Cassels-Catalan Curves

In this chapter we prove that the function field of a Cassels-Catalan curve not containing√

−1 can only have Pythagoras number two if its base field is hereditarily Pythagorean.

We first give a proof of geometric flavor, before we take a look at a very early version of the proof of more combinatorial flavor. In the second section of this chapter, we give a purely geometric proof for the special case of conic curves.

Remark 4.1. A field is hereditarily Pythagorean if and only if it is real and√

−1 is contained in each of its nonreal finite extensions. See [Bec78, III, Thm. 1].

Remark4.2. An ordering on a fieldK extends toK(√

a) fora∈K^× if and only if a is positive at this ordering. See [PD01, 1.2.3]. In particular,K(√

a) is nonreal if and only if−a∈P K^×2.

Proposition 4.3. Let C be a smooth integral curve over a non hered-itarily Pythagorean field K. If P ∈ C is a closed point with K(P) nonreal and√

−1∈/ K(P), thenp(K(C))≥3.

Proof. If K(P)=OC,P/m_C,P is nonreal and √

−1 ∈/ K(P), the as-sertion follows with (2.26), as P ∈ C is a point of codimension one,

wherebyOC,P is a discrete valuation ring.

1. The general case

The proof of the following (4.4) uses a small observation on scalar restrictions obtained in the third chapter. At the end of the current section, I present my first proof for (4.4), which omits all geometry and uses only combinatorial arguments.

Proposition 4.4. Let K be an infinite field andL/K be a finite sep-arable extension such that L is not Pythagorean. Then there exists ξ∈L such that L=K(ξ²) andξ²+ 1∈/ L^×2. Moreover, there exists σ∈P

L^×2\L^×2 such that L=K(σ) and σ+ 1∈/L^×2.

45

46 4. CASSELS-CATALAN CURVES

Proof. Fixz∈Lwithz²+ 1∈/L^×2. Forν∈L^×, consider the terms α= ^ν_z²2,β=ν²+z²,γ=^(z2_ν⁺¹⁾2 ² +z²,δ= ^(z_z²2⁺¹⁾ν²² and= ^z2_ν⁺¹2 . In terms of elementary geometry, these are rational functions inνover L. Let G = {x ∈ L | K(x) = L}. This is a K-Zariski open subset of L, as it is the complement of the finitely many subspaces ofLthat correspond to the finitely many intermediate extensions of L/K. By (3.5), the preimage ofGunder any nonconstantK-rational function on L is nonempty. Moreover, it is K-open in L. As the intersection of finitely many nonemptyK-open subsets ofLis nonempty, there exists ν ∈L^× such thatα, β, γ, δ, ∈G.

Note that,¹ ∈PL^×2\L^×2. If+ 1∈/ L^×2, we setσ=. Otherwise we have that ¹⁺ = ¹ + 1∈/L^×2, and we setσ= ¹.

Note that α∈L^×2. Ifα+ 1∈/ L^×2, chooseξ= ^ν_z. Assume now that α+ 1∈L^×2. Then β ∈L^×2. Ifβ+ 1∈/ L^×2, chooseξ ∈L such that ξ²=β. Assume now thatβ+ 1∈L^×2. Then ν²+z²+ 1∈L^×2 and ν²+z²∈L^×2. It follows that^(z2_ν⁺¹⁾2 ²+z²+1∈/ L^×2, sincez²+1∈/L^×2. Recall thatδ=^(z_z²2⁺¹⁾ν²². Ifδ+1∈/ L^×2, chooseξ=^z2_zν⁺¹. Ifδ+1∈L^×2, then γ ∈ L^×2 and γ+ 1 ∈/ L^×2, and we can choose ξ ∈ L such that

ξ²=γ.

We denote by±K^×2the setK^×2∪ −K^×2, for any fieldK.

Lemma 4.5. Letu∈K^×\ ±K^×2 andr≥1. Letγ∈Kalg be such that γ^2r =u. ThenK^×∩K(γ)^×2=K^×2∪uK^×2.

Proof. As −u /∈K^×2, and thus −u /∈4K^×4, the polynomial T^2r−u is irreducible by [Lan02, Chap. VI, (9.1)]. Write d =γ², L =K(d) and M =K(γ). Note thatM/L is a quadratic extension. AsT^2r−1−u is the minimal polynomial of d over K, the norm of d with respect to L/K is ±u. As u /∈ ±K^×2, it follows that K^× ∩dL^×2 = ∅. As L^×∩M^×2=L^×2∪dL^×2, we have that

K^×∩M^×2=K^×∩(L^×2∪dL^×2) =K^×∩L^×2.

The statement thus follows by induction onr.

Corollary 4.6. Suppose−1∈/ K^×2. Letu∈K^×\ ±K^×2 andn∈N. There existsx∈Kalg with xⁿ =uandK^×∩K(x)^×2⊆K^×2∪uK^×2. Proof. If n is odd, we can choose x∈ Kalg with xⁿ = usuch that [K(x) :K] is odd, wherebyK^×∩K(x)^×2⊆K^×2. Assume now thatn is even. If u /∈K^×2, then we write n= 2^rmwith m odd and r ≥1, and apply (4.5) together with the previous case.

1. THE GENERAL CASE 47

Corollary4.7. Suppose−1∈/K^×2. Letv∈K^×\ −K^×2 andm∈N. There existsy∈Kalg such that y^m=v and−1∈/K(y)^×2.

Proof. Letr∈Nbe maximal such that 2^r|mandv∈K^×2r. Let u∈K be such thatu^2r =v. We writem=n2^r. Ifnis odd, then we can choosey ∈Kalg such thaty^m =v and [K(y) : K] is odd, and it follows trivially that−1∈/K(y)^×2.

Assume thatnis even. Thenu /∈K^×2by the maximality ofr. Further-more, we claim thatu /∈ −K^×2. Ifr= 0 we have that u=v /∈ ±K^×2. If r > 0 then u /∈ −K^×2 by the maximality of r and the fact that (−u)^2r = v. Using (4.6), we choose y ∈ K_alg such that yⁿ = u and K^×∩K(y)^×2 ⊆K^×2∪uK^×2. Theny^m=v and−1∈/ K(y)^×2, since

u /∈ −K^×2.

Proposition4.8. LetKbe a field with−1∈/ K^×2. Letu∈K^×\±K^×2 and v ∈ K^×\(−K^×2∪ −uK^×2). Let n, m≥ 1. Then there exists a finite extension M/K such that −1 ∈/ M^×2, with x, y ∈ M such that xⁿ =uandy^m=v, and such that M =K(x, y).

Proof. We choose x∈ Kalg such that xⁿ = uand K^×∩K(x)^×2 ⊆ K^×2∪uK^×2. Then−1,−v /∈K(x)^×2. By (4.7) there existsy∈Kalg

such thaty^m=v and−1∈/K(x, y)^×2. SetM =K(x, y).

Corollary4.9. LetL/K be a finite field extension such thatLis real and not Pythagorean. Leta, b∈K such thata, b∈L^×2∪ −L^×2. For integersn, m≥1, there exists a finite extensionM/L, such that −1∈/ M^×2, and with x, y∈M such that1 =axⁿ+by^m andM =K(x, y).

If moreovernor mis even, then we can chooseM to be nonreal.

Proof. By (4.4) there exists ξ ∈ L with ξ²+ 1∈ P

L^×2\L^×2 and L=K(ξ²), and furtherσ∈P

L^×2\L^×2withL=K(σ) andσ+ 1∈ PL^×2\L^×2.

In the case wherea, b∈L^×2, setu=−_aσ¹ andv= ¹_b(1 +_σ¹). Thenu /∈

±L^×2and−v /∈L^×2∪uL^×2, as−uv= _ab^{1 σ+1}_σ2 . Moreover, 1 =au+bv.

In the case where −a,−b ∈ L^×2, set u = ^ξ2_a⁺¹ and v = ^−ξ_b². Then u /∈ ±L^×2and−v /∈L^×2∪uL^×2. Moreover, 1 =au+bv.

In the case where−a, b ∈ L^×2 set u= ^σ+1_a and v = ^−σ_b . Then u /∈

±L^×2 and−v /∈L^×2∪uL^×2. Moreover, 1 =au+bv.

In the case wherea,−b ∈ L^×2 set u= ^−σ_a and v = ^σ+1_b . Then u /∈

±L^×2 and−v /∈L^×2∪uL^×2. Moreover, 1 =au+bv.

In each case, by (4.8), there existx, y∈Lalg such thatxⁿ=u,y^m=v and √

−1 ∈/ L(x, y). Moreover, since u ∈ L(x, y) and K(u) = L, it

48 4. CASSELS-CATALAN CURVES

follows thatL(x, y) =K(x, y). Obviously 1 =axⁿ+by^mas 1 =au+bv.

SetM =L(x, y). Now suppose that normis even. By symmetry, we can assume that nis even. Then xⁿ =uis both a square inM and, by the choices ofuin each case, a negative sum of squares inM. Thus

M is nonreal.

Theorem 4.10. Let K be a field not containing √

−1. Let a, b∈K^× andn, m∈Nsuch that char(K) does not dividenm. Let F/K denote the function field of the plane affine curveCdefined by1 =aXⁿ+bY^m. Thenp(F) = 2 implies that K is hereditarily Pythagorean.

Proof. Note that the affine curve C is smooth, by (3.42). We first show that K is real. Assume that K is nonreal. Then, by (4.3), we have that √

Let us first consider the case wherenis odd. ThenF is clearly an odd degree extension of the rational function fieldK(X). Thenp(K(X))≤ p(F) ≤ 2 by Springer’s theorem (5.1). Thus p(K(X)) = 2, and the claim follows from (1.5). Thus assume thatnis even.

Suppose there exists a finite real extension L/K that is not Pythago-rean. We can assume that a, b∈L^×2∪ −L^×2 since at least one of the four extensionsL(√

±a)(√

±b) is real, and not Pythagorean by (1.3).

By (4.9) there exists a pointP∈Csuch that−1∈/K(P)^×2andK(P) is nonreal. Again (4.3) yields the contradictionp(F)≥3.

A combinatorial proof. The core of the proof of the main re-sult (4.10) of this chapter was (4.4). The previously presented proof of (4.4) is of geometric nature. In the following, the author’s firstly ob-tained proof for (4.4) is presented, which is of combinatorial nature. In particular, the geometric preparation on scalar restriction of the third chapter is not needed.

LetK be an infinite field andL/K a finite separable extension that is not Pythagorean. We will letGdenoteL\SI={x∈L|K(x) =L}, the set of primitive elements forL/K.

1. THE GENERAL CASE 49

Lemma4.11. Let c∈L^×,d∈L andν∈G. Then

(a) we have thatc(ν+x)²+d∈Gfor all but finitely manyx∈K.

(b) for each except finitely manyα∈Kthere are infinitely manyx∈K with _(ν+x)^cα22 +d∈G.

Proof. LetI ={N/K|N (L}, the finite set of intermediate fields ofL/K that are different fromL.

(a) ForN ∈I, let AN =

As the Vandermonde matrix on the left has nonvanishing determinant (x₃−x₂)(x₂−x₁)(x₁−x₃), we can multiply across by its inverse matrix,

(b) Assume that there are infinitely many elementsα∈K^× for which the set Bα = {x ∈ K | _(ν+x)^cα22 +d ∈ G} is finite. Set n := |I| and distinctx∈C⁰, the corresponding fieldsK_xmay not coincide. However, applying the pigeon-hole principle again, we see that there existsN∈I for whichC={x∈C⁰ |K_x=N} is infinite. In particular, forx∈C,

contradicting the result by (a), that _c(α^(ν+x)2 ²

`−α²_m) ∈G for all but finitely

manyx∈K.

50 4. CASSELS-CATALAN CURVES

We recall (4.4) and present a geometry free proof.

Proposition 4.12. There exists ξ∈Lsuch that L=K(ξ²)and ξ²+

2. CONICS 51

Assume now thatα∈K^× is such thatMαis not cofinite in K. Then the set

K\(Hα∪Mα) =

x∈K\Hα

α²(z²+ 1)²

z²(ν+x)² + 1∈/L^×2

is infinite. It follows from (4.11.(a)) that_α^z2²(z^(ν+x)2+1)²² ∈Gfor somex∈K\ (Hα∪Mα), whereby we can letξ= ^α(z_z(ν+x)²⁺¹⁾, completing the proof.

2. Conics

In this section, we give a more conceptual proof for (4.10) in the special case of function fields of conics . No result of the previous section is used, instead we use a result obtained in the third chapter on generic splitting properties of quadrics and Severi-Brauer varieties.¹

Proposition4.13. LetCbe a regular projective conic over an infinite fieldK, and K(C)its function field. LettingL/K be a finite separable extension, we have thatLis the residue fieldK(P)of a pointP ∈C if and only ifCL has a rational point.

Proof. A smooth projective conic is a one dimensional regular pro-jective quadric. Hence (3.38) yields the result.

Remark4.14. In the case whereKis perfect andC/Ka smooth conic, this result yields that the property‘beeing the residue field of point on C’ is a hereditary property, that is, if L is a finite field of K with K(P)⊂Lfor someP ∈C, thenL=K(P⁰) for someP⁰ ∈C.

The same is not true if we replace ‘smooth conic’ by an arbitrary smooth projective curve overK. The Fermat curveX⁶+Y⁶=Z⁶, for example, has the ‘trivial’Q-rational points (0 : 1 : 1) and (1 : 0 : 1). In [Aig57], it was shown that (0 : 1 : 1) and (1 : 0 : 1) are also the onlyL-rational points of this curve for any quadratic field extension L/Q, whereby L6=Q(P) for any pointP onX⁶+Y⁶=Z⁶.

Theorem 4.15. Let K be a field not containing √

−1. Let F/K be the function field of a conic over K. Then p(F) = 2 implies that K is hereditarily Pythagorean, or even hereditarily Euclidean in the case whereF is nonreal.

Proof. Note that char(K)6= 2, by the assumption that −1 ∈/ K^×2. The conic C is given by Z² = aX²+bY² where a, b ∈ K^×. First assume that K is finite. Then every quadratic form of dimension at

1J. Van Geel and A. Wadsworth helped me establish this connection.

52 4. CASSELS-CATALAN CURVES

least 3 over K is isotropic, whereby the conic has a rational point and thus F/K is the rational function field K(X). Hence −1 +X² is a sum of 3 but not 2 squares by the second representation theorem ([Lam05, IX.2.1.]). Suppose that K is infinite and nonreal. Then at least one of the four biquadratic extensions L = K(√

±a,√ is a separable field extension. If√

a∈M, then (1 : 0 :√ -rational point. In any case,C has anM-rational point, whereby (4.13) implies thatCM is rational. ThusChas a pointP withK(P) =M by (3.37). Since−1∈/ M^×2, it follows thatp(F)≥3, by (4.3).

Finally, assume thatK is a real field but is not hereditarily Pythago-rean. Let L be a finite real extension of K that is not Pythagorean.

F is the function field of a smooth projective conic C over K. The conic C is given by Z²=aX²+bY², where a, b∈K^× can be chosen such thatabis positive at a fixed ordering onL. This ordering extends either toL(√

a,√

b) or toL(√

−a,√

−b), whereby at least one of these field extensions is real. Neither of them are Pythagorean, by (1.3). We can therefore assume, without loss of generality, that eithera, b∈L^×2 or −a,−b∈L^×2. C. In any case,Chas anM-rational point, andCthus isM-rational by (4.13). By (3.37),C has a pointP withK(P) =M. Since√

−1∈/M it follows that p(F)≥3 by (4.3).

HenceKis hereditarily Pythagorean. IfF is nonreal, in addition, then a, b ∈ −P

K^×2. SinceK is Pythagorean, the corresponding conic is thus Y² =−(X²+ 1) and by (iii) ⇒(i) of (1.4) it follows thatK is

hereditarily Euclidean.

CHAPTER 5

Im Dokument Sums of Squares in Algebraic Function Fields (Seite 55-63)