Lower bounds for resultants II.

Transcription

1 Lower bounds for resultants II. [Page 1] Jan-Hendrik Evertse Abstract. Let F X, Y ), GX, Y ) be binary forms in Z[X, Y ] of degrees r 3, s 3, respectively, such that F G has no multiple factors. For each matrix U = a b c d ) GL 2 Z), define F U X, Y ) = F ax + by, cx + dy ), and define G U similarly. We will show that there is a matrix U GL 2 Z) such that for the resultant RF, G) of F, G we have RF, G) C HF U ) s HG U ) r ) 1/718, where HF U ), HG U ) denote the heights maxima of absolute values of the coefficients) of F U, G U, respectively, and where C is some ineffective constant, depending on r, s and the splitting field of F G. A slightly weaker result was announced without proof in [3] Theorem 3). We will also prove a p-adic generalisation of the result mentioned above. As a consequence, we will obtain under certain technical restrictions a symmetric improvement of Liouville s inequality for the difference of two algebraic numbers. In our proofs we use some results from [4], [5], and the latter were proved by means of chlickewei s p-adic generalisation of chmidt s ubspace theorem Mathematics ubject Classification: 11J68, 11C Introduction. Let F X, Y ) = a 0 X r + a 1 X r 1 Y + + a r Y r, GX, Y ) = b 0 X s + b 1 X s 1 Y + + b s Y s be two binary forms with coefficients in some field K of characteristic 0. The resultant RF, G) of F and G is defined by the determinant of order r + s, RF, G) = a 0 a 1 a r, 1.1) b 0 b 1 b s a 0 a 1 a r a 0 a 1 a r b 0 b 1 b s b 0 b 1 b s of which the first s rows consist of coefficients of F and the last r rows of coefficients of G. Both F, G can be factored into linear forms with coefficients in the algebraic

2 2 Jan-Hendrik Evertse closure of K, i.e. and we have F X, Y ) = α i X + β i Y ), GX, Y ) = RF, G) = j=1 γ j X + δ j Y ), j=1 α i δ j β i γ j ). 1.2) Hence RF, G) = 0 if and only if F, G have a common linear factor. Further, if for a matrix U = a b c d ) with determinant det U 0 we define F U X, Y ) := F ax + by, cx + dy ) and similarly G U, it follows that RF U, G U ) = det U) rs RF, G). 1.3) Now assume that F, G have their coefficients in Z. For a polynomial P with coefficients in Z, we define its height HP ) to be the maximum of the absolute values of the coefficients of P. From 1.1) and Hadamard s inequality it follows that RF, G) r + 1) s/2 s + 1) r/2 HF ) s HG) r. On the other hand, there are some results in the literature on lower bounds for RF, G) which have been obtained by applying Diophantine approximation techniques. To state these results, we need some terminology. A binary form is called square-free if it is not divisible by the square of any non-constant binary form. The splitting field over a field K of a binary form with coefficients in K is the smallest extension of K over which this binary form factors into linear forms. By C ineff 1 ), C ineff 2 ),... we denote ineffective positive constants depending only on the parameters between the parentheses. Improving on a result of Wirsing [14], chmidt [12] proved that if r, s are integers with r > 2s > 0 and if F is a square-free binary form of degree r in Z[X, Y ] without irreducible factors of degree s, then for every binary form G Z[X, Y ] of degree s which is coprime with F one has RF, G) C ineff 1 r, s, F, ε)hg) r 2s ε for ε > 0, 1.4) where the dependence of C 1 on F is unspecified. From Theorem 4.1 of Ru and Wong [9] it follows that 1.4) holds true without the constraint that F have no irreducible factors of degree s. Győry and the author [5], Theorem 1) proved that for each pair of binary forms F, G with coefficients in Z such that deg F = r 3, deg G = s 3, F G has splitting field L over K and F G is square-free one has RF, G) C2 ineff r, s, L, ε) s r ) 1 DF ) r 1 DG) s 1 17 ε for ε > 0, 1.5) where DF ), DG) denote the discriminants of F, G. We recall that if F X, Y ) = r α ix + β i Y ) then DF ) = 1 i<j r α iβ j α j β i ) 2. Győry and the author showed also in [5] that if r 2 or s 2 or if we allow the splitting field of F G to vary, then DF ), DG) may grow arbitrarily large while RF, G) remains

3 Lower bounds for resultants II 3 bounded. For more information on lower bounds for resultants and on applications we refer to [4], [5]. Our aim is to derive instead of 1.5) a lower bound for RF, G) which is a function increasing in both HF ) and HG). In general such a lower bound does not exist. Namely, 1.3) implies that RF U, G U ) = RF, G) for U GL 2 Z), 1.6) where GL 2 Z) = { a b c d ) : a, b, c, d Z, ad bc = ±1}) while HF U ), HG U ) may be arbitrarily large for varying U. However, assuming that r 3, s 3, we can show that there is an U GL 2 Z) such that RF, G) is bounded from below by a function increasing in both HF U ), HG U ). The next result, with exponent 1/760 instead of 1/718, was stated without proof in [3], Theorem 3. Theorem 1. Let r 3, s 3, and let F, G) be a pair of binary forms with coefficients in Z such that deg F = r, deg G = s, F G is square-free and F G has splitting field L over Q. Then there is an U GL 2 Z) such that RF, G) C ineff 3 r, s, L) HF U ) s HG U ) r) 1/ ) Remark. imilarly as for 1.5), the conditions r 3, s 3, as well as the dependence of C 3 on L, are necessary. Namely, the discriminant of a binary form F of degree r is a homogeneous polynomial of degree 2r 2 in the coefficients of F, and for U GL 2 Z) one has DF U ) = DF ). Therefore, there is a constant cr) such that DF ) cr){inf U GL2Z) HF U )} 2r 2. Now, by the result from [5] mentioned above, if r 2 or s 2 or if we allow the splitting field of F G to vary, then DF ), DG), and hence inf U GL2Z) HF U ), inf U GL2Z) HG U ) may grow arbitrarily large while RF, G) remains bounded. The proof of Theorem 1 ultimately depends on chmidt s ubspace theorem, which explains the ineffectivity of the constant C 3. It would be a remarkable breakthrough to obtain an effective lower bound for RF, G) which is a function increasing in both HF U ) and HG U ) for some U GL 2 Z). We also prove a p-adic generalisation of Theorem 1. To state this, we have to introduce some further terminology. Let K be an algebraic number field. Denote by O K the ring of integers of K. The set of places M K of K consists of the isomorphic embeddings σ : K R which are called real infinite places; the pairs of complex conjugate isomorphic embeddings {σ, σ : K C} which are called complex infinite places; and the prime ideals of O K which are called finite places. We define absolute values v v M K ) normalised with respect to K as follows: v = σ ) 1/[K:Q] if v = σ is a real infinite place; v = σ ) 2/[K:Q] = σ ) 2/[K:Q] if v = {σ, σ} is a complex infinite place; v = N ) ord )/[K:Q] if v = is a finite place, i.e. prime ideal of O K,

4 4 Jan-Hendrik Evertse where N = #O K / ) denotes the norm of and ord x) is the exponent of in the prime ideal decomposition of x), with ord 0) =. These absolute values satisfy the Product formula v M K x v = 1 for x K. For any finite extension L of K, we define absolute values w w M L ) normalised with respect to L in an analogous manner. Thus, if w M L lies above v M K, then the restriction of w to K is equal to v [Lw:Kv]/[L:K], where K v, L w denote the completions of K at v, L at w, respectively. We will frequently use the Extension formula x w = N L/K x) 1/[L:K] v for x L, v M K so in particular w v x w = x v for x K, v M K, w v where the product is taken over all places w M L lying above v. Now let be a finite set of places on K, containing all real and complex) infinite places. The ring of -integers and its unit group, the group of -units, are defined by O = {x K : x v 1 for v / }, respectively, where v / means v M K \. We put Thus, x := v O = {x K : x v = 1 for v / }, x v for x K. x > 1 for x O, x 0, x / O, x = 1 for x O. 1.8) We define the truncated height H by H x) = H x 1,..., x n ) = v max x 1 v,..., x n v ) for x = x 1,..., x n ) K n. For a polynomial P with coefficients in K we put H P ) := H p 1,..., p t ), where p 1,..., p t are the coefficients of P. By 1.8) we have H x) 1 for x O n \{0}, 1.9) H ux) = H x) for x O n \{0}, u O. 1.10) Further, one can show that for every A > 0 the set of vectors x O n with H x) A is the union of finitely many O -cosets {uy : u O } with y On fixed. 1.3) and 1.8) imply that for binary forms F, G with coefficients in O we have RF U, G U ) = RF, G) for U GL 2 O ), 1.11)

5 b Lower bounds for resultants II 5 where GL 2 O ) = { a c d ) : a, b, c, d O, ad bc O }. We prove the following generalisation of Theorem 1: Theorem 2. Let r 3, s 3, and let F, G) be a pair of binary forms with coefficients in O such that deg F = r, deg G = s, F G is square-free and F G has splitting field L over K. Then there is an U GL 2 O ) such that RF, G) C ineff 4 r, s,, L) H F U ) s H G U ) r) 1/ ) In the proof of Theorem 2 we use a lower bound for resultants in terms of discriminants from [5] which has been proved by means of chlickewei s p-adic generalisation [10] of chmidt s ubspace theorem [11], a lower bound for discriminants in terms of heights from [4] which follows from Lang s p-adic generalisation [6] Chap. 7, Thm. 1.1) of Roth s theorem [8], and also a semi-effective result on Thue-Mahler equations, stated below, which follows also from the p-adic generalisation of Roth s theorem. Theorem 3. Let F X, Y ) O [X, Y ] be a square-free binary form of degree r 3 with splitting field M over K and let A 1. Then every solution x, y) O 2 of satisfies F x, y) = A 1.13) H x, y) C ineff 5 r,, M, ε) H F ) A) 3 r +ε for every ε > ) Using the techniques from the paper of Bombieri and van der Poorten [1] it is probably possible to derive instead of 1.14) an upper bound H x, y) C ineff 6 r,, M, ε) H F ) cr,ε) A 1 r 2 +ε for every ε > 0, where cr, ε) is a function increasing in r, ε 1. We derive from Theorem 2 a symmetric improvement of Liouville s inequality. The absolute) height of an algebraic number ξ is defined by hξ) = max1, ξ v ), v M K where K is any number field containing ξ. By the Extension formula, this height is independent of the choice of K. Let K be an algebraic number field and ξ, η numbers algebraic over K with ξ η. Put L = Kξ, η). Further, let T be a finite set of places on L not necessarily containing all infinite places). By the Product formula we have w T ξ η w max1, ξ w ) max1, η w ) = w / T max1, ξ w ) max1, η w ) ξ η w ) hξ) 1 hη) 1

6 6 Jan-Hendrik Evertse 1 2 hξ)hη) ) 1, 1.15) where as usual, the absolute values w are normalised with respect to L. The latter is known as Liouville s inequality. Under certain hypotheses we can improve upon the exponent 1. Assume that L = Kξ, η); [Kξ) : K] 3, [Kη) : K] 3; 1.16) [L : K] = [Kξ) : K][Kη) : K], i.e. Kξ), Kη) are linearly disjoint over K. Further, let T be a finite set of places on L such that if is the set of places on K lying below those in T then 1 W := max [L w : K v ] < 1 v [L : K] 3, 1.17) where for each place v, the sum is taken over those places w T that lie above v. w T w v Theorem 4. Assuming that ξ, η, L, T satisfy 1.16), 1.17) we have ξ η w max1, ξ w ) max1, η w ) Cineff 7 L, T ) hξ)hη) ) 1+δ w T with δ = W 1 + 3W. 1.18) For instance, suppose that L, ξ, η satisfy 1.16) with K = Q and that T is a subset of the set of infinite places on L, satisfying 1.17) with K = Q and with consisting of the only infinite place of Q. Inequality 1.18) has been stated in terms of absolute values normalised with respect to L and we will renormalise these to Q. Each w T is either an isomorphic embedding of L into R and then L w = R; or a pair of complex conjugate embeddings of L into C and then L w = C. Therefore, the union of all places w T is a collection Σ of isomorphic embeddings of L into C such that with an isomorphic embedding also its complex conjugate belongs to Σ and moreover, the quantity W of 1.17) is precisely #Σ/[L : Q]. We recall that if w = σ is real then w = σ ) 1/[L:Q] while if w = {σ, σ} is complex then w = σ ) σ ) ) 1/[L:Q]. This implies that the left-hand side of 1.18) equals σ Σ σξ η) / max1, σξ) ) max1, ση) ) ) 1/[L:Q]. For an algebraic number ξ, we define Hξ) to be the maximum of the absolute values of the coefficients of the minimal polynomial of ξ over Z. Then hξ) deg ξ c Hξ) where c depends only on the degree of ξ cf. [6], Chap. 3, 2, Prop. 2.5). Thus, Theorem 4 implies the following: Corollary. Let ξ, η be algebraic numbers of degrees r 3, s 3, respectively, such that the field L = Qξ, η) has degree rs. Further, let Σ be a collection of isomorphic embeddings of L into C such that if σ Σ then also σ Σ, and such

7 that W := #Σ/[L : Q] < 1 3. Put δ = 1 σ Σ Lower bounds for resultants II 7 1 3W W. Then σξ η) max1, σξ) ) max1, ση) ) Cineff 8 L) Hξ) s Hη) r ) 1 δ. 1.19) For instance, assume that L R and take Σ = {identity}. Then [L : Q] = rs 9 and hence W 1 9. o by 1.19) we have ξ η w max1, ξ w ) max1, η w ) Cineff 9 L) s r Hξ) Hη) ) ) If L C, L R then with Σ = {identity, complex conjugation} we have W 2 9 and so 1.19) gives ξ η ) 2 w C ineff 10 L) s r Hξ) Hη) ) ) max1, ξ w ) max1, η w ) Results similar to 1.20), 1.21) with better exponents were derived in [3] Corollary 3, i)). For an inequality of type 1.18) with δ > 0 to hold it is certainly necessary to impose some conditions on ξ, η, L, T but 1.16), 1.17) are probably far too strong. Using for instance geometry of numbers over the adeles of a number field one may prove a generalisation of Dirichlet s theorem of the sort that for a number field M, a number η of degree 2 over M and a finite set of places T on L := Mη) satisfying some mild conditions, there is a constant c = cη, M, T ) such that the inequality w T ξ η w max1, ξ w ) chξ) 1 has infinitely many solutions in ξ M. Thus, for an inequality of type 1.18) to hold it is probably necessary to assume that [L : Kξ)] 3, [L : Kη)] 3. The following example shows that the condition W < 1 is necessary. Assume that W = 1. Then there is a place v on K such that T contains all places on L lying above v. Fix two elements ξ 0, η 0 of L such that L = Kξ 0, η 0 ), [Kξ 0 ) : K] 3, [Kη 0 ) : K] 3 and [L : K] = [Kξ 0 ) : K][Kη 0 ) : K]. Let γ 1, γ 2,... be a sequence of elements from K such that lim i γ i v =. By the strong approximation theorem, there exists for every i an α i K such that α i γ i v < 1 and α i v 1 for every place v v on K. Now put ξ i := ξ 0 + α i, η i := η 0 + α i for i = 1, 2,.... Then for all places w M L lying outside a finite collection depending only on ξ 0, η 0 we have ξ i w 1, η i w 1, while for the remaining places on L not lying above v we have ξ i w 1, η i w 1 for i = 1, 2..., where the constants implied by, depend only on ξ 0, η 0. Further, for w M L lying above v we have ξ i w α i w γ i w, η i w γ i w for i sufficiently large. Therefore, by the Extension formula, hξ i ) w v max1, ξ i w ) γ i v for i and similarly, hη i ) w v max1, η i w ) γ i v for i, where the products are taken over the places w M L lying above v. Moreover, since

8 8 Jan-Hendrik Evertse ξ i η i = ξ 0 η 0 we have ξ i η i w max1, ξ i w ) max1, η i w ) ξ 0 η 0 w max1, ξ i w ) max1, η i w ) w T w v hξ i )hη i ) ) 1 for i = 1, 2, Proof of Theorem 3. As in ection 1, K is an algebraic number field and a finite set of places on K containing all infinite places. Further, F X, Y ) is a square-free binary form of degree r 3 with coefficients in O and A a real 1. We assume that F X, Y ) = α i X + β i Y ) with α i, β i O for i = 1,..., r. 2.1) This is no loss of generality. Namely, suppose that F has splitting field M over K. Thus, F X, Y ) = r α i X + β i Y ) with α i, β i M. Let L be the Hilbert class field of M/K and T the set of places on L lying above those in. Then for i = 1,..., r, the fractional ideal with respect to O T generated by α i, β i is principal and since F has its coefficients in O this implies that F can be factored as in 2.1) but with α i, β i O T. From the Extension formula it follows that for x, y) O 2 we have F x, y) T = F x, y), H T x, y) = H x, y) and that also H T F ) = H F ), where T = w T w, H T,..., ) = w T max w,..., w ). o, if we have proved that for all x, y) OT 2 with F x, y) T = A and all ε > 0 we have H T x, y) C ineff 11 r, T, L, ε) H T F )A ) 3 r +ε, then Theorem 3 readily follows, on observing that T, L are uniquely determined by, M. In the proof of Theorem 3 we need some lemmas. The first lemma is fundamental for everything in this paper: Lemma 1. Let x 0,..., x n be non-zero elements of O such that x x n = 0, x i 0 for each proper nonempty subset I of {0,..., n}. i I Then for all ε > 0 we have H x 0,..., x n ) C12 ineff n 1+ε K,, ε) x i. Proof. Lemma 1 in this form appeared in Laurent s paper [7]. It is a reformulation of Theorem 2 of [2]. For n = 2, Lemma 1 follows from the p-adic generalisation of Roth s theorem [6] Chap. 7, Thm. 1.1) and for n > 2 from chlickewei s p- adic generalisation [10] of chmidt s ubspace theorem [11]. The constant C 11 i=0

9 Lower bounds for resultants II 9 and also each other constant in this paper) is ineffective because the ubspace theorem is ineffective. In fact, we need Lemma 1 only for n = 2 in which case the non-vanishing subsum condition is void. However, Lemma 1 with n > 2 has been used in the proof of a result from [5] which we will need in the present paper. For a polynomial P with coefficients in K and for v M K we define P v := max p 1 v,..., p t v ) where p 1,..., p t are the coefficients of P. Lemma 2. Let F X, Y ) = r α ix + β i Y ) with α i, β i O for i = 1,..., r. There is a constant c depending only on r and K such that H α i, β i ) H F ) c H α i, β i ). 2.2) c 1 Proof. According to, for instance [6], Chap. 3, 2, we have for any polynomials P 1,..., P r K[X 1,..., X n ], v M K that c 1 v P 1 P r v P 1 v P r v c v P 1 P r v if v is infinite, P 1 P r v = P 1 v P r v if v is finite, where each c v is a constant > 1 depending only on r, n, K. Now Lemma 2 follows by applying this with P i X, Y ) = α i X + β i Y for i = 1,..., r and any v, and then taking the product over v. We complete the proof of Theorem 3. Let F X, Y ) be a square-free binary form of degree r 3 satisfying 2.1) and let ε > 0. Put ε := ε/10. In what follows, the constants implied by will be ineffective and depending only on K,, r, ε. Define ij := α i β j α j β i for i, j = 1,..., r. We will use that ij v max α i v, β i v ) max α j v, β j v ) for v M K 2.2) whence, on taking the product over v, ij H α i, β i )H α j, β j ). 2.3) Pick three distinct indices i, j, k from {1,..., r} and define the linear forms A 1 = jk α i X + β i Y ), A 2 = ki α j X + β j Y ), A 3 = ij α k X + β k Y ). Thus, Further, A 1 + A 2 + A 3 = ) ij jk ki X = ki β j A 1 jk β i A 2, ij jk ki Y = ki α j A 1 + jk α i A )

10 10 Jan-Hendrik Evertse Let x, y) O 2 be a pair satisfying 1.13). Put a h := A h x, y) for h = 1, 2, 3. From 2.5) and 2.2) it follows that for v, ) ij jk ki v max x v, y v ) max α p v, β p v ) max a 1 v, a 2 v ). p {i,j,k} By taking the product over v we get ij jk ki H x, y) p {i,j,k} By Lemma 1 and 2.4) we have H a 1, a 2 ) H a 1, a 2, a 3 ) ij jk ki ) H α p, β p ) H a 1, a 2 ). p {i,j,k} By combining these inequalities we obtain ) H x, y) ij jk ki ε H α p, β p ) p {i,j,k} p {i,j,k} p {i,j,k} α p x + β p y ) 1+ε. α p x + β p y ) 1+ε H α p, β p ) α p x + β p y ) ) 1+3ε in view of 2.3). By taking the product over all subsets {i, j, k} of {1,..., r} we get, using Lemma 2 and r α ix + β i y = A which is a consequence of 2.1), 1.13), that H x, y) r 3) This proves Theorem 3. H α i, β i ) α i x + β i y ) ) r 1 2 )1+3ε ) H F ) A ) r 3) 3 r +ε). 3. Proof of Theorem 2. Let again K be an algebraic number field and a finite set of places on K containing all infinite places. We recall that the discriminant of a binary form F X, Y ) = r α ix + β i Y ) is given by DF ) = 1 i<j r α iβ j α j β i ) 2. This implies that DF U ) = DF ) for U GL 2 O ). We need some results from other papers. Lemma 3. Let F be a square-free binary form of degree r 3 with coefficients in O and with splitting field M over K. Then there is an U GL 2 O ) such that DF ) C ineff 13 r, M, )H F U ) r 1 21.

11 Lower bounds for resultants II 11 Proof. This follows from Theorem 2 of [4]. The proof of that theorem uses Lemma 1 mentioned above with n = 2 and a reduction theory for binary forms. I would like to mention here that the reduction theory for binary forms developed in [4] is essentially a special case of a reduction theory for norm forms which was developed some years earlier by chmidt [13] for a totally different purpose). I apologize for having overlooked this in [4]. Lemma 4. Let F, G be binary forms of degrees r 3, s 3, respectively, with coefficients in O such that F G is square-free and F G has splitting field L over K. Then RF, G) C14 ineff r, s, L,, ε) DF ) s r 1 DG) r s 1 ) 1 17 ε for ε > 0. Proof. This is Theorem 1A of [5]. The proof of that theorem uses Lemma 1 with n > 2. We now prove Theorem 2. We assume that DF ) s r 1 DG) r s 1 3.1) which is clearly no loss of generality. Let U GL 2 O ) be the matrix from Lemma 3. We will show that 1.12) holds with this U. Let M be the Hilbert class field of L/K, and T the set of places on M lying above those in. Thus, we have F U X, Y ) = α i X + β i Y ), G U X, Y ) = γ j X + δ j Y ) with α i, β i, γ j, δ j O T for i = 1,..., r, j = 1,..., s. 3.2) The height H T and the quantity T are defined similarly to H, but with respect to the absolute values w w T ). In what follows, the constants implied by, will be ineffective and depending only on r, s, L, and ε, where ε is a positive number depending only on r, s which will later be chosen sufficiently small. j=1 Note that by Lemma 4, 3.1), our choice of U, and Lemma 3 we have RF, G DF ) s r 1 DG) s ) 1 r 1 17 ε DF ) s r ε) H F U ) s ε 21 ). 3.3) We estimate H G U ) from above. By 1.2), 3.2) we have RF U, G U ) = j=1 α i δ j β i γ j ) = F U δ j, γ j ), j=1

12 12 Jan-Hendrik Evertse and together with 1.11) and the Extension formula this implies that RF, G) = RF U, G U ) T = F U δ j, γ j ) T. 3.4) Further, using that H F U ) = H T F U ), H G U ) = H T G U ) by the Extension formula, we have H G U ) H T γ j, δ j ) j=1 H T γ j, δ j ) by 3.2), Lemma 2, 3.5) j=1 H F U ) F U δ j, γ j ) T ) 3 r +ε for j = 1,..., s by Theorem 3, 3.6) where both Lemma 2, Theorem 3 have been applied with M, T replacing K,. Now 3.4), 3.5), 3.6) together imply H G U ) In combination with 3.3) this gives H F U ) s RF, G) ) 3 r +ε. H F U ) s H G U ) r H F U ) s4+rε) RF, G) 3+rε RF, G) 4+rε) ε 21 ) 1 +3+rε RF, G) 718 for ε sufficiently small. This implies 1.12), whence completes the proof of Theorem Proof of Theorem 4. As before, let K be an algebraic number field and a finite set of places on K containing all infinite places. For a matrix U = a b c d ) with entries in K we define U v := max a v, b v, c v, d v ) for v M K, H U) = v U v. We need the following elementary lemma: Lemma 5. Let F X, Y ) be a square-free binary form of degree r 3 with coefficients in O and U GL 2 O ). Then for some constant c depending only on r and the splitting field of F over K we have H U) c H F )H F U ) ) 3/r. 4.1)

13 Lower bounds for resultants II 13 Proof. We prove 4.1) only for binary forms F such that F X, Y ) = α i X + β i Y ) with α i, β i O for i = 1,..., r. 4.2) This is no restriction. Namely, in general F has a factorisation as in 4.2) with α i, β i O T where T is the set of places lying above those in on the Hilbert class field of the splitting field of F over K. Now if we have shown that H T U) c H T F )H T F U ) ) 3/r then 4.1) follows from the Extension formula. From 4.2) it follows that F U X, Y ) = b αi X + βi Y ) with αi, βi ) = α i, β i )U for i = 1,..., r. 4.3) Let U = a c d ). Pick three indices i, j, k from {1,..., r}. Then a, c, b, d, 1, 1, 1) is a solution to the system of six linear equations α i β i 0 0 α x i α i β i βi x α j β j αj x α j β j 0 βj x 0 4 =. 4.4) 0 α k β k αk x α k β k 0 0 βk x 6 0 x 7 4.4) can be reformulated as x 5 αi, β i ) = α i, β i )X, x 6 αj, β j ) = α j, β j )X, x 7 αk, β k ) = α k, β k )X, with X = x1 x 3 x 2 x 4 ). It is well-known that up to a constant factor, there is at most one 2 2-matrix mapping three given, pairwise nonproportional vectors to scalar multiples of three given other vectors. Therefore, the solution space of system 4.4) is one-dimensional. One solution to 4.4) is given by 1, 2,..., 7 ) where p is the determinant of the matrix obtained by removing the p-th column of the matrix at the left-hand side of 4.4). Therefore, there is a non-zero λ K such that U = λ ). Note that 1,..., 4 contain the fifth, sixth, and seventh column of the matrix at the left-hand side of 4.4). Therefore, each of 1,..., 4 is a sum of terms each of which is up to sign a product of six numbers, containing one of α p, β p for p = i, j, k and one of αp, βp for p = i, j, k. Consequently, U v = λ v max 1 v,..., 4 v ) c v λ v max αp v, β p v ) max αp v, βp v ) ) for v M K, 4.5) p=i,j,k where for infinite places v, c v is an absolute constant and for finite places v, c v = 1. Let v /. Then since U GL 2 O ) we have det U v = 1, whence U v = 1. Further, α p, β p, αp, βp O for p = i, j, k, therefore, these numbers have v-adic absolute value 1. It follows that λ v 1 for v /, and together with the Product formula this implies λ = v λ v 1. Now 4.5) implies, on taking

14 14 Jan-Hendrik Evertse the product over v, H U) c 1 λ c 1 p=i,j,k p=i,j,k H α p, β p )H α p, β p) ) H α p, β p )H α p, β p) ), where c 1 depends only on K. By taking the product over all subsets {i, j, k} of {1,..., r}, on using 4.2), 4.3), Lemma 2, we obtain H U) r 3) c2 H F )H F U ) ) r 1 2 ), where c 2 depends only on K, r. This implies 4.1). Lemma 6. Let M be an extension of K of degree r and T the set of places on M lying above those in. Denote by x x i) i = 1,..., r) the K-isomorphisms of M. i) Let F X, Y ) = r αi) X + β i) Y ), where α, β O T. Then F O [X, Y ] and H F ) 1/r H T α, β). ii) Let ξ M with ξ 0. Then there are α, β O T such that ξ = α/β and such that for the binary form F X, Y ) = r αi) X + β i) Y ) we have H F ) 1/r hξ). Here the constants implied by, depend only on M. Proof. i) F has its coefficients in O since O T is the integral closure of O in M. Let M be the normal closure of M/K and T the set of places on M lying above those in T. By the Extension formula, we have H T α, β) = H T α, β). Further, by the Extension formula and Lemma 2 we have H F ) = H T F ) H T α i), β i) ). Now M /K is normal, hence if w 1,..., w g are the places on M lying above some v M K then for i = 1,..., r, the tuple of absolute values i) wj : j = 1,..., g) is a permutation of wj : j = 1,..., g). Therefore, H T α i), β i) ) = H T α, β) = H T α, β) for i = 1,..., r. This implies i). ii) The ideal class of 1, ξ) the fractional ideal with respect to O M generated by 1, ξ) contains an ideal, contained in O M, with norm 1. This implies that there are α, β O M with ξ = α/β such that the ideal α, β) has norm 1. It follows that w / T max α w, β w ) 1. Now by the Product formula we have hξ) = w M M max1, ξ w ) = w M M max α w, β w ) and so hξ) w T max α w, β w ) = H T α, β). Together with i) this implies ii). We now complete the proof of Theorem 4. Let L = Kξ, η), r = [Kξ) : K], s = [Kη) : K]. Then 1.16) implies that r 3, s 3, [L : K] = rs. Further, let T be a finite set of places on L such that 1.17) holds and the set of places on K lying below those in T. We add to all infinite places on K that do not belong

15 Lower bounds for resultants II 15 to. Thus, contains all infinite places and the places lying below those in T. There might be places in above which there is no place in T but then 1.17) still holds. Denote by T 1 the set of places on L lying above the places in. Note that T is a proper subset of T 1. In what follows, the constants implied by, depend only on L,. We mention that constants depending on some subfield of L may be replaced by constants depending on L since L has only finitely many subfields. Denote by x x i) i = 1,..., r) the K-isomorphisms of Kξ) and by y y j) j = 1,..., s) the K-isomorphisms of Kη). From part ii) of Lemma 6 applied with M = Kξ), M = Kη), respectively) it follows that there are α, β, γ, δ such that ξ = α β, η = γ δ, where α, β belong to the integral closure of O in Kξ) and γ, δ to the integral closure of O in Kη) and such that for the binary forms F X, Y ) = α i) X + β i) Y ), GX, Y ) = γ j) X + δ j) Y ) 4.6) we have j=1 H F ) 1/r hξ), H G) 1/s hη). 4.7) The forms F, G have their coefficients in O, and deg F = r 3, deg G = s 3. Further, since Kξ), Kη) are linearly disjoint over K, the numbers ξ and η are not conjugate over K and so F G is square-free. Hence all hypotheses of Theorem 2 are satisfied. The splitting field of F G is the normal closure of L over K. By Theorem 2 there is a matrix U GL 2 O ) such that RF, G) H F U ) s H G U ) r) ) By 4.6) we have F U X, Y ) = α ) i) X + β ) i) Y ), G U X, Y ) = with α, β ) = α, β)u, γ, δ ) = γ, δ)u. We define the following quantities: γ ) j) X + δ ) j) Y ), ξ η w Λ w := max1, ξ w ) max1, η w ) = αδ βγ w for w T 1, max α w, β w ) max γ w, δ w ) Λ α δ β γ w w := max α w, β w ) max γ w, δ for w T 1, w ) j=1 H := H F ) 1/r H G) 1/s, H := H F U ) 1/r H G U ) 1/s. Thus, 4.7) and 4.8) translate into H hξ)hη), RF, G) 1/rs H ) ) Note that we have to estimate from below w T Λ w. For matrices A = a b c d ) and places w on L we put A w = max a w,..., d w ). Let v and w T 1 a place lying above v. Using that the restriction of w to K

16 16 Jan-Hendrik Evertse is [Lw:Kv]/[L:K] v and also that αδ βγ = det U 1 α δ β γ ), max α w, β w ) U 1 w max α w, β w ), max γ w, δ w ) U 1 w max γ w, δ w ), we obtain Λ w det U 1 w U 1 2 w det U Λ 1 v w = U 1 2 v ) [Lw :Kv ] [L:K] Λ w. Note that by Lemma 5 we have H U 1 ) H F )H F U ) ) 3/r and H U 1 ) H G)H G U ) ) 3/s. Hence H U 1 ) H H ) 3/2. We take the product over w T. Using 1.17), det U 1 v / U 1 2 v 1 for v and det U O we get v det U 1 ) [Lw :Kv ] v [L:K] U 1 2 v v w T w v det U 1 v U 1 2 v H H ) 3W. ) W = det U 1 H U 1 ) 2 ) W Hence w T Λ w H H ) 3W w T Λ w. 4.10) We need also lower bounds for w T 1 Λ w, w T 1 Λ w. Note that since [L : K] = [Kξ) : K][Kη) : K] = rs we have RF, G) = j=1 α i) δ j) β i) γ j) ) = N L/K αδ βγ). Together with the Extension formula this implies RF, G) 1/rs v = w v αδ βγ w for v M K, and by applying part i) of Lemma 6 and 4.9) we obtain RF, G) 1/rs Λ w = H T1 α, β)h T1 γ, δ) RF, G) ) 1/rs RF, G) 1/rs = H F ) s H G) r H w T 1 H ) H ) Completely similarly we get, in view of 1.11), Λ w RF U, G U ) H w T 1 1/rs = RF, G) 1/rs H H ) )

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