QUADRATIC FORMS, BASED ON (A COURSE IN ARITHMETIC BY SERRE)

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1 QUADRATIC FORMS, BASED ON (A COURSE IN ARITHMETIC BY SERRE) HEE OH 1. Lecture 1:Introduction and Finite fields Let f be a olynomial with integer coefficients. One of the basic roblem is to understand if f = m has an integral solution for any given m, or more secially, if f = 0 has an integral solution. Suose f = 0 has an integral solution. Then the equation f 0 (modr) should be solvable for all integer r = k 1 1 k s s. Hence f 0 mod k i i should be solvable for each rime i and ositive integer k i. These last two are in fact equivalent, which can be seen using the Chinese Remainder Theorem. The last equation can be always decided, since this is a finite tye roblem. However it is not a sufficient condition. Now consider an integral quadratic form, that is, f(x 1,, x n ) = a ij x i x j where a ij Z and a ij = a ji. Note that f(sx) = s 2 f(x), that is f is homogeneous of degree 2 and f(0) = 0. Hence the existence of a non-trivial integral solution to f = 0 is equivalent to that of a non-trivial rational solution to f = 0. Then the Hasse-Minkowski theorem says: Theorem 1.1. The following are equivalent: f = 0 has a non-trivial rational solution. f = 0 has a non-trivial real solution and f = 0(mod k ) has a non-trivial solution (that is, a solution that is not divisible by k ) for all rime and k. f = 0 has a non-trivial real solution and f = 0 has a non-trivial -adic solution. One of the main goal of this course is to understand the above Hasse-Minkowski theorem for quadratic forms. We remark that such a theorem is not already true for cubic forms (examle??). We will first talk about the solutions of f = 0(mod ), that is the solvability of an equation over a finite field F. Then in trying to understand when a solution to f = 0(mod ) can be lifted to a solution f = 0 mod 2, and to f = 0 mod 3 etc, we will see how the notion of -adic integers Z aears, as well the -adic field Q. 1

2 2 HEE OH 1.1. Finite fields. If K is any field, let i be the homomorhism from Z to K defined by i(1) = 1. Then i(z) is either Z or Z/mZ for some m, since any subgrou of Z is of the form e or mz. Moreover since k is a field and has no zero divisor, m has to be a rime. The characteristic of K is defined to be 0 if i(z) = Z and if i(z) = Z/Z. In other words, char(k) is the smallest ositive integer such that 1 = 0, unless it is 0. We denote by F the field given by Z/Z. Fix an algebraically closed field Ω F. The following is the classification of all fields F K Ω, and hence the classification of all finite fields u to isomorhism. Theorem 1.2. Set F r := {x Ω : x r x = 0}. Then F r is the unique subfield of Ω of order r. Proof. Let q = r. Note that σ : Ω Ω given by x x is an automorhism. It is a homomorhism since (x + y) = x + y. It is surjective since Ω is algebraically closed, and injective since x y = (x y). Hence σ r is also an automorhism. Therefore its fixed field F q := {x Ω : x q = x} is a field. Since the olynomial f(x) = x q x has the formal derivative f (x) = qx q , it follows that #F q = q (Lang, P. 179). Now, to show the uniqueness, let K be a field extension of F. If x K, then x q 1 = 1, since the order of K = q 1. It follows that K F q. By comaring the order, we have K = F q. Theorem 1.3. The multilicative grou F q is cyclic. Proof. Since F q is a finite abelian grou of order q 1, using the classification of finite abelian grous (the elementary divisor theorem) we have F q = Z/m i Z Z/m s Z. Here q 1 = m s and m i m i+1. It follows that every element in F q has an order at most m s, i.e., x m s = 1. This imlies that F q {x Ω : x ms = 1}. Since x ms has at most m s roots and hence q 1 m s, we have s = 1 and q 1 = m 1. Hence F q is cyclic. Remark In the textbook, there is a roof using the Euler φ-function, instead of using the classification of finite abelian grous.

3 LECTURE NOTES Equations over a finite field. Let K be a finite field, say F r. Consider a olynomial f K[x 1,, x n ] and the equation f = 0. Suose that f(x 1,, x n ) = a i x i, a i 0 i.e., a linear form. If n = 1, and f(x) = ax, then the number of solutions is 1 = q 0. If n 2, then the number of solution is q n 1. More generally, when the number of variables is bigger than the degree of f, we have the following Chevalley-Warning theorem concerning the number of solutions in K n of f = 0: Theorem 1.4 (Chevalley-Warning). Suose n > deg(f). Set V K := {X K n : f(x) = 0}. Then #V K 0(mod ). Corollary 1.5. If f K[x 1,, x n ] is a form of degree d in n variables and n > d, then f 0 has a non-trivial solution. In articular, any quadratic form over a finite field of at least 3 variables has a non-trivial zero. Remark If f is a quadratic form over K of 2 variables, is it true or not that f = 0 has a non-trivial solution? Proof. Since #V K 1 (0 V K ), it is at least by the above theorem. In order to give a roof for the Chevalley-Warning theorem, we need the following simle observation about the average of monomials: Lemma 1.6. Let 0 s < q 1. Then x s = 0(mod ). x K Here we agree 0 0 = 1. Hence if 0 s i < q 1 for some i, then x s1 1 x s n n = 0(mod ). (x 1,,x n) K n Proof. If s = 0, then the above sum is equal to #K = q, hence ok. If s 0, and s < q 1, there exists y K such that y s 1, for otherwise K {x Ω : x s = 1}, contradiction. Now x s y s = y s ( x s ). x s = x K x K Since y s 1, this roves the claim. Proof of Chevalley-Warning theorem Note that #V K = X K n χ VK (X)

4 4 HEE OH where χ VK denotes the characteristic function of V K in K n. So we need to understand the average of χ VK over K n. Observe that χ VK (X) = 1 f q 1 (X). This is clear, if X V K. If X / V K and hence f(x) K, then f(x) q 1 = 1. Hence It suffices to show that X K f q 1 (X) 0(mod ) to rove the theorem. n Note that f q 1 (X) is a linear combination of monomials x s 1 1 x sn n. Since d < n, we have s i (q 1)d < (q 1)n. Hence there exists s i < q. It remains to aly the above lemma.

5 LECTURE NOTES 5 2. Lecture 2: Quadratic residues modulo Last time, as a consequence of Chevalley-Warning theorem, we obtained that any quadratic form over finite field with at least 3 variables has a non-trivial solution. To comlete the icture, we state the following theorem. (note that x 2 + xy + y 2 is not considered as an integral binary quadratic form, since the coefficient of xy and yx must be symmetric integers: Let f(x, y) = ax 2 + 2bxy + cy ( 2. The ) discriminant d is defined to be ac b 2. That a b is, if we write f(x, y) = (x, y) (x, y) b c t, then the discriminant is recisely the determinant of the symmetric matrix corresonding to f. Theorem 2.1. The congruence f(x, y) 0(mod ) has a non-trivial solution if and only if d is a square (including 0) modulo. Proof. If a = c = 0(mod ), then d = b 2 and hence a square modulo, then it is easy to see the claim. Suose a 0(mod ). Hence there exists the inverse, say a 1, mod. In the same way we do in order to find a solution for a usual quadratic equation, we try to comlete the square above by or ax 2 + 2bxy + cy 2 = a(x + a 1 by) 2 + (c a 1 b 2 )y 2. Hence solving the above quadratic congruence amounts to solving a 2 (x + a 1 by) 2 ( ac + b 2 )y 2, dy 2 (ax + by) 2. Easy to see that if (x 0, y 0 ) is a non-trivial solution, then y 0. Hence y 0 has the inverse mod. Hence d y0 2 (ax 0 + by 0 ) 2 F 2. If d F 2, then clearly we have a non-trivial solution. (Do we need to use that 2 in the above roof?) Therefore the solvability of the quadratic congruences is reduced to the solvability of y 2 r(mod ) for any y. Definition 2.2. Any integer y such that y is called a quadratic residue modulo if x 2 y(mod ) has a solution, in other words, y is a square modulo (assuming that y is not divisible by ). So we meet the secific question about which elements a of F is a square of an element of F, or when is a (F ) 2?. Remark If = 2, clearly every element in F 2 is a square. Hence we assume that 2 below.

6 6 HEE OH Definition 2.3 (Legendre symbol). Let 2. Let x F. The Legendre symbol ( ) x is defined to be 1 if x (F ) 2 and 1 otherwise. ( ) ( ) For an integer x not divisible by, the symbol is understood as x(mod ). Proosition 2.4 (Euler s criterion). Let x F. Then ( ) x x ( 1)/2 (mod ). Proof. Let x F. Then x 1 = 1 and hence x ( 1)/2 = ±1. Let y Ω such that y 2 = x. If y F that is if x (F ) 2, then y 1 = 1 and hence x ( 1)/2 = 1. If y / F, then y 1 1, otherwise, it imlies that y F = {z Ω : z 1 = 1}. Hence x ( 1)/2 1 and hence 1. This roves the claim. The above can be formulated to saying that (F ) 2 is recisely the kernel of the homomorhism F F given by x x ( 1)/2. Since F is cyclic of order 1 and the image of this ma has to be recisely {±1(mod )} which has 2 elements since is odd, it follows that [F : (F ) 2 ] = 2. As a corollary of the Euler s criterion, we see that the Legendre symbol defines a homomorhism from F C, in fact from F onto {±1}. Corollary 2.5. ( a ) ( ) b = x ( ) ab Proof. By the Euler criterion, we have ( ) ab (ab) ( 1)/2 (mod ). ( ) ( ) a b Hence it is congruent to mod. Since it is ±1 and is odd rime, the congruence imlies equality as desired. Corollary 2.6. Let a = r 1 1 r k k Z. We have ( ) a = ( ) ri i. By the above corollary, to understand the solvability of the ( equation ) x 2 a(mod ) in general, it suffices to understand the Legendre symbol for for rimes and q. Theorem 2.7 (Quadratic recirocity law). Let, q be rimes not equal to 2. Then ( ) ( ) q = ( 1) ( 1)/2 (q 1)/2. q q

7 LECTURE NOTES 7 Indeed, the quadratic recirocity law rovides an algorithm which ( ) reduces the calculation of Legendre symbol for any integer to the calculations of for q = ±1, 2 (is there a systematic roof for this statement?). We remark that in addition to its role in determining the solvability of the quadratic congruence, it roves to be the key tool, often unexectedly, for solving many basic number theoretic roblems (for instance,??). Exercise 2.8. ( ( 7 13) = 13 ) ( 7, which is same as 6 ) ( 7 which is same as 2 ) ( 7 3 ) 7. However ( ) ( ) ( ) = = ( ) x 2.1. Comutation for. Let 2. Since is an odd rime, we have mod 4, ( ) ±1. Clearly we have = 1. 1 Theorem ( 2.9. ) TFAE: 1 = 1 ( 1) mod 4. a = 1 for any a Z. ( ) 1 Proof. (1) is clear. Using the Euler criterion = ( 1) 1/2, since 1 is even iff 2 ( ) ( ) a 1 mod 4 (instead of 3 mod 4), it follows. Lastly 1 = in view of the last corollary. Exercise Consider the equation x 2 19 mod 23. Observe that ( ) ( ) ( ) ( ) = = = since mod 4 we have Hence the above equation cannot be solved. Here is a much less trivial theorem in deciding 2 is a square mod : recall that mod 8, ±1, ±5. ( ) 2 Theorem We have = 1 iff ±1 mod 8. ( ) Proof. We have = 2 ( 1)/2 = y 1 if y satisfies y 2 = 2(mod ). We claim that 2 y = α + α 1 where α denotes a rimitive 8-th root of unity in Ω. It is so, since α 4 = 1 and y 2 = α α 2 = 2 + α 2 (α 4 + 1) = 2. Now we need to understand y 1. Note that y = (α + α 1 ) = α + α since characteristic of Ω is. Then if ±1 mod 8, then y = y, and if ±5 mod 8, then since α 4 = 1, y = y. q

8 8 HEE OH Therefore y 1 = ±1, and it is 1 exactly when ±1 mod 8. We can now finish the revious exercise 2.8: Exercise ( ( 7 13) = 2 ) ( 7 1 ) 7. Since 7 3 mod 4, and 7 1 mod 8, ( ) ( 1 7 = 1 and 2 ) ( 7 = 1. Hence 7 13) = 1. Therefore x 2 = 7 mod 13 is solvable. We note that the above methods tells only when the congruence x 2 a mod is solvable or not, but when there is a solution, it doesn t give any information on how to find one. Theorem 2.13 (Quadratic recirocity law). Let, q be rimes not equal to 2. Then ( ) ( ) q = ( 1) ( 1)/2 (q 1)/2. q Let w be a rimitive q-th root of unity in Ω, which is the algebraic closure of F. Then the following so-called Gauss sum is well defined: y = ( ) x w x Ω q x F q Lemma ( y 1 = q ). ( Note that by Lemma (1), (2) finishes the roof. Proof. y 2 ( 1) (q 1)/2 q(mod ). q ) q ( 1)/2 = ( 1) ( 1)/2 (q 1)/2 y ( 1). Hence Lemma y 2 = ( ) xz w x+z q x,z = ( ( ) ) t(u t) w u q u t = u w u = ( 1) (q 1)/2 u ( 1 ut ( 1) (q 1)/2 1 q t F q C u w u since t(u t) = t 2 (1 ut 1 ) and we have ut C u = t ) ( ) 1 ut 1. q

9 LECTURE NOTES 9 Note that if u = 0 then C u = q 1, and u F q, then 1 ut 1 runs over F q 1. Hence C u = ( ) t. q t F q 1 Since [F q : (F q) 2 ] = 2, we have ( ) ( ) t 1 = 0 = + C u. q q t F q ( ) Hence C u = = 1. Hence ( 1) q 1/2 y 2 = u F q C u w u = q 1 ( u F wu ) = q 1 q q. since r = ( u F q w u ) = 1 + w w q 1 = 0 (because wr = r). (If y C, then what is the meaning of y mod??) For the second claim, ) w x y = x F q ( x q = ( ) z 1 w z q z F q ( ) 1 = y q ( ) = y q We remark that the methods using the quadratic recirocity comletely determines whether x 2 = a(mod ) is solvable or not. However when it is solvable, it does not yield any information on the solution. Another roof of the quadratic recirocity law given by Einsenstein makes use of transcendental functions such as sin.

10 10 HEE OH 3. Lecture 3:On the number oints of an algebraic variety in a finite rime field 3.1. Main Theorem. A olynomial F (x 1,, x n ) is called absolutely irreducible if F cannot be written as gh where g, h C[x 1,, x n ]. Theorem 3.1. Let F (x 1,, x n ) be an absolutely irreducible olynomial with integral coefficients. Let N(F, ) be the number of solutions of the congruence f 0(mod ). Then N(F, ) n 1 < C(F ) n 1 1/2 where C(F ) deends only on F, but not. Corollary 3.2. There exists 0 such that N(F, ) 0 for all > 0. We remark that all known roofs of the above corollary go by way of the above theorem. The first roof of Corollary was given by Weil 1948 by roving a weaker version of the above theorem. For n = 1, Corollary is trivial since then F has to be linear. But already for n = 2, the roof is quite involved and requires some dee results in algebraic geometry. The theorem 3.1 was given by Lang and Weil. Note that N(F, ) = X F n χ V (X) where V is the set of solutions to F 0(mod ). We use the following notation: e(x) = ex(2πix). Note that e( 1 ) is a rimitive -th root of unity. Lemma 3.3. We have Proof. We first claim that N(F, ) = n χ V (X) = 1 x F X F n x F e( e( xf (X) ). xf (X) ). Note that x F e( xu ) = if u 0(mod ) and 0 otherwise. If X V and hence F (X) = 0, then the right hand side is 1. Otherwise it is 0. Since N(F, ) = X F χ n V (X), we have N(F, ) = 1 e(0) + xf (X) e( ) = n xf (X) e( ). X F n x F X F n x F X F n

11 LECTURE NOTES 11 Therefore we must show that as increases the remaining term increases in absolute value slower than the main term n 1, which is where all the difficulties lie Sums of owers. We will now rove a secial case of Theorem 3.1 when F (x,, x n ) = a 1 x a n x 2, a i 0(mod ). Indeed, we will rove: Theorem 3.4. N(F, ) n 1 C( 1) n/2 1. Examle 3.5. F is absolutely irreducible if and only if n 1, 2. Note that though the statement of Theorem is true for n = 1, 2, it has no content in those cases. We will see that in this case, everything boils down to obtaining an uer bound for the Gaussian sum (or the exonential sum) for the character given by the Legendre symbol. Note that the above is not meaningful for n = 1, 2. Remark For the case when F (x,, x n ) = a 1 x r a n x rn, the roof is almost identical, since we will be roving uer bound for a general Gaussian sum. Proosition 3.6. Set Then τ a (χ LS ) := t F ( t N(F, ) = n x F i=1 ) e( at ). n τ ai x(χ LS ). We have seen that N(F, ) = n = n x(a1x2 e( x F x 1,,x n x F i=1 n e( x i xaix2i ) a n x 2 n) ) Hence it suffices to have the following lemma in order to rove the roosition. Lemma 3.7. e( by2 ) = ( ) t e( bt ) =: τ bt(χ LS ). y F t F

12 12 HEE OH Proof. Note that by2 y e( ) = x m(x)e( bx ) where m(x) is the number of solutions of y 2 x(mod ). We have m(0) = 1. For x F, we can easily check that ( ) x m(x) = 2 = 1 + Hence the above is equal to (1 + e(bx )) + ( ) x e( bx ). x F x F Note that the sum of ( first ) two terms is 0, since b and hence we rove the claim, 0 since we agree that = 0. Theorem 3.8. For any a such that a, we have τ a (χ LS ) = The above theorem finishes the roof of Theorem 3.4, since we have N(F, ) n 1 C( 1) n/2 1. Examle 3.9. The same roof works for the case when we consider F c = F c. Note that ( N(F c, ) = n n ) xaix2i e( ) e( cx ) = n and that the modulus of e( cx x F x F ) is 1. ( ) i=1 x i τai x(χ LS )e( cx ) 3.3. Gaussian sum. For a, t t e( at ) is an examle of Gaussian sum. More generally it is defined as follows: Let χ : F S 1 be a multilicative character. We set χ(0) = 0. Then the sum τ a (χ) := t F χ(t)ζ at is well defined, and is called a Gaussian sum. We have: Theorem Let 1 d. Let a and χ 1, i.e., the unit of the trivial character. Then τ a (χ) 2 =. Let F be the vector sace of all comlex functions defined on F. F is clearly a comlex vector sace of dimension, since the ma f (f(0),, f( 1)) is a bijection between F and C.

13 Consider the inner roduct on F given by (f, g) = 1 f(x)g(x). LECTURE NOTES 13 x The functions f a (x) = e( ax) on F for a F form an orthonormal basis, since (f a, f b ) = 1 x F e( (a b)(x) ) is 1 if a = b and otherwise 0. Lemma χ = 1 τa (χ)f a and τ a (χ) is constant for all a F and τ 0 (χ) = 0. Note that (χ, χ) = 1 χ 2 = 1. Hence assuming the above lemma, we have and hence τ a (χ) =. 1 1 = 1 2 τ a (χ) 2 We now rove the above lemma: Write χ = a (χ, f a)f a. Then α a := (χ, f a ) = x χ(x)ζax = 1τ a(χ), and hence the first claim is roved. Hence we need to understand the coefficient α a. We claim that the modulus α a are all constants for a F. We will do this by comaring α ac with α a for any c such that c. Note that χ(cx) = χ(c)χ(x) = χ(c) a α a f a (x) = a χ(c)α ac f ac (x) = a χ(c)α ac f a (cx). On the other hand, χ cx = α a f a (cx). Hence α a = χ(c)α ac. Since χ(c) = 1, α 0 = 0 and α c = α 1. That is the modulus of α a is constant for any a F. Remark Note that in understanding N(F, ), the crucial thing was to understand the size of the exonential sum or the Gaussian sum. Having a meaningful estimate on the exonential sum means that there is some hidden cancelation in the sum. In a sense, what we exloited here is that the Gaussian sum aears as a Fourier coefficient of a character in a finite dimensional comlex vector sace. And hence, it has to satisfy certain relations among themselves {τ a (χ)}. Indeed, for understanding N(F ), the number of integral solutions for F = 0, we come u with a roblem of understanding Fourier coefficient of modular forms in an infinite dimensional sace. In dealing with F (X) = a i x r i i, we use the characters defined below, beyond the Legendre symbol:

14 14 HEE OH Examle Let d be such that 1 d, and let ɛ C be a rimitive d-th root of unity. For 0 s d 1, let χ s (x) = ɛ ks if x g k (mod ) for some generator of F (i.e., a rimitive root modulo ). Note that k is unique modulo 1. Hence χ s is a multilicative character of F. Note that χ 0 is the unit character, and for d = 2, χ 1 (x) = symbol. For a general χ 1, is k determined by the condition x (F ) k??? By doing some similar stes as before, we will have: Lemma Let N(F, ) be the number of congruence F (X) = a i x r i i where a i 0(mod ). Then where d i = (r i, 1). N(F, ) = n n d i 1 x F i=1 s=1 τ ai x(χ i,s ) ( x ) is the Legendre 0(mod ), In the above, χ i,s is defined as in the above examle with resect to d i. Hence the modulus of the Gaussian sum yields the main theorem for the above F, since N n 1 is then bounded above by 1 ( 1) (d i 1) = ( 1) n/2 1 n (d i 1). i=1

15 LECTURE NOTES Lecture 4: On -adic numbers Consider the congruence x 2 2 mod 7 n. For n = 1, we have x = ±3 mod 7. If x 2 2 mod 7 2, then x 2 2 mod 7, and hence x has to be the form x = ±3 + 7t. Let x = 3 + 7t. Plugging this, we have (3 + 7t) 2 = t 2 mod 7 2 and hence 6t 1 mod 7 and hence t = 1. So x = is a solution for x 2 2 mod 7 2. Again, a solution of x 2 2 mod 7 3 must satisfy the congruence mod 7 2. Hence a solution must be of the form x = t... By continuing this rocess, we obtain a sequence {x 1,, x n } of the form x n = a 0 + a a a a n 1 7 n 1, 0 a i 6, a solution of x 2 2 mod 7 n. And x n x n 1 mod 7 n 1. Definition 4.1. The -adic integer is a sequence of the form {x 1 = a 0, x 2 = a 0 + a 1, x 3 = a 0 + a 1 + a } where 0 a i 1}. More formally, this can be defined as the inverse limit (or rojective limit) of finite rings A k = Z/ k Z. Let φ n : A n A n 1 be the surjective ring homomorhism. The sequence A 1 φ2 A 2 forms a rojective system. Definition 4.2. The rojective (or inverse) limit of (A n, φ n ), denoted by lim A n is defined to be the subset {(x 1, x 2,, ) : φ n (x n ) = x n 1 } of the roduct ring X = n=1 A n. Note that X is a ring where the addition and the multilication is defined coordinate by coordinate. We can easily check that lim A n is also a subring. Set Z := lim A n, and call it the ring of -adic integers. Moreover, if we ut a discrete toology on each A n, then the roduct sace is comact by Tychonoff. Lemma 4.3. Z is comact as well. Proof. It suffices to show that it is closed. Let z / Z. Then for some n, φ n (z n+1 ) z n. Let O be the subset of X defined by the condition {(Y ) X : φ(y n+1 ) y n }. Clearly this is a oen neighborhood of Z disjoint from Z. Note that an element of Z is uniquely determined by the formal series i=0 a i i, 0 a i 1. The subring n Z consists of elements corresonding to i=n a i i, or the sequence (x 1, x 2,...) where x i 0 for all 1 i n and x n+1 = a n n, x n+2 = a n+1 n+1,... Lemma 4.4. The subring n Z is oen and closed in Z. They in fact form a basis around 0.

16 16 HEE OH The sequence 0 Z n Z ɛn A n 0 defined by ɛ n (X) = x n is an exact sequence. Hence Z / n Z is isomorhic to Z/ n Z. Proof. We only rove the second art. Let X = (x 1, x 2, ) = 0. Then if x i = i 1 j=0 a j j, then x i = 0 mod i imlies that a j = 0 for all j < i. Hence x j = 0 for all j < i. Hence the multilication by n is an injection. Note that x n 0 mod n means that all x i = 0 mod i for i n Hence X corresonds to i=n a i i. Hence X = n (a n, a n + a n+1, ) n Z. In view of the above lemma, X Z is divisible by n iff its n-th comonent x n is congruent to 0 mod n. In order to define a metric on Z as well as its field of fractions Q, we need the following theorem: An invertible element in Z is called a unit. We denote by U the subring of units of Z. Theorem 4.5. Any non-zero element x in Z is uniquely written as n u where u U. To rove this, Lemma 4.6. Let x = (x 1, x 2, ) Z Then x U if and only if x 1 0(mod ). Proof. If x U, there is y such that yx = (1, 1,...). Hence y 1 x 1 1(mod ), roving x 0(mod ). Suose x 1 0 (mod ). Then there is y 1 Z/Z such that x 1 y 1 1(mod ). Let s find y 2 = y 1 + a such that x 2 y 2 1 mod 2. Since x 2 = x 1 + b, we need to solve x 1 y 1 + (ax 1 + by 1 ) 1 mod 2. Since x 1 y 1 = 1 + c, we only need to solve ax 1 + by 1 + c 0(mod ). Since x 1 y 1 1(mod ), we have a y 1 (by 1 + c). This rocess can be continued.. Since Z/ m Z/Z/ m Z is isomorhic to Z/Z, if x 1 0 and hence x m / (Z/ m Z), then there is an element y m in Z/ m Z (which is an inverse mod ). Hence x m y m = 1 z m for some z Z/ m Z. Since 1 z m has the inverse 1 + z + (z) m 1, x m is invertible. This roves the claim. We now rove the theorem: Let x = (x 1, x 2,...) Z. If x is a unit, then n = 0. And this is a necessary and sufficient by the above lemma. Let k be the smallest index such that x k+1 0. Then by the lemma 4.4, x k Z and x / k+1 Z. Hence u := k x Z is a unit since does not divide u. It follows that Z is an integral domain from the theorem and hence we can talk about a field of fraction: Definition 4.7. Now the field of fraction of the ring Z is given by Z [ 1 ] = { m Z : m Z}. We denote this field by Q and call it the -adic field.

17 LECTURE NOTES 17 Before introducing the -adic valuation and the -adic absolute value, let s recall those notions for an arbitrary field: Definition 4.8. An absolute value on K is a function : K R + such that x = 0 iff x = 0 and xy = x y and x + y x + y. If it additionally satisfies x + y max{ x, y }, then it is called non-archimedean and otherwise archimedean. If we have an absolute value, then it defines a metric on K by d(x, y) = x y. A non-archimedean absolute value defines an ultra metric, that is, d(x,, y) max{d(x, z), d(z, y)}. Definition 4.9. A valuation of a field K is a surjective ma v : K Z which satisfies v(xy) = v(x) + v(y). If it also satisfies v(x + y) min(v(x), v(y), then it is called non-archimedean. If we have a valuation v on K, then we can define an absolute value x on K by x = e v(x) for a non-zero x and 0 = 0. Note that a non-archimedean valuation gives a non-archimedean absolute value, and hence a ultra metric. Note that any non-zero element in Q is of the form m u where m Z and u U are uniquely determined. Definition 4.10 (-adic valuation). For x = n u Q, the -adic valuation v (x) is defined as n. Lemma The above is indeed a valuation, i.e., the ma v : Q Z satisfies v (xy) = v (x) + v (y) and v (x + y) min(v (x), v (y)). Proof. Note that if x = m u and y = k v where u and v are units, and m k, then x+y = m (u+ k m v) and u+ k m v Z. Hence v (x+y) m = min{v (x), v (y)}. Definition 4.12 (-adic absolute value). For x Q, x := v (x) or x = e v (x). This defines an ultra metric by d(x, y) = x y. (Indeed, for any x y, x + y = max{ x, y }. Examle Hence n 0 as n and n 0 as n. Theorem Any non-archimedean valuation on Q is given by v for some. Any non-trivial absolute value on Q is either the usual absolute value or, u to equivalence (Ostrowski). Here two absolute values are called equivalent if they define the same toology, or if they define the same notion of convergence.

18 18 HEE OH Note that Z = {x Q : x 1} and U = {x Q : x = 1}. Using this metric, the ideals m Z = {x Q : x m } = {x Q : x < y} (where y is any element between m and m+1 ), m = 0, 1, 2, are closed and oen subsets in Q. Observe that the toology on Z induced by this metric coincides with the rofinite toology on Z, and hence Z is comact, and hence Z is a comlete metric sace. Hence Q is locally comact, and is comlete as well, since if {X n } is cauchy, we have X n is bounded, and hence for some k, k X n Z. Since k X n is cauchy and Z is comlete, it converges and so does {X n }. Examle Show that Q Q. Lemma Z and Q are dense in Z and Q resectively. We have a notion of continuous function from Q n Q. For instance, clearly olynomials in Q [x 1,, x n ] are continuous. Theorem A sequence X n Q converges if and only if X n X n+1 0. Proof. One direction is clear. Suose X n X n+1 0. It follows from the ultrametric inequality that {X n } is a cauchy and hence it converges. Corollary Any series i= k a i i, 0 a i 1 converges to an element in Q. Indeed, Q is the comletion of Q with resect to the -adic absolute value.

19 LECTURE NOTES Lecture 5: Solutions over Z Theorem 5.1. Let F Z [x 1,, x n ]. Let F k be the olynomial with coefficients in A k = Z/ k Z. Then F k 0 mod k is solvable for all k 1 iff F = 0 is solvable over Z. This is again equivalent to saying F 0 mod k i is solvable for an infinite sequene=ce k i. Proof. Suose F (X) = 0 for X Z n. This imlies that mod k, we have a congruence solution. Now suose that we have F k (x k 1,, x k n) 0 mod k. First choose a convergent subsequence x k1 i 1 for x k 1 in Z. Among x k1 i 2, choose a convergent subsequence x k2 i 2, etc. Then we have a subsequence X l i = (x l i 1,, x l i n ) which is a solution mod l i and x l i j α j Z as i. Since F (X l i ) = l i u i u i Z, we have F (X l i ) 0 and since F is continuous, we have F (α 1,, α n ) = 0. Note that the above is again equivalent to F li 0 mod l i is solvable for some infinite sequence l i. Theorem 5.2. Let F Z[x 1,, x n ] be a form. Then F 0 mod k has solutions not divisible by for all k 1 iff F = 0 has a rimitive solution over Z. Proof. Suose X = (x 1,, x n ) Z n be a rimitive solution. Then min(v (x 1 ),, v (x n )) = 0. Hence for some i, v (x i ) = 0. Then ɛ k (x i ) = k j=1 αj j with α 0 0. And F (ɛ k (x 1 ),, ɛ k (x n )) = 0 mod k. Now, conversely, suose (x k 1,, x k n) are rimitive solutions mod k. Then for some i, there is an infinite subsequence x k j i not divided by, and converge to α a unit in Z. By taking a suitable subsequence among (x k j 1,, x k j n ), we have a limit in Z n which is not divisible by. Theorem 5.3. Let F Z [X 1,, X n ], and (Z 1,, Z n ) Z n. F (Z) = 0 mod m and 2 min 1 i n {v ( F X i (Z))} < m v (F (Z)). Or equivalently, min v { F X i (Z))} = k < m/2. Then there is a Y Z n such that F (Y ) = 0 and Z Y mod m k. Suose that Examle 5.4. Can we do better, that is, can the condition k < m/2 be weakened?? Lemma 5.5. Let F Z [X]. Let X Z be such that F (X) = 0 mod m and v (F (X)) = k < m/2. Or equivalently 2v (F (x)) < v (F (X)) or F (X) 2 < F (X). Then there is a Y Z such that F (Y ) = 0 mod m+1 and v (F (Y )) = k and X Y mod m k.

20 20 HEE OH Proof. Write Y = X + t m k. By the taylor formula (for olynomials, easy to check), we have f(y ) = f(x) + m k tf (X) + 2m 2k a with a Z. We want to solve f(y ) 0 mod m+1. Write f(x) = m c and f (X) = k d. Then m (c + td + m 2k a) 0 mod m+1 is same as c + td = m 2k a mod. Since d is a unit, we can find t. Only need to check that v (f (Y )) = k. Note that f (Y ) = k d + m k r where r Z. Since d is a unit, and m k > k, v (f (Y )) = k. Proof of the theorem For some i, v ( F X i (Z))) = k. Without loss of generality, let i = n. Then, set f(x n ) := F (Z 1,, Z n 1, X n ). Let W 0 = Z n. By alying the above lemma to f, we have W 1 Z such that f(w 1 ) 0 mod m+1 and v (f (W 1 )) = k and W 1 W 0 modulo m k. Alying the same to W 1 recursively, we obtain a sequence W q Z such that f(w q ) 0 mod m+q and W q = W q 1 modulo m+q k. Note that v (W q+l W q ) m + q + 1 k and hence W q+l W q (m+q+1 k) for all l > 1. Hence W q is a Cauchy sequence and hence converges to α Z. Then f(α) = 0, and hence F (Z 1,, Z n 1, α) = 0. Note that since α W 0 mod m k, we have (Z 1,, Z n 1, α) (Z 1,, Z n ) mod m k. Corollary 5.6. Let F Z [X 1,, X n ]. Let F (Z 1,, Z n ) 0(mod ). If F X i (Z)) is not divisible by for some i, then F = 0 is solvable over Z. Corollary 5.7. Let 2. Let f(x 1,, X n ) = a ij X i X j be a quadratic form (a ij = a ji ) with coefficients in Z. Suose that determinant of (a ij ), i.e, the discriminant of f is not divisible by. Then any rimitive solution to f 0(mod ) lifts to a Z solution. Proof. Suose that Z is a zero of f mod and all artial derivatives f X i (Z)) are divisible by. Since f (5.8) (Z)) = 2 a ij Z j X i and (a ij ) is invertible mod, it follows that (Z 1,, Z n ) cannot be rimitive. Corollary 5.9. = 2. Let f be as above and let det(a ij ) 1 mod 2. Then a rimitive solution Z of f(x) 0 mod 2 3 = 8 can be lifted to a Z 2 zero. That is, if f(x) 0 mod 8 has a solution whose coordinates are not all even, then it lifts to a solution over Z 2. Proof. The det condition imlies that the artial derivatives of f at Z are not divided by 4 by 5.8. Hence min i {v ( f X i (Z))} 1. Hence take m = 3 and k = 1, to aly the revious theorem.

21 LECTURE NOTES 21 Corollary Let f(x) = a ij X ij be a quadratic form (a ij = a ji ) with coefficients in Z and at least 3 variables Then f = 0 is solvable over Z for all but finitely many. Proof. By Chevalley-Warning, f has a rimitive solution for all. Let be any odd rime not dividing the determinant of a ij. Then aly the above theorem. For instance, if f(x) = n i=1 x2 i and n 3, then f(x) = 0 is solvable for all Z for all odd rime. For = 2, x x x 2 3 = 0 or x x x x 2 4 = 0 has no solution over Z 2 (since mod 8 has no solutions such that at least one of which is odd). Remark We will see later that for any quadratic form f over Z of at least 5 variables, f 0 has a non-zero solution for all Z. This imlies that for any quadratic form f over Q, SO n (f, Q ) cannot be comact,i.e., isotroic, if n 5.

22 22 HEE OH 6. Lecture 6 More generally, we have the following fact: Theorem 6.1 (BS, P. 44). Let F Z[X 1,, X n ] be an absolutely irreducible olynomial. Then F 0 can be solvable over Z for all excet for finitely many. The above follows from the fact that for all sufficiently large, the number of solutions F 0(mod ) is greater than the number of solutions of the system F (X 1,, X n ) 0(mod ) and F Xn (X 1, X n ) 0(mod ). Indeed, the number of the solutions to the former behave bigger than n 1 C F n 1 1/2, and the latter is bounded above by D n 2 where D deends only on F. Hence there has to be some zero of F 0 mod whose artial derivatives are not divided by. Examle 6.2. Fill out the details!! At least check the case for f = a i x r i i Squares in Q. Let x Q. When is x a square in Q? If x = y 2 Q 2, v (x) has to be even. Hence x = 2m u where u is a unit in Z. Hence this roblem reduces to when a unit u is a square. Lemma 6.3. Let 2. A unit u is a square in Z iff u is a square mod. Proof. If f(x) = x 2 u has a solution y mod, does not divide y. f (y) = 2y 0(mod ), hence y can be lifted to a solution over Z. If u = y 2 for y Z, clearly mod u is a square. Corollary 6.4. Let 2. Let x = k u Q. Then x is a square if and only if k is even and u((mod )) is a quadratic residue mod. Hence Q /(Q ) 2 is isomorhic to Z/2Z Z/2Z. Proof. The ma x = k u (k mod 2, umod /(F ) 2 ) is an isomorhism onto Z/2Z F /(F ) 2. The latter is isomorhic to Z/2Z. Now let = 2. Lemma 6.5. A unit u in Z 2 is a square iff u 1 mod 2 3 = 8. Proof. Suose u = y 2. So that u = y 2 mod 8. Note that y mod 8 is an odd integer, since otherwise u = 0 mod 2, contradiction to the unit. However, the square of any odd integer is 1 mod 8: (2k + 1) 2 = 4k 2 + 4k + 1. If k is even, clear. If k is odd, Check. Suose u = 1 mod 8. If F (x) = x 2 u, F (1) = 0 mod 8 and F (1) = 2. Hence 2v 2 (F (1)) = 2 < 3 v 2 (F (1)). Hence there is a Z -lift y such that F (y) = 0. Hence [Q 2 : (Q 2) 2 ] = 8, and since Q 2/(Q 2) 2 is an abelian grou with every element having an order 2, it is isomorhic to Z/2Z Z/2Z Z/2Z.

23 LECTURE NOTES Quadratic forms. Let A be a commutative ring (e.g., A = Z, A = Z, a field) and V a module over A. Definition 6.6. A function Q : V A is a quadratic form if Q(ax) = a 2 Q(x) and if we set (x, y) := (Q(x + y) Q(x) Q(y)), then (, ) is a bilinear form. In the following let k be a field of characteristic not 2 and V be an n-dimensional vector sace over k. We will relate quadratic forms to symmetric bilinear forms on V V and to symmetric matrices in M n (k). Let Q be the sace of quadratic forms on V and S be the sace of symmetric bilinear forms on V. Let M be the symmetric matrices in M n (k). Theorem 6.7. For a quadratic form Q on V, consider the symmetric bilinear form (x, y) Q = 1 (Q(x + y) Q(x) Q(y)) (also called as a scalar roduct). 2 This is a bijection between Q and S. Fix a basis (e i ) of A. Then for Q Q, set A to be the matrix defined by (e i, e j ) Q. This gives a bijection and the inverse is given by And Q( x i e i ) = ij a ij x i x j. ( x i e i, y i e i ) = (x 1,, x n )A(y 1,, y n ) t. Consider the equivalence classes as [Q 1 ] = [Q 2 ] if and only if there is a vector sace isomorhism f : V V such that Q 1 (x) = Q 2 (f(x)) for all x V. Let A, B M be equivalent if and only if A = XBX t for some X GL n (k). Then the above ma induces a bijection. Proof. The first two claims are clear. If [Q 1 ] = [Q 2 ] then (x, y) Q1 = (f(x), f(y)) Q2. Hence if f i = x ij e j, (e i, e j ) Q1 = (f(e i ), f(e j )) Q2 = X(e i, e j ) Q2 X t. Hence the ma is well defined. If A i are symmetric matrices such that Q i (v) = va i v t, and A 1 = XA 2 X t, then define f : V V by f(e i ) = x ij e j. Then Q 1 (v = v i e i ) = va 1 v t = vxa 2 X t v t = Q 2 (f(v)). We would like to understand the classification of quadratic forms u to equivalence. For instance, note that understanding Q(V ), i.e, what numbers are reresented by Q does not deend on the equivalence class, since f : V V is an automorhism. One of the imortant invariants is this: Definition 6.8. The discriminant of Q is an element of k/(k ) 2 reresented by the determinant of the symmetric matrix corresonding to Q with resect to some basis. If d(q) 0, then Q is called non-degenerate.

24 24 HEE OH Lemma 6.9. TFAE Q is non-degenerate. its radical rad(v ) := {w V : (w, v) Q = 0 for all v V } = 0. the ma Q V : V V given by l v (w) = (v, w) Q is an isomorhism. Proof. If Q is degenerate, for some v = v i e i V, Av = 0. Hence (w, v) Q = 0 for any w V, and vice versa. Note that Q V is a linear ma and the kernel of Q V is radv. Hence Q is non-degenerate if and only if Q V is injective. Since we have dim(v ) = dim(v ), the claim is roved. We would like to rove that any quadratic form has an orthogonal basis, or equivalently, the corresonding symmetric matrix is equivalent to a diagonal matrix. Definition Let U V be a subsace. The orthogonal comlement U 0 is defined as {v V : (u, v) Q = 0 for all u U}. V = U i if V = U i and U i are airwise orthogonal to each other, i.e., (u i, u j ) Q = 0 for i j. Note that Q = Q Ui The last one follows since (x, y) = 0 = 1/2(Q(x + y) Q(x) Q(y)) imlies Q(x + y) = Q(x) + Q(y).

25 LECTURE NOTES Lecture 7 Lemma 7.1. Let (V, Q) be a quadratic module and U be any subsace such that V = U radv. Then Q U is non-degenerate and V = U radv. Lemma 7.2. Let (V, Q) be a non-degenerate quadratic module. Then for a subsace U V, we have U 00 = U, radu = U U 0 = radu 0 and dimv = dimu + dimu 0. Proof. Clearly U U OO. Since V is nondegenerate, the ma V V U is a surjection and the kernel is U 0. Hence dim(v ) = dimu + dimu 0. Since dimu = dimu, we have dimv = dimu + dimu 0. Now alying the same argument to U 0 gives dimv = dimu 0 + dimu 00. Hence U = U 00 by comaring the dimension. Corollary 7.3. Let (V, Q) be as above and U < V. TFAE U is non-degenerate. U 0 is non-degenerate. U U 0 = 0. V = U U 0. That is, any non-degenerate subsace U of V has an orthogonal comlement. Theorem 7.4. Any quadratic module has an orthogonal basis. That is, there exists a basis e 1,, e n for V such that (e i, e j ) Q = 0 whenever i j. It follows that Q( x i e i ) = n i=1 a ix 2 i where a i = Q(e i ). In other words, the symmetric matrix ((e i, e i ) Q )) corresonding to Q wrt {e i } is a diagonal matrix. Proof. First we have V = radv U for some non-degenerate subsace U. Note that any basis for radv is orthogonal to each other, and orthogonal to any choice of basis for U. Hence it suffices to rove the claim for Q U, and hence non-degenerate. Hence there exists e 1 V be such that Q(e 1 ) 0. Then U = ke 1 is a non-degenerate subsace, we have V = ke 1 U 0 and Q U 0 is non-degenerate. Hence we roceed by induction. Examle 7.5. Give an examle of degenerate subsace of non-degenerate (V, Q). Let V = R 2 and Q((x, y)) = x 2 y 2. Then on U = R(1, 1), Q U = 0 and hence U is degenerate. Another useful decomosition for quadratic form is obtained via hyerbolic lanes. Definition 7.6. An element x (V, Q) is isotroic if Q(x) = 0. That is, Q contains a non-zero isotroic vector if and only if Q reresents 0. A two dimensional subsace U is called hyerbolic lane if it has ( a basis ) u 1 and u which are isotroic and (u 1, u 2 ) = 1. Hence Q U has the matrix with resect 1 0 to u 1 and u 2. Lemma 7.7. Let (V, Q) be non-degenerate and x 0 an isotroic element. Then there is a subsace U of V which is hyerbolic lane and contains x.

26 26 HEE OH Proof. Since V is nd, for some z V, (x, z) 0 and hence we may assume (x, z) = 1. Note that z is not a multile of x since (x, z) = 0. Let y = dz+cx. To have (x, y) = 1, we have d = 1 Then(y, y) = (z, z) + 2c(z, x) = 0 imlies that y = z 1 (z, z)x. 2 Corollary 7.8. Let (V, Q) be as above. Then V = H W where H is the orthogonal sum of hyerbolic lanes and H 0 = W is a subsace with no non-zero isotroic element. Corollary 7.9. If (V, Q) is non-degenerate, and contains a non-zero isotroic vector, then Q(V ) = k. Proof. Let x and y be a basis consisting of isotroic elements for the hyerbolic lane. Then for any a k, Q(x+a/2y) = a, since a/2 = (x, a/2y) = 1(Q(x+a/2y) Q(x) 2 Q(a/2y)). Hence a/2 = 1/2Q(x + a/2y) Reresentation of 0 by -adic quadratic form. Theorem Any quadratic form over Q in at least 5 variables always reresents 0 (by a non-zero element). Why not for n = 3, 4? For instance,if f(x) = n i=1 x2 i and n 3, then f(x) = 0 is solvable for all Z for all odd rime (since we have congruence solutions mod and the discriminant is 1 and hence invertible mod ). For = 2, x x x 2 3 = 0 has no solution over Z 2 : if so, there is congruence solution mod 4 one of which is odd, say, x 1. Then x 2 1 = 1 mod 4. But the sum any two squares cannot be 3 mod 4 (it is either 0 when both are even, 2 when both are odd, and 1 if one is even and the other is odd). Similarly x x x x 2 4 = 0 has no solution over Z 2 : if so, let x 1 be odd and (x 1,, x 4 ) be a congruence solution mod 8. Then x 2 1 = 1 mod 8. But the sum of three squares cannot be 7 mod 8 (as it was in the hw). Hence 5 variables is the best we can hoe for. Corollary If Q is a quadratic form over Q with at least 4 variables, Q(V ) = Q. Proof. Let a Q and f(x, x n+1 ) = Q(x) ax 2 n+1. Then by the above theorem, f has a non-zero isotroic vector. If x n+1 0, Q(Xx 1 n+1) = a. If x n+1 = 0, then Q has a non-zero isotroic vector. Hence Q(V ) = Q by Corollary 7.9. In the following, we may assume that Q(X) = n i=1 α ix 2 i with α i 0 and n 5. Let k = Q. Then we can furthermore have r n F (X) = ɛ i x 2 i + ( ɛ i x 2 i ) = F 0 (X 1,, X r ) + F 1 (X r+1,, X n ) i= where ɛ i are -adic units. i=r+1

27 LECTURE NOTES 27 We would like to understand when Q(X) = 0 has a non-zero solution over Q, or equivalently over Z. Without loss of generality, we assume r n r, by relacing Q by Q if necessary. Corollary Any quadratic form with at least 5 variables reresents 0 for all Q for any odd rime. Proof. We have F (X) = F 0 + F 1 as before. We may assume r n r, and hence r 3. Hence F 0 (x 1,, x r ) = ɛ i x 2 i always reresents 0 over Z for all odd rime, since mod it has a solution and determinant is invertible mod. Hence it lifts to a zero over Q. Though we do not need this, we will rove the following: Theorem F reresents 0 if and only if F 0 or F 1 reresents 0. Proof. Suose F (X) = 0. We may assume X Z n and X. If (X 1,, X r ) is not divisible by, i.e., X i for some 1 i r, then F 0 (X) 0 mod, and F 0 X i (X) = 2ɛ i X i 0(mod ). Hence F 0 has a zero over Z. If X i for all 1 i r, then 2 F 0 (X 1,, X r ) and hence mod 2, (F 1 (X r+1,, X n )) = 0. Hence F 1 (X r+1,, X n ) = 0(mod ) and (X r+1,, X n ). Hence F 1 = 0 reresents 0 over Z. To see the other direction, if F 0 reresents 0, then consider (X 1,, X r, 0,, 0) etc. Examle Counterexamle to the above for = 2?

28 28 HEE OH 8. Lecture 8 We begin by clarifying differences between two notions quadratic module and quadratic form: A quadratic module (over k) is a air (V, Q) where V is a k vector sace and Q is a quadratic form on V. A quadratic form over k in n variables is a function on k n given as f(x 1,, x n ) = a ij x i x j 1 i,j n where a ij = a ji. These two are related as follows: For a given quadratic form f over k in n variables, the air (k n, f) is a quadratic module. Conversely, if dim(v ) = n. and we fix any basis {e i }, then for a given quadratic module ( V, Q), we can associate a quadratic form over k in n variables as f(x 1,, x n ) := Q( x i e i ). Then f(x 1,, x n ) = a ij x i x j where a ij = (e i, e i ) Q. 1 i,j n Definition 8.1. f is non-degenerate if the quadratic module (k n, f) is nondegenerate. We say f 1 and f 2 are equivalent if (k n, f 1 ) and (k n, f 2 ) are isometric, that is, for some linear isomorhism A : k n k n, f 1 (X) = f 2 (AX). f reresents a k over k if there is a non-zero x k n such that f(x) = a. Hence f reresents 0 over k if (k n, f) has a non-zero isotroic vector. Lemma 8.2. If two quadratic forms f 1, f 2 are equivalent, then f 1 reresents a k if and only if f 2 reresents a. Now the theorem saying that any quadratic module admits an orthogonal basis can be translated as follows: Theorem 8.3. Let f be a quadratic form in n variables. Then f is equivalent to g where g(x 1,, x n ) = r i=1 a ix 2 i where all a i 0 and r n. Here, r is called the rank of f. Hence f is non-degenerate if and only if r = n. Any quadratic form f is equivalent to g where g(x 1,, x n ) = h(x 1,, x r ) where h is a non-degenerate quadratic form of r variables. The only thing to check is that the rank of f deends only on the equivalence class of f. It is because that r is recisely the rank of the symmetric matrix corresonding to (k n, f) (i.e., the maximum number of linearly indeendent columns of the matrix). Equivalent symmetric matrices have the same rank. Remark We note that studying the quadratic form via the quadratic module is a very owerful tool. Many of the imortant theorems on quadratic forms are roven using the theory of quadratic module, for instance, the diagonalization.

29 LECTURE NOTES 29 Theorem 8.4. Any quadratic form Q over Q with rank n 5, or equivalently any non-degenerate quadratic form in at least 5 variables, always reresents 0. Proof. First, any such Q is equivalent to s i=1 ɛ ix 2 i + ( n s+1 ɛ ix 2 i ) where ɛ i are units in Z and s n s. Hence we may assume Q is given as above. Then since s 3 and hence s i=1 ɛ ix 2 i reresents 0 over Q for any odd, so does Q. we only need to consider the case = 2. We consider the case when s = n and s < n. Let r = n, and may assume that 5 f = ɛ i x 2 i. i=1 We only need to show that f 0 mod 8 has a solution x 1,, x 5 which are not all even. Note that ɛ 1 +ɛ 2 and ɛ 3 +ɛ 4 are 0 or 2 mod 4. If both are 2 mod 4, then the sum 4 i=1 ɛ i = 4c for some c Z 2. Hence f(1, 1, 1, 1, 2c) = 4c(1 + ɛ 5 c) 0 mod 8. Now suose that ɛ 1 +ɛ 2 = 0 mod 4. Then ɛ 1 +ɛ 2 = 4c for c Z. Then f(1, 1, 0, 0, 2c) 0 mod 8. Let s < n. Then take 3 f = ɛ i x 2 i + 2ɛ n x 2 n. i=1 We only need to show that f 0 mod 8 has a solution (x 1, x 2, x 3, x n ) such that one of x 1, x 2, x 3 is odd. Note that ɛ 1 +ɛ 2 = 2c for c Z. Then ɛ 1 +ɛ 2 +2ɛ n c 2 = 2c(1+cɛ 2 n) = 0 mod 4. Hence ɛ 1 + ɛ 2 + 2ɛ n c 2 = 4β with β Z 2. Then f(1, 1, 2β, c) = 4β + 4β 2 ɛ 3 = 4β(1 + βe 3 ) 0 mod 8. Corollary 8.5. Any quadratic form Q over Q with rank n 4 reresents all nonzero elements in Q. Or Q(Q n ) = Q Witt s theorem. Let (V, Q) be a non-degenerate quadratic module. An isometry from V to V is called autometry. The set Aut(V, Q) becomes a grou with resect to the comosition oeration, and denoted by O(V ). Theorem 8.6 (Witt). Let U 1, U 2 be subsaces of V and let s : U 1 U 2 be an isometry. Then s extends to an autometry V V. Definition 8.7 (symmetry). Let v V be such that Q(v) 0. Then the symmetry τ v id defined as follows: Since kv is non-degenerate, we have V = kv U 0. Set τ v (kv + w) = kv + w where w U 0. Hence τ v is identity on U 0. Clearly τv 2 = id. In fact, 2(w, v) τ v (w) = w (v, v) v. Lemma 8.8. Let u, v V and Q(u) = Q(v). If Q(u v) 0, then τ u v (u) = v. Proof. Comutation!

30 30 HEE OH Corollary 8.9. If Q(u) = Q(v) 0, then there is an autometry S : V V such that S(u) = v. Moreover S is either a symmetry or the roduct of two symmetries. In articular, it follows that O(V ) acts transitively on {v V : Q(v) = c} for any c k. Proof. Note that Q(u + v) + Q(u v) = 2(Q(u) + Q(v)) = 4Q(u) 0. Hence one of Q(u + v) or Q(u v) is non-zero. Suose Q(u v) 0, then set S = τ u v. If Q(u + v) 0, then τ u+v (u) = v. Hence set S = τ v τ u+v. Proof of Witt s theorem for U 1 non-degenerate We use induction: Let dim(u 1 ) = 1. Then for some u U 1, Q(u) 0. Set v = s(u) and aly the above corollary. Since U 1 is non-degenerate, we have for some u U 1, Q(u) 0. By alying S as above, we may assume that s(u) = u. Let W be the orthogonal comlement of ku in V. Set W i = U i W. The restricted isometry s : W 1 W 2 extends to S W : W W by induction. Now on V = ku W, we have id S W which extends s. Proof of Witt s theorem for a general case It suffices to show that for the case when U 1 is degenerate, we can extend s to a subsace V 1 which contains U 1 as a hyerlane. The idea is to look for an isotroic element z / U 1. Let u radu 1 and l U 1 be such that l(u) = 1. Using the isomorhism V V given by v l v, there is v V such that (v, u) = l(u) for all u U 1. Set z = v 1 2 (v, v)x. Check that z is isotroic. Since (v, x) = 1, v / U 1 and hence z / U 1. Similarly, let v be such that (v, u) = l (u) where l U 2 such that l (s(x)) = 1. Then we find an isotroic element z / U 2 as above. Now define S : U 1 kz U 2 kz by s and sending z z. We can check that this is indeed an isometry.

31 LECTURE NOTES Lecture 9 We now resent some corollaries of Witt theorem. Recall: Theorem 9.1 (Witt). Let (V, Q) be a non-degenerate quadratic module. Let U 1, U 2 be subsaces of V and let s : U 1 U 2 be an isometry. Then s extends to an autometry V V. Theorem 9.2. Let (V i, Q i ) be nondegenerate quadratic modules which are isometric to each other. Let U i be a subsace of V i and let s : U 1 U 2 be an isometry. Then s extends to an isometry V 1 V 2. Proof. If Φ : V 2 V 1 such that Q 1 Φ = Q 2, then Φ s : U 1 φ(u 2 ) is an isometry such that Q 1 (Φ s) = Q 1. Hence it extends to S : V 1 V 1. Set S = Φ 1 S. Definition 9.3. A quadratic form f(x, y) is called hyerbolic if the quadratic module (k 2, f) is hyerbolic. Note that a hyerbolic quadratic form is equivalent to xy, or x 2 y 2 or 2xy. ( ) ( ) ( ) /2 It is so, since the matrices S 1 = S = S = 1/2 0 are all equivalent. ( We have) S 2 = A t S 3 A where A = diga(1, 1/2) and S 1 = A t S 2 A for A = 1/2 1/ If g and h are quadratic forms in r and s variables, f = g + h is defined as the quadratic form in r + s variables such that f(x 1,, x r+s ) = g(x 1,, x r ) + h(x r+1,, x r+s ). Theorem 9.4 (Cancellation theorem). Let f i be non-degenerate quadratic forms in n variables and f i = g i + h i. If f 1 f 2 and g 1 g 2 then h 1 h 2. Proof. Set V i = (k n, f i ) = U i W i and U i = (k r i, g i ) where r i is the number of variables and W i = (k n r i, h i ). Then aly the above theorem. Theorem 9.5. Let f be a non-degenerate quadratic form over k in n variables. If f reresents 0, then f g + h where g is hyerbolic and f reresents all elements in k. Theorem 9.6. Let f be a non-degenerate quadratic form in n variables. Then f g g m + h where g i s are hyerbolic and h does not reresent 0. Moreover this decomosition is unique u to equivalence. That is, m i=1 g i + h m i=1 g i + h imlies m = m and h h. Proof. The module (k n, f) has a decomosition as the orthogonal sum of hyerbolic lanes and an anisotroic subsace. Hence by a change of basis, f can be written as above. Now the uniqueness follows from the above.