Algebraic Number Theory part: 12: Solutions to Problems

If you face any problem to loading this page, 
 click here for PDF file of this chapter.


Chapter 1
Section 1.1
1. Multiply the equation by an−1 to get
a
−1 = −(cn−1 + · · · + c1an−2 + c0an−1) ∈ A.
2. Since A[b] is a subring of B, it is an integral domain. Thus if bz = 0 and b = 0, then
z = 0.
3. Any linear transformation on a finite-dimensional vector space is injective iff it is
surjective. Thus if b ∈ B and b = 0, there is an element c ∈ A[b] ⊆ B such that bc = 1.
Therefore B is a field.
4. Since P is the preimage of Q under the inclusion map of A into B, P is a prime ideal.
The map a + P → a + Q is a well-defined injection of A/P into B/Q, since P = Q ∩ A.
Thus A/P can be viewedas a subring of B/Q.
5. If b + Q ∈ B/Q, then b satisfies an equation of the form
xn + an−1xn−1 + · · · + a1x + a0 = 0, ai ∈ A.
By Problem 4, b+Q satisfies the same equation with ai replacedb y ai +P for all i. Thus
B/Q is integral over A/P.
6. By Problems 1-3, A/P is a fieldif andonly if B/Q is a field, and the result follows.
(Note that B/Q is an integral domain (because Q is a prime ideal), as required in the
hypothesis of the result just quoted.)

Section 1.2
1. If x /∈M, then by maximality ofM, the ideal generated byMand x is R. Thus there
exists y ∈ M and z ∈ R such that y + zx = 1. By hypothesis, zx, hence x, is a unit.
Take the contrapositive to conclude that M contains all nonunits, so R is a local ring by
(1.2.8).
2. Any additive subgroup of the cyclic additive group of Z/pnZ must consist of multiples
of some power of p, andit follows that every proper ideal is containedin (p), which must
therefore be the unique maximal ideal.
3. The set of nonunits is M = {f/g : g(a) = 0, f(a) = 0}, which is an ideal. By (1.2.8),
R is a local ring with maximal ideal M.
4. S−1(g ◦ f) takes m/s to g(f(m))/s, as d oes (S−1g) ◦ (S−1f). If f is the identity on
M, then S−1f is the identity on S−1M.
5. By hypothesis, g◦f = 0, so (S−1g)◦(S−1f) = S−1(g◦f) = S−10 = 0. Thus im S−1f ⊆
ker S−1g. Conversely, let y ∈ N, s ∈ S, with y/s ∈ ker S−1g. Then g(y)/s = 0/1, so for
some t ∈ S we have tg(y) = g(ty) = 0. Therefore ty ∈ ker g = imf, so ty = f(x) for some
x ∈ M. We now have y/s = ty/st = f(x)/st = (S−1f)(x/st) ∈ im S−1f.
6. The sequence 0 → N → M → M/N → 0 is exact, so by Problem 5, the sequence
0 → NS → MS → (M/N)S → 0 is exact. (If f is one of the maps of the first sequence,
the corresponding map in the second sequence is S−1f.) It follows from the definition of
localization of a module that NS ≤ MS, andb y exactness of the secondsequence we have
(M/N)S
∼=
MS/NS.

Section 2.1
1. A basis for E/Q is 1, θ, θ2, and
θ21 = θ2, θ2θ = θ3 = 3θ − 1, θ2θ2 = θ4 = θθ3 = 3θ2 − θ.
Thus
m(θ2) =


0 −1 0
0 3 −1
1 0 3


andw e have T(θ2) = 6, N(θ2) = 1. Note that if we hadalread y computedthe norm of θ
(the matrix of θ is
m(θ) =


0 0 −1
1 0 3
0 1 0


and T(θ) = 0, N(θ) = −1), it wouldb e easier to calculate N(θ2) as [N(θ)]2 = (−1)2 = 1.
2. The cyclotomic polynomial Ψ6 has only two roots, ω andits complex conjugate ω. By
(2.1.5),
T(ω) = ω + ω = eiπ/3 + e
−iπ/3 = 2 cos π/3 = 1.
3. We have min(θ,Q) = X4 − 2, min(θ2,Q) = X2 − 2, min(θ3,Q) = X4 − 8, and
min(
√
3θ,Q) = X4 − 18. (To compute the last two minimal polynomials, note that
(θ3)4 = (θ4)3 = 23 = 8 and(
√
3θ)4 = 18.) Therefore all four traces are 0.
4. Suppose that
√
3 = a + bθ + cθ2 + dθ3. Take the trace of both sides to conclude
that a = 0. (The trace of
√
√ 3 is 0 because its minimal polynomial is X2 − 3.) Thus
3 = bθ + cθ2 + dθ3, so
√
3θ = bθ2 + cθ3 + 2d. Again take the trace of both sides to get
d = 0. We now have
√
3 = bθ + cθ2, so
√
√ 3θ2 = bθ3 + 2c. The minimal polynomial of
3θ2 is X2 − 6, because(
√
3θ2)2 = 6. Once again taking the trace of both sides, we get
c = 0. Finally,
√
3 = bθ implies 9 = 2b4, andw e reach a contradiction.

Section 2.2
1. By the quadratic formula, L = Q(
√
b2 − 4c). Since b2 − 4c ∈ Q, we may write
b2 − 4c = s/t = st/t2 for relatively prime integers s and t. We also have s = uy2 and
t = vz2, with u and v relatively prime andsquare-free. Thus L = Q(
√
uv) = Q(
√
m).
2. If Q(
√
d) = Q(
√
e), then
√
d = a + b
√
e for rational numbers a and b. Squaring both
sides, we have d = a2 + b2e+2ab
√
e, so
√
e is rational, a contradiction (unless a = 0 and
b = 1).
3. Any isomorphism of Q(
√
d) and Q(
√
e) must carry
√
d into a+b
√
e for rational numbers
a and b. Thus d is mappedto a2 + b2 + 2ab
√
e. But a Q-isomorphism maps d to d, and
we reach a contradiction as in Problem 2.
4. Since ωn = ω2
2n, we have ωn ∈ Q(ω2n), so Q(ωn) ⊆ Q(ω2n). If n is odd, then n+1 = 2r,
so
ω2n = −ω2r
2n = −(ω2
2n)r = −ωrn
.
Therefore Q(ω2n) ⊆ Q(ωn).
5. Q(
√
−3) = Q(ω) where ω = −1
2 + 1
2
√
−3 is a primitive cube root of unity.
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 .
9. Write xi =

n
k=1 aikyk, and take the inner product of both sides with xj to conclude
that aij = (xi, xj ).

Section 2.3
1. The first statement follows because multiplication of each element of a group G by a
particular element g ∈ G permutes the elements of G. We can work in a Galois extension
of Q containing L, andeac h automorphism in the Galois group restricts to one of the σi
on L. Thus P + N and PN belong to the fixedfieldof the Galois group, which is Q.
2. Since the xj are algebraic integers, so are the σi(xj ), as in the proof of (2.2.2). Thus P
and N, hence P + N and PN, are algebraic integers. By (2.2.4), P + N and PN belong
to Z.
3. D = (P − N)2 = (P + N)2 − 4PN ≡ (P + N)2 mod4. But any square is congruent
to 0 or 1 mod4, andthe result follows.
4. We have yi =

n
j=1 aijxj with aij ∈ Z. By (2.3.2), D(y) = (d etA)2D(x). Since D(y)
is square-free, detA = ±1, so A has an inverse with coefficients in Z. Thus x = A−1y, as
claimed.
5. Every algebraic integer can be expressedas a Z-linear combination of the xi, hence of
the yi by Problem 4. Since the yi form a basis for L over Q, they are linearly independent
andthe result follows.
6. No. For example, take L = Q(
√
m), where m is a square-free integer with m ≡ 1
mod4. By (2.3.11), the fieldd iscriminant is 4m, which is not square-free.

Section 3.1
1. We may assume that I is not containedin the union of any collection of s − 1 of the
Pi’s. (If so, we can simply replace s by s−1.) It follows that elements of the desired form
exist.
2. Assume that I ⊆ P1 and I ⊆ P2. We have a1 ∈ P1, a2 /∈
P1, so a1+a2 /∈
P1. Similarly,
a1 /∈
P2, a2 ∈ P2, so a1 + a2 /∈
P2. Thus a1 + a2 /∈
I ⊆ P1 ∪ P2, contradicting a1, a2 ∈ I.
3. For all i = 1, . . . , s − 1 we have ai /∈
Ps, hence a1 · · · as−1 /∈
Ps because Ps is prime.
But as ∈ Ps, so a cannot be in Ps. Thus a ∈ I and a /∈ P1 ∪ ·· ·∪Ps.

Section 3.2
1. The product of ideals is always contained in the intersection. If I and J are relatively
prime, then 1 = x + y with x ∈ I and y ∈ J. If z ∈ I ∩ J, then z = z1 = zx + zy ∈ IJ.
The general result follows by induction, along with the computation
R = (I1 + I3)(I2 + I3) ⊆ I1I2 + I3.
Thus I1I2 and I3 are relatively prime. Continue in this manner with
R = (I1I2 + I4)(I3 + I4) ⊆ I1I2I3 + I4
andso on.
2. We have R = Rr = (P1 + P2)r ⊆ Pr
1 + P2. Thus Pr
1 and P2 are relatively prime for all
r ≥ 1. Assuming inductively that Pr
1 and Ps
2 are relatively prime, it follows that
Ps
2 = Ps
2R = Ps
2 (Pr
1 + P2) ⊆ Pr
1 + Ps+1
2
so
R = Pr
1 + Ps
2
⊆ Pr
1 + (Pr
1 + Ps+1
2 ) = Pr
1 + Ps+1
2
completing the induction.
3. Let r be a nonzero element of R such that rK ⊆ R, hence K ⊆ r−1R ⊆ K. Thus
K = r−1R. Since r−2 ∈ K we have r−2 = r−1s for some s ∈ R. But then r−1 = s ∈ R,
so K ⊆ R andconsequen tly K = R.
Section 3.3
1. By (2.1.10), the norms are 6,6,4 and9. Now if x = a + b
√
−5 and x = yz, then
N(x) = a2 + 5b2 = N(y)N(z). The only algebraic integers of norm 1 are ±1, andthere
are no algebraic integers of norm 2 or 3. Thus there cannot be a nontrivial factorization
of 1 ±
√
−5, 2 or 3.
2. If (a+b
√
−5)(c+d
√
−5) = 1, take norms to get (a2 +5b2)(c2 +5d2) = 1, so b = d = 0,
a = ±1, c = ±1.
3. By Problem 2, if two factors are associates, then the quotient of the factors is ±1,
which is impossible.
4. This is done as in Problems 1-3, using 18 = (1 +
√
−17)(1 −
√
−17) = 2 × 32.
5. By (2.2.6) or (2.3.11), the algebraic integers are of the form a + b
√
−3, a, b ∈ Z, or
(u/2) + (v/2)
√
−3 with u and v odd integers. If we require that the norm be 1, we only
get ±1 in the first case. But in the secondcase, we have u2 +3v2 = 4, so u = ±1, v = ±1.
Thus if ω = ei2π/3, then the algebraic integers of norm 1 are ±1, ±ω, and ±ω2.

Section 3.4
1. 1 −
√
−5 = 2 − (1 +
√
−5) ∈ P2, so (1+
√
−5)(1 −
√
−5) = 6 ∈ P2
2 .
2. Since 2 ∈ P2, it follows that 4 ∈ P2
2 , so by Problem 1, 2 = 6 − 4 ∈ P2
2 .
3. (2,1+
√
−5)(2,1+
√
−5) = (4, 2(1+
√
−5), (1+
√
−5)2), and(1 +
√
−5)2 = −4+2
√
−5.
Therefore each of the generators of the ideal P2
2 is divisible by 2, hence belongs to (2).
Thus P2
2
⊆ (2).
4. x2+5 ≡ (x+1)(x−1) mod3, which suggests that (3) = P3P
3, where P3 = (3,1+
√
−5)
and P
3 = (3, 1 −
√
−5).
5. P3P
3 = (3, 3(1+
√
−5), 3(1−
√
−5), 6) ⊆ (3), because each generator of P3P
3 is divisible
by 3. But 3 ∈ P3 ∩P
3, hence 9 ∈ P3P
3, andtherefore 9−6 = 3 ∈ P3P
3. Thus (3) ⊆ P3P
3,
andthe result follows.
Section 4.1
1. The kernel is {a ∈ A : a/1∈MS−1A} = A ∩ (MS−1A) =M by (1.2.6).
2. By hypothesis, M∩S = ∅, so s /∈M. By maximality of M we have M+ As = A, so
y + bs = 1 for some y ∈M, b ∈ A. Thus bs ≡1 mod M.
3. Since 1−bs∈M, (a/s)−ab = (a/s)(1−bs)∈MS−1A. Therefore (a/s)+MS−1A =
ab +MS−1A = h(ab).
Section 4.2
1. By the Chinese remainder theorem, B/(p) ∼=

i B/Pei
i . 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 ∈ Pe−1 \ Pe andobserv e
that (x + Pe)e = xe + Pe is zero in B/Pe.
2. The minimal polynomial of a nilpotent element is a power of X, andthe result follows
from (2.1.5).
3. Let β =

n
i=1 biωi with bi ∈ Z. Then, with T denoting trace,
T(A(βωj)) = T(
n
i=1
biA(ωiωj)) =
n
i=1
biT(ωiωj) ≡0 modp.
If β /∈ (p), then not all the bi can be 0 mod p, so the determinant of the matrix (T(ωiωj )),
which is the discriminant d by (2.3.1), is 0 mod p. Therefore, p divides d.
4. This follows from the Chinese remainder theorem, as in Problem 1. The fields Fi all
have characteristic p because p annihilates B/(p).
5. The Ti are nondegenerate by separability, and

i Ti is nondegenerate by orthogonality,
that is, πi(x)πj(y) = 0 for i = j.
6. Since Fi/Fp is a finite extension of a finite field, it is a Galois extension, so all embeddings
are actually automorphisms. Thus for any z ∈ Fi, the endomorphism given by
multiplication by z has trace TFi/Fp (z) = Ti(z). Since B/(p) is, in particular, a direct
sum of the Fi, the result follows.

Section 4.3
1. Factoring (2) is coveredb y case (c1) of (4.3.2), andw e have (2) = (2,1 +
√
−5)2.
Factoring (3) is coveredb y case (a1), and x2 + 5 ≡ (x + 1)(x − 1) mod3. Therefore
(3) = (3,1 +
√
−5) (3, 1 −
√
−5).
2. We have (5) = (5,
√
−5)2, as in case (b). To factor (7), note that x2 + 5 factors mod7
as (x + 3) (x − 3), so (7) = (7,3 +
√
−5) (7, 3 −
√
−5), as in case (a1). Since -5 is not a
quadratic residue mod 11, we are in case (a2) and 11 remains prime.
3. Mod5 we have x3−2 ≡ x3−27 = x3−33 = (x−3)(x2+3x+9) = (x+2)(x2+3x−1).
Thus
(5) = (5, α + 2)(5, α2 + 3α − 1)
where α = 3
√
2.
Section 5.3
1. We have r2 = 1 and n = 2, so the boundis (4/π)(2/4)
   
|d| = (2/π)
   
|d|. The
discriminant may be calculated from (2.3.11). We have d = 4m for m = −1,−2, and
d = m for m = −3,−7. The largest |d| is 8, andthe corresponding boundis 4
√
2/π,
which is about 1.80. Thus all the class numbers are 1.
2. We have r2 = 0 and n = 2, so the boundis
   
|d|/2. We have d = 4m for m = 2, 3, and
d = m for m = 5, 13. The largest |d| is 13, andthe corresponding boundis
√
13/2, which
is about 1.803. Thus all the class numbers are 1.
3. The discriminant is -20 and the Minkowski bound is 2
√
20/π, which is about 2.85.
Since 2 ramifies [see (4.3.2), case (c1)], there is only one ideal of norm 2. Thus the class
number is at most 2. But we know that Q(
√
−5) is not a UFD, by the exercises for Section
3.3. Therefore the class number is 2.
4. The discriminant is 24 and the bound is
√
24/2 =
√
6, which is about 2.45. Since 2
ramifies [see (4.3.2), case (b)], the argument proceeds as in Problem 3. Note that Q(
√
6) is
not a UFD because −2 = (2+
√
6)(2−
√
6). Note also that 2+
√
6 and2 −
√
6 are associates,
because (2 +
√
6)/(2 −
√
6) = −5 − 2
√
6, which is a unit [(−5 − 2
√
6)(−5 + 2
√
6) = 1].
5. The discriminant is 17 and the bound is
√
17/2, which is about 2.06. Since 2 splits
[(4.3.2), case (c2)], there are 2 ideals of norm 2. In fact these ideals are principal, as can
be seen from the factorization −2 = [(3 +
√
17)/2] [(3 −
√
17)/2]. Thus every ideal class
contains a principal ideal, so the ideal class group is trivial.
6. The discriminant is 56 and the bound is
√
56/2 =
√
14, which is about 3.74. Since 3
remains prime [(4.3.2), case (a2)], there are no ideals of norm 3. (The norm of the principal
ideal (3) is 32 = 9.) Since 2 ramifies [(4.3.2), case (b)], there is only one ideal of norm 2.
This ideal is principal, as can be seen from the factorization 2 = (4 +
√
14)(4 −
√
14). As
in Problem 5, the class number is 1.

7
7. This follows from the Minkowski bound(5.3.5) if we note that N(I) ≥ 1 and2 r2 ≤ n.
8. By a direct computation, we get a2 and
an+1
an
= π
4
(n + 1)2n+2
n2n
1
(n + 1)2 = π
4
(1 +
1
n
)2n.
By the binomial theorem, an+1/an = (π/4)(1 + 2 + positive terms) ≥ 3π/4. Thus
|d| ≥ a2
a3
a2
· · · an
an−1
≥ π2
4
(3π/4)n−2,
andw e can verify by canceling common factors that (π2/4)(3π/4)n−2 ≥ (π/3)(3π/4)n−1.
9. By Problem 8,
log |d| ≥ log π
3
+ (n − 1) log
3π
4
= log π
3
− log
3π
4
+ n log
3π
4
andthe result follows.
10. This follows from the boundgiv en in Problem 8.

Section 6.1
1. Since x, hence jx, as well as ei, hence jbi ei, all belong to H, so d oes xj . We have
xj ∈ T because jbi − jbi ∈ [0, 1).
2. We have x = x1 +

r
i=1
bi ei with x1 ∈ H ∩ T andthe ei ∈ H ∩ T. Since H ∩ T is a
finite set, there are only finitely many choices for x1. Since there are only finitely many
ei, H is finitely generated.
3. There are only finitely many distinct xj andinfinitely many integers, so xj = xk for
some j = k. By linear independence of the ei, we have (j − k)bi = jbi − kbi for all i,
andthe result follows.
4. By the previous problems, H is generatedb y a finite number of elements that are linear
combinations of the ei with rational coefficients. If d is a common denominator of these
coefficients, then d = 0 and dH ⊆

r
i=1 Zei. Thus dH is a subgroup of a free abelian
group of rank r, hence is free of rank at most r.
5. Since dH
∼=
H, H is free, andsince H ⊇

r
i=1 Zei, the rank of H is at least r, and
hence exactly r.
Section 6.3
1. m = 2 ⇒ 2 × 12 = 12 + 1, so the fundamental unit u is 1 +
√
2 andw e stop at step
t = 1.
m = 3 ⇒ 3 × 12 = 22 − 1, so u = 2+
√
3 and t = 1.
m = 5 ≡1 mod 4⇒ 5 × 12 = 12 + 4, so u = 1
2 (1 +
√
5) and t = 1.
m = 6 ⇒ 6 × 22 = 52 − 1, so u = 5 + 2
√
6 and t = 2.
m = 7 ⇒ 7 × 32 = 82 − 1, so u = 8 + 3
√
7 and t = 3.
m = 10 ⇒ 10 × 12 = 32 + 1, so u = 3+
√
10 and t = 1.
m = 11 ⇒ 11 × 32 = 102 − 1, so u = 10 + 3
√
11 and t = 3.
m = 13 ≡1 mod 4⇒ 13 × 12 = 32 + 4, so u = 1
2 (3 +
√
13) and t = 1.
u = 1
2 (a + b
√
m), we compute
8u3 = a(a2 + 3b2m) + b(3a2 + b2m)
√
m.
Now a2 − b2m = ±4, andif we add4 b2m to both sides, we get
a2 +3b2m = 4b2m±4 = 4(b2m±1). Since m ≡1 mod 4, m must be odd, and since b is
also odd, b2m ± 1 is even, so 4(b2m ± 1) is divisible by 8. Similarly,
3a2 + b2m = 4a2 − (a2 − b2m) = 4a2 ± 4, which is also divisible by 8 because a is odd. It
follows that u3 ∈ B0.
4. If u2 ∈ B0, then u2 is a positive unit in B0, hence so is (u2)−1 = u−2. Therefore
u = u3u−2 ∈ B0. But a and b are odd, so u /∈ Z[
√
m], a contradiction.
5. When m = 5, we have u = 1
2 (1 +
√
5), so 8u3 = 1+3
√
5 + (3 × 5) + 5
√
5. Thus
u3 = 2+
√
5. Also, 4u2 = 6 + 2
√
5, so u2 = (3+
√
5)/2. When m = 13, we have
u = 1
2 (3+
√
13), so 8u3 = 27+27
√
13+(3×3×13)+13
√
13. Therefore u3 = 18+5
√
13.
Also, 4u2 = 22 + 6
√
13 = (11 + 3
√
13)/2.
Note that the results for u3 in Problem 5 are exactly what we wouldget by solving
a2 − mb2 = ±1. For m = 5 we have 5 × 12 = 22 +1, so a = 2, b = 1. For m = 13 we have
13 × 52 = 182 + 1, so a = 18, b = 5.

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 andits 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 mappedto -1, then −ζ is mappedto 1. But 1 is also mappedto 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 andonly if p ≡ 1 mod4. If p = 2, the sign will be positive iff r > 2.

Section 8.1
1. If τ ∈ I(Q) and x ∈ B, then
στσ
−1(x) − x = σ(τσ
−1(x) − σ
−1(x)) ∈ σ(Q)
so σI(Q)σ−1 ⊆ I(σ(Q)). Conversely, let τ ∈ I(σ(Q)), x ∈ B. Then τ = σ(σ−1τσ)σ−1,
so we must show that σ−1τσ ∈ I(Q), in other words, σ−1τσ(x) − x ∈ Q. Now we have
τσ(x)−σ(x) ∈ σ(Q), so τσ(x)−σ(x) = σ(y) for some y ∈ Q. Thus σ−1τσ(x)−x = y ∈ Q,
the desired result.
2. Since G is abelian, σD(Q)σ−1 = σσ−1D(Q) = D(Q), so by Problem 1 and(8.1.2),
all the decomposition groups are the same. The decomposition groups depend only on P
because P determines the unique factorization of PB into prime ideals of B. The analysis
is the same for the inertia groups.

Section 9.1
1. This follows from (6.1.5) andthe observation that a root of unity must have absolute
value 1.
2. If the characteristic is p = 0, then there are only p integers, andthe result follows from
(9.1.7).
3. Assume the absolute values equivalent. By nontriviality, there is an element y with
|y|1 > 1. Take a = log |y|2/ log |y|1. For every x there is a real number b such that
|x|1 = |y|b
1. Finda sequence of rational numbers s/t converging to b from above. Then
|x|1 = |y|b
1 < |y|s/t
1 , so |xt/ys|1 < 1. By hypothesis, |xt/ys|2 < 1, so |x|2 < |y|s/t
2 . Let
s/t → b to get |x|2 ≤ |y|b
2. But by taking a sequence of rationals converging to b from
below, we get |x|2 ≥ |y|b
2, hence |x|2 = |y|b
2. To summarize,
|x|1 = |y|b
1
⇒ |x|2 = |y|b
2.
Taking logarithms (if x = 0), we have log |x|2/ log |x|1 = a, hence |x|a1
= |x|2.

Section 9.2
1. Let a = ±

pri
i , hence a ∞ =

pri
i . If p is one of the pi, then a p = p
−ri
i , and
if p is not one of the pi, then a p = 1. Thus only finitely many terms of the product
are unequal to 1, andthe infinite prime cancels the effect of the finite primes. The result
follows.

Section 9.3
1. For each i = 1, . . . ,n, choose yi, zi ∈ k such that |yi|i > 1 and |zi|i < 1. This is
possible by (9.3.2). Take xi = yi if i ≤ r, and xi = zi if i > r. By (9.3.3), there is an
element a ∈ k such that |a − xi|i < < for all i. (We will specify < in a moment.) If i ≤ r,
then
|yi|i ≤ |yi − a|i + |a|i < < + |a|i
so |a|i > |yi|i − <, andw e need0 < < ≤ |yi|i − 1. On the other hand, if i > r, then
|a|i ≤ |a − zi|i + |zi|i < < + |zi|i
so we need0 < < ≤ 1 − |zi|i. Since there are only finitely many conditions to be satisfied,
a single < can be chosen, andthe result follows.
2. Since | |1 is nontrivial, there is an element b such that |b|1 < 1. Then we have
|a|1 = 1 ⇒ |anb|1 < 1 ⇒ |anb|2 < 1 ⇒ |a|n2
|b|2 < 1 ⇒ |a|2 < 1/|b|1/n
2 .
Let n→∞ to get |a|2 ≤ 1. Similarly, |a−1|2 ≤ 1, so |a|2 = 1. Finally,
|a|1 > 1 ⇒ |a
−1|1 < 1 ⇒ |a
−1|2 < 1 ⇒ |a|2 > 1.
Note also that the implication |a|1 ≥ 1 ⇒ |a|2 ≥ 1 is sufficient for equivalence (consider
the cntrapositive).

Section 9.4
1. The condition stated is equivalent to v(a/b) ≥ 0.
2. The product is 4+2p+4p2+p3+p4. But 4 = 1+3 = 1+p and4 p2 = p2+3p2 = p2+p3.
Thus we have 1 + 3p + p2 + 2p3 + p4 = 1 + 2p2 + 2p3 + p4.
3. We have −1 = (p−1)−p = (p−1)+[(p−1)−p]p = (p−1)+(p−1)p−p2. Continuing
inductively, we get
−1 = (p − 1) + (p − 1)p + (p − 1)p2 + · · · .
The result can also be obtainedb y multiplying by -1 on each side of the equation
1 = (1 − p)(1 + p + p2 + · · · ).
4. Since n! = 1·2 · · · p · · · 2p · · · 3p · · · , it follows that if rp ≤ n < (r+1)p, then |n!| = 1/pr.
Thus |n!| → 0 as n→∞.
5. No. Although |pr| = 1/pr → 0 as r →∞, all integers n such that rp < n < (r + 1)p
have absolute value 1. Thus the sequence of absolute values |n| cannot converge, hence
the sequence itself cannot converge.
6. We have |an| = |1/n| = pv(n), where v(n) is the highest power of p dividing n. Thus
pv(n) ≤ n, so v(n) ≤ log n/ log p andconsequen tly v(n)/n → 0. We can apply the root
test to get lim sup |an|1/n = lim pv(n)/n = 1. The radius of convergence is the reciprocal
of the lim sup, namely 1. Thus the series converges for |x| < 1 andd iverges for |x| > 1.
The series also diverges at |x| = 1 because |1/n| does not converge to 0.
7. Since n/pi ≤ n/pi and
∞
i=1 1/pi = (1/p)/(1 − 1/p) = 1/(p − 1), the result follows.
8. By Problem 7,
v[(pm)!] = pm
p
+ pm
p2 + · · · + pm
pm = 1+p + · · · + pm−1 = pm − 1
p − 1 .
9. We have 1/|n!| = pv(n!) ≤ pn/(p−1) by Problem 7. Thus |an|1/n ≤ p1/(p−1). Thus
the radius of convergence is at least p−1/(p−1). Now let |x| = p−1/(p−1) = (1/p)v(x), so
v(x) = 1/(p − 1). Taking n = pm, we have, using Problem 8,
v(xn/n!) = v[xpm
/(pm)!] = pmv(x) − v[(pm)!] = pm
p − 1
− pm − 1
p − 1
=
1
p − 1.
Since 1/(p−1) is a constant independent of m, xn/n! does not converge to 0, so the series
diverges.
Note that 0 < 1/(p − 1) < 1, andsince v is a discrete valuation, there is no x ∈ Qp
such that v(x) = 1/(p−1). Thus |x| < p−1/(p−1) is equivalent to |x| < 1. But the sharper
boundis useful in situations where Qp is embedded in a larger field that extends the
p-adic absolute value.

Section 9.5
1. Take F(X) = Xp−1 − 1, which has p − 1 distinct roots mod p. (The multiplicative
group of nonzero elements of Z/pZ is cyclic.) All roots are simple (because deg F = p−1).
By (9.5.3), the roots lift to distinct roots of unity in Zp.
2. Take F(X) = X2 − m. Since p does not divide m and p = 2, F andits derivative are
relatively prime, so there are no multiple roots. By (9.5.3), m is a square in Zp iff m is a
quadratic residue mod p.
3. Successively find a0, a1, . . . , such that (a0 + a1p + a2p2 + · · · )2 = m in Zp. If we take
p = 5, m = 6, then the first four coefficients are a0 = 1, a1 = 3, a2 = 0, a3 = 4. There is
a secondsolution, the negative of this one. When computing, don’t forget the carry. For
example, (1+3×51 +a2 ×52 +· · · )2 = 1+1×51 yields a term 6×51 = 1×51 +1×52,
so the equation for a2 is 2a2 + 10 (not 9) ≡ 0 mod5, so a2 = 0.

If you face any problem to loading this page,  click here for PDF file of this chapter.

You Might Also Like

Posted in: Linear Algebra, Solutions to Problems