)
E
z3 is a
solution to the equation there exists another integer solution of
smailer height (H(P) = m a x { ~ l , M , ~ > ) .Contiming in this fashion (descent) would
result in 8k + 7
= Xr
2
+
+ zr 2,
a contradiction.
The method of descent seen above was h s t introduced by Fermat.
We can now
effortlessly derive Lagrange's result .
Theorem 3.4 (Lagrange's Four Square Theorem)
Any positive integer n is a sum of squares of four integers.
R o o f . If n is not of the form 4'(8k + 7), it is the sum of 3 squares. Otherwise,
4-n = -1 (mod 8) and thus 4-n - 1 = 6(mod 8) so n - 4r is a sum of three squares.
Thus n
=
(n - 4')
+ 4'
is the s u m of four squares.
Chapter 4
The Hilbert Symbol
It is natural at this point to introduce the Hilbert symbol which will make our
statements more concise.
4.1 The Hilbert Symbol
Since any quadratic form is equivalent to a diagonal form, we may confine our
attention to quadratic forms which are in the diagonal form.
In particular any
d + by + a2
ternary quadratic form with coefficients in K is equivalent to
(a,b, c E K) which represents zero if and only if z2 - ar'- b y (a,b
zero.
E
K) represents
The representability of zero in the field K by a quadratic form in three
variables can then be expressed with the Hilbert symbol of a and b relative to K
( U , ~ ) K=
if z2 - ax2 - b$ = O has a non-trivial solution in K
otherwise
Proposition 4.1
~(6).
The equation z2 - ax2 - bfl = O has a
non-trivial solution in P if and only if a is the norm of an element in ~(6)'.
Let a,b E K" be square £iee and let
Roof Let a = Z+
~nbe a typicd nonzero element of
fi)" and a = N(a) its
is a quadratic extension of K. There exists
norm. Since b is not a square, ~(6).
The equation 12-m2-by' = O then has the
Y , Z E K " such that a = Z 2 - b p .
solution (x,y,z) = ( 1 , Y,Z ) and so (a,b ) =~1. Conversely, i f (a,b ) =~ 1, the equation
z2
- ax2 - b?
=0
has a non-trivial solution in
K3. Observe that
x t O or else b
would be a square so a = 22 - b p where Z = z/x and Y = ylx. We see that a is then
the n o m of Z + J ~ Y .
By Proposition 4.1, one can rewrite the Hilbert symbol in terms of norms
i f a = N(a) ; a
otherwise
E
~(6).
From the multiplicativity of the norrn, it follows that
(a,b)(a,b l ) = (a,66').
In particular (a,b)
=
1
(a,bb')
=
(a,b').
Furthermore, since the definition of the
Hilbert symbol depends on the solvability of z2 - d
- b$
= O, and since
(xo,yo,zo)
is a solution of z2 - ax2 - bbyZ = O then (xo,zo,y o ) is a solution of z2 - bx2 - ay2 = 0.
Thus the Hilbert symbol is symmetric, that is
( a b ) = @,a).
Combining these last two properties one gets the bilinearity of the Hilbert symbol
Now, the equation z2 - pi2 + a?
9-tu2-y2+&=0
=O
has solutions x
= y,
z = O, the equation
has solutions x = y = z and z 2 - a x 2 t c 2 y 2 = 0 bas the
solutions (O, y, z = kcy) also (a,-a) = (a,-l)(a, a).
W e thus get the foliowing
properties
(a,*) = (a,c2) = ( a , l - a ) = 1
and
The Hilbert symbol also has that property
(a,b ) = (a,-ab)
@,a) = (a,-1).
For later
use, we introduce an explicit means by which to calculate the Hilbert
symbol of o and b relative to the padic field
Qp,
the proof of which is herein
ornitted.
Theorem 4-1
Let a,b E Qp where a = ~ p ~ pand
( ~ b) = v p ' ~ ( ~U),;v
Hilbert symbol over
R mf
Qp
E
Zi, p
f
2. The values of the
for dinerent primes p are given by:
See [24] pp. 2 1.
Corollary 4-1
Let u,b E Q where a = upn and b = vpm; u7v E z;.
F o r p f 2 , q ifp ,f ab then the
Hilbert symbol (a,b)Qp= 1 . For p = 2 if 2 1 ab and u or v is congruent t o 1 modulo
4 then ( a , b ) ~=, 1 .
Corollary 4.2
Let p and q be odd primes, p
+ q.
Then
1
-1
ifp = l ( m d 4 )
ifp = 3(mod4)
4.2 Hilbert's Tenth Problem
10. Determination of the solvability of a Diophantine equation
"Given a Diophantzne equation with any number of unknowns and with
rational integer coeffiaents. Devise a process, which w u l d detennine by a
finite number of operations whether the equatim i s solvable i n rational
integers ." [34]
Corollary 4.1 impiies that there is a method to test whethcr or not z2 - ax2- b y = O
has a solution in Q,,, p < 00 (in a finite number of steps) since we need only consider
the primes which divide a or b. By the Hasse-Minkowski Theorem, this means that
there is a method to test, in a finite number of steps, whether or not a given ternary
quadratic form with rational coefficients has a rational zero. For p = a the method
is even more simple.
This answers Hilbert's Tenth ptoblem for Diophantine
equations of degree 2 in three variables. The Hasse-Minkowski Theorem provides a n
algorithm to determine the solvability of any quadratic form over a field of
characteristic other than 2. The general Hilbert's tenth problem has been shown to
be impossible to solve by Matijasevic in 1970.
Chapter 5
Proof of the HasseMinkowski Theorem
We now have all the necessary tools to prove the Hasse-Minkowski Theorem (stated
below). Most important to keep in mind is that even though we assumeci solutions
in Q,, @
#
2) at the omet, once we have reduced to the diagonal form with
relatively prime, square fiee nonzero ai, this is equivalent to having a solution in IFp
(p
#
2) because of Hemel's Lemma.
5.1 Proof for Ternary Quadratic Forms (n = 3)
Since all ternary quadratic forms with rational coefficients can be transformed into
the form flx,y,z)
= z2
- ux2 - b y ,
we can focus our attention on those forms and
rewrite the Hasse-Minkowski Theorem of Chapter 2 as
Theorem 5.1 (Hasse-Minkowski)
The quadratic form flx,y,z) = 9 - ax2- b y = O, where a,b
E
Q, has non-trivial
solutions in Q if and only if it has nomtrivial solutions in Qpfor each p
< m.
Awf First, some preliminary remarks. We may assume that a and b are integers.
If not, for instance if we have z2- (l/s)x2- c$
then have z2-sxo
2
- cy2 = O to work
= O, simply substitute x = sxo and we
with. We may also assume that a and b are
square kee by absorbing the square factor into one of the unknowns. We may look
for x, y,z
E
Z since j(x, y,z)
=O
can be solved in Q if and only if it can be solved in Z
(because f is homogeneous). Without loss of generality we may also suppose that
5 161 and that x,y,z have no cornmon factor. Let us now begin the proof, which is
by induction on m =
+ 161.
The general case a
and b
=O
#
O, which covers the cases m = 1 s a
= 0,
Ibl = 1 and
m = 2 where a = O and 161 = 2, reduces to the question: If b is a square modulo p
for al1 p 5 ao is b a squa~ein Q? This question was answered by the Global Square
Theorem (Theorem 1.1).
If m
= 2, where
la1 = 161 = 1 , we have the four poçsibilities:
The first three equations have solutions in Q thus trivially in all the padic fields Q.,
The last equation has no solutions in W so it clearly has no solutions in Q, i.e.
( 1 , l ) ~= -1
s
( 1 , l ) Q=
l -1.
Now suppose m 2 3 and that for b l + lbl < k and that (a,b)%
implies (a,b)q
=
=
1 for all p 5 cc
1 . Write
where the pi are distinct primes.
Claim: a is a square modulo pi for all pi(b.
If a = O(mod pi) then a is a square modulo pi.
If otO(modpi), then since (a,b)% = 1, 3 x,y,z
z2
-
4ir2
-b y
= O.
E Qpl
-
not d zero such that
This can be reduced to the congruence z2
&(modpi).
If a is
not a square modulo pi then x = O ( m d p i ) which in turn implies z2 = O(modpi).
Hence pi divides x,y and z, a contradiction.
Now, a is a square modulo pi for all pilb, so by the Chinese Rernainder Theorem
,a
is a square modulo b. There thus e s t integers t, and b'such that;
t2 = a + bb'
(5.1)
B y inspection of (5.1), z2 -a?- bb'y = O has the solution ( l , l , t )= (x,y,z), that is
to say (a,bb')Q = 1 which trivially impües that (a,bb')qp= 1 V p < a. Choose t such
that # 5
y.
Then,
+ 16'1 < k and
since Ibl 2 2. So
(a,bl)% = 1 V p 5 m
a,
thus, by the hypothesis, we have the implication
( ~ , b '=) 1~. Applying the properties of the Hilbert symbol
seen in the previous chapter, we have (a,b)qp= I V p 5
00
and (a,bb')qP= 1Vp 5 m
thus (a,bl)% = 1 V p < m which implies (a,b l ) p= 1 by hypothesis. Finaily, we have
(~,bb')~=
and
l ( ~ , b ' ) ~ =so
l , (a,b)q=l.
This ends the proof of the
Hasse-Minkowski Theorem for ternary quadratic forms.
5.2 Proof for n > 3
The idea behind the proof of the Hasse-Minkowski Theorem for four variables is to
construct
2
g = alxl
an integer r which
+<
r 2 ~ and
~ ~
is simultaneously represented
h = -a3x32 - acr42
(ai E Q )
resulting
by the forms
in
a
rational
representation of zero by the indefinite form
XXI,x2,x3,xd) = aixl2 + ~ 2 x22+ a3x3 2 + ~
4
-
q
~
We will see that the construction of such an integer follows fiom the padic
solvabiüty off = O at all primes. The existence of the integer r with the required
properties will then be proved using Dirichlet's Theorem on primes in arithmetic
progressions.
Theorem 5.2 (Dirichlet's Theoiem)
If a and m are relatively prime integers greater than O, then there exist infinitely
many primes p such that p
= a(modm).
Theorem 5.3
Let K be a field with more than five elements. If a i x 1 2+ ... + onxn2 = O (ai E K) has
a non-trivial solution in K then it has solution in K for which all the variables take
non-zero values.
R w f [6] pp.394.
Theorem 5.4
The quaternaxy quadratic form with rational coefficients
flxi,xz,x3,x4) = o p l
2
+ a2x22 + a3x32 + am2
=O
where the ais are square f i e , is solvable in Q if and only if it is solvable in
Qp
for all
p Ia.
R o o f (Ref. [6] pp.68). Since the form f is solvable in
IR,it is indefinite, Say a 1 > O
and a4 < O. We will consider the forms
g = o 1 x l 2 + a z ~ 2 2 and
h=-a3x32 -a4x4 2
Let S denote the set of distinct odd primes which divide the coefficients a 1, a2, a3
and ad, together with the prime 2 and the infinite prime. Since f = O is solvable in
Qp
for all p 5
00,
it is in particular solvable for all p
choose a representation of zero for each prime p
where ad
#
O. Fix bP = a 1
and g represent zero in
+ a2822=
E
E
S. By Theorem 5.3, we can
S
+ 0 4 8 ~ ~If)some
.
b,
- ( t z j ~ ~ ~
=
O, then h
Qp,hence they represent all numbers by Theorem 2.3. So
we may assume that our representation is such that each bp is a nonzero padic
integer divisible by a t most the k s t power of p.
Consider the system of congruences
The integer r
* O,
which we may choose to be positive, is uniquely determinecl
2
moduio n = 16p12--p~ by the Chinese Remainder Theorern. Since bpi is divisible
by at most the first power of pi, then Ibp,r-'l,, = 1 (Le. bP,r-l is a pi-adic unit) and
-
= l(modpi). Hence bp,r-' is a quadratic residue modulo p z p and thus bp,r-' is
a pi-adic square (Corollary 1.1) . Also, 22 X 62, so b2r1 l(mod8Z2)and b 2 r 1is a
b#'
2-adic square (Corollary 1.2). Since bp and r dways ciiffer by a square factor in all
Q p for p
E
S, both the forms
- a s 3 2 -a&
represent zero in Q p for al1 p
2
E
-rx,,2
and
a i q 2 +a2232 - q2
S. Since we assumed ai < O and a4 > O, both forms
are indefinite and thus have a solution in W.
If p e
S and p ,f r then p does not
divide any of the coefficients of the ternary forms and thus the forms represent zero
in Q p (Corollary 4.1). If there is one more prime q
S such that q
1 r,
we can use
the fact that if a quadratic form with rational coefncients f represents zero in all
fields Q p for all p 5
except possibly for Q , for some prime q, then f &O represents
zero in Q,. The reason why this is true will become clear in Chapter 9. Now, by
the ternary case of the Hasse-Minkowski Theorem proved above, there exists
rationai numbers ci,c2,
o and e4 such that r = oicl2 + ~
2
2 ~ =2 -a3c3
~ '
- a4c42 , so
that
To prove the existence of such an integer r, we may use Dirichlet's Theorem on
prime numbers in arithmetic progressions. Let s be a number satisfymg the system
of congruences above and let d = gcd(s,m). Then sld and mld are relatively prime
and Dirichlet's Theorem implies that there is a positive integer k such that
sld + km/d = q is prime. We can then let r = s + km = dq. The result follows.
For five variables we repeat the argument above, £inding an integer r by Dirichlet's
'+ =lx2
Theorem which is represented by g = o 1x1
2
and h = -(a3x3
2
+ a& 2 + asx52).
Then by Hasse-Minkowski Theorem for three and four variables, g and h represents
r in Q, giving us a rational representation of zero by
alxl
2
+ a2x22 + a3x32 + a4x42 + asx52 .
It is easy to see that repeated application of this process will produce a proof of the
n variable Hasse-Minkowski.
Theorem 5.5
Any quadratic form in five or more variables with coefficients in Qp always has a
non-trivial solution in Qp.
R o o f See [6] pp.51.
In combination with the Hasse-Minkowski Theorem, Theoreu 5.5 implies the result
Theorem 5.6
Every indefinite quadratic form f in five or more variables has a non-trivial rational
zero.
The success in applying padic numbers to the theory of quadratic forms made it
natural to consider applying them to forms of higher order. Artin had conjectureci;
Artin's Conjecture
,, ...,x.]
All homogeneous equations in Qp[x
of degree r in at least 9 + 1 variables
have a non-trivial solution in Q.
Artin's conjecture holds for r = 2 by virtue of Theorem 5.5. The conjecture has also
been proved for r = 3, (Refer to [27]). The general Artin Conjecture remained open
for approximately thirty years until a counterexample of degree r = 4 in 18 variables
was discovered in 1966 by G. Terjanian [28]. Terjanian first notes that it sunices to
have a homogeneous polynomial of degree 4, flxi,...,x9)
property that J x l ,
...,x9) = O(mod4)
E
Z[xi, ...,x93 with the
implies al1 the xj are divisible by 2 to have a
counterexample. Indeed the polynomial
would then be a counterexample because if it had a zero in Q , it would have a
primitive zero in Z2, which is not possible. To verify the 1 s t statement, suppose
there is a primitive zero in &. The form flxi, ..x9) is then congruent to O modulo 4
which impiies 2 divides xi for i = 1, ..., 9. So 16 divides (5.2) which implies that 4
divides j(xio, ...,x l s ) and hence 2 divides xi for i = 10, ..., 18.
Terjanian then
constructs the following polynomial:
which bas the desired property. It is easily verified that if x,y and z are divisible by
2, then n
= O(mod4). If
Consequently flx 1, ...,x9)
-
any one of x, y or z is not divisible by 2 then n
= 3(mod4).
O(mod4) implies t hat all xi's are divisible by 2 and we
have a counterexample by the discussion above.
There is a conciliatory theorem [Z] which states that
Theorem 5.7
For any natural number r, there exists a h i t e set S of primes such that any form of
degree r over Q, represents zero non-trivially for every p not in S, provided that the
number of variables is at least
6 + 1.
However, there is no known method for determining the set S of exceptionai prime
numbers ([24] pp.39).
Theorems 5.5 and 5.7 are part of a f d y of theorems in
Number Theory which read, as Borevich-Shafarevich put it: "A11 is well as lmg as
the number of variables és sufficiently large" [6] pp.57.
In 1957, B.J. Birch
proved in his article Homogeneous f m s of odd degree in a large nurnber o f
variables [5] that:
Given an odd number d, there exists an n such that every
homogeneous equation of degree d in n variables has a non-trivial rational solution.
5.4 Projecting a Conic onto a Line
The Hasse-Minkowski Theorem does not help us find a rational solution, it only tells
us whether or not o u - search would be in vain.
If our quadratic equation has a
rational solution, it has infinitely many. Once we have explicitly found a rational
point P = (xo,yo) on a rational conic f ; we can derive ail others by drawing the
rational line L = y - y0 = t(x - xo) where t is rational and projecting the conic onto
the line from point P.
Consider the classicai problem of parametrizing the
Pythagorean triples. Integer solutions to
P+P=ZZ
where X,Y and Z have no common factors correspond tc rational solutions t o the
unit circle
x2 +y2 = 1
where x = X/Z and y = Y/Z are in lowest term. We will project £rom the point (-1,O)
onto the y-axis. Let (O, t ) be the yintercept and let L denote the l h e through (-1.0)
and (O,?). Armed with the knowledge of x and y, we may get t easily since the dope
of L is
t =
2
E Q. The equation of L is then y = t(x + 1). If the point
1 +x
(x,y) is
on the circle and the line L we obtain the relation
(1
+ x)(l -x)
= 1 - x 2 = y2 = t 2 ( l +x)2
Cancelling out an (x + 1), solving for x and ushg the relation y
= t(1
+ x ) to solve for
y one gets the familiar parametrization of the unit circle as the set
Clearly x and y are rational if and only if t is rational. AU rational points on the
unit circle x2 + y 2 = 1 may be found by simply plugging in arbitrary rational
numbers in place of t. This technique dows us to describe the infinite amount of
rational points on any conic completely given one ratiorial point.
Hence, the
Diophantine theory of conics, dominated by the Hasse Principle, is well-understood.
Chapter 6
Equivalence of Quadratic Forms
Mathematicians of the 18th century were greatly preoccupied by the problem of
determining conditions for the existence of soiutions to binary quadratic equations
+ bxy + c$ + dr + ey +f = O
with integer coefficients and to find algorithms to
find ail such solutions. It was determined early on that one need only consider
equations of the form (a,b, c)
= ux2
+ bxy + cy2 = n
where a,b and c are relatively
prime Le. primitive forms. When faced with the problem of finding integer solutions
for such an equation, it is normal to try to simplifv it by doing a change of
variables.
For instance performing the transformation xo = 30x + 43y, y0
lSx2 + 43xy + 32y2 = 223 &es
us the simpler equation:
elementary calculation yieids the solutions xo
2
xo
+ 7 lyo2
= 13380.
= S î , y0 = f8.
comesponding solutions for our original equation, x
=y
= (1/30)(x0
on
An
However, the
- 43yo), y = YO, are
not integers.
The need to find invertible transformations led Lagrange to lay the foundation for
the classification of quadratic forms. Let K be an arbitrary field of characteristic
=x
n
different from 2. Let f
agXiXj, where
= aji (av E
K) . Let K be an arbitrary
field of characteristic different £rom 2. The determinant d of the quadratic form f is
the determinant of A = (ar/), the matrix of the quadratic form j: If d = O the form f
is said to be singular. Otherwise, it is c d e d nonsingular. Two quadratic forms f
and g are calleci equivalent (denoted by f-g) if there is a nonsingular linear change
of variables, which takes one form to the other.
Equivalent forms represent the
same numbers. The following theorems were found in the Algebraac Supplement of
Borevich-Shsfarevich [6]. The reader is referred t o this text for the proofs.
Theorem 6.1
If two quadratic forms over the field K are equivalent over K, then their
determinants differ by a nonzero factor which is a square in K.
Definition 6.1
Let f and g be two noasingular quadratic forms. The form f + g denotes l(x)
+ g(y)
where x and y are independent sets of variables (Le. we are taking the direct sum of
the two quadratic spaces.)
Theorem 6.2 (Witt's Theorem)
Let Ag, h be nonsingular quadratic forms over K.
If the forms f + g and f+ h are
equivalent over K then g and h are equivalent over K.
Theorem 6.3
If a quadratic form f in n variables with coefficients in a field K represents a
#
O
then f is equivalent to ux2 + g(x2, ...,x,,) over K.,where g is a quadratic form in n - 1
variables with coefficients in K.
6.2
A Local-Global Principle
Hasse applied Hensel's p a d i c methods to the problem of equivalence of quadratic
forms which resulted in a variant of the Hasse-Minkowski Theorem.
Hasse's
Habilitation dissertation (1923) contained a proof of the following
Theorem 6.4 (Hasse-Minkowski)
Two nonsingular quadratic forms with rational coefncients are equivalent over the
field of rational numbers, if and only if they are equivalent over W and al1 the padic
fields Qp.
Roof
The proof 1 give is fkom [6] pp.70. It is by induction on n, the number of
variables. If n = 1, f= ax2-g = b*2 in Q, for all p 5
oo
if and only if d b is a square
in Q, for all p 5 m which is true if and only if a/b is a square in Q (Theorem 1.1). If
n > 1, let r
E
Q be represented by f in a l l
Q,.
Since f-g in all Qp,r is also
represented by g in all Qp.By the Hasse-Minkowski Theorem r b then represented
by f and g in Q. Applying Theorem 6.3
where fi and gl are quadratic forms in n - 1 variables. By Witt's Theorem, since f-g
in al1 Q,,, then f i -gi in all
Qp.By the induction hypothesis,
-gl in Q, so f-g in
This result was also printed in the Crelle journal of 1923 in the article Ober die
Aquiualenz quadratischer F m e n irn K&pr
der rationalen Zahlen ( A h t the
equivalence of quadrutic f m s mer the field of rational numbers) [18].Minkowski
had once again previously stated an equivalent result.
See [13], pp.315, for a
statement of Minkowski's result.
The importance of the Hasse Principle is that both the representability of a number
by a given quadratic form and whether two quadratic forms are @valent
determined by local information.
can be
This is interesting because questions of
equivalence, like representability of a rational nurnber over local fields, are
manageable. For instance, Sylvester's Law states that (Ref. 1201 pp. 42)
Theorem 8.5 (Sylvester's Law)
Let ri be the number of positive terms in the diagonalization of a form and r z be the
number of negative terms. Two quadratic forms over IR are equivalent if and only if
t hey have the same dimension n = r 1 + r2 and the same signature ri
- r2.
Chapter 7
Failure of the Hasse Principle
The result for quadratic forms with rational coefficients certainly looks encouraging.
Unfortunately, a general equation having solutions in HP and all padic fields Qp, by
no means guarantees solutions in Q.
That the Local-Global R-inclple fails for
certain curves was seen early on. The fîrst explicit counterexample, X" - 17 = ZP,
was offered by Reichardt in 1942. We will see in detail that A? - 17 = 2Y2 has
solutions everywhere locaily, but not globally in the proof of Theorem 7.4. Ernst S.
Selmer later gave a large number of Diophantine equations for which the Hasse
Principle fails, including the now famous 3P + 4Y3 + SZ3 = O. Selmer's example will
also be discussed at length.
7.1
A Simple Counterexample
First , we wiU present the considerably simpler counterexample to the Hasse
Principle. The equation
f l X ) = (Xl - 2 ) ( P - 17)(X2
has a root in Qp for all p 5
Roof
ab,
- 34)
=O
but has no mot in Q.
It is clear that f has a solution in
IR and no solution in Q since the numbers
2, 17 and 34 are not squares of rational numbers. It remains to be shown that f = O
has a solution in all completions of Q.
Case 1 p # 2,l7
If either (or both)
in
(5)
= 1 or
(F)
= 1, then
-
a2 2 or a2 = I7(modp) is solvable
mpZ. Thus there exists an a E WpZ such that fla) = O(modp).
The
derivat ive 2u*O(mod p) since p + 2 and a+O(mod p), so solutions lift by Hemel's
multiplicativity of the Legendre symbois.
Case3 p = 2
Here, 17 = 1(mod 8) so 17 is a 2-adic square by Corollary 1.2.
Hence,
(p- 2)(* - 17)(* - 34) = O d o e ~not have a solution in
Q even though it
has a solution in al1 padic fields Q.
So we see that, in general, there is no analogue of Hasse's principle for c u v e s of
genus g 2 1 or even for general nonsingular cubic c w e s , although there are
interesting classes of Diophantine cubic equations for which it is known to hold. For
instance, we will see in Chapter 10 that Selmer worked out in 1953 sufficient criteria
for the Hasse Principle to hold for the cubic
(a,b,c,d E Z, x,y,z, w
E
Q).
equation u3+ by3 + a3+ dw3 = 0,
There is no known method to determine in a finite
number of steps whether a given rational cubic h a . a rational point.
Moreover,
when there are rational solutions, there may not be infinitely many.
As we have seen in the previous section, the idea of looking modulo p for all primes
p is not sufficient for cubic c w e s . Let us take a closer look at the counterexample
due to Selmer (1951)
f l ~ , ~ , z=
) 3 P + 4 P +5z3
where X, Y,Z E Q.
(7-1)
Since the form is homogeneous we may look for X,Y.2e Z.
Selmer's cuve has padic solutions for aii p, but no rational solution. Selmer proved
the global insolubility of 3 s
+ 4Y3 + Sz3= O by considering factorization in a cubic
x+6y
5
(4
83 = 3 For a
3sketch of Selmer's argument see [7],pp.206. We will go about it in quite a dinerent
field Q(S). Selrner rewrites the equation as N
=
3
way. There is a theorern which asserts
Theorem 7.1
A c u v e of genus 1 over IFp has a point defined over IFp and thus over Q,.
Roof See [IO],pp.119.
In our particular example, all primes p except for the primes p
=
2,3 and 5 have
BY
good reduction (since they do not divide the discriminant D = -2 123"S 2 )
Theorem 7.1, the form thus hss a zero in
Qp
for all p
+ 2,3,5.
Note that for p
the form clearly has a zero. For p = 2, the form (7.1) has the zero
This
zero
is
in
Q2
since
3
is
the
inverse
ol
3
4 5 is a 3-adic cube because 4 / 5
=
3 9 = Y3(mod5) has
f '(1,2,0) = 9
8
so
3, note that
= (4)(-16) = lO(mod27) (since -16 is the inverse
of 5) and 10 = 1 + 9 is a 3-adic cube. So (7.1) has the 3-aàic zero (O, 1,
p=5,
(dm,
O, 1).
modulo
-5/3 = (-5)(3) = -15 = l(m0d8) hence -513 is a Zadic cube. For p
= oo
the
solution
X = 1,Y=2.
The
da).
For
derivative
- lZ~O(mod5)so the solution to 3 p + 4]A + 5Z3 = O lifts to a padic
solution by Hemel's Lemma.
Now to show that 3 P + 4p + 5Z3 = O has no rational solutions, we will need the
following t heory.
Lemma 7.1
Let p3 = 1, p
A,B
E
#
1 and Q(p) = K.
Let the Galois group Gal(Q(p)/Q)
=
G.
If
f l are distinct Galois conjugate points on the c w e u3 + v' + d = O where
u,v E Q, d E
Z,then the iine joining A and B is rational and intersets the cwve in
a third point with rationd coefficients.
Roof
p2. Let A = (u1,vi)
The Galois group G is generated by the automorphism O : p
and B
=
(uz,v2) be such that ou1 = u2 and
be denoted by L : v
= mu
+ bo.
ml
The slope of
= v2. The line joiaing A and B
L is m =
(VI
( ~
-VI)
-2 u1)
Applying
-
O
will
to the
Applying a to the
intersect bo = V I
- mu* we get ab0 = v2 - mu2 = bo,
thus bo E Q. Let C = (u3,v3) be
the third point of intersection. Substituting the line in the cuve
-
0 = u3 + (mu + bo)3+ d = (m3 + l ) ( u - u ~ ) ( u
- UZ)(U u t )
where u1,u2,ui
are the zeros.
Comparing the coefficients of u2 results in
ui +uz +u3 = (3m2bo)/(m3+ 1). Now ul +u2 and (3m2bo)/(m3+ 1) are rational thus
u3 is rational. Plugging back into the equation of the line,
v3
is also rational and
hence C is a rational point.
0
Theorem 7.2
Let a,b,c > 1 be distinct integers such that d = abc is cube kee. Suppose there are
integers x,y,z not a l l zero such tbat ax3 + by'
U,V,Wwith W # O such that
+ d = O.
Thm there exists integers
u +v3+dW3 = O.
R w f ([IO],pp.86) Let p3 = 1,p + 1. Bear in mind that 1 + p + p2 = O. Write
C = ax3 + pby3 + p2cz)
and
=ar'+p2b9+pcr'
p25 + Pq = 369
and
p{
+ p2q = 3
d
where y = - 3 9 .
The two points A = (e,pq,y) and B = (q,p2& y ) are solutions to the Selmer curve
U.' +
+dl@ = O. The points A and B are Galois conjugates over Q, since
6
and q
are conjugates, pq and p2P are conjugates and y is rational. First observe that the
line L joining A and B will intersect U3+ Y3 + dW' = O in a third point C which is
not (1,-1,O).
If it was, it would imply that a = b, but a,b, c are distinct integers.
Dehomogenizing with respect to W, we may apply Lemma 7.1. The line L is rational
and will intersect u3 + v3 + d3 = O in a third point with rational coordinates. Putting
C in homogeneous coordinates by clearing the denominators we get integers U,V, W
with W # O such that U3+
+dW3 = 0.
0
Lemma 7.2 The only rational point of X'
A mf (Sketch
+
+ 602)= O is (1, -1,O)
[IO],pp.86 )
Let u = UIW, v = V I K The Selmer c w e u3 + v3 = d is birationdy equivalent to
the Weierstrass curve 9 = x3 - î 4 33dt under the transformation
for which the inverse transformation is
In our example, d = -60 so X'
+ r +602 = 0
is birationally e q u i v d e ~ ~to
t
Y2 = X3 - 2433602. Let E(Q) denote the group of rational points of the eiiiptic curve.
The group E(QP)IZE(Q) for P = X
- 2433602is trivial. For a proof, see [IO],pp.53.
The group
(pinot necessarily distinct) and
Then E(Q)/2E(Q) = (Z/2Z)Hs where r is the rank and s is the number of 2-torsion
points. If E(Q)/ZE(Q) is trivial then E(Q) has no 2-torsion points (and the rank is
O).
A cdculation with PARI with the function elltors shows that the c w e has no
ptorsion for p odd (it actually shows there is no 2-torsion as weli). So E(Q) is
trivial.
O
Theorem 7.3
3X'
+ 4Y' - 5Z3 = O has no rational solutions.
+ 4r + 5Z3 = O had a rational point then
that x3 +y3 + 60z3 = 0, but by Lemma 7.2,
R o o f By the two previous results, if 3 s
there would exist x,y,z
E
Z,z # O such
no such point exists.
O
We now turn our attention to Reichardt's equation;
P - 17 = 2p.
Theorem 7.4
X - 17
= 2Y2 has padic solutions for
Roof
Let flx Y) = p - 17 - 2F = O. Since the reduction of the curve has genus 1
dl primes p 5 00, but has no rational solution.
for ail p except for p = 2 and p = 17, we need only take a closer look at what
happens to the curve a t those primes. Observe that 17
= l(mod 16) so 17 a tadic
fourth power by the discussion foilowing Theorem 1.4.
2-adic solution
(m,O).
- 17
=
2Y2 thus has a
X - 17 - 2Y2 = p - 2 p = O(mod 17)
Furthermore,
o a2 = 2(mod 17) where a = XIY, Y z O. The congruence has a non-trivial solution
since
(2
1 7 ) = 1.
17-adic
For example X = Y = 6 .
solution
by
Hensel's
We can then lift this solution to a
Lemma
since
f16,6) = O(mod 17)
where
@7aX(6,6)$O(rnod 17). Hence our equation has local solutions everywhere.
Now to show that A?
Number Theory.
K
- 17 = 2p has no global solutions, we will use some Algebraic
The following is not the argument used by Reichardt.
= Q(Jlf).
Suppose
- 17 = 2Y2
Let
has a rational solution. Set X = a/c in lowest
terms. The equation
where a,b and c are rational integers would then have an integral solution. Note
that if a prime p divides a and b this implies that p divides c. Similarly if p divides
both b and c it divides a. So gcd(a,c) = gcd(b,c) = gcd(a, b) = 1 . Both a and c are
odd, since the right hand side of (7.2) is even and both a and c cannot be even since
gcd(a, c ) = 1 . Since K has integral basis (1,
('
+
2
and
>, this makes a2+c2J17
2
Now note that
ideal p
a 2 + c 2 m and
2
a2
-' 17
2
are relatively prime, since if a prime
(m)divides both, then p divides a and c. This contradicts the tact that
= 1. If (m)
divides both, then (,/if)divides a2 so (17) divides a4. By
#
gcd(a,c)
equation (7.2), (17) divides b2 which is a contradiction (since (17) would then
divides all of (7.2)). Now
a 4 - 17c4 =
4
Since both
a2 f c 2 m
2
are algebraic integers,
b2
is a rational integer.
2
Hence b is
even, say b = 2bo. So
The class nurnber of
K is 1 we thus have unique factorkation. In the field K,2 splits
into
Since
a2 fc2m
by one of
2
are relatively prime and conjugate to each other, each is divisible
( '*2m ).
~e can thus mite
for some unit q. The left hand side and its norm are positive, as are ( 5 * 2 m ) ,
p2 and their n o m (N( 5 ' 2 m=)2).
unit in K is
EO = 4
+ m. Every unit in K
hindamental unit (Le. q
power even, thus
-
r)
= *(eo)").
= 5u2(md17)
A fundamental
N(q)> O .
is thus plus or minus some power of the
If n oàd N((eO)") < O since N(Eo) = -1.
a square and can be absorbeci in pz.
This is not
Hence 4a2 = 5(u2 f 3 4 w + 17v2).
4a2
So q > 0 thus
( 2 ~ / u=
) ~S(mod 17) and
(*)
possible over
Put q
SO
= 1,
QI, because
= -1.
O
7.4 Representation of Integers by Quaciratic For=
A question naturaliy mises: If a quadratic fmfi)&th integer coefficients has
a non-trivial zero in
Zp for aM p
5 a, does Ax) have a non-trivial zero in Z? In
marked contrast to the question of solutions in rational numbers, the answer is NO.
This point is articulateci by J.W.S. Cassels, in the introduction to his survey article
Diophantine equations with special reference to elliptic curves:
''The field Q of
ratiaal numkr is already less recalcitrant (because less basic?) t h n the ring Z
of integers [. ..]"
Consider the equation x2
p
* 3.
If p
- 3 4 9 = -1.
= 3, the equation has the solution (3/5,1/5) in
does not have a solution in
field K =
The solution (x,y) = (5/3,1/3) is in
Z3.Yet,
Z. To see this, look at x2 - 3 4 3 = -1
5
for ail
x2 - 34y2 = -1
over the quadratic
Q(m).
The fundamental unit in K is &a = 35 + 6
m where the norm of
ao is 1. Therefore, -1 is not the n o m of an algebraic integer (i-e.of a unit in
OK).
We thus have no nice equivalent to the Hasse-Minkowski Theorem for the
representation of an integer by a quadratic form with integer coefficients, although
many results in this area still rely heavily on the Local-Global Rinciple. A striking
example of such a result due to Conway and Schneeberger (1993) is the following
theorem [3]
Theorem 7.5 (The Fifteen Theorem)
If a positivedefinite quadratic form having integer matrix represents every positive
integer up to 15 then it represents every positive integer.
The criterion is indeed necessary since it is easily seen that a2 + 2b2 + Sc2 + 5d2
represents the integers O through 14, but it does not represent 15 because
a2 + 2b2 = O(mod 5) e
(* ) = 1 which is false.
The original proof of the Fifteen
Theorem was never published, but the recent proof by Manjui Bhargava does rely
on some Local-Global concepts. For more on this fascinating subject, refer to [12]
and [3].
Chapter 8
Finite Extensions of Q,
Up to this point, Local Field meant the padic field
referred to the field of rationd numbers Q.
Qp
or l
k and Globd Field
However, these are the simplest
occurrences of Local and Globai Fields. In general a Local Field is a finite, algebraic
extension of
IR, Qp or
the field of forma1 power series over some finite field.
A
Globai Field is a h i t e , algebraic extension of Q (Le. an algebraic number field) or
the field of algebraic functions in one variable over a finite field. We saw that the
behaviour of a problem in Q can sometimes be deduced hom it's behaviour in
Qp
and W. In this chapter, we lay the foundation necessary to generalize this idea to
finite algebraic extensions of
Qp.
8.1 The p-adic Fields
Let L be a local field of degree n over Qp.
i t can be shown that there exists a unique
absolute value 1 IL on L which extends 1 1, on Qp given by 1331:
I*IL =
( I I v U Q ~ ( ~ )lhI ~ )
(X E
Now let K be am arbitrary algebraic number field. A f i t e prime ideal p induces a
non-archimedean absolute value 1 jp on K which restricts to the usud padic absolute
value if pp. The completion of the algebraic number field K with respect to 1 1, is a
local field (proof see [I5],pp.55) c d e d the padic number field. This field is denoted
Kp. Elements of K p are c d e d p-adic numbers. Pictorially,
By the discussion above, the padic absolute value 1 lp of Qpis extended t o a unique
absolute value
1 1,
d e h d by:
where np = [Kp : QJ
If cr is an embedding a : K
4
@, it is called an intinite prime and it induces an
archimedean absolute value on K (given by laIo = laal). An archimedean absolute
value is an extension of the archimedean absolute value
1 1,
on Q.
The
comesponding cornpletion Kp is W or @. Hilbert was the fiist to introduce the formal
concept of innnite primes [13], pp.239.
K
=
~ ( a The
) . Galois
Let us laok a t an explicit example. Take
group is Gal(K@) = { l , ~ )where a :
1 1 + a l ~ = l + nand ~ l + n l , = n - l .
(3 + n)(3
- f i )= p,p,'.
13 + n l p ,
=
117,
P- n l p ,
fi
+
-f i .
The prime 7 splits in K
So
into
Calculating the various absolute values we find
= 1,
P+
alP,<
= 1, and P - filp,?
Local fields have many of the same properties as
=
ln.
op. If 1 lp is a non-archimedean
blp 5 1) forms the ring of p-adic integers
denoted Op and the group of units is OP = (x E Kp : BIp = 1). Techniques which
involve proving theorems about the ground field K by exploring a l l its embeddings in
its completions Kp and W (or @) are called Local methods.
absolute value, the subset (x e K p :
8.2 Hasse-Minkowski for Algebraic Number Fields
Using Takagi's work on class field theory and Hilbert's ideas, Hasse fuily solved
Hilbert's eleventh problem by extending his Local-Global A-inciple to algebraic
number fields in 1924. The problem of representability and equivalence of quadratic
forms for an arbitrary nuniber field was thus completely solved. Instead of looking
for rational solutions in the usual sense of the word we may look for solutions in a
general algebraic number field. The ring of integers of K (i.e. the ring of elements of
K integral over Z) will be denoted OK. The more general Hasse-Minkowski Theorem
then reads as follows;
Theorem 8.1
Let j ( ~ i , . . . , x ~be
) a quadratic form with coefficients in some algebraic number field
K.
Suppose f l x !,...,x.)
finite prime ide&
=O
p of
has a non-trivial solution in aU p-adic fields Kp for ad
OK and in all the completions corresponding to the
archimedean absolute values. Then f = O has a non-trivial solution in K.
We will see in Chapter 10 that the prwf of Theorem 8.1 follows easily from a
beautiful theorem of Hasse's known as the N o m Theorem.
Most Local-Global
results concerning the relationship between Q and the padic fields Qpgeneralize to
algebraic number fields and the corresponding local fields, iike the Hasse-Minkowski
Theorem. The proofs are often more sophisticated.
The objective of Local Class Field Theory is to describe d abelian extensions of
Local Fields.
Unlike
IR, Qp possesses
many such extensions.
The problem of
c l a s s e n g the abelian extensions of Q is known as the Kronecker-Weber Theorem.
A famous theorem of Kronecker and Weber states that: Every abelian extension of
Q is contained in a cgdotomic extension. This theorem can be proved from the
corresponding theorem for the local fields
Qp,[9] pp.151, although historicaily, the
local result was proved using the global theorem.
Chapter 9
The Product Formula and Quadratic
Reciprocity
Euler (1707-1783) was led to the discovery of the celebrated law of quadratic
-
reciprocity in the years 1744-46 after examining many individual cases of t h e
solutions of the congruences x2
= p(modq) and x2 q(modp). Euler noticed that if p
and q are both primes of the form 4k+3, then only one of the congruences is
solvable. On the other hand, if one or both p and q are of the form 4k + 1, then both
or neither congruences are solvable. Legendre independently rediscovered this result
in 1785. Using the Legendre symbol the quadratic reciprocity law then becomes the
familiar
As Weil mentions in a letter to Simone Weil [32],the quadratic reciprocity is a
startling result because it means that the prime numbers p for which m is a
quadratic residue are precisely the ones that belong to a certain arithmetic
progression moduio 4n1. For instance 1 =
(*)
=
(31)
=
(*)
=
...
This is
quite an astounding statement in light of the fact that the distribution of primes in
arithmetic progression seems to be random.
Legendre attempted to prove
reciprocity, but his proof was flawed in six of the eight cases.
Gauss rediscovered quadratic reciprocity when he was just eighteen years of age. He
declared it to be Wlie Gdden nieorena of Number Ine@.
He gave the £ ù s t
complete proof after one year of intense work in 1795. "It t d u r e d me for a whde
gear" said Gauss "and eluded m y most strenuuus efforts befme finallp 1 p
t the
proo f expIained
[.ln.
A bitter animosity between Legendre and Gauss was sparked
when Gauss exposed his proof in his 1801 text Disquisiticmes Arithmetzcae
attributing the law to himself, making only a passing remark about Legendre's
contribution. Gauss went back to the law of quadratic reciprocity nwnerous times
during his illustrious career, proving it in eight drastically different ways.
His
ultimate goal was to £ind an approach which would generalize to higher powers.
This QU& led to the introduction of Cyclotomy and Gaussian integers by Gauss and
eventualiy was the motivation behind the development of the Theory of Algebraic
Numbers.
Gauss' prooh can
be found in many Number Theory books. Our goal will be to
prove the quadratic reciprmity in Local-Global fashion. This is accomplished by
k t proving Hilbert's Product Formula, for a,b E Q,
n (a,b)Q
P.9
that it is equivalent to quadratic reciprocity.
P
=
1 and showing
The source of al1 such product
formulas is the padic product formula
Theorem 9.1
For any nonzero x
Roof Let x
E
E
Q,
2. We can write x = f
n p " ~ ( ~ )The
.
sign of
F
general result for a rational x follows.
x is
-a-. Thus
kl.0
a
Note that in Theorem 9.1, Lxl, = 1 for h a r t a.U primep in Q.
9.1 Euler's Criterion
Proving Hiibert's Product Formula will require us to become f h d i a r with some
preliminary notions.
Theorem 8.2 (Euleri Criterion)
A padic unit u E Zi is a square if and only if u e l ) n= I(modp)
Roof [
e
lLet p be an odd prime. There exista a
C*')R = -l(modp) and {0,C,C2, ...,{PI>
6 E Z such that CFp-'= I(modp),
natural number k where O ( k 5 p - 1 we may write u
and only if u * ~ )=~(-l)k(modp).
f l x ) = x2 - u.
f (4")
Thus
= 24"+O(modp)
#
-u
-
O(modp)
For some
Ck(modp) which is h u e if
Suppose k = 2m (Le.
Aem) = 5'
since p
- -
is a complete residue system.
and
l(modp)) and let
the
derivat ive
2. We can apply Hensel's Lemma to find z
E
Z,,
such thatAz) = 9 - u = O. Thus u is a square.
[cl Suppose u = $ for some z E Z,.
We have z = tm(modp) for some m
Recall that in Chapter 1 we saw that if p is an odd prime and u a padic unit, then
it is necessary and sunicient for u to be a square that
(f ) = 1.
wit h Euler's criterion, this result becornes
Lemma 9.1
Let p be an odd prime and let a be an integer such that p X a. Then
(p)= 0@-~'L(m0dp)
In conjunction
Euler's Criterion is extremely useful in pmving various properties of the Legendre
symbol. The following are easy consequences of Euler's criterion.
Corollary 9.1 Let p be an odd prime and let a and b be integers such that p
p
a,
1 b. Then
The &st complementary Law of quadratic reciprocity, which is a consequence of
Euler's Criterion,
was a result already known by Pierre de Fermat.
Simply stated
the law says that the only prime divisors of x2 + 1 are 2 and primes of the form
4 k + 1. The second law states that 2 is a square in
F, if and only i f p = 8 k f
1.
Corollary 9.2 (Complementary Law 1) Let p be an odd prime, then
($1
1 ifp = l(mod4)
-1 i f p = 3(mod4)
= (-1)WIQ =
($-) = 1 e (-l)@lIn = 1(modp) o
Note that clearly, i f p = 2 then ($) = 1
R w f Applying Euler's Criterion to -1,
O
-
= (p- 1)/2(mod2) o p
I(mod4).
0
Corollary 9.3 (Complementary Law 2) Let p be an odd prime, then
(5)
f
=(+Y-M =
R o o f From [24], pp.7.
1 ifp*fl(modS)
-1 ifp =î3(mod8)
First observe that if 2 has x as a square root x in an
algebraic closure of Fp then 2 = x2(modp) o
(5)= 2*lIn
= xP' (moâp). Let 5. be
a primitive 8" mot of unity in an algebraic closure of iFp. We have that
c4 = -1
and
9.2 Proof of the Product Formula
We are now in a position to prove Hilbert's product formula.
Theorem 9.3 (The Product Formula)
Let a, b
E
Q. Then
Roof The following proof is kom [24] pp.23.
The properties of the Hilbert symbol allows us to reduce the proof of the general
product formula to three special cases a = b = - 1 , a = - 1 , b
=
1 (1 a prime) and a = 1,
b = Ir (1 and 11 distinct primes). The following Hilbert symbols have been calculateci
using Theorem 4.1.
Case 1. a
= b = -1.
has (-l,-l)q, = -1
For p
* 2 . q vp(-1) = O so we have that ( - l , - l ) q p
=
and, since a and b are both negative, (a,b)n = -1.
1. One
Hence
Case 2 a = -1, b = I a prime nurnber.
If I = 2 and p + 2 ,
a,
then v , ( - l ) = v p ( 2 ) = O so (-1,2)qP = 1 .
Moreover, since
9 + x2 = 2 9
has
the
(-1,2)n = 1 = (-l,2)Q,
=
non-trivial
solution
1. Hencen (-1,2)$
( 11 1
=,
we
get
;+ 2
where
y ,
= 1.
P**
I f 1 # 2 and p = 2, v2(-1) = vI(i) = O , so ( - 1 , l ) ~=
~ (-l)(I-I)%
Now if p = i * 2, then v2(-1) = O and ~ ( l=) 1, so
p s 1, we have (-l,oQp = 1.
(-1, r ) ~ =
,
(+)=
(-1)('-ly2.
I f 1 O 2, p
The product
n (-l,i)pp is thus equd to 1.
P*rn
Case 3 a = 1, b = f f .
I f 1 = 11, b y properties of the Hilbert symbol we £ind that
Hence (1,I)Qp= (-l,oQp for a l l p.
( 1=( 1 , ) =( 1 - l ) ( , ( .
with in Case 2. So assume that 1 + 1. I f 1'
thus (1,2)% = 1.
If
i'
= 2 and
) = 1-
and p
#
(+) = ( - ~ ) ( l ' - ~ ~If. i
For p = 2 we obtain (1,P)Q, = (-1)(1-1)(1'-i)'4
Finally, since v1(l1)= 1
W e thus obtain
n (1,1'
PP'
= vif (l)
this results in (1,
= (- 1)('2-1)'4
)
2,l then vp(l) = vP(l1)
=O
p = 2, (1,2)q, = (-1)(12-'@. I f
vl(0 = 1 and vr(2) = 0 so ( & 2 ) ~=,
( 11
=2
QP
(1)(
) = 1.
=
This was dealt
#
11 =
2 and p = 1 # 2,
/', 1, l'p # 2, we get
since v2(l) = v2(11)= O.
($1 and (l,l!)qt
=
The general result follows.
O
9.3 Quadratic Reciprocity
A direct consequence of the prduct formula is that for m y quadratic form f over Q,
the number of primes p, including eo, for which f has no zero over Q p is always even.
As previously mentioned, the product formula is equivalent to the classic law of
quadratic reciprocity.
From the proof of the product formula, we see that in
general, if p,q, i are distinct primes not equal to 2, we have (p, qIp, = 1.
then @ , q ) ~ =
, (-l)*i)(q-1v4
becornes
and @,q)Qp =
(%).
I f i = 2,
The product formula then
The product formula
n ( ~ , b )1 ~is ~quite useful since if ne know all but one
=
P."
value of the Hilbert symbol we can earrily determine the missing value. In particular
Lemma 9.2
If a quadratic form with rational coefficients f in three variables represents zero in
all fields Q, for all p
<
except possibly for
Qq
for some prime q, then f
&O
represents zero in Q,.
For instance Borevich and Shafarevich prove the HaaseMinkowski Theorem in their
text (61 noting that one never needs to verify the solvability at p = 2 in order to
deduce global solvability.
9.4 HasseysProduct Formula
There is an analogue of the Product Formula for algebraic number fields. Let K be
a finite extension of Q and let a be an element of P.Similarly to Theorem 9.1 we
find that
where p goes through all fînite and infinite primes- of K. Hilbert had defined the
norm residue symbol for quadratic and Kummer extensions of Q, but had verifîed
the product formula only for special cases ([l3], pp.262-265). Hasse proved Hilbert's
product formula and then generaiized the norm residue symbol t o abelian
extensions. To accomplish this, H a s e worked in p-adic fields using the adapted
Hilbert norm residue. Given a field K.a prime ideal p of K and v , p
E
O Ksuch that
p is not a square in OKwe can define the quadratic Hiibert norm residue
To be consistent with our prior notation, we will denote this n o m residue as (v, u ) ~ ~ .
There is a correspondhg nom residue aymbol for the infinite primes p,.
If K has r
real embeddings a,...or, then for each real embedding w e associate an infinite prime
PQ, and
As in the case K = Q, we see that ( v , u ) ~=
, 1 if v is the n o m of an element in
K,(fi).
Hasse then proved the general prduct formula
taken over all primes p of the field K. To work comfortably in these fields, Hasse
had to prove theorems about padic fields which were analogues of Takagi's results
in class field theory.
These results paved the way for Local Class Field Theory.
Hilbert had managed to prove the above product formula only for a few special
cases, such as when K is a field with class number 1 or 2 and with less than two
classes or when K is an imaginary extension with an odd class nurnber ([13],
pp.238-241). In these cases, Hilbert constructed an extension L of K he cailed a class
field. He t hen proved various conjectures concerning t hese special fields.
Chapter 10
The Hasse Norm Theorem
10.1 The Hilbert Norm Theorem
In part III of his report Zahlbericht on algebraic number theory to the Deutache
Mathemcrtiker Ve~einigung(German Mathematical Society), Hilbert r e h u l a t e d a
theorem of Gauss ([Ml pp.225)
Theorem 10.1 (Hilbert Norm Theorem)
Let b be a square £ree integer and a a nonzero integer. If ( ~ ~= 61 for
) all
~ primes
~
p5
a, then
kom
That is to
a = N(x) is solvable in the field Q(6).
Say
a is a global norm
Q(G)
if and only if it is a local n o m everywhere.
We saw in Chapter 4 that z2 - ar2 - by2 = O has a solution in
only if a is the norm of an element in
%(fi).
Since a
Qp
for all p 5 a, if and
is a norm in
Q(6)
if and
only if z2 - ax2 - by2 = O has a solution in Q, we see that the Hilbert Norm theorem
is completely equivalent to the Hasae-Minkowski Theorem for rational quadratic
forms in three variables.
10.2 The Hasse Norm Theorem for Cyclic Extensions
Let K be a number field. In 1924 Hasse proved that v is the norm of an elemant in
K ( F ) if and only if v is a local norm everywhere. Hasse later (1930) extended the
Norm Theorem from quadratic extensions of K to al1 cyclic extensions of K. He thus
obtained the celebrated Hasse Norm Theorem. The following is the general Norm
Theorem that Hilbert had proved only for special cases.
Theorem 10.2 (Hasse Norm Theorem )
Let LIK be a cyclic extension of number fields. An element a in K" is the norm of
an element in Lx (Le. a
E
N(LX))if and only if a is the norm of an element in the
correaponding local extensions L d K p for al1 prime divisors p in K,including infinite
primes (i.e. a
E N ~ ~ (L$)v~).
, K ~
Roof.See [lg], Theorem 4.5, p:156. Note: Analogues of the product formula seen
in Chapter 8 play a crucial role in the proof of the Hasse Norm Theorem.
In Theorem 10.2, by cyclic extension we mean a Galois extension with cyclic Galois
group. In the context of ration& the Hasse Norm Theorem reads as follows: Let
HQ be
a qclic extension.
A nmzero rational n u m h r n is a norm from K if and
a l y if it is a local n m at eue*
particular, for L
=
prime of
~ ( na degree
) two
K
induding infinite primes.
In
cyclic extension of Q, the Hasse Norm
Theorem reduces to the Hilbert Norm Theorem. It has been observed that one can
leave out one particular prime, in the application of the Hasse Norm Theorem (Ref.
~91)-
Theorem 10.3
If L/Kis a cyclic extension of global fields and a
E
P is a local norm at all primes of
K with the possible exception of one particular prime, then a is a local n o m at that
prime as well.
Rwf. See [19]pp.190. Note: The proof follows hom the product formula for the
local Artin maps (p:189 of [NI).
10.3 Abelian Extensions
fiasse had a t first conjectureci that his N o m Theorem for cyclic extensions would
hold for abelian extensions. Hasse found the first counterexample to this conjecture
himself. The biquadratic field K
for all p
E
x
(,/x,
Q, but 3 is not a nom in K.
Q(m,
Jlf ).
L=
2/22
=Q
n)has the property that (3,K)p
P
= 1
Another well-known counterexample is
The extension L is Galois over Q with non-cyclic Galois group
2 / 2 2 = (1,6,r,&) = G, where
The norm of an element of a is
m +d m
where a E G. Hence the n o m of an element a = a + bm + c
from
Q to L is NUQ(a)= a6(a)r(a)Ss(a). I t can be shown that -1 is not a norm of any
a E L. We will now show that -1 is a local n o m everywhere, Le.
completions LI. If p = a o , Lm = P and -1
is a norm in L
in each
If p z 13,17. The
extension Lp is an unramified extension of Q where ILp : QJ 5 2. By the following
result of local claas field theory
Theorem 10.4
Let L, be an u n r d e d extension of
with u E U,,m
E
2. Then
Qp
of de-
f over Q,.
Let
fl = pmu E Q;,
fl E NLPmp(L;)
if and only if f 1 m. In particular, every unit
of Q is the norm of a unit of L,.
The number -1 is a n o m in L, for al1 p
I f p = 13, Li3 =
+ 13,17 since -1
is a unit of Q, p
~,,(m,
m)= Q13(m)
since 17 = 4(mod 13) Le.
#
E
13,17.
Q13and
-1 is the n o m of a = 3 - m E L I 3 .Similarly, 13' r (4)8
= l(modl7) thus 13 is a
2
square in
Qi7 thus
Q,,(n,m)
=Q17(m)
and
-1
is the
norm of
4 + f l E Q 17(Ji7). We then see that ali local conditions are satisfied.
10.4 Some Applications
Using the Hasse N o m Theorem for cubic cyclic extensions of an algebraic number
field K, Selmer proved the foilowing theorem in his 1953 article Sujficient
m g m e n c e conditions for
the existence of rational points on certain n<bic
surfaces
Theorem 10.5
+ a, b,c,d E Z cube-free integers
ar3+ by3 + cz3 + &v3 = 0 has non-trivial solutions
Given
O
such
that
ad = bc
then
in Q if and only if it has
non-trivial solutions in Qp for all p 5
R m f (Idea of
x3 +n?v) = n(u3
Selmer's
++a,
with
(m,n) = 1). Let
proof).
Selmer considers the
cubic
equation
integer, cubekee m aad n (but not necessarily
K be an algebraic number field.
Selmer boils down the question,
by an argument about the ide& of the field, to an application of Hasse's Norm
Theorem where K ( p ) is hi8 ground field (p a complex cube mot of unity) and
R = K(p)(m ln) is the cyclic extension over K(p). For more detail, see [25].
Now recall the statement of the generalized Hasse-Minkowski Theorem [13]pp.263,
Theorem
Let A x 1, ...,xn) be a quadratic form with coefficients in some algebraic n u m k field
,,
K. Suppose j(x ...,x,)
= O has a non-trivial solution
in all p-adic fields Kp for al1
finite prime ide& p of OK and in all the completions corresponding to the
archimedean absolute values. Then f = O has a non-trivial solution in K.
(n = 2,3) The two variable case n = 2 is a direct consequence of the
theorem: a E K is a square in K which is true if and only if a is a square in Kp for all
A-oof
p and in each r d embeddings; the proof of which is similar to that of Theorem 1.1.
The 3 variable case n
Ax)
=d
=
3 is a particular case of the Hasse Norm Theorem since
+ 6 9 + cz2 = O haa a non-trivial solution if and only if -ab
element of ~(j-bc).
Moreover, f l x )
=O
is a norm of an
haa non-trivial solutions in all Kp and in IR
implies that -ab is a local norm of an element in K,(J Zwith
) respect to Kp. The
general case is obtained in the ssme manner as the previous case for Q (see the proof
of Theorem 5.4).
There is a finite set of primes which must be taken into consideration when applying
the Hasse Nom Theorem.
Using the Hasse Norm Theoram, Voncenzo Acciaro
constructed an algorithm in order to m e r the following question: Let L = Q(a) be
a cyclic extension of the rationals of prime degree q and let a
E
Qx.
Does the
equation
NJ&)
=a
admit any solution in L? (Le. 1s a the norm of some element in L. It is worth
noting that we are not asking how to h d such a 1, but only whether a solution
exists. Ses Voncenzo Acciaro's Doctoral thesis for a detailed account see [l]Chapter
4.)
Although the N o m Theorem gives a satisfktory criterion for establishing
whether or not a given element of K is the norm of an element kom the cyclic
extension L of K, it does not help in identifying what integers of K are norms of
integers of L.
In particular, there is no systematic way to determine which
quadratic extensions have a unit of n o m -1.
Refer to the discussion about the
representation of integers by quadratic forms in Section 7.4.
Chapter 11
Measuring the Failure
Even when the Hasse Principle d m not hold, there some+imes remains a close
interaction between the local and the global. We have already seen that the Hasse
Principle fails for
3x3 + 4 P +SZ3 = O
It has been conjectured that although the Hasse Principle Eails for eIliptic curves, it does
not fail that badly, that is the measure of the obstruction to the Hasse Pirinciple is
finite. Since studying the extent of the obstruction to the Hasse Principle is an active
area of reseazch in Number Theory, this chapter aims to give a brief glimpse at the rich
theory involved. Hasse's principle may be expressed cohorn~logically~This formulation,
is essential in measuring the m u r e of the Local-Global Principles. We thus begin by
getting reacquainted with some definitions and theory kom the cohomology of groups
and Galois cohomology. The content of this chapter relies heavily on the references
[IO],1261 and some useful notes on the subject mitten up by David S. Dummit.
11.1 First Cohomology Group
Let G be a finite group and let A be a Gmodule written additively (Le. A is an abelian
group on which G acts on the left). The action is written o(a) where a
The group G acts on itself by ~ ( r=)mu-' where a,r
E
E
G and a
G. A 1-cayde is a map
E
A.
which satisfies the cocycle identity
for all o,r E G.
If f and g are 1-cocycles then f + g is also a 1-cocycle where
(f+ g)(a) =fia) + g(a).
The 1-cocycles thus form a group denoted Z1(G,A). The group
G acts trivially on A if and only if o(a) = a for aU o E G and all a
and G acts trivially on A, then a, = a*
E
A. I f f
+ nar = au + ar which is true if
E
Z1(G,~)
aiid only if f E
Hom(G,A). The map f is called a I - o o b o u h ~if and only if there exists aa element
b
EA
such that
for all a
E
G. A coboundary is a cocycle because
a,
-b
= m(b) - o ( b ) + a(b) - b
= a(r(b) - b ) + o ( b ) - b
= m(b)
+
= Car aa
The coboundaries in fact form a subgroup of Z1(G,A)since the surn of two coboundaries
is again a coboundary. The group of coboundaries is denoted B1(G,A). Note that if G
acts trivially on A then B' (G,A) = {O).
'
The quotient group Z1(G,A)/B(G,A) is c d e d
the first cohomology group and is denoted by H'(G, A). If G acts trivially on A
The cohomology group H"(G,A) is the set of elements of A fixed by G,
{a
E A laa
= a V a E G), denoted
There are higher order cohomology groups
ff (G,A), but we will not need them. If G is a topological group and A is a topological
A-module (Le. A is a topological group and G acts continuously on A) then the
continuous cohomology is defined as above except that the cocycles are continuous
hinctions (as are the coboundatie~).
11.2 Hasse Principle
Given an elliptic c w e E defined over Q, we can look at E over
g, over
which it
acquires points. Write
G = GuI(Q/Q)
and let
Gp
=
G4Qp/Qp)
be the decomposition group of the prime p. We can view Gp as a subgroup of G. The
embedding of fields
0
induces the embedding of groups
- op
-
GP
G
The diagram lwks like
Let E(Q) be the group of rational points of E. The elliptic curve E will be identified
with its group of algebraic points which will be denoted by
E = E(Q)
By restricting all cocycles in Z'(G,E) to Gp we get a canonical group homomorphism
-
H1(G,E) ffl(Gp,E)
called a localization. If P is a point on the c u v e E it gives rise to a coboundary
o -. a P - P
E
-
Bi(G,E) c B1(Gp,E). The map
ECO)
E(QJ
induces the sequence
where jp is the localization map. If a cocycle f is a coboundary we say f is trivial (in the
trivial class). We wodd like to answer the question: If a cocycle f when restricted to
Gp is an element of B 1(Gp,E) for all p, isf an element of B'(G, E)? In other words, if f is
locally trivial for al1 p (E has a local point) isf globally trivial (E has a global point).
11.3 Principal Homogeneous Spaces
Now the elliptic curves over Q which become isomorphic to E over 0 are c d e d twists of
E. To each element of H1(G,E), we can associate a certain twist of E called a f i n c i p l
Homogeneous S'ce. The concept of Principal Homogeneous Spaces was introduced by
Weil [IO]. The following two definitions are taken kom [26].
Definition 11.1
Let E/Q be an elliptic c w e . A principal homogeneous space for E/Q consists of a pair
(C,p), where C/Q is a smwth cuve and a morphism
p:CxE+C
d e h d over Q by &O,
P ) = po
+ P with the following properties:
i) p o + O = p o for allpo EC.
ii) ( p o + P ) + Q = p o + ( P + Q ) for allpo E C, P,Q E E.
iii) For all po,q
E
C there is a unique P E E satisfying po
+ P = q.
Definition 11.2
Two homogeneous spaces C and V for EIQ are equivalent if there is an isomorphiam 9 :
C-r 27 which preserves the action of E on
have 0(po + P) = 8(po) + P.
C Le. such that for
all po E C
and P
E
E we
Note that E is a principal homogeneous space for E and that any other principal
homogeneous space is in the same class as E, that is in the trivial class, if and only if it
has a point in Q.
Theorem 11.1
Let E/Q be an elliptic c w e . There is a natural bijection
Set of equivalence classes of
H1(G,E) *
Principal Homogeneous Spaces
for E/Q which have rational points in
defined as follows:
Let C/Q be a homogeneous space and choose a point po
{O
+
E C.
Then
u p o -PO>-, WQ>
where { ) indicates an equivalence clws and
op0
-po
means (iiij in definition 11.1. For
a prwf see [26], pp.291.
11.4 The Tate-Shafarevich Group
The group H1(G,E) is called the Weil-Chatelet group which we will denote WC. The
trivial class of WC corresponds to the dass of Principal Homogeneous Spaces having a
rational point. So the problem of checking the triviality of the Weil-Chatelet group is
equivalent to answering the Diophantine equation of whether a given c w e has a
rational point. We have the exact sequence
0
+
E(Q)/mE(Q)
H1(G,Em)
[H1
(G,
El 1,
+
0
where En is the group of elements of E ( 0 ) of order exactly m and
[...lm denotes
the
subgroup of elements of E(Q) of order dividing m. The principal homogeneous spaces
come into the computation of the weak Mordell-Weil group E(Q)/mE(Q). This problem
may be reduced to determining whether each associateci homogeneous space has a
rationalpoint. One can either explicitly find such a point or show that there is some
local completion of Q for which it has no point.
The difnculty is that there are
homogeneous spaces that have points locdy everywhere, yet have no rational point,
that is the Hasse Principle fi. So we are interested in the elements of
HI (G, [an)= Sm, the m-Selma group, which are images of E(Q)/mE(Q). Since the
sequence is exact, these are precisely the elements of the kernel of the map
Being in the kernel of the latter map meana that there is a point on C defineci over Q.
Let the Russian letter iii (sha) denote the elements of WC for which there is a point on
C everywhere locdy (elements of WC for which f restricted to Gp is an elernent of
BI (Gp,E) for aii p, Le. having a gppoint for ail p). These elements t o m a group called
the Tate-Sbfarevich Group
It counts the number of equivalence classes of
homogeneous spaces of C which have points in all local fields. The group
intersection of the kernels of all localization maps j,.
Lü
is the
We get the exact sequence
The Tate-Shafarevich group then measures the obstruction to the Local-Global
Prhciple for the eUiptic curve E, that is, it measures the gap between rational
solvability and everywhere local solvability. It was conjectured in the 60's that the
TateShafarevich group is h i t e . The conjecture rernains unproved. Establishing the
finiteness of the Tate-Shafarevich group is tantamount to proving that the Hasse
Principle holds up to a h i t e obstruction for c w e s of genus 1. The problem is that
there are no algorithms for computing iii for a curve kom the initial data. Rubin with
help of some idem by Francsico Thaine gave the first concrete examples of elliptic
curves over number fields with a TateShafarevich p o u p proved to be finite [22]. It is
now known that if E(Q) is finite then 1 üi1 < 00. For more on the conjectured finiteness
theorems concerning Tate-Shafarevich look at Ju. 1. Man$
moduLar curves, Russian Mathematical Surveys 26 (lWl), 7-78.
Cyclotomic fields and
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