Algebraic Number Theory HW 2 Solutions PDF

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Algebraic Number Theory homework 2 Solutions set....

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Algebraic Number Theory (Homework 2) Due March 11th SOLUTIONS Part A 1. Find O√ K where K is: √ (a) Q( 15); (b) Q( 56); (c) Q(α) where α solves 2x2 − 3x + 2 = 0. Solution: √ (a) OK = Z[ √15] since 15 6≡ 1 (mod 4). (b) OK = Z[ 14] since 56 = 4 · 14 and 14 6≡ 1 (mod 4). √ 1+ −7 (c) OK = Z[ 2 ], since (−3)2 − 4 · 2 · 2 = −7 ≡ 1 (mod 4). √ √ √ √ 2. Let L = Q( 4 2), K = Q( 2), α = 4 2 and β = α2 = 2. Calculate the following: (a) TrL/Q (α); NL/Q (α); (b) TrL/Q (β); NL/Q (β); (c) TrL/K (α); NL/K (α); (d) TrL/K (β); NL/K (β); (e) TrL/Q (α + β); NL/Q (α + β); Solution: Before we do the calculations, we note that L has 4 Q-embeddings given by σj (α) = (−i)j α where i2 = −1. Similarly, we note that L has 2 K-embeddings, given by σj (α) = (−1)j α. (a) TrL/Q (α) = 0, NL/Q (α) = −2, since mα,Q (x) = x4 − 2. (b) TrL/Q (β) = β + (−i)2 β + (−1)2 β + (i)2 β = 0. NL/Q (β) = (NL/Q (α))2 = 4. √ √ (c) TrL/K (α) = 0, NL/K (α) = −√ 2, since mα,K (x) = x2 − 2. (d) TrL/Q (β) = β + (−1)2 β = 2 2, NL/K (β) = (NL/K (α))2 = 2 (e) TrL/Q (α + β) = TrL/Q (α) + TrL/Q (β) = 0 NL/Q (α + β) = (α + β)(−iα − β)(−α + β)(iα − β) = (−α2 + β 2 )(−(iα)2 + β 2 ) = (−α4 + β 4 ) = (−2 + 4) = 2

3. (Thinking about Thm 3.2.2) Let L/K be an extension of number field. (a) Let α ∈ L. Suppose NL/K (α) ∈ OK , can we conclude that α ∈ OL ? Give a proof or counter-example. (b) Let α ∈ L. Suppose TrL/K (α) ∈ OK , can we conclude that α ∈ OL ? Give a proof or counter-example. (c) Let α ∈ L. Suppose NL/K (α) ∈ OK and TrL/K (α) ∈ OK , can we conclude that α ∈ OL ? Give a proof or counter-example. (d) Let α ∈ L. Suppose NL/Q (α) = ±1, can we conclude α ∈ OL×? Give a proof or counter-example. (e) Let α, β ∈ OL . Suppose NL/Q (α)|NL/Q (β) in OK . Can we conclude α|β in OL ? Solution: √ (a) This is not true, for example, take L = Q( L = √ −2), K = Q, then we know O √ / OL . Then NL/K (α) = ( 31)2 + Z[ −2], OK = Z. Consider α = 31 (1 + 2 −2) ∈ 2( 23 )2 = 1 ∈ Z. √ (b) This = Q( −2), K = Q, then we know OL = √ is not true, for example, take L1 √ Z[ −2], OK = Z. Consider α = 1 + 3 −2 ∈ / OL . Then TrL/K (α) = 2 ∈ Z. (c) This is still not true, but we now need to go to a cubic field to find a counter example. Let α be a root of x3 − 12 x + 1, note that this polynomial is irreducible 1

over Q[x] (since 2x3 − 1 + 2 has no roots in Z). Let L = Q(α), and since α ∈ / O, α ∈ / OL (since mα,Q (x) ∈ / Z[x]). On the other hand, NL/Q (α) = −1, while TrL/Q (α) = 0 (using the coefficients of mα,Q (x)). (d) Note that the counterexample to the first part √ works for this question as well. (e) This is not true, for example take L√ = Q( −5), K √= Q, then we know OL = √ Z[ −2], OK = Z. Consider α = 1+ 2 −5 and β = 1− 2 −5 . Then α ∤ β and β ∤ α but NL/K α = NL/K β = 6. √ 4. Let K = Q( d), where d < 0 and d ≡ 1 (mod 4). We show a different approach to √ 1+ dZ. finding the units of O√K , by using the integral basis OK = Z ⊕ 2 (a) Let α = x + y 1+2 d ∈ K, with x, y ∈ Q. Prove that NK/Q (α) = x2 + xy + y 2 d−1 . 4 √ 1− d Solution: We first note that the conjugate of α = x + y 2 , hence √ ! √ ! 1− d 1+ d x+y NK/Q (α) = x + y 2 2 √ √ √ √ 1+ d+1− d 1+ d− d−d 2 2 =x + y xy + 4 2 1−d 2 = x2 + xy + y 4 ≥ 0 for all x, y ∈ Q. (b) By completing the square, show that x2 + xy + y 2 1−d 4 2 y2 2 2 1−d 2 = (x + y/2)2 − dy4 . Solution: We have x + xy + y 4 = (x + y/2) − 4 + y 2 1−d 4 Since d < 0, we have the sum of two squares, which is always positive. = 1. (c) Find all units of OK , i.e., find all x, y ∈ Z such that x2 + xy + y 2 1−d 4 dy2 2 Solution: We solve the equation (x + y/2) − 4 = 1. If d < −3, then as d ≡ 1 2 (mod 4), we have d ≤ −7 and − dy4 > 1 for all y ≥ 1. So y − 0 and we must have x = ±1. 2 If d = −3, we analyse (x + y/2)2 + 3y4 = 1. Again, we see |y| ≤ 1. When y = 1, we have (x + 21 )2 = 41 , i.e., x + 12 = ± 21 , giving either x = 0 or x = −1. When y = −1, we need to solve x − 21 = ± 12 , i.e., either x = 0 or x = 1. When y = 0, we have x = ±1. √ √ √ 1+ 3 −1+ 3 1+ 3 , 0− = , −1+ This gives us the 6 solutions we expected, {±1, 0+ 2 2 2 √ √ √ 1+ 3 1+ 3 −1− 3 , 1 − 2 = 2 }. 2 5. Prove that a quadratic number field is uniquely determined by its discriminant. Write down all the quadratic number fields with discriminant between −10 and 10. Which integers between −10 and 10 is not the discriminant of a quadratic number field? Solution: We know that √ for any quadratic field, K, there exists a unique square-free d 6= 1 such that K = Q( d). We also know that disc(K) = d if d ≡ 1 (mod 4) and disc(K) p = 4d if d ≡ 2, 3 (mod 4). Hence, if disc(K) √ = D, if D ≡ 0 (mod 4) then K = Q( D/4) and if D ≡ 1 (mod 4) then K = Q( D). We start by listing all the integers between −10 and √ 10 which √ are 0, 1 mod √ 4. These are √ {−8, −7, −4, −3, 0, 1, 4, 5, 8, 9}. We see that Q( −2), Q( −7), Q( −1) and Q( −3) will give the first four discriminant. Now 0/4 = 0 is not square free, so there are no quadratic fields of discriminant 0. Similarly, we don’t have √ a quadratic √ field of discriminant 1 or 4 (since 4/4 = 1 and d 6= 1.) We have Q( 5) and Q( 2) gives the next two discriminant, and there are no quadratic fields of discriminant 9. So {−10, −9, −6, −5, −2, −1, 0, 1, 2, 3, 4, 6, 7, 9, 10} are all integers which are not the discriminant of quadratic number fields. Part B 1. Let K = Q[α], where α is a root of the equation x3 − 6x2 + 11x − 7. 2

(a) (3 marks) Is it true that {1, α, α2 } form an integral basis for Z[α]? Justify. Solution: We first show that the polynomial x3 − 6x2 + 11x − 7 is irreducible modulo 2. We have x3 − 6x2 + 11x − 7 ≡ x2 + x + 1 = p(x) (mod 2), and we can check that p(0) = p(1) = 1. Hence, this cubic doesn’t have any linear factor and must be irreducible. Hence x3 − 6x2 + 11x − 7 is the minimal polynomial of α over Q. Therefore, α is of degree 3 over the rationals and we see that α ∈ O. Hence by Proposition 3.1.2 (part 2), we have that {1, α, α2 } is an integral basis for Z[α]. (b) (3 marks) Define θ = α − 2. Prove that Q(α) = Q(θ) and Z[α] = Z[θ]. Is it true that {1, θ, θ 2 } form an integral basis for Z[θ]? Justify. Solution: The assertions Q(α) = Q(θ) and Z[α] = Z[θ] are immediate as α = θ + 2. The minimal polynomial of θ can be found using the minimal polynomial of α. It is mθ,Q (x) = mα,Q (x + 2) = (x + 2)3 − 6(x + 2)2 + 11(x + 2) − 7

= x3 + 6x2 + 12x + 8 − 6x2 − 24x − 24 + 11x + 22 − 7 = x3 − x − 1.

Since θ ∈ O, we conclude that {1, θ, θ 2 } form an integral basis for Z[θ]. (c) (2 marks) Prove OK = Z[θ]. Solution: We start by calculating the discriminant of Z[θ] using Lemma 3.4.2. We have disc(Z[θ]) = −(4 · (−1)3 + 27 · (−1)2 ) = −23. As −23 is square-free, we conclude that OK = Z[θ]. (d) (4 marks) Prove that Z[ 2α] is not finitely generated. Solution: We have OK = Z[α] by the previous part. Since 2α = 0·1+ 12 · α + 0 · α2 , we conclude α2 ∈ / Z[α]. Hence α2 ∈ / O, and so by Proposition 3.1.2, it’s minimal polynomial is not in Z[x] and by Lemma 2.3.5, Z[ α2 ] is not finitely generated. 2. (3 Marks) Prove Lemma 3.2.5: Let K be a number field of degree n. Suppose that α1 , . . P . , αn is a basis for K/Q and C = (cij ) a n × n matrix with entries in Q. Let wi = j cij αj . Then disc(w1 , . . . , wn ) = det(C)2 disc(α1 , . . . , αn ). Solution: Let A = (σi (αj ))i,j and B = (σi (wj ))i,j . For ease of notation, let wj = P k cjk αk . Note that X X σi (wj ) = σi (cjk αk ) = cjk σi (αk ). k

k

Hence B = CAT , using the fact that AT = (σj (αi ))i,j . Therefore, since det(AT ) = det(A), we have disc(w1 , . . . , wn ) = det(B 2 ) = det(C 2 ) det(A2 ) = det(C)2 disc(α1 , . . . , αn )

3. (5 Marks) Prove Lemma 3.2.6: Let α be an algebraic number of degree n and K = Q(α). Let α = α1 , α2 , . . . , αn be the roots of mα,Q (x) and let m′ (x) be the derivative of mα,Q (x). Then Y (αr − αs )2 = ±NK/Q (m′ (α). disc(1, α, . . . , αn−1 ) = 1≤r...