Answers and Hints to Exercises - Springer LINK

Answers and Hints to Exercises

Chapter 1

EXERCISE 1.1. Oa = (0 + O)a = Oa + Oa.

EXERCISE 1.Z. 0 = I:::? Oa = 1a = a, Va E K:::? K = {O} which contradicts the definition of the field (Definition 1.1).

EXERCISE 1.3. 0 = (1 + (-1»(-1) = 1(-1) + (-1)(-1) = (-1) + (-1)( -1). Add 1 on both ends.

EXERCISE 1.4. For k E I\J, put Sk(n) = 1k + Zk + ... + nk. Use Sk(n + 1) = Sk(n) + (n + II and Sl(n) = n(n + l)/Z.

EXERCISE 1.5. Think of the definition of the baldness.

EXERCISE 1.6. If a > 1 is not prime, then a = be, 1 < b ~ e < a. Hence a =be~b2.

EXERCISE 1.7. Za + Zb + Ze = (Za + Zb) + Ze = Za + (Zb + Ze).

EXERCISE 1.8. an+2 = an + an+l:::? (an' an+l) = (an+l> an+2)' Induction works because (ao, at) = (0, 1) = 1. an+l = an-l + an ~ Zan·

EXERCISE 1.9. Note first that the series I(x) converges for every x; hence one can differentiate I(x) as many times as one wants. From an+2 = an + an+l we find that y = I(x) satisfies the differential equation y" = y + y'; hence y = AeA.e + Bel'X. As A and Il are solutions of X2 = 1 + X, we have A = (1 + V5)/Z and Il = (1- V5)/Z. Putting x = 0 in l(x)=AeAx+Bel'X and f'(x)=AAeAx+Bllel'X, we find O=A+B and 1 = AA + BIl; hence A = 1/0 and B = -1/V5. Values of an are obtained by the equality eX = I:~=o (xn In!).

EXERCISE 1.10. Use Exercise 1.6 to check each P(x), 0 ~x ~ 39, is prime.

EXERCISE 1.11. One proves an+l = ala2' .. an + 1 by induction. Then we find (an+l> ai) = 1 for all i, 1 ~ i ~ n; hence (ai' aj) = 1 for i =1= j. Since

195

196 Answers and Hints to Exercises

aj = 2, i;:: 1, if we take any prime pj dividing aj, we get infinitely many primes.

EXERCISE 1.12. Similar to Exercise 1.11.

EXERCISE 1.13. (95,432) = 1, (1986,304) = 2.

EXERCISE 1.14. (-42,191).

EXERCISE 1.15. (-93,62, -1).

EXERCISE 1.16. V2 = [0] + (V2)0 = 1 + (0 -1), ql = 1. 1/(0-1)=0+1=2+(V2-1), q2=2. Thenqj=2, i;::2.

EXERCISE 1.17. For n = 1,2,3, see text. For general n, the induction goes as follows:

[AI> ... , An+d = [AI> ... , An-I> An + l!An+l]

Pn(Al,"" An-I' An +~) (An +~)Pn-l +Pn- 2 =

Qn( AI, ... , An-I' An + An1+J (An + An1+JQn-l + Qn-2

An+l(AnPn- 1 + Pn- 2) + Pn- 1 An+lPn + Pn- 1 Pn+1

= An+l(AnQn-l + Qn+2) + Qn-l = An+1Qn - Qn-l Qn+l'

EXERCISE 1.18. PnQn-l - Pn-1Qn = (AnPn- 1 + Pn- 2)Qn-l - Pn-1(AnQn-l + Qn-2) = Pn- 2Qn-l - Pn- 1Qn-2 = -(Pn- 1Qn-2 - Qn-lPn-2) = ... = (-It-1(P1Qo - PoQl) = (-It. From this we find that polynomials Pn and Qn are mutually prime.

EXERCISE 1.19. PnQn-2 - Pn- 2Qn = (AnPn- 1 + Pn- 2)Qn-2 - Pn- 2(AnQn-l + Qn-2) = A.n(Pn- 1Qn-2 - Pn-2Qn-l) = An( -It-1•

EXERCISE 1.20. a = lim [ql> ... , qn-l, qN] = lim [ql> ... , qn-l> N~oo N~oo

1 1 = lim ql +-----1---= ql + ------1----

N_oo

EXERCISE 1.21.

1 qn-l +-----

lim [qn, ... , qN] N_oo

1 VS 0-1 1 1 £1'=-+-=1+--=1+--=1+--

2 2 2 2 0+1 VS-1 2

1 = 1 + 0 _ 1 '" = [1, 1, 1, ... ].

1+--2

Answers and Hints to Exercises 197

Since Qo = 0, Ql = 1, and Qn = qnQn-l + Qn-Z = Qn-l + Qn-Z, {Qn} is the Fibonacci sequence.

EXERCISE 1.22. By Proposition 1.2, we have n = qm + r, 0 ~ r < m. Then e = an = a qm+T = (am)qaT = aT. By the minimality of m, r = O.

EXERCISE 1.23. If aa# = a#a = 1, then a*aa# = a# = a* which proves the uniqueness of the inverse of a when it exists. The rest is routine.

EXERCISE 1.24. Routine.

EXERCISE 1.25. Easy.

EXERCISE 1.26. Let a = [ql' ... , qn-l, an] and a' = [q~, ... , q;"-l> a;"]. Note that an = a;" =? an+k = a;"+k' for all kEN.




EXERCISE 1.30. By the assumption aZ' == -1 (p), we have az·+1 == 1 (p). Here in the group rr; = (lL/ p lL) X, the order of [a] must divide 2n+t, i.e., ord[a] = 2m, O~k~n+1. If k~n, then aZ'==l (p) and so -1 == 1 (p), a contradiction (because p =f=. 2). Hence k = n + 1. On the other hand, aP-l == 1 (p) which implies that 2n + l 1 p - 1.

EXERCISE 1.31. Let a = 2 and n = 5 in Exercise 1.30. If p 1 Fs, then 26 1 p - 1, i.e., p = 1 + k26. Try k until 10:

k 1 2 3 4 5 6 7 8 9 10

1+k26 65 129 193 257 321 385 449 513 577 641

prime? no no yes yes no no yes no yes yes

Here 641 is the first prime which divides Fs = 4294967297. In fact, the prime decomposition of Fs = 641 x 6700417 (try!).

EXERCISE 1.32. paz = 2 P- l -1 = (2(p-l)/Z + 1)(2(p-l)/Z - 1) =? 2(p-l)12

+ 1 = bZ or 2(p-l)/2 - 1 = bZ, bEN. Since b is odd, write b = 2c + 1, c ::=:0.

Case 1: 2(p-l)12 = bZ - 1 = 4c(c + 1) =? c = 1 =? b = 3 =? p = 7.

Case 2: 2(p-l)12 = bZ + 1 =2(2cz+2c+ 1) =?c= O=? b = 1 =? p = 3. EXERCISE 1.33. pe = aP-l - 1 = (a(p-l)/Z + l)(a(p-l)12 - 1) =? a(p-l)/Z

-1 = 1 (because the two factors of aP - l -1 above are relatively prime) =?a=2, p = 3.




EXERCISE 1.36. Given at> ... , a/ E 7L, we must show that the system

has a solution. Let m = ml ... m/ = mimi. Hence (mi' mi) = 1, mj I mi (i =1= j). Let Xi be a solution of mixi == 1 (mi). Then x = almtxl + ... + a/mIx/ is a solution to the system.


EXERCISE 1.38. Induction on n = degA (for A =1= 0). If n < m, put Q = 0 and R = A. If n;::: m, then let A' = A - B(anlbm)Xn- m, where an (resp. bm) is the coefficient of xn (resp. xm) of A (resp. B). Since degA' <n, we get A'=Q'B+R, R=O or degR<m and so A=QB+R, with Q = Q' + (anlbm)Xn- m.

EXERCISE 1.39. Similar to Theorem 1.1.

EXERCISE 1.40. We can take r = 2 as a primitive root mod 13.

x 1 2 3 4 5 6 7 8 9 10 11 12

7: 2 4 8 3 6 12 11 9 5 10 7 1

a 1 2 3 4 5 6 7 8 9 10 11 12

indz a 0 1 4 2 9 5 11 3 8 10 7 6

{2, 6,11, 7} is the set of all primitive roots mod 13. Solutions of three equations are x == 10, x == ±5, and x == 8, 11, 7 (mod 13), respectively.


EXERCISE 1.42. For k, 1 $. k $. P - 1, put

= (P) =p(p -1) ... (p - k + 1) N k k! .

Then p I k!N. As p f k!, we have piN.


EXERCISE 1.44. Obvious when x = e. For x =1= e, write x = gk, where g is a generator of G. Hence n f k. If we put Xl(g) = 1;;, then Xl(X) = I;;k =1=


1. Therefore

S = L X(x) = L (XIX)(X) = XI(X)S = ° xeG x

because Xl(X) *0. EXERCISE 1.45. The pair {17,2} is an example which satisfies the

condition.

EXERCISE 2.1. Easy.

EXERCISE 2.2.

Chapter 2

1. !ex = gh, g, h E K[X], 0< deg g, deg h < degfa => ° = fa( a) = g( a)h( a) => g( a) = ° or h( a) = ° => fa I g or fa I h, a contradiction.

2. Suppose fa = (X - y)2g, YEO, g E a[X]. Then the derivative f~ = 2(X - y)g + (X - y)2g , E a[X] and d = (fa, f~) is a nonconstant polynomial E a[X]. On the other hand, as d is obtained from fa, f~ by the Euclidean algorithm in K[X], we have dE K[X] which contradicts that fa is irreducible in K[X].

3,4. By Theorem 1.14, cp induces an isomorphism cp': K[X]/(fa) = K[a]. By the argument in the proof of Theorem 1.12 (¢:) (applied to K[X] in place of Z), we see that K[X]/(fa) is a field; hence K[a] = K(a). If n = degfa, then, by the minimality of fa, {I, a, ... , a"-l} is independent over K, which implies [K(a): K] = n.

EXERCISE 2.3 .. By Note B, we know that C is algebraically closed; hence for f E a[X] c qX] there exists a E C such that f(a) = 0. We claim that a E a. As is seen easily, we can assume that

a; E 1:.

Now, since each ll'i satisfies a relation of the form

ll'7' + cil ll'7i - 1 + ... + Ci,n-lll'i + Cin = 0,

if we call M the submodule of C generated by finitely many complex numbers ll'jll'~I ... ll'~n, 0:$ j:$ n - 1, 0:$ k i :$ n; -1, 1:$ i:$ n, then we have ll'M c M and hence ll' E 7L (Proposition 1.20).

EXERCISE 2.4. Easy.

EXERCISE 2.5. There are I-I primitive Ith roots of 1, i.e., ~;, 1::5 i ::5/- 1. For a E G(a/Q), we have (~ay = (~l)a = 1, hence ~a = ~j(a), j( a) E Z, is a primitive root of 1. Thus, K a = Q(~) = Q( ~j(a») = Q(C) = K which proves (1). (2) This is a famous theorem with many proofs. The shortest proof uses Eisenstein's criterion: If for a polynomial P(X) = anXn + an_1Xn- 1 + ... + ao E Z[X] there is a prime p such that


(a) p.J-an , (b) p I ai' O;§;i;§;n-l, and (c) p2.J-ao, then P(X) is irreducible in Q[X]. (Proof of this statement is a good exercise in itself.) Now the question of the irreducibility of fl;(X) is equivalent to that of P(X) = fl;(X + 1) = [(X + W -1]IX = X I- 1 + (DXI- 2 + ... + (DX + I. By Exercise 1.42, we see that (a), (b), and (c) hold for P(X) with p = I. (3) The mapping a~ j(a) induces a homomorphism G(KIQ)~ (7L/l7L) X

whose kernel is trivial because A. generates KIQ. To see the surjectivity, let a E 7L be such that (a, l) = 1. Then 1;a shares the same minimal polynomial with 1; [by (2)] and so there is a a E G(KIQ) such that a == j(a) (I) [Proposition 2.2(4)] which proves (3). (4) By definition 1.20, the Gauss sum takes value in K. Use Proposition 1.25. (5) Since G(KIQ) = IFf is cyclic of order 1-1, an even number, it has only one subgroup of order 2 (Proposition 1.14). By Theorem 2.3, [K: Gb] = 2 and Gb is the only one subfield of K with this property. Since I '* 2 we have 1; '* 1;-1 = ~ and so a: 1; ~ ~ is the unique element of order 2 in G(KIQ) and we find that k = 10(1; + 1;-1) = Gb.


EXERCISE 2.7. f,AX) = ll?=1 (X - a(i)~ f~(a(i) = n""i (a(i) - aU».

EXERCISE 2.8. Easy.

EXERCISE 2.9. Use Exercise 2.6. To see that NK1Q(1; -1) = I, notice thatfl;_1(X) = fl;(X + 1).



EXERCISE 2.12. Tedious but easy.

EXERCISE 2.13. If 0 is a generator of K X , then 00 = O(qf- 1)/(q-1) is a generator of kX.

EXERCISE 2.14. Jr!(KX) = H1(KX) = 1.

EXERCISE 2.15. Consider the matrices in GLn(7L) of the following three types:

1

1 1';= -1

1

1


i j

1

1 o 1

1 Qi,j=

1 1 o j

1

1

j

1

R- ·(A) = 1,1 i, ii=j, A E Z.

1

For X EMn(Z) 1. X ~ P;X: change the sign of the ith row of X,

X ~ XP;: change the sign of the ith column of X. 2. X ~ Qi,jX: exchange the ith row of X with the jth row.

X ~ XQi,j: exchange the ith column of X with the jth column. 3. X ~ RiA A)X: add A times the jth row to the ith row.

X~XRi,j(A): add A times the ith column to thejth column. By X ~ Y, X, Y E Mn(Z), we mean that Y is obtained from X by applying finitely many operations of types (1), (2) and (3). Prove that for any A E Mn(Z) one can find D, a diagonal matrix, such that

d 1 0

A~D= d r

r= rank A, di E N, di I di+l. 0

0 0

EXERCISE 2.16. Any a E k satisfies a 2 - (tkIOa)a + NklOa = O.

EXERCISE 2.17. Use Exercise 2.16.


EXERCISE 2.18. We know that DkIO(~) = (_1)(1-1)1211-2 (Exercise 2.9) and Dk/Ci~) = dk[Ok: ..£igW (Proposition 2.19). Therefore, for this Exercise, it is enough to prove that Ok = 1"[ n So assume that 0k"* 1"[ n As [Ok:1"[~]] is a power of I, there is an W E Ok' W It. 1"[~] such that

XiE1".

Since 1"[1 - ~] = 1"[ n we can also write Iw = Yo + Y1 (1 - ~) + ... + YI-2(1 - ~)1-2, Yi E 1". Then 1= (1 - ~)1-1 A implies Yo = (1 - ~)Jlo, Jlo E Ok and so II Yo as Nk,o(l - ~) = I. Similarly, we get, in tum, II Yb ... , I/YI-2; this means that W E 1"[1 - ~] = 1"[ n a contradiction.



EXERCISE 2.21. If 0 = Ok, then put c = ah-\ h being the class number of k. If 0"* Ok> let lJb ... , lJl be all distinct prime factors of o. First find Jl E a such that Jllt. alJi' for all i, 1::5 i ::51. To do this, put rn = alJ1 ... lJ/J rni = rnlJi\ find Jli E rni such that Jli It. rn, and put Jl = Jl1 + ... + JlI. Next, let c be an ideal COk such that (Jl) = ac. Verify that c has the required property.

EXERCISE 2.22. By Exercise 2.18, we have Ok = 1"[n Use this ~ in place of () in Theorem 2.17.

EXERCISE 2.23. Use Exercise 2.17 and Theorem 2.17.



EXERCISE 2.26. When Ok = [Wb ... , Wn], d k = dklO( Wb ... , Wn) = dk(W1, ... , Wn) = det(tk(wiWj» (Exercise 2.25). Now, {Wb ... , Wn} forms a basis of A = Ok/(P) over IFp. For a E Ok, (aWb ... , aWn) = (Wb . .. , wn)M(a), M(a) E Mn(1") and so (a-Wb ... , a-wn) = (Wb ... ,wn)M(a-), M(a-) = M(a) E Mn(lFp). Since tA(a-) = tr(M(a-» = tr(M(a» = tk(a), we get dA(Wb ... , Wn) = det(tA(wiWj» =

det(tk(wiWj)) = !:,.k. EXERCISE 2.27. Easy.




EXERCISE 2.31. Define a mapping CPn: Io(I)--;.IF('!(lFit so that for a prime P "* I we have CPn(P) = [p] mod(lFrt. Check that Ker CPn = Hn(I).

EXERCISE 2.32. 1. (x,N)=(y,N)=l, x==y(N)::3;>(forp I~) (x,p)=(y,p)=l,

x == Y (p)::3;> Ap(X) = Ap(y)::3;> T/N(X) = T/N(Y)·


2. Let N = I1p p and M = I1q q where p and q are prime. Then TJM(N)TJN(M) = I1p Ap(M) I1q Aq(M) = I1p,q Ap(q )Aq{p) and use the reciprocity for Legendre characters.

3,4. These can be proved similarly.

EXERCISE 2.33. 1. llk == 1 (4). x E N, (x, llk) = 1, x = 2ay, Y odd, Y = I1p. For

brevity, write Xk=X and llk=ll. X(x) =X(2,)X(Y) = TJ\6,i2')I1p X(p) = TJ\6.\(2£1') I1p Ap(ll) = TJ\6.\(2')TJy(ll). Now

TJy(ll) = TJy(sgn ll) Ill!) = TJy(sgn ll)TJy(lll!)

= TJY( sgn II )TJ\6.\(Y)( -1 )[(y -1)I2H(l6.1-1)/2]

{ TJ\6.\(Y)( -1 )I(y-1)l2j '[(6.-1)l2j = TJ\6.\(Y) = TJY( -1)TJ\6.\(Y)( _1)[(y-1)l2j .[(-6.-1)l2j if II > 0,

= TJ\6.\(Y)( _1)[(y-1)/2j .[1-(M1)l2j = TJ\6.\(Y) if II < O.

Hence X(x) = TJ\6.\(2')TJ\6.\(Y) = TJ\6.\(x). 2. Similar to (1). 3. Case 1, II == 1 (4). Write Illl = pg, p a prime, g E N, (p, g) = 1.

Find a E 7l.. such that (~) = -1. Find x E N such that x == a (p),

x == 1 (g). Then, using (1), we get X(x) = -1. Use (2) to handle Case 2, ll==O (4).

4. Case 1, II == 1 (4), is easy. Case 2, II == 0 (4), requires a little more effort.

Chapter 3

EXERCISE 3.1. Let M = {1/2 v, v = 0, 1,2, ... } c IR. M is discrete but not finite.

EXERCISE 3.2. Let N c M = [U1' ... ,un]. If n = 0, the matter is trivial. If n = 1, then M = 7l.. and the statement is nothing but Theorem 1.1. Assume that the statement is true for a submodule of M* = [Uv ... , un-d. If N cM* then the statement follows from the induction hypothesis. If N¢M*, there is a Vn=b1U1+···+bn-1Un-1+bnunEN with bn EN; we assume Vn is such that the natural number bn is minimal. Let v = a1u1 + ... + anUn be any element of N and let an = bnq + r, 0:;;; r < bn. Since v - qVn EN, we have v - qVn E M* by the minimality of bn and so v = U + qVn with U EN n M* c M*. By induction N n M* is a free module of rank:;;; n - 1. As v = U + qVn = 0 => U = 0, q = 0, N is a free module of rank:;;; n.


EXERCISE 3.3. Let cp: ~n~ ~n be the linear mapping defined by cp(e;) = Vi' 1:s; i :s; n, where {ei} is the standard basis of ~n. Notice that A = Jrp = the Jacobian matrix for cpo

EXERCISE 3.4. Easy.

EXERCISE 3.5. Passing to polar coordinates (xS+i = lje i8i) we have vol (B) = 2S (2trYI and 1= 4-tp nJ where J is a Dirichlet integral with the value

r(lyr(2yr(l) 1 J - --- -

r(s+2t+l) n!

[E. T. Whittaker and G. N. Watson, A Course of Modern Analysis, 4th ed., p. 258, Cambridge University Press, London (1958)].

EXERCISE 3.6. Easy.

EXERCISE 3.7. Since n=2, r2=1, we find Mk=(2/tr)V-l!!..k< 0.63663V-l!!..k. We have Mk <2 for -m = 1,2,3,7, and hk = 1 in these cases. There remain five more cases to check: -m = 11, 19, 43, 67, 163. Note that m == 1 (4) and I!!..k = m for these cases. Let us explain the case -m = 43. Suppose hk ~ 2 and take an ideal a C Ok such that Na:s; Mk < 0.63663\143 < 4.174663. Hence Na = 2, 3, 4.

1. Na = 2. Then a = p, Np = 2, X(2) = 1. On the other hand, X(2) = (_1)(432-1)12 = (_1)231 = -1, so this case does not happen.

2. Na = 3. In this case, a = p, Np = 3, X(3) = 1. On the other hand,

X(3) = (~) = (-;3) = (~1) = -1, impossible.

3. No = 4 = 22. If 0 = p, then since 2 is unramified for k/IJJ, p = 2 - 1 which contradicts 0 + 1. Hence 0 = p2 or 0 = pq, P =1= q. If 0= p2, then 22 = No = (Np)2, i.e., Np = 2 which is impossible by (1). If 0 = pq, then 22 = No = NpNq and so Np = Nq = 2 which is impossible again.

EXERCISE 3.8. Since n = 2, r2 = 0, we have

Hence hk = 1 whenever I!!..k < 16; this is the case for m = 2, 3, 5, 13.

EXERCISE 3.9. As m = -ab == 2, 3 (4), we have I!!..k = 4m and Ok = [1, Ym]. Verify that a = (a, Ym) is not principal.



EXERCISE 3.11. Generalize Exercise 2.5(1), (2) for arbitrary cyclotomic field.

EXERCISE 3.12. Let m be the number of roots of unity in k. Then, on the one hand, 211 m but, on the other hand, q;(m):s; 1 - 1.


EXERCISE 3.14. Y E Wk ~ yN = 1 ~ (y(i»N = 1 ~ ly(i)1 = 1 for all i. Conversely, suppose that ly(i)1 = 1 for all i. Since X(Ok) is a lattice in £"''2 = [Rn (Proof of Theorem 3.2), x(y) must be in a finite set of [Rn. As x is injective, it follows that there are only a finite number of y's such that I y(i) I = 1. Since for any j E N, absolute values of all conjugates of y are 1 like y, we must have ya = yb for a> b, a, bEN, and so ya-b = 1, i.e., y E Wk'

EXERCISE 3.15. O~:::> (')0: is obvious. One sees easily that (') n 0: = 1 because 1 * 2. It remains to show that 0 ~ C ( ') 0:. For any E E o~, its complex conjugate E is also in o~. Since G(K/O) is abelian, EW=(E)(i). It follows that 1(E/E)(i)l=l; hence E/EEWK={±,V} (Exercise 3.12, Exercise 3.14). As E/E = ±'v = ±,±V+1, we may assume that v = 2g, g E 7L. Put Eo = ,gE E o~. Then Eo = ,-gE = ,-g(±1),2gE = ±,g E = ±Eo. If Eo = Eo, then we are done. Suppose, on the contrary, that Eo= -Eo. Write Eo= co+ C1' + ... + CI_2,-(1-2) (Exercise 2.18). Then Eo = Co + C1,-1 + ... + CI_2,-(1-2). Let {= (1- '), this being the prime ideal in OK such that (I)K = £1- 1 (Exercise 2.22). As ,== 1 (0, ,-1 == 1 (0, we have Eo == Eo = Co + ... + CI-2 (0; but Eo = - Eo implies 2Eo == 0 (0, a contradiction (because 1 * 2).

EXERCISE 3.1Q. From the inequalities

In+ 1 dx 1 In dx -<-< -s s s' n X n n-1 X

we get

iN+ 1 dx N 1 iN dx

-<2:-<1+ -. 1 XS n=1ns 1 XS

(*)

Since

r ~ = [ X 1-

s ]00 =_1 J1 x 1- s 1 s -1

is finite (s > 1), 'o(s) = E:=11/ns converges for s > 1. Next, letting N-H/J in (*), we have

1 1 s -1 < 'o(s) < 1 + s -1 '

hence 1 < (s -l),o(s) <s and so lim (s -l),o(s) = 1. 8-+1+0



EXERCISE 3.18. Let no n 7L = a7L, a > O. Then one verifies that x(a)ECnr, i.e., cnr*0.

EXERCISE 3.19. As C1 = C n VI = {x E C; IF(x)1 :51}, it is enough to prove that S = {x E C1; IF(x) I = 1} is bounded.

EXERCISE 3.20. Let

We find (u,).,) = 0, orthogonality. Let r = [u, l#(El), ... ,1#(Er )], a lattice in w,+r2 • Verify that V(r) = Rk •





EXERCISE 3.25. I is the only prime ramified for klQ and 1=£1- 1

(completely ramified, Exercise 2.22). Hence h = g[ = 1 and T(s) = l. EXERCISE 3.26. m = 4 and there is only one 1:> such that 1:> 14, i.e., the

1:> with 2 = 1:>2. For this 1:> we have 12 = g2 = 1 and so T(s) = l. EXERCISE 3.27. By Exercise 3.26, we have ~k(S) = ~Q(s)L(s, X).

Compare coefficients of Dirichlet series on both sides.

EXERCISE 3.28. For any E > 0, take no so that 11nno < E/2B. If n ~ no, s ~ So, then 1lns :5 11nno < E 12B. For M > N ~ no, we have

M M A A M A M-l A 2: av = 2: v - v-I = 2: ~ - 2: v

v=NV v=N V v=NV v=N-l(v+1Y

Hence

~ <B -+ -- +- <-<E 1M a I (1 M-I ( 1 1) 1) 2B V~NV - N S V~N v (v+1Y M S - N S •

EXERCISE 3.29.

OX> 1 OX> 1 1 1 IR(s,x)I<2: 2:-;;:;:52: 2:--;:;=2:-2-<2:( 1)2<b.

p n=2P p n=2P p P - P p P-

Answers and Hints to Exercises


EXERCISE 4.2. Easy.

207

Chapter 4

EXERCISE 4.3. Let Ix be the characteristic function of the one point set {x}, x E IFp, and let A be the matrix determined by the relation

( ... , TIx, ... ) = ( ... ,/y, .. . )A,

Applying Tix on ;, we find A = (~xY) by using 1jJy(x) = ~xy. Then we find that

l o ... ···0 0 0 .. . ... l

A2= 0 o ... l 0

0 l 0

and so detA = ±il(/-1)12lll2. It remains to show that detA = il(/-1)12 x (a positive number). To do this, use the Vandermonde theorem to get

detA = n (~y - ~X) = n eni(x+Y)'ln (eni(y-x)ll-ni(x-y )ll)

Osx<ys/-1 x<y x<y

and verify that the first factor in the right-hand side equals 1 and the second factor equals i l(/-1)12 X (a positive number).

EXERCISE 4.4., Easy.

EXERCISE 4.5. Easy.

EXERCISE 4.6. First derive that

Next, compute 1-1

Sx = L X(;) 10g(l- ~-/;). /;=1

Case 1, X = an even character. Apply the change of variable ; 1-+ -;.

As X( -1) = 1, we have Sx = E~:'\ X(;) 10g(1 - t;l;) and so

2Sx = 2 % X(;) log (2 sin (Jli) ) because Ex (;) = o.


Case 2, X = an odd character. By the change of variable ~ ~ - ~ we have, this time, Sx = - ~ X(~) log(1- ,(;) and then 2Sx = 2 ~ X(~)Jti(!~/I). Use ~ X(~) = 0, again.

EXERCISE 4.7. Since 199=7 (8), hk=~O<{;<99.5A199(~)=54-45=9.

EXERCISE 4.8. It is enough to prove that me [a, b + cro]. So take any x + yro E m and write y = qc + r, 0:5 r < c. Then x + yro -q(b + cro) = (x - qb) + rro E m, and, by the minimality of c, we have r=O, i.e., y=qc. Now, x+yro=x+qcro=x-qb+q(b+cro)~ x - qb Em n £: = £:a ~ x = sa + qb, s E £: ~ x + yro = sa + qb + qcro = sa + q(b + cro) E [a, b + cro].

EXERCISE 4.9. Easy. EXERCISE 4.10. If m is an ideal, then aro E m and so c I a by the

minimality of c. As for c I b, start with ro(b + cro) E m and consider the casesm=2,3 (4)andm=1 (4),separately.

EXERCISE 4.11. (~) N(b+ro)=(b+ro)(b+ro')Emn£:=£:a~ a I N(b + ro). (¢:) It is enough to prove that aro, (b + ro)ro E m. Now, aro = -ab + ab + aro = -ab + a(b + ro) E m. Next, (b + ro)ro = (b + ro)(b +s) -aa' Em where s = ro + ro' and N(b + ro) = aa'.

EXERCISE 4.12. Use Exercise 4.11. EXERCISE 4.13. Easy. EXERCISE 4.14. We may assume that a is primitive: a = [a, b + ro] =

(a, b + ro). Hence a' = [a, b + ro'] = (a, b + ro'). Then aa' = (a, b + ro)(a, b + ro') = (a 2, a(b + ro), a(b + ro'), N(b + ro» = a(a, b + ro, b + ro', d)=afJ. with N(b+ro)=ad. Nowa2=(Naf=NaNa'=N(aa')= N(afJ) =a2NfJ~NfJ = 1~b = Ok. Hence aa' = (a) = (Na).

EXERCISE 4.15. a I N(b + ro)~ai I N(b + ro)~ai = [a;, b + ro] is a primitive ideal, i = 1,2. Verify that a1a2 c a and compare norms of both sides.

EXERCISE 4.16. Notice first that Mk < I. (~) Assume that P(b), 0:5 b :51 - 2, are all prime. For p E Sk (Proposition 4.3), we have p=N!,Jp, !,Jp=[p,b+ro], O:5b<p. As p:5Mk <l, we have O:5b:5 /-2. 1~[P(b), b+ro] and p IN(b+ro)=P(b)~P(b)=p. Hence 1 ~ [p, b + ro] = !,Jp, p E Sk> which implies that hk = 1. (¢:) Notice that P(b) = P(l- 2) < /2. Hence if P(bo) is not prime for abo, 0:5 bo:5/ - 2, and if p is the smallest prime dividing P(bo), then we have p :5/- 1 < / by P(bo) < r. Verify that !,J = [p, bo + ro] is a prime ideal which is not principal.

EXERCISE 4.17. Easy. EXERCISE 4.18. Routine. EXERCISE 4.19. For a EN Ib put b = Ok + a + a1+a + ... +

1+a+···+0"-2 = [k. "'] a , n .1IaI!.


EXERCISE 4.20. k = Q(Y221).

EXERCISE 4.21. hk = hi; = 2t - 1h%. For m = 1- 2225 == 1 (4), Nw = 2223 (223 is prime), ll.k = m, and t = 14. Let a = [2, w]. As a + 1, we find 2231 hk and hence 213 X 223 = 18268161 hk •


EXERCISE 4.23. (1) 1 + Vi, (2) 2 + \13, (3) 5 + 2'16, (4) 8 + 3V7.

EXERCISE 4.24. (1)!(1 + 0), (2) !(3 + V13). EXERCISE 4.25. Just do!

EXERCISE 4.26.

b c b' c' b b' x2 +-x +-= (x - a)(x - a') =x2 +-x +-~- = -,

a a a' a' a a'

EXERCISE 4.27. Let

Then

pa+q al=--,

ra+s

A=ra+s,

al 2" ( bl) ta-JAa-l=Al = b

l ~(al> I)Al(~l)

2 Cl

= alai + blal + Cl = O~detAI

c c' a a'·

= detA = b2 - 4ac = D(a) = D(al) ~ al = a(a) E X 2(D).


EXERCISE 4.29. One way is obvious. Conversely, assume that a E R2(ll.). Then, by (4.59) and (4.60), a + a' = -b/a > 0 and all" = (b2-ll.)/4a2 = c/a < O.

EXERCISE 4.30. Similar to Exercise 4.26. EXERCISE 4.31. Easy. EXERCISE 4.32. Use (4.63).

Notes

A. Peano Axioms

In Section 1.1, we denoted by N the set {1, 2, 3, ... } of all natural numbers. We did so under the assumption that everybody can feel the totality of natural numbers intuitively. As N is only a letter, what we did is merely a matter of convenience. To develop the theory of natural numbers less intuitively, one usually starts with the following axioms due to Peano (1858-1932):

I. 1 E N, II. there is a mapping cp: N~ N,

III. cp is injective, IV. 1 ~ cp(N), V. if a subset A of N satisfies the following two conditions:

(a) 1 EA, (b) cp(A) cA, then A = N.

Once axioms are stated, it is natural to ask the existence and the uniqueness of objects (the set N satisfying I-V). In our case, one can prove the uniqueness.t However, to prove the existence of the set N seems to be extremely difficult; the set N is infinite and studies such as cosmic physics, cerebral physiology, or metaphysical ontology seem to be helpless for such an attempt. If one wants to avoid the vicious circles of analyzing a product of human thoughts such as N by using human thoughts, the best way must be to believe in its existence.

From N we pass to Q (rationals) via Z (integers) purely algebraically. We obtain ~ (reals) by completing Q with respect to the distance in Q defined by the absolute value. Of course, the best way is to assume that

tE.g., p. 176 of S. Iyanaga, System of Numbers, Vol. 1, Iwanami, 1972 (in Japanese).

210

Notes 211

everybody can feel the totality ~ of real numbers as the straight line. The passage from ~ to C (complex numbers) is again purely algebraic.

Our standpoint in this book is quite naive; think of 71. as a subset of ~ (the straight line) and extract its properties as a commutative ring but not a field.

B. Fundamental Theorem of Algebra

This theorem asserts that C is algebraically closed, i.e., for any polynomial f(X) E qX], degf ~ 1, there is an a E C such that f(a) = O. It was proved by C. F. Gauss (1777 4/30-1855 2/23) in the fall of 1797. While he was a student at University of Gottingen (1795-1798) Gauss made important discoveries in mathematics including the famous Disquisitiones arithmeticae. However, his first proof of the fundamental theorem of algebra was submitted to the University of Helmstedt as Dissertation and published in 1799 under the title Demonstratio nova theorematis omnen functionem algebraicam rationalem integram unius variabilis in factores reales primi vel secundi gradus revolvi posse. Actually, Gauss's first proof is the first proof of the theorem and so the word nova in the title is misleading. The polynomial (integral rational function) f(X) in the Dissertation has real coefficients: f(X) E ~[X]. However, it is easy to extend the result for any f(X) E qX]; in fact, write f(X) = g(X) + ih(X), g, h E ~ [X], and consider

r(X) = f(X)/(X) = g(X? + h(X? E ~[X].

Then, for a E C, r( a) = O::} f( a) = 0 or f( it) = 0 which proves the theorem for any f(X) E qX]. After the Dissertation, Gauss published his second and third proof in 1816 and his fourth proof in 1850.

One of the shortest proof we have today depends on a theorem in the theory of complex variables due to Liouville (1809-1882): a bounded integral function is a constant. In fact, if f(X) E qX], degf ~ 1, has no roots in C, then l/f is integral and bounded as f(z)-700 when Z-700,

hence l/f or f is a constant, a contradiction. Anyway, it is delightful to know that one reaches the algebraic

closure of ~ by a quadratic extension C = ~ (i) where the ground field ~ is far from being algebraically closed; e.g., ±1 are the only roots of 1 in ~.

212 Notes

c. Zorn's Lemma

Let X be a set. X is an ordered set if there is defined a relation -< (order) among elements in X satisfying the following conditions:

1. x -<x, Vx EX, 2. x -< y, y -< x ~ x = y, 3. x-<y, y-<z~x-<z.

If, in addition to (1)-(3), we have

4. for any x, y E X, either x -< y or y -< x,

then X is a totally ordered set. An ordered set X is called an inductively ordered set if every totally

ordered subset Y of X has an upper bound, i.e., for any y E Y, there is a b E X such that y -< b.

ZORN'S LEMMA. Every inductively ordered set X '* 0 has at least one maximal element m, i.e., an element m such that m -< x, x E X ~ X = m.

REMARK. Although the statement is called "lemma" it is rather an "axiom" because we have no intention of proving it.

As an example, let us prove that any vector space V over a field K has a basis. First of all, a subset S in V is called independent if any set of finitely many vectors in S is independent. Let X be the family of all independent subsets of X. In X, define the relation -< by

S-<S'~ScS', S, S' EX.

Clearly, X is an ordered set. Let Y = {1).;). E A} be any totally ordered subset of X. Then T = U). 1). belongs to X and, as 1). -< T for all 1). in Y, X is inductively ordered. By Zorn's lemma, there is a maximal element Min X. We claim that M is a basis for V, i.e., M is independent and any vector in V is a linear combination of finitely many vectors in M. In fact, if some v E V fails to be such a linear combination, then S = {M, v} EX with M -< S, M '* S, which contradicts the maximality of M.

D. Quadratic Fields and Quadratic Forms

Fixing once and for all a quadratic field k = IlJl(Ym), consider the set

Q(dk ) = {f = ax2 + bxy + cy2; a, b, c E 71, b2 - 4ac

= d k, f > 0 if dk < O},

Notes 213

where f is a binary quadratic form with integral coefficients a, b, and c, with discriminant b2 -4ac, andf>O~f(x,y»O, for all (x,y),*(O, 0). As we shall see later, Q(~k) '* 0. For f, g E Q(~k)' put

def f±g ~g(z)=f(yz), Y E SLz(£:) = {a' E GLz(£:), deta' = 1},

this being an equivalence relation in Q(~k). Let Q(~k) denote the quotient space Q(~k)/ ±. Now, let a = [a', {:J] be a fractional ideal of k. We agree to orient the basis so that

if ~k >0,

For example, the order of the canonical basis of a primitive ideal a = [a, h + co] is the right one. In general, for a fractional ideal a = [a', {:J] put

1 fa = -N(xa' + y{:J).

Na

Then fa E Q(~k). It follows that

where ± for ideals is the one defined in Section 4.6. We then find that the mapping a ~ fa induces a bijection ik :

Since the left-hand side is a group, the right-hand side becomes a group, too. This is the "group" of Gauss; in Disquisitiones arithmeticae, he defined in Q(~k) a product called composition and proved that Q(Dok) forms a finite abelian group.

(group)

(character)

ik

_

Ht -=-. Q(Dok) ("group")

x7~ /'I/J; ("character")

{±1}

214 Notes

In Theorem 4.9, we associated with each i, 1:5 i:5 t, t being the number of distinct prime factors of !J.k , the character xi of H7;, and proved that

t n Ker xi = (H7;)2. ;=1

Each coset of Hid(H7;)2 is a genus, each genus consists of hZ = [(H7;)2] classes, and h7; = 2t - 1hZ- All these results are in Disquisitiones in the language of quadratic forms and we have only translated them into the language of ideals via our dictionary ik •

Although we did not touch it at all, we can define the notion of genera of quadratic forms by using p-adic numbers as a classification of quadratic forms weaker than the classification by f ± g. This method can be generalized almost endlessly to classify polynomial mappings and leads us to the idea of the adelization of algebraic groups. t

t See T. Ono, Algebraic Groups and Number Theory, Sugaku Expositions, American Mathematical Society Vol. 1 (1988), pp. 95-112.

List of Mathematicians

This list includes names and dates of birth and death for mathematicians quoted in the text, except those still living.

Euclid (330?-275? B.C.) Diophantus (246?-330?) Fibonacci (1174?-1250?) Mersenne (1588-1647) Fermat (1601-1665) Pell (1611-1685) Goldbach (1690-1764) Euler (1707-1783) Lagrange (1736-1813) Legendre (1752-1833) Fourier (1768-1830) Germain (1776-1831) Gauss (1777-1855) J\bel (1802-1829) Jacobi (1804-1851) Dirichlet (1805-1859) Liouville (1809-1882) Kummer (1810-1893)

215

Galois (1811-1832) Catalan (1814-1894) Eisenstein (1823-1852) Kronecker (1823-1891) Riemann (1826-1866) Dedekind (1831-1916) Weber (1842-1913) Frobenius (1849-1917) Peano (1858-1932) Hurwitz (1859-1919) Hilbert (1862-1943) Minkowski (1864-1909) Takagi (1875-1960) J\rtin (1898-1962) Hasse (1898-1979) Herbrand (1908-1932) Chevalley (1909-1984)

Bibliography

Here are some books, mainly on number theory, chosen from my own library and, with the exception of the last two general references, arranged in alphabetical order by author's name.

1. E. Artin, in: The Collected Papers of Emil Artin (S. Lang and J. Tate, eds.), Addison-Wesley, Reading (1965).

2. A. Baker, A Concise Introduction to the Theory of Numbers, Cambridge University Press, Cambridge (1984).

3. Z. I. Borevich and I. R. Shafarevich, Number Theory, Academic Press, New York (1966).

4. J. Cassels and A. Frohlich (eds.), Algebraic Number Theory, Thompson Book Co., Washington D.C. (1967). (Reprint by Academic Press, New York, 1986.)

5. R. Courant and H. Robbins, What is Mathematics?, Oxford University Press, Oxford (1941).

6. H. Davenport, The Higher Arithmetic, 5th ed., Cambridge University Press, Cambridge (1982).

7. L. E. Dickson, Introduction to the Theory of Numbers, Dover, New York (1957).

8. L. E. Dickson, History of the Theory of Numbers, 3 volumes, Chelsea, New York (1971).

9. P. G. L. Dirichlet, Mathematische werke, I and II, Chelsea, New York (1969).

10. P. G. L. Dirichlet, Vorlesungen iiber Zahlentheorie, Chelsea, New York (1968).

11. M. Eichler, Introduction to the Theory of Algebraic Numbers and Functions, Academic Press, New York (1966).

12. G. Fujisaki, Introduction to Algebraic Number Theory, 2 volumes, Shokabo, Tokyo (1957) (in Japanese).

216

Bibliography 217

13. C. F. Gauss, Disquisitiones arithmeticae, Yale University Press, New Haven (1966).

14. L. Garding, Encounter with Mathematics, Springer, New York (1977).

15. G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Oxford University Press, Oxford (1960).

16. G. H. Hardy, Ramanujan, Cambridge University Press, Cambridge (1940).

17. H. Hasse, Bericht fiber neuere Untersuchungen und Probleme aus der Theorie der algebraischen Zahlkorper, I, la, II, Jber. Deutsch. Math. Verein., Leipzig (1926, 1927, 1930).

18. H. Hasse, Zahlentheorie, Akademie-Verlag, Berlin (1949). 19. H. Hasse, Vorlesungen fiber Zahlentheorie, Springer, Berlin (1950). 20. H. Hasse, Uber die Klassenzahl Abelscher Zahlkorper, Akademie

Verlag, Berlin (1952). 21. E. Heeke, Vorlesungen fiber die Theorie der Algebraischen Zahlen

(English translation, Springer, Chelsea, New York 1970). 22. E. Heeke, Mathematische Werke, Vandenhoeck & Ruprecht,

Gottingen (1959). 23. D. Hilbert, Gesammelte Abhandlungen, 3 volumes, Springer, Berlin

(1970). 24. L. K. Hua, Introduction to Number Theory, Springer, (1982). 25. K. Ireland and M. Rosen, A Classical Introduction to Modern

Number Theory, Springer, Berlin (1982). 26. M. Ishida, Introduction to Algebra, Jikkyo, Tokyo (1978) (in

Japanese). 27. M. Ishida, Algebraic Number Theory, Morikita, Tokyo (1974) (in

Japanese). 28. K. Iwasawa, Theory of Algebraic Functions, Iwanami, Tokyo (1952)

(in Japanese). 29. K. Iwasawa, Local Class Field Theory, Iwanami, Tokyo (1980) (in

Japanese). 30. S. Iyanaga (ed.), Number Theory, North-Holland, Amsterdam (1975). 31. S. Iyanaga, System of Numbers, 2 volumes, Iwanami, Tokyo (1972,

1978) (in Japanese). 32. G. J. Janusz, Algebraic Number Fields, Academic Press, New York

(1973). 33. K. Katayama, Introduction to Number Theory, Jikkyo, Tokyo (1975)

(in Japanese). 34. Y. Kawada, Algebraic Number Theory, Kyoritsu, Tokyo (1957) (in

Japanese). 35. Y. Kawada, Number Theory, 3 volumes, Iwanami, Tokyo (1983) (in

Japanese).

218 Bibliography

36. F. Klein, Vorlesungen fiber die Entwicklung der Mathematik im 19. lahrhundert, Chelsea, New York (1956).

37. M. Kuga, Galois's dream, Nippon Hyoron sha, Tokyo (1968) (in Japanese).

38. L. Kronecker, Mathematische Werke, 5 volumes, Chelsea, New York (1968).

39. T. Kubota, Introduction to Number Theory, Asakura, Tokyo (1971) (in Japanese).

40. E. Landau, Vorlesungen fiber Zahlentheorie, 2 volumes, Chelsea, New York (1950, 1955).

41. S. Lang, Algebra, Addison-Wesley, Reading (1967). 42. S. Lang, Introduction to Diophantine Approximations, Addison

Wesley, Reading (1966). 43. S. Lang, Algebraic Number Theory, Addison-Wesley, Reading

(1970). 44. R. L. Long, Algebraic Number Theory, Marcel Dekker, New York

(1977). 45. D. A. Marcus, Number Fields, Springer, New York (1977). 46. G. B. Mathews, Theory of Numbers, Chelsea, New York (1982). 47. H. Minkowski, Gesammelte Abhandlungen, I and II, Chelsea, New

York (1967). 48. H. Minkowski, Geometrie der Zahlen, Chelsea, New York (1953). 49. L. J. Mordell, Diophantine Equations, Academic Press, New York

(1969). 50. T. Nagell, Introduction to Number Theory, Chelsea, New York

(1964). 51. K. Nakamura, Modern History of Mathematics, Nippon Hyoron sha,

Tokyo (1980) (in Japanese). 52. W. Narkiewicz, Elementary and Analytic Theory of Numbers, Polish

Sci. Pub!., Warszawa (1974). 53. C. D. Olds, Continued Fractions, Random House, New York

(1963). 54. T. Ono, Variations on a Theme of Euler, Jikkyo, Tokyo (1980) (in

Japanese). 55. O. Perron, Die Lehre von den Kettenbrfichen, Chelsea, New York

(1950). 56. G. Polya, How to Solve it, Doubleday, Garden City (1957). 57. P. Ribenboim, 13 Lectures on Fermat's Last Theorem. Springer, New

York (1979). 58. B. Riemann, Gesammelte Mathematische Werke, Dover, New York

(1953). 59. P. Samuel, Theorie Algebrique des nombres, Hermann, Paris (1967). 60. J.-P. Serre, Corps locaux, Hermann, Paris (1962). 61. J.-P. Serre, A Course in Arithmetic, Springer, New York (1970).

Bibliography 219

62. D. Shanks, Solved and Unsolved Problems in Number Theory, Chelsea, New York (1978).

63. K. Shoda, A Manual of Number Theory, Kyoritsu, Tokyo (1949) (in Japanese).

64. G. Shimura and Y. Taniyama, Complex multiplication of abelian varieties and its application to number theory, Publ. Math. Soc. Japan 6, Tokyo (1961).

65. C. L. Siegel, Analytische Zahlentheorie, I, II, G6ttingen, Berlin (1963-1964).

66. C. L. Siegel, Gesammelte Abhandlungen, I, II, III, Springer, Berlin (1966).

67. H. J. S. Smith, Report on the Theory of Numbers, Chelsea, New York (1965).

68. D. E. Smith, A Source Book in Mathematics, 2 volumes Dover, New York (1959).

69. H. M. Stark, An Introduction to Number Theory, The MIT Press, Cambridge (1978).

70. D. J. Struik, A Concise History of Mathematics, Dover, New York (1948).

71. J. Suetsuna, Analytic Number Theory, Iwanami, Tokyo (1950) (in Japanese).

72. T. Takagi, An Introduction to Analysis, Iwanami, Tokyo (1961) (in Japanese).

73. T. Takagi, Lectures on Algebra, Kyoritsu, Tokyo (1983) (in Japanese).

74. T. Takagi, Lectures on Elementary Number Theory, Kyoritsu, Tokyo (1983) (in Japanese).

75. T. Takagi, Algebraic Number Theory, Iwanami, Tokyo (1971) (in Japanese).

76. T. Takagi, A Historical Story of Modem Mathematics, KYOritSll, Tokyo (1984) (in Japanese).

77. B. L. van der Waerden, Algebra, 2 volumes, Frederick Ungar, New York (1970).

78. B. A. Venkov, Elementary Number Theory, Wolters-Noordhoff, Groningen (1970).

79. I. M. Vinogradov, Elements of Number Theory, Dover, New York (1954).

80. H. Wada, World of Numbers, Iwanami, Tokyo (1984) (in Japanese). 81. L. C. Washington, Introduction to Cyclotomic Fields, Springer, New

York (1982). 82. H. Weber, Lehrbuch der Algebra, 3 volumes, Chelsea, New York

(1891, 1894, 1896). 83. A. Weil, Number Theory for Beginners, Springer, New York (1985). 84. A. Weil, Basic Number Theory, Springer, New York (1967).

220 Bib60grapby

85. A. Weil, Oeuvres Scientifiques, Collected Papers, 3 volumes, Springer, New York (1979).

86. A. Weil, Number Theory, Birkhauser, Basel-Boston (1984). 87. H. Weyl, Algebraic Theory of Numbers, Princeton University Press,

Princeton (1940). 88. H. Weyl, Gesammelte Abhandlungen, Springer, Berlin (1968). 89. E. T. Whittaker and G. N. Watson, A Course of Modern Analysis,

4th ed., Cambridge University Press, Cambridge (1958). 90. D. B. Zagier, ZetaJunktionen und quadratische Korper, Springer,

Berlin (1981). 91. O. Zariski and P. Samuel, Commutative Algebra, 2 volumes, D. Van

Nostrand, New York (1958,1960). 92. Encyclopedic Dictionary of Mathematics, Mathematical Society of

Japan, The MIT Press, Cambridge (1977). 93. Mathematical Reviews (Monthly), American Mathematical Society,

Providence.

Comments on the Bibliography

1. As I said before, these books are chosen from my bookshelf; most of them I myself bought since the 1950s but some are gifts from authors. Of course, there are many more books on number theory and the above list is only a small and perhaps one-sided collection.

2. The following are some classic publications: 9, 10, 13, 22, 23, 38, ~,~,~,~,~,~,~,~,~,~d~.

3. The following are those books I always kept at the side of my desk for reference while writing this book: 3, 13, 15, 21, 25, 27, 30, 34, 41, 42, 43, 53, 69, 72, 74, 75, 76, 83, and 92.

4. The following have been adopted as textbooks in classes taught by myself and by my colleagues: 2, 6, 40, and 83 (Elementary Number Theory); 3, 4, 43, 45, and 84 (Algebraic Number Theory); 41, 77, and 90 (Algebra).

Index

Algebra, 86 Algebraic closure, 36 Artin map, 96 Artin symbol, 93

Basis integral, 70 normal,73

Beta function, 57

Central symmetric, 109 Character(s), 37

of abelian groups, 37. Jacobi, 101 Kronecker, 101 Legendre, 39

Class field theory, 96 Class number, 76, 95

of Q(.Ji*), 155 Compact, 106 Cone, 119 Conjugate, 45 Constructible, 59 Continued fraction, 11

periodic, 182 Coset, 22

Degree, 28, 46, 79 Direct product, 27 Dirichlet

L-function, 136 theorem on arithmetical progressions,

103, 140 unit theorem, 118

Discrete, 106 Discriminant, 71 Divisor, 4, 28

Embedding, 46 Endomorphism, 63 Equivalence relation, xi, 18 Euclidean algorithm, 8 Euler's function, 24 Euler-Lagrange theorem, 184 Exact sequence, 64 Extension

abelian, 92 field,45 Galois, 45

Fermat's theorem, 24 Fibonacci sequence, 6, 16 Field(s),2

algebraic number, 44 automorphisms of, 45 characteristic of, 68 cyclotomic, 132 finite, 25, 68 Hilbert class, 95 Ith cyclotomic, 52 quadratic, 52

First supplementary law, 40 Fourier expansion, 145 Free, 17 Frobenius automorphism, 69, 92 Fundamental set, 126 Fundamental theorem

of algebra, 211

221

222

Fundamental theorem (Cont.) of finite abelian groups, 134 of ideal theory, 77 of number theory, 7

Gamma function, 57 Gauss reciprocity law, 41, 100 Gauss sum, 41, 55, 144 Greatest common divisor, 6, 29 Group(s), 16

abelian, 16 acting on a set, 18 automorphism, 45 character, 37 cohomology, 64 commutative, 16 cyclic, 22 decomposition, 89 free abelian, 95, 107 Galois, 45 general linear, 17 ideal class, 75, 94 inertia, 91 quotient, 22 transformation, 17 unimodular, 17 of units, 94

Herbrand quotient, 66 Hilbert theorem 90, 67 Homomorphism, 25, 26 Hurwitz lemma, 75

Ideal(s), 23 class, 75 class group, 77, 94 fractional, 93 prime, 74 primitive, 160 principal, 74 principal fractional, 93

Index, 32 Integer(s), 2

algebraic, 34 of k, 70

Invertible element, 17 Irreducible, 28 Isomorphism, 25, 26, 46

Jacobi sum, 56

Kernel, 25, 26 Kronecker-Weber theorem, 133 Kummer's theorm, 83

Lattice, 106 lth cyclotomic field, 52

Artin map of a subfield of, 97

Index

class number of a subfield of, 152 decomposition of rational primes in, 85,

133 discriminant of, 72 Galois group of, 52 group of units of, 119 integral basis of, 72 maximal real subfield of, 52 normal basis of, 73 roots of unity in, 115 subfields of, 60 zeta function of, 139

Mathematical induction, 3 Minkowski constant, 110 Minkowski's theorem, 109 Module, 5, 17

G-,63 Multiple, 4, 28

Norm, 53, 78 nth power residue, 33 Number(s)

algebraic, 34 complex, xi, 2 Fermat, 8, 25 geometry of, 105 irrational, 11 natural, xi, 3 prime, 4 quadratic irrational, 182 rational, xi, 2 real, xi, 2 transcendental, 34

Order, 17,22

Parallelotope, 107 Peano's axiom, 3, 210 Polynomial, 28

minimal,46 monic, 29 normalized, 29

Index

Prime(s),4 Fermat, 59 Sophie Germain, 142 twin, 142

Primitive element, 48 Primitive root, 32

Quadratic field(s), 52 another proof of Gauss reciprocity law,

100 Artin map of, 100 canonical basis of an ideal of, 160 cohomology of, 166 decomposition of rational primes in, 86 discriminant of, 72 fundamental unit of real, 156, 179, 191 Galois group of, 52 genus theory of, 173,212 ideal class group of, 159 ideal class group in the narrow sense of,

167 integral basis of, 72 Kronecker character of, 101 normal basis of, 73 primitive ideal of, 160 Rabinovitch's theorem, 166 roots of unity in, 114

Quadratic form, 212 Quadratic residue, 33 Quotient, 4, 29 Quotient set, 18

Ramification index, 82, 83 relative, 83

Ramified, 82 totally, 85

Reduced, 183 Regulator, 128 Relatively prime, 6, 29 Remainder, 4, 29 Residue, 120 Ring(s),2

of algebraic integers, 36 endomorphism, 63 of integers of k, 70 noetherian, 74 polynomial, 28 quotient, 23

Second supplementary law, 40 Subgroup, 21

normal, 22 Sublattice, 106

Trace, 53

Unramified, 82

Zeta function Dedekind, 123 pre-, 119 Riemann, 120

Zorn's lemma, 47, 212

223

Answers and Hints to Exercises - Springer LINK

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