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The Logarithmic Derivative on the Elliptic Curve

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The Logarithmic Derivative on the Elliptic Curve The Logarithmic Derivative Homomorphism on an Elliptic Curve De…ned over Constants: Addendum Phyllis J. Cassidy City College of CUNY November 16, 2007 1 Introductory remarks At the Friday meeting, the logarithmic derivative map came into play, since it is the analogue for elliptic curves de…ned over constants of the Manin homo- morphism. Let be a di¤erentially closed di¤erential …eld, with derivation d G operator = dt , and with (algebraically closed) constant …eld of character- istic 0. Note that is also algebraically closed. Let x; y; z Cbe di¤erential G indeterminates. For any subset S of P2( ), and sub…eld of , set S ( ) equal to the set of points in that are rationalG over . Let E L= E (G ) be theL elliptic L L G curve in P2( ) de…ned by the a¢ ne Legendre equation G 2 y = (x e1)(x e2)(x e3); e1; e2; e3 distinct elements of : C We equip E with the Kolchin topology, making it a di¤erential algebraic group. # Let Etors be the torsion group of E, and denote by E the Kolchin closure of # Etors . It is clear that E = E( ). Let ` : E Ga ( ) be the logarithmic derivative map. Dating back toC Kolchin’s 1953! paper,G the fact that ` is a surjective di¤erential rational homomorphism with kernel E# has been crucial in the development of the Galois theory of di¤erential …elds. An essential hypothesis is that E be de…ned over constants. The Manin homomorphism is its replacement in the case where E does not descend to constants. In this addendum to Friday’stalk, I will show that ` is everywhere de…ned and surjective, and that its kernel is E#. For the fact that it is a group homomorphism, I refer you to Kovacic, “On the inverse problem in the Galois theory of di¤erential …elds: II”, Annals of Mathematics, 93, 269-284. The logarithmic derivative map is the dual of the invariant di¤erential on the elliptic curve. However, I would like to discuss it ab initio without reference to this – and in the language of di¤erential algebraic groups. mainly because it is a good exercise in de…ning a global section of the di¤erential algebraic variety E by patching di¤erential rational functions on an a¢ ne open cover. I hope to give a proof in a later note that ` is an invariant derivation on E, thus 1 proving that it is a homomorphism of di¤erential algebraic groups. The proof of surjectivity in this language is very easy. 2 The kernel of ` In this section, we will de…ne ` on the a¢ ne open yz = 0, and assume that is everywhere de…ned on E. We will show that ker ` = E6 #. The de…ning equation of E = E ( ) is G 2 y = (x e1)(x e2)(x e3); e1; e2; e3 distinct elements of : C The associated homogeneous equation is 2 y z = (x e1z)(x e2z)(x e3z): Let U be the a¢ ne open subset yz = 0 of P2 ( ). We identify the a¢ ne open subset z = 0 of the projective plane6 with the a¢G ne (x; y)-plane. So, U is the complement6 of the x-axis. We de…ne the logarithmic derivative map ` : E (U) Ga ( ) ! G by the formula x `(x; y) = 0 : y The constant trace E ( ) of E is a Zariski dense connected di¤erential alge- braic subgroup of E ( ). AsC we remarked in the introduction, it is the Kolchin # G closure E of Etors . We will show that ker (` U) = U ( ). j C 2 y = (x e1)(x e2)(x e3): 3 2 = x (e1 + e2 + e3)x + (e1e2 + e1e3 + e2e3) x e1e2e3: Let 3 2 P (x) = x (e1 + e2 + e3)x + (e1e2 + e1e3 + e2e3) x e1e2e3: y2 = P (x) : Set F (x; y) = y2 P (x): @F @F dP = 2y: = : @y @x dx @F dP y = x @y 0 dx 0 2 Let (x; y) U. Since E is a smooth algebraic curve, either @F or dP fails to 2 @y dx vanish at (x; y). @F = 2y = 0 y = 0 @y () P (x) = 0 () x = e1; e2; e3: () So, the fact that the roots of P (x) are distinct guarantees the smoothness of the elliptic curve! We have eliminated them from U. @F (x; y) U, = 0 at (x; y): 8 2 @y 6 (x; y) U; x0 = 0 y0 = 0: 8 2 () x ` (x; y) = 0 : y ker (` U) = U ( ) : j C U ( ) is an open subset of the Kolchin closed subgroup E (C). Suppose ` is everywhereC de…ned on E, hence on E ( ). It follows that since it vanishes on a dense open subset of E(C) it will vanishC on all of E( ): Can it vanish on a point (x; y; z) not in E(C)? What are the points in theC complement of U? If z = 0, then (x; y; z) = (0; 1; 0), the unique point at . If y = 0, then x = e1 1 or e2 or e3. These 4 points are all in E(C). Thus, ker ` = E( ). C 3 ` is an everywhere de…ned di¤erential ratio- nal function on E. To show this, we must extend the de…nition to an open cover of E. So, we cover E by a …nite number of a¢ ne open subsets on which the logarithmic derivative map is de…ned by everywhere di¤erential rational functions that agree on the intersections. I decided to transform the Legendre form of the de…ning equation of E to the Weierstrass form, which is easier, since it is missing the second degree term in the cubic P (x). So, we choose a so that the translation x x a, which …xes y, removes the degree 2 term.2C This transformation does not7! change the formula for the logarithmic derivative in (x; y) coordinates. It su¢ ces to show that ` is a global section of the di¤erential structure sheaf on E, now de…ned by the Weierstrass equation 2 3 2 3 y = x g2x g3; g2; g3 ; 4g2 27g3 = 0: 2 3 2 3 2 C 6 y z = x g2xz g3z : 3 2 3 In Weierstrass form,the inequation 4g2 27g3 = 0, which tells us that the discriminant of the cubic vanishes nowhere on the6 curve, guarantees that its roots are distinct, and the elliptic curve is smooth, as we saw in the …rst section. 3.1 The a¢ ne open yz = 0: 6 As usual, we de…ne ` : E E ( ), by the formulae ! G x ` (x; y) = 0 ; y zx0 xz 0 ` (x; y; z) = yz …rst in a¢ ne coordinates, then in homogeneous coordinates, on the a¢ ne open U1 : z = 0; y = 0. ` is everywhere de…ned on U1. ` is not everywhere de…ned on the6 entire6 open subsetz = 0 of E. We have not covered the points (x; y), y = 0. We need an open neighborhood6 of these 3 points on E. 6 y = 0 x is a root of the cubic () 3 Q (x) = x g2x g3: dQ 2 Since all the roots are simple roots, = 3x g2 does not vanish at any of dx these points in E. Let U2 be the a¢ ne open subset of the open a¢ ne subset 2 z = 0, de…ned by the inequation 3x g2 = 0. De…ne 6 6 2y0 ` (x; y) = 2 ; 3x g2 2 (zy0 yz0) ` (x; y; z) = 2 2 ; 3x g2z on U2 What’smissing? We havent shown that the log derivative is de…ned on an open neighborhood of the unique point at in…nity on E = (0; 1; 0), which we get by setting z = 0 in the homogenized equation. Now,1 set y = 1 in the homogeneous equation for E. We now have a¢ ne coordinates (x; z), and we get the de…ning a¢ ne equation and its homogenization 3 2 3 z = x g2xz g3z ; 2 3 2 3 y z = x g2xz g3z : satis…ed by = (0; 1) : The de…ning polynomial of E in the (x; z) plane is 1 3 2 3 P (x; z) = z x g2xz g3z : We follow the technique used to capture the roots of the cubic: We use the fact that E is smooth. Therefore, one of the partial derivatives of P (x; z) must be nonzero at . Indeed, although @P vanishes at = (0; 0) ; 1 @x 1 @P = 1 + 2g xz + 3g z2 @z 2 3 4 does not vanish. So, to capture , we use the same trick we used to de…ne ` 1 at the 3 roots of the cubic. Let U3 be the a¢ ne open de…ned by the inequation 2 y = 0; 1 + 2g2xz + 3g3z = 0. De…ne 6 6 2x0 ` (x; z) = 2 ; 1 + 2g2xz + 3g3z 2(yx0 xy0) ` (x; y; z) = 2 2 : y + 2g2xz + 3g3z The union of the three a¢ ne open sets equals E = E ( ). The di¤erential rational functions de…ning the logarithmic derivative on theseG open subsets are everywhere de…ned. However, we must verify agreement on the intersections U1 U2;U1 U3;U2 U3. We will then have a global di¤erential rational map\ on E.\ Note that\ although for ease of exposition I have avoided it, the logarithmic derivative function is actually a di¤erential polynomial function on E. 3.1.1 U1 U2 \ 2 2 We must show that if z = 0; y = 0; 3x g2 = 0; the formulae for ` agree. On the one hand, 6 6 6 x ` (x; y) = 0 : y On the other hand, 2y0 ` (x; y) = 2 3x g2 But, recall 2 dQ 2 y = Q (x) ; and = 3x g2: dx 2 3 Di¤erentiate y = Q(x); where Q(x) = x g2x g3 dQ 2yy = x : 0 dx 0 So, 2 3x g2 x0 = 2yy0; and the two di¤erential rational functions agree on the intersection.
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