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».

Thus IX - ß, IX - Y are squares. Similarly ß - IX, ß - y, y - IX, Y - ß are squares. This leads to - I as a square, a contradiction.

CHAPTER 1, §5

1. A cusp occurs if az + af!4 = O. A double point occurs if and only if(az + aU4) > O. Theseeonditionsareeasilyobtainedfromy = x(-a1/2 ± Jx + (a 2 + aU4». The grapht; below indicate the forms ofthe eurve for a 1 < O.

.. One checks that the quartic equation has no solution in Q, by Eisenstein's irreducibility criterion. Appendix. Guidc to thc Excrcises 319

x x

(i) a2 > 0 (double point) (ii) a2 < 0, a2 + aV4 > 0 (double point)

y y

"", y = -ta,x .-.",,--'"

x (0,0) is an isolated point .

(iii) a2 + aV4 = 0 (iv) a2 + aV4 < 0

CHAPTER 2, §1 1. There are (q + 1) points on a line and (q2 + q + 1) lines in P1(k). There are (q' _ l) ... (q' _ q'-I) (q' -l) ... (q' _ q'-I)

s-dimensional subspaces in P,(k).

2. Any dimension from mu(O, SI + S2 - r)to min(sl' 52) can occur.

3. Each (r + 1)-dimensional subspac:e, M, or Pn(k) corresponds to M+ ~ kn+1 containing Mri and of dimension (r + 2). Such a subspace is specified by that vector 'E M+ n (Mri).l (unique up to scalar multiplicatlon). However (Mri).l has dimension (n + 1) - (r + 1) = n - r. So, we parametrize the (r + 1)-dimensional subspaces of P.(k) containing Mo by vectors in kn-, up to scalar multiplication, i.e., by elements of a projective space of dimension n - r - 1. 320 Appendix. Guide to the Exercises

CHAPTER 2, §2

1. This is a straightforward calculation.

2. From Exercise 1, H~ is five-dimensional. Conics which pass through PI' P2 , P3 have equations with six coefficients (up to seal ar multiplication) which must satisfy three' relations. This leaves us with two de~ees of freedom. If (1, 0, 0), (0, 1,0), (0,0, 1) are the given points, then the possible conics are axy + bwx + cwy = ° with a, b, c specifying the conic (up to scalar multiplication). The subfamily of S containing w': x': y' is orte-dimensional as long as at most one of x', y', w' is zero. The subfamily of S containing w':x':y' and w":x":y" is: two-dimensional if

(w':x':y'), (w":X":y")E {(I, 0, 0), (0,1,0), (0, 0, I)};

one-dimensional if precisely one of (w':x':y'), (w":.x:':y") has two zero components or w = w' = °or x = x' = °or y = y' = 0; zero-dimensional (i.e., one conic) otheFWise.

CHAPTER 2, §3

1. Similar to Proposition (3.1).

2. We give a sketch ofthe proof. Suppose ABCDEF is the hexagon, and A = (1,0,0), B = (0, 1,0), C = (0, 0, 1). Let R = AB n DE, Q = AF n DC, P = BC n EF. Then if D = (dl> d2, d3), etc.,

R = (d t e3 - d3e.l , d2e3 - d3e2, 0),

P = (0, ed2 - e2fl' ed3 - eJitl, and

Since

i l f2(ed3 - e3ftlR + dd3(dt e3 - d3e t )P + (d l e3 - d3etl(ft e3 - f3etlQ = ° thus P, Q, Rare collinear. 3. The same techniques as in Exercise 2 are used. This is somewhat more straight• forward than Exercise 2.

CHAPTER 2, §4

1. Follows from the definitions in a straightforward way.

2. Family of curves through P with order :Sr in H~ has dimension t(r + l)(r + 2). The subfamily of H~ consisting of curves through P; with order sr; at P; for i = 1, 2, ... , t has dimension at least L::=I (t(rj + t)(r; + 2)) - !m(m + 3). 3. Transform so that (w, x, y) is at (1, 0, 0). Then (I, 0, 0) has order t on Cf and s on Cg • TlHis, am , has terms of degree ;?; r + m' - ltI only waen m' = m - r + 1, ... , m. Appendix. Guide to the Exercises 321

Similarly forthebi' So R(j; g) is homogeneous in w, x and of degree mn and divisible by x". Hence result. 4. Use Exercise 3. 5. Apply Exercise 4 to 1'. For the last part, the case in which we get n lines through

one point, gives only one rl , namely n; and m = n. 6. Co.1sider the family of curves of degree rn-I (containing 1') and the subfamilies ofthose eurves of degree rl - 1 at PI' Thus subfamily has dimension trl(rl + 1) and the original family has dimension t(m - l)(m + 2). 7. If f is redueible, then f = gh where g, h are of degree 1. Recall the formula Vf = hVg + gVh. So Vf is a linear combination of Vg, Vh. Note that Vg, Vh are eonstant vectors. Thus of/ox I' iJiliJX2' iJflcx 3 satisfy a linear relation with eonstant coefficients, namely Vf' (Vg /\ Vh) = O. Hence (f-f/ox i oXj with i = 1,2,3 satisfy a

comm~ relation for all j, and thll.s det(02flox l OXj) = O. Conversely, suppose det(02f/oXI'O:I) = O. Let ßij = 02flj).~,.oxj' Now diagonalize ßij (possible since ßu = ßj ;). One of the diagonal elements is zero, since the deter• minan! is zew, Thus we get a diagonal form o 0) ßZ2 0 . o O. However, by aversion of Euler's theorem for second derivatives,

2f = 'i. ßijxiXj i,i

= ßIIXt + ß'2X~ = (,/jj;;x 1 + J -ß22X2)(~Xl - J -ß22X2)' This factorization shows that f is reducible. 8. The assertion does not hold in general for characteristics p unless we impose a condition on f, e.g., it has degree :5 (p - 1). 9. Same answer as for Exercise 8.

10. If Ais the matrix of 02floxi oXj then det Ais homogeneous of degree 2:: 3 (it cannot be identically 0 by Exercise 7). Thus det A = 0, f = 0 has at least one common solution, and hence we get at least one Ilex by Exercise 9.

ApPENDIX TO CHAPTER 2 Exercises I, 2, and 3 are straightforward verifications. 4. If f has a repeated root in an extension field k' of k then f, I' have a common root in k' and so RU, 1') = 0 since f, I' have a common factor in k'(X). Thus DU) = O. Conversely, if DU) = 0, then RU, 1') = 0 and so f, I' have a common faetor gE k[X] with og > O. Let k' be an extension of k over which 9 splits completely. Then f, I' have a common linear faetor over k'[X]. So f, I' have a common root in k', i.e., f has a repeated root -in k'. 322 Appendix. Guide to the Exercises

5. A straightforward calculation givcs

D(ax2 + bx + c) = -a(b2 - 4ac) and D(x2 + px + q) = 27q2 + 4p3.

CHAPTER 3, §1 These qucstions are quite tedious to do completeJy. They just consist of many cases. For Exercise 2, we consider 0, P, Q, PP, PQ, P + P, P + Q, pcp + Q), (P + P)Q. The latter two are to j)e shown equal. Using Theorem (3.3), (P + P)Q is one of the other eight points. We must eliminate all other possibilities. For example, (P + P)Q = P implies P + P = PQ. Tben P(P + Q) = prO' PQ) = P(O'(P + P» = P(PP) = P, as required.

CHAPTER 3, §2 80th· follow by straightforward algebra. Substitute the relations in (2.3) into the equation in (2.4).

CHAPTER 3, §3

1. Discriminant is 27c2 - 18abc + 4a3c + 4b 3 - a2b2 (obtained by evaluating a 5 x 5 determinant).

2. Discriminant is (Oll - 1(2)2(1X2 - 1(3 )2(0I3 - IXd2• This is obtained as follows. It is a homogeneous polynomial of degree 6 which vanishes whenever two of OI,'S coincide. Thus it is divisible by (lXI - 0(2)(012 - 0(3)(1X3 - lXil. It is zero only if two OI,'S coincide, giving D(f) = C(lX t - 0(2)2(1X2 - 1(3)2(0I3 - IXI)2 some constant c. Use the special case f = x 3 - X to get c = 1. 3. Evaluation of a 7 x 7 determinant gives discriminant

256c3 - 51b4 - 20a3b2 + 80ab2c + 96a 2c2 - 32a4c -16a3c + 16ab3c - 48a 2bc2• 4. (a) j = -4906/11, A = -11. (b) j = 4096/43, A = -43. (c) j = (27/37)(4096), A = 37. (d) f = (27/91)(4096), A = -91. (e) j = (27 x W/79634), A = -28. (f) j=O,A= _39 •

5. General formula: 16b 2(a 2 - 4b). So we get (a) 80; (b) -48; (c) 128; (d) 230400.

CHAPTER 3, §5

1. A = -16D. To simplify algebra, shifting to give a2 = 0 is convenient. Then D = 27a~ + 4al alld A = 9b2 b4 b6 - 27bt, - 8bl-: bibs reducing to -16D. Appendix. Guide to the Exercises 323

2. A suitable elhptic curve is y2 = xJ - x. It has A = -1, j = O. In F3 , it has graph:

2

o * o 2

CHAPTER 3, §6

1. It is found that E3 , E4 have four points while Es has eight points. So E3 *Es while EJ ;;:; E4 under a shift.

2. The only part that needs checking is invt!rses. In F16, the relation given leads to (a + bvrl = (a' + b'v) where

a' = (b - 1)j(wb2 + ab - a2 ),

b' = bj(wb2 + ab - a2 ), and wb2 + ab - a2 "# 0, as can be easily verified. A slightiy more complicated version ofthis holds in F256.

3. Going back to (2.4) gives us that for an isomorphism betweeu E3 and Es, t 2 + t + v3 = 0 must be soluble. In F2S6 , t = w is a solution. In F16 , there are no such solutions.

4. This question is best answered by writing down (2.4) with ai = ai in each of the cases Ei (i = 1,2,3,4,5) separately. We then find that EI' E 2 give automorphism groups Zj2Z over any field of characteristic 2, of the identity and x=x } y=x+Y

However E3 , E4 , Es are more complicated, and the different k fields must be

considered separately. In fact, EJ , E4 give an identical set of equations and thus the same automorphism groups,

over F2 , F4 : Zj4Z,

over F16 , F2S6: 24-element group. Also Es gives automorphism groups,

over F2 : Zj2Z,

over F4 , F16• F2S6 : 24-element group, G. In this latter case, (2.4) gives: x = vx + r, y = y + r2 vx + t, 324 Appendix. Guide to the Exercises

where v = u2, S = r2• Here, U = 1, w, w' and either r=O, t = 0,1 or

reFt, t = w,w'.

Let av•r •• be the above transformation. Then

Ci v. r. t 0 (J.v',,' .t' =" (luv'. ur'+r,t+t'+,l,'v'

When v = I, we get a subgroup of eight elements isomorphie to H (the quaternion group {± I, ± i, ±j, ± k}). In this isomorphism:-

j -+ a t •w•w - j -+ a t •w•w•

-k -+ 1X1 ...... and

a 1.0,1 0:% V,r,t --av.r,t+l - -av,r,t oa 1.0.1"

In fact, the group of 24 elements in the Es cas~ is isomorphie to {± 1, ±i, ±j, ±k, i(± 1 ± i ± j ± lc)} (all are units). This ean be seen as folIows. If

I = HI + i + j + k) then I has order 6. Also a •. o.OOat.r .• oa;;:~.o = a t .•r .•. Thus the subgroup,. H, of our 24-element group G isomorphie to H, given by H = {at.r..• lr = 0, te{O, I} or reFt and te{w, w'}}, is anormal subgroup of G. Now r1j[ = -k can easily be verified. Thus eonjugation by [2 permutes i, j, k eyelieally, and.we ean thus eorrespond

[2 ...... a ..•• o•o

and since a;.r•• = IXv'.(v+IJr.vr3, thus [ean eorrespond to a w•o.o or to a w•o. t ' Thus G is a semidirect product of H with Z/3Z. There are four 3-Sylow subgroups of G; namely those generated by

!( - 1 + i + j + k), t( - 1 + i - j - k), i( - 1 - i - j + k), t( - 1 - i + j - k). These are eorresponded to

(lw'.o,O I%w',w',w and t( - 1 - i - j - k), t( - 1 - i + j + k), t( - 1 + i + j - k), t( -1 + i - j + k) are their squares (= conjugates) and are thus eorr.esponded to

Clw,W'.w' IXw,w,w' (Xw,I,w' Appendix. Guide to tbe Exercises 325

So, in the fuH eorrespondence,

1 +-+ IXI.O,o

- j +-+ IX I , w, w' -k+-+IXI ,,..,,.,

t(1 + i + j + k)+-+lXw,O,1 t( -1 - i'-j - k)+-+IX,.,o,o t(1 + i + j - k)+-+ IX,.., I,w' t( -1 - i - j + k) +-+ IX,.., I,,. t(1 + i - j + k)+-+IX,."w',w' t( -1 - i + j - k)+-+lXw',,.,,,. t(1 + i - j - k)+-+lXw,,.,,,. t( -1 - i + j + k)+-+IX,.,w',w' t

5. Tbe elliptie eurves over F4 up to ilOmorphism over F4 are: j=O: y2+ Y =X1, y2 + Y = x3 + w, r + y = x 3 + X + w, y2 + Y =x3 + x, y2+ y =.x3 +wx, y2 + Y = x3 + wx + w, ". + Y = x3 + w'x, r + y = x3 + w'x + w, y2 + wy =x3, y2 + wy = x3 + w, y2 + w'y = x3, r + w'y = x3 + w',· j = 1: r + xy = x3 + 1, y2 + xy = x3 + wx2 + 1. j = w: y2 + "y = x 3 + w', y2 + xy -= x3 + wx2 +·w'. j=w': y2+ xY =X3 +w, y2 + xy = x3 + wx2 + w. 326 Appendix. Guide to tbe Exercises

Over F16, these reduee to: j = 0: y2 + Y = x 3 (isomorphie to all above eurves with j = 0, a3 = 1),

y2 + wy = x 3 , y2 + wy = x 3 + ~, y2 + w'y = x 3, y2 + w'y = Xl + w'. j = 1: y2 + xy = x 3 + 1 (isomorphie to y2 + xy = Xl + wx2 + 1). j = w: y2 + xy = x 3 + w' (isomorPhie to y2 + xy = x 3 + wx2 + w'). j = w': y2 + xy = Xl + W (isomorphie to y2 + xy = Xl + wx2 + w).

6. The elliptie eurves over Fl up to isomorphism over Fl are:

j= 0: y2 = Xl + X} 2 3 isomorphie over F9 (16 elements), y =x -x

y2 = Xl - X + I} 2 3· isomorphie over F9 (7 elements) y =X ~-1

j = 1: y2 = Xl _ x 2 + I} 2 l 2 isomorphie over F9 (15 elements), y=x+x-l j = -1: y2 = x 3 + x 2 + I} 2 3 2 isomorphie over F9 (12 elements). y=x-x-l

7. We find that when k = Fl Z/2Z for eurves with j = 1, Autk(E) = { Z/2Z for y2 = Xl + X, Sl for y2 = Xl - X + Cl (Cl = 0, ± 1),

and with the notation V4 = Z/2Z x Z/2Z

Z/2Z for j "" 0, Z/4Z for y2 = x 3 "+ x, 3 AutF.(E) = jV4 x Z/3Z for y2 = x - x, Z/6Z for y2 = x 3 - X + 1, Z/6Z for y2 = x 3 - x-I.

8. We think of F8 as a eubic extension of F2 formed by adjoining w such that 3 w + w + 1 = O.We find that noncycIie groups occur only over F 16, FZS6 in this example. For

i = 1: F8 gives Z/14Z} F16 gives a 16-element group with a subgroup Z/8Z i = 2: F8 gives Z/4Z "of { 0, (0, 1), (w)1, w' '.( 2, 1)w' ., (w, ,I)} w for E.

i = 3: F8 gives Z/5Z} F16 gives.a 25-element group, with subgroup of i=4: F8 gives 0 order 5 gJVen by {O, (0, 0), (0, 1), (1, 0), (1, I)}. Appendix. Guide to the Exercises 327

; ~" F. g;,~ Z/9Z; F" g;'" {(O, 0), (0, " C:' ) ,0 l

This exercise can become a bit tedious: we only need to check Fs, F16 here. 9. The elliptic curves over Fs are: j=O: y2 = x3 + 1 (Z/6Z), y2 = x 3 + 2 (Zj6Z). y2 = Xl + X (Zj2Z x Zj2Z), y2 = x 3 + 2x (Zj2Z), y2 = x 3 + 3x (ZjlOZ), y2 = x 3 +.4x (Zj2Z x Zj2Z). j = 1: y2 = x 3 + X + 2 (Zj4Z), y2 = x 3 + 4x + 1 (Zj8Z). j = 2: y2 = Xl + 4x + 2 (Zj3Z), y2 = x 3 + X + 1 (Zj9Z). j=4: y2 = x 3 + 2x + 4 (Zj7Z), y2 = x 3 + 3x + 2 (Zj5Z).

CHAPTER 4, §1

These exercises are routine algebra, using the definitions of the bj , Ci' Ö, and j.

CHAPTER 4, §2

1. Here p, q are clearly inverses. Thus, one only needs to check that q = 2p. This 'is done by showing that the tangent at p cuts the elliptic curve at a tripIe point (namely p).

CHAPTER 5, §1 1. It is easy to verify that C is nonsingular over Q and that C is singular. A suitable pair P, P' is

P' = (p, p2, 1),

and then L is w + X = (p + p2)y. Thus i is w + X = 0 and alI the conditions are satisfied.

CHAPTER 5, §2

1. The required condition is that

ordp(ö) + min(O, ordp(j» < 12 + 12152P + M3p 328 Appendix. Guide to the Exercises

for all primes p. It is easily verified that

A = -16(4a3 + 27b2 ),

j = (3 3 . 28 )a 3 j(4a 3 + 27b 2 )

for y2 = x 3 + ax + b. Thus, for y2 = x 3 + ax, A = _26 a3 , and j = 33. 26. This gives the conditions for minimality:

ord2(a) < 6

ord3 (a) < 6 ordp(a) < 4,

For y2 = x 3 + a, A = _24 . 33a2, j = O. This gives thecondition forminimality:

ord2(a) < 10

ord3 (a) < 8 ordp(a) < 6,

CHAPTER 5, §3 The first four questions in this section are all very similar. We therefore give only the answers, and in Exercise 4(c) we give a complete solution. 1. (a) Never get bad reduction at prime p, unless a == 0 mod p. So the prim~s p at which bad reduction occurs~re those which divide Q. (b) Bad reduction occurs at primes p "# 3 as long as p divides a; and occurs at p = 3 always. (c) Bad reduction occurs at p = 13. (d) Bad reduction occurs at p = 3, 11.

2. (a) p = 5. (b) p = 3. (c) We never get bad reduction (d)p = 3,5.

3. (a). p = 37. Modulo 2 gives E3 in 3(6.4) and modulo 3 gives y2 = x 3 - X + 1. (b) p = 43. Modulo 2 gives E] in 3(6.4) and modulo 3 gives y2 = x 3 + x 2 + 1. 3 . (c) p = 91. Modulo 2 gives E3 in 3(6.4) and modulo 3 gives y2 = x + X + 1. (d»)' = 3. Modulo 2 gives E's. (e) p = 2, 53. Modulo 3 gives y = x 3 - X + 1. 3 (f) p = 5, 17,31. Modulo 2.give E l and modulo 3 gives y2 = x + 1. 4. (a) p = 3, 7 give good reduction and p = 2, 5 give bad reduction. (h) p = 2, 3 give good reduction and p = 5, 7 give bad reduction. (c) Here y2 + xy + Y = x 3 - x 2 - 3x + 3. Thus

(2y + x + I)y' = 3x2 - 2x - 3 - y and bad.reduction at an odd prime p requires

2y + x + 1 = 0, i.e., y = -tex + 1) and

3x2 - 2x - 3 = y. Appendix. Guide to the Exercises 329

This is never satisfied for odd p, and (x, y) on the curve. So we consider p = 2. This gives y2 + x'y = X'3 where x' = x + 1. Thus bad reduction occurs at p = 2 and good reduction occurs for p = 3, 5, 7. (d) p = 2, 5, 7 give good reduction and p = 3 gives bad reduction. (e) p = 5 gives good recluction and p = 2, 3, 7 give bad reduction. 5. The subgroup, G, generated by (0, 0) has image under r2 which is {(O, 0), (1,1), (1; 0), (0,1), O}. 6. The discriminant is 5077 and this is easily seen to be prime.

CHAPTER 5, §4

1. These fo11ow very simply from the relation for x + x' + x" given in (4.3) since ordp(x + x' + x") ;;:: 3n as a 1 = O. 2. It is quite clear that Rp is a maximal ideal in R iff R(p) forms an additive group, and thus R(p) = R. Uniqueness of this maximal ideal follows by assuming land J = Rp to be two distinct maximal ideals. Thus I 't J, J 't I. So thereexists ao e I such that ordpao = 0, and then Va e k, if and only if ordpa;;:: 1. But for a11 aeR.

So I = R because ao is invertible, a contradiction. It is now easy to show that I + Rp !:; R*. We then get k*jR* ..... Z, a + R* ..... max {ordp(a + r)}. rER·

Reduction mod p maps R~ ..... k(p)* with kernel 1 + Rp. The canonical map R ..... RjRp composed with the map:

1 + Rp" ..... R} 1 + ap" ..... a produces a map 1 + Rp" ..... RjRp. The kernel consists of 1 + Rp"+l. Thus (1 + Rp")j(I + Rp"+l) ~ RjRp by the first isomorphism theorem: 3. This is very straightforward from the definitions.

CHAPTER 5, §5

1. For the last part, note that good reduction occurs mod p far all odd primes other than 3. When E is reduced mod 5, we get a group of order 6. Since E(Q),ors has no elements of order 2, thus all elements must be of ordl:r 3. Hence E(Q),ors ~ Zj3Z. 2. Modulo 2and modulo 3 give no singularities. So there are injections from E(Q),ors into Zj3Z, Zj7Z. Hence E(Q)'on = O. 330 Appendix. Guide to the Exercises

3. Modulo 2, E contains just O. Modulo 3 we get no singularities and the group is Z/4Z. So E(Q),ors is thus 0 or Z/2Z. However, (2, -1) has order 2, and so E(Q),ors = Zj2Z.

4. Hefe, E(Q),ors = Z/7Z since modulo 3 gives Z/7Z and (1, 0) has order 7.

5. Here GL.(Z) denotes the group ofn x n matrices with entries in Z and determinant

± I. Thus GL.(Z) maps into GL.(ZjqZ) under rq • Showing that (I. + p·X)· == I. + np' X mod p.+b+l. Thus we find that C) p.r is divisible by p.+b+I, whenever r ~ 2.

If G ~ GL.(Z) is finite and A e G (") ker r p , then A has finite order. Thus,

as Aeker rp , A = I + X, and Am = I + mX mod pb+l. So X = 0 since the Am cannot all be distinct. This gives G (") ker r, = I, 'r/p > 2. For p = 2, A"' = I. + 2mX mod· pordp(m)~2. and so G (") ker'4 = 1 for p = 2.

6. It is easily seen that p = 3, 5 give good reduction, and groups Z/7Z (p = 3) and a group of order 10 (p = 5). So ß(Q),ors = O.

CHAPTER 5, §6

1. It is easily computed that E has discriminant 3. So, any torsion point has y = 0, ± 1, ±3. Thus, 0, (0, 0), (I, I), (1, -1) are the torsion points (Z/4Z). The curve has good reäuction modulo 5 and E(Fs) i~ja group of eight points: {O, (0, 0), (1, ± 1), (2, ± 1), (-2, ± 2)}. 2 The torsion points are 0, (0, 0). The curve has good reduction both modulo 3 and

modulo 5, and E(F3 ) = Z/6Z, E(Fs) = Z/4Z; In fact E(F3 ) = {O, (0, 0), (± 1, ± I)}, E(Fs) = {O, (0,0), (-2, ± I)}.

3. Here E(F3 ) =Z/6Z = {O, (0, 0), (± I, ± I)}. Also E(Q),o" = E(F3 ) and so both are Z/6Z.

4. To prove this, we use the fact that - 2P is not 11 torsion point, so that the tangent line at P cuts the curve at an integer point. This is used to determine that the slope of this line must be integral.

CHAPTER 6, §4

1. The first part is quite easy, as x + (f) generates Rf as an algebra. If f factors as distinct linear factors, n?=, (x - IX;) then we can map Rf to k· by evaluating at the IXi · When f factors as n~=, (x - IX;)", we can map Rf to a direct sum of R",r,. This then gives thestructure of Rf . 2. We use (); as our three maps which produce a tripie for each PeE. This tripie is converted into an element of Rf using g(r;). Thus, Jmg ~ Rj,I/(Rj)2.

CHAPTER 9, §1

I. The proof is by induction on n, and so we assume that it holds for (n - 1). For n = 2, it obviously holds. A discrete subgroup r of R" Jives. r' = rf(Rw t ) ~ Appendix. Guide to tbe Exera- 331

R"J(ftllJ 2 ) ~ Rn-l where w t er has minimal nonzero absolute value. (This holds unless r = {O}). Thus r = Zw. + ... + Z"', sinee r' = Z"', + ... + Zw; by indue• tive assumption, and by disereteness of r. The eondition for compactness ofR"jr follows sinee Rnjr is Rn-, x T. where T,. is torus. 2. Trus follows from the last exereise.

3. SupPose f is analytic, so that f, g: C --> C and fog = gof = id. We ean extend to maps t --> t where t = C U { oo} by

f(oo) == g(oo) = 00.

Hence f(z) = (az + b)!(ez + d). Sinee f( 00) = 00, C = 0, and so f(z) = az + b with a # O. Since f i~ a homomorphism, f(z, + Z2) = f(z tl + f(Z2)' Thus b = O;and so f(z) = .l.z some .l. eC

CHAPTER 9, §3 1. We consider djdz«((z + w) - (z)), and show that it vanishes for aH z. Thus, (z + w) - (z) is eonstant, independent of z, and is ,,(w) so me function IJ of w.

2. Integrate ( around a rectangle from Xo to Xo + w" Xo + W l + Wz, uno Xo + W2' Then we get 21[; by the eaIculus of residues. Integrating around pairs of opposite sides will give the result (using Exereise 1). 3. To show that u is entire, we only need to show that In(1 - zjUJ) + zlw + z2j2w2 forms a eonvergent sum over (OE L - {O}.

4. In this exereise u(z + wJ = u(z)e"= Ai some eonstant Ai from Exereises 1 and 3. Also u is odd, and so Ai = _e-·,,,,,/2. Henee result.

5. Since any elIiptic function has only a finite number of zeros and poles ai , bj say, we have " u(z - b·) f(z) [1 • i~l u(z - a,) eontaining no zeros or poles and also periodic. Thus, we get a constant function, as required.

CHAPTER 9, §4

1. Since h is a group isomorphism and A + B = - A· B, therefore (1, p(zd, p'(zd), (1, P(Z2)' P'(Z2))' -(1, p(Zt + Z2)' t,)'(z, + Z2)) are coHinear. This gives the result. 2. To check the first result, we note that both si des, as a function of z" are periodic with poles when z 1 E L. These poles are double poles with equal coeffieients of Ijzf near 0 (namely.l). So, they can dilTer at most by a constant. At z 1 = Z2, hoth sides agree. Hence the resuIt r.. uows. As Z2 -> Z l, we obtain the second part. 3. DilTerentiate the first resuIt in Exereise 2. 332 Appendix. Guide to the Exercises

4. Apply Exercise 3 with ZI' Z2 interchanged,and add to the result of Exercise 3. Simplifying gives the result. To derive the addition formulae, we need only derive (I).

CHAPTER 9, §5 1. We have

r(s)r(1 - s) = (L"" x·-Ie-x dX) (L"" x-'e-X dX)

= (L"" 2x 2.-1e- x2 dX) (L"" 2y-2·+1e-,2 dY)

("" (Kf2 = J0 J0 4e-"(cos 0)2'-I(sin 0)1-2'r dO dr

putting r = x 2 + y 2 = 2 LK/2 (tan 0)1-2. dO

= Ir> le du/(I + u) (u = tan 2 0).

lntegrating u-'/(l + u) around the contour below, since 0< Re(s) < 1 thus the integrals around circ1es of radii e, R about 0 tend to 0 as e -+ 0, R -+ 00. The only . enc10sed pole is at -1, with residue e-ttis• Thus

21rie- .is = r(s)r(1 - s)(1 _ e-2i",)

and so r(s)r(l - s) = n/sin ns. Q.E.D.

-1 -f.

Questions 2, 3, and 4 follow straight from-the definitions of the hypergeometrie function. Bibliography

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Abelian L-functions, 296-303 Bombieri,247 Abelian variety, 229 Boundary function, 146 Absolute values, 136,137 Addition fommla, 172 Additive group, 40, 41 Canonica1 height, see Height Additive reduction, see Reduction of an Cebotarov density theorem, 281 clIiptic curve Character, 298-303 Admissiblc change of variables, 65 Characteristic polynomial, 230, 259, 261, Aigebraic extension, 139, 140 271, 277,280 Aigebraie group homomorphism, 301 Chinese remainder theorem, 210 Aigebraie Hecke character, 301-303 Chord-tangent composition law, 12-15,22- Analytic continuation, 290, 293, 295, 298, 32 299,303,305 Cla.:s field thcory, 301 Arclength of an ellipse, 182 Class number, 234, 310 Artin, E., 243 one.233 Artin, M., 245 Coates, J., 17: 19,307 Associative law for an elliptic curYe, 63 Cohomologous one-cocycles, 144 Automorphism group, 68, 71, 73, 75-77, Cohomology c\ass, 144, 146 150,207,323-326 Cohomology group, 142-147 as a Gal (k/k)-module, 150,237 Cohomology set, 143-147 ofTate module, 259 Complex analytic isomorphism, 1163 Complcx multiplication, 230-234, 302- 303 Bad reduction, see Reduction of an elliptic Conductor of an elliptic curve, 265, 273, curve 274 Bezout's theorem, 48, 54, 55 Congruence subgroup, 206 Bilinear pairing, 311 Connecting homomorphism, 146 Birch arid Swinnerton-Dyer conjecture, 17, Cubic curve, 13 35,306-313 Cubic equation, 45 346 Index

Curve, 45 Division point, 224, 228. 253-255, 258, elliptic, see Elliptic curve 270,271 modular, see Modular curve Divisor, 168,229 twists of. 150, 151 associated to a differential, 273 Cusp, 39,77-80 associated to a function, 168 of a congruence subgroup, 209, 212 degree, 168 Cuspform, 214 degree O. 168 for a congruence subgroup, 304--306 of a differential, 273 Cyclotomic representation, 275 evaluated at a function. 229 group (Div), 168 principal. 168 Decomposition group, 278 Divisorclassgroup, 168 Dedekind zeta function, 297, 307 Oouble point, 39,77-80 Definite quaternion algebra, 258 Dual isogeny, 223, 226, 228 Degree of an algebraic field extension, 139 of a curve, 18,45 Eichler-Shimura, 304, 305 of a divisor, 227, 228 Eigenfunctions, 304, 305 of an endomorphism, 223, 228 Ei~nsteil) series. 170--172, 184-185 ofanisogeny,223,224,228 Elimination theory, 55 of a map, 130, 227 Ellipse, 182 of amorphism, 227, 228 EIliptiC curve, I, 15 of multiplication-by-m map, 228 in characteristic two and three, 72-77, Degree map is positive definite, 228 113, 114, 323-327 Deligne, P., 243, 245, 248 defined over k, 15 Density of a subset, 281 families, 81-89 Descent, 155-161 group of rational points 0ver via two-isogeny, 155-158 complex numbers, 21 Deuring, M., 258, 302, 303 finite fjelds, 242-261 Duering normal form, 248, 251 local fields, 262-271 Deuring polynomial, 249, 251 number fields and rational numbers, Differential equation, 169-171 28-35 Differential form, 65, 67-77 real numbers, 19--21 associated to a cusp form, 215 k-isomorphism classes, 150 invariant, 65, 67-77, 251, 258 Elliptic functions, 161, 163-1"14, 185, Diophantine equation, 17 191, 192 Diophantine geometry, 17 field, 167, 168, 226 Diophantus, 7, 8 as functions of Weierstrass functions, DiriChlet, G.L., 298 167, 168 Dirichlet class number formula. 310 no poles implies it is constant, 164 Dirichlet L-function, 298 as product of a-functions, 169 Dirichlet series. 220 Elliptic integral, 178-182 Dirichlet unit theorem. 127 e .. -pairing or Weil pairing. 226, 229. 230 Discrete valuation, 57, 100 N~' 224, 226 ring. 57. 100 Endomorphism ring, 223, 228-234. 236, Discriminant, 67,70,84, 1I6, 322 237, 255-258 minimal, 103-105,273,274 classification of, 23 I as a modular form, 214 is an integral domain, 230 of a polynomial, 61, 69, 70, 72. 321 e.-pairing or Weil pairing. 226, 229, 230 Divisible element, 55 Euler characteristic, 221 Index 347

Euler product, 217-220 Good reduction, see Reduction of an Exponential of a formal group, 240, elliptic curve 241 Greenberg, R., 17, 308, 309 . Extended upper half-plane, 209 Gross, B., 309, 31I-313 Grössencharacters, 299-301, 303 Grotbendieck, A., 245, 248 Factorial rings, 55 Group cohomology, 141-147; see also Faltings, G., 9,17,272,288 Galois cohomology Families of elliptic curves, 81-90, 204- non-abelian, 143-146 208 Group Iaw on an elliptic curve, 14, 63 Fermat, P., 7, 9 Group scheme, 264-267, 271 Fermat descent, 122 Fermat equation, 9, 29, 30, 122, 123 Fermat's last theorem, 9 H", 146 Fiber, 265, 266 H', 144, 146 Finitely generated group, 120, 121 of Aut(E), 150-151 Fixed fjeld, 139 of E, 152 Formal addition law, 238, 239 of ...E, 158-161 Formal group, 237-241, 255, 258, 263 of...,E, 155-158, 161

Fourier coefficients, 183-187,213,214, of GLn , 148 216 of Isom(E), 152-155 Fourier series or q-expansions, 183-187 ofk*, 148, 149 Fractional ideal, 232 of ILm' 149 Frobenius element, 281 Hasse, H., 7, 17,231,243,258 Frobenius morphism, 243, 244, 248, 254, Hasse invariant, 248; see also Ordinary, 258,272 Supersingular characteristic polynomial, 243 Hasse-Minkowski theorem, 7 degree of, 243 Hasse-Weil conjecture, 290 is njlrely inseparable, 254 Hasse-Weil L-function, 290, 292-2% Function fjeld, 226, 227 Hasse-Weil zeta function, 292 map induced on, by rational map, 226, Hecke L-series, 215, 216, 299 227 Hecke operator, 220, 221 Functional equation, 293-299 Heegner, K., 312 ' Fundamental parallelogram, 164- Heegner point, 312, ~:3 Height, 130-137, 311 on an abelian 'group, 135 G-invariant element, 142 behavior under maps, 131-133 G-module, 142-147 canonical height, 133-137 exact sequence of, 146-'147 on an elliptic curve, 133-135 homomorphism, 142 finitely many points with bounded, 135 set, 141 .on P" (Q), 130-133 Galois cohomology, 147-151,270,271, on projective space, 136, 137 279 Hessian family, 84-86, Il8 Galois group, 139 Hilbert theorem ninety, 148 Galois theory, 138-141 Homogeneous coordinates, 3, 44 Gamma function, 174-176, 332 Homogeneous space, 143-146, 152-155 Gauss, C.F., 58 associated with a quadratic extension, General linear group, 148 153-156 Genus, 18, 212 Homothetic lattices, 163, 203 Goldfeld, 0., 310, 311 Honda, 261 348 Index

Hurwitz-Riemann genus formula, 211 f-adic representation, 274--289 Hypergeometric function, 162, 176, 177, compatibility, 283 179, 180,250 image of, 288, 289 Hyperplane, 44 integral, 282 Hypersurface, 45 rational, 282 unramified, 279 C-adic Tate module, see Tate module Ideal dass group, 232 C-adic Weil pairing, see Weil pairing or Identity component, 265 em-pairing Inertial group, 279, 280 L-series, 215-217, 293-295, 298, 299; action on ,.E and Te (E), 270 303, 306-310, 312-313; see also Infinite descent , 160 Birch and .Swinnerton-Dyer Inflection point, 53 conjecture Intersection multiplicity , 52 ~ce, 163, 164, 223-226 'Intersection theory, 48-50, 52 Lcgendre normal fonn, 81, 84,117,249- Invariant differential, 65,67-77,251, 258 254 of a formal group, 240 Legendre theorem, 7 Irreducible, 56 Lie group, 19, 20 Isogenous eIliptic curves, 163, 203 Une, 2-4, 6, 11-15 Isogeny, 163,203,223,224,226-230 at infinity, 3 associated moduli problem, 206-208 Uouville's theorem, 164 of degreetwo, 91-96 Locus, 1,2 Frobenius, see Frobenius morphism Lutz, 112 given by an analytic map, 163 kerne I of, 163, 254, 255 Weil pairing on kernei, 226, 230 Mass fonnula, 253 Isomorphism Mazur, B., 16, 316 complex analytic, 163, 203, 206, 207 Mellin transform, 215, 305 of curves, 62, 65, 68, 71, 73, 75-77 Minimal discriminant, see Discriminant Isomorphism group, 71, 73, 75, 77, 86 Minkowski, H., 7 may be non-abeJian, 324, 325 Modular curve, 208-212, 304, 305 cusps,209 genus, 212 j-invariant, 67-69, 71-77, 83, 85,150, Modillar form, 213-215 151,187,232-234 algebra of, 214 of some CM elIiptic curves, 233 associated L-series, 215-217 integral implies potential good reduction, ·for. a corigruence subgroup, 304, 305 117,118 Fourier series, 186, 214 j = 0 andj = 1728,33-36,71,73-78, Modular fUfl(;tion, 202, 213 324, 325 tield of, 234 Jacobi, c.G.J., 14, 166 Modular groups, 202, 206, 210, 211 Jacobian, 248 Moduli space, 204-208 Monsky, 247 MordelI, L.J., 15 Katz, 245, 247 MordelI con jecture, 17 Kummer sequence, 149' MordelI theorem or MordelI-Weil theorem, Kummer theory, 9, 149 . 10, 13, 15, 120 Multiplication by m !llap, 36-38, 123 f-adic cohomology, 248 Multiplicative group, 42 Index 349

Nagell, T., 112 Quadratic imaginary field, 231-234 Neron, A., 264 dass field, 232 Neron model, 264-267, 311 dass number one, 233 Neron-Ogg':"Safarevic criterion, 269-271, Quasilinear maps, 127-130 277-279 Quasiparallelogram law, 128 Node, see Double point Quasiquadratic maps, 127-130 Non-abelian cohomology, 145 Quaternion algebra, 256, 258 Nonsingular Quaternion group, 325 curve, 12, 51 Quotient of a curve by a finite group, 256 hypersurface, 51 point, 51 reduction, see Reduction (good) of an Rank of an eIliptic curve, 16 elliptic curve Rational Weierstrass equation, 68 conic, 5, 13 Nonn, 121, 122, 133-135 cubic, 10-16 Nonnal fonn, 22, 64, 272-274 curve, 4, 17 NullstelIensatz, 131, 132 line, 4 point, I, 4, 5, 11 plane curve, 2, 4 Ogg, A., 16 Reduction map, lQO, 106, 107, 263 Order of a point, 51 is injective 'on torsion, 112 Ordinary, 258; see also Supersingular kernel of, 107-109, 112 Reduction modulop, 99-102 Reduction of an clliptic curve, denoted E. Parallelogram law, 128, 129 105, 263, 328-330 Parametrized by modul~r functions, 305 additive, 116, 265, 271 Periods of an elliptic curve, 178-180 bad, 105, 116, 265, 271 Petersson inner product, 221 everywhere good, 118 Picard group (Pic), 232 no cvcrywhere good reduction over Q. Plane curve, I, 43 118 Poincare, 15 good (stable), 105, 106,265,271, T/8 Points at infinity, 3 morphism, 106 Point of finite order, see Torsion point multiplicative (semistable), 116, 117, Potential good reduction, see Reduction of 265, 271 an elliptic curve potential good, 118 Principal homogeneous space, see Residue of a function, 165 Homogeneous space Residuc theorem, 165 Product formula, 136 Rest;ltant, 59 Profinitc group, 140 Riemann hypothesis, 242-245, 247, 248 Projective Riemann-Roch theorem, 66. 199 plane, 2-4 Riemann surfacc. 18 space, 43, 44, 136 Riemann zeta function, 217, 292 Proper function, 127 Rohrlieh, D., 308, 309 p-torsion in characteristic p, 258 Roots of unity, 234 Pythagoras, 7 Rosati involution, 223 Pythagorean tripie, 7-10 Rubin, K., 308

q-expansion, see FourieI series Scheme, 264-266 Quadratic form, 3 i I Seimer, E.,I4 350 Index

Selmer group, 160-161 Torsion point, 21, 39, 81-91, 96-98 Serre, J-P., 289, 314 associated moduli problem, 206-209 Safare\ i~, I.R., 287 integrality conditions, 114-116 Safarevi~-Tate grotip. 160, 161,306, 311, Torsion subgroup, 16, 96-98, 112 313 Torus, 162-164, 189-191, 203 Sheaf eohomology, 66 Trace of Frobenius, 246 Shimura, G., 16,305 TunnelI, J., 9, 10 Siegel, C.L., 18, 19,310 Siegel's theorem, 18 .. 19, 310 is not effective, 310 Uniformizing element (local), 100 Sigma funetion, see Weierstrass u-funetion Unique factorization domain, see Factorial Sign of the funetional equation, 295 ring Singular point, 12, 39, 50, 77-80 Unit group, 127 Silverman, J., 222, 242, 262 Unramified, 279, 280

SL2(Z), 204, 206, 210 Smooth, see Nonsingular Spee (R), 260 Valuation, 57, 58, 100 Special fiber, 266 Variety over a finite field, 248 Standard absolute values, 136 Supersingular, 248~258; see also Ordinary Weber, H., 233 eharaeterizations, 258 Weierstrass, K., 166 number of eurves, 252, 253 Weie'tstrass p·function, 166-174, 331 points of order p, 253, 254 algebraic relation, 170 Sympleetie pairing, 224-226, 235 generates elliptie funetion field, 167 Laurent series, 170 relation with u, 174 Tangent line, 13 Weierstrass u-funetion, 169, 174, 331 Taniyama. 272 Weierstrass ~-funetion, 167, 169, 174 Taniyama-Weil eonjecture, 305, 306 Weil, A .. 244, 247 Tate curve. 192-195,200. 201 Weil eonjectures, 247, 248 Tate, ,1.,264 ,287 Weil pairing, 288, 230, 235 Tate module, (jenoted Tl (E), 234-:237, Weil reciprocity law, 229 259-261 Weil-Taniyama eonjeeture, see Taniyama• action of Gal (hk), 259 Weil conjeeture . action of inertia, 270 Wiles, A., 17, 308

Tp (E) in characteristie p, 258 Wronskian determinant, 250 Tate normal form, 8~ Tate's algorithm, 26~9 Tate-Safarevi~ group, ~ee Safarevi~-Tate Zagier, D., 309, 311-313 group Zeta funetion Theorem 90, see Hilbert theorem ninety of a eurve;245-248, 291 Theta functions, 183, t87-192. 197-200 of an ellip\ie curve, 244 coastants. 190 of a variety, 248