Quadratic Forms Over Global Fields [Lecture notes]

16 Pages·2017·0.351 MB·English

by  Pete L. Clark

QUADRATIC FORMS OVER GLOBAL FIELDS
PETEL.CLARK
Contents
1. The Hasse Principle(s) For Quadratic Forms Over Global Fields 1
1.1. Reminders on global fields 1
1.2. Statement of the Hasse Principles 2
2. The Hasse Principle Over Q 3
2.1. Preliminary Results: Reciprocity and Approximation 3
2.2. n≤1 6
2.3. n=2 6
2.4. n=3 6
2.5. n=4 8
2.6. n≥5 9
3. The Hasse Principle Over a Global Field 9
3.1. n=2 10
3.2. n=3 10
3.3. n=4 11
3.4. n≥5 12
4. Some Applications to Integral Forms 12
4.1. The Aubry-Davenport-Cassels Lemma 12
4.2. Two, Three and Four Squares 14
4.3. More on Euclidean Forms and ADC Forms 15
References 16
1. The Hasse Principle(s) For Quadratic Forms Over Global Fields
All quadratic forms and quadratic spaces will be assumed nondegenerate unless
otherwise noted.
1.1. Reminders on global fields.
A number field is a field K which is a finite-dimensional field extension of Q.
Let k be a field. A function field over k is a field K which can be expressed as
a finite degree separable extension of the rational function field k(t). (By the Sep-
arable Noether Normalization Theorem, it follows that the class of function
fields over k is not enlarged if the word “separable” is omitted.)
A global field is a field which is either a number field or a function field over
ThankstoFranceDacar,TimoKellerandDannyKrashenforhelpfulcomments.
1
2 PETEL.CLARK
a finite field F.
To any global field K we associate the set Σ of places of K, namely equiva-
K
lence classes of absolute values on K.
BIG OSTROWSKI
1.2. Statement of the Hasse Principles.
Let q be a quadratic form over the global field K, and let v be a place of K.
We write q for the base change of q to a quadratic form over K .
v v
Now we can state one of the most important and influential results in quadratic
forms theory and number theory.
Theorem 1. (Hasse Principle) For a quadratic form q over the global field K, the
following are equivalent:
(i) q is isotropic.
(ii) q is locally isotropic: for all places v of K, q is isotropic.
v
It will take us most of the chapter to give a complete proof of Theorem 1. For now
we assume it and derive further “Hasse Principles”.
For a quadratic space (V,B) over a global field K and v ∈ Σ , write V for
K v
(V ⊗ K ,B⊗ K ).
K v K v
Theorem2. (HassePrincipleforIsometricEmbedding)ForquadraticspacesV,W
over a global field K, TFAE:
(i) There is an isometric embedding V (cid:44)→W.
(ii) For all v ∈Σ , there are isometric embeddings V (cid:44)→W .
K v v
Proof. (i) =⇒ (ii) is immediate.
(ii) =⇒ (i): We go by induction on dimV. The case dimV = 0 being trivial,
suppose n ≥ 1, that the result holds in dimension n−1, and write V = (cid:104)α(cid:105)⊕V(cid:48).
For v ∈ Σ , we have (cid:104)α(cid:105) (cid:44)→ V (cid:44)→ W , i.e., W represents α. By the First
K v v v v
Representation Theorem, the space W ⊕(cid:104)−α(cid:105) is locally isotropic, so by the Hasse
Principle it is isotropic, and thus we may write W = (cid:104)α(cid:105)⊕W(cid:48). However, for all
v ∈σ , since V is a nondegenerate subspace of W , we may write
K v v
(cid:104)α(cid:105)⊕W(cid:48) =W =V ⊕X =(cid:104)α(cid:105)⊕V(cid:48)⊕X .
v v v v v v
ApplyingWittCancellation,wegetisometricembeddingsV(cid:48) (cid:44)→W(cid:48) forallv ∈Σ .
v v K
We are done by induction. (cid:3)
The case of dimV =dimW is important enough to be stated separately.
Theorem 3. (Hasse Principle for Isomorphism) For quadratic forms q,q(cid:48) over
the global field K, the following are equivalent:
(i) q ∼= q(cid:48).
K
(ii) q and q(cid:48) are locally isomorphic: for all places v of K, q ∼= q(cid:48).
v Kv v
For a quadratic form q and v ∈ Σ , we define the v-adic Hilbert symbol
/K K
simply as the Hilbert symbol of the base change: H (q)=H(q ).
v v
QUADRATIC FORMS OVER GLOBAL FIELDS 3
Corollary 4. (ClassificationofQuadraticFormsOverGlobalFields)Forquadratic
forms q,q(cid:48) over a global field K, the following are equivalent:
(i) q ∼=q(cid:48),
(ii) All of the following hold:
(a) dimq =dimq(cid:48).
(b) discq =discq(cid:48).
(c) For all places v ∈Σ , H (q)=H (q(cid:48)).
K v v
Proof. ThisfollowsimmediatelyfromtheHassePrincipleforIsomorphismandthe
fact that for all v ∈Σ , quadratic forms over K are classified by their dimension,
K v
discriminant and Hasse invariant (XXX). (cid:3)
Exercise: ShowthatCorollary4givesanexplicitprocedurefortestingisomorphism
of quadratic forms over global fields.
Recall that we say a rational quadratic form q is indefinite if qR is isotropic:
i.e., neither positive nor negative definite.
Corollary5. (Meyer)Forn≥5,anindefiniten-aryquadraticformqQ isisotropic.
Proof. This follows from the Hasse Principle together with the fact that for all
primes p, every quadratic form in at least five variables over Q is isotropic. (cid:3)
p
Corollary 6. For a global field K, the following are equivalent:
(i) The u-invariant of K is 4.
(ii) K has no real places.
Exercise: Prove Corollary 6. (Don’t forget to show that u(K) ≥ 4, i.e., that K
admits an anisotropic quaternary form.)
Exercise: Prove (that the Hasse Principle implies) the Hasse Principle for Hy-
perbolicity: if q is a quadratic form over a global field K, then q is hyperbolic iff
q is hyperbolic for all v ∈Σ .
v K
2. The Hasse Principle Over Q
2.1. Preliminary Results: Reciprocity and Approximation.
Let a,b ∈ Q×. Then for all v ∈ ΣQ we have the Hilbert symbol (a,b)v. One
of the key ideas in this subject is to examine relations among the various (a,b) .
v
Lemma 7. Let a,b∈Q×.
a) For all odd primes p(cid:45)ab, (a,b) =1.
p
b) In particular, {v ∈ΣQ | (a,b)v =−1} is finite.
Exercise: Prove Lemma 7.
Exercise: Let a,b ∈ Q×. Show that if the quadratic form ax2 + by2 − z2 = 0
is isotropic, (a,b)v =1 for all v ∈ΣQ.
Theorem 8. (Hilbert Reciprocity Law) For a,b∈Q×, we have
(cid:89)
(1) (a,b) =1.
v
v∈ΣQ
4 PETEL.CLARK
Proof. Because of the bilinearity of Hilbert symbols, it suffices to prove the result
when each of a and b is either −1 or a prime number. Moreover, of course what we
are trying to establish is that in each case, #{v ∈ΣQ | (a,b)v =−1} is even.
Case 1: a=b=−1. We have (−1,−1) =−1 iff v ∈{2,∞}.
v
Case 2: a = −1,b = 2. Since −(1)2+2(1)2−(1)2 = 0, by Exercise X.X we have
(−1,2)v =1 for all v ∈ΣQ.
Case 3: a = −1, b = (cid:96) an odd prime. We have (−1,(cid:96)) = 1 except possibly for
v
v ∈{2,(cid:96)} and (−1,(cid:96))2 =(1,(cid:96))(cid:96) =(−1)(cid:96)−21.
Case4: a=b=pisapime(possibly2). ByX.X,forallv ∈ΣQ,(a,a)v =(−1,a)v,
so we are reduced to Cases 2 and 3.
Case 5: a = 2, b = (cid:96) an odd prime. We have (2,(cid:96)) = 1 except possibly for
v
v ∈{2,(cid:96)}. Further, we have
(2,(cid:96)) =1 ⇐⇒ (cid:96)≡±1 (mod 8),
2
(cid:18) (cid:19)
2
(2,(cid:96)) =1 ⇐⇒ =1 ⇐⇒ (cid:96)≡±1 (mod 8),
(cid:96) (cid:96)
where we have used the second supplement to the Quadratic Reciprocity Law.
Case 5: a = (cid:96) , b = (cid:96) are distinct odd primes. Then ((cid:96) ,(cid:96) ) = 1 except possibly
1 2 1 2 v
for v ∈{2,(cid:96) ,(cid:96) }. Further:
1 2
((cid:96)1,(cid:96)2)2 =(−1)((cid:96)12−1)((cid:96)22−1),
(cid:18) (cid:19)
(cid:96)
((cid:96) ,(cid:96) ) = 2 ,
1 2 (cid:96)1 (cid:96)
1
(cid:18) (cid:19)
(cid:96)
((cid:96) ,(cid:96) ) = 1 .
1 2 (cid:96)2 (cid:96)
2
Thus the fact that ((cid:96) ,(cid:96) ) ((cid:96) ,(cid:96) ) ((cid:96) ,(cid:96) ) = 1 follows from – indeed, is – the
1 2 2 1 2 (cid:96)1 1 2 (cid:96)2
Quadratic Reciprocity Law. (cid:3)
Exercise: If you haven’t realized it already, verify that Theorem 8 is equivalent to
quadratic reciprocity together with its first and second supplements. However, for
thestudyofquadraticformsoverglobalfieldsitisanespeciallygracefulformulation
of these classical results: as we will see later, Hilbert’s Reciprocity Law extends
verbatim to any global field.
Theorem 9. (Global Existence Theorem) Let a ,...,a ∈Q×. For each 1≤i≤
1 N
N and v ∈ΣQ, let (cid:15)i,v ∈{±1}. The following are equivalent:
(i) There is α∈Q× such that for all 1≤i≤N and all v ∈ΣQ,
(α,a ) =(cid:15) .
i v i,v
(ii) All of the following hold:
(a) {(i,v) | (cid:15) =−1} is finite.
i,v
(cid:81)
(b) For 1≤i≤N, we have (cid:15) =1.
v∈ΣQ i,v
(c) For all v ∈ΣQ, there is αv ∈Q×v such that (αv,ai)v =(cid:15)i,v.
Proof. (i) =⇒ (ii): Condition (a) follows from Lemma 7, condition (b) follows
from Theorem 8, and condition (c) is obvious: take α =α for all v.
v
(ii) =⇒ (i): It is no loss of generality to assume that ai ∈Z for all i. Let S ⊂ΣQ
consist of 2,∞ and all primes dividing at least one ai, and let T ⊂ΣQ be the set of
all places such that (cid:15) =−1 for some i: these are finite sets.
i,v
Case 1 (S∩T =∅):
QUADRATIC FORMS OVER GLOBAL FIELDS 5
(cid:81) (cid:81)
Let a= (cid:96) and m=8 (cid:96). Since S and T are disjoint, a and m
(cid:96)∈T\{∞} (cid:96)∈S\{2,∞}
are coprime, so by Dirichlet’s Theorem on Primes in Arithmetic Progression there
is a prime number p ≡ a (mod m) and p ∈/ S ∪T. We may take α = ap. The
verification of this is left to the reader as an exercise (or see [Se, pp. 25-26]).
Case 2 (General Case):
Forv ∈ΣQ,Q×v2isopeninQ×v. ByArtin-WhaplesApproximation,thereisα(cid:48) ∈Q×
such that for all v ∈S, α(cid:48) ∈Q×2 and thus
αv v
(α(cid:48),a ) =(α ,a ) =(cid:15) ∀v ∈S.
i v v i v i,v
Letη =(cid:15) (α(cid:48),a ) ;thenthefamily(η )satisfiesconditions(a),(b)and(c)and
i,v i,v i v i,v
η = 1 for all v ∈ S. By Case 1 above, there is β ∈ Q× such that (β,a ) = η
i,v i v i,v
for all i and all v ∈ΣQ. We may take α=βα(cid:48). (cid:3)
Exercise: Fill in the details of Case 1 in the proof of Theorem 9.
Let R be a Dedekind domain with fraction field R. Then the discrete valuations
on K with valuation ring containing R are precisely the p-adic valuations v for
p
p a nonzero prime ideal of R. For each such p, let |·| denote a corresponding
p
non-Archimedean absolute value on R, say x(cid:55)→e−vp(x).
Theorem 10. (Dedekind Approximation Theorem) Let R be a Dedekind domain
with fraction field K. Let P be a finite set of nonzero prime ideals of R. For each
p∈P we give ourselves n ∈Z and x ∈K. Then there is x∈K such that:
p p
(i) For all p∈P, v (x−x )=n and
p p p
(ii) v (x)≥0 for all nonzero prime ideals q∈/ P.
q
Proof. Step 1: For p ∈ P, the set U of elements y ∈ K such that v (y−x ) =
p p p p
n is nonempty and (cl)open. Applying Artin-Whaples Approximation to the set
p
{|·| } , we get y ∈K such that v (y−x )=n for all p∈P.
p p∈P p p p
Step2: LetQbethefinitesetofprimeidealsq∈/ P suchthatv (y)<0. IfQ=∅,
q
then we’re done – take x = y – so assume Q (cid:54)= ∅. For each p ∈ P, there exists a
positive integer ap such that for all z ∈R with z ≡1 (mod pap), vp(yz−xi)=np.
ApplyingtheChineseRemainderTheoremtotheset{pap |p∈P}(cid:96)Qofpairwise
comaximal ideals of R, we get z ∈R such that for all p∈P v (yz−x )=n and
p p p
for all q∈Q, v (yz)=0. So we may take x=yz. (cid:3)
q
Theorem 11. Let P be a finite subset of ΣQ. Suppose given tp ∈ Q×p2 for all
v ∈P. Then there is t∈Q× such that
(i) t≡t (mod Q×2) for all v ∈P, and
p p
(ii) |t| =1 for all but possibly at most one finite prime p∈/ P.
p
Proof. Let P = P \{∞}. Let (cid:15) ∈ {±1} have the same sign as p if ∞ ∈ P; if
f ∞
∞∈/ P, put (cid:15)=1. Put
(cid:89)
β =(cid:15) pvp(tp).
p∈Pf
Thenforallp∈P ,wemaywriteβ =u t forsomeu ∈Z×;if2∈/ P,putu =1.
f p p p p 2
BytheDedekindApproximationTheorem,thereisz ∈Zsuchthatz ≡u (mod p)
p
for all odd p ∈ P and z ≡ u (mod 8). By the Local Square Theorem, z ≡ u
f 2 p
(mod Q×2) for all p ∈ P . Since gcd(z,8(cid:81) p) = 1, by Dirichlet’s Theorem
there ispa prime p ≡z (mfod 8(cid:81) p). Wep∈mPafy take t=p β. (cid:3)
0 p∈Pf 0
6 PETEL.CLARK
2.2. n≤1.
These cases are absolutely trivial, as every quadratic form over a field in at most
one variable is anisotropic!
2.3. n=2.
Recall that a binary form q over any field K is isotropic iff −discq ∈K×2.
Let q be a binary form over Q – we may assume that q has Z-coefficients – such
that qv = q/Qv is isotropic for all v ∈ ΣQ. By the above remark, for all v ∈ ΣQ,
−discq ∈ Q×2. For v = ∞ this says −discq > 0. For v = p a finite prime it cer-
v
tainly implies that v (−discq) is even. By unique factorization in Z, this implies
p
that −discq =n2 for some n∈Z, so discq ≡−1 (mod Q×2) and q is isotropic.
2.4. n=3.
TheHassePrincipleforternaryformsoverQisequivalenttothefollowingclassical
theorem of Legendre. Let qQ be a locally isotropic ternary quadratic form. Via a
change of variables, we may take q to have the form
q(x,y,z)=ax2+bx2+cx2,
1 2 3
with a,b,c nonzero squarefree integers. If a,b,c were all positive or all negative,
q/R would be anisotropic. Thus, up to relabeling and multiplying through by −1,
we may – and shall – assume that a is positive and b and c are negative.
Finally, wemayreducetothecaseinwhicha,b,carecoprimeinpairs, orequiv-
alently that abc is squarefree. We leave this as a simple but enlightening exercise
for the reader. Thus we are led to consider the Legendre equation
(2) ax2+by2+cz2 =0,
with a>0, b,c<0 and abc squarefree.
Wecan“removethep-adicnumbers”viathefollowingobservation: qQ isisotropic
p
forallprimespiffqZp isisotropicforallprimespiffqZ/pnZ isisotropicforallpand
n iff q/Z/nZ is isotropic for all n∈Z+.
We claim that if q(x,y,z) is isotropic, then −bc is a square modulo a, −ac is
a square modulo b, and −ab is a square modulo c. Indeed, suppose there are
x,y,z ∈ Q, not all zero, such that ax2 +by2 +cz2 = 0. By rescaling, we may
assume that (x,y,z)∈Z3 and gcd(x,y,z)=1.
Let p be a prime dividing a. Reducing 2 modulo a gives
by2+cz2 ≡0 (mod p).
If y and z were both divisible by p, then since ax2+by2+cz2 = 0, p | ax2. Since
p | a and gcd(a,c) = 1, p | x2 and thus p | x, contradicting gcd(x,y,z) = 1. So we
may assume that at least one of y and z is invertible modulo p; with no real loss of
generality we assume y is invertible modulo p. Then by2 ≡−cz2 (mod p), so
(cid:18)cz(cid:19)2
−bc≡ (mod p),
y
QUADRATIC FORMS OVER GLOBAL FIELDS 7
i.e., −bc is a square modulo p. Since this argument holds for every prime dividing
the squarefree integer a, by the Chinese Remainder Theorem −bc is a square mod-
ulo a. And of course a perfectly symmetrical argument shows that −ac is a square
modulo b and that −ab is a square modulo c.
This was easy. Remarkably, Legendre showed that these easy necessary condi-
tions are also sufficient for the existence of a nontrivial solution to 2.
Theorem 12. (Legendre) The Legendre Equation
q(x,y,z)=ax2+by2+cz2 =0
has a nontrivial integer solution iff −bc is a square modulo a, −ac is a square
modulo b and −ab is a square modulo c.
We will give a geometry of numbers proof of Theorem 12. For an even more
elementary proof, see [NT, §17.2].
Lemma 13. Let m ∈ Z+ and let (cid:15) ,(cid:15) ,(cid:15) ∈ R>0 be such that (cid:15) (cid:15) (cid:15) ≥ m. Let
1 2 3 1 2 3
(cid:96)(x,y,z) = αx+βy +γz ∈ Z[x,y,z] be any linear polynomial. Then there are
(x,y,z)∈(Z3)• such that
(3) (cid:96)(x,y,z)≡0 (mod m)
and |x|≤(cid:15) , |y|≤(cid:15) , |z|≤(cid:15) .
1 2 3
Exercise: ProveLemma13. (Suggestion: showthat(3)definesasublatticeΛ⊂Z3
of index dividing m, and apply Minkowski’s Linear Forms Theorem [GoN, §9.2].)
We now begin the proof of Legendre’s Theorem. First, we may assume that b
and c are not both −1. Indeed, if b = c = −1, then the condition −bc is a square
modulo a gives that −1 is a square modulo a and thus a is a sum of two integer
squares, yielding a nontrivial solution to ax2−y2−z2 =0.
We claim that our congruence conditions force q(x,y,z) are necessary and suffi-
cient for the existence of linear forms L (x,y,z),L (x,y,z)∈Z[x,y,z] such that
1 2
q(x,y,z)≡L (x,y,z)L (x,y,z) (mod abc).
1 2
Since a,b,c are coprime in pairs, it is sufficient to show the factorization of q into
linear forms modulo a, modulo b and modulo c; then by the Chinese Remainder
Theorem we may choose L ,L ∈Z[x,y,z] which reduce modulo a, b and c to the
1 2
linearfactorsofq. So: letr besuchthatr2 ≡−bc (mod a),andletc(cid:48) besuchthat
cc(cid:48) ≡1 (mod a). Then
q(x,y,z)=ax2+by2+cz2 ≡by2+cz2 ≡cc(cid:48)(by2+cz2)≡c(cid:48)(bcy2+c2z2)
≡c(cid:48)(c2z2−r2y2)≡c(cid:48)(cz+ry)(cz−ry)≡L (x,y,z)L (x,y,z) (mod a).
1 2
By symmetry similar arguments can be made modulo b and c. So we get
q(x,y,z)≡L (x,y,z)L (x,y,z)=(αx+βy+γz)(α(cid:48)x+β(cid:48)y+γ(cid:48)z) (mod abc).
1 2
(cid:112) (cid:112) (cid:112)
Now apply Lemma 13 with m=abc, (cid:15) = |bc|, (cid:15) = |ac|, (cid:15) = |ab|: there are
1 2 3
(x ,y ,z )∈(Z3)• with
1 1 1
√ √ √
|x |≤ bc, |y |≤ ac, |z |≤ ab
1 1 1
and
L (x ,y ,z )≡0 (mod abc).
1 1 1 1
8 PETEL.CLARK
Since q ≡L L (mod abc), this implies
1 2
q(x ,y ,z )≡0 (mod abc).
1 1 1
Note that we have
x2 ≤bc, y2 ≤−ac, z2 ≤−ab.
1 1 1
In fact, since bc is squarefree and greater than 1, we must have x2 <bc. Similarly,
1
if y2 =−ac then a=1 and c=−1, and if z2 =−ab then a=1 and b=−1, so at
1 1
least one of the two inequalities must be strict and thus
−2abc<by2+cz2 ≤ax2+by2+cz2 ≤ax2 <abc.
1 1 1 1 1 1
Thus either q(x ,y ,z ) = 0 – great! – or q(x ,y ,z ) = −abc. In the latter case,
1 1 1 1 1 1
the ternary form q represents −disc(q) hence is isotropic by [NCA, Cor. 95].1 If
one wants to avoid this result, here is a completely elementary finish: put
x =−by +x z ,
2 1 1 1
y =ax +y z ,
2 1 1 1
z =z2+ab.
2 1
Then
q(x ,y ,z )=ab(ax2+by2+cz2)+z2(ax2+by2+cz2)+abcz2+a2b2c
2 2 2 1 1 1 1 1 1 1 1
=ab(−abc)−abcz2+abcz2+a2b2c=0,
1 1
so (x ,y ,z ) satisfies 2. If z = z2+ab = 0 then a = 1, b = −1, and (1,1,0) is a
2 2 2 2 1
nontrivial solution of 2.
Before moving on we record the following strengthening of the ternary Hasse Prin-
ciple over Q.
Theorem 14. Let q be a ternary rational quadratic form, and let p0 ∈ΣQ. If for
all p∈ΣQ\{p0}, qp is isotropic, then q is isotropic.
Proof. Since an/isotropy is not affected by scaling, we may assume q = (cid:104)a,b,−1(cid:105),
so for any v ∈ ΣQ, qv is isotropic iff (a,b)v = 1. By Hilbert Reciprocity we have
(cid:81)
1 = (a,b) = (a,b) . Thus q is also isotropic, i.e., q is locally isotropic,
v∈ΣQ v v0 v0
and by the ternary Hasse Princile q is isotropic. (cid:3)
2.5. n=4.
After consulting the literature, we were not able to choose between two differ-
entproofsofthen=4case. Wewillgivebothofthem: thefirstistakenfrom[Se],
the second from [C] and [G].
First proof: Let q =(cid:104)a,b(cid:105)−(cid:104)c,d(cid:105) be locally isotropic. By the Isotropy Criterion,
for each v ∈ΣQ, since qv is isotropic, there is αv ∈Q×v which is Qv-represented by
both ax21+bx22 and cx23+dx24. By [QF4, Lemma 9], for all v ∈ΣQ,
(α ,−ab) =(a,b) , (α ,−cd) =(c,d) .
v v v v v v
We may now apply Theorem 7 to get α∈Q× such that
(α,−ab)v =(a,b)v, (α,−cd)v =(c,d)v ∀v ∈ΣQ.
1IamindebtedtoDannyKrashenforpointingoutthissimplificationoftheendoftheproof.
QUADRATIC FORMS OVER GLOBAL FIELDS 9
By [QF4, Lemma 9] the binary form (cid:104)a,b(cid:105) Kv-represents α for all v ∈ ΣQ; equiv-
alently, the ternary form (cid:104)a,b,−α(cid:105) is locally isotropic and thus isotropic by the
previous section; equivalently, (cid:104)a,b(cid:105) represents α. Similarly, (cid:104)a,b(cid:105) repesents α, and
thus by the Isotropy Condition q is isotropic.
Second proof: We may take q of the form (cid:104)a ,a ,a ,a (cid:105), with each a a square-
1 2 3 4 i
free integer. Let P = {p | p | 2discq} ∪ {∞}. Then for all p ∈/ P, we have
(ai,aj)p = 1 for all 1 ≤ i,j ≤ 4. For p ∈ ΣQ, since qp is isotropic, by the Isotropy
Criterion there is t ∈ Q× such that (cid:104)a ,a (cid:105) and (cid:104)a ,a (cid:105) both Q -represent t .
p p 1 2 p 3 4 p p p
Apply Theorem 11 to get t ∈ Z and a prime number p . Then (cid:104)a ,a ,−t(cid:105) and
0 1 2 p
(cid:104)a ,a ,−t(cid:105) areisotropicforallp∈P. Further, forallp∈/ (P∪{p }), (cid:104)a ,a ,−t(cid:105)
3 4 p 0 1 2 p
and (cid:104)a ,a ,−t(cid:105) are isotropic since their entries are p-adic units. Thus the ternary
3 4 p
forms (cid:104)a ,a ,−t(cid:105) and (cid:104)a ,a ,−t(cid:105) are locally isotropic except possibly at p , so by
1 2 3 4 0
Theorem 14 they are both isotropic. Then (cid:104)a ,a (cid:105) and (cid:104)a ,a (cid:105) both Q-represent t,
1 2 3 4
so q =(cid:104)a ,a (cid:105)−(cid:104)a ,a (cid:105) is isotropic.
1 2 3 4
2.6. n≥5.
SinceeveryquadraticforminatleastfivevariablesoveraCDVRwithfiniteresidue
field is isotropic, in this case the Hasse Principle for isotropy amounts to: if q is
indefinite, thenq isisotropic. Anindefiniteforminmorethanfivevariableshasan
indefinite subform in exactly five variables, so it suffices to treat the case n=5.
Now write h = (cid:104)a ,a (cid:105)−(cid:104)a ,a ,a (cid:105). Let V be ∞ together with the set of primes
1 2 3 4 5
dividing 2a ···a . For each v ∈ V, h is isotropic, so by the Isotropy Criterion
1 5 v
there is α ∈ Q× such that (cid:104)a ,a ,−α (cid:105) and (cid:104)a ,a ,a ,−α (cid:105) are both isotropic.
v v 1 2 v 3 4 5 v
In particular, for each v ∈V there are b ,b ∈Q such that
1,v 2,v v
α =a b2 +a b2 .
v 1 1,v 2 2,v
By Artin-Whaples Approximation, there are b ,b ∈Q× such that for all v ∈V,
1 2
a b2+a b2 ≡α (mod Q×2) ∀v ∈V.
1 1 2 2 v v
Put α=a b2+a b2. It follows that (cid:104)a ,a ,a ,−α(cid:105) is Q -isotropic for all v ∈V. It
1 1 2 2 3 4 5 v
also Q -isotropic for all other v – indeed, the ternary subform (cid:104)a ,a ,a (cid:105) already
v 3 4 5
has this property – so by the n = 4 case (cid:104)a ,a ,a ,−α(cid:105) is Q-isotropic: there are
3 4 5
x ,x ,x ,y ∈Q, not all 0, such that
3 4 5
a x2+a x2+a x2 =αy2 =a (b y)2+a (b y)2,
3 3 4 4 5 5 1 1 2 2
and thus h is Q-isotropic.
3. The Hasse Principle Over a Global Field
We now wish to prove the Hasse Principle for Isotropy for quadratic forms over an
arbitrary global field K. Here are the main ideas of the proof: the case n ≤ 1 is
stilltrivial, ofcourse. Eachofthecasesn=2andn=3turnsouttofollowfroma
big theorem in algebraic number theory, as we will explain (and then outsource to
standardworks). Ontheotherhand, viaadirtytrickitwillturnoutthatknowing
the n=3 case of the Hasse Principle for all global fields allows for an easier proof
of the n=4 case of the Hasse Principle over our fixed global field K! Finally, the
case of n≥5 goes exactly as in the case of K =Q.
10 PETEL.CLARK
3.1. n=2.
Letq beabinaryformoverK. AsinthecaseK =Q, itcomesdowntothefollow-
ing: given that −discq ∈K×2 for all v ∈Σ , we must show that −discq ∈K×2.
v K
The following theorem accomplishes this.
Theorem 15. (Global Square Theorem) For a∈K×, the following are equivalent:
(i) a is a square in K: there exists b∈K with b2 =a.
(ii) a is a local square: for all v ∈Σ , there is b ∈K with b2 =a.
v v v v
Proof. (i) =⇒ (ii) is immediate.
√
¬(i) =⇒ ¬(ii): supposethata∈K×\K×2: thenL=K( a)isaquadraticfield
extension (and even a Galois extension, since we are “globally” excluding the case
of characteristic 2). We may therefore apply a case of the celebrated Cebotarev
Density Theorem: the set of finite places v of K which split in the extension L/K
hasdensityequalto 1 . Inparticular,thereareinfinitelymanyplacesvwhichdo
[L:K] √
not split in L, which means that K ⊗ L=K ( a) is a quadratic field extension
v K v
of K , so a∈K×\K×2. (cid:3)
v v v
Exercise: Let K be a global field. Show that the natural map
(cid:89)
K×/K×2 → K×/K×2
v v
v∈ΣK
is injective.
Exercise: Letq beabinaryquadraticform,andletS ={v ∈Σ |q is isotropic }.
/K q K v
a) Show that if q is isotropic, S =Σ .
q K
b) Show that if q is anisotropic, S has density 1.
q 2
3.2. n=3.
Proposition 16. For a global field K, the following are equivalent:
(i) The Hasse Principle holds for ternary quadratic forms over K.
(ii) The Hasse Principle holds for Hilbert Symbols over K: for a,b∈K×, we have
(a,b)=1 iff (a,b) =1 for all v ∈Σ .
v K
(iii) The Hasse Principle holds for plane conics over K: if C is a smooth plane
/K
conic such that C(K )(cid:54)=∅ for all v ∈Σ , then C(K)(cid:54)=∅.
v K
(iv) The Hasse Principle holds for quaternion algebras over K: if B is a quater-
/K
nion algebra such that B =B⊗ K ∼=M (K ) for all v ∈Σ .
v K v 2 v K
(v) The Hasse Principle holds for quaternary quadratic forms over K of square
discriminant.
Proof. First observe that a form q is isotropic iff any similar form αq (for α∈K×
is isotropic.
(i) =⇒ (ii): Thisisimmediate,sincetheHilbertsymbol(a,b)tracksthean/isotropy
of the ternary form ax2+by2−z2 =0.
(ii) =⇒ (i): Every ternary quadratic form is similar to some ax2+by2−z2 =0.
(i) ⇐⇒ (iii) is immediate.
(iii) ⇐⇒ (iv)followsfromtheequivalencebetweenconicsandquaternionalgebras.
(iv) ⇐⇒ (v): Aquadraticformisquaternaryofsquarediscriminantiffitissimilar
to the norm form of a quaternion algebra. (cid:3)