Algebraic number theory LTCC Solutions to Problem Sheet 2

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1 Algebraic number theory LTCC 008 Solutions to Problem Sheet ) Let m be a square-free integer and K = Q m). The embeddings K C are given by σ a + b m) = a + b m and σ a + b m) = a b m. If m mod 4) then R K = Z + Z m by Lemma 3.6, so β =, β = m is a Z-basis of RK. Hence ) = 4m. σ β d K = det ) σ β ) σ β ) σ β ) ) m = det m If m mod 4) then R K = Z+Z + m by Lemma 3.6, so β =, β = + m is a Z-basis of R K. Hence d K = det σ β ) σ β ) σ β ) σ β ) ) = det + m m Therefore we have shown that { 4m if m mod 4), d Q m) = m if m mod 4). ) = m. ) Let a Z be such that a m mod ), and consider the ideals P =, m + a) and P =, m a) of R Q m). Claim: P P = ) Proof of claim. We have P P =, m + a), m a), m a ). It is clear that, m+a), m a) ). Furthermore m a ) because a m mod ). This shows that P P ). To show the converse we first observe that P P and a = m + a) m a) P P. Since m and a m mod ), it follows that a. Furthermore is odd, so a. Therefore a and are corime, so there exist u, v Z such that u a + v =. Multilying this by gives = u a + v P P. This imlies ) P P. Claim: P and P are rime ideals of R Q m) Proof of claim. Using Lemmas 8.5 and 8. we have NP )NP ) = NP P ) = N)) = [K:Q] =. Furthermore it is easy to see that NP ) = NP ) observe that the automorhism τ of K mas P onto P and therefore induces an isomorhism R K /P = RK /P ). It follows that NP ) = NP ) =. By Question 3)a) this imlies that P and P are rime ideals. 3) a) Assume that A is a non-zero ideal of R K which is not a rime ideal. If A = R K then NA) =, i.e. in this case NA) is not a rime number. If A R K then by Theorem 4.7 we can write A = P P n with n where P,..., P n are non-zero rime ideals of R K. By Lemma 8.5 we obtain NA) = NP ) NP n ). Now for all i we have NP i ) N and NP i ) because P i R K. This shows that in this case NA) is a comosite number, i.e. again NA) is not a rime number.

2 Algebraic number theory, Solutions to Problem Sheet, LTCC 008 b) Let K = Q ) and A = 3), i.e. A is the rincial ideal of the ring R K generated by 3. We claim that A is a rime ideal of R K such that NA) is not a rime number. By Lemma 8. we have NA) = 3 [K:Q] = 9, so NA) is not a rime number. To show that A is a rime ideal, we must show that if a + b ) c + d ) A where a + b, c + d R K then a + b A or c + d A. Now a+b ) c+d ) A imlies a+b ) c+d ) = 3 u+v ) for some u+v R K. Since a+b ) c+d ) = ac+bd)+ad+bc) it follows that ac + bd = 3u, ad + bc = 3v. If b 0 mod 3) then these two equations imly ac ad 0 mod 3). Therefore either a 0 mod 3) and so a + b A, or c d 0 mod 3) and so c + d A. Now assume that b 0 mod 3). From the two formulas ac+bd = 3u and ad + bc = 3v we can deduce bc + bd 0 mod 3). This imlies c + d 0 mod 3). Therefore c d 0 mod 3) and hence c d 0 mod 3). This shows that c + d A. 4) If z > is a real number then ζz) = n z < + z n= comare the comutation in the notes after Definition 7.). Also ζz) = n z > x z dx = z. n= These two inequalities imly < z )ζz) < z for all z >. Letting z + gives hence lim z )ζz) =. z + lim z )ζz) lim z =, z + z + 5) Let m >. The olynomial X m has roots ζm i for i = 0,,..., m, therefore X m = m i=0 X ζi m). Dividing this equation by X gives m X m + X m + + X + = X ζm). i Letting X = shows that m = ζm). i m i= Write n = a ar r where,..., r are distinct rime numbers and a N. Alying the above formula to m = n gives n n = ζn). i i= i=

3 Algebraic number theory, Solutions to Problem Sheet, LTCC Alying the above formula to m = a a a = i= ζ n/ a n i and using that ζ a ) = ζn) = ζ n/a n gives where the roduct is over those {,..., n } for which ζ n has order a ower of. Hence n = a a r r = ζ n) where the roduct is over those {,..., n } for which ζn has rime ower order. It follows that = ζn) where the roduct is over those {,..., n } for which ζn is not of rime ower order. Since n has at least two distinct rime factors this roduct contains the factor ζ n. Hence ζ n ) = ζ n) where the roduct is over those {,..., n } for which ζn has not rime ower order. Since ζ n R Qζn) and ζ n ) = ζ n) R Qζn), it follows that ζ n is a unit in R Qζn ). 6) Let n N. We recall that [Qζ n ) : Q] = φn) where φ is Euler s φ-function. Indeed, for n = this is clear, for n > and n mod 4) this is Theorem 0..), and for n mod 4) it follows from the other cases because we have using that n/ is odd) [Qζ n ) : Q] = [Qζ n/ ) : Q] = φn/) = φn). Claim: If n > is an even integer then µ Qζn ) = {ζ i n : 0 i n }. Proof. The inclusion {ζ i n : 0 i n } µ Qζn ) is clear. Conversely let ε µ Qζn). Let m N be the order of ε. Then ε = ζ i m for some integer i with i, m) =. It follows that ζ m µ Qζn ) because if u i + v m = for u, v Z then ζ m = ζ m = ζ i m) u ζ m m) v = ε u ). Now let l = lcmm, n). Then l/m, l/n) =, so there exist x, y Z such that = x l/m + y l/n. It follows that ζ l = ζl = ζ l/m l ) x ζ l/n l ) y = ζm x ζn y Qζ n ). Thus Qζ l ) Qζ n ). The inclusion Qζ n ) Qζ l ) is obvious because n l, hence Qζ n ) = Qζ l ) and therefore φn) = φl). Since n l and n is even, this imlies that n = l and thus m n. Hence ζ m = ζn for some Z. It follows that ε = ζm i = ζn i. Thus we have shown that µ Qζn ) {ζn i : 0 i n }. Now let n > be an integer such that n mod 4). If n is divisible by 4 then µ Qζn) = {ζ i n : 0 i n } by the claim. If n is odd then µ Qζn) = µ Qζn) = {ζ i n : 0 i n } = {±ζ i n : 0 i n } where for the second equality we use the claim and for the first and third equalities we use that ζ n = ζ n ) n+)/. Thus we have shown that { {±ζn i : 0 i n } if n is odd, µ Qζn) = {ζn i : 0 i n } if n is divisible by 4 as required.

4 4 Algebraic number theory, Solutions to Problem Sheet, LTCC 008 7) a) By Theorem 0. we now that Qζ 5 ) + = Qζ 5 + ζ5 ). Note that ζ 5 = + i. 4 8 Since ζ5 = ζ 5 it follows that 5 ζ 5 + ζ5 =. From this it is obvious that Qζ 5 ) + = Q 5). b) The grou of units R Q is generated by {±} and a fundamental 5) unit. To find a fundamental unit ε = a + b 5 of Q 5) we can use the method from Question 7) on Problem Sheet. Since 5 mod 4), we must try a =,, 3,... until we find an a for which there exists a b such that a+b 5 R Q, i.e. a+b 5 R 5) Q 5) and Na+b 5) = ±. For a = we find N + b 5) = 4 5b and this is equal to for b =. Therefore ε = + 5 is a fundamental unit of Q 5). It follows that R Qζ 5 ) = R + Q = {±} + ) Z 5 5) εz = {±}. c) By definition the grou C + of cyclotomic units of Qζ 5 ) + is generated by and by the units ξ a for all a Z with a, 5) =. An easy comutation shows that ξ a+5 = ξ a and ξ a = ξ a. Furthermore ξ =. It follows that C + is generated by and ξ. We have Hence ξ = ζ )/ 5 ζ 5 ζ 5 = ζ 5 ζ 5 + ) = ζ 5 + ζ 5) =... = + 5. C + = {±} ξ Z = {±} + ) Z 5. d) By arts b) and c) we have R Qζ 5 ) = C +, therefore it follows from + Theorem.4 that h Qζ5 ) + = [R Qζ 5) + : C + ] =. 8) Let be a rime number and fx) = X Z [X]. We will use Hensel s lemma to show that the equation fx) = 0 has solutions in Z. Note that f X) = )X. For every a Z there exists an ã Z such that a ã mod ). It easily follows that Z /Z = Z/Z, so Z /Z is a field with elements. Now let α Z /Z be a non-zero element. Choose a Z Z such that a reduces to α in Z /Z. Since α 0 it follows that a and therefore by Euler s theorem a mod ). Hence a satisfies fa ) 0 mod ). Furthermore f a ) = )a 0 mod ) since 0 mod ) and a 0 mod ). Therefore by Hensel s lemma Theorem 4.) there exists a unique a Z such that fa) = 0 and a a mod ). The last condition imlies that a reduces to α in Z /Z because a reduces to α.

5 Algebraic number theory, Solutions to Problem Sheet, LTCC We have shown that for every non-zero α Z /Z there exists a unique solution a Z of fx) = 0 which reduces to α. As there are choices for α, we have therefore found solutions of the equation fx) = 0. 9) Existence of a: We will construct a sequence a 0, a, a,... in Z such that for all i 0 we have i) fa i ) 0 mod M++i ), ii) a i a i mod M+i ) if i. Clearly the given a 0 Z satisfies i) and ii). Now suose that we have constructed a 0, a,..., a i satisfying i) and ii). We want to find a i+ Z for which i) and ii) hold. In order for ii) to be satisfied we must have a i+ = a i + λ M++i for some λ Z. We will show that there exists λ such that fa i + λ M++i ) 0 mod M++i ). Note that fa i + λ M++i ) fa i ) + f a i )λ M++i mod M++i ). Hence we will have fa i + λ M++i ) 0 mod M++i ) if and only if fa i ) + f a i )λ M++i 0 mod M++i ). By assumtion fa i ) 0 mod M++i ). Since a i a 0 mod M+ ) we have f a i ) f a 0 ) mod M+ ), so in articular f a i ) 0 mod M ). It follows that the revious congruence is equivalent to fa i ) M++i + f a i ) λ 0 mod ). M Now f a i ) 0 mod M+ ), hence f a i ) 0 mod ). Therefore f a i ) is M M a unit in Z, so there exists a λ Z such that the revious congruence is satisfied. It follows that a i+ = a i + λ M++i satisfies condition i). By ii) the sequence a 0, a, a,... is a Cauchy sequence in Z. Hence we can define a = lim i a i Z. Then again by ii)) we have a a 0 mod M+ ). Furthermore fa) = flim i a i ) = lim i fa i ) = 0 where the last equality comes from i). This shows the existence of a Z with the required roerties. Uniqueness of a: Suose that a Z satisfies fa ) = 0 and a a 0 mod M+ ). We must show that a = a. First we mae the following observation. We showed above that for every i 0 there exists a i+ Z such that conditions i) and ii) are satisfied. It follows from the above that a i+ must be of the form a i+ = a i + λ M++i and that λ is unique modulo. Therefore a i+ is unique modulo M++i. Now we claim that for all i 0 we have a a i mod M++i ). For i = 0 this is true by assumtion. Suose that we have shown a a i mod M++i ) for some i N {0}. Since fa ) 0 mod M++i ), it follows that a satisfies conditions i) and ii) for i +, hence by the uniqueness result stated in the revious aragrah it follows that a a i+ mod M++i ). From a a i mod M++i ) for all i it follows that a = lim i a i = a, as required. 0) Let fx) = X )X 7)X 34). Claim : The equation fx) = 0 has solutions in R. Proof. Clearly the solutions of fx) = 0 in R are X = ±, ± 7, ± 34. Claim : The equation fx) = 0 has solutions in Q 7.

6 6 Algebraic number theory, Solutions to Problem Sheet, LTCC 008 Proof. Let gx) = X. Then a = 6 Z Z 7 satisfies ga ) = 34 0 mod 7) and g a ) = 0 mod 7). Therefore by Hensel s lemma Theorem 4.) there exists a Z 7 such that ga) = 0. This imlies that fa) = a )a 7)a 34) = 0, i.e. fx) = 0 has a solution in Z 7 Q 7. Claim 3: The equation fx) = 0 has solutions in Q for every rime number with and 7. ) ) ) 34 7 Proof. We have =, therefore at least one of the Legendre ) ) ) 7 34 symbols,, is equal to. Let c {, 7, 34} be such that ) = and let gx) = X c. Then by the definition of the Legendre c symbol there exists a Z Z such that ga ) = a c 0 mod ). Furthermore g a ) = a 0 mod ) because c 0 mod ) imlies a 0 mod ). Therefore by Hensel s lemma Theorem 4.) there exists a Z such that ga) = 0. This imlies that fa) = a )a 7)a 34) = 0, i.e. fx) = 0 has a solution in Z Q. Claim 4: The equation fx) = 0 has solutions in Q. Proof. Let gx) = X 7. Let a 0 = Z Z and M = N {0}. Then ga 0 ) = 6 0 mod M+ ), g a 0 ) = 0 mod M ) and g a 0 ) = 0 mod M+ ). Therefore by the generalisation of Hensel s lemma stated in Question 9) there exists a Z such that ga) = 0. This imlies that fa) = a )a 7)a 34) = 0, i.e. fx) = 0 has a solution in Z Q. Claim 5: The equation fx) = 0 has no solutions in Q. Proof. In the roof of Claim we listed all solutions of fx) = 0 in R. Clearly all of these solutions are irrational, therefore fx) = 0 has no solutions in Q.