aa r X i v : . [ m a t h . N T ] A p r Zeros of Pairs of Quadratic Forms
D.R. Heath-BrownMathematical Institute, Oxford
Let Q ( X , . . . , X n ) and Q ( X , . . . , X n ) be a pair of quadratic forms definedover Q , or more generally over a number field k . This paper will be concernedwith the existence of non-trivial simultaneous zeros of the two forms, over Q or k as appropriate. In particular one may hope for a local-to-globalprinciple. Indeed one may also ask whether the global points are dense inthe ad`elic points. When n ≥ k , and from [6, Theorem A(ii), page 40] thatthe weak approximation principle holds if the intersection Q = Q = 0is nonsingular. Moreover they proved [6, Theorem C, page 38] that when n ≥ k istotally imaginary.Colliot-Th´el`ene, Sansuc and Swinnerton-Dyer [7, Section 15] give a num-ber of examples showing that such results would be false for forms in suffi-ciently few variables. In particular the example Q = X X − ( X − X ) , Q = ( X + X )( X + 2 X ) − ( X − X )due to Birch and Swinnerton-Dyer [3], gives a smooth intersection violatingthe Hasse principle, while the example Q = X X − ( X + X ) , Q = (4 X − X )(4 X − X ) − ( X + X )due to Colliot-Th´el`ene and Sansuc [5], gives a smooth intersection for whichweak approximation fails. Colliot-Th´el`ene, Sansuc and Swinnerton-Dyer con-jecture [7, Section 16] that the Hasse principle should hold for nonsingular1ntersections as soon as n ≥
6. The goal of the present paper is the corre-sponding result for forms in 8 variables.
Theorem 1
Let Q ( X , . . . , X ) and Q ( X , . . . , X ) be two quadratic formsover a number field k such that the projective variety V : Q ( X¯ ) = Q ( X¯ ) = 0 is nonsingular. Then the Hasse principle and weak approximation hold for V . For pairs of diagonal forms this was shown by Colliot-Th´el`ene, in un-published work. The proof used the corresponding special case of Theorem2, which he established using the argument ascribed to him in the proof ofLemma 7.2.Perhaps it is as well to explain precisely what we mean by the variety Q ( X , . . . , X n ) = Q ( X , . . . , X n ) = 0 being nonsingular over a general field K . By this it is meant that the matrix ∂Q ∂x . . . ∂Q ∂x n ∂Q ∂x . . . ∂Q ∂x n ! has rank 2 for every non-zero vector x¯ ∈ K n . It follows automatically fromthis that if n ≥ Q = Q = 0 is absolutelyirreducible (see Lemma 3.2) and is not a cone.The papers by Colliot-Th´el`ene, Sansuc and Swinnerton-Dyer suggest aline of attack for the 8 variable case (see [7, Remark 10.5.3]). The mainobstacle to this plan, which they failed to handle, was a purely local prob-lem concerning the forms Q and Q . This we now resolve in the followingtheorem. Theorem 2
Let k v be the completion of a number field k at a finite place v . Let F v be the residue field of k v and assume that F v ≥ . Supposethat Q ( X , . . . , X ) and Q ( X , . . . , X ) are quadratic forms over k v suchthat the projective variety Q ( X¯ ) = Q ( X¯ ) = 0 is nonsingular, and assumefurther that the forms have a non-trivial common zero over k v . Then thereis a form Q in the pencil a Q + a Q (with a , a ∈ k v ) containing (at least)3 hyperbolic planes.
2e will deduce Theorem 1 from Theorem 2, but our argument differssomewhat from that sketched by Colliot-Th´el`ene, Sansuc and Swinnerton-Dyer.While Theorem 2 can be appropriately extended to singular intersections,it is unclear whether one can prove a corresponding global statement. Thusthe methods of the present paper seem insufficient to handle a version ofTheorem 1 for singular intersection of two quadrics in P .The next section will describe some basic facts and terminology from thetheory of quadratic forms, but we should stress at the outset that much ofour analysis involves the reduction of integral forms from k v down to F v .It follows that if χ ( F v ) = 2 we are forced to consider quadratic forms incharacteristic 2, which may be less familiar to some readers than the case ofodd (or infinite) characteristic.Throughout the paper we will write k v for the completion of a numberfield k . When v is a non-archimedean valuation we will write O v for thevaluation ring of k v , and F v for the residue field. We will tend to use uppercase Q and L (and other letters) for quadratic and linear forms over k or k v ,and similarly lower case q and ℓ (and other letters) for forms over the finitefield F v . In this section we recall some basic facts about the theory of quadratic forms.In the case of characteristic 2 the reader may wish to treat some of ourstatements as exercises.For any field K we will think of a quadratic form q ( X , . . . , X n ) over K as a polynomial of the shape q ( X , . . . , X n ) = X ≤ i ≤ j ≤ n q ij X i X j . (2.1)It is usual to represent quadratic forms using coefficients q ij for every pair i, j ≤ n , subject to the symmetry constraint q ij = q ji . However this is inap-propriate in characteristic 2 and we therefore depart from this convention.We define rank( q ) to be the least integer m such that there is a lineartransformation M ∈ GL n ( K ) for which q ( M X¯ ) is a function of X , . . . , X m alone. If K is contained in a larger field K ′ then the rank over K ′ will be thesame as the rank over K . Moreover rank( q ) < n if and only if q (X¯ ) = 0 has3 singular zero over K , that is to say a zero x¯ = 0¯ for which ∇ q (x¯) = 0¯. Thisis exactly the case in which the projective variety Q : q = 0 is singular, orequivalently, is a cone. Indeed the set of vertices for Q (as a cone in projectivespace) is a linear space V of codimension rank( q ). If P is a nonsingular pointon Q with tangent hyperplane H , then V ⊆ H , and one sees that all pointsof < P, V > are vertices for Q ∩ H as a cone. The linear space < P, V > hascodimension rank( q ) − H . Thus if we project Q ∩ H to P n − , so as toproduce a quadric hypersurface, the corresponding quadratic form will haverank at most rank( q ) − q ) − Lemma 2.1
Let q ( X , . . . , X n ) be a quadratic form of rank r , and suppose X m = 0 is tangent to q = 0 at a nonsingular point. Then q ( X , . . . , X n − , has rank r − . For a quadratic form in the shape (2.1) we may define an n × n matrix M ( q ) with M ( q ) ij = q ij , i < j, q ij , i = j,q ji , j < i. (2.2)When χ ( K ) = 2 this will have diagonal entries equal to zero. We notethat ∇ q (X¯ ) = M ( q )X¯ . Moreover if T ∈ GL n ( K ) and q T (X¯ ) = q ( T X¯ ) then M ( q T ) = T t M ( q ) T , where T t denotes the transpose of T .We also define the determinant det( q ) asdet( q ) := det( M ( q )) , so that det( q T ) = det( T ) det( q ). Our definition, which is designed to help inthe case of characteristic 2, differs by powers of 2 from that which the readermay have expected. It should be observed that det( q ) will vanish identicallyif χ ( K ) = 2 and n is odd. In this case one may use the “half-determinant”of q which we denote det / ( q ). Although one can define this in completegenerality (see Leep and Schueller [10]) we shall need it only for the case inwhich K is the residue field F v of a completion k v for some number field k ,with respect to a non-archimedean valuation v . Then, if q (X¯ ) ∈ F v [X¯ ] is thereduction of some form Q (X¯ ) ∈ O v [X¯ ], one can verify that det( Q ) ∈ O v .We then define det / ( q ) = det( Q ) , θ denotes the reduction to F v of θ ∈ O v ). One may verify that thisis indeed independent of the choice of the lift Q of q , and that det / ( q T ) =det( T ) det / ( q ). A quadratic form is nonsingular if and only if its determi-nant (or in the case in which n is odd and χ ( K ) = 2, its half-determinant)is non-zero.If the form q has rank at most r the corresponding matrix M ( q ) willalso have rank at most r . When r < n it follows that all ( r + 1) × ( r + 1)submatrices of M ( q ) are singular. In the case in which χ ( K ) = 2 and r iseven we can say slightly more. Suppose that q is a quadratic form in r + 1variables obtained by setting to zero n − r − q . Then q will be singular and det / ( q ) = 0. Thus not only do all the ( r + 1) × ( r + 1)minors vanish, but also the “central” ( r + 1) × ( r + 1) half-determinants.We shall express these conditions in general by saying somewhat looselythat all the ( r + 1) × ( r + 1) minors of q vanish. Here one should bear inmind that if r is even and χ ( K ) = 2 then this refers to the half-determinantin the case of “central” minors.Conversely, the vanishing of all the ( r +1) × ( r +1) minors of q implies thatrank( q ) ≤ r . To prove this it is enough to verify that if the ( r + 1) × ( r + 1)minors of q all vanish then so do those of q T for any elementary matrix T .These facts are enough to show thatrank (cid:0) q ( X , . . . , X m ) + X m +1 X m +2 (cid:1) = rank (cid:0) q ( X , . . . , X m ) (cid:1) + 2 . (2.3)If q (x¯) = 0 has a nonsingular zero over K there is a linear transformation T ∈ GL n ( K ) for which q ( T X¯ ) takes the form X X + q ′ ( X , . . . , X n ). Wethen say that q “splits off a hyperbolic plane”. If q splits off m hyperbolicplanes then the form q ′ will split off m − K is finite, as is often the case of interest to us, the Chevalley–Warning theorem implies that q has a nonsingular zero over K as long asrank( q ) ≥
3. It follows that there is a matrix T such that q ( T X¯ ) = X X + . . . + X s − X s + cX s +1 for some c = 0, if q has odd rank equal to 2 s + 1, or such that q ( T X¯ ) takesone of the shapes q ( T X¯ ) = X X + . . . + X s − X s or q ( T X¯ ) = X X + . . . + X s − X s − + n ( X s − , X s )5f rank( q ) = 2 s is even. Here n ( X, Y ) is an anisotropic form. We note that,when K is finite, if n ( X, Y ) and n ′ ( X, Y ) are two anisotropic forms there willbe a linear transform in GL ( K ) taking n to n ′ . Moreover there is a lineartransform in GL ( K ) taking n ( X , X ) + n ′ ( X , X ) to X X + X X . Weshall be somewhat lax in our notation for anisotropic forms, writing n ( X, Y )for a generic form of this type, not necessarily the same at each occurrence.Moreover we shall sometimes use the notation N ( X, Y ) for a binary formover O v whose reduction to F v is anisotropic.The following result will be useful in recognizing when a form over O v splits off hyperbolic planes. Lemma 2.2
Let π be a uniformizing element for k v and suppose that thequadratic form Q ( X . . . , X n ) ∈ O v [ X¯ ] satisfies Q ( X¯ ) ≡ X X + . . . + X s − X s + e Q ( X s +1 , . . . , X n )+ π s X i =1 X i L i ( X , . . . , X n ) ( mod π ) for some quadratic form e Q over O v . Then there exists T ∈ GL n ( O v ) suchthat Q ( T X¯ ) = X X + . . . + X s − X s + Q ( X s +1 , . . . , X n ) with Q ≡ e Q ( mod π ) .In particular, if Q splits off at least s hyperbolic planes over F v , then sodoes Q over k v . Indeed if Q has rank at least 7 then Q splits off at least 3hyperbolic planes. Unfortunately if Q and Q are forms in 8 variables with coefficients in O v it is possible that every linear combination aQ + bQ has rank at most 6,even if the variety Q = Q = 0 is nonsingular over k v . This is where thedifficulty in proving Theorem 2 lies.We may establish Lemma 2.2 by adapting the argument for Hensel’sLemma. We show inductively that for every positive integer h there is a T h ∈ GL n ( O v ) and linear forms L ( h ) i ( X , . . . , X n ) over O v such that Q ( T h X¯ ) ≡ X X + . . . + X s − X s + Q h ( X s +1 , . . . , X n )+ π h s X i =1 X i L ( h ) i ( X , . . . , X n ) (mod π h +1 )6ith Q h ≡ e Q (mod π ). Once this is established one may choose a convergentsubsequence from the T h and the lemma will follow.To prove the claim we observe that the case h = 1 is immediate. Generallyif U h ∈ GL n ( O v ) corresponds to the substitution X i − → X i − + π h L ( h )2 i ( X , . . . , X n ) , (1 ≤ i ≤ s ) X i → X i + π h L ( h )2 i − ( X , . . . , X n ) , (1 ≤ i ≤ s ) X i → X i , (2 s < i ≤ n ) , then it suffices to take T h +1 = U h T h .There is one further result which will be useful in finding forms whichsplit off three hyperbolic planes. Lemma 2.3
Let Q ( X , . . . , X ) be a quadratic form over k v whose determi-nant is not a square in k v . Then Q contains at least three hyperbolic planes. The form Q must be nonsingular. However any nonsingular form in 8variables over k v will split off two hyperbolic planes, leaving a form in 4variables, so that Q = X X + X X + Q ′ ( X , . . . , X ). Then det( Q ′ ) =det( Q ) is a non-square, which implies that Q ′ is isotropic, and hence splitsoff a further hyperbolic plane.In view of Lemma 2.3, it would suffice for the proof of Theorem 2 to findany form aQ + bQ whose determinant is not a square. Unfortunately it ispossible that all forms in the pencil have square determinant, as the example Q = X − X + X − X + X − X + X − X Q = X X + X X + X X + X X over Q shows. These forms have a common zero at (2 , , , , , − , , Q of quadratic forms Q , defined over O v . The argument will require a greatmany invertible linear transformations of variables. In this context we willuse the fact that any element of GL n ( F v ) can be lifted to GL n ( O v ). Thuswhenever we refer to a “change of variables” or a “substitution” among thevariables of such a form over F v , we mean that one applies an appropriateelement of GL n ( O v ) which reduces to the relevant mapping over F v .7ince we will use a succession of changes of variables it will become cum-bersome to use different notation for all the different forms that arise. We willtherefore abuse notation by saying for example that the change of variables X → X + X transforms q ( X , X ) = X into q ( X , X ) = X + 2 X X X ,rather than producing q ′ ( X , X ) = X + 2 X X X , say. Thus we may havea number of forms, all denoted by “ q ”, which are not actually the same. Wetrust that this will not cause confusion. Given two quadratic forms q ( X , . . . , X n ) , q ( X , . . . , X n ) defined over a field K we may consider the pencil P = < q , q > = < q , q > K , which is the setof all linear combinations aq + bq with a, b ∈ K , not both zero. We willalso consider P ∗ = < q , q > K , where K is the algebraic closure of K . Thereare three notions of rank that we attach to such a pencil. We define R ( P ) asthe least integer m such that there is a linear transformation T ∈ GL n ( K )for which q ( T X¯ ) is a function of X , . . . , X m alone, for every q ∈ P ∗ . Thus m is the least integer such that there are variables X , . . . , X m which sufficeto represent every form in the pencil. We define a second number r ( P ) asmax rank( q ), where q runs over all forms in P ∗ . The third notion of rankwhich we shall use is r min ( P ) defined similarly as min rank( q ), where q runsover all forms in P ∗ .For a pencil P generated by q and q we will write R ( q , q ) = R ( P ), r ( q , q ) = r ( P ) and r min ( q , q ) = r min ( P ). Clearly one has r min ( P ) ≤ r ( P ) ≤ R ( P )in every case. We remark that if rank( tq + q ) ≤ r for every value of t thenthe ( r +1) × ( r +1) minors of tq + q all vanish identically as polynomials in t ,whence the minors of aq + bq also vanish identically, yielding r ( q , q ) ≤ r .As an example of these notions of rank one may consider the forms q ( X , X , X ) = X X and q ( X , X , X ) = X X , which generate a pencil P with r min ( P ) = r ( P ) = 2 and R ( P ) = 3.When the variety q = q = 0 is nonsingular we can describe these ranksprecisely. Lemma 3.1
Suppose that Q ( X , . . . , X n ) and Q ( X , . . . , X n ) are quadraticforms over a field K of characteristic not equal to 2. Define the binary form F ( x, y ) = F ( x, y ; Q , Q ) = det (cid:0) xM ( Q ) + yM ( Q ) (cid:1) (3.1)8 f degree n . Then if the variety Q = Q = 0 is nonsingular the form F does not vanish identically, and has distinct linear factors over the algebraiccompletion K . Moreover r ( Q , Q ) = n and r min ( Q , Q ) = n − . This follows from Heath-Brown and Pierce [8, Proposition 2.1], for example.We now give a general condition for an intersection of quadrics to beabsolutely irreducible of codimension 2.
Lemma 3.2
Let Q ( X , . . . , X n ) and Q ( X , . . . , X n ) be two quadratic formsover an algebraically closed field K , such that r min ( Q , Q ) ≥ and r ( Q , Q ) ≥ . Then the projective variety V : Q = Q = 0 is absolutely irreducible of codimension 2. Moreover if n ≥ and V is non-singular then it is absolutely irreducible of codimension 2. If V fails to be absolutely irreducible of codimension 2 then it must containeither a quadric of codimension 2, or a linear space of codimension 2. Supposefirstly that V contains a quadric of codimension 2, given by the simultaneousvanishing of Q (X¯ ) and L (X¯ ) say. Then we may write Q i = c i Q + LM i forsuitable constants c i and linear forms M i . We now set a i = c i for i = 1 , c = c = 0, in which case we take a = 1 and a = 0. Then a Q − a Q is a multiple of L and hence has rank at most 2, contradictingour hypotheses. Now consider the second case, in which V contains a linearspace of codimension 2, given by the simultaneous vanishing of L (X¯ ) and L (X¯ ) say. Then Q and Q must take the shape Q i = L M i + L M i for i = 1 and 2, where M ij are suitable linear forms. Hence every quadraticin the pencil < Q , Q > will have rank at most 4. This contradicts ourhypothesis that r ( Q , Q ) ≥ V is nonsingular and n ≥ r min ( Q , Q ) = n − ≥
3, so that V cannot contain a quadric of codimension2. Moreover if V contained the linear space L = L = 0, so that we canwrite Q i = L M i + L M i for i = 1 and 2, there would be singular pointswherever L = L = M M − M M = 0 . r ( P ) < R ( P ). In this context we have the following result. Lemma 3.3
Suppose that P is a pencil of quadratic forms in n variablesover a field F , and assume that F ≥ n . Write r ( P ) = r and R ( P ) = R and suppose that r < R . Assume further either that χ ( F ) = 2 , or that r is even and F is the residue field of the completion of a number field undera non-archimedean valuation. Then we can choose a basis for P such that rank( q ) = r . For any such basis q , q there is an invertible change ofvariables over F so that q ( X , . . . , X n ) = q ′ ( X , . . . , X r ) and q ( X , . . . , X n ) = q ′ ( X , . . . , X R − ) + X r X R for certain quadratic forms q ′ , q ′ defined over F . If q and q generate P and r ( q , q ) = r , the r × r minors of a linearcombination q + xq cannot all vanish identically in x . Since at least one ofthese minors is a non-zero polynomial of degree at most r in x there can beat most r values of x for which rank( q + xq ) < r . Thus if F ≥ n ≥ R > r there will be a linear combination q ′ = q + xq defined over F with rank r .Thus it will be possible to choose a basis q ′ , q for P in which rank( q ′ ) = r .Now, assuming that rank( q ) = r , we make an appropriate change ofvariables so that only X , . . . , X R appear in the forms in P ∗ . After a furtherchange of variables we can then write q = q ′ ( X , . . . , X r ) and q = q ( X , . . . , X r ) + R X i = r +1 X i ℓ i ( X , . . . , X r ) + q ( X r +1 , . . . , X R ) (3.2)say, with appropriate linear forms ℓ i . With the notation (2.2) the matrix for q + xq then takes the form M ( q ′ + xq ) xℓxℓ t xM ( q ) ! . (3.3)10ince q + xq has rank at most r for every value of x we deduce that all the( r +1) × ( r +1) minors vanish identically in x (with the obvious interpretationfor the central minors when r is even and χ ( F ) = 2). Suppose now that q contains a term cX i X j with r < i ≤ j ≤ R , so that c = ( q ) ij . If i = j ,or if χ ( D ) = 2, we consider the minor formed from rows 1 to r and i , andcolumns 1 to r and j . This will be a polynomial in x , in which the coefficientof x is c det( q ′ ). Since the polynomial vanishes identically we deduce that c det( q ′ ) = 0. Similarly if i = j and χ ( F ) = 2 (so that r is even) the half-determinant corresponding to the r + 1 variables X , . . . , X r and X i mustvanish. However this half determinant is a polynomial in x with linear term c det( q ′ ) x , and again we deduce that c det( q ′ ) = 0. However since q hasrank r , and we have avoided the case in which χ ( F ) = 2 and r is odd, wewill have det( q ′ ) = 0, so that c = 0. We therefore deduce that all coefficientsof q vanish.We see now that (3.2) simplifies to q = q ( X , . . . , X r ) + R X i = r +1 X i ℓ i ( X , . . . , X r ) . If all the forms ℓ r +1 , . . . , ℓ R were identically zero we would have R ( P ) ≤ r ,contrary to our assumption. We therefore suppose that ℓ R , say, is not identi-cally zero. Without loss of generality we may assume that ℓ R involves X r withnon-zero coefficient. Indeed, after a change of variable among X , . . . , X r wecan then assume that ℓ R ( X , . . . , X r ) = X r . It then follows that q and q are of the shape given in Lemma 3.3.For the rest of the present section, the pairs of forms q , q will be definedover a field F v , which will be the residue field of the completion k v of anumber field k v with respect to a non-archimedean valuation. Some of theresults we prove will in fact be valid in a more general setting, but this willsuffice for our needs.We now examine the remaining case, in which χ ( F v ) = 2 and r is odd. Byextending the argument above we will prove the following structure result. Lemma 3.4
Let F v and P be as in Lemma 3.3, except that χ ( F v ) = 2 andthat r is odd. Then there is a basis < q , q > for P in which rank( q ) = r .For any such basis there is an invertible change of variables over F v suchthat q ( X , . . . , X n ) = q ′ ( X , . . . , X r )11 nd q is in one of the shapes q ( X , . . . , X n ) = q ′ ( X , . . . , X R − ) + X r X R , (3.4) or q ( X , . . . , X n ) = q ′ ( X , . . . , X r ) + X r +1 , (3.5) or q ( X , . . . , X n ) = q ′ ( X , . . . , X r ) + X r +1 + X r X r +1 . (3.6) When r ≤ R − we can always take q to be of the shape (3.4). We start the proof by choosing generators q and q for P as before, andexpressing q in the shape (3.2). As before the ( r + 1) × ( r + 1) minorsof (3.3) must all vanish identically in x . We begin by considering an “off-diagonal” term, cX i X j say, in q ( X r +1 , . . . , X R ). Thus we will assume that r +1 ≤ i < j ≤ R . Choose any sets I , J ⊂ { , . . . , r } of cardinality r − I = I ∩ { i, j } and J = J ∩ { i, j } . If I = J then the I, J minor of (3.3) isa determinant which is a polynomial in x , which must vanish identically. Thecoefficient of x will be − c times the I , J minor of q ′ , so that this productmust vanish. Similarly if I = J the I, I minor of (3.3) is a half-determinantwhich is again a polynomial in x . The coefficient of x will similarly be − c times the I , I minor of q ′ , which must also vanish. If all the ( r − × ( r − q ′ were to vanish we would have rank( q ′ ) ≤ r −
2, contrary tohypothesis. We therefore conclude that c = 0. Thus q will be a diagonalform. Since F v is a finite field of characteristic 2 there is therefore a changeof variable reducing q to the shape cX r +1 . Now, if c = 0, or if any of theforms ℓ r +2 , . . . , ℓ R is not identically zero, we can proceed as in the proof ofLemma 3.3, so as to put q into the form (3.4). Moreover if ℓ r +2 , . . . , ℓ R allvanish identically we see that we must have R = r + 1.Thus it remains to consider the case in which R = r + 1 and q ( X , . . . , X n ) = q ′ ( X , . . . , X r ) + cX R + X R ℓ ( X , . . . , X r )with c = 0. Since every element of F v is a square we may replace X R by c − / X R so as to reduce to the case c = 1. Then, if ℓ vanishes identicallywe obtain a form of type (3.5), while if ℓ does not vanish identically we canmake a change of variables so as to replace ℓ ( X , . . . , X r ) by X r , giving us aform of type (3.6). This completes the proof of Lemma 3.4.We can say a little more about the structure of our forms when q hasthe shape (3.4). 12 emma 3.5 Let P be a pencil with r ( P ) = r < R = R ( P ) , and supposeeither that r is even or that r ≤ R − . Then r ≥ . Moreover if rank( q ) = r we can make a change of variable so that q = q ( X , . . . , X R − ) + X R − ℓ ( X , . . . , X R − ) and q = q ( X , . . . , X R − ) + X R − X R with rank( q ) = r ( q , q ) = r − and R ( q , q ) = R − or R − . In particularif r = 2 the forms q and q vanish identically. We begin the proof by observing that, according to Lemmas 3.3 and 3.4,we can put q into the shape (3.4). We may then rewrite the forms as q = q ′′ ( X , . . . , X r − ) + X r ℓ ( X , . . . , X r )and q = q ′′ ( X , . . . , X r − , X r +1 , . . . , X R − ) + X r (cid:0) X R + ℓ ( X , . . . , X R − ) (cid:1) . We then replace X R by X R + ℓ ( X , . . . , X R − ) and re-number the variablesso as to interchange X r and X R − . This puts q and q into the shape givenin the lemma. For any x we then have xq + q = xq + q + X R − (cid:0) xℓ ( X , . . . , X R − ) + X R (cid:1) . Replacing X R by X R + xℓ ( X , . . . , X R − ) we see from (2.3) thatrank( xq + q ) = rank( xq + q ) + 2for all x , whence we must have r ( q , q ) ≥ r ( q , q ) = r −
2, as required.Moreover it is evident that r = rank( q ) ≤ rank( q ) + 2, so that rank( q ) = r −
2. Clearly we have R ( q , q ) ≤ R −
2, and if it were possible to write q and q using only R − q and q given inthe lemma would allow us to write q and q with only R − R ( q , q ) ≥ R − Lemma 3.6
Let q and q be quadratic forms in variables, defined overa field F v as above. Then if r ( q , q ) < the forms will have a singularcommon zero over F v ; that is to say there is a non-zero vector x¯ ∈ F v with q ( x¯ ) = q ( x¯ ) = 0 and ∇ q ( x¯ ) and ∇ q ( x¯ ) proportional.
13t should be observed that when χ ( F v ) = 2 this result may fail for forms in n = 4 variables. For example, when n = 2 the forms q = X and q = X have r = 1 < n , but have no singular common zero over F v (indeed theyhave no non-trivial common zero). Equally, when n = 6 the forms q (X¯ ) = X X + X X + X and q (X¯ ) = X X + cX X + X X + X have no singular common zero over F v provided that c ∈ F v is chosen so thatthe polynomial T + T + c is irreducible.We can choose x¯ = (0 , , ,
1) when q is of the shape (3.4). Accordingto Lemmas 3.3 and 3.4 it therefore remains to consider the case in which χ ( F v ) = 2 and r = 3, and q takes one of the forms (3.5) or (3.6).We begin by examining the first case, in which q ( X , . . . , X ) = q ′ ( X , X , X ) and q ( X , . . . , X ) = q ′ ( X , X , X ) + X . We begin by changing variables so that q ′ ( X , X , X ) = q ( X , X ) + cX . We can now write c = d and replace X by X + dX to get q ( X , . . . , X ) = q ( X , X ) + X . Then (0 , , ,
0) will be a singular common zero if q ′ (0 , ,
1) = 0. Otherwisewe may take q ( X , . . . , X ) = q ( X , X ) + ℓ ( X , X ) X + eX , say, with e = 0. We now choose x , x ∈ F v , not both zero, such that ℓ ( x , x ) = 0. There then exist x , x ∈ F v with ex = q ( x , x ) and x = q ( x , x ), so that q (x¯) = q (x¯) = 0. However we have arranged that thethird and fourth entries of ∇ q i (x¯) vanish for i = 1 and 2. Moreover ∇ q (X¯ )is automatically orthogonal to X¯ , for any quadratic form q over a field ofcharacteristic 2. Hence x ∂q (x¯) ∂x + x ∂q (x¯) ∂x = 0 , and x ∂q (x¯) ∂x + x ∂q (x¯) ∂x = 0 . ∇ q (x¯) and ∇ q (x¯) are proportional.In the case of (3.6) we write q (X¯ ) = q ( X , X ) + X ℓ ( X , X , X )and q (X¯ ) = q ( X , X ) + X (cid:0) ℓ ( X , X , X ) + X (cid:1) + X , whence det( q + xq ) = x det( q + tq ). However det( q + xq ) vanishesidentically since r ( q , q ) <
4, and we therefore see that det( q + xq ) alsovanishes identically, yielding r ( q , q ) ≤
1. Since χ ( F v ) = 2, we know thatevery element of F v is a square, and it follows that we can write q (X¯ ) = ℓ ( X , X ) + X ℓ ( X , X , X )and q (X¯ ) = ℓ ( X , X ) + X ( ℓ ( X , X , X ) + X ) + X = ( ℓ ( X , X ) + X ) + X ( ℓ ( X , X , X ) + X ) . We may then complete the proof of the lemma by choosing a non-zero vectorx¯ ∈ F v such that x = ℓ = ℓ + x = 0 . v -Adically Minimized Pairs of Forms Since our analysis is based on reduction to F v we begin by multiplying theforms Q , Q in Theorem 2 by a suitable scalar so as to give them coefficientsin O v . For such integral forms we will write q = Q and q = Q for thereductions to F v . We retain this notation through to the end of Section 11.With these conventions we can now explain the fundamental idea behindour proof. We begin by choosing a suitable model for the pencil < Q , Q > over O k . For example, one may divide Q by a suitable power of π so as toreach a form for which q does not vanish identically. This choice of modelwill depend on a v -adic minimization technique from the work of Birch,Lewis and Murphy [2]. We then consider a large number of cases, dependingprincipally on the values of r = r ( q , q ) and R = R ( q , q ). If our model issuitably chosen it turns out that small values of r and R cannot occur. Inthe remaining cases we prove that some form in the pencil < Q , Q > splitsoff three hyperbolic planes. Often we will be able to do this by showing that15ome form aq + bq over F v splits off three hyperbolic planes, so that we canapply Lemma 2.2. However other cases will require more work.There is one particularly easy case. If r ( q , q ) ≥
7, and assuming that F v ≥
8, there will be a form aq + bq with a, b ∈ F v whose rank is at least7. Then we can conclude from Lemma 2.2 and that Q splits off 3 hyperbolicplanes, as required for Theorem 2.We record this observation as follows. Lemma 4.1 If Q and Q are defined over O v and have r ( q , q ) ≥ , thenthere is a form in the pencil < Q , Q > which splits off three hyperbolicplanes. We now describe the v -adic minimization process given by Birch, Lewisand Murphy [2]. We start with two quadratic forms Q ( X , . . . , X n ) and Q ( X , . . . , X n ) defined over k v , the completion of a number field k withrespect to a non-archimedean valuation v . We will assume that the variety Q = Q = 0 is nonsingular, in the sense specified in Section 1. We willretain this hypothesis throughout the paper, without further comment.Since the variety Q = Q = 0 is nonsingular, Lemma 3.1 tells us that F ( x, y ) does not vanish identically, and has no repeated factors. We now set∆( Q , Q ) = Disc( F ( x, y ; Q , Q )) (4.1)which will be a non-zero element of k v . Indeed if the forms Q and Q aredefined over O v then ∆( Q , Q ) will also be in O v .For any matrices U ∈ GL ( k v ) and T ∈ GL n ( k v ) we define actions onpairs of quadratic forms Q , Q by setting( Q , Q ) U = ( U Q + U Q , U Q + U Q )and ( Q (X¯ ) , Q (X¯ )) T = ( Q ( T X¯ ) , Q ( T X¯ )) . Notice that Q = Q = 0 defines a smooth variety with a non-trivial pointover k v if and only if the same is true for the forms ( Q , Q ) UT . Similarly thepencil defined over k v by Q , Q contains a form which splits off 3 hyperbolicplanes if and only if the same is true for the quadratics ( Q , Q ) UT . Thus wemay transform our pair Q , Q in this way in the hope of producing formsof a convenient shape. We shall say that a pair of integral forms Q , Q is16minimized” if there are no transforms U and T such that ( Q , Q ) UT is alsointegral and | ∆(( Q , Q ) UT ) | v > | ∆( Q , Q ) | v . It is clear that there is always a pair of transforms U and T such that( Q , Q ) UT is minimized. Indeed it may happen that there are many quitedifferent pairs U, T which may be used. Since one can compute in generalthat ∆(( Q , Q ) UT ) = (det( U )) n ( n − (det( T )) n − ∆( Q , Q ) , the condition for a pair of integral forms Q , Q to be minimized is that thereare no matrices U, T for which ( Q , Q ) UT are integral and such that | det( U ) | nv | det( T ) | v > . We observe for future reference that if | det( U ) | nv | det( T ) | v = 1 (4.2)and Q , Q is a minimized pair of integral forms, then ( Q , Q ) UT will also beminimized, provided of course that the resulting forms are integral.From now on we will restrict to the case n = 8, for which the abovecondition says that there are no suitable T, U with | det( U ) | v | det( T ) | v > . (4.3)Continuing with the notation above we have the following simple criteria fora pair of forms not to be minimized. Lemma 4.2
Suppose that the projective variety q = q = 0 contains a linearspace of projective dimension at least 4 defined over F v . Then the pair Q , Q is not minimized. In particular, if R ( q , q ) ≤ then the pair Q , Q is notminimized. A convenient condition equivalent to the existence of a linear space ofprojective dimension at least 4 is that the forms q and q can be written as q (X¯ ) = ℓ (X¯ ) λ (X¯ ) + ℓ (X¯ ) λ (X¯ ) + ℓ (X¯ ) λ (X¯ ) q (X¯ ) = ℓ (X¯ ) µ (X¯ ) + ℓ (X¯ ) µ (X¯ ) + ℓ (X¯ ) µ (X¯ ) (4.4)17or suitable linear forms ℓ i , λ i , µ i defined over F v .To prove the lemma we assume for simplicity that ℓ , ℓ and ℓ are lin-early independent. In the alternative case we will have a similar argumentinvolving fewer linear forms. We apply a transform T ∈ GL ( O v ) so as toreplace ℓ i (X¯ ) by X i for 1 ≤ i ≤
3. If π is a uniformizing element for k v wecan then write Q ( T X¯ ) = X L (X¯ ) + X L (X¯ ) + X L (X¯ ) + πQ (X¯ )for suitable linear forms L i and a quadratic form Q , all defined over O v .There is also an analogous expression for Q . We now define T = diag( π, π, π, , , , , T = T T , so that | det( T ) | v = | π | v . We then see that both Q ( T X¯ )and Q ( T X¯ ) are divisible by π . Hence if we take U = diag( π − , π − ) then( Q , Q ) UT is a pair of integral forms. However since | det( U ) | v = | π | − v thecondition (4.3) is satisfied. Thus the pair Q , Q is not minimized.When R ( q , q ) ≤ q and q in terms of just 3 variables X , X , X so that the 4-plane X = X = X = 0 is contained in the variety q = q = 0. Lemma 4.2 then follows.There is a further instance in which one can see that a pair of forms isnot minimized, given by the following lemma. Lemma 4.3
Suppose that R ( q , q ) = R ≤ and that the forms Q , Q takethe shape Q i ( X , . . . , X ) = G i ( X , . . . , X R ) + π R X j =1 X j L ( i ) j ( X R +1 , . . . , X )+ πH i ( X R +1 , . . . , X ) (4.5) for i = 1 , , with appropriate quadratic forms G i , H i and linear forms L ( i ) j ,all defined over O v . Then if H and H have a common zero over F v the pair Q , Q is not minimized. To prove this we make a change of variables among X R +1 , . . . , X so asto suppose that H (0 , . . . , ,
1) = H (0 , . . . , ,
1) = 0 . One then sets T = diag( π, . . . , π,
1) and U = diag( π − , π − ). These satisfy(4.3) and produce a pair of integral forms ( Q , Q ) UT . Thus Q and Q cannotbe minimized. 18 The Case r ≤ Lemma 4.2 shows that a minimized pair of forms cannot have R ( q , q ) ≤ R ( q , q ) and r ( q , q ). In this section we examine thepossibility that r ( q , q ) ≤
4. We begin by proving the following result.
Lemma 5.1
Suppose that Q , Q is a minimized pair of quadratic forms in8 variables, with a common non-trivial zero over k v . Then R ( q , q ) = 4 . For the proof we argue by contradiction. If R ( q , q ) = 4 there is a changeof variables in GL ( O v ) so that q and q are functions of X , . . . , X only.Thus we may write Q and Q in the shape (4.5) with R = 4. In view ofLemma 4.3 we know that H and H cannot have a common zero over F v . Wenow claim similarly that the reductions G and G cannot have a commonzero over F v . If they did, then after a change of variables in GL ( O v ) wecould assume that G (1 , , ,
0) = G (1 , , ,
0) = 0 . Taking T = diag(1 , π, π, π, , , ,
1) it would then follow that both the forms Q ( T X¯ ) and Q ( T X¯ ) are divisible by π . Hence if U = diag( π − , π − ) wewould find that ( Q , Q ) UT is a pair of integral forms. However the condition(4.3) is satisfied, since det( T ) = | π | v and det( U ) = | π | − v . Since Q and Q are minimized this contradiction proves our claim.We therefore conclude that if the pair Q , Q is minimized then neither G = G = 0 nor H = H = 0 can have a non-trivial solution over F v .However it is then easy to see that Q = Q = 0 cannot have a non-trivialzero over k v , giving us the desired contradiction. This completes our proofof Lemma 5.1.We are now ready to show that the case r ( q , q ) ≤ Lemma 5.2
Suppose that F v ≥ and that Q , Q is a minimized pair ofquadratic forms in 8 variables, with a common non-trivial zero over k v . Then r ( q , q ) ≥ . We argue by contradiction. We know from Lemmas 4.2 and 5.1 that R = R ( q , q ) ≥
5. Thus if r = r ( q , q ) ≤ r < R with r even, or r ≤ R −
2. We may therefore deduce from Lemma 3.5 that q = q ( X , . . . , X R − ) + X R − ℓ ( X , . . . , X R − )19nd q = q ( X , . . . , X R − ) + X R − X R with r ( q , q ) = r − ≤ R ( q , q ) ≤ R −
2. If R ( q , q ) ≤ q = q ′ ( X , X ) + X R − ℓ ( X , . . . , X R − )and q = q ′ ( X , X ) + X R − X R . These are in the form (4.4) and hence the pair Q , Q cannot be minimized,by Lemma 4.2. This contradiction shows that R ( q , q ) ≥ q and q . Since r ′ = r ( q , q ) ≤ R ′ = R ( q , q ) ≥ r ′ < R ′ with r ′ even, or r ′ ≤ R ′ −
2. In either case we deduce from Lemma3.5 firstly that q can be put into the shape (3.4), then that r ′ = 2 and finallythat q = X R ′ − ℓ ′ ( X , . . . , X R ′ − ) and q = X R ′ − X ′ R . It follows that q and q can be put into the form (4.4) (indeed with two terms on the right, ratherthan three). As before, Lemma 4.2 implies that Q , Q cannot be minimized,giving the required contradiction. This proves the lemma. R = 5 So far we have shown that certain small values of R ( q , q ) or r ( q , q ) areimpossible. For large values we will show that the pencil < Q , Q > containsa form which splits off three hyperbolic planes. We have already remarkedin Lemma 4.1 that this is the case when r ( q , q ) ≥
7. In this section we dealwith the case R = 5. Lemma 6.1
Suppose that F v ≥ . Then if R ( q , q ) = 5 there is at leastone form in the pencil < Q , Q > which splits off three hyperbolic planes. We begin by writing our forms in the shape (4.5) with R = 5. We knowfrom Lemma 5.2 that if R = 5 we will also have r = 5, so that if g i = G i for i = 1 ,
2, then r ( g , g ) = 5.We claim that if h i = H i for i = 1 ,
2, then R ( h , h ) = r ( h , h ) = 3.To prove this, suppose firstly that R ( h , h ) ≤
2. Then h , h would havea common zero over F v and Lemma 4.3 would contradict the minimality of20he pair Q , Q . On the other hand, if R ( h , h ) = 3 and r = r ( h , h ) ≤ r is even or r ≤ R ( h , h ) −
2. We may therefore apply Lemma3.5. Here the forms corresponding to q and q must vanish identically since r ≤
2. It follows that h and h have a common zero at (0 , , R ( h , h ) = r ( h , h ) = 3, as claimed.Since r ( g , g ) = R ( g , g ) = 5, the form g + tg cannot be singular for allvalues of t . However its determinant (or half-determinant) is a polynomialof degree at most 5, and we therefore see that g + tg is singular for at most5 values of t . Similarly we find that h + th is singular for at most 3 valuesof t . Since F v > g + tg has rank5 and h + th has rank 3. Let τ be any lift of t to O v , and let Q = Q + τ Q so that Q ( X , . . . , X ) = G ( X , . . . , X ) + π X i =1 X i L i ( X , X , X ) + πH ( X , X , X )with rank( g ) = 5 and rank( h ) = 3. We may then write g ( X , . . . , X ) = X X + X X + cX and h ( X , X , X ) = X X + c ′ X after a suitable change of variables.Thus Q is in the appropriate shape to apply Lemma 2.2 with s = 2 and e Q ( X , . . . , X ) = cX + πX L ( X , X , X ) + πH ( X , X , X )for a suitable linear form L ( X , X , X ). We see from Lemma 2.2 that Q splits off two hyperbolic planes and so it remains to show that Q splits offat least one hyperbolic plane. Let J ( X , . . . , X ) = π − Q ( πX , X , X , X ).Then J has coefficients in O v and satisfies J ≡ H (mod π ). Now, since h ( X , X , X ) = X X + c ′ X we are able to make a second application ofLemma 2.2 to show that J splits off a hyperbolic plane. Lemma 6.1 thenfollows. R = 8 In this section we prove the following result.21 emma 7.1
Suppose that F v ≥ . Let Q , Q be a minimized pair of formsin 8 variables, with a common zero over k v . Then if R ( q , q ) = 8 there isat least one form in the pencil < Q , Q > which splits off three hyperbolicplanes. The result is an immediate consequence of Lemma 4.1 unless r ( q , q ) ≤ r = r ( q , q ). ThenLemma 3.5 will apply, giving us representations q = q ( X , . . . , X ) + X ℓ ( X , . . . , X )and q = q ( X , . . . , X ) + X X with rank( q ) = r ( q , q ) = r − ≤ ≤ R ( q , q ) ≤
6. Hence we mayapply Lemma 3.5 a second time, leading to expressions q = q ( X , . . . , X ) + X ℓ ′ ( X , . . . , X ) + X ℓ ( X , . . . , X )and q = q ( X , . . . , X ) + X X + X X with rank( q ) = r ( q , q ) ≤
2. If R ( q , q ) ≥ r ( q , q ) = 2 and producing q = X ℓ ′′ ( X , X ) + X ℓ ′ ( X , . . . , X ) + X ℓ ( X , . . . , X )and q = X X + X X + X X . However the forms q , q would then be in the shape (4.4), so that, accord-ing to Lemma 4.2, the pair Q , Q could not have been minimized. Thiscontradiction shows that R ( q , q ) ≤
2, allowing us to write q = q ( X , X ) + X ℓ ′ ( X , . . . , X ) + X ℓ ( X , . . . , X )and q = q ( X , X ) + X X + X X , after a suitable change of variables. It is now apparent that, in order to have R ( q , q ) = 8, the linear forms X , X , X , X , X , X , ℓ ′ ( X , . . . , X ) and ℓ ( X , . . . , X )22ust be linearly independent. We may therefore write q = q ( X , X ) + X X + X X q = q ( X , X ) + X X + X X , (7.1)after a further change of variables.To complete the argument we now call on the following result, which willbe used repeatedly in the rest of the proof of Theorem 2. Lemma 7.2
Let Q ( X , . . . , X ) and Q ( X , . . . , X ) be forms over O v , notnecessarily minimized. Suppose that Q ( X , X , X , X , X , , , ≡ X X + X X ( mod π ) and that π ∤ Q (0 , , , , , , , . Then there is at least one form in the pencil < Q , Q > which splits off threehyperbolic planes. We shall prove the lemma in a moment, but first we demonstrate howit may be used to complete our proof of Lemma 7.1. It is not possible forboth the forms q and q in (7.1) to vanish identically, since R ( q , q ) = 8.Moreover, if both q and q were merely multiples of X X at least one of q or q would be a sum of three hyperbolic planes, in which case an applicationof Lemma 2.2 completes the proof. We may therefore assume that q , say,contains a non-zero term in X . We now replace Q by Q + cQ for a suitable c ∈ O v and substitute X and X for X + cX and X + cX respectively.This enables us to assume that q (1 ,
0) = 0. Then, re-labelling the variables,we may write q = X X + X X + q ( X , X ) q = X X + X X + q ( X , X ) , with q (1 ,
0) = 0 and q (1 , = 0. Thus Lemma 7.2 applies to Q , Q , andcompletes the proof of Lemma 7.1It therefore remains to establish Lemma 7.2. It will be convenient towrite S i ( X , X , X , X , X ) = Q i ( X , X , X , X , X , , , , ( i = 1 , .
23e claim that there exists λ ∈ O v and T ∈ GL ( O v ) such that T (0 , , , ,
1) = (0 , , , , S − λS )( T X¯ ) = X X + X X . (7.2)In particular S − λS will be singular. Since O v and GL ( O v ) are compactit will suffice to show that for every positive integer f there are suitable λ and T such that( S − λS )( T X¯ ) ≡ X X + X X (mod π f ) . (7.3)We will prove this by induction on f , the case f = 1 being handled by thehypotheses of the lemma.We therefore suppose that (7.2) holds for some particular f and showhow to derive a corresponding statement with exponent f + 1. It will beconvenient to write S (X¯ ) = ( S − λS )( T X¯ )= X X + X X + π f S ′ ( X , . . . , X )+ π f L ( X , . . . , X ) X + π f cX and U (X¯ ) = S ( T X¯ ) = U ( X , X , X , X ) + M ( X , . . . , X ) X + dX , Since T (0 , , , ,
1) = (0 , , , ,
1) we have U (0 , , , ,
1) = S (0 , , , , π ∤ d .We now examine S − π f cd − U , which will have coefficients in O v . Byconstruction this form will have no term in X . Let( S − π f cd − U )( X , X , X , X ,
0) = V ( X , X , X , X )= X X + X X + π f S ′ ( X , . . . , X ) − π f cd − U ( X , X , X , X , , say, so that V ( X , X , X , X ) ≡ X X + X X (mod π ). We therefore seethat Lemma 2.2 applies, producing a transform in T ∈ GL ( O v ) such that V ( T X¯ ) = X X + X X . Thus there is an admissible T ∈ GL ( O v ) with( S − π f cd − U )( T X¯ ) = X X + X X + π f L ′ ( X , . . . , X ) X , L ′ ( X , . . . , X ) = L (cid:0) T ( X , . . . , X ) (cid:1) − cd − M (cid:0) T ( X , . . . , X ) (cid:1) . We now make a further change of variable, of the shape X i → X i + π f γ i X (1 ≤ i ≤ S − π f cd − U into the form( S − π f cd − U )( T X¯ ) = X X + X X + π f c ′ X . Since 2 f > f this establishes (7.3) with exponent f + 1, and with a new value λ + π f cd − in place of λ . We have therefore completed the induction step,thereby establishing the claim that we can choose λ so that (7.2) holds.Returning to the statement of Lemma 7.2 we now see that there is achange of variable putting Q − λQ into the shape X X + X X + X L ( X , . . . , X ) + X L ( X , . . . , X ) + X L ( X , . . . , X ) . If rank( Q − λQ ) = 8 then X must appear in at least one of the forms L i .In this case after a further change of variable we can represent Q − λQ as X X + X X + X X + X L ′ ( X , . . . , X ) + X L ′ ( X , . . . , X ) . Another change of variable, of the form X i → X i + µ i X + ν i X , (1 ≤ i ≤ Q − λQ = X X + X X + X X + X L ′′ ( X , X ) + X L ′′ ( X , X ) , giving us a form which splits off three hyperbolic planes.It remains to consider the possibility that Q − λQ is singular. Here weuse an argument shown to the author by Colliot-Th´el`ene. We recall fromLemma 3.1 that the form F ( x, y ) = det( xM ( Q ) + yM ( Q )) cannot vanish,and will have distinct factors. However F (1 , − λ ) = det( Q − λQ ) = 0 , F ( x, y ) = ( λx + y ) G ( x, y )for some form G ( x, y ) ∈ k v [ x, y ] with G (1 , − λ ) = 0. Suppose that we have v ( G (1 , − λ )) = e , so that π − e G (1 , − λ ) is a unit in O v . Then if r is a largeenough integer we will have v ( G (1 , π r − λ )) = e . Moreover F (1 , π r − λ ) = π r G (1 , π r − λ ) , whence v ( F (1 , π r − λ )) will be odd for any sufficiently large integer r ofopposite parity to e . In particular F (1 , π r − λ ) is not a square in k v forsuch a choice of r , so that Q + ( π r − λ ) Q is nonsingular and containsthree hyperbolic planes, by virtue of Lemma 2.3. This completes the proofof Lemma 7.2.We end this section by giving a useful corollary to Lemma 7.2. Lemma 7.3
Let Q ( X , . . . , X ) and Q ( X , . . . , X ) be forms over O v , notnecessarily minimized. Suppose that Q ( X , X , X , X , X , , , ≡ X X + X X ( mod π ) and Q ( X , X , X , X , X , , , ≡ Q ′ ( X , X , X , X ) ( mod π ) for some quaternary form Q ′ . Suppose further that π | Q (0 , , , , , , , and π ∤ Q (0 , , , , , , , Then there is at least one form in the pencil < Q , Q > which splits off threehyperbolic planes. If π ∤ Q (0 , , , , , , ,
0) this follows at once from Lemma 7.2. Otherwisewe define ( Q , Q ) UT = ( V , V ) with T = diag( π, π, π, π, , π , π , π ) and U = diag( π − , π − ). Then V , V are integral forms. Moreover we have V (X¯ ) ≡ X X + X X + L ( X , X , X , X ) X + cX (mod π )26or some linear form L and some c ∈ O v , while V (X¯ ) ≡ dX (mod π )for some unit d ∈ O v . A suitable substitution X i → X i + c i X for 1 ≤ i ≤ V so that V (X¯ ) ≡ X X + X X + c ′ X (mod π )say, while leaving V (X¯ ) ≡ dX (mod π ). We may then apply Lemma 7.2 tothe forms V − c ′ d − V and V , and Lemma 7.3 follows. rank( q ) ≤ — First Steps In the next two sections we examine the case in which q has small rank.This turns out to be an important prelude to our treatment of pairs forwhich R ( q , q ) = 6. Our goal is the following result. Lemma 8.1
Suppose that F v ≥ . Let Q , Q be a minimized pair of formswith a common zero over k v . Then if rank( q ) ≤ there is at least one formin the pencil < Q , Q > which splits off three hyperbolic planes. However in the present section we will content ourselves with the followingintermediate statement.
Lemma 8.2
Suppose that F v ≥ . Let Q , Q be a minimized pair of formsover k v such that rank( q ) ≤ . Suppose indeed that q ( X¯ ) = q ( X , X ) .Then either there is at least one form in the pencil < Q , Q > which splitsoff three hyperbolic planes, or q is an anisotropic form of rank 2, and thereis a linear change of variable in GL ( O v ) , involving only X , . . . , X whichmakes q ( X¯ ) = X X + X X + n ( X , X ) . (8.1)We begin by observing that we must have rank( q ) = 2. Indeed we willbe able to write q (X¯ ) = n ( X , X ) for some anisotropic form n , after asuitable change of variables. To see this we notice that in all other cases wecan write q as a product of linear factors over F v . Thus, after a change ofvariable we can take q (X¯ ) = X ℓ (X¯ ). The transforms T = diag( π, , . . . , U = ( π − ,
1) then make ( Q , Q ) UT integral, but would satisfy (4.3),contradicting the minimality of the pair Q , Q .In view of Lemma 4.1 we can assume that r = r ( q , q ) ≤
6. Similarly,in view of Lemmas 4.2, 5.1 and 6.1 we may suppose that R = R ( q , q ) ≥ q may take. With q (X¯ ) = n ( X , X ), we write q (X¯ ) = q ( X , X ) + X ℓ ( X , . . . , X ) + X ℓ ( X , . . . , X ) + q ( X , . . . , X )for appropriate quadratic forms q i and linear forms ℓ i . Let rank( q ) = m ,say, and change variables so as to write q (X¯ ) = q ( X , X ) + X ℓ ′ ( X , . . . , X m +2 ) + X ℓ ′ ( X , . . . , X m +2 )+ X ℓ ′′ ( X m +3 , . . . , X ) + X ℓ ′′ ( X m +3 , . . . , X )+ q ( X , . . . , X m +2 ) (8.2)with rank( q ) = m . Our analysis now splits into 3 cases in which the linearforms ℓ ′′ , ℓ ′′ both vanish, or are linearly dependent but do not both vanish,or are linearly independent. If ℓ ′′ and ℓ ′′ both vanish identically then R ( q , q ) ≤ m + 2. Since we areassuming that R ≥ m ≥
4. On the other hand, if χ ( F v ) = 2,or m is even, the determinant of q + tq (considered as a quadratic form in m + 2 variables) is a polynomial in t of degree at most m + 2, in which theterm in t has coefficient det( n ) det( q ) = 0. It follows that r ( q , q ) ≥ m + 2.Similarly if χ ( F v ) = 2 and m is odd, the half-determinant of q + tq is apolynomial in which the coefficient of t is det( n )det / ( q ) = 0, and again weconclude that r ( q , q ) ≥ m + 2. We are supposing that r ≤
6, so we musthave m ≤ m = 4, in which situation achange of variables will allow us to put q into one of the shapes X X + X X or X X + n ( X , X ). We can now use further linear transformations, of thetype X i → X i + λ i ( X , X ) (3 ≤ i ≤ q into one of the shapes q (X¯ ) = q ( X , X ) + X X + X X (8.3)28r q (X¯ ) = q ( X , X ) + X X + n ( X , X ) . (8.4)We now call on the following lemma, which we shall prove shortly. Lemma 8.3
Let s ( X, Y ) and s ( X, Y ) be quadratic forms over a finite field F . Suppose that s and s have no common factor and that r ( s , s ) = 2 .Then there are at least ( F − pairs a, b ∈ F , not both zero, for which as + bs is a hyperbolic plane, and at least ( F − such pairs for whichit is anisotropic of rank 2. It is clear that r ( q , n ) = 2, because rank( n ) = 2. Moreover, since n cannot have a linear factor, the forms q and n are coprime unless q is amultiple of n . Suppose firstly that q is not a multiple of n . In Lemma 8.3we must have ( F v − > F v − F v ≥
9, and we therefore concludethat there are linear combinations aq + bn with a = 0 which are hyperbolicplanes, and also which are anisotropic of rank 2. We may then deduce thatthere is a linear combination q ′ = q + cq taking one of the forms X X + X X + X X , or n ( X , X ) + X X + n ( X , X ) , in the two cases given by (8.3) and (8.4). As explained in Section 2, if q ′ (X¯ ) = n ( X , X ) + X X + n ( X , X ) then a change of variable will make q ′ a sum of 3 hyperbolic planes, and so in either case Lemma 8.2 follows fromLemma 2.2.In the alternative situation in which q is a multiple of n , there is a linearcombination q ′ = q + cq taking one of the forms X X + X X , or n ( X , X ) + X X + n ( X , X ) , in the two cases given by (8.3) and (8.4) respectively. As above, in the secondof these cases q ′ is a sum of 3 hyperbolic planes, and again we may completethe proof of Lemma 8.2. We may therefore reduce our considerations to thecase in which q (X¯ ) = X X + X X , and q (X¯ ) = n ( X , X ) . However in this situation we may apply Lemma 7.2 after re-ordering thevariables, since π ∤ n (1 , F = N temporarily. Let S be the set of quadruples ( a, b, x, y ) ∈ F forwhich ( a, b ) = (0 ,
0) and ( x, y ) = (0 ,
0) and such that as ( x, y )+ bs ( x, y ) = 0.Since s and s have no factor in common there is no ( x, y ) = (0 ,
0) such that s ( x, y ) = s ( x, y ) = 0. Thus there are N − a, b ) for each admissiblepair ( x, y ). It follows that S = ( N − N − . On the other hand, if ( a, b ) is a pair such that as ( X, Y ) + bs ( X, Y ) is ahyperbolic plane then there are 2( N −
1) corresponding pairs ( x, y ), while if as ( X, Y )+ bs ( X, Y ) is anisotropic of rank 2 there are none. Moreover, in theremaining case, in which as ( X, Y ) + bs ( X, Y ) has a repeated linear factor,there are N − x, y . Suppose that the hyperbolic plane, anisotropic,and repeated factor cases occur N h , N a and N r times each respectively. Then N h + N a + N r = N − N − N h + ( N − N r = S = ( N − N − . (8.6)However the determinant of as ( X, Y ) + bs ( X, Y ) is a binary quadratic formin a and b which does not vanish identically, since r ( s , s ) = 2. It followsthat N r ≤ N − N h = N − − N r ≥ N − − N − N a = 2( N − − N h − N r ) = N − − N r ≥ N − − N − . These inequalities suffice for the lemma.
The next case is that in which ℓ ′′ and ℓ ′′ are not both zero but are linearlydependent. After a change of variable between X and X we may supposethat ℓ ′′ = 0, and then, after a further change of variable among X m +3 , . . . , X we may assume that ℓ ′′ = X m +3 . It follows that R ( q , q ) ≤ m + 3. Since we30re assuming that R ≥ m ≥
3. The representation (8.2)then becomes q (X¯ ) = q ( X , X ) + X ℓ ( X , . . . , X m +2 ) + X ℓ ( X , . . . , X m +2 )+ X X m +3 + q ( X , . . . , X m +2 ) . (8.7)(Our numbering of forms q i and ℓ i will be independent of that used forCase 1.) Thus if the coefficients of X in q and q are a and b = 0 respectively,we may write q + tq as q + tq = q ( X , . . . , X m +2 ) + X ( X m +3 + ℓ ( X , . . . , X m +2 ; t )) (8.8)for some linear form ℓ ( X , . . . , X m +2 ; t ) depending on t , where q = ( a + tb ) X + X ℓ ( X , . . . , X m +2 ) + q ( X , . . . , X m +2 ) . Then rank( q ) ≥ rank( q ) = m and so r ( q , q ) ≥ rank( q ) + 2 ≥ m + 2, by(2.3). It follows that m ≤
4. In fact, if m = 4 then the half-determinant of q is a polynomial in t whose linear term has a coefficient b det( q ) = 0. Thusthere is some value of t for which q has rank 5. This however would implythat q + tq , given by (8.8), has rank 7, contradicting our assumption that r ( q , q ) ≤
6. We must therefore have m ≤
3, and indeed we can deduce that m = 3, since 6 ≤ R ( q , q ) ≤ m + 3.After a change of variable amongst X , X , X we can now take q ( X , X , X ) = X X + X , whence (8.7) becomes q (X¯ ) = q ( X , X ) + X ℓ ( X , X , X ) + X ℓ ( X , X , X )+ X X + X X + X . Further transforms of the type X → X + λX , X → X + µX and X → X + ℓ ( X , . . . , X ) simplify this to q (X¯ ) = cX + dX X + X X + X X + X for certain c, d ∈ F v . Now, with the same notation b for the coefficient of X in q = n ( X , X ) we consider q = q − cb − q , which takes the shape q = X ( X + ℓ ( X , X )) + dX X + X X + X . X + ℓ ( X , X ) by X we finally see that we can replace ouroriginal forms Q , Q by a pair Q, Q for which, after a change of variablesin GL ( O v ), we have q = X X + dX X + X X + X and q = n ( X , X ) . We may now re-label the variables so that q = X X + X X + dX X + X and q = n ( X , X ). Thus we can apply Lemma 7.2 to complete the proof. If ℓ ′′ and ℓ ′′ are linearly independent we may take them to be X m +3 and X m +4 respectively, after an appropriate change of variable. The representation (8.2)then becomes q (X¯ ) = q ( X , X ) + X ℓ ′ ( X , . . . , X m +2 ) + X ℓ ′ ( X , . . . , X m +2 )+ X X m +3 + X X m +4 + q ( X , . . . , X m +2 ) , which simplifies to q (X¯ ) = X X m +3 + X X m +4 + q ( X , . . . , X m +2 )after a substitution of the form X m +3 → X m +3 + ℓ ( X , . . . , X m +2 ) , X m +4 → X m +4 + ℓ ( X , . . . , X m +2 ) . It is clear that R ( q , q ) ≤ m + 4. Since we are assuming that R ≥ m ≥
2. Thus q will split off a hyperbolic plane unless m = 2and q is anisotropic. Thus either q is a sum of three hyperbolic planes, orwe have q (X¯ ) = X X + X X + n ( X , X ) . Thus one or other of the conclusions in Lemma 8.2 will hold. rank( q ) ≤ We proceed with the proof of Lemma 8.1. Clearly we may assume that q takes the form (8.1) and that q (X¯ ) = n ( X , X ). We begin by setting S i ( X , X ) = π − Q i (0 , , , , , , X , X ) , ( i = 1 , . (9.1)32hese forms are defined over O v since q and q contain no terms in X , X .Suppose firstly that S does not vanish modulo π . In this case we can makea change of variables in X and X so as to suppose that π ∤ S (1 , λ = S (1 , S (1 , − , we replace Q by Q = Q − λQ , which willcontain no term in X . We have q (X¯ ) = X X + X X + n ( X , X ) − λn ( X , X ) , so that a change of variable, of the type X → X + ℓ ( X , X ) , X → X + ℓ ′ ( X , X )puts q (X¯ ) into the shape X X + X X + n ( X , X ) while having no ef-fect on q (X¯ ) = n ( X , X ). These manoeuvres allow us in effect to as-sume that Q contains no term in X . If we now relabel the variables sothat X , X , X , X , X become X , X , X , X , X respectively we see thatLemma 7.3 applies, showing that the pencil contains a form splitting off threehyperbolic planes.We can therefore assume that S vanishes modulo π . Next, unless S isanisotropic of rank 2 it will have a zero over F v , so that Lemma 4.3 may beapplied. Since this would contradict the minimality of the pair Q , Q wemay therefore assume that S ( X , X ) ≡ N ( X , X ) , S ( X , X ) ≡ π ) . We now consider the pair of forms ( Q ′ , Q ′ ) = ( Q , Q ) UT , where T = diag( π, π, , . . . ,
1) and U = diag(1 , π − ) . Then Q ′ and Q ′ have coefficients in O v , and T and U satisfy (4.2). Sincethe pair Q , Q is minimized we then deduce that Q ′ , Q ′ is also minimized.However q ′ = n ( X , X ) so that Lemma 8.2 applies. We therefore concludethat the pencil generated by Q ′ and Q ′ contains a form which splits off 3hyperbolic planes, except possibly when q ′ takes the shape X ℓ + X ℓ + n ( ℓ , ℓ ) (9.2)for linearly independent linear forms ℓ i ( X , X , X , X , X , X ). However q ′ contains no terms in X or X , by construction. Since n is anisotropic we seethat ℓ and ℓ must be independent of X and X , and then we deduce that33 and ℓ must also be independent of X and X . Similarly we observe that q ′ contains no quadratic terms in X and X , by construction, whence ℓ and ℓ cannot involve X or X . Thus ℓ and ℓ are functions of X , . . . , X ,while ℓ and ℓ are functions of X and X alone. Moreover ℓ and ℓ arelinearly independent.We are finally in a position to complete the treatment of Lemma 8.1. Theconsiderations above have shown that if q ′ is given by (9.2) we may assumethat our forms take the shape Q (X¯ ) ≡ X X + X X + N ( X , X )+ π { c Q ( X , . . . , X ) + X M ( X , . . . , X )+ X M ( X , . . . , X ) + N ( X , X ) } (mod π )and Q (X¯ ) ≡ N ( X , X ) + π { X L ( X , . . . , X ) + X L ( X , . . . , X )+ X L ( X , . . . , X ) + X L ( X , . . . , X )+ N (cid:0) L ( X , X ) , L ( X , X ) (cid:1) } (mod π ) , where L , . . . , L are lifts of ℓ , . . . , ℓ . According to our hypotheses there isa non-trivial point x¯ ∈ k v for which Q = Q = 0. By a suitable rescalingwe may suppose that x¯ has entries in O v , at least one of which is a unit.Since π | Q (x¯) we deduce that π | N ( x , x ), whence π | x , x . Then, since π | Q (x¯) we deduce that π | N ( x , x ), leading to π | x , x . We next usethe fact that π | Q (x¯), which shows that π | N (cid:0) L ( X , X ) , L ( X , X ) (cid:1) .This implies π | L ( x , x ) , L ( x , x ), and since the forms L and L areindependent modulo π we find that π | x , x . Finally, since π | Q (x¯) weobtain π | N ( x , x ), leading to π | x , x . We therefore have π | x¯, contraryto our assumption. Hence it is impossible for q ′ to take the shape (9.2). Wetherefore conclude that in every case the pencil contains a form which splitsoff 3 hyperbolic planes. This concludes the proof of Lemma 8.1.
10 The Case R = 6 We next suppose that R ( q , q ) = 6, and prove the following result. Lemma 10.1
Suppose that F v ≥ . Let Q , Q be a minimized pair offorms with a common zero over k v . Then if R ( q , q ) = 6 there is at leastone form in the pencil < Q , Q > which splits off three hyperbolic planes.
34n view of Lemma 5.2 we may assume that r ( q , q ) ≥
5. Since we have R ( q , q ) = 6 we may write q i (X¯ ) = q ′ i ( X , . . . , X ) for i = 1 ,
2. We nowconsider the forms S i given by (9.1). These will have coefficients in O v , sothat we may define s i as the reduction S i over F v . The forms s and s cannot have a common zero over F v by Lemma 4.3.We first consider the possibility that s and s are proportional. Then byconsidering appropriate linear combinations of Q and Q we may supposeindeed that s = 0. We now examine the forms ( V , V ) = ( Q , Q ) UT where T = diag( π, . . . , π, ,
1) and U = diag( π − , π − ). The forms V , V will beintegral, while T and U satisfy (4.2). Thus V and V are minimized. More-over, if v is the reduction of V to F v then rank( v ) ≤
2. It then followsfrom Lemma 8.1 that the pencil < V , V > contains a form which splits offthree hyperbolic planes, and the required conclusion then follows, subject toour assumption that s and s are proportional.If r ( s , s ) = 1, then every non-trivial linear combination has rank atmost one, and hence has a zero. Otherwise, since we are now supposingthat s and s are not proportional and have no common zero, Lemma 8.3shows that there are at least ( F v − non-trivial linear combinations as + bs with a non-trivial zero. In contrast, at most 6( F v −
1) non-triviallinear combinations aq + bq can have rank smaller than r ( q , q ), as onesees by considering the corresponding determinant, or half determinant, asa binary form in a and b . It follows that if F v >
13 then there is at leastone form aq + bq with rank at least 5, such that as + bs has a non-trivial zero. By a change of variable between X and X we may arrangethat as + bs vanishes at (1 , a = 0. Then s (1 ,
0) cannot vanish, since (1 ,
0) cannot be a common zero of s and s .Thus, on replacing Q by aQ + bQ we may suppose that rank( q ) ≥
5, that s (1 ,
0) = 0 and that s (1 , = 0. Since rank( q ) ≥ q splitsoff at least two hyperbolic planes, whence we may make a change of variableso as to put q into the form X X + X X + q ( X , X ). On interchangingthe variables X and X we now see that our forms are in the correct shapefor an application of Lemma 7.3, which provides a suitable form splitting off3 hyperbolic planes. 35 R = 7 We turn now to the final case, in which R ( q , q ) = 7. Lemma 11.1
Suppose that F v ≥ . Let Q , Q be a minimized pair offorms with a common zero over k v . Then if R ( q , q ) = 7 there is at leastone form in the pencil < Q , Q > which splits off three hyperbolic planes. Our argument will depend on the following result, which we prove at theend of this section.
Lemma 11.2
Let q = Y Y + s ( Y , Y , Y ) and q = Y Y + s ( Y , Y , Y ) bequadratic forms over a finite field F v with F v ≥ . Suppose that q (0 , , and q (0 , , are not both zero. Then either the forms have a singular com-mon zero over F v , or there are at least F v − non-zero pairs ( a, b ) ∈ F v for which aq + bq is a sum of two hyperbolic planes. We now present our proof of Lemma 11.1. In view of Lemmas 4.1 and 5.2we may restrict attention to the cases r = 5 and r = 6. Hence Lemma 3.5shows that we may take q (X¯ ) = q ( X , . . . , X ) + X ℓ ( X , . . . , X )and q (X¯ ) = q ( X , . . . , X ) + X X with rank( q ) = r ( q , q ) = r − R ( q , q ) = 4 or 5.We begin by considering the easy case in which R ( q , q ) = 5. Herewe may make a second application of Lemma 3.5 and a further change ofvariables so as to write q (X¯ ) = q ( X , X , X ) + X ℓ ′ ( X , . . . , X ) + X ℓ ( X , . . . , X )and q (X¯ ) = q ( X , X , X ) + X X + X X with rank( q ) = r ( q , q ) = r − R ( q , q ) = 3 Lemma 3.5 wouldshow that q and q have a common linear factor. This however would allow36 and q to be written in the form (4.4), contradicting the minimality of thepair Q , Q . Thus R ( q , q ) ≤
2, so that we can write q (X¯ ) = q ( X , X ) + X ℓ ′ ( X , . . . , X ) + X ℓ ( X , . . . , X )and q (X¯ ) = q ( X , X ) + X X + X X . If we now set T = diag( π, π, , π, , π, ,
1) and U = diag( π − , π − ) we seethat U and T satisfy (4.2), and that the forms ( Q , Q ) UT = ( V , V ), say, areintegral. Thus V and V are also minimized. However one readily checksthat if v and v are the reductions of V and V to F v then R ( v , v ) ≤ v and v contain no terms in X and X . Thus our previous resultsshow that the pencil < V , V > contains a form splitting off three hyperbolicplanes, and this suffices for the lemma.We may therefore restrict our attention to the case in which R ( q , q ) = 4and r ( q , q ) = 3 or 4. Thus we may assume that q (X¯ ) = q ( X , . . . , X ) + X ℓ ( X , . . . , X )and q (X¯ ) = q ( X , . . . , X ) + X X . Since R ( q , q ) = 7 the variable X must genuinely occur in ℓ ( X , . . . , X )and so we can make a change of variable to obtain q (X¯ ) = q ( X , . . . , X ) + X X and q (X¯ ) = q ( X , . . . , X ) + X X . We proceed to write Q (X¯ ) = Q ( X , . . . , X ) + X X + π { Q ( X , . . . , X ) + X i =5 X i L (1) i ( X , . . . , X ) } and similarly Q (X¯ ) = Q ( X , . . . , X ) + X X + π { Q ( X , . . . , X ) + X i =5 X i L (2) i ( X , . . . , X ) } , Q , . . . , Q and linear forms L ( j ) i suchthat the reductions Q and Q are q and q respectively. We note that π cannot divide both Q (0 , , ,
1) and Q (0 , , , G , G ) = ( Q , Q ) UT , where T = diag( π, π, π, π, , π, ,
1) and U = diag( π − , π − ) . (11.1)Then G (X¯ ) = πQ ( X , . . . , X ) + H ( X , . . . , X ) + π X i =5 X i M (1) i (X¯ )and G (X¯ ) = πQ ( X , . . . , X ) + H ( X , . . . , X ) + π X i =5 X i M (2) i (X¯ )for appropriate linear forms M ( j ) i , with H ( X , . . . , X ) = X X + Q ( X , , X , X )and H ( X , . . . , X ) = X X + Q ( X , , X , X ) . Thus G and G will be integral forms, but since | det( U ) | v | det( T ) | v < G , G will not be minimized. If we denote the reductions of H and H by h and h respectively we see that they are in the right shape for anapplication of Lemma 11.2, after a re-labelling of the variables. In view ofthe alternative conclusions of the lemma there are now two cases to consider. If h and h have a singular common zero over F v , there will be a change ofvariable putting them into the shape h i (X¯ ) = c i X ℓ ( X , X , X ) + w i ( X , X , X ) ( i = 1 , . A further substitution allows us to write h i (X¯ ) = c i X X + w ′ i ( X , X , X ) = X λ i ( X , . . . , X ) + w ′ i (0 , X , X )38or i = 1 , λ , λ . Hence if we apply thetransforms T = diag(1 , , , , , , π, π, π ) and U = diag( π − , π − ) (11.2)the forms ( G , G ) UT = ( V , V ) say, will be integral. However if we label thetransforms (11.1) as T and U , and the transforms (11.2) as T and U , wesee that T T and U U satisfy (4.2), so that V , V must be minimized. Onthe other hand one sees that R ( V , V ) ≤
6. Thus Lemmas 4.2, 5.1, 6.1 and10.1 show that the pencil < V , V > contains a form which splits off threehyperbolic planes, which suffices for the present lemma. Suppose next that there are at least 5( F v −
1) non-zero pairs ( a, b ) ∈ F v for which ah + bh is a sum of two hyperbolic planes. Since r ( q , q ) = 3 or4 there are at most 4( F v −
1) non-zero pairs ( a, b ) ∈ F v for which aq + bq has rank less than 3. Thus there is at least one pair such that ah + bh is asum of two hyperbolic planes and aq + bq splits off a hyperbolic plane. Nowtake a ′ , b ′ ∈ O v to be arbitrary lifts of a and b , and consider G = a ′ G + b ′ G .This will take the shape G (X¯ ) = πG ( X , . . . , X ) + G ( X , . . . , X ) + π X i =5 X i M i (X¯ ) , where G splits off a hyperbolic plane, and G is a sum of two hyperbolicplanes.According to Lemma 2.2 we may make a change of variable so that G becomes X X + X X + πG ( X , . . . , X )with G = G . A second application of Lemma 2.2 then shows that G splitsoff a hyperbolic plane, so that the pencil < G , G > contains the form G which splits off three hyperbolic planes. This establishes Lemma 11.1 in Case2. We end this section by establishing Lemma 11.2. We suppose for the proofthat r ( q , q ) = 4 since otherwise Lemma 3.6 provides a singular commonzero. Without loss of generality we will assume that q (0 , , = 0. We may39hen replace q by q + aq with a suitable value of a , and substitute Y + aY for Y so as to produce a form in which q (0 , ,
1) = 0. We can then write q ( Y , . . . , Y ) = Y (cid:0) Y + λ ( Y , Y , Y ) (cid:1) + Y ℓ ( Y , Y ) . We proceed to substitute Y + λ ( Y , Y , Y ) for Y so that q ( Y , . . . , Y ) = Y Y + Y ℓ ( Y , Y ) . This transformation puts q into the shape q ( Y , . . . , Y ) = Y Y + q ( Y , Y , Y )for some new quadratic form q with q (0 , ,
1) = q (0 , , = 0. Now, forany a ∈ F we have q + aq = Y ( Y + aY ) + q ( Y , Y , Y ) + aY ℓ ( Y , Y )and on substituting Y + aY for Y this becomes Y Y + q ( Y − aY , Y , Y ) + a ( Y − aY ) ℓ ( Y − aY , Y )= (cid:0) Y + µ ( Y , Y , Y ) (cid:1) Y + q ( − aY , Y , Y ) − a Y ℓ ( − aY , Y ) . If det( q + aq ) = 0 this is a sum of two hyperbolic planes provided thatthe binary form q ( − aY , Y , Y ) − a Y ℓ ( − aY , Y ) has a linear factor over F . Since r ( q , q ) = 4 we need to exclude at most 4 values of a for whichdet( q + aq ) vanishes.We claim that, unless the forms q and q have a singular common zeroover F v , there are at least 9 values of a such that the polynomial f a ( U ) = q ( − a, , U ) − a ℓ ( − a, U )has a root u ∈ F v . There will then be at least 5 values with the additionalproperty that det( q + aq ) = 0, and then bq + abq will be a sum of twohyperbolic planes for any non-zero b ∈ F v . Thus the claim suffices for theproof of Lemma 11.2.The coefficient of U in f a ( U ) is q (0 , , = 0 so that f a ( U ) is quadraticin U for every value of a . Since there are at most two roots u for any valueof a it will therefore suffice to show that the curve X : f ( U, V ) = f V ( U ) = 040as at least 18 affine points over F v .We first consider the case in which ℓ does not vanish identically, so that X is a curve of degree 3. If X is absolutely irreducible the number of projectivepoints over F v is at least F v + 1 − √ F v ≥
21, by the Hasse–Weil bound.At most three of these can be at infinity, so that there are at least 18 affinepoints, as required. If X contains a line defined over F v there are at least F v + 1 ≥
33 projective points and we have the same conclusion. Thereremains the possibility that X splits into three cubic conjugate lines. Howeverthis case cannot arise since F ( U, V ) contains no term in U and a non-zeroterm q (0 , , U . This completes the proof in the case in which ℓ does notvanish identically.In the alternative case in which ℓ vanishes identically, q reduces to Y Y and f ( U, V ) becomes q ( − V, , U ). If rank( q ) = 3, the curve X has F v + 1projective points over F v , of which at most two lie at infinity. We may thencomplete the argument as before. Finally, if rank( q ) ≤ q = Y Y + q (cid:0) ℓ ( Y , Y , Y ) , ℓ ( Y , Y , Y ) (cid:1) and q = Y Y for suitable linear forms ℓ , ℓ over F v . If we choose a non-zero point y¯ over F v , such that y = ℓ ( y , y , y ) = ℓ ( y , y , y ) = 0we then see that y¯ is a singular common zero for q and q . This completesthe proof of the lemma in this second case.
12 Global Forms Splitting Off 3 HyperbolicPlanes
Our task now is to deduce Theorem 1 from Theorem 2. In doing this we willbe inspired by the plan outlined by Colliot-Th´el`ene, Sansuc and Swinnerton-Dyer [7, Remark 10.5.3]. However our argument looks somewhat differentfrom that which they proposed.We begin by producing a global statement related to Theorem 2.
Proposition 1
Let k be a number field for which every prime ideal has ab-solute norm at least 32, and let Q ( X , . . . , X ) and Q ( X , . . . , X ) be twoquadratic forms such that the projective variety Q ( X¯ ) = Q ( X¯ ) = 0 is non-singular and has a point over every completion k v of k . Then there exist , b ∈ k such that aQ + bQ has rank 8 and splits off at least 3 hyperbolicplanes. Our first move will be to establish a variant of Theorem 2 for archimedeanvaluations.
Lemma 12.1
Let Q ( X , . . . , X ) and Q ( X , . . . , X ) be quadratic formsover R , such that the projective variety Q ( X¯ ) = Q ( X¯ ) = 0 is nonsingular.Then there is a non-trivial linear combination aQ + bQ which splits off atleast 3 hyperbolic planes. We prove this by adapting an argument of Swinnerton-Dyer [12, § F θ ( z ) = det (cid:0) (sin θ ) Q + (cos θ ) Q − zI (cid:1) . This is a polynomial with real roots. For any open interval I ⊆ R let n I ( θ )denote the number of roots of F θ ( z ), counted according to multiplicity, lyingin I . If I = ( a, b ) is a finite interval, and F θ ( z ) is non-zero at z = a and z = b , we will have n ( a,b ) ( θ ) = 12 πi Z Γ F ′ θ ( z ) F θ ( z ) dz, where Γ is the path from a − i to b − i to b + i to a + i and back to a − i . Thisformula makes it clear that there is a neighbourhood of θ on which n ( a,b ) ( φ )is constant, whenever F θ ( a ) and F θ ( b ) are non-zero.We now write n + ( θ ) , n − ( θ ) , n ( θ ) for the number of roots of F θ ( z ) whichare positive, negative, or zero, respectively. It follows from the above that n + and n − are locally non-decreasing. Moreover, since the variety Q = Q = 0is nonsingular Lemma 3.1 shows that n ( θ ) = 0 or 1. We also observe that n + ( θ ) = n − ( θ + π ) and n − ( θ ) = n + ( θ + π ), so that either n + ( θ ) or n + ( θ + π )must be at least 4. Suppose that n + ( θ ) ≥ θ = sup { ξ ∈ [ θ, θ + π ] : n + ( ξ ) ≥ } . If n + ( θ ) ≥ θ = θ + π , since n + is locally non-decreasing tothe right of θ . This however is impossible since n + ( θ + π ) = n − ( θ ) ≤ − n + ( θ ) ≤ . On the other hand, if n + ( θ ) ≤ n − ( θ ) ≥
3. Thus there is aninterval ( θ − δ, θ ] on which n − ≥
3. However it follows from the definition42f θ as a supremum that there is a point φ ∈ ( θ − δ, θ ] such that n + ( φ ) ≥ (cid:0) n + ( φ ) , n − ( φ ) (cid:1) ≥
3. Thus (sin φ ) Q + (cos φ ) Q has at least 3positive eigenvalues, and at least 3 negative ones, so that it will split off atleast 3 hyperbolic planes over R . This completes the proof of Lemma 12.1.Moving now to our treatment of Proposition 1, we begin by replacing Q and Q by suitable scalar multiples, so that they are defined over O k . Wethen define a set B consisting of all infinite places, all places above 2, and allplaces corresponding to prime ideals dividing ∆( Q , Q ), as given by (3.1)and (4.1).We now claim that for a place v not belonging to B , every nontrivial linearcombination Q = aQ + bQ has rank( Q ) ≥
7, where Q is the reduction to F v ,as usual. Then Q must split off three hyperbolic planes over k v , by Lemma2.2.To prove this latter claim we suppose for a contradiction that rank( Q ) ≤ Q = aQ + bQ with a , say, a unit in O v . After a suitable change of variablein GL ( O v ) we can arrange that Q is a function of X , . . . , X only. Thendet( xQ + yQ ) will be divisible by y . Thus if F ( x, y ) is given by (3.1) andif π is a uniformizing element for k v then F ( x, y ) will have a repeated factormodulo π , contradicting our assumption that π ∤ ∆( Q , Q ). This suffices forthe proof of our claim.We can now prove Proposition 1. Since every prime ideal of O k hasabsolute norm at least 32 we have F v ≥
32 for every finite place v of k .Then, according to Theorem 2 and Lemma 12.1, for every v ∈ B there isa non-trivial pair a v , b v of elements of k v such that a v Q + b v Q splits off3 hyperbolic planes in k v . If v is a finite place and Q = a v Q + b v Q thenthere is a change of variables in GL ( k v ) transforming Q into X X + X X + X X + Q ′ ( X , X ) with Q ′ defined over O v . According to Lemma 2.2, anyother quadratic form congruent to this modulo π also splits off 3 hyperbolicplanes over k v . We therefore deduce that there is a positive real ε v such that aQ + bQ splits off 3 hyperbolic planes in k v whenever | a − a v | < ε v and | b − b v | < ε v . We may obtain the analogous conclusion for infinite placesby using the local non-decreasing property for the functions n + ( θ ) , n − ( θ )introduced above.By weak approximation, there are suitable elements a, b ∈ k satisfyingthe additional condition that det( aQ + bQ ) = 0. Then to complete theproof we merely note that a quadratic form splits off three hyperbolic planes43ver k if and only if it does so over every completion k v .
13 The Hasse Principle in the Absence ofSmall Prime Ideals
Our next goal is the following major result, which establishes the Hasseprinciple for fields which have no prime ideals with norm less than 32.
Proposition 2
Let k be a number field for which every prime ideal has ab-solute norm at least 32, and let Q ( X , . . . , X ) and Q ( X , . . . , X ) be twoquadratic forms such that the projective variety V : Q ( X¯ ) = Q ( X¯ ) = 0 is nonsingular and has a non-trivial point over every completion k v of k .Then there is a non-trivial point over k . We take Q to be the form aQ + bQ given by Proposition 1. Assumingthat a = 0, as we may by symmetry, the forms Q and Q then generate thepencil < Q , Q > k . The proof of Proposition 2 now depends on the followinglemma. Lemma 13.1
There is a change of variable in GL ( k ) such that Q ( X¯ ) = Q ( X , . . . , X ) + X L ( X¯ ) + X L ( X¯ ) and Q ( X¯ ) = Q ( X , . . . , X ) + X L ( X¯ ) + X L ( X¯ ) with rank( Q ) = 4 and rank( Q + αQ ) ≥ for every α ∈ k . Moreover Q and Q have a nonsingular common zero over every completion k v of k . We will prove this in the next section, but first we show how it suffices forProposition 2.We make a change of variable so that Q depends only on X , . . . , X .Then if Q ( X , X ) = Q (0 , . . . , , X , X ) has rank less than 2, or is a hyper-bolic plane, it will have a non-trivial zero x , x over k , whence V will have anon-trivial point at (0 , . . . , , x , x , , Q is anisotropic over k . We then have Q ( x , x ) = 0 for a point ( x , x ) defined44nly over a quadratic extension of k , so that the variety Q = Q has a pairof conjugate singular points. We now apply the following result, which is aconsequence of Theorem 9.6 of Colliot-Th´el`ene, Sansuc and Swinnerton-Dyer[7]. Lemma 13.2
Let Q ( X , . . . , X ) and Q ( X , . . . , X ) be quadratic formsover a number field k , such that the projective variety Y : Q = Q = 0 isabsolutely irreducible of codimension 2, and is not a cone. Suppose that Y has a pair of conjugate singular points over k , and assume further that thepencil < Q , Q > k does not contain two independent forms of rank 4. Thenif Y has nonsingular points over every completion of k it will have a pointover k . We proceed to investigate the possibility that Y might be a cone, ormight fail to be absolutely irreducible of codimension 2. It is trivial that Y has a point over k if Y is a cone. Moreover if Y is not absolutely irreducibleof codimension 2 then, according to Lemma 3.2, either r min ( Q , Q ) ≤ r ( Q , Q ) ≤
4, in the notation of §
3. In the first case we would have a Q + a Q = R ( X , X ) say, and hence a Q + a Q = R ( X , X ) + X L ′ ( X , . . . , X ) + X L ′′ ( X , . . . , X ) , say. This however has rank at most 6 in contradiction to Lemma 3.1. Thuswe cannot have r min ( Q , Q ) ≤
2. On the other hand, if r ( Q , Q ) ≤ < k -point, given by taking x = . . . = x r = 0 and x R = 1 in the notation of the lemma.Thus in any cases in which Lemma 13.2 is not applicable we will auto-matically have a point on Y defined over k . Thus Proposition 2 follows fromLemma 13.2.
14 Proof of Lemma 13.1
Lemma 13.1 requires us to control both local solvability and ranks, and webegin by investigating the local solvability condition. Our first result is thefollowing. 45 emma 14.1
Let S ( X , . . . , X n ) and S ( X , . . . , X n ) be quadratic formsover k such that r min ( S , S ) ≥ and r ( S , S ) ≥ . Then there is a fi-nite set B = B ( S , S ) of places of k such that for any v B , and any linearform L ( X , . . . , X n ) defined over k , the variety V L : S = S = L = 0 has a nonsingular point over k v . Without loss of generality we may assume that S , S and L are definedover O k . Moreover, if L ( X , . . . , X n ) = P ℓ i X i we can multiply L by asuitable constant so that the ideal generated by ℓ , . . . , ℓ n has norm at most c k , the Minkowski constant. We shall require B to include all the infiniteplaces, together with all finite places corresponding to prime ideals of normat most c k .When v B at least one coefficient, ℓ n say, of L is a unit in O v . We maythen replace the variety V L by S ,L ( X , . . . , X n − ) = S ,L ( X , . . . , X n − ) = 0,where S i,L ( X , . . . , X n − )= S i (cid:0) X , . . . , X n − , − ℓ − n ( ℓ X + . . . , ℓ n − X n − ) (cid:1) , ( i = 1 , . and ask whether the variety S ,L = S ,L = 0 has a nonsingular point in k v .It follows from the Lang–Weil theorem [9] that there is a constant C depending only on n , such that the variety V L : S ,L = S ,L = 0has a nonsingular point over F v provided firstly that it is absolutely irre-ducible of codimension 2, and secondly that F v ≥ C . By Hensel’s Lemmathese conditions then suffice for the existence of a nonsingular point on V L over k v . We therefore require B to include all places corresponding to primeideals of norm at most C , and all places such that there is some linear form L for which the variety V L fails to be absolutely irreducible of codimension2 over F v . In view of Lemma 3.2 it is sufficient to add places for which thereis some linear from L having either r min ( S ,L , S ,L ) ≤ r ( S ,L , S ,L ) ≤ . (14.2)46hus V L will always have a nonsingular point over k v when v B , and itremains to prove that B is finite. We begin by considering those v for which(14.1) holds for some L . Thus there exist a, b ∈ F v , not both zero, and forms q, ℓ , ℓ over F v , such that a S ,L + bS ,L = ℓ ℓ . We then deduce that there is a third form ℓ such that a S + bS = ℓ ℓ + Lℓ . Thus the 5 × M ( XS + Y S ), which are binary formsin X and Y , all vanish at X = a, Y = b . However the 5 × M ( XS + Y S ) cannot have a common zero, since r min ( S , S ) ≥
5. Thusthere are two of them which have a non-zero resolvent Res ∈ O v say. Thisgives us an element depending only on S and S . However the minors canonly have a common zero over F v when Res = 0, and this can only happenfor a finite set of places v .In the alternative case for which (14.2) holds, any linear combination a S + bS differs from the corresponding form a S ,L + bS ,L by a multipleof L , so thatrank (cid:0) a S + bS (cid:1) ≤ rank (cid:0) a S ,L + bS ,L (cid:1) + 2 ≤ a , a ∈ F v . Hence all 7 × M ( XS + Y S ) mustvanish identically. However the 7 × M ( XS + Y S ) cannot allvanish identically, since r ( S , S ) ≥
7. Thus there is some non-zero coefficient µ of some minor, such that (14.2) can hold only when µ = 0. This too canhold only for a finite set of places v . We have therefore shown that we cantake B to be finite, thereby completing the proof of Lemma 14.1.In our next result we handle the remaining places. Lemma 14.2
Let S ( X , . . . , X n ) and S ( X , . . . , X n ) be quadratic formsover k such that the variety S = S = 0 has a nonsingular point overevery completion k v . Suppose that r min ( S , S ) ≥ and r ( S , S ) ≥ , andthat S splits off two hyperbolic planes.If P is a nonsingular projective point on S = 0 , defined over k , let L P ( X , . . . , X n ) = 0 be the tangent hyperplane to S = 0 at P . Then there isa Zariski-dense set of such P for which the variety V ( P ) : S ( X , . . . , X n ) = S ( X , . . . , X n ) = L P ( X , . . . , X n ) = 047 as a nonsingular point over every completion k v . Although the lemma requires S = S = L P = 0 to have nonsingular solu-tions for every place v it is clear that Lemma 14.1 may be applied, producinga finite set B of places outside which a nonsingular solution is guaranteedfor any linear form L P .For each v ∈ B we have a nonsingular point x¯ v = ( x v , . . . , x nv ) ∈ k nv on S = S = 0. Since x¯ v is nonsingular we have M ( S )x¯ v = 0¯. We now considerthe codimension 2 quadric defined over k v by Q : S (X¯ ) = X¯ t M ( S )x¯ v = 0 . Since S splits off two hyperbolic planes over k we may write it as S (X¯ ) = X X + X X + S ( X , . . . , X n ) , after a change of variable. We then choose non-zero pointsy¯ = ( a, b, , . . . ,
0) and z¯ = (0 , , c, d, , . . . , k v , both lying on the hyperplane X¯ t M ( S )x¯ v = 0. Thus both y¯ and z¯lie on the quadric Q . Moreover ∇ S (y¯) = (0 , , a, b, . . . ,
0) and ∇ S (z¯) = ( c, d, , . . . , . Since these cannot both be proportional to M ( S )x¯ v we deduce that Q hasat least one nonsingular point, w¯ v say, over k v . When an irreducible quadrichas a nonsingular point over an infinite field such points are automaticallyZariski-dense. Thus in our case the available points w¯ v cannot be restrictedto a line. We may therefore suppose that our point w¯ v is chosen so that M ( S )w¯ v does not lie on the line through M ( S )x¯ v and M ( S )x¯ v .We therefore have S (x¯ v ) = S (x¯ v ) = S (w¯ v ) = 0 , w¯ tv M ( S )x¯ v = 0 , ∇ S (w¯ v ) = 0¯ , and rank ∂S ( x¯ v ) ∂X . . . ∂S ( x¯ v ) ∂X n ∂S ( x¯ v ) ∂X . . . ∂S ( x¯ v ) ∂X n ∂S ( w¯ v ) ∂X . . . ∂S ( w¯ v ) ∂X n = 3 . (14.3)48hus P v = w¯ v is a nonsingular point on S = 0, and x¯ v is a nonsingular pointon the variety V ( P v ) defined in the lemma.For each place v there is a 3 × v , w¯ v ) formed from thematrix (14.3) such that ∆(x¯ v , w¯ v ) = 0. If | ∆(x¯ v , w¯ v ) | v = δ v then there isa positive real ε v such that | ∆(x¯ v , w¯ ) | v = δ v whenever | w¯ − w¯ v | < ε v . Wenow use weak approximation on the quadric Q = 0 to choose a point P = w¯suitably close to w¯ v for each v ∈ B . Since w¯ v is a nonsingular point on Q = 0the resulting point P will also be nonsingular. We then claim that, if w¯ issufficiently close to w¯ v , the variety V ( P ) will have a nonsingular point over k v . This will follow from Hensel’s Lemma, using the starting value x¯ = x¯ v ,for which | ∆(x¯ v , w¯ ) | v = δ v > v . Moreover wehave Q (x¯ v ) = Q (x¯ v ) = 0, while L P (x¯ v ) can be made sufficiently small forHensel’s Lemma to apply, merely by taking w¯ suitably close to w¯ v . Thus wedo indeed obtain a point on V ( P ), and since the lifting argument for Hensel’sLemma preserves the value of | ∆(x¯ , w¯ ) | v we see that the resulting point isnonsingular. This establishes Lemma 14.2, since we can use any point w¯sufficiently close to w¯ v for each v ∈ B .We are now ready to prove Lemma 13.1, which will require us to useLemma 14.2 twice. The hypotheses of Lemma 14.2 are satisfied for the forms S = Q and S = Q , since r min ( Q, Q ) ≥
7. Let us write temporarily I forthe variety S = 0, and W for the variety S = S = 0, both considered in P .Each of these varieties is nonsingular. We now proceed to consider their dualvarieties I ∗ and W ∗ . The reader should recall that the dual of a variety Y isthe closure of the set of hyperplanes which are tangent at a nonsingular pointof Y . For any variety Y ⊂ P m , the dual Y ∗ is a proper subvariety of P m ∗ , andis irreducible. Moreover we have ( Y ∗ ) ∗ = Y . In our case I ∗ will be a quadrichypersurface. We claim that I ∗ cannot be contained in W ∗ . Indeed since I ∗ is an irreducible hypersurface and W ∗ is an irreducible proper subvariety of P the only situation in which one could have I ∗ ⊆ W ∗ is when I ∗ = W ∗ .However this would imply that I = ( I ∗ ) ∗ = ( W ∗ ) ∗ = W . The claim then follows since
I 6 = W .It now follows from Lemma 14.2 that we may choose our point P so thatthe hyperplane L P = 0 is not tangent to W . Thus the variety V ( P ) will benonsingular. Since P is defined over k the hyperplane L P = 0 is also defined49ver k . We may therefore make a change of variables in GL ( k ) so that thehyperplane L P = 0 is X = 0. The forms S and S then become S i (X¯ ) = T i ( X , . . . , X ) + X L i ( X , . . . , X ) , ( i = 1 , T ) = 6 by Lemma 2.1. The conclusions of Lemma 14.2 implythat the forms T and T have a nonsingular common solution over everycompletion k v . Moreover the variety T = T = 0 will be nonsingular since V ( P ) is nonsingular. In particular Lemma 3.1 implies that r min ( T , T ) = 6and r ( T , T ) = 7. Moreover, as S = Q splits off three hyperbolic planes wededuce that T splits off two hyperbolic planes.We therefore see that the forms T and T satisfy the hypotheses for asecond application of Lemma 14.2. Since r ( T , T ) = 7 there are at most7 linear combinations W = T + αT with α ∈ k such that rank( W ) < W , . . . , W m , say. For each of these the variety X i : W i = 0 isdistinct from X : T = 0 so that, by the argument above, their duals, whichare irreducible quadric hypersurfaces, are different.We may therefore choose P so that the hyperplane L P = 0 belongs to X ∗ but to none of the sets X ∗ i . We now repeat the manoeuvres above. We makea change of variables in GL ( k ) so that the hyperplane L P = 0 is X = 0.The forms T , T then become T i ( X , . . . , X ) = U i ( X , . . . , X ) + X M i ( X , . . . , X ) , ( i = 1 , U ) = rank( T ) − U and U have a nonsingular common solution over everycompletion k v .We claim now that rank( U + αU ) ≥ α ∈ k . Since our changeof variables has transformed ( T + αT )( X , . . . , X ) into( U + αU )( X , . . . , X ) + X L α ( X , . . . , X ) (14.4)for a suitable linear form L α , we see that rank( U + αU ) ≥ rank( T + αT ) − α . This verifies the claim except when T + αT is one of the forms W i above. Suppose then that T + αT = W i . Since r min ( T , T ) ≥ W i ) = 6. However, if rank( U + αU ) ≤ λ ( X , . . . , X ) , . . . , λ ( X , . . . , X ) say, such thatone can write U + αU as a form in λ , . . . , λ alone. Thus (14.4) produces W i = U ( λ , . . . , λ ) + X L ( λ , . . . , λ , X ) , W i ) = 6 the linear form L ( λ , . . . , λ , X ) must properlycontain at least one of λ or λ . One then sees from (14.4) that the variety X i : W i = 0 has a nonsingular point with λ = . . . = λ = X = 0 and L = 0.The tangent hyperplane at this point would be X = 0, which is also thetangent hyperplane of X : T = 0 at P . This however is impossible, since P was chosen so that the hyperplane L P = 0 was in none of the dual varieties X ∗ i . This completes the proof of our claim.We now see that the effect of our two applications of Lemma 14.2 is toproduce a change of variables putting Q and Q into the shape Q (X¯ ) = U ( X , . . . , X ) + X L (X¯ ) + X L (X¯ )and Q (X¯ ) = U ( X , . . . , X ) + X L (X¯ ) + X L (X¯ )for suitable linear forms L (X¯ ) , . . . , L (X¯ ), all defined over k . Moreover wehave arranged that the variety U = U = 0 has nonsingular points in everycompletion k v of k , that rank( U ) = 4 and that rank( U + αU ) ≥ α ∈ k . This therefore suffices for Lemma 13.1, and thereby also completesour demonstration of Proposition 2.
15 Deduction of Theorem 1
In this final section we first show how to remove the condition that all primeideals of O k have norm at least 32, and then explain why the weak approxi-mation property holds automatically in our situation.We begin by choosing a prime q ≥ k : Q ],and we proceed to construct a number field Q ( θ ) of degree q over Q , suchthat every prime ideal of O Q ( θ ) has norm at least 32. Clearly it suffice thatevery rational prime p ≤
31 is inert in Q ( θ ) / Q . For each such prime wechoose a monic polynomial f p ( X ) ∈ Z [ X ] of degree q which is irreduciblemodulo p . We then use the Chinese Remainder Theorem to produce a monicpolynomial f ( X ) ∈ Z [ X ] of degree q , with f ( X ) ≡ f p ( X ) (mod p ) for every p ≤
31. Then f is irreducible over Q since it is irreducible modulo 2, forexample. We claim that Q ( θ ) will be a suitable field, where θ is a root of f ( X ). Let p ≤
31 be prime. Then f p is irreducible modulo p , since it hasno repeated factors over F p . Thus p ∤ Disc( θ ), so that we may apply theKummer–Dedekind theorem to deduce that p is inert in Q ( θ ). It follows51hat N ( P ) ≥
32 for every prime ideal of O Q ( θ ) , as required. Indeed we have N Q ( θ ) / Q ( I ) ≥
32 for every non-trivial integral ideal I .We now consider the field k ′ = k ( θ ). If P is a prime ideal of O k ′ then N k ′ / Q ( P ) = N Q ( θ ) / Q (cid:0) N k ′ / Q ( θ ) ( P ) (cid:1) ≥ . We also note that [ k ′ : Q ] = [ k ′ : Q ( θ )][ Q ( θ ) : Q ], and [ k ′ : Q ] = [ k ′ : k ][ k : Q ],so that both [ Q ( θ ) : Q ] = q and [ k : Q ] divide [ k ′ : Q ]. It follows that q [ k : Q ]divides [ k ′ : Q ], since we chose q to be coprime to [ k : Q ]. On the otherhand it is clear that [ k ( θ ) : k ] ≤ q , whence [ k ′ : Q ] ≤ q [ k : Q ]. We thereforeconclude that [ k ′ : Q ] = q [ k : Q ], so that [ k ′ : k ] = q .Since V has points over every completion of k it will also have points overevery completion of k ′ . Thus Proposition 2 is applicable, and shows that V has a point x¯ = ( x , . . . , x ) say, over k ′ . It follows that the quadratic form Q (X¯ ) + T Q (X¯ ), which is defined over the function field k ′ ( T ), has a non-trivial point at x¯. However Q (X¯ ) + T Q (X¯ ) is also defined over the subfield k ( T ), and [ k ′ ( T ) : k ( T )] = [ k ′ : k ] = q , which is odd. It therefore followsfrom a result of Springer [11] that Q (X¯ ) + T Q (X¯ ) has a zero over k ( T ).Finally, we apply the Amer–Brumer Theorem [1, §
3, Satz 7], [4], which westate as follows.
Lemma 15.1
Let S ( X , . . . , X n ) and S ( X , . . . , X n ) be quadratic forms de-fined over a field K of characteristic not equal to 2. Then if S + T S hasa non-trivial zero over the function field K ( T ) , the forms S and S have asimultaneous zero over K . We therefore conclude that our forms Q and Q have a simultaneous zeroover k .It remains to show that V has the weak approximation property. Thishowever is a direct consequence of Theorem 3.11 of Colliot-Th´el`ene, Sansucand Swinnerton-Dyer [6]. References [1] M. Amer,
Quadratische formen ¨uber funktionenk¨orpern , (Thesis, Mainz,1976).[2] B.J. Birch, D.J. Lewis and T.G. Murphy, Simultaneous quadratic forms,
Amer. J. Math. , 84 (1962), 110–115.523] B.J. Birch and H.P.F. Swinnerton-Dyer, The Hasse problem for rationalsurfaces,
J. Reine Angew. Math. , 274/275 (1975), 164–174.[4] A. Brumer, Remarques sur les couples de formes quadratiques,
C. R.Acad. Sci. Paris S´er. A-B,
286 (1978), no. 16, A679–A681.[5] J.-L. Colliot-Th´el`ene and J.-.J. Sansuc, La descente sur une vari´et´e ra-tionnelle d´efinie sur un corps de nombres,
C. R. Acad. Sci. Paris S´er.A , 284 (1977), 1215–1218.[6] J.-L. Colliot-Th´el`ene, J.-.J. Sansuc, and H.P.F. Swinnerton-Dyer, In-tersections of two quadrics and Chˆatelet surfaces. I,
J. Reine Angew.Math. , 373 (1987), 37–107.[7] J.-L. Colliot-Th´el`ene, J.-.J. Sansuc, and H.P.F. Swinnerton-Dyer, In-tersections of two quadrics and Chˆatelet surfaces. II,
J. Reine Angew.Math. , 374 (1987), 72–168.[8] D.R. Heath-Brown and L.B. Pierce, Simultaneous integer values of pairsof quadratic forms, to appear.[9] S. Lang and A. Weil, Number of points of varieties in finite fields,
Amer.J. Math. , 76 (1954), 819–827.[10] D.B. Leep and L.M. Schueller, A characterization of nonsingular pairsof quadratic forms,
J. Algebra Appl. , 1 (2002), 391–412.[11] T.A. Springer, Sur les formes quadratiques d’indice z´ero,
C. R. Acad.Sci. Paris , 234 (1952), 1517–1519.[12] H.P.F. Swinnerton-Dyer, Rational zeros of two quadratic forms,
ActaArith. , 9 (1964), 261–270.Mathematical Institute,24–29, St. Giles’,OxfordOX1 3LBUK [email protected]@maths.ox.ac.uk