REVIEW SHEET FOR LECTURE 10/20, 10/25 AND 10/27
MING-LUN HSIEH
1. Quadratic forms 1.1. Preliminaries.
Definition 1.1. Let k be a field. Let V be a vector space over k of dimension n. A quadratic form is a function Q : V → k such that
(1) Q(ax) = a2Q(X) for x∈ V and a ∈ k×.
(2) The pairing (x, y)7→ Q(x + y) − Q(x) − Q(y) is bilinear.
If the characteristic of k is not 2, then we associate a k-valued symmetric bilinear form ( , )Q on V×V to Q defined by
(x, y)Q:= 1
2· (Q(x + y) − Q(x) − Q(y)).
It is clear that (x, x)Q = (x). This gives e a bijection between quadratic forms and symmetric bilinear forms
From now on, we assume char k ̸= 2. In our later application, k = Q or Qp. Let e ={ei} be a k-basis of V . The n×n matrix Q[e] of Q associated to e is defined by
Q[e] := ((ei, ej)Q)1≤i,j≤n.
It is easy to see that if e′ is another basis of V , then there exists J ∈ GLn(k) such that Q(e′) = Jt· Q(e) · J.
We introduce an important invariant attached to Q.
Definition 1.2. We define the discriminant
d(Q) := det Q(e)∈ k×/(k×)2. This definition is independent of the choice of a basis of V . Definition 1.3. Let (V, Q) be a quadratic space. Then
(1) Orthogonal: x, v∈ V are if (x, y)Q = 0
(2) Orthogonal complement U⊥:={x ∈ U | (x, U)Q = 0} . (3) Radical V⊥
(4) (V, Q) is non-degenerate if V⊥= 0.
(5) Orthogonal sum: V = U ⊕ W such that (U, W )Q = 0.
(6) x∈ V is called isotropic if Q(x) = 0.
(7) We call (V, Q) a hyperbolic plane if V = kx + ky such that x, y are isotropic and (x, y)Q = 1.
Date: December 25, 2011.
1
Lemma 1.4. Suppose (V, Q) is non-degenerate. For a isotropic vector x, there exists y ∈ V such that U := kx + ky is a hyperbolic plane.
Proposition 1.5. If (V, Q) is non-degenerate, then there exists a orthogonal basis e ={e1, e2,· · · , en}.
In other words,
Q[e] = diag(a1, a2,· · · , an) with ai ∈ k×.
1.2. Quadratic forms over local fields. Let k = Qp and (V, Q) be a non-degenerate qua- dratic space. Let e be a orthogonal basis and let Q[e] = diag(a1,· · · , an). Then d(Q) = a1a2· · · an∈ k×/(k×)2. We define the Hasse invariant h(Q) by
h(Q) := ∏
1≤i<j≤n
(ai, aj)p ∈ {±1} .
Theorem 1.6. The Hasse invariant h(Q) does not depend on the choice of a orthogonal basis.
2. Solutions of the equation of quadratic forms over local fields To a quadratic space (V, Q), we can associate a quadratic form of rank n
f (x1,· · · , xn) := ∑
1≤i,j≤n
(ei, ej)Q· xixj
for a choice of basis e ={ei} of V . Conversely, given a (symmetric) quadratic form of rank n f (x1,· · · , xn) = ∑
1≤i≥j≤n
ai,jxixj (ai,j = aj,i),
we define a quadratic space (kn, Qf) by Qf(
∑n i=1
aiei) = f (a1,· · · , an).
For f, f′ two quadratic forms of rank n, we say f ∼ f′ (f is equivalent to f′) if (kn, Qf) ≃ (kn, Qf′).
We say f is non-degenerate if Qf is non-degenerate or equivalently d(Qf) ̸= 0. In what follows, we assume f is a non-degenerate quadratic form of rank n.
Definition 2.1. For a ∈ k, we say f represents a in k if f(x) = a has a nonzero solution x∈ kn. Equivalently,
f represents a
⇐⇒ Qf(v) = a for some nonzero v ∈ V = kn
⇐⇒ f ∼ f′+ aXn2 for some quadratic form f′ of rank n− 1.
Lemma 2.2. (1) If f represents 0, then f represents every element in k.
(2) Let a∈ k×. Then f represents a if and only f − aXn+12 represents 0.
Let k = Qp. Let d(f ) := d(Qf)∈ k×/(k×)2 be the determinant and h(f ) := h(Qf) be the Haase invariant of f .
Example 2.3. If f ∼ f′ + aXn2 with f′ quadratic form of rank n− 1, then d(f ) = d(f′)· a and h(f) = h(f′)· (a, d(f′))p.
Theorem 2.4. f represents 0 in k if and only if (1) n = 2 and d(f ) =−1.
(2) n = 3 and (−1, −d(f))p = h(f ).
(3) n = 4 and d(f )̸= 1 or d(f) = 1 and h(f) = (−1, −1)p. (4) n≥ 5.
Example 2.5. x2+ 2y2 = 7 has no solution in Q2.
Corollary 2.6. Let a∈ k×. Then f represents a in k if and only if (1) n = 1 and d(f ) = a.
(2) n = 2 and (a,−d(f))p = h(f ).
(3) n = 3 and either a̸= −d(f) or a = −d(f) and (−1, −d(f))p = h(f ).
(4) n≥ 4.
Theorem 2.7 (Classification of quadratic forms over local fields). Two quadratic forms f and f′ of rank n are equivalent if and only if d(f ) = d(f′) and h(f ) = h(f′).
3. Hasse-Minkowski theorem and applications
Theorem 3.1. Let f be a quadratic form of rank n over Q. Then f represents 0 in Q if and only f represents 0 in Qp for all p (including p =∞).
Proof. We sketch a proof here.
The case n = 2: Since f represents 0 in R, we may assume f = X2− aY2 with a > 0 ∈ Q.
Then it is not difficult to see that f represents 0 in Qp for all primes p if and only if the p-adic valuation vp(a) is even for all p. This is equivalent to saying that a∈ (Q×)2, and hence f represents 0 in Q.
The case n = 3: We may assume f = X2− aY2− bZ2. We may assume that a and b are co-prime square-free integers with |a| ≤ |b|. Let m = |a| + |b| ≥ 2. We prove the theorem by induction on m. In other words, we will show that if f represents 0 in Qp for all p, then f represents 0 in Q. If m = 2, then a, b = ±1, then f = X2 ± Y2 ± Z2. It can be checked directly that f represents 0 in Q except for f = X2+ Y2+ Z2 which do not represent 0 in Q∞= R.
We suppose that m > 2. Write b =±p1· · · ps, where pi are distinct primes. For each pi|b, f represents 0 in Qpi, so X2− aY2− bZ2 = 0 has a solution X, Y ∈ Z×pi and Z ∈ Zpi, which in particular implies that a∈ (Z×pi)2. By Chinese remainder theorem, we conclude that a (mod b) is a square. Thus we may write
t2 = a + bb′ for some b′ ∈ Z, |t| ≤ b 2. In particular, bb′ is a norm of Q(√
a). Thus we can deduce that
f represents 0 in k if and only if X2− aY2− b′Z2 represents 0 in k, where k can be Q or Qp. On the other hand,
|b′| = t2− a
b
≤ |b|
4 + 1 <|b| .
Write b′ = u2b′′ for some u∈ Z and square-free b′′. Then
X2− aY2− b′Z2 represents 0 in k if and only if X2− aY2− b′′Z2 represents 0 in k.
We thus reduced the problem to X2− aY2− b′′Z2 with |b′′| ≤ |b′| < |b|. The usual induction argument completes the proof for the case n = 3.
The case n = 4: We may assume f = g− ℓ, where g = a1X2+ a2X22 and ℓ = a3X32+ a4X42 with a1, a4 > 0. For k = Q or Qp, we know that f represents 0 in k if and only if there exists x ∈ k× such that x is represented by g and ℓ. Let N = 4p1p2· · · ps, where pi are odd prime divisors of the product a1a2a3a4. We have assumed that f represents 0 in Qp for all p. In particular, for p|N, there exists bp ∈ Zp − p2Zp which is represented by g and ℓ in Qp. By Chinese remainder theorem, there exists a0 ∈ Z≥0 such that
(3.1) a0 ≡ b2(mod 16), a0 ≡ bpi(mod p2i) for i = 1,· · · s.
Let m = N2 = 16p21p22· · · p2s. Any a satisfying the congruence relation must be of the form a = a0+ mk for k∈ Z, and
b−12 a≡ 1 (mod 8), b−1pi a ≡ 1 (mod pi) for all i = 1,· · · s.
In particular, b−1p a∈ (Z×p)2 for all p|N. We recall the following theorem of Dirichlet:
Theorem (Dirichlet). Let a, b be two co-prime integers. Then there exist infinitely primes of the form a + bn, n∈ Z.
Let d = (a0, m) be the G.C.D. of a0 and m. Then by the above theorem, there exists a prime q = ad0 +md · k for some k ∈ Z>0 and q- N. Let a := dq = a0+ mk. Consider two rank 3 quadratic forms:
g′ := g− aX52, ℓ′ = ℓ− aX52.
Then g′ and ℓ′ both represent 0 in Qp for all p|N. On the other hand, if p - Nq, then the coefficients of g′ and ℓ′ all belong to Z×p. It follows that g′ and h′ represent 0 in Qp for all p- Nq by Theorem 2.4 (2). To sum up, we have proved that g′ and ℓ′ both represent 0 for all Qp with p ̸= q. Again by Theorem 2.4 (2) combined with the quadratic reciprocity law, we conclude that g′ and ℓ′ represent 0 in Qq as well, and hence g′ and ℓ′ represent 0 in Q by the solution to Hasse-Minkowski’s theorem for n = 3, which is equivalent to saying that g and ℓ represent a in Q. This completes the proof for the case n = 4.
The case n≥ 5: The proof is similar to the case n = 4 as we discussed in the class. I leave
it to you as an exercise.
4. Exercise
Exercise 4.1. Show that 3x2+ 5y2 = z2 has no non-zero solution in Q.
Exercise 4.2. For which a∈ Z, the quadratic form x2 + 2y2− az2 represents 0 in Q.
Exercise 4.3. Let a, b, c∈ Z be three square-free integers. Suppose that a, b, c are relatively prime to each other and abc < 0. Let
f (x, y, z) = ax2+ by2+ cz2.
Then f represents 0 in Q if and only if the following congruence equations x2 ≡ − bc (mod a)
x2 ≡ − ca (mod b) x2 ≡ − ab (mod c) have solutions in Z.
Department of Mathematics, National Taiwan University, Taipei, Taiwan E-mail address: mlhsieh@math.ntu.edu.tw
