數論中的局部-全域原則

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(1)國立臺灣師範大學數學系碩士班碩士論文. 指導教授：夏. 良. 忠. 博士. The Local-Global Principle in Number Theory. 研究生：藍. 茂. 愷. 中華民國一 ○ 三年六月.

(2) 致謝在這份論文完成之後，代表著我的碩士階段即將告一段落。兩年的碩士生涯中，對於那些指導過我的老師及支持我的朋友們，我十分地感謝他們無私的幫助。首先，特別感謝我的指導教授夏良忠老師。在討論數學之際，總是以相當謹慎且有耐心的態度來為我解惑，同時提醒我面對問題時該如何突破窘境來化解危機。感謝老師以富有吸引力的自身觀點攜我進入代數數論這一門令人驚嘆的領域。謝謝老師您的指導！感謝姚為成老師以及李華介老師來擔任我的口試委員，在口試時問了許多先前我沒有考慮過的問題，以及對於論文中敘述不周的地方給予建議，讓我受益良多。感謝林延輯老師和張毓麟老師在複分析中的教導，為我在學習代數數論的過程中鋪路；感謝劉容真老師在交換代數這方面給予的指導，幫助我快速掌握日後處理問題的工具；感謝李華介老師時常熱情地與我討論代數數論中重要結果的證明細節，著實增加我的體會；最後感謝劉家新老師曾經帶我做過群分類的問題以及過去每周一次的討論，正是這一段時期，奠定了我代數方面的基礎。感謝 M410 研究室的朋友們長久以來的陪伴，讓我在研究室中備感溫馨。感謝大學部讀書會每晚的問題討論，讓我可以轉換心情藉此紓壓。僅將這份論文獻給所有關心我的師長、朋友們，以及我的家人！茂愷 2014.6.

(3) 摘要在這篇論文中，我們探討算術當中最重要的課題之一：局部-全域原則。作為第一個例子，我們將呈現著名的有理域 ℚ 上之哈瑟-閔可夫斯基定理及其幾個應用。第二個例子則是乘冪上的哈瑟原則。第三個例子我們欲討論當 𝑝 不等於 2 時，函數域 𝔽𝑝 (𝑡) 上之哈瑟-閔可夫斯基定理。最後，我們將研究局部-全域原則的幾個反例。關鍵字：局部-全域原則，哈瑟原則，二次互反律，希爾伯特符號，哈瑟-閔可夫斯基定理，平方和。.

(4) Abstract. In this thesis, we investigate one of the most important topics in arithmetic: the local-global principle. As a first example, we will present the well-known Hasse-Minkowski Theorem for ℚ and give some of its applications. The second example will be the Hasse principle for powers. The third one, we would like to discuss the Hasse-Minkowski Theorem for 𝔽𝑝 (𝑡) with 𝑝 ≠ 2. Finally, we will study some counterexamples to the local-global principle. Key Words: Local-Global Principle, Hasse Principle, Quadratic Reciprocity Law, Hilbert Symbol, Hasse-Minkowski Theorem, Sum of Squares..

(5) The Local-Global Principle in Number Theory. By Mao-Kai Lan. Advisor Liang-Chung Hsia. Department of Mathematics, National Taiwan Normal University Taipei, Taiwan June, 2014.

(6) Contents 1. 2. 3. 4. 5. 6.. Introduction……….................................................................................................1 Preliminaries……………………..……………………………....…………………….3 Basic Results on Quadratic Spaces and Quadratic Forms…………….6 Hilbert Symbol……………………………………………………………………….10 Hasse-Minkowski Theorem for ℚ and Its Applications..................16 Hasse Principle for Powers…………………………………………………......22. 7. Hasse-Minkowski Theorem for 𝔽𝑝 (𝑡) with 𝑝 ≠ 2..............……......23 8. Some Counterexamples to the Hasse Principle…………………………29 9. References……………………………………………………………………………..31.

(7) The Local-Global Principle in Number Theory Mao-Kai Lan June 27, 2014. 1. Introduction. Suppose that we are now given a polynomial f 2 Q[X1 ; :::; Xn ] and if f has a solution in Q, that is, there exists an n-tuple (r1 ; :::; rn ) 2 Qn such that f (r1 ; :::; rn ) = 0. Then we can deduce immediately that when viewing f as a polynomial with coe¢ cients in the completion Qv of Q at the place v, it has a solution in Qv ; it is easy to see this, since for any place v of Q, Q can always be embedded into Qv . We can conclude in the above observation that if a polynomial with coe¢ cients in Q has a solution in Q, then it also has a solution in every completion of Q. This suggests the following question: is the converse still true? More explicitly, given a polynomial f over Q (i.e. with coe¢ cients in Q) in n variables, if f has a solution in every completion of Q, dose f necessarily posses a solution in Q? Unfortunately, the converse may fail in some cases. This problem has reached the face of the so-called local-global principle, also known as the Hasse principle, which states that if a statement concerning an algebraic number …eld K holds true for every completion of K, then this statement holds true for K as well. Roughly speaking, this principle says that in some sense we can collect all the "local" information and piece together to say something about the "global" situation. From the beginning, we will survey a classical example: the Hasse-Minkowski Theorem for Q, which states that a quadratic form (i.e. homogeneous polynomial of degree 2) over Q has a nontrivial solution in Q if and only if it has a nontrivial solution in every completion of Q. In the above terminology, the Hasse-Minkowski Theorem for Q is equivalent to saying that a nontrivial-solubility of a quadratic form over Q satis…es the local-global principle. As an application, we list several results on sum of squares of integers. The second example presented here is the Hasse principle for powers. It says that in a number …eld K, the statement " an element in K is a pe th power " satis…es the localglobal principle (here p being an odd prime). Moreover, this statement still holds true for K provided that it is true for almost all completions of K (i.e. with a …nite number of exceptions). The proof of this example make use of the Chebotarev Density Theorem, a deep result from Class Field Theory. The third example is the Hasse-Minkowski Theorem for the function …eld Fp (t) (p being an odd prime). This part is very parallel to the …rst example, with only some proofs should be modi…ed. As the end, the failure of the local-global principle is indicated. Three conunterexamples are demonstrated here; the …rst two simply require the knowledge of the basic number theory and consequences of the Hilbert symbol while the last one needs the support of Weil’s Theorem. Served as a background, we will recall the materials needed in the basic theory of quadratic spaces and quadratic forms, including the useful facts (Chevalley-Warning Theorem) about quadratic forms over …nite …elds. Then we introduce the concept of the Hilbert symbol, which dominates the proof of the Hasse-Minkowski Theorem (both for Q and for 1.

(8) Fp (t)). With the help of the Hilbert symbol, we are able to study the quadratic forms over the completion of Q (and of Fp (t)) and come to understand two astonishing behaviors of the Hilbert symbol: the Product Formula and the Existence Theorem. After that, the way to the main theorem is reached. The main references we follow in this thesis are [2] and [8]. For the materials of algebraic number theory, we follow [4] and [6].. 2.

(9) 2. Preliminaries. 2.1. Reciprocity Laws and the Dirichlet’s Theorem. Let p be an odd prime. It is known that the multiplicative group G = (Z=(p)) of Z=(p) is cyclic, and there is a natural epimorphism from G onto f 1g with kernel the squares G2 in G. This map is denoted by ( p ) and for a 2 Z with gcd(a; p) = 1 the image of a 2 G under ( p ) is called the Legendre symbol of a (with respect to p). Note that we only de…ne the values on the integers relatively prime to p and what the Legendre symbol describes is the sovablility of the equation X 2 a (mod p). More explicitly, this equation is solvable if and only if a lies in G2 if and only if ( ap ) = 1. Now we list the well-known result about the computation of the Legendre symbol. Theorem (Gauss). Let p be an odd prime. Then 1. (. 1 ) p. = ( 1). 2. ( p2 ) = ( 1). p 1 2. p2 1 8. 3. ( pq )( pq ) = ( 1). ,. , and. p 1 q 1 2 2. for any odd prime q 6= p.. Proof. See Page 48 in [4]. The third equation stated in the theorem is usually called the Quadratic Reciprocity Law. We give an interesting consequence of this theorem: there are in…nitely many primes of the form 4n + 1. We prove this by contradiction. Suppose that there is only a …nite number of primes of the form 4n + 1, say p1 , p2 , ..., pk . Thus the integer N = (2p1 p2 pk )2 + 1 is never a prime and has no prime divisors of the form 4n + 1 (otherwise pj divides 1 for some j). Therefore N is divisible by a prime p of the form 4n + 3, that is, (2p1 p2 pk )2 1 2 (mod p) and X 1 (mod p) is solvable, which contradicts to the fact that p 3 (mod 4). Remarkably, there is a more general phenomenon. Theorem (Dirichlet’s Theorem). Let a and m be relatively prime integers. Then there are in…nitely many primes p with p a (mod m). Proof. See Page 138 in [4]. One advantage of this nontrivial theorem is that we can always select a prime of particular type that avoids a given …nite set of prime numbers.. 2.2. Facts on Algebraic Number Theory. Now we recall some facts in algebraic number theory. Let k be any …eld. A map j j de…ned on k into the real numbers is called a valuation if for each a; b 2 k, it satis…es the following conditions: 1. jaj > 0 and jaj = 0 if and only if a = 0. 2. jabj = jajjbj. 3. ja + bj 6 jaj + jbj. 3.

(10) We call a valuation j j nonarchimedean if it satis…es the stronger assumption: ja + bj 6 maxfjaj; jbjg for each a; b 2 k; and archimedean otherwise. Notice that, in particular, a valuation is a metric. Two valuations are equivalent if they induce the same (metric) topology and inequivalent otherwise. This de…nes an equivalence relation on the set of all valuations on k, and an equivalence class is called a place of k. If v is a place of k, we select any valuation in v and denotes the completion of k with respect to this valuation by kv , which is independent of the choice of the valuation since all valuations in the same place induce the same topology. We also call kv the completion of k at the place v. If j j is a nonarchimedean valuation on k. Then A = f 2 k j j j 6 1g is a local ring with M = f 2 k j j j < 1g as its maximal ideal, U = f 2 k j j j = 1g as its unit group and k as its …eld of fractions. We call A the valuation ring of k. We now come to a concrete example. Let A be a Dedekind domain and k its …eld of fractions. Fix a nonzero prime ideal P of A, we have seen that the localization of A at P is a discrete valuation ring (DVR), i.e., a local PID that is not a …eld, with maximal ideal PAP . If 2 A is a nonzero element, we let vP (a) be the power of PAP occurred in the factorization of the principal ideal (a) of AP ; this map can be extended to k by setting vP ( ) = vP ( 1 ) if 2 k but 2 = AP . Hence j jP = cvP ( ) (where c < 1 is any positive number) de…nes a nonarchimedean valuation on k. Let kP be the completion of k with respect to j jP , then j jP can be extended to a valuation on kP which is still nonarchimedean, cP of kP is still a DVR we also denote this extended map by j jP . The valuation ring A n cP =PA [P and the unit group UP . Moreover, A [P is isomorphic to with maximail ideal PA [P and PAP can be generated by AP =(PAP )n as rings for any positive integer n; also, PA the same element, called the uniformizer. If A = Z and k = Q. Let P = (p) where p is a prime number, we denote vP ( ) by cP and Up for UP , so vp ( ) and j jP = (1=p)vp ( ) by j jp . We write Qp for kP , Zp for A Zp =pn Zp ' Z(p) =pn Z(p) ' Z=(pn ). If p and q are distinct prime numbers, then j jp and j jq are inequivalent. Ostrowski’s theorem states that every valuation on Q is equivalent to j jp for some prime p or j j1 (the usual absolute value). Hence the set of places of Q is just the set of all prime numbers together with the symbol 1 (which denotes the place containing j j1 ). Let p be a prime. If A = Fp [t] and k = Fp (t). Let P = (P ) where P is a monic irreducible polynomial, we denote vP ( ) by vP ( ) and j jP = (1=pdeg(P ) )vP ( ) by j jP . There is an additional (nonarchimedean) valuation: j j1 = j jP = (1=p)vP ( ) where P = (t 1 ) Fp [t 1 ], we denote the uniformizer t 1 by . Similar consequence as for Q, every valuation on Fp (t) is equivalent to j jP for some monic irreducible polynomial P or j j1 . Hence the set of places of Fp (t) is the set of monic irreducible polynomail P with the symbol 1.. 2.3. Approximation Theorem and Hensel’s Lemma. We now present a useful tool that helps us to esimate the distance with respect to a …nite number of pairwise inequivalent valuations simultaneously. Theorem (Approximation Theorem). Let j j1 , ..., j jn be nontrivial pairwise inequivalent valuations on k and 1 ,..., n 2 k. Then for any " > 0, there is an element 2 k such that j j jj < " for each j.. Proof. See Page 109 in [4].. Let k be a complete …eld with respect to a nonarchimedean valuation with its valuation ring A a DVR. Let M be the unique maximal ideal of A. The following lemma gives 4.

(11) su¢ cient conditions under which we may determine a factorization of a polynomial f (X) 2 A[X] from a factorization of f (X) 2 (A=M )[X]. Lemma 2.1 (Hensel’s lemma). If f (X) is monic and f (X) = G(X)H(X) in (A=M )[X] where G(X) and H(X) are monic and relatively prime in (A=M )[X]. Then f (X) = g(X)h(X) in A[X] for some monic g(X); h(X) 2 A[X] with g(X) = G(X), h(X) = H(X) in (A=M )[X] and deg(g) = deg(G), deg(h) = deg(H). Proof. See Page 38 in [4]. We can use Hensel’s lemma to characterize the nonzero squares in Qp where p is a prime. First we let p be an odd prime. From the canonical epimorphism Zp Zp =pZp , the units in Zp are mapped into the unit group in Zp =pZp and (Zp =pZp ) ' (Z=(p)) ' Fp . Hence this induces an injection i : Up =Up2 ,! Fp =Fp2 . If 2 Up with i( ) is a square in Fp , then this implies that f (X) = X 2 2 Zp [X] has a root in Zp =pZp . Since f 0 (X) = 2X and p is odd, we …nd that f has no multiple roots in Zp =pZp . Now apply Hensel’s lemma, f (X) has a root in Zp , in other words, is a square in Up , proving surjectivity of i. Since Up =Up2 ' Fp =Fp2 and Fp is a cyclic group, we have Up =Up2 ' Z=(2). Since every element 2 Qp is uniquely expressed as pn u where n = vp ( ) 2 Z and u 2 Up , we see that Qp ' Z Up so that Qp =Qp2 ' Z=(2) Up =Up2 ' Z=(2) Z=(2). This also tells us that is a square in Qp if and only if vp ( ) is even and u 1 mod pZp (or u 1 (mod p)). Next we let p = 2. We claim that U2 =U22 ' (Z=(8)) . If is a square in U2 , then = (1 + 2 )2 where 2 Z2 and 2 + 2 2Z2 so = 1 + 4( 2 + ) 2 1 + 8Z2 . From U2 ,! Z2 Z2 =8Z2 we have U2 =U22 ,! (Z2 =8Z2 ) . If 2 U2 with 2 1 + 8Z2 . Write = 1 + 8 where 2 Z2 . In order to be a square in U2 , we have to …nd 2 Z2 such that (1 + 2 )2 = 1 + 8 or 2 + = 2 . Consider this time f (X) = X 2 + X 2 2 Z2 [X], 0 f (X) always has a root modulo 2Z2 since 0 + 0 1 + 1 0 (mod 2), and f (X) = 1 modulo 2Z2 , can indeed be selected by Hensel’s lemma. Thus U2 =U22 ' (Z2 =8Z2 ) ' (Z=(8)) ' Z=(2) Z=(2). The same argument as above, Q2 =Q22 ' Z=(2) Z=(2) Z=(2). And = 2n u 2 Q2 is a square in Q2 if and only if v2 ( ) is even and u 1 mod 8Z2 (or u 1 (mod 8)). We summarize this. Proposition 2.1. Qp =Qp2 ' Z=(2) Remark 2.1. Let holds:. ;. 2 Q. j. Z=(2) for p 6= 2 and Q2 =Q22 ' Z=(2). Z=(2). Z=(2).. If j j is any valuation on Q and the following condition j < " for su¢ ciently small ".. Then j = 1j < "=j j = "0 for "0 small enough. If j j = j j1 , then this condition implies that = is positive, in other words, = is a square in R . If j j = j jp for some odd prime p, then this condition implies that = 1 2 pZp or = 2 1 + pZp . Therefore = is a square in Qp . If j j = j j2 , then this condition implies that = 1 2 8Z2 or = 2 1 + 8Z2 , which shows that = is a square in Q2 as well. We conclude that: if v is a place of Q, then j jv < " for su¢ ciently small " implies that = is a square in Qv .. 5.

(12) 3. Basic Results on Quadratic Spaces and Quadratic Forms. 3.1. Basic Properties. Let k be a …eld with char(k) 6= 2 and V be a vector space over k of dimension n. We say that a map q : V ! k is a quadratic form on V if q satis…es the following two conditions: 1. q(ax) = a2 q(x) for all a 2 k and x 2 V , and 2. The map b : V form on V .. V ! k de…ned by b(x; y) = (q(x + y). q(x). q(y))=2 is a bilinear. Two things should be noted: we assume from the beginning that k is of characteristic distinct from 2, so the term 1=2 occurred in the condition (2) is meaningful; and the map b de…ned above is clearly symmetric, that is, b(x; y) = b(y; x) for all x; y 2 V . Let q be a quadratic form on V and b the bilinear form on V associated to q. If = (e1 ; :::; en ) is an ordered basis for V , we call Q = (b(ei ; ej )) 2 Mn (k) the matrix of q relative to ; it is symmetric since b is symmetric. For any vector x 2 V , [x] denotes the coordinate of x with respect to , i.e. [x] = (c1 ; :::; cn )t 2 k n where the ci are given P t by x = 16i6n ci ei , then we have the relation: for each x; y 2 V , b(x; y) = [x] Q[y] . t Conversely, if A 2 Mn (k) is such that b(x; y) = [x] A[y] for any x; y 2 V , then A = Q. If is another ordered basis for V and Q0 is the matrix of q relative to , then we have [x] = P [x] for any x 2 V where P = [idV ] (it is invertible of course), thus b(x; y) = [x]t Q0 [y] = [x]t P t Q0 P [y] for each x; y 2 V and Q = P t Q0 P . This shows that det(Q) = det(Q0 ) det(P )2 , in other words, the discriminants of these two matrices di¤er by a nonzero square in k. In the above discussion, we know that if the discriminant of the matrix of q relative to some basis is zero, then the same is true for any other basis. Otherwise, up to nonzero squares, the element det(Q) is unique. Hence, in any case, we call det(Q) the discriminant of q, denoted by d(q). Note that the d(q) is independent of the choice of the bases; it is undoubtedly an element in k, sometimes we also view it as the class containing det(Q) in k =k 2 if det(Q) 6= 0. The rank of q is de…ned to be the rank of Q, which is well-de…ned (note that P is of full rank): Once we have a quadratic form q on V , we call this pair (V; q) the quadratic space; if x; y 2 V , we denote b(x; y) by x y. A linear map f from (V; q) into another quadratic space (V 0 ; q 0 ) is called a morphisim if f (x) f (y) = x y for all x; y 2 V . Two quadratic spaces are said to be isomorphic if there is a morphism that is bijective between them. Two elements x and y in V are orthogonal if x y = 0; for any subset S of V , S ? is the set of all elements of V that are orthogonal to every element of S. We say that V is the orthogonal direct sum of two subspaces U and W if V = U W and U W ? , denoted by V = U W . Let q be a quadratic form on V , then q induces a linear map from V into its dual space V given by x 7 ! x b where x b(y) = x y. If is injective, we say that q is nondegenerate (or (V; q) is nondegenerate). In other words, q is nondegenerate if and only if the only element x 2 V with x y = 0 for all y 2 V is x = 0, or equivalently, if and only if V ? = f0g. As we have seen before, the condition x y = 0 for all y 2 V can be translated into the following statement: If is some ordered basis for V and Q the matrix of q relative to , then [x]t Q[y] = 0 for all y 2 V , or equivalently, Q[x] = 0. Thus q is nondegenerate is equivalent to saying Q[x] = 0 only if x = 0, that is, Q is invertible (and d(q) 6= 0). We summarize this as: 6.

(13) Proposition 3.1. Let (V; q) be a quadratic space. Then q is nondegenerate , V ? = f0g , d(q) 6= 0. Proposition 3.2. Let (V; q) be a nondegenerate quadratic space. 1. If U is a subspace of V , then (U; q) is nondegenerate , (U ? ; q) is nondegenerate , V = U 2. If V = U. U ?:. W , then (U; q) and (W; q) are nondegenerate.. Proof. See Page 287 in [2]. De…nition 3.1. An element x 2 V is called isotropic if q(x) = 0. Corollary 3.1. If (V; q) is nondegenerate and contains a nonzero isotropic element, then q(V ) = k: Proof. See Page 288 in [2].. 3.2. Orthogonal Bases. We say that a basis = (e1 ; :::; en ) of a quadratic space (V; q) is orthogonal if the ei are pairwise orthogonal. We now give the following fundamental result. Theorem. Every quadratic space has an orthogonal basis. Proof. See Page 288 in [2]. Now the information is easily seen: when we have an orthogonal basis for a quadratic space (V; q), the matrix of q relative to this basis is a diagonal matrix, hence d(q) is the product of the diagonal entries and the rank of q is the number of nonzero diagonal entries.. 3.3. Translation. If V = k n and e = (ei ) is the standard basis of V , we can view a quadratic form q on V as a homogeneous polynomial of degree 2 over k in n variables by the formula X q(X1 ; :::; Xn ) = Qij Xi Xj ; 16i<j6n. where the Qij are the entries of the symmetric matrix Q. We will simply speak of a quadratic form with coe¢ cients in k (or a quadratic form over k). Conversely, if q is given as a homogeneous polynomial of degree 2 as above, we can consider q as a quadratic form on k n in a natural way. De…nition 3.2. Two quadratic forms q and q 0 are equivalent if the quadratic spaces (k n ; q) and (k n ; q 0 ) are isomorphic, and will be denoted by q q 0 .. 7.

(14) If q(X1 ; :::Xn ) and q 0 (X1 ; :::; Xm ) are two quadratic forms in n and in m variables respectively, the form q q 0 is the quadratic form in n + m variables de…ned by (q q 0 )(X1 ; :::; Xn+m ) = q(X1 ; :::; Xn ) + q 0 (Xn+1 ; :::; Xm ). We also set q q 0 = q ( q 0 ). We say that q represents c 2 k if q(x) = c for some x 2 k n ; moreover, we say that q represents 0 in k if q(x) = 0 for some nonzero x 2 k n . In the above de…nition, we see that two equivalent quadratic forms represents the same elements, that is, if c 2 k is represented by a quadratic form q, then it is represented by any quadratic form equivalent to q as well. Proposition 3.3. If q is nondegenerate and represents 0 in k, then q represents all elements of k. Proof. This is just Corollary 3.1. Corollary 3.2. Let q be a nondegenerate quadratic form in n variables and c 2 k . The following conditions are equivalent: 1. q represents c in k. 2. There exists a quadratic form q 0 in n 3. q. 1 variables such that q. q0. cZ 2 .. cZ 2 represents 0 in k.. Proof. See Page 292 in [2]. Corollary 3.3. Let q 0 and q 00 be two nondegenerate quadratic forms and let q = q 0 The followings are equivalent:. q 00 .. 1. q represents 0 in k. 2. There exists c 2 k represented by both q 0 and q 00 . 3. There exists c 2 k such that both q 0. cZ 2 and q 00. cZ 2 represent 0.. Proof. See Page 292 in [2]. Corollary 3.4. Every quadratic form is equivalent to some diagonal quadratic form. Proof. This is another language of the previous theorem (Page 6).. 3.4. Quadratic Forms over Finite Fields. If k is a …eld and f is a polynomial over k in n variables. It is di¢ cult to check whether or not f has a solution in k. Fortunately, there is a way to …nding some information. Suppose that the coe¢ cients of f lies in some subring A of k and suppose we could choose a suitable maximal ideal M of A for which the residue …eld A=M is …nite. If f (the polynomial f with coe¢ cients modulo M ) has a solution in the …nite …eld A=M , then this solution may have the chance to be lifted to a solution in A by Hensel’s lemma (undoubtedly the assumptions of Hensel’s lemma should be satis…ed). Therefore we have to seek a method to judge the solvability of polynomials over …nite …elds (in …nite …elds). Let p be an odd prime and q a power P of p. Fq denotes a …nite …eld with q elements. If f 2 Fq [X1 ; :::; Xn ], we denote S(f ) by x2Fnq f (x). 8.

(15) Lemma 3.1. Let be a positive integer. Then the sum S(X ) = if is divisible by q 1; it is equal to 0 otherwise.. P. x2Fq. x is equal to. 1. Proof. If is divisible by q 1, then 0 = 0 and x = 1 for x 2 Fq ; hence S(X ) = q 1 = 1. If is not divisible by q 1, then there P exists y 2Pk such that y 6= 1 since Fq is cyclic of order q 1, and it follows that S(X ) = x2Fq x = x2Fq y x = y S(X ) or (1 y )S(X ) = 0, which gives us S(X ) = 0. Theorem (Chevalley-Warning). Suppose that f1 ; :::fr 2 Fq [X1 ; :::; Xn ] are polynomials such P that 16j6r deg fj < n, and let V be the set of their common solutions in Fnq . Then we have Card(V ) 0 (mod p). Yr Proof. Let P = (1 fjq 1 ) and let x 2 Fnq . If x 2 V , then P (x) = 1 since fj (x) = 0 for j=1. all j; if x 2 = V , then P (x) = 0 since fj (x) 6= 0 for some j so that fj (x)q 1 =P1. It follows that Card(V ) S(P ) (mod p). We claim that S(P ) = 0: The hypothesis 16j6r deg fj < n e1 en e implies Pthat deg P < n(q 1), thus P is a linear combination of monomials X = X1 :::Xn with 16i6n ei < n(q 1), and ei < q 1 for some i. It su¢ ces to show that for such a monomial X e , we have S(X e ) = 0, but this follows directly from Lemma 3.1. This completes the proof. P Corollary 3.5. If 16j6r deg fj < n and if the fj have no constant terms, then the fj have a nontrivial common solution. Proof. If the fj have no nontrival common solutions, then V = f0g and Card(V ) would not be divisible by p, a contradiction. Corollary 3.6. All quadratic forms in at least 3 variables over Fq have a nontrivial solution. Proof. Apply Corollary 3.5 (note that quadratic forms are of degree 2). Hence we deduce the following: Proposition 3.4. A quadratic form over Fq of rank > 2 (resp: of rank > 3) represents all elements of Fq (resp: of Fq ). Proof. By Corollary 3.6, a quadratic form over Fq represents 0 in Fq . By Proposition 3.3 we get the desired result.. 9.

(16) 4. Hilbert Symbol. Let k be a …eld. We start to study quadratic forms over k. Let q be a nondegenerate quadratic form over k. We are interested in what characterizes the representability of 0 by q in k (note that degenerate forms always represent 0, so we need only focus on the nondegerate case). For simplicity, we assume at …rst that q is in diagonal form (remember that equivalent forms represent the same elements). When q is in 1 variable, then q is of the form aX 2 where a 2 k . Thus q never represents 0 in k. When q is in 2 variables, q is of the form aX 2 + bY 2 where a; b 2 k . If q represents 0 in k, say ax2 + by 2 = 0 for some nontrivial solution (x; y) in k. Clearly x and y are both nonzero, and a=b = y 2 =x2 or ab = y 2 b2 =x2 2 k 2 , in other words, d(q) is a square in k . Conversely, if d(q) is a square in k , say ab = c2 for some c 2 k . Then aX 2 + bY 2 has (b; c) as a solution so that q represents 0 in k. We conclude that for q is of rank 2, q represents 0 in k if and only if d(q) is a square in k , i.e., d(q) characterizes the representability of 0 by q in k. When q is in 3 variables, we write q as aX 2 + bY 2 + cZ 2 where a; b; c are nonzero. We face an obstacle this time. It seems that there is no simple manner to see what’s important in this situation. But we realize that aX 2 + bY 2 + cZ 2 represents 0 in k if and only if acX 2 bcY 2 c2 Z 2 represents 0 in k if and only if acX 2 bcY 2 Z 2 represents 0 in k. This motivates the study of the Hilbert symbol.. 4.1. Basic Properites. In this subsection, k denotes any …eld. De…nition 4.1. Let a,b 2 k . We set (a; b) = 1 if aX 2 +bY 2 = Z 2 has a nontrivial solution in k; and (a; b) = 1 otherwise. The number (a; b) is called the Hilbert symbol of a and b relative to k (when there is no ambiguity, we may just speak of the Hilbert symbol of a and b). From the de…nition we see that (a; b) does not change when a or b is multiplied by a nonzero square in k, thus we can view the Hilbert symbol as a map from k =k 2 k =k 2 into f 1g. Next we give the following proposition, which turns the problem of determination of the value of the Hilbert symbol into that of …eld norms. p Proposition 4.1. Let a,b 2 k . We have (a; b) = 1 if p and only if a 2 N (k( b) ), in other words, if and only if a is the norm of an element of k( b) . p 2 2 2 b) as a solution so that (a; b) = 1, and Proof. If b is a square, then aX +bY Z has (0; 1; p k( b) = k, thus this proposition follows in this case. If b is not a square and if (a; b) = 1, then ax2 + by 2 = z 2 for some nontrivial solution (x; y; z) in p k. Then x 6= 0 (for otherwise b wouldpbe a square) and a is the norm of (z=x) + (y=x) b. Conversely, if a is a norm from k( b) , then a is of the form u2 bv 2 where u; v 2 k, which implies that (a; b) = 1 by de…nition. We also list the needed properties of the Hilbert symbol; all the letters involved below are assumed to be nonzero elements of k. 1. (a; b) = (b; a) and (a; c2 ) = 1. The …rst is clear from the de…nition and aX 2 + c2 Y 2 = Z 2 has (0; 1; c) as a solution.. 10.

(17) 2. (a; a) = 1 and (a; 1. a) = 1.. aX 2 aY 2 = Z 2 (resp. aX 2 + (1 solution.. a)Y 2 = Z 2 ) has (1; 1; 0) (resp. (1; 1; 1)) as a. 3. (a; b) = 1 implies (aa0 ; b) = (a0 ; b).. p p 0 (a; b) = 1 implies that a is the norm from k( b) , hence aa 2 N (k( b) ) if and only p if a0 2 N (k( b) ).. 4. (a; b) = (a; ab) = (a; (1. a)b).. (a; b) = (b; a) = ( ab; a) = (a; ab) by (1); (2), and (3). (a; (1 a)b).. Similar for (a; b) =. For simplicity, we call above four results the basic properties of the Hilbert symbol.. 4.2. Computation of (a; b) relative to Qv. Now we restrict k to be the …eld Q of rational numbers. Recall that if v is a place of Q, Qv denotes the completion of Q at v. Hence Qv denotes either the …eld R of real numbers or the …eld Qp of p-adic numbers. We use ( ; )v to denote the Hilbert symbol relative to Qv (or simply ( ; ) when v is speci…ed). We now state the Explicit Formula of the Hilbert symbol (relative to Qv ) in terms of the Legendre symbol, which helps us to derive the bilinearity and the nondegeneracy of the Hilbert symbol. We begin with the following lemma. For 1 mod 8Z2 (denoted by u 2 U2 , the symbol ( u2 ) is de…ned as ( u2 ) = 1 if and only if u u 1 (mod 8)). Lemma 4.1. Let u 2 Up . If the equation pX 2 + uY 2 = Z 2 has a nontrival solution in Qp , it has a solution (x; y; z) such that x 2 Zp and y,z 2 Up . Proof. Let (x; y; z) be a nontrivial solution in Qp of this equation. After dividing by pe with e = minfvp (x); vp (y); vp (z)g, we may assume that x; y; and z are in Zp with at least one of them in Up . If y 2 = Up , then y 0 (mod p) and z 0 (mod p). Thus px2 0 (mod 2 2 p ) and x 0 (mod p), which implies that x 0 (mod p), contradicting the fact that one of x; y; and z are in Up . Hence y 2 Up and z 2 Up as well. Theorem (Explicit Formula). Let a,b 2 Qv . For v = 1. We have (a; b)1 = 1 if a < 0 and b < 0 and (a; b)1 = 1 otherwise. For v = p. If we write a = p ua , b = p ub with ua and ub in Up , we have (a; b)p = ( 1) (a; b)2 = ( 1). (ua. p 1 2. 1)(ub 4. (. ua u b ) ( ) p p. 1). (. ua ub ) ( ) 2 2. if p 6= 2 if p = 2.. Proof. For v = 1, it is clear that aX 2 + bY 2 = Z 2 has no nontrivial solutions in R unless a or b is positive. Now let v = p. Assume that p 6= 2. Notice that the Hilbert symbol (a; b)p depends only on a and b modulo squares, so we may reduce this problem to the three cases: Case 1: = = 0. (, (ua ; ub )p = 1) 11.

(18) Since quadratic forms over …nite …elds with rank n > 3 represents all elements by Proposition 3.4, we know that ua X 2 + ub Y 2 = Z 2 has a nontrivial solution modulo p, so z 2 (mod p). there exist x; y; z 2 Zp with x; y; z not both zero such that ua x2 + ub y 2 Without loss of generality, say x nonzero; consider f (X) = X 2 ub y 2 =ua z 2 =ua 2 Zp [X]. 0 0 Since f (x) = 0 and f (x) = 2x 6= 0, by Hensel’s lemma f (x0 ) = 0 for some nonzero x0 2 Zp so that (a; b)p = 1. Case 2: = 1, = 0. (, (pua ; ub )p = ( upb )) From Case 1 we see that (ua ; ub )p = 1 so that (a; b)p = (pua ; ub )p = (p; ub )p by the basic properties of the Hilbert symbol. If (p; ub )p = 1, then by de…nition pX 2 + ub Y 2 = Z 2 has a nontrivial solution in Qp and thus has a solution (x; y; z) in Zp with y; z 2 Up by Lemma 4.1. Hence ub (z=y)2 (mod p) and ( upb ) = 1. If ( upb ) = 1, then (p; ub )p = 1, for otherwise pX 2 + ub Y 2 = Z 2 would again have a solution (x; y; z) in Zp with y; z 2 Up and ( upb ) = 1, a contradiction. Hence (pua ; ub )p = ( upb ) in this case. p 1. Case 3: = = 1. (, (pua ; pub )p = ( 1) 2 ( upa )( upb )) Using the basic properties of the Hilbert symbol, we have (a; b)p = (pua ; pub )p = (pua ; p2 ua ub )p = (pua ; ua ub )p p 1 ua ub u a ub ) = ( 1) 2 ( )( ), = ( p p p where the …rst equality in the second row is deduced from Case 2. Assume that p = 2. We still consider the same three cases. (ua 1)(ub 1) 4 Case 1: = = 0. (, (ua ; ub )2 = ( 1) ) If ua 1 (mod 4), then ua 1 (mod 8) or ua 5 (mod 8). If ua 1 (mod 8), then ua is a square in Q2 so that (ua ; ub )2 = 1. If ua 5 (mod 8), then since ua + 4ub 1 (mod 8), ua + 4ub becomes a square in Q2 , and hence ua X 2 + ub Y 2 = Z 2 has (1; 2; w) a solution where w 2 Z2 is some square root of ua + 4ub , i.e. (ua ; ub ) = 1. Similarly, if ub 1 (mod 4), we also have (ua ; ub )2 = 1. This shows that (ua ; ub )2 = 1 if ua 1 (mod 4) or ub 1 (mod 4). If ua 1 (mod 4) and ub 1 (mod 4). Suppose that ua X 2 + ub Y 2 = Z 2 has a 3 nontrivial solution (x; y; z) 2 Q2 . We may assume that x; y; z 2 Z2 with at least one of them in U2 . Then from ua x2 + ub y 2 = z 2 we obtain x2 + y 2 + z 2 0 (mod 4), which implies that x, y, and z are all nonunits, a contradiction. Hence (ua ; ub )2 = 1 in this case. (ua 1)(ub 1) 4 ( u2b )) Case 2: = 1, = 0. (, (2ua ; ub )2 = ( 1) ub We claim that (2; ub )2 = ( 2 ) …rst. If (2; ub )2 = 1, then there exist x 2 Z2 and y; z 2 U2 such that 2x2 +ub y 2 = z 2 . Since y,z 2 U2 , y 2 z 2 1 (mod 8) so ub 1 2x2 (mod 8) 1 ub ub (mod 8). Thus ( 2 ) = 1. Conversely, if ( 2 ) = 1, then ub 1 (mod 8). If ub 1 (mod 8), then ub is a square in Q2 and (2; ub )2 = 1; if ub 1 (mod 8), then ub 1 (mod 8) and ub is a square so that (2; ub )2 = (2; 1)2 = 1 since 2X 2 Y 2 = Z 2 has (1; 1; 1) a solution. If (2; ub )2 = 1, then (2ua ; ub )2 = (2; ub )2 (ua ; ub )2 ; if (ua ; ub )2 = 1, we get the same conclusion. So we consider the situation (2; ub )2 = (ua ; ub )2 = 1. (2; ub )2 = 1 implies that ub 3 (mod 8) and (ua ; ub )2 = 1 implies that ua ub 1 (mod 4). Thus, ub 3 (mod 8) and ua 1; 3 (mod 8). It follows that 2ua + ub 1 (mod 8) so 2ua + ub is a square and (2ua ; ub )2 = 1. This proves that (2ua ; ub )2 = (2; ub )2 (ua ; ub )2 . (ua 1)(ub 1) 4 Case 3: = = 1. (, (2ua ; 2ub )2 = ( 1) ( u2a )( u2b )) (2ua ; 2ub )2 = (2ua ; ua ub )2 = (2; ua ub )2 (ua ; ua ub )2 = (2; ua ub )2 (ua ; ub )2 = ( ( 1). (ua. 1)(ub 4. 1). ( u2a )( u2b ). 12. ua ub )( 2. 1). (ua. 1)(ub 4. 1).

(19) From this theorem it is now easy to deduce the following results. Corollary 4.1. The Hilbert symbol relative to Qv is a nondegenerate bilinear form on the F2 -vector space Qv =Qv2 . Proof. The bilinearity of the Hilbert symbol relative to Qv is deduced from the Explicit Formula. We claim the nondegeneracy. Let v = 1. If [a] 2 R =R 2 is not the identity class, then a < 0 and we …nd that ( 1; a) = 1, proving this result in this case. Let v = p. We …rst consider the case p 6= 2. Thus we may choose f1; n; p; npg as the representatives of Qp =Qp2 where n 2 Z is a quadratic nonresidue modulo p. If [a] 2 Qp =Qp2 is not the identity class, then [a] = [n], [p], or [np], and (n; p) = ( np ) = 1 by the Explicit Formula, hence the result. Next in the case p = 2, we choose f 1; 5; 2; 10g as representatives of Q2 =Q22 and if [a] is not the identity class in Q2 =Q22 , from (2; 5) = ( 52 ) = 1 and the bilinearity of the Hilbert symbol relative to Q2 , we can choose the corresponding b 2 Q2 so that (a; b) = 1. This completes the proof. Next we are concerned with the problem of representability of 0 by quadratic forms in 3 variables over Qp . Corollary 4.2. Let q = aX 2 + bY 2 + cZ 2 be a nondegenerate quadratic form over Qv . Set " = "(q) = (a; b)(a; c)(b; c) and let d = d(q) = abc be the discriminant of q. Then q represents 0 in Qv if and only if ( 1; d) = ". Proof. Note that q represents 0 in Qv if and only if the form cq does if and only if acX 2 bcY 2 = Z 2 has a nontrivial solution in Qv , that is, ( ac; bc) = 1. By bilinearity of the Hilbert symbol relative to Qv this condition implies that 1 = ( ac; bc) = ( 1; 1)( 1; b)( 1; c)(a; 1)(a; b)(a; c)(c; b)(c; c); which is equal to the condition ( 1; abc) = (a; b)(a; c)(b; c) and the result follows. Corollary 4.3. Let c 2 Qv and let q = aX 2 + bY 2 be a nondegenerate quadratic form over Qv . Then q represents c in Qv if and only if (c; ab) = (a; b). Proof. q represents c in Qv if and only if aX 2 + bY 2 cZ 2 represents 0 in Qv . By Corollary 4.2, this holds true if and only if ( 1; abc) = (a; b)(a; c)(b; c) = (a; b)(ab; c) if and only if (a; b) = ( 1; abc)(ab; c) = ( 1; abc)(ab; c)(c; c) = ( 1; abc)(abc; c) = ( c; abc) = (c; c)(c; ab) = (c; ab). Remark 4.1. The key fact in the proofs of last two results is that the Hilbert symbol relative to Qv is bilinear. Later, we will also de…ne the Hilbert symbol relative to the completion of Fp (t) with p 6= 2. Once we have proved the Hilbert symbol (in the latter sense) is bilinear, the analogues of above two corollaries follow immediately. We prove the following lemma, which is essential for the characterization of representability of 0 by a quadratic form over Qp . Lemma 4.2. Let p be a prime number. 1. We have jQp =Qp2 j = 2r with r = 2 if p 6= 2 and r = 3 if p = 2. 13.

(20) 2. Let a 2 Qp =Qp2 and " = 1. We de…ne H" (a) = fx 2 Qp =Qp2 j(x; a) = "g. Then jH1 (1)j = 2r , H 1 (1) = and jH" (a)j = 2r 1 if a 6= 1. 3. Let a,a0 2 Qp =Qp2 , and ","0 equal to 1; assume that H" (a) and H"0 (a0 ) are nonempty. Then H" (a) \ H"0 (a0 ) = if and only if a = a0 and " = "0 . Proof. (1) is explained in Proposition 2.1. For (2), it is clear that jH1 (1)j = 2r and jH 1 (1)j = 0 follows from the de…nition of the Hilbert symbol; if a 6= 1, then H1 (a) is the kernel of the linear map from Qp =Qp2 into f 1g and the nondegeneracy of the Hilbert symbol relative to Qp implies surjectivity of this map, we have jH1 (a)j = 2r 1 (and hence jH 1 (a)j = 2r 1 ) by the isomorphism theorem. Finally, since H" (a) and H"0 (a0 ) are both nonempty, each of whom has at least 2r 1 elements by (2). Thus H" (a) \ H"0 (a0 ) = amounts to saying that H" (a) is the complement of H"0 (a0 ), so " = "0 . Without loss of generality, we assume " = 1. Then H1 (a) = H1 (a0 ) (since H1 (a0 ) is the complement of H"0 (a0 )), it follows that (x; a) = (x; a0 ) for all x 2 Qp =Qp2 . By the nondegeneracy of the Hilbert symbol, a = a0 and the proof is completed.. 4.3. Quadratic Forms over Qp. We are now in a position to study general quadratic forms over Qp . Let q be a quadratic form in n variables over Qp . Since degenerate quadratic forms represent 0 nontrivially, we will always assume that P q is nondegenerate. Up to equivalence, we may assume that q is in diagonal form as q = 16i6n ai Xi2 where ai 2 Qp . We set "(a1 ; :::; an ) =. Y. (ai ; aj ),. 16i<j6n. where (ai ; aj ) is the Hilbert symbol relative P to Qp , so that " = "(a1 ; :::; an ) = 1. But q may be expressed as another diagonal form 16i6n a0i Xi2 , and we obtain "0 = "(a01 ; :::; a0n ). However, it can be shown that this value does not depend on the expression of q (see Page 301 in [2]), in which the proof is mainly based on the bilinearity of the Hilbert symbol relative to Qp . That is, " = "0 . We denote this common value by "(q), which is an invariant of q. Later, we will face this situation again. After presenting the bilinearity of the Hilbert symbol relative to Fp (t)v , the analogues of the invariance of "(q) for a nondegenerate form q over Fp (t)v follows. In the following two results, we let q be a nondegenerate quadratic form of rank n over Qp ; set d = d(q) and " = "(q). Theorem. q represents 0 in Qp if and only if one of the following holds: 1. n = 2 and d =. 1.. 2. n = 3 and ( 1; d) = ". 3. n = 4 and either d 6= 1 or d = 1 and ( 1; 1) = ". 4. n > 5. Proof. Since d and " are invariants. In our discussion, we assume that q is in diagonal form. When n = 1 and n = 2, explanations are placed in Page 9. When n = 3, the result has been done before by Corollary 4.2. 14.

(21) When n = 4, we write q as the form q1 q2 where q1 = a1 X 2 + a2 Y 2 and q2 = (a3 Z 2 + a4 W 2 ). Hence by Corollary 3.3 q represents 0 in Qp if and only if there exists c 2 Qp which is represented simultaneously by q1 and q2 , and this condition is equivalent to that there exists c 2 Qp such that (c; a1 a2 ) = (a1 ; a2 ) and (c; a3 a4 ) = ( a3 ; a4 ) by Corollary 4.3. De…ne A = fc 2 Qp =Qp2 j(c; a1 a2 ) = (a1 ; a2 )g and. B = fc 2 Qp =Qp2 j(c; a3 a4 ) = ( a3 ; a4 )g,. thus q does not represent 0 in Qp if and only if A \ B = . Since a1 2 A and a3 2 B, A and B are nonempty. Thus A \ B = if and only if a1 a2 = a3 a4 and (a1 ; a2 ) = (a3 ; a4 ) by Lemma 4.2. a1 a2 = a3 a4 is equivalent to d = 1, and if d = 1, (a1 ; a2 ) = (a3 ; a4 ) tells us that ( 1; 1) = ". Combine these facts together, q represents 0 in Qp if and only if A \ B 6= if and only if d 6= 1 or d = 1 and ( 1; 1) = ". Finally, we show that a nondegenerate form q in 5 variables represents 0 in Qp . We have seen that a nondegenerate form q 0 = aX 2 + bY 2 represents c 2 Qp in Qp if and only if (c; ab) = (a; b) (Corollary 4.3 ). Consider the set S = fc 2 Qp =Qp2 j(c; ab) = (a; b)g. Since a 2 S, S is nonempty and thus jSj > 2 by Lemma 4.2. That is to say, the number of elements in Qp =Qp2 represented by q 0 is at least 2, and the situation holds true for q as well. Therefore we can choose c 2 Qp =Qp2 with c 6= d(q) such that it is represented by q. It follows that there exists a quadratic form q 00 in 4 variables such that q is equivalent to q 00 cZ 2 (Corollary 3.2 ); since equivalent forms have the same discriminant, we have d(q) = cd(q 00 ) and d(q 00 ) 6= 1 due to the fact c 6= d(q). From what we just proved (the case n = 4), q 00 represents 0 in Qp , and so does q. Corollary 4.4. Let c 2 Qp =Qp2 . following holds:. Then q represents c in Qp if and only if one of the. 1. n = 1 and c = d. 2. n = 2 and (c; d) = ". 3. n = 3 and either c 6=. d or c =. d and ( 1; d) = ".. 4. n > 4. Proof. Consider the quadratic form q 0 = cZ 2 q. We see that d0 = d(q 0 ) = cd and "0 = "(q 0 ) = ( c; d)". Apply the above theorem to q 0 and the bilinearity of the Hilbert symbol relative to Qp , we can easily check that d0 = 1 if and only if c = d; ( 1; d0 ) = "0 if and only if (c; d) = ", and the conditions (3).. 15.

(22) 5. Hasse-Minkowski Theorem for Q and Its Applications. After studying in great detail quadratic forms over Qp , we are now going to prove the Hasse-Minkowski Theorem for Q. As a preparation, we stop for a while to prove two results concerning the Hilbert symbol relative to Qv .. 5.1. Product Formula for the Hilbert Symbol relative to Qv. Let V be the set of all places of Q, i.e. the set of all prime numbers together with the symbol 1. The Product Formula seems to tell us that there is a balance among the solubility of two given nonzero rational numbers in every completion of Q. In the following proof, we use the Explicit Formula of the Hilbert symbol relative to Qp frequently. For the sake of convenience, we put that result here again. (Explicit Formula) Let p be a prime and a; b 2 Qp . If a = p ua , b = p ub with ua and ub in Up , we have p 1 2. (a; b)p = ( 1) (a; b)2 = ( 1) where ( u2 ) = ( 1). u2 1 8. (ua. 1)(ub 4. (. u a ub ) ( ) p p. 1). (. ua ub ) ( ) 2 2. if p 6= 2 if p = 2.. for u 2 U2 (that is, ( u2 ) = 1 , u. 1 (mod 8)).. Theorem (Product Formula). If a; b 2 Q , then (a; b)v = 1 for almost all v 2 V and Y (a; b)v = 1: v2V. Proof. Since the Hilbert symbol relative to Qv is bilinear, after multiplying nonzero squares in Q, we can just consider particular cases (where ` and `0 below are distinct odd primes): Case 1: a = b = 1. ( 1; 1)2 = 1 and clearly ( 1; 1)1 = 1; ( 1; 1)v = 1 for v 6= 2; 1. Case 2: a = 1, b = `. If ` = 2. Then ( 1; 2)2 = ( 21 ) = 1 and ( 1; 2)v = 1 for v 6= 2. ` 1 ` 1 If ` 6= 2. Then ( 1; `)2 = ( 1) 2 and ( 1; `)` = ( `1 ) = ( 1) 2 ; ( 1; `)v = 1 for v 6= 2; `. Case 3: a = 2, b = `. `2 1 (2; `)2 = ( 2` ) = ( 1) 8 = ( 2` ) = (2; `)` ; (2; `)v = 1 for v 6= 2; `. Case 4: a = b = `. (`; `)v = ( 1; `)v for each place v, so this case reduces to Case 2. Case 5: a = `, b = `0 . ` 1 `0 1 0 (`; `0 )2 = ( 1) 2 2 , (`; `0 )` = ( `` ) and (`; `0 )`0 = ( ``0 ); (`; `0 )v = 1 for v 6= 2; `; `0 . Apply Quadratic Reciprocity Law we have (`; `0 )2 = (`; `0 )` (`; `0 )`0 .. 16.

(23) 5.2. Existence Theorem for Q. Y Let a 2 Q . From the Product Formula we have shown above, we know that (a; b)v = v2V 1 for any b 2 Q . Suppose now that for each v 2 V , we have "v , 1 or 1, such that the product of these "v is 1. Thus we may ask: under what situation can we …nd an element x 2 Q such that (a; x)v = "v for each v? The following Existence Theorem, somewhat like a converse of the Product Formula, suggests an answer. Theorem (Existence Theorem). Let (ai )i2I be a …nite set of elements of Q and let ("i;v )i2I;v2V be a set of numbers equal to 1. There exists x 2 Q such that (ai ; x)v = "i;v for all i 2 I and all v 2 V if and only if the following conditions are satis…ed: 1. Almost all the "i;v are equal to 1: Y 2. "i;v = 1 for each i 2 I: v2V. 3. For each v 2 V there exists xv 2 Qv such that (ai ; xv )v = "i;v for all i 2 I. Proof. The necessity of conditions (1) and (2) follows from the Product Formula for the Hilbert symbol, and (3) is obtained by taking xv = x. Our goal is to show that these conditions are su¢ cient. After multiplying the ai by nonzero squares, we may assume that these ai are integers. De…ne two sets as follows: S = fv 2 V j ai is a nonunit in Qv for some ig [ f2; 1g, T = fv 2 V j "i;v = 1 for some ig. Note that S and T are both …nite sets. First of all, we put the additional assumption S \ T = and we set Y Y v. v and m = 8 a= v2S;v6=2;1. v2T;v6=2;1. The integers a and m are relatively prime since S \ T = ; there exists a prime number p a (mod m) such that p 2 = S [ T by Dirichlet’s Theorem. We claim that the positive integer x = ap is the desired candidate by considering two cases: Case 1: v 2 S We have "i;v = 1 for all i since v 2 = T . Since x > 0, (ai ; x)1 = 1 = "i;1 . And since 2 x ap a (mod m), it follows that x a2 (mod 8) so that x is a square in Q2 and (ai ; x)2 = 1 = "i;2 . If v = ` is an odd prime. Since a is a unit in Q` , the fact that x a2 (mod `) implies that x is a square in Q` by Hensel’s lemma and (ai ; x)` = 1 = "i;` . Hence (ai ; x)v = "i;v for all i, …nishing the theorem in this case. Case 2: v 2 =S In this case v = ` where ` is an odd prime, and ai is a unit in Q` for all i. By the Explicit Formula of the Hilbert symbol, we have ai (ai ; b)` = ( )v` (b) for all b 2 Q` . ` If ` 2 = T [ fpg, then ` does not divide x and x is a unit in Q` (i.e. v` (x) = 0), hence (ai ; x)` = 1 = "i;` , the latter equality comes from ` 2 = T . If ` 2 T , then v` (x) = 1 and condition (3) implies that there exists x` 2 Q` such that (ai ; x` )` = "i;` for all i. Moreover, 17.

(24) "i;` = 1 for some i since ` 2 T , hence v` (x` ) 1 (mod 2) by the Explicit Formula, so for each i 2 I we have ai (ai ; x)` = ( ) = (ai ; x` )` = "i;` . ` If ` = p, then we deduce that Y Y (ai ; x)p = (ai ; x)v = "i;v = "i;p , v6=p. v6=p. where the …rst equality comes from the Product Formula; the second from the discussion we have made, and the condition (2) implies the third one. This proves the theorem provided that S \ T = . Back to the general case. From Approximation Theorem, we know that there exists x0 2 Q such that x0 =xv 2 Qv2 for all v 2 S by Remark 2.1. Thus (ai ; x0 )v = (ai ; xv )v = "i;v for all v 2 S (note that x0 =xv 2 Qv2 implies the …rst equality). For each i and each v, we set i;v = (ai ; x0 )v "i;v . Then i;v satis…es the conditions (1), (2), and (3) (note that (ai ; x0 xv )v = i;v ). De…ne T 0 = fv 2 V j i;v = 1 for some ig, we …nd that i;v = 1 if v 2 S and S \ T 0 = . By the special case we have shown above, there exists y 2 Q with (ai ; y)v = i;v for all i and all v. Now x = yx0 has the required properties (note that (ai ; yx0 )v = i;v (ai ; x0 )v = "i;v ).. 5.3. The Main Theorem for Q. We now state Hasse-Minkowski Theorem for Q and prove this by dividing into individual cases. Theorem (Hasse-Minkowski for Q). Let q be a quadratic form in n variables over Q: Then q represents 0 in Q if and only if it represents 0 in every completion of Q. If q represents 0 in Q, there is no doubt that it represents 0 in each Qv , so the part "only if" is clear. Moreover, every degenerate quadratic form represents 0, so in the proof we assume that q is nondegenerate. Finally, to simplify this problem, we assume more that q is in diagonal form (thus d(q) is easily read). The case n = 2 Proof. Multiply suitable nonzero squares in Z if necessary, we let q = X 2 aY 2 where a 2 Z is nonzero. Y First, a > 0 since q represents 0 in R. Express a as the form pvp (a) where the p product is taken over all prime numbers p, we see that for each p, a is a square in Qp since q represents 0 in Qp , hence all the vp (a) are even, and a is a square in Q, which shows that q represents 0 in Q: The case n = 3 Proof. Multiply suitable nonzero squares in Z, we assume that q = X 2 aY 2 bZ 2 where a,b 2 Z are nonzero with b squarefree and jaj 6 jbj. We prove this theorem by induction on m = jaj + jbj. When m = 2, we see that jaj = jbj = 1 and thus q = X 2 Y 2 Z 2 . Since q represents 0 in R, q can not be X 2 + Y 2 + Z 2 , and the other possibilities all represent 0 in Q. 18.

(25) Qr When m > 2, jbj > 1 since jaj 6 jbj, we may write b = j=1 pj with pj the distinct prime divisors of b (remember that b is squarefree). Fix a prime divisor p of b, we now claim that a is a square modulo p. If a 0 (mod p), there is nothing to prove; if otherwise, a is thus a unit in Qp , since q represents 0 in Qp , X 2 = aY 2 + bZ 2 has a nontrivial solution in Qp , say (x; y; z). We may as usual assume that x; y; z are in Zp with at least one of them in Up . Now x2 ay 2 (mod p), we deduce that y is a unit in Qp , for otherwise x2 0 (mod p) and x 0 (mod p) so that bz 2 x2 ay 2 0 (mod p2 ) and z 2 0 (mod p) (since b is squarefree and b 0 (mod p)), which contradicts to the fact that one of x, y, and z are in Up . Hence a is a square modulo p. Since p is arbitrary, by the Chinese Remainder Theorem and the fact that b is squarefree, we know that a is a square modulo b. Therefore k 2 = a + bb0 for some integers k and b0 with jkj 6 jbj=2 (if k is well-chosen). bb0 = k 2 a p tells us that bb0 is a norm from the extension k( a)=k where k = Q or each Qv . By the basic properties of the Hilbert symbol, this implies that q represents 0 in Q or in any Qv if and only if the same is true for the form q 0 = X 2 aY 2 b0 Z 2 . By the assumption, q represents 0 in each Qv , so q 0 does as well. Moreover, we have jb0 j = j. k2. a b. j6j. k2 a jbj j+j j6 + 1 < jbj. b b 4. Here jaj 6 jbj, jkj 6 jbj=2, and jbj > 2 are used. By the induction hypothesis, we see that q 0 represents 0 in Q. Hence q represents 0 in Q as well. This completes the proof. The case n = 4 Proof. We may assume that q = a1 X 2 + a2 Y 2 a3 Z 2 a4 W 2 . Let v be a place of Q: By Corollary 3.3 The assumption that q represents 0 in Qv implies that there is some cv 2 Qv which is represented both by a1 X 2 + a2 Y 2 and by a3 Z 2 + a4 W 2 . Therefore (cv ; a1 a2 )v = (a1 ; a2 )v and (cv ; a3 a4 )v = (a3 ; a4 )v for each place v by Corollary 4.3. Apply the Product Formula for the Hilbert symbol and the Existence Theorem we …nd that there exists c 2 Q such that (c; a1 a2 )v = (a1 ; a2 )v and (c; a3 a4 )v = (a3 ; a4 )v for all v. Since (c; a1 a2 )v = (a1 ; a2 )v for each v implies that the form a1 X 2 + a2 Y 2 cT 2 represents 0 in each Qv (Corollary 4.3 and Corollary 3.2 ) and thus it represents 0 in Q by the Hasse-Minkowski Theorem for n = 3, that is, a1 X 2 + a2 Y 2 represents c in Q. Likewise, a3 Z 2 + a4 W 2 represents c in Q, so q represents 0 in Q. The case n = 5 Proof. We assume that q = g h with g = a1 X12 + a2 X22 and h = (a3 X32 + a4 X42 + a5 X52 ). Let S be the subset of V consisting of 1; 2; and all the prime divisors p of ai for i = 3; 4; 5; which is a …nite set. Let v 2 S. There exists av 2 Qv represented by g and h since q represents 0 in Qv , so there exist xvi 2 Qv for i = 1; :::; 5 such that g(xv1 ; xv2 ) = av = h(xv3 ; xv4 ; xv5 ): We now claim that we can …nd x1 ; x2 2 Q such that if a = g(x1 ; x2 ), one has a=av 2 Qv2 for all v 2 S. We see that for each v 2 S, g is continuous on the product space Q2v whose topology is induced by the metric topology of Qv . This means that for each v 2 S and " > 0, there is a v > 0 such that jg(y1 ; y2 ) av jv < " provided that (y1 ; y2 ) 2 Q2v and dv ((y1 ; y2 ); (xv1 ; xv2 )) < v where dv is the metric on Q2v de…ned by dv ((x; y); (x0 ; y 0 )) = (jx x0 j2v + jy y 0 j2v )1=2 . Let be the least positive number of these v . By Approximation 19.

(26) Theorem, there exist x1 ; x2 2 Q such that jxi xvi jv is small enough for i = 1; 2 and v 2 S so that d((x1 ; x2 ); (xv1 ; xv2 )) < . Hence if a = g(x1 ; x2 ), then for each v 2 S we have ja av jv < ". We can choose " su¢ ciently small such that ja av jv < " implies a=av is a square in Q2v , proving this claim. Consider the form f = aZ 2 h: If v 2 S, then h represents a in Qv since it represents av in Qv and a=av 2 Qv2 ; hence f represents 0 in Qv : If v 2 = S, since the ai (i > 3) are units in Qv , so is the discriminant d(h) of h, and also "(h) = 1 by the Explicit Formula and v 6= 2. It follows that h represents a in Qv by Corollary 4.4 and f represents 0 in Qv . In all cases, f represents 0 in Qv and thus f represents 0 in Q by the Hasse-Minkowski Theorem for n = 4. Now q represents 0 in Q since both g and h represent a in Q. The case n > 5 P Proof. We let q = q1 q2 with q1 = 16i65 ai Xi2 and a1 > 0 and a5 < 0 (we can always choose such ai since q represents 0 in R). Since q1 represents 0 in R, it represents 0 in Q by Hasse-Minkowski Theorem for n = 5 and thus q represents 0 in Q by choosing the remaining variables equal to zero. Hence the proof of Hasse-Minkowski Theorem for Q is …nished.. 5.4. Some Applications. In this subsection, we use Hasse-Minkowski Theorm for Q to illustrate a few examples. Most of the time, in number theory, we are more interested in expressing a positive integer as a sum of squares in integers (no longer in rationals); in another point of view, we are now concerned with the problem that when a positive integer n is represented by a quadratic form in Z. However, what Hasse-Minkowski Theorem involves is merely a quadratic form q over Q. So we need a criterion to determine when we can assure that n is represented by q in Z by just veri…ng it is represented by q in Q. Let q be a quadratic form over Q in k variables and b the bilinear form associated to q, we say that q is positive-de…nite if q(x) > 0 for x 6= 0 and is matrix-integral if for any x; y 2 Zk , we have b(x; y) 2 Z. The following lemma achieves the preceding idea. Lemma 5.1. Let q be a positive de…nite and matrix-integral quadratic form over Q in k variables. Suppose that for each x 2 Qk , there exists y 2 Zk such that q(x. y) < 1.. (*). Then n 2 Z is represented by q in Z if it is represented by q in Q. Proof. See Page 311 in [2]. Remark 5.1. An easy consequence of Lemma 5.1 is derived: if a positive integer n is represented by the quadratic form q = X 2 + Y 2 (resp. q 0 = X 2 + Y 2 + Z 2 ) in Q, then it is also represented by q (resp. q 0 ) in Z. To see this, we have to show that q and q 0 both satisfy the assumptions of the lemma and we check ( ) now. For any x = (r; s) 2 Q2 , we choose y = (h; k) 2 Z2 such that jr hj 6 1=2 and js kj 6 1=2, we have q(x y) 6 1=2 < 1. The similar argument works for q 0 as well (here we have q 0 (x y) 6 3=4 < 1). Corollary 5.1 (Fermat). Let n 2 N and q be as in Remark 5.1. The following conditions are equivalent: 20.

(27) 1. n is represented by q in Z. 2. n is represented by q in Q. 3. For each prime divisor p of n with p. 3 (mod 4), 2 divides vp (n).. Proof. (1) implies (2) is clear. And (2) implies (1) is explained earlier. We claim that (2) implies (3). Assume that q represents n in Q, then it represents n in Qv for each place v of Q, which amounts to saying that (n; 1)p = (1; 1)p = 1 for each prime p by Corollary 4.3. Hence for each prime p with p 3 (mod 4), 1 = (n; 1)p = ( 1)vp (n) so that 2 divides vp (n). Conversely, if the condition (3) is satis…ed, then for each odd prime p, q represents n in Qp by Corollary 4.3 and the Explicit Formula; also, (3) implies that n=2v2 (n) 1 (mod 4) so q represents n in Q2 , and q represents n in R is clear. Hence q nZ 2 represents 0 in each Qv , so it represents 0 in Q by Hasse-Minkowski Theorem for Q and (2) is obtained. Lemma 5.2. Let q 0 be as in Remark 5.1. Then n 2 Q is represented by q 0 in Q if and only if n > 0 and n is not a square in Q2 . Proof. Note that the form q 0 has discriminant d(q 0 ) = 1 and "(q 0 ) = 1. We see that q 0 represents n in Q if and only if n > 0 and q 0 represents n in Qp for each prime number p by Hasse-Minkowski Theorem for Q. The latter statement is equivalent to that for each p, n is not a square in Qp , or n is a square in Qp and ( 1; 1)p = 1 by Corollary 4.4. By the Explicit Formula of the Hilbert symbol relative to Qp , we know that for each odd prime p, we always have ( 1; 1)p = 1 and ( 1; 1)2 = 1. So we need to require that n is not a square in Q2 to guarantee that n is represented by q 0 in Q2 , hence the result. Corollary 5.2 (Gauss). Let n 2 N. The following conditions are equivalent: 1. n is represented by q 0 in Z. 2. n is represented by q 0 in Q. 3. n is not of the form 4a (8k. 1) for a > 0 and k > 1.. Proof. The equivalence of condition (1) and (2) follows from Lemma 5.1. We see that n is of the form 4a (8k 1) if and only if n is a square in Q2 . Combine with Lemma 5.2, the equivalence of condition (2) and (3) follows. Corollary 5.3 (Lagrange). Every positive integer is a sum of 4 squares of integers. Proof. Express a positive integer n as the form 4a (8k + m) where a > 0 and 1 6 m < 8. If m 6= 7, then n is a sum of 3 squares of integers by Corollary 5.2 ; if m = 7, then n 4a = 4a (8k + m 1) is also a sum of 3 squares, hence the result.. 21.

(28) 6. Hasse Principle for Powers. In this section, we present the second example of the local-global principle. Let K be an algebraic number …eld and be an element in K. For an odd prime p and a positive integer e, we claim that is a pe th power in K provided that the same is true in every completion of K. Moreover, we allow there are …nitely many exceptions locally. In order to prove this result, we should know some knowledge in algebraic number theory such as places of an algebraic number …eld, Frobenius automorphisms, and the Dirichlet density of a set of primes in an algebraic number …eld, and so on. These terminologies can be found, for instance, in the …rst four chapters of [4]. Also, we need a nontrivial theorem in Class Field Theory listed below. Theorem (Chebotarev Density Theorem). Let L=K be a …nite extension of algebraic number …elds. Suppose that L=K is Galois with Galois group G and 2 G has c conjugates in G. Then the set of primes of K which have a prime divisor in L whose Frobenius automorphism is has Dirichlet density c=jGj. Proof. See Page 182 in [4]. Theorem. Let K be an algebraic number …eld, p be an odd prime number, and e 1 be an integer. Let S be a …nite set of places of K and 2 K . Assume that for each v 2 = S, the e e element is a p th power in Kv . Then is a p th power in K. Proof. Let L = K( ) where is a primitive pe th root of unity in L. Consider the extension e M = L( 1=p ) of L. We see that M=L is a normal extension, being a splitting …eld over L e of the polynomial X p , and is obviously a separable extension. Hence M=L is Galois with Galois group G = Gal(M=L) isomorphic to a subgroup of Z=pe Z via the inclusion map i : G ,! Z=pe Z de…ned by i( ) = k mod pe where k is chosen to be any integer with ( ) = k (note that k is unique modulo pe ). Thus G is cyclic, say G = h i. By the Chebotarev Density Theorem, there are in…nitely many primes of L whose Frobenius automorphism is . Since the number of primes of L rami…ed in M is …nite. We can choose a …nite place v0 of L not above any place of S, unrami…ed in M , such that whose Frobenius automorphism of v0 is , that is, a generator of G. Let v be any place of K below v0 . Since v 2 = S (by our choice of v0 ), it follows by e assumption that the element is a p th power in Kv as well as Lv0 since Kv Lv0 . Hence e for any place w0 of M above v0 , we always have Mw0 = Lv0 ( 1=p ) = Lv0 , which tells us that v0 splits completely in M , being the consequences of the facts that v0 is unrami…ed and that [Mw0 : Lv0 ] = 1. The completely-split of v0 implies that the Frobenius automorphism of v0 is the identity element so G is a trivial group. Combine with the fact that M=L is Galois, e M = L. This shows that 1=p lies in L and is a pe th power in L. e Let = p for some 2 L . Let H = Gal(L=K). This time we embed H into Gal(Q( )=Q) by restricting elements of H to Q( ). We have seen that Gal(Q( )=Q) is isomorphic to the multiplicative group of Z=pe Z, which is cyclic (the assumption that p is odd is used here). Thus H is cyclic and we set be its generator. Let = ( )= , we …nd e e e that p = ( p )= p = ( )= = 1 since 2 K and …xes K. Also, by the de…nition of , NL=K ( ) = 1. Since L=K is a cyclic extension, there is an element in L such that = ( )= by the Hilbert’s 90 theorem. Let = = , then ( )= = = = 1 by direct e e e computation and hence 2 K . Finally, p = p = p = and is a pe th power in K. It completes the proof.. 22.

(29) Hasse-Minkowski Theorem for Fp(t) with p 6= 2. 7. Throughout this section, p denotes an odd prime number, k = Fp (t) and V the set of all places of k, i.e., the set of all monic irreducible polynomial in A = Fp [t] together with the symbol 1. If v 2 V , Uv is the unit group of the valuation ring Ov of kv and ( ; )v denotes the Hilbert symbol relative to kv . When we write v = P or Q, P and Q are understood to be (distinct) monic irreducible polynomials in A tacitly.. 7.1. Computation of (a; b) relative to Fp (t)v. Lemma 7.1. Let v be a place of k and u 2 Uv . If the equation X 2 + uY 2 = Z 2 has a nontrival solution in kv , it has a solution (x; y; z) such that x 2 Ov and y,z 2 Uv (here = P if v = P and = if v = 1). Proof. This proof is parallel to Lemma 4.1, since in kv (v being arbitrary), every nonzero element can be uniquely expressed as n u with u 2 Uv , which is similar to the property of Q` . The characteristic of k has nothing to do with the whole argument. Let P 2 A be an irreducible polynomial, we know that A=P A is a …nite …eld with pdeg(P ) elements. Let G = (A=P A) be the unit group of A=P A, which is cyclic of even order, hence there is a canonical isomorphism G=G2 ' f 1g. If a 2 A and is relatively prime to P , then a 2 G; we de…ne ( Pa ) = 1 if a 2 G2 and 1 otherwise. It can be shown that p 1 for a 2 Fp , ( Pa ) = adeg(P ) 2 (see Page 24 in [7]). The symbol ( u ) and ( Pu ) occurred in the following theorem are de…ned similarly as above. For example, for u 2 U1 , we set ( u ) = 1 if u 2 (O1 = O1 ) 2 and 1 otherwise. Theorem (Explicit Formula). Let a; b 2 kv . If v = 1, we write a = ua ; ub 2 U1 . Then p 1 ua ub (a; b)1 = ( 1) 2 ( ) ( ) .. ua , b =. ub with. If v = P , we write a = P ua , b = P ub with ua ; ub 2 UP . Then (a; b)P = ( 1). deg(P ) p 2 1. (. u a ub ) ( ) . P P. Proof. This process is similar to the proof of Explicit Formula (see Page 10). We sketch this now. Assume v = 1 …rst. We consider three cases again. Case 1: = = 0. As before, by Proposition 3.4 ua X 2 + ub Y 2 = Z 2 has a nontrivial solution modulo since O1 = O1 ' Fp [ ]=( ) ' Fp is a …nite …eld. By Hensel’s lemma, this solution lifts to a solution in O1 . Here p 6= 2 is used. Hence (ua ; ub )1 = 1. Case 2: = 1; = 0. Since (ua ; ub )1 = 1, so ( ua ; ub )1 = ( ; ub )1 . The rest of the proof is the same as Case 2 (see Page 11). Case 3: = = 1. 2 Then ( ua ; ub )1 = ( ua ; ua ub )1 = ( ua ; ua ub )1 = ( ua ub ) = ( 1 )( ua )( ub ) = p 1 ( 1) 2 ( ua )( ub ). Now assume v = P . We still consider the same three cases with the third case should be modi…ed slightly. p 1 The di¤erent term is ( P1 ) = ( 1)deg(P ) 2 . 23.

(30) Corollary 7.1. The Hilbert symbol relative to Fp (t)v is a nondegenerate bilinear form on the F2 -vector space Fp (t)v =Fp (t)v 2 . Proof. The bilinearity is easily seen. And for the nondegeneracy, notice that in the proof of Corollary 4.1 we pick particular representatives in Qp =Qp2 , which works in this proof as 2 . well (here kv =kv 2 is of order 4). For instance, if v = 1, we have k1 =k12 ' Z=(2) U1 =U1 2 We may take f1; ; n; n g as representatives of k1 =k1 where n 2 U1 is not a square in O1 (i.e. ( n ) = 1). If [a] 2 k1 =k12 is not the identity class, say [a] = [n]; [ ], [n ], respectively. Then we choose b 2 k1 with [b] = [ ]; [n], [n], respectively. From ( ; n)1 = ( n ) = 1 and (n; n)1 = 1 (since n 2 U1 ), the nondegeneracy follows. The case for v = P is the same, so we omit the details. Once the bilinearity of the Hilbert symbol relative to Fp (t)v is established, the following two corollaries are obtained, see Remark 4.1 for a reason. Corollary 7.2. Let q = aX 2 + bY 2 + cZ 2 be a nondegenerate quadratic form over Fp (t)v . Set " = "(q) = (a; b)(a; c)(b; c) and let d = d(q) be the discriminant of q. Then q represents 0 in Fp (t)v if and only if ( 1; d) = ". Corollary 7.3. Let c 2 Fp (t)v and let q = aX 2 + bY 2 be a nondegenerate quadratic form over Fp (t)v . Then q represents c in Fp (t)v if and only if (c; ab) = (a; b). We have an analogue of Lemma 4.2. Lemma 7.2.. 1. We have jFp (t)v =Fp (t)v2 j = 4.. 2. If a 2 Fp (t)v =Fp (t)v2 and " = 1. De…ne H" (a) = fx 2 Fp (t)v =Fp (t)v2 j(x; a) = "g. Then jH1 (1)j = 4, H 1 (1) = and jH" (a)j = 2 if a 6= 1. 3. Let a,a0 2 Fp (t)v =Fp (t)v2 , and ","0 equal to 1; assume that H" (a) and H"0 (a0 ) are nonempty. Then H" (a) \ H"0 (a0 ) = if and only if a = a0 and " = "0 . Proof. (1) is explained in the Preliminaries. For (2) and (3), arguments involved are bilinearity and the nondegeneracy of the Hilbert symbol relative to kv (see the proof of Lemma 4.2 ).. 7.2. Quadratic Forms over Fp (t)v. Let q be a nondegenerate quadratic form P over Fp (t)v in n variables. Up to equivalence, we assume that q is in diagonal form q = 16i6n ai Xi2 where the ai 2 Fp (t)v . As before, Y we can also de…ne "(a1 ; :::; an ) = (ai ; aj ) (( ; ) denotes the Hilbert symbol relative to i<j. Fp (t)v ). This is an invariant of q (see Page 13). We have the following analogues. In the following, d = d(q) and " = "(q). Corollary 7.4. q represents 0 in Fp (t)v if and only if one of the following holds: 1. n = 2 and d =. 1.. 2. n = 3 and ( 1; d) = ". 3. n = 4 and either d 6= 1 or d = 1 and ( 1; 1) = ". 4. n > 5. 24.

(31) Proof. (1) is true in general (see Page 9) and (2) is Corollary 7.2. For (3) and (4), similar to the parallel version for Q (Page 13), are derived mainly by Lemma 7.2. Corollary 7.5. Let c 2 Fp (t)v =Fp (t)v2 . Then q represents c in Fp (t)v if and only if one of the following holds: 1. n = 1 and c = d. 2. n = 2 and (c; d) = ". 3. n = 3 and either c 6=. d or c =. d and ( 1; d) = ".. 4. n > 4. Proof. This is a direct consequence of Corollary 7.4, see the proof of Corollary 4.4 for comparison.. 7.3. Product Formula for the Hilbert Symbol relative to Fp (t)v. For convenience, we list the Explicit Formula of the Hilbert symbol relative to Fp (t)v . Let a; b 2 kv . Then. If v = 1, we write a = (a; b)1 = ( 1). p 1 2. (. ua , b = ua. ) (. ub. ub with ua ; ub 2 U1 .. ) .. If v = P , we write a = P ua , b = P ub with ua ; ub 2 UP . Then (a; b)P = ( 1). deg(P ) p 2 1. (. ua u b ) ( ) . P P. Before proving the Product Formula, we state the Reciprocity Law in function …elds. Theorem (Reciprocity Law). Let P ,Q 2 Fp [t] be two distinct monic irreducible polynomials with degree m and n, respectively. Then p 1 Q P ( )( ) = ( 1) 2 mn . P Q. Proof. See Page 25 in [7]. Theorem (Product Formula). If a; b 2 Fp (t) , then (a; b)v = 1 for almost all v 2 V , and we have the product formula Y (a; b)v = 1. v2V. Proof. We consider the following cases: Case 1: a = b = 1. This is clear from the Explicit Formula. Actually, the quadratic form X 2 + Y 2 + Z 2 represents 0 in Fp (and hence in kv ) by Proposition 3.4, so by de…nition of the Hilbert symbol ( 1; 1)v = 1 for each v. Case 2: a = 1; b = P . p 1 We know that ( 1; P )v = 1 for v 6= P; 1, and ( 1; P )P = ( P1 ) = ( 1)deg(P ) 2 while p 1 ( 1; P )1 = ( 1; deg(P ) uP )1 = ( 1 )deg(P ) = ( 1)deg(P ) 2 where uP 2 U1 (in fact, uP is a square in U1 since P is monic). Hence the result holds in this case. 25.

(32) Case 3: a = b = P . By the basic properties of the Hilbert symbol, (P; P )v = ( 1; P )v for each place v. Case 4: a = P; b = Q. P ) and (P; Q)Q = ( Q ). (P; Q)1 = (P; Q)v = 1 for v 6= P ,Q since 1 2 Uv ; (P; Q)P = ( Q P p 1. ( deg(P ) uP ; deg(Q) uQ )1 = ( deg(P ) ; deg(Q) )1 = ( 1) 2 deg(P ) deg(Q) where uP ,uQ 2 U1 (they p 1 P )( Q ) = ( 1) 2 deg(P ) deg(Q) by the Reciprocity are also squares, so can be ignored). Since ( Q P Law, the product formula holds in this case.. 7.4. Existence Theorem for Fp (t). Remenber that in proving the Existence Theorem for Q, we use the Dirichlet’s Theorem. In the following, we also need a parallel version in function …elds. For our purpose, a more stronger result. Theorem (Strong Dirichlet’s Theorem). Let a; m 2 Fp [t] be relatively prime polynomials. Given a positive number N; consider the set SN (a; m) of all monic irreducible polynomials N=2 1 pN + O( p N ) where in Fp [t] with P a (mod m) and deg(P ) = N . Then #SN (a; m) = (m) N (m) is the order of (A=mA) and O is the Landau big-oh notation. Proof. See Page 40 in [7]. This theorem indicates, if N is su¢ ciently large, the number #SN (a; m) > N=2. N=2. 1 pN (m) N. N=2. C p N = p N ( p (m) C) for some …xed constant C, hence #SN (a; m) > 0. In particular, we can always pick an irreducible polynomial P with P a (mod m) and deg(P ) is controled to be odd or even (if we wish). Theorem (Existence Theorem). Let (ai )i2I be a …nite set of elements of Fp (t) and let ("i;v )i2I;v2V be a set of numbers equal to 1. There exists x 2 Fp (t) such that (ai ; x)v = "i;v for all i 2 I and all v 2 V if and only if the following conditions are satis…ed: 1. Almost all the "i;v are equal to 1: Y 2. "i;v = 1 for each i 2 I: v2V. 3. For each v 2 V there exists xv 2 Fp (t)v such that (ai ; xv )v = "i;v for all i 2 I: Proof. The same as the proof of the Existence Theorm for Q, we sketch this process now. We assume that these ai are in A. De…ne two sets as follows: S = fv 2 V j ai is a nonunit in kv for some ig [ f1g, T = fv 2 V j "i;v = 1 for some ig. First, we assume that S \ T =. and we set Y a= v and m = v2T;v6=1. Y. v.. v2S;v6=1. We see that a and m are relatively prime; there exists a monic irreducible polynomial P a (mod m) such that P 2 = S [ T and deg(P ) deg(a) (mod 2) by the Strong Dirichlet’s Theorem in function …elds version. We still consider x = aP : 26.

(33) Case 1: v 2 S We have "i;v = 1 for all i. And since x is monic and is of even degree, x is a square in Fp (t)1 so that (ai ; x)1 = 1. If v = Q is a monic irreducible polynomial, then x aP a2 (mod Q) and by Hensel’s lemma x is a square in kQ so that (ai ; x)Q = 1. Hence (ai ; x)v = "i;v for all i. Case 2: v 2 =S In this case v = Q where Q is a monic irreducible polynomial, and ai is a unit in kQ for all i. By the Explicit Formula, we have ai (ai ; b)Q = ( )vQ (b) for all b 2 kQ . Q If Q 2 = T [ fP g, then x is an unit in kQ , hence (ai ; x)Q = 1 = "i;Q . If Q 2 T , then vQ (x) = 1 and condition (3) implies that there exists xQ 2 kQ such that (ai ; xQ )Q = "i;Q for all i. We know that "i;` = 1 for some i so v` (x` ) is odd, hence for each i 2 I we have ai (ai ; x)Q = ( ) = (ai ; xQ )Q = "i;Q . Q If Q = P , then we deduce that (ai ; x)P =. Y. (ai ; x)v =. v6=P. Y. "i;v = "i;P .. v6=P. This proves the theorem under the assumption S \ T = . Now everything goes similar to the proof of the Existence Theorem for Q.. 7.5. The Main Theorem for Fp (t). Theorem (Hasse-Minkowski for Fp (t)). Let q be a quadratic form in n variables over Fp (t): Then q represents 0 in Fp (t) if and only if it represents 0 in every completion of Fp (t). As before, we now assume that q is nondegenerate and is in diagonal form. The case n = 2 Q Proof. We let q = X 2 aY 2 where a 2 A is nonzero. Express a as the form a0 P P vP (a) where a0 2 Fp is the leading coe¢ cient of a and the product is taken over all monic irreducible polynomials P . Since q represents 0 in k1 , a is a square in k1 , and a0 is thus a square. And for each P , q represents 0 in kP so a is also a square in kP , which implies that vP (a) is even. Hence a is a square in A, and q represents 0 in k. The case n = 3 Proof. Let q = X 2 aY 2 bZ 2 where a; b 2 A are nonzero with b squarefree and deg a 6 deg b. We prove this theorem by induction on m = deg a + deg b. When m = 0, we see that deg a = deg b = 0. Since a; b 2 Fp and q is in 3 variables, q represents 0 in Fp as well as in k. When m = 1, we have deg a = 0 and deg b = 1. Thus q is of the form X 2 aY 2 (ct+d)Z 2 where c; d 2 Fp . Since q represents 0 in k1 , from the Explicit Formula we get 1 = (a; b)1 = ( a ), so a is a square in k1 and thus a is a square in Fp . Hence q represents 0 in k. 27.