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Problems For Section 2.2
1. Let L = Q(α), where α is a root of the irreducible quadratic X 2 + bX + c, with b, c ∈ Q. √ Show that L = Q( m) for some square-free integer m. Thus the analysis of this section covers all possible quadratic extensions of√ . Q 2. Show that the quadratic extensions Q( m), m square-free, are all distinct. 3. Continuing Problem 2, show that in fact no two distinct quadratic extensions of Q are Q-isomorphic. Cyclotomic fields do not exhibit the same behavior. Let ωn = ei2π/n , a primitive nth 2 root of unity. By a direct computation, we have ω2n = ωn and
n+ −ω2n 1 = −eiπ(n+1)/n = eiπ eiπ eiπ/n = ω2n ....
Problems For Section 3.1
This problem set will give the proof of a result to be used later. Let P1 , P2 , . . . , Ps , s ≥ 2, be ideals in a ring R, with P1 and P2 not necessarily prime, but P3 , . . . , Ps prime (if s ≥ 3). Let I be any ideal of R. The idea is that if we can avoid the Pj individually, in other words, for each j we can find an element in I but not in Pj , then we can avoid all the Pj simultaneously, that is, we can find a single element in I that is in none of the Pj . The usual statement is the contrapositive of this assertion....
Problems For Section 3.4
We will now go through the factorization of an ideal in a number field. In the next chapter, we will begin to develop the necessary background, but some of the manipulations are accessible to us now. By (2.3.11), the ring B of algebraic integers of the number field √ √ Q( −5) is Z[ −5]. (Note that −5 ≡ 3 mod 4.) If we wish to factor the ideal (2) = 2B of B , the idea is to factor x2 + 5 mod 2, and the result √ x2 + 5 ≡ (x + 1)2 mod 2. is √ Identifying x with −5, we form the ideal P2 = (2, 1 + −5), which turns out to be 2 prime. The desired factorization is (2) = P2 . This technique works if B = Z[α], where √ the number field L is Q( α). √ 2 1. Show that 1 − −5 ∈ P2 , and conclude that 6 ∈ P2 . 2 2 2. Show that 2 ∈ P2 , he√ e (2) ⊆ P2 √ nc . 2 2 3. Expand P2 = (2, 1 + −5)(2, 1 + −5), and conclude that P2 ⊆ (2). 4. Following the technique suggested in the above problems, factor x2 + 5 mod 3, and √ conjecture that the prime factorization of (3) in the ring of algebraic integers of Q( −5) is (3) = P3 P3 for appropriate P3 and P3 . 5. With P3 and P3 as found in Problem 4, verify that (3) = P3 P3 ....
Proof. Adjoin a root θi of hi to produce the field Fp [θi ] ∼ Fp [X ]/hi (X ). The assignment = θ → θi extends by linearity (and reduction of coefficients mod p) to an epimorphism λi : Z[θ] → Fp [θi ]. Since Fp [θi ] is a field, the kernel of λi is a maximal, hence prime, ideal of Z[θ] = B . Since λi maps fi (θ) to hi (θi ) = 0 and also maps p to 0, it follows that Pi ⊆ ker λi . We claim that Pi = ker λi . To prove this, assume g (θ) ∈ ker λi . With a...
Problems For Section 4.3
1. In the exercise√ for Section 3.4, we factored (2) and (3) in the ring B of algebraic s integers of L = Q( −5), using ad hoc techniques. Using the results of this section, derive the results rigorously. 2. Continuing √ roblem 1, factor (5), (7) and (11). P √ 3. Let L = Q( 3 2), and assume as known that the ring of algebraic integers is B = Z[ 3 2]. Find the prime factorization of (5)....
Remarks
Note that G, a finite cyclic group, has a generator, necessarily a primitive root of unity. Thus G will consist of all tth roots of unity for some t, and the field L will contain only finitely many roots of unity. This is a general observation, not restricted to the quadratic case...
Assume that [K L : Q] = mn. Let σ be an embedding of K in C and τ an embedding of L in C. Then there is an embedding of K L in C that restricts to σ on K and to τ on L. Proof. The embedding σ has [K L : K ] = n distinct extensions to embeddings of K L in C, and if two of them agree on L, then they agree on K L (because they coincide with σ on K ). This contradicts the fact that the extensions are distinct. Thus we have n embeddings of K L in C with distinct restrictions to L. But there are only n embeddings of L in C, so one of them must be τ , and the result follows. ♣...
/ We use the notation QK or the Frobenius automorphism. The behavior of the Frobenius under conjugation is similar to the behavior of the decomposition group as a whole (see the exercises in Section 8.1)....
Problems For Section 9.4
1. Show that a rational number a/b (in lowest terms) is a p-adic integer if and only if p does not divide b. 2. With p = 3, express the product of (2 + p + p2 ) and (2 + p2 ) as a p-adic integer. 3. Express the p-adic integer -1 as an infinite series. 4. Show that the sequence an = n! of p-adic integers converges to 0. 5. Does the sequence an = n of p-adic integers converge? ∞ 6. Show that the p-adic power series for log(1 + x), namely n=1 (−1)n+1 xn /n, converges in Qp for |x| < 1 and diverges elsewhere. This allows a definition of a p-adic logarithm: logp (x) = log[1 + (x − 1)]. In Problems 7-9, we consider the p-adic exponential function. 7. ∞ ecall from elementary number theory that the highest power of p dividing n! is R i i=1 n/p . (As an example, let n = 15 and p = 2. Calculate the numb er of multiples of 2, 4,and 8 in the integers 1-15.) Use this result to show that the p-adic valuation of n! is at most n/(p − 1). 8. Show that the p-adic valuation of (pm )! is (pm − 1)/(p − 1). ∞ 9. Show that the exponential series n=0 xn /n! converges for |x| < p−1/(p−1) and diverges elsewhere....
Section 2.1
1. A basis for E /Q is 1, θ, θ2 , and θ2 1 = θ2 , θ2 θ = θ3 = 3θ − 1, θ2 θ2 = θ4 = θθ3 = 3θ2 − θ. Thus 0 m(θ2 ) = 0 1 and we have T (θ2 ) = 6, N (θ2 ) = 1. Note that (the matrix of θ is 0 m(θ) = 1 0 −1 3 0 0 −1 3...
There√ re Q(ω2n ) ⊆ Q(ωn ). fo √ 5. Q( −3) = Q(ω ) where ω = − 1 + 1 −3 is a primitive cube root of unity. 2 2 6. If l(y ) = 0, then (x, y ) = 0 for all x. Since the bilinear form is nondegenerate, we must have y = 0. 7. Since V and V ∗ have the same dimension, the map y → l(y ) is surjective. 8. We have (xi , yj ) = l(yj )(xi ) = fj (xi ) = δij . Since the fj = l(yj ) form a basis, so do the yj . n 9. Write xi = k=1 aik yk , and take the inner product of both sides with xj to conclude that aij = (xi , yj )....
i 1. By the Chinese remainder theorem, B /(p) ∼ B /Piei . If p does not ramify, then = ei = 1 for all i, so B /(p) is a product of fields, hence has no nonzero nilpotents. On the other hand, suppose that e = ei > 1, with P = Pi . Choose x ∈ P e−1 \ P e and observe that (x + P e )e is a nonzero nilpotent in B /P e . 2. The minimal polynomial of a nilpotent element is a power of X , and the result follows from (2.1.5). n 3. let β = i=1 bi ωi with bi ∈ Z. Then, with T denoting trace, T (A(β ωj )) = T ( in
=1...
7 7. This follows from the Minkowski bound (5.3.5) if we observe that N (I ) ≥ 1 and 2r2 ≤ n. 8. By a direct computation, we get a2 and an+1 1 π (n + 1)2n+2 π 1 = = (1 + )2n . an 4 n2n (n + 1)2 4 n By the binomial theorem, an+1 /an = (π /4)(1 + 2 + positive terms) ≥ 3π /4. Thus |d| ≥ a2 a3 an π2 ··· ≥ (3π /4)n−2 , a2 an−1 4...
Section 7.1
1. The missing terms in the product defining the discriminant are either squares of real numbers or occur as a complex number and its conjugate. Thus the missing terms contribute a positive real number, which cannot change the overall sign. 2. Observe that (c − c)2 is a negative real number, so each pair of complex embeddings contributes a negative sign. 3. We have 2r2 = [Q(ζ ) : Q] = ϕ(pr ) = pr−1 (p − 1), so the sign is (−1)s , where, assuming pr > 2, s = pr−1 (p − 1)/2. To show that there are no real embeddings, note that if ζ is mapped to -1, then −ζ is mapped to 1. But 1 is also mapped to 1, and (assuming a nontrivial extension), we reach a contradiction. Examination of the formula for s allows further simplification. If p is odd, the sign will be positive if and only if p ≡ 1 mod 4. If p = 2, the sign will be positive iff r > 2....
Index
m-n means chapter m, page n absolute value, 9-1 on the rationals, 9-4 AK LB setup, 2-5 algebraic integer, 1-2 approximation theorem, 9-5, 9-6 archimedean absolute value, 9-1 Artin symbol, 8-5 Artin-Whaples, see approximation theorem Cauchy sequence, 9-7 characteristic polynomial, 2-1 class number, 5-7 coherent sequence, 9-8 completion of a field with an absolute value, 9-7 conjugates of an element, 2-3 of a prime ideal, 8-2 contraction of an ideal, 4-1 cyclotomic extension, 2-5, 2-7, 6-5, 7-1, 8-6 polynomial, 7-1 decomposition field, 8-6 group, 8-2 Dedekind domain, 3-1 Dedekind’s lemma, 2-4 denominator of a fractional ideal, 3-3 Dirichlet unit theorem, 6-1, 6-3, 6-4 discrete valuation, 9-1 discrete valuation ring, 4-3, 9-2, 9-3 discriminant, 2-8, 7-3, 7-4 divides means contains, 3-6 DVR, see discrete valuation ring embedding, canonical, 5-4 complex, 5-4 logarithmic, 6-1 real, 5-4 equation of integral dependence, 1-2 1...
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