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Guide to the Exercises Ty Ruthlawrence Oxford University APPENDIX Guide to the Exercises ty RuthLawrence Oxford University The purpose of this appendix is to give, for each exercise, a comment, a hint, a sketch, or in a few cases, a complete solution. Pure algebra has not been worked. Exercises which are merely a matter of applying techniques given in the text to particular examples for the purpose of drill have not generally been solved fully here-the answers only being given. When there is -a batch of somewhat similar questions a representative question has been selected to be completely solved and answers only have been given to the rest. A few questions involve rather tedious numerical calculations not reducible by technique. Readers who have c:arried these out can, by checking their answers against the given ones, gain confidence. They mayaiso be comforted to discover that they have not missed some subtle point or ingeniously simple route to a solution. In a few cases a link is suggested to oonnected problems which the interested reader may like to pursue. I hope that this appendix will fulfill its purpose of helping readers who experience any difficulties with the problems to overcome them; thus gaining the maximum understanding and insight which the author intended by his carefully chosen incorporation into the text. I wish to acknowledge the help received from Professor Husemöller whilst compiling this appendix, by way of some useful discussions and suggestions. CHAPTER 1, §1 1. TedioUl calculation gives: 7P = ( - 5/9, 8/27), 8P = (21/25, -69/125), 9P = (-20/49, -435/343), 10P = (161/16, -2065/64), 316 Appendix. Guide to the Exercises -7P = (-5/9, -35/27), 8P = (21/25, -56/125), -9P = (-20/49,97/343), -lOP = (161/16, 2001/64). Mazur's theorem then implies P has infinite order. For, 7P, 8P, 9P, lOP # 0 and none of the above eight points coincide, i.e., p's order does not divide any integer 14 to 20 inclusive. Thus P cannot have order 2, 3, 4, 5, 6, 7, 8, 9, 10, or 12 (the laUer since 6P # - 6P). 2. Very straightforward. Here - 2P = p. P = (0, 2), i.e., - 2P = Pas required. 3. As in Exercise 2, we work out - 2P = p. P = (0, 0). Clearly, (0,0) has order 2, and so P has order 4. 4. It is found that 2P = (0, 1), 3P = (-1,0). Thus 3P has order 2. So p's order is 2 or 6 (divides 6 and does not divide 3). As 2P # 0, thus p's order is 6. 5. Here - 2P = (24, 108), and 2P = (24, - 108). From P, 2P we obtain - 3P. It is found that - 3P = (24, -108) = 2P. So P has order s. 6. A suitable subgroup is {O, (0, 0), (1, 0),( - 9,0)}. The calculations for nP (n = -7, ... , + 7) get very tedious! Here are the first few: -14161 -466004) P = (-1,4), 2P = (25 195) 3P= ( ------ 16' 64 ' 1681 '68921 ' 14421367921 - 67387812533791) 4P= ( 4090881600' 6381775296000 ' -P=(-l, -4), _ 2P = (25 - 195)" 3 (-14161 466004) 16' 64 ' - P = l681' 68921 ' -4P = (14421367921 67387812533791). 4090881600 ' 6381775296OöO Now use Mazur's theorem. Since P has infinite order, no finite subgroup contains P. It helps in the calculations to realize that if PI' P2 , P3 are collinear points on the curve, then XI + X 2 + x 3 = (slope of line)2 - 8, and thus for any Q, R on the curve, Q. R (and hence Q + R) can be determined. CHAPTER 1, §2 1. Answers: 6P = ( - 2/9, - 28/27), 7 P = (21, -99), 8P = (11/49, 20/343), - 6P ~ ( - 2/9, 1/27), - 7 P = (21, 98)' - 8P = (11/49, - 363/343). Use Mazur's theorem-very similar to §1, Exercise 1. 2. Straightforward algebra gives u = (w/3x)(y + 4), u = (w/3x)(5 - y). 3. For c = 1, the group is {O, (12, 36), (12, -36)}, ie., Z/3Z. For c = 2, the group is {O, (3, On, i.e., Z/2Z. The complexit) increases rapidly for c ~ 8. The interested reader may Iike to consider c = 3, 4, .... Appendix. Guide to the Exercises 317 4. Here 2P = (9, -18), 4P = (0, 0). So 4P = -4P, i.e., P has order 8. 5. One easily computes 2P = (-1, -2), -3P = (3, 6), -4P = (3,.-6). Thus P has order 7. 6. Multiples are 0, (3, 8),( - 5, -16),(11, - 32),(11,32),( - 5, 16),(3, -8)in that order. Thus (3, 8), (11, 32) both have order 7, since (11, 32) = 4(3, 8) and 4,7 are coprime. 7. Here E(Q) = {(O, 0), (0, -I)} (i.e., Z/2Z). 8. The giyen cubic in P follows since P = (x, y) gives - P = (x, y') where y, y' are two roots, for Y, of y 2 + (a\x + a3)Y - (x3 + a2x2 + a4 x + a;;) = 0 and so -(y + y') = (a\x + a3). Thus 2P = 0 iff y = -t(a1x + a3) and thi.. reduces to the given condition on x. Thus there are zero, one, or·three solutions for x. In every case 0 is a solution of 2P = O. So, the group consists of one, two, or four elements. In the fir;;t two cases, the groups must be 0, Z/2Z, respectively. In the last case, there are three elements of order 2, and so we get (Zj2Z x Z/2Z). CHAPTER t §3 1. We get (-1,0) (order 2); (0,1), (0, -1) (order 3); and (2, 3), (2, -3) (order 6). The laUer two are possible r:enerators. 2. (a) The condition is that there are three points of order 2. Hence· we need x2 + ax + b to have two roots in k, i.e., a2 - 4b must be a square. (b) The condition is that there exists a point, P, on the curve, of order 4. Thus 2P has order 2 and is thus (cx, 0) some cx. Since a2 - 4b is not a square, so x 2 + ax + b #- 0 for an x E k. Therefore cx = 0, and so 2P = (0, 0). This gives us a condition that there exists A such that y = AX intersects the cubic curve in a double point. So 4b = (a - A2 )2 and so b = c2 with a - A2 = ±2c. So one of a ± 2c must be a square. (c) If a + 2c, a - 2c are squares (with b = c2 ), then from (b), (0,0) = 2P some P. Hence the group E(k) contains a subgroup generated by P and (cx i, 0) where CXi are the two roots of x 2 + ax + b = O. Thus Zj4Z x Z/2Z s; E(k). Conversely, if Z/4Z x Z/2Z s; E(k), then (0,0) E 2E(k) gives, from (b), that one of a ± 2c must be a square. From (a), a2 ._ 4b = (a + 2c)(a - 2c) is a square. So a + 2c and a - 2c are squares. Note that in this question, the extra conditions, which are apparently asymmetric, are required since·the equation of the cubic curve has fixed (0, 0) as a point of order 2. 3. Points of infinite order are (l, 2), (2, 3~ (~1, 1), (3, 5), respectively. To check this, use Theorem (3.2). 4. Consider the canonicaI map E(k).!. E(k)/2E(k). Then, if P, Q a;e Iinearly dependent, say nP = mQ with m, nEZ, we get n'[P] = m'[Q], 318 Appendix. Guide to the Exercises where [P] is the equivalence c1ass of'P (i.e., O(P» and n' == n(mod 2). Thus, if P = (3,41. Q = (15, 58), then P, Q have infinite order. We thus only need to check that [P], [Q!O are alI distinct. In fact Q - P = (313/36, -5491j216). ;So, we now evaluate 2(x, y). The condition that 2(x, y) = (IX, ß) on y2 = x 3 - 11 is that 0 = x 4 - 4IXX3 + 88x + 44IX. SO we use IX = 3, 15,313/36 to show that P, Q, Q - Prt2E(k).* Thus [P], [Q], 0 are aII distinct. Hence nP '" mQ, "Vm, neZ, not both zero. 5. This is similar to Exercise4. If P = (0, 2), Q = (I, 0), R = (2, 0) then we check: (i) P, Q, R have infinite order. (ii) [P], [Q], [R], 0 are aII distinct. (iii) [P + Q + R] '" O. CHAPTER I, §4 1. This is very straightforward: define K ;= P' - P and then show that K e .A. 2. In (4.1) the conditions are that -IX, -ß are squares. In §3, 2(c) the conditions are IXß = c2, some.c, and 2c - (IX + ß) or - 2c - (IX + ß) is a square. Here ±2c - (IX + ß) = ±2~ - IX - C2/IX = -(0: + C)2/a.. SO -IX (and hence also -ß) is a square. 3. When 2E(k) = 0, 4E(k) = o. When zE(k) = Z/2Z x Z/2Z, 4E(k) is Z/2Z x Z/2Z or Z/4Z x Z/2Z or Z/4Z x Z/4Z. When 2E(k) = Z/2Z, 4E(k) is Z/2Z or Z/4Z. The numbers of elements of order 4 are: 0; 0, 4, 12; 0, 2, respectively. 4. Suppose otherwise, that (Z/4Z)2 s; E(Q), by an inc1usion map j. Then the elements of order 2 are j«2, 0», j«(O, 2)), j«2, 2». Thus in normal form we get y2 = (x - IX)(X - ß)(x - y), where (IX, 0) = j«2, 0», ete. Hence (IX, 0) = 2P where P = j«I, 0», j«(1, 2», j«3, 0» or j«3, 2».
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