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Abstract Algebra
Chapter 13 - Field Theory
David S. Dummit & Richard M. Foote
Solutions by positrón0802
positron0802@mail.com
Contents
13 Field Theory 1
13.1 Basic Theory and Field Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
13.2 Algebraic Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
13.3 Classical Straightedge and Compass Constructions . . . . . . . . . . . . . . . . . . . . . 9
13.4 Splitting Fields and Algebraic Closures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
13.5 Separable and Inseparable Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
13.6 Cyclotomic Polynomials and Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
13 Field Theory
13.1 Basic Theory and Field Extensions
Exercise 13.1.1.
p(x) = x3 + 9x+ 6 is irreducible in Z[x] by Eisenstein Criterion with p = 3. By Gauss Lemma, then it
is irreducible in Q[x]. To find (1 + θ)−1, we apply the Euclidean algorithm (long division) to p(x) and
1 + x. We find
x3 + 9x+ 6 = (1 + x)(x2 − x+ 10)− 4.
Evaluating at θ, we find (1 + θ)(θ2 − θ + 10) = 4. Therefore
(1 + θ)−1 =
θ2 − θ + 10
4
.
Exercise 13.1.2.
Let f(x) = x3 − 2x − 2. f is irreducible over Z by Eisenstein Criterion with p = 2, hence over Q by
Gauss Lemma. Now, if θ is a root of f , then θ3 = 2θ + 2. Hence
(1 + θ)(1 + θ + θ2) = 1 + 2θ + 2θ2 + θ3 = 3 + 4θ + 2θ2.
For computing
1 + θ
1 + θ + θ2
, first we compute (1+θ+θ2)−1. Applying the Euclidean algorithm, we obtain
x3 − 2x− 2 = (x2 + x+ 1)(x− 1)− 2x− 1,
and
x3 − 2x− 2 = (2x+ 1)(x
2
2
− x
4
− 7
8
)− 9
8
.
Evaluating at θ, from this equalities we obtain
(θ2 + θ + 1)(θ − 1) = 2θ + 1 and (2θ + 1)−1 = 8
9
(
θ2
2
− θ
4
− 7
8
).
Combining these two equations we obtain
8
9
(θ2 + θ + 1)(θ − 1)(θ
2
2
− θ
4
− 7
8
) = 1.
1
13.1 Basic Theory and Field Extensions
So,
(θ2 + θ + 1)−1 =
8
9
(θ − 1)(θ
2
2
− θ
4
− 7
8
) = −2θ
2
3
+
θ
3
+
5
3
,
where we used θ3 = 2θ + 2 again. Therefore,
1 + θ
1 + θ + θ2
= (1 + θ)(−2θ
2
3
+
θ
3
+
5
3
) = −θ
2
3
+
2θ
3
+
1
3
.
Exercise 13.1.3.
Since 03 + 0 + 1 = 1 and 11 + 1 + 1 = 1 in F2, then x3 + x+ 1 is irreducible over F2. Since θ is root of
x3 + x+ 1, then θ3 = −θ − 1 = θ + 1. Hence, the powers of θ in F2(θ) are
θ, θ2, θ3 = θ + 1, θ4 = θ2 + θ, θ5 = θ2 + θ + 1, θ6 = θ2 + 1, and θ7 = 1.
Exercise 13.1.4.
Denote this map by ϕ. Then
ϕ(a+ b
√
2 + c+ d
√
2) = a+ c− b
√
2− d
√
2 = ϕ(a+ b
√
2) + ϕ(c+ d
√
2),
and
ϕ((a+ b
√
2) · (c+ d
√
2)) = ϕ(ac+ 2bd+ (ad+ bc)
√
2)
= ac+ 2bd− (ad+ bc)
√
2
= (a− b
√
2)(c− d
√
2)
= ϕ(a+ b
√
2)ϕ(c+ d
√
2),
hence ϕ is an homomorphism. Moreover, if ϕ(a+b
√
2) = ϕ(c+d
√
2), then a−b√2 = c−d√2, hence (since√
2 6∈ Q) a = b and c = d, so ϕ is injective. Also, given a+ b√2 ∈ Q(√2), then ϕ(a− b√2) = a+ b√2,
so ϕ is surjective. Therefore, ϕ is an isomorphism of Q(
√
2) with itself.
Exercise 13.1.5.
Let α = p/q be a root of a monic polynomial p(x) = xn + · · · + a1x + a0 over Z, with gcd(p, q) = 1.
Then
(
p
q
)n + an−1(
p
q
)n−1 + · · ·+ a1 p
q
+ a0 = 0.
Multiplying this equation by qn one obtains
pn + an−1pn−1q + · · ·+ a1pqn−1 + a0qn = 0
⇒ q(an−1pn−1 + · · ·+ a1pqn−2 + a0qn−1) = −pn.
Thus, every prime that divides q divides pn as well, so divides p. Since gcd(p, q) = 1, there is no prime
dividing q, hence q = ±1. The result follows.
Exercise 13.1.6.
This is straightforward. If
anα
n + an−1αn−1 + · · ·+ a1α+ a0 = 0,
then
(anα)
n + an−1(anα)n−1 + anan−2(anα)n−2 + · · ·+ an−2n a1(anα) + an−1n a0
= annα
n + an−1n an−1α
n−1 + an−1n an−2α
n−2 + · · ·+ an−1n a1α+ an−1n a0
= an−1n (anα
n + an−1αn−1 + an−2αn−2 + · · ·+ a1α+ a0) = 0.
Exercise 13.1.7.
If x3 − nx + 2 is reducible it must have a linear factor, hence a root. By the Rational Root Theorem,
if α is a root of x3 − nx+ 2, then α must divide its constant term, so the possibilities are α = ±1,±2.
If α = −1 or 2, then n = 3; if α = −1, then n = 1; and if α = 2, then n = 5. Therefore, x3 − nx+ 2 is
irreducible for n 6= −1, 3, 5.
2
13.1 Basic Theory and Field Extensions
Exercise 13.1.8.
We subdivide this exercise in cases and subcases.
If x5 − ax − 1 is reducible then it has a root (linear factor) or is a product of two irreducible
polynomials of degrees 2 and 3.
Case 1. If x5− ax− 1 has a root, then, by the Rational Root Theorem, it must be α = ±1. If α = 1
is a root, then a = 0. If α = −1 is a root, then a = 2.
Case 2. Now, suppose that there exists f(x) and g(x) irreducible monic polynomials over Z of
degrees 2 and 3 respectively, such that x5 − ax − 1 = f(x)g(x). Write f(x) = x2 + bx + c and
g(x) = x3 + rx2 + sx+ t, where b, c, r, s, t ∈ Z. Then
x5 − ax− 1 = (x2 + bx+ c)(x3 + rx2 + sx+ t)
= x5 + (b+ r)x4 + (br + c+ s)x3 + (bs+ cr + t)x2 + (bt+ cs) + tc.
Equating coefficients leads to
b+ r = 0
br + c+ s = 0
bs+ cr + t = 0
bt+ cs = −a
ct = −1.
From ct = −1 we deduce (c, t) = (−1, 1) of (c, t) = (1,−1), which give us two cases.
Case 2.1. First suppose (c, t) = (−1, 1). Then the system of equations reduces to
b+ r = 0
br − 1 + s = 0
bs− r + 1 = 0
b− s = −a.
Now, put b = −r into second and third equations to obtain −r2 − 1 + s = 0 and −rs− r + 1 = 0, that
is, r2 + 1 − s = 0 and rs + r − 1 = 0. Adding these last two equations we obtain r2 + rs + r − s = 0.
Thus r2 + rs+ r+ s = 2s, so (r+ 1)(r+ s) = 2s. Now, from r2 + 1− s = 0 we have r2 = s− 1, so then
r2 + rs + r − s = 0 becomes rs + r = 1, that is, r(s + 1) = 1. Hence, r = 1 and s = 0, or r = −1 and
s = −2. If r = 1 and s = 0, then (r + 1)(r + s) = 2s leads to 2 = 0, a contradiction. If r = −1 and
s = −2, it leads to 0 = −4, another contradiction.
Therefore, (c, t) = (−1, 1) is impossible. We now pass to the case (c, t) = (1,−1).
Case 2.2. Suppose that (c, t) = (1,−1). The system of equations reduces to
b+ r = 0
br + 1 + s = 0
bs+ r − 1 = 0
−b+ s = −a.
Adding the second and third equation we obtain b(r + s) + r + s = 0, so that (b+ 1)(r + s) = 0. Then
b = −1 or r = −s, so one more time we have two cases. If r = −s, then br + 1 + s = 0 becomes
br+ 1− r = 0. Hence, b = −r and br+ 1− r = 0 gives r2 + r− 1 = 0. By the Rational Root Theorem,
this equation has no roots on Z. Since r ∈ Z, we have a contradiction. Now suppose b = −1. From
b = −r we obtain r = 1, so, from br + 1 + s = 0 we obtain s = 0. Finally, from −b + s = −a we
obtain a = −1. Therefore, the solution (b, c, r, s, t) = (−1, 1, 1, 0,−1) is consistent and we obtain the
factorization
x5 − ax− 1 = (x2 + bx+ c)(x3 + rx2 + sx+ t) = (x2 − x+ 1)(x3 + x2 − 1).
3
13.2 Algebraic Extensions
13.2 Algebraic Extensions
Exercise 13.2.1.
Since the characteristic of F is p, its prime subfield is (isomorphic to) Fp = Z/pZ. We can consider F as
a vector space over Fp. Since F is finite, then [F : Fp] = n for some n ∈ Z+. Therefore
|F| = |Fp|[F:Fp] = pn.
Exercise 13.2.2.
Note that g and h are irreducible over F2 and F3. Now, is θ is a root of g, then F2(θ) ∼= F2/(g(x))
has 4 elements and F3(θ) ∼= F3/(g(x)) has 9 elements. Furthermore, is θ2 is a root of h, then F2(θ2) ∼=
F2/(h(x)) has 8 elements and F3(θ2) ∼= F3/(h(x)) has 27 elements.
The multiplication table for F2/(g(x)) is
· 0 1 x x+ 1
0 0 0 0 0
1 0 1 x x+ 1
x 0 x x+ 1 x
x+ 1 0 x+ 1 x x
The multiplication table for F3/(g(x)) is
· 0 1 2 x x+ 1 x+ 2 2x 2x+ 1 2x+ 2
0 0 0 0 0 0 0 0 0 0
1 0 1 2 x x+ 1 x+ 2 2x 2x+ 1 2x+ 2
2 0 2 1 2x 2x+ 2 2x+ 1 x x+ 2 x+ 1
x 0 x 2x 2x+ 1 1 x+ 1 x+ 2 2x+ 2 2
x+ 1 0 x+ 1 2x+ 2 1 x+ 2 2x 2 x 2x+ 1
x+ 2 0 x+ 2 2x+ 1 x+ 1 2x 2 2x+ 2 1 x
2x 0 2x x x+ 2 2 2x+ 2 2x+ 1 x+ 1 1
2x+ 1 0 2x+ 1 x+ 2 2x+ 2 x 1 x+ 1 2 2x
2x+ 2 0 2x+2 x+ 1 2 2x+ 1 x 1 2x x+ 2
In both cases, x is a generator of the cyclic group of nonzero elements.
Exercise 13.2.3.
Since 1 + i 6∈ Q, its minimal polynomial is of degree at least 2. We try conjugation, and obtain
(x− (1 + i))(x− (1− i)) = x2 − 2x+ 2,
which is irreducible by Eisenstein with p = 2. Therefore, the minimal polynomial is x2 − 2x+ 2.
Exercise 13.2.4.
First, note that (2+
√
3)2 = 4+4
√
3+3 = 7+4
√
3. Let θ = 2+
√
3. Then θ2−4θ = 7+4√3−8−4√3 = −1,
hence θ is a root of x2−4x+1. Moreover, x2−4x+1 is irreducible over Q (because θ 6∈ Q), so x2−4x+1
is the minimal polynomial of 2 +
√
3. Therefore, 2 +
√
3 has degree 2 over Q.
Now, let α = 3
√
2 and β = 1 + α+ α2. Then β ∈ Q(α), so Q ⊂ Q(β) ⊂ Q(α). We have [Q(α) : Q] =
[Q(α) : Q(β)][Q(β) : Q]. Note that [Q(α) : Q] = 3 since α has minimal polynomial x3 − 2 over Q, so
[Q(β) : Q] = 1 or 3. For a contradiction, suppose [Q(β) : Q] = 1, that is, Q(β) = Q so β ∈ Q. Then
β2 = (1 + α+ α2)2 = 1 + 2α+ 3α2 + 2α3 + α4 = 5 + 4α+ 3α2,
where we used α3 = 2. So
β2 − 3β = 5 + 2α+ 3α2 − 3(1 + α+ α2) = 2− α,
hence α = −β2 + 3β + 2 ∈ Q(β) = Q, a contradiction. Therefore, [Q(β) : Q] = 3.
4
13.2 Algebraic Extensions
Exercise 13.2.5.
Since the polynomials have degree 3, if they were reducible they must have a linear factor, hence a root
in F . Note that every element of F is of the form a+bi, where a, b ∈ Q. The roots of x3−2 are 3√2, ζ 3√2
and ζ2 3
√
2, where ζ is the primitive 3’rd root of unity, i.e., ζ = exp(2pii/3) = cos(2pi/3) + i sin(2pi/3) =
− 12 +
√
3
2 . Since
√
3 6∈ Q, none of this elements is in F , hence x3− 2 is irreducible over F . Similarly, the
roots of x3 − 3 are 3√3, ζ 3√3 and ζ2 3√3, and by the same argument non of this elements is in F . Hence
x3 − 3 is irreducible over F .
Exercise 13.2.6.
We have to prove that F (α1, . . . , αn) is the smallest field containing F (α1), . . . , F (αn). Clearly F (αi) ⊂
F (α1, . . . , αn) for all 1 ≤ i ≤ n. Now let K be a field such that F (αi) ⊂ K for all i. If θ is an
element of F (α1, . . . , αn), then θ is of the form θ = a1α1 + · · · + anαn, where a1, . . . , an ∈ F . Every
aiαi is in K, hence θ ∈ K. Thus F (α1, . . . , αn) ⊂ K. Therefore, F (α1, . . . , αn) contains all F (αi)
and is contained in every field containing all F (αi), hence F (α1, . . . , αn) is the composite of the fields
F (α1), F (α2), . . . , F (αn).
Exercise 13.2.7.
Since
√
2 +
√
3 is in Q(
√
2,
√
3), clearly Q(
√
2 +
√
3) ⊂ Q(√2,√3). For the other direction we have to
prove that
√
2 and
√
2 are in Q(
√
2 +
√
3). Let θ =
√
2 +
√
3. Then
θ2 = 5 + 2
√
6, θ3 = 11
√
2 + 9
√
3 and θ4 = 37 + 15
√
6.
So √
2 =
1
2
(θ3 − 9θ),
√
3 =
1
2
(11θ − θ3) and θ4 = 49 + 20
√
6.
Therefore
√
2 ∈ Q(θ) and √3 ∈ Q(θ), so Q(√2,√3) ⊂ Q(√2 + √3). The equality follows. Hence,
[Q(
√
2 +
√
3) : Q] = [Q(
√
2,
√
3) : Q] = 4.
We also have
θ4 − 10θ2 = (49 + 20
√
6)− 10(5 + 2
√
6) = −1, so θ4 − 10θ2 + 1 = 0.
Since [Q(
√
2 +
√
3) : Q] = 4, then x4 − 10x2 + 1 is irreducible over Q, and is satisfied by √2 +√3.
Exercise 13.2.8.
The elements of F (
√
D1,
√
D2) can be written in the form
a+ b
√
D1 + c
√
D2 + d
√
D1D2, where a, b, c, d ∈ F.
We have
[F (
√
D1,
√
D2) : F ] = [F (
√
D1,
√
D2) : F (
√
D1)][F (
√
D1) : F ].
Since [F (
√
D1) : F ] = 2, then [F (
√
D1,
√
D2) : F ] can be 2 or 4. Now, [F (
√
D1,
√
D2) : F ] = 2 if and
only if [F (
√
D1,
√
D2) : F (
√
D1)] = 1, and that occurs exactly when x2 − D2 is reducible in F (
√
D1)
(i.e., when
√
D2 ∈ F (
√
D1)), that is, if there exists a, b ∈ F such that
(a+ b
√
D1)
2 = D2, so that a2 + 2ab
√
D1 + b
2D21 = D2.
Note that ab = 0 as ab 6= 0 implies √D1 ∈ F , contrary to the hypothesis. Then a = 0 or b = 0. If
b = 0, then D2 is a square in F , contrary to the hypothesis. If a = 0, then b2D1 = D2, and thus
D1D2 = (
D2
b )
2, so D1D2 is a square in F . So, x2 −D2 is reducible in F (
√
D1) if and only if D1D2 is a
square in F . The result follows.
Exercise 13.2.9.
Suppose
√
a+
√
b =
√
m +
√
n for some m,n ∈ F , then a + √b = m + n + 2√mn. Since b is not a
square in F , this means
√
b = 2
√
mn. We also have
√
a+
√
b−√n = √m, so
√
b = 2
√
n(
√
a+
√
b−√n).
5
13.2 Algebraic Extensions
Hence, √
b = 2
√
n(a+
√
b)− 2n
⇒ (
√
b+ 2n)2 = 4n(a+
√
b)
⇒ b+ 4n
√
b+ 4n2 = 4n(a+
√
b)
⇒ b+ 4n2 − 4na = 0
⇒ n = 4a±
√
16a2 − 16b
8
⇒
√
a2 − b = ±2n
a
.
Therefore, since a and n are in F ,
√
a2 − b is in F .
Now suppose that a2− b is a square in F , so that √a2 − b ∈ F . We prove that there exists m,n ∈ F
such that
√
a+
√
b =
√
m+
√
n. Let
m =
a+
√
a2 − b
2
and n =
a−√a2 − b
2
.
Note that m and n are in F as char(F ) 6= 2. We claim
√
a+
√
b =
√
m+
√
n. Indeed, we have
m =
(a+
√
b) + 2
√
a2 − b+ (a−√b)
4
= (
√
a+
√
b+
√
a−√b
2
)2,
and
n =
(a+
√
b)− 2√a2 − b+ (a−√b)
4
= (
√
a+
√
b−
√
a−√b
2
)2.
Thus
√
m =
√
a+
√
b+
√
a−√b
2
and
√
n =
√
a+
√
b−
√
a−√b
2
.
Therefore,
√
m+
√
n =
√
a+
√
b+
√
a−√b
2
+
√
a+
√
b−
√
a−√b
2
=
√
a+
√
b,
as claimed.
Now, we this to determine when the field Q(
√
a+
√
b), a, b ∈ Q, is biquadratic over Q. If a2 − b
is a square in Q and b is not, we have Q(
√
a+
√
b) = Q(
√
m +
√
n) = Q(
√
m,
√
n), so by last exercise
Q(
√
a+
√
b) is biquadratic over Q when a2 − b is a square in Q, and neither b, m, n or mn are squares
in Q. Since
mn =
a+
√
a2 − b
2
a−√a2 − b
2
=
b
4
,
then mn is never a square when b isn’t. Thus, Q(
√
a+
√
b) is biquadratic over Q exactly when a2 − b
is a square in Q and neither b, m nor n is a square in Q.
Exercise 13.2.10.
Note that
√
3 + 2
√
2 =
√
3 +
√
8. Recalling last exercise with a = 3 and b = 8, we have a2−b = 9−8 = 1
is a square in Q and b = 8 is not. Hence, we find (m = 2 and n = 1 from last exercise)
√
3 +
√
8 =
√
2+1.
Therefore, Q(
√
3 + 2
√
2) = Q(
√
2) and the degree of the extension Q(
√
3 + 2
√
2) over Q is 2.
Exercise 13.2.11.
(a) First, note that the conjugation map a+bi→ a−bi is an isomorphism of C, so it takes squares roots
to square roots, and maps numbers of the first quadrant to the fourth (and reciprocally). Since
√
3 + 4i
is the square root of 3+4i in the first quadrant, its conjugate is the square of root of 3−4i in the fourth
quadrant, so is
√
3− 4i. Hence √3 + 4i and √3− 4i are conjugates each other. Now, we use Exercise
9 again. Note that
√
3 + 4i =
√
3 +
√−16. With a = 3 and b = −16, we have a2 − b = 25 is a square
in Q and b = −16 is not. Hence, we find m = 1 and n = −4 and thus √3 + 4i = 1 + √−4 = 1 + 2i.
Furthermore, we find
√
3− 4i = 1− 2i. Therefore, √3 + 4i+√3− 4i = 4, i.e., √3 + 4i+√3− 4i ∈ Q.
6
13.2 Algebraic Extensions
(b) Let θ =
√
1 +
√−3 +
√
1−√−3. Then
θ2 = (
√
1 +
√−3 +
√
1−√−3)2 = (1 +√−3) + (2√1 + 3) + (1−√−3) = 6.
Since x2 − 6 is irreducible over Q (Eisenstein p = 2), then θ has degree 2 over Q.
Exercise 13.2.12.
Let E be a subfield of K containing F . Then
[K : F ] = [K : E][E : F ] = p.
Since p is prime, either [K : E] = 1 or [E : F ] = 1. The result follows.
Exercise 13.2.13.
Note that, for all 1 ≤ k ≤ n, we have [Q(α1, . . . , αk) : Q(α1, . . . , αk−1)] = 1 or 2. Then [F : Q] = 2m
for some m ∈ N. Suppose 3√2 ∈ F . Then Q ⊂ Q( 3√2) ⊂ F , so [Q( 3√2) : Q] divides [F : Q], that is, 3
divides 2m, a contradiction. Hence, 3
√
2 6∈ F .
Exercise 13.2.14.
Since α2 ∈ F (α), clearly F (α2) ⊂ F (α). Thus we have to proveα ∈ F (α2). For this purpose, consider
the polynomial p(x) = x2 − α2, so that p(α) = 0. Note that α ∈ F (α2) if and only if p(x) is reducible
in F (α2). For a contradiction, suppose p(x) is irreducible in F (α2), so that [F (α) : F (α2)] = 2. Thus
[F (α) : F ] = [F (α) : F (α2)][F (α2) : F ] = 2[F (α2) : F ],
so [F (α) : F ] is even, a contradiction. Therefore, p(x) is reducible in F (α2) and α ∈ F (α2).
Exercise 13.2.15.
We follow the hint. Suppose there exists a counterexample. Let α be of minimal degree such that F (α)
is not formally real and α having minimal polynomial f of odd degree, say deg f = 2k+1 for some k ∈ N.
Since F (α) is not formally real, then −1 can be express as a sum of squares in F (α) ∼= F [x]/((f(x))).
Then, the exists polynomials p1(x), . . . , pm(x), g(x) such that
−1 + f(x)g(x) = (p1(x))2 + · · ·+ (pm(x))2.
As every element in F [x]/((f(x)) can be written as a polynomial in α with degree less than deg f ,
we have deg pi < 2k + 1 for all i. Thus, the degree in the right hand of the equation is less than
4k + 1, so deg g < 2k + 1 as well. We prove that the degree of g is odd by proving that the degree of
(p1(x))
2 + · · ·+ (pm(x))2 is even, because then the equation −1 + f(x)g(x) = (p1(x))2 + · · ·+ (pm(x))2
implies the result. Let d be the maximal degree over all pi, we prove that x2d is the leading term of
(p1(x))
2 + · · · + (pm(x))2. Note that x2d is a sum of squares (of the leading coefficients of the pi’s of
maximal degree). Now, since F is formally real, 0 can’t be expressed as a sum of squares in F . Indeed,
if
∑l
i=1 a
2
i = 0, then
∑l−1
i=1(ai/al)
2 = −1. Therefore x2d 6= 0, so the degree of (p1(x))2 + · · ·+ (pm(x))2
is 2d, as claimed. Hence, the degree of g must be odd by the assertion above. Then g must contain an
irreducible factor of odd degree, say h(x). Since deg g < deg f , we have degh < deg f as well. Let β be
a root of h(x), hence a root of g(x). Then
−1 + h(x)f(x)g(x)
h(x)
= (p1(x))
2 + · · ·+ (pm(x))2,
so −1 is a square in F [x]/((h(x)) ∼= F (β), which means F (β) is not formally real. Therefore, β is a root
of an odd degree polynomial h such that F (β) is not formally real. Since degh < deg f , this contradicts
the minimality of α. The result follows.
Exercise 13.2.16.
Let r ∈ R be nonzero. Since r is algebraic over F , there exist an irreducible polynomial p(x) =
a0 + a1x + · · · + xn ∈ F [x] such that p(r) = 0. Note that a0 6= 0 since p is irreducible. Then
r−1 = −a−10 (rn−1 + · · ·+ a1). Since ai ∈ F ⊂ R and r ∈ R, we have r−1 ∈ R.
7
13.2 Algebraic Extensions
Exercise 13.2.17.
Let p(x) be an irreducible factor of f(g(x)) of degree m. Let α be a root of p(x). Since p is irreducible,
then [F (α) : F ] = deg p(x) = m. Now, since p(x) divides f(g(x)), we have f(g(α)) = 0 and thus g(α)
is a root of f(x). Since f is irreducible, this means n = [F (g(α)) : F ]. Note that F (g(α)) ⊂ F (α).
Therefore,
m = [F (α) : F ] = [F (α) : F (g(α))][F (g(α)) : F ] = [F (α) : F (g(α))] · n,
so n divides m, that is, deg f divides deg p.
Exercise 13.2.18.
(a) We follow the hint. Since k[t] is an UFD and k(t) is its field of fractions, then, by Gauss Lemma,
P (X)−tQ(X) is irreducible in k((t))[X] is and only if it is irreducible in (k[t])[X]. Note that (k[t])[X] =
(k[X])[t]. Since P (X)− tQ(X) is linear in (k[X])[t], is clearly irreducible in (k[X])[t] (i.e., in (k[t])[X]),
hence in (k(t))[X]. Thus, P (X)− tQ(X) is irreducible in k(t). Now, x is clearly a root of P (X)− tQ(X)
since P (x)− tQ(x) = P (x)− P (x)
Q(x)
Q(x) = P (x)− P (x) = 0.
(b) Let n = max{degP (x), degQ(x)}. Write
P (x) = anx
n + · · ·+ a1x+ a0 and Q(x) = bnxn + · · ·+ b1x+ b0,
where ai, bi ∈ k for all i, so at least one of an or bn is nonzero. The degree of P (X)− tQ(X) is clearly
≤ n, we prove is n. If an or bn is zero then clearly deg (P (X)− tQ(X)) = n. Suppose an, bn 6= 0. Then
an, bn ∈ k, but t 6∈ k (as t ∈ k(x)), it cannot be that an = tbn. Thus (an − tbn)Xn 6= 0, so the degree of
P (X)− tQ(X) is n.
(c) Since P (X) − tQ(X) is irreducible over k(t) and x is a root by part (a), then [k(x) : k(t)] =
degP (X)− tQ(X), and this degree equals max{degP (x), degQ(x)} by part (b).
Exercise 13.2.19.
(a) Fix α in K. Since K is (in particular) a commutative ring, we have α(a + b) = αa + αb and
α(λa) = λ(αa) for all a, b, λ ∈ K. If, in particular, λ ∈ F , we have the result.
(b) Fix a basis for K as a vector space over F . By part (a), for every α ∈ K we can associate a
F -linear transformation Tα. Denote by
(
Tα
)
the matrix of Tα with respect to the basis fixed above.
Then define ϕ : K → Mn(F ) by ϕ(α) =
(
Tα
)
. We claim ϕ is an isomorphism. Indeed, if α, β ∈ K,
then T(α+β)(k) = (α + β)(k) = αk + βk = Tα(k) + Tβ(k) for every k ∈ K, hence T(α+β) = Tα + Tβ .
We also have T(αβ)(k) = (αβ)(k) = αkβk = Tα(k)Tβ(k) for every k ∈ K, so T(αβ) = TαTβ . Thus
ϕ(α + β) = ϕ(α) + ϕ(β) and ϕ(αβ) = ϕ(α)ϕ(β) (since the basis is fixed), so ϕ is an homomorphism.
Now, if ϕ(α) = ϕ(β), then αk = βk for every k ∈ K, so letting k = 1 we find that ϕ is injective.
Therefore, ϕ(K) is isomorphic to a subfield of Mn(F ), so the ring Mn(F ) contains an isomorphic copy
of every extension of F of degree ≤ n.
Exercise 13.2.20.
The characteristic polynomial of A is p(x) = det(Ix − A). For every k ∈ K, we have (Iα − A)k =
αk −Ak = αk − αk = 0, so det(Iα−A) = 0 in K. Therefore, p(α) = 0.
Now, consider the field Q( 3
√
2) with basis {1, 3√2, 3√4} over Q. Denote the elements of this basis by
e1 = 1, e2 = 3
√
2 and e3 = 3
√
4. Let α = 3
√
2 and β = 1 + 3
√
2 + 3
√
4. Then α(e1) = e2, α(e2) = e3 and
α(e3) = 2e1. We also have β(e1) = e1 + e2 + e3, β(e2) = 2e1 + e2 + e3 and β(e3) = 2e1 + 2e2 + e3. Thus,
the associated matrices of the their linear transformations are, respectively,
Aα =
0 0 21 0 0
0 1 0
 and Aβ =
1 2 21 1 2
1 1 1
 .
The characteristic polynomial of Aα is x3 − 2, hence is the monic polynomial of degree 3 satisfied by
α = 3
√
2. Furthermore, the characteristic polynomial of Aβ is x3 − 3x2 − 3x − 1, hence is the monic
polynomial of degree 3 satisfied by β = 1 + 3
√
2 + 3
√
4.
8
13.3 Classical Straightedge and Compass Constructions
Exercise 13.2.21.
The matrix of the linear transformation "multiplication by α" on K is found by acting of α in the basis
1,
√
D. We have α(1) = α = a+ b
√
D and α(
√
D) = a
√
D + bD. Hence the matrix is
(
a bD
b a
)
. Now
let ϕ : K →M2(Q) be defined by ϕ(a+ b
√
D) =
(
a bD
b a
)
.
We have
ϕ(a+ b
√
D + c+ d
√
D) =
(
a+ c (b+ d)D
b+ d a+ c
)
=
(
a bD
b a
)
+
(
c dD
d c
)
= ϕ(a+ b
√
D) + ϕ(c+ d
√
D),
and
ϕ((a+ b
√
D) · (c+ d
√
D)) =
(
ac+ bdD (ad+ bc)D
ad+ bc ac+ bdD
)
=
(
a bD
b a
)
·
(
c dD
d c
)
= ϕ(a+ b
√
D)ϕ(c+ d
√
D),
so ϕ is an homomorphism. Since K is a field, its ideals are {0} and K, so ker(ϕ) is trivial or K. Since
ϕ(K) is clearly non-zero, then ker(ϕ) 6= K and thus ker(ϕ) = {0}. Hence, ϕ is injective. Therefore, ϕ
is an isomorphism of K with a subfield of M2(Q).
Exercise 13.2.22.
Define ϕ : K1 × K2 → K1K2 by ϕ(a, b) = ab. We prove that ϕ is F -bilinear. Let a, a1, a2 ∈ K and
b, b1, b2 ∈ K2. Then
ϕ((a1, b) + (a2, b)) = ϕ(a1 + a2, b) = (a1 + a2)b = a1b+ a2b = ϕ(a1, b) + ϕ(a2, b),
and
ϕ((a,1 b) + (a, b2)) = ϕ(a, b1 + b2) = a(b1 + b2) = ab1 + ab1 = ϕ(a, b1) + ϕ(a, b2).
We also have, for r ∈ F , ϕ(ar, b) = (ar)b = a(rb) = ϕ(rb). Therefore, ϕ is a F -bilinear map. Hence, ϕ
induces a F -algebra homomorphism Φ : K1 ⊗F K2 → K1K2. We use Φ to prove both directions. Note
that K1 ⊗F K2 have dimension [K1 : F ][K2 : F ] as a vector space over F .
First, we suppose [K1K2 : F ] = [K1 : F ][K2 : F ] and prove K1 ⊗F K2 is a field. In this case
K1 ⊗F K2 and K1K2 have the same dimension over F . Let L = Φ(K1 ⊗F K2). We claim L = K1K2,i.e. Φ is surjective. Note that L contains K1 and K2. Since L is a subring of K1K2 containing K1 (or
K2), then L is a field (Exercise 16). Hence, L is a field containing both K1 and K2. Since K1K2 is
the smallest such field (by definition), we have L = K1K2. Therefore Φ is surjective, as claimed. So,
Φ is an F -algebra surjective homomorphism between F -algebras of the same dimension, hence is an
isomorphism. Thus, K1 ⊗F K2 is a field.
Now suppose that K1 ⊗F K2 is a field. In this case Φ is a field homomorphism. Therefore, Φ
is either injective or trivial. It is clearly nontrivial since Φ(1 ⊗ 1) = 1, so it is injective. Hence,
[K1 : F ][K2 : F ] ≤ [K1K2 : F ]. As we already have [K1K2 : F ] ≤ [K1 : F ][K2 : F ] (Proposition 21 of
the book), the equality follows.
13.3 Classical Straightedge and Compass Constructions
Exercise 13.3.1.
Suppose the 9-gon is constructible. It has angles of 40◦. Since we can bisect an angle by straightedge and
compass, the angle of 20◦ would be constructible. But then cos 20◦ and sin 20◦ would be constructible
too, a contradiction (see proof of Theorem 24).
Exercise 13.3.2.
Let O,P,Q and R be the points marked in the figure below.
9
13.3 Classical Straightedge and Compass Constructions
Then α = ∠QPO, β = ∠RQO, γ = ∠QRO, and θ is an exterior angle of 4PRO. Since 4PQO is
isosceles, then α = ∠QPO = ∠QOP . Since β is an exterior angle of 4PQO, it equals the sum of the
two remote interior angles, i.e., equals ∠QPO+∠QOP . This two angles equals α, hence β = 2α. Now,
since 4QRO is isosceles, then β = γ. Finally, since θ is an exterior angle of 4PRO, equals the sum of
the two remote interior angles, which are α and γ. Therefore, θ = α+ γ = α+ β = 3α.
Exercise 13.3.3.
We follow the hint. The distances a, b, x, y and x− k are marked in the figure below.
From the figure, using similar triangles for (a), (b) and (c), and Pythagoras Theorem for (d), the 4
relations are clear. Hence, we have
y =
√
1− k2
1 + a
, x = a
b+ k
1 + a
,
y
x− k =
√
1− k2
3k
and (1− k2) + (b+ k)2 = (1 + a)2.
So,
√
1− k2 = y(1 + a) = 3kyx−k implies 3k = (x − k)(1 + a). From the equation for x above, we find
3k = (a(b+a)1+a − k)(1 + a) = a(b + k) − k(1 + a), so b + k = 4k+kaa . Using this in the last equation and
reducing, we get
(1− k2) + (b+ k)2 = (1 + a)2
⇒ (1− k2) + (4k + ka
a
)2 = (1 + a)2
⇒ a2(1− k2) + (4k + ka)2 = a2(1 + a)2
⇒ a2 − (ka)2 + (4k)2 + 8k2a+ (ka)2 = a2 + 2a3 + a4
⇒ a4 + 2a3 − 8k2a− 16k2 = 0.
We let a = 2y to obtain
h4 + h3 − k2h− k2 = 0.
We find h = k2/3, hence a = 2k2/3. From b = 4k+kaa −k, we find b = 2k1/3. Therefore, we can construct
2k1/3 and 2k2/3 using Conway’s construction.
Exercise 13.3.4.
Let p(x) = x3 + x2 − 2x− 1 and α = 2 cos(2pi/7). By the Rational Root Theorem, if p has a root in Q,
it must be ±1 since it must divide its constant term. But p(1) = −1 and p(−1) = 1, so p is irreducible
over Q. Therefore, α is of degree 3 over Q, hence [Q(α) : Q] cannot be a power of 2. Since we can’t
construct α, the regular 7-gon is not constructible by straightedge and compass.
Exercise 13.3.5.
Let p(x) = x2 + x − 1 = 0 and α = 2 cos(2pi/5). By the Rational Root Theorem, if p has a root in Q,
it must be ±1. Since p(1) = 1 and p(−1) = −1, p is irreducible over Q. Hence, α is of degree 2 over
Q, so it is constructible. We can bisect an angle by straightedge and compass, so β = cos(2pi/5) is also
constructible. Finally, as sin(2pi/5) =
√
1− cos2(2pi/5), sin(2pi/5) is also constructible. Therefore, the
regular 5-gon is constructible by straightedge and compass.
10
13.4 Splitting Fields and Algebraic Closures
13.4 Splitting Fields and Algebraic Closures
Exercise 13.4.1.
Let f(x) = x4 − 2. The roots of f are 4√2, − 4√2, i 4√2 and −i 4√2. Hence, the splitting field of f is
Q(i, 4
√
2). So, the splitting field of f has degree [Q(i, 4
√
2) : Q] = [Q(i, 4
√
2) : Q( 4
√
2)][Q( 4
√
2) : Q] over Q.
Since 4
√
2 is a root of the irreducible polynomial x4 − 2 over Q, then [Q( 4√2) : Q] = 4. Furthermore,
since i 6∈ Q( 4√2), then x2 + 1 is irreducible over Q( 4√2) having i as a root, so [Q(i, 4√2) : Q( 4√2)] = 2.
Therefore, [Q(i, 4
√
2) : Q] = 8.
Exercise 13.4.2.
Let f(x) = x4 + 2. Let K be the splitting field of f and let L be the splitting field of x4 − 2, that is,
L = Q(i, 4
√
2) (last exercise). We claim K = L, so that [K : Q] = 8 by last exercise. Let ζ =
√
2
2 + i
√
2
2 .
First we prove ζ ∈ L and ζ ∈ K, then we prove K = L.
We prove ζ ∈ L. This is easy. Let θ = 4√2. Since θ ∈ L, then θ2 = √2 ∈ L. We also have i ∈ L, so√
2, i ∈ L implies ζ ∈ L.
We prove ζ ∈ K. We have to prove i ∈ K and √2 ∈ K. Let α be a root of x4 + 2, so that α4 = −2.
Let β be a root of x4 − 1, so that β4 = 1. Then (αβ)4 = α4β4 = −2, hence αβ is also a root of x4 + 2.
Since the roots of x4 − 1 are ±1,±i, the roots of x4 + 2 are ±α and ±iα. Since K is generated over
Q by there roots, then iα/α = i ∈ K. Now let γ = α2 ∈ K. Since γ2 = α4 = −2, then γ is a root of
x2 + 2. Since the roots of x2 + 2 are i
√
2 and −i√2, then γ is one of this roots. In either case γ/i ∈ K,
which implies
√
2 ∈ K. Therefore, ζ ∈ K.
Now we prove L = K proving both inclusions.
Let α be a root of x4 + 2 and θ be a root of x4 − 2. Then α4 = −2 and θ4 = 2. Note that ζ2 = i,
so ζ4 = −1. Hence, (ζθ)4 = ζ4θ4 = −2, so ζθ is a root of x4 + 2. Then, as we proved earlier, the roots
of x4 + 2 are ±ζθ and ±iζθ. We also have (ζα)4 = ζ4α4 = 2, so ζα is a root of x4 − 2. Then, by last
exercise, the roots of x4− 2 are ±ζα and ±iζα. Now, since ζ and α are in K, we have ζα ∈ K. We also
have i ∈ K, so all roots of x4 − 2 are in K. Since L is generated by these roots, then L ⊂ K. Similarly,
ζ and θ are in L, so ζθ ∈ L; since i ∈ L, then all roots of x4 + 2 are in L. Since K is generated by these
roots, we have K ⊂ L. Therefore, K = L, so [K : Q] = 8.
Exercise 13.4.3.
Let f(x) = x4 + x2 + 1. Note that f(x) = (x2 + x + 1)(x2 − x + 1), so the roots of f are ± 12 ± i
√
3
2 .
Let w = 12 − i
√
3
2 , so this roots are w,−w,w,−w, where w denotes the complex conjugate of w (i.e.,
w = 12 + i
√
3
2 ). Hence, the splitting field of f is Q(w,w). Since w + w = 1, then Q(w,w) = Q(w).
Furthermore, w is a root of x2 − x+ 1, that is irreducible over Q since w 6∈ Q. Therefore, the degree of
the splitting field of f is [Q(w) : Q] = 2.
Exercise 13.4.4.
Let f(x) = x6 − 4. Note that f(x) = (x3 − 2)(x3 + 2). The roots of x3 − 2 are 3√2, ζ 3√2 and ζ2 3√2,
where ζ is the primitive 3’rd root of unity, i.e., ζ = exp(2pii/3) = cos(2pi/3) + i sin(2pi/3) = − 12 +
√
3
2 .
Furthermore, the roots of x3 + 2 are − 3√2, −ζ 3√2 and −ζ2 3√2. Therefore, the splitting field of f is
Q(ζ, 3
√
2). Then [Q(ζ, 3
√
2) : Q] = [Q(ζ, 3
√
2) : Q( 3
√
2)][Q( 3
√
2) : Q]. We have that 3
√
2 is a root of the
irreducible polynomial x3− 2 over Q, so 3√2 has degree 3 over Q. Furthermore, ζ is a root of x2 +x+ 1,
irreducible over Q( 3
√
2), so ζ has degree 2 over Q( 3
√
2). Hence, the degree of the splitting field of f is
[Q(ζ, 3
√
2) : Q] = 6.
Exercise 13.4.5.
We follow the hint. First suppose that K is a splitting field over F . Hence, there exists f(x) ∈ F [x]
such that K is the splitting field of f . Let g(x) be an irreducible polynomial in F [x] with a root α ∈ K.
Let β be any root of g. We prove β ∈ K, so that g splits completely in K[x]. By Theorem 8, there
is an isomorphism ϕ : F (α) ∼−→ F (β) such that ϕ(α) = β. Furthermore, K(α) is the splitting field for
f over F (a), and K(β) is the splitting field for f over F (β). Therefore, by Theorem 28, ϕ extends to
an isomorphism σ : K(α) ∼−→ K(β). Since K = K(α), then [K : F ] = [K(α) : F ] = [K(β) : F ], so
K = K(β). Thus, β ∈ K.
Now suppose that every irreducible polynomial in F [x] that has a root in K splits completely in
K[x]. Since [K : F ] is finite, thenK = F (α1, . . . , αn) for some α1, . . . , αn. For every 1 ≤ i ≤ n, let pi
be the minimal polynomial of αi over F , and let f = p1p2 · · · pn. Since every αi is in K, every pi has a
11
13.5 Separable and Inseparable Extension
root in K, hence splits completely in K. Therefore, f splits completely in K and K is generated over
F by its roots, so K is the splitting field of f(x) ∈ F [x].
Exercise 13.4.6.
(a) Let K1 be the splitting field of f1(x) ∈ F [x] over F and K2 the splitting field of f2(x) ∈ F [x] over
F . Thus, K1 is generated over F by the roots of f1, and K2 is generated over F by the roots of f2.
Therefore, f1f2 splits completely in K1K2 and K1K2 is generated over F by its roots, hence is the
splitting field of f1f2(x) ∈ F [x].
(b)We follow the hint. By last exercise, we have to prove that every irreducible polynomial in F [x] that
has a root in K1 ∩K2 splits completely in (K1 ∩K2)[x]. So, let f(x) be an irreducible polynomial in
F [x] that has a root, say α, in K1∩K2. By last exercise, f splits completely in K1 and splits completely
in K2. Since K1 and K2 are contained in K, by the uniqueness of the factorization of f in K, the roots
of f in K1 must coincide with its roots in K2. Hence, f splits completely in (K1 ∩K2)[x].
13.5 Separable and Inseparable Extension
Exercise 13.5.1.
Let f(x) = anxn + · · · + a1x + a0 and g(x) = bmxm + · · · + b1x + b0 be two polynomials. Suppose,
without any loss of generality, that n ≥ m. Thus, we can write g(x) = bnxn+ · · ·+ b1x+ b0, where some
of the last coefficients bi could be zero. We have f(x) + g(x) = (an+ bn)xn+ · · ·+ (a1 + b1)x+ (a0 + b0),
so
Dx(f(x) + g(x)) = n(an + bn)x
n−1 + · · ·+ 2(a2 + b2)x+ (a1 + b1) = Dx(f(x)) +Dx(g(x)).
Now we prove the formula for the product. Let cn =
∑n
k=0 akbn−k, so that
f(x)g(x) = (
n∑
k=0
akx
k)(
n∑
k=0
bkx
k) =
2n∑
l=0
(
l∑
k=0
akbl−k)xl =
2n∑
l=0
clx
l.
Hence,
Dx(f(x)g(x)) = Dx(
2n∑
l=0
clx
l) =
2n−1∑
l=0
(l + 1)cl+1x
l,
so the coefficient of xl in Dx(f(x)g(x)) is (l + 1)cl+1.
Now, we have Dx(f(x)) = nanxn−1 + · · · + 2a2x + a1, and Dx(g(x)) = nbnxn−1 + · · · + 2b2x + b1.
So (recall product of polynomials, page 295 of the book)
Dx(f(x))g(x) = (
n∑
k=1
kakx
k−1)(
n∑
k=0
bkx
k)
= (
n−1∑
k=0
(k + 1)ak+1x
k)(
n∑
k=0
bkx
k) =
2n−1∑
l=0
(
l∑
k=0
(k + 1)ak+1bl−k)xl,
and
f(x)Dx(g(x)) = (
n∑
k=0
akx
k)(
n∑
k=1
kbkx
k−1)
= (
n∑
k=0
akx
k)(
n−1∑
k=0
(k + 1)bk+1x
k) =
2n−1∑
l=0
(
l∑
k=0
ak(l − k + 1)bl−k+1)xl.
12
13.5 Separable and Inseparable Extension
Therefore, the coefficient of xl in Dx(f(x))g(x) +Dx(g(x))f(x) is
(
l∑
k=0
(k + 1)ak+1bl−k) + (
l∑
k=0
(l − k + 1)akbl−k+1)
= (l + 1)al+1b0 + (
l−1∑
k=0
(k + 1)ak+1bl−k) + (
l∑
k=1
(l − k + 1)akbl−k+1) + (l + 1)a0bl+1
= (l + 1)al+1b0 + (
l∑
k=1
kakbl−k+1) + (
l∑
k=1
(l − k + 1)akbl−k+1) + (l + 1)a0bl+1
= (l + 1)al+1b0 + (
l∑
k=1
(l + 1)akbl−k+1) + (l + 1)a0bl+1
= (l + 1)(
l+1∑
k=0
akbl−k+1) = (l + 1)cl+1.
Since all their coefficients are equal, we conclude Dx(f(x)g(x)) = Dx(f(x))g(x) +Dx(g(x))f(x).
Exercise 13.5.2.
The polynomials x and x + 1 are the only (non-constant, i.e. 6= 0, 1) polynomials of degree 1 over F2;
they are clearly irreducible. A polynomial f(x) ∈ F2[x] of degree 2 is irreducible over F2 if and only
if it does not have a root in F2, that is, exactly when f(0) = f(1) = 1. Hence, the only irreducible
polynomial of degree 2 over F2 is x2 + x + 1. Now, for a polynomial f(x) ∈ F2[x] of degree 4 to be
irreducible, it must have no linear or quadratic factors. We can also apply the condition f(1) = f(0) = 1
to discard the ones with linear factors. Furthermore, f must have an odd number of terms (or it will
be 0), and must have constant term 1 (or x will be a factor). We are left with
x4 + x3 + x2 + x+ 1 x4 + x3 + 1
x4 + x2 + 1 x4 + x+ 1.
For any of this polynomials to be irreducible, it can’t be factorized as two quadratic irreducible factors.
Since x2+x+1 is the only irreducible polynomial of degree 2 over F2, only (x2+x+1)2 = x4+x2+1 of
this four is not irreducible. Hence, the irreducible polynomials of degree 4 over F2 are x4+x3+x2+x+1,
x4 + x3 + 1 and x4 + x+ 1.
Now, since x+ 1 = x− 1 in F2, we have (x+ 1)(x4 + x3 + x2 + x+ 1) = x5 − 1. We also calculate
(x2 + x+ 1)(x4 + x+ 1)(x4 + x3 + 1) = x10 + x5 + 1. So, the product of all this irreducible polynomials
is
x(x+ 1)(x2 + x+ 1)(x4 + x+ 1)(x4 + x3 + 1)(x4 + x3 + x2 + x+ 1)
= x(x5 − 1)(x10 + x5 + 1) = x16 − x.
Exercise 13.5.3.
We follow the hint. Suppose d divides n, so that n = qd for some q ∈ Z. Then xn − 1 = xqd − 1 =
(xd − 1)(xqd−d + xqd−2d + . . .+ xd + 1). So xd − 1 divides xn − 1.
Conversely, suppose d does not divide n. Then n = qd + r for some q, r ∈ Z with 0 < r < d. Thus
xn−1 = (xqd+r−xr)+(xr−1) = xr(xqd−1)+(xr−1) = xr(xd−1)(xqd−d+xqd−2d+. . .+xd+1)+(xr−1).
Since xd − 1 divides the first term, but doesn’t divide xr − 1 (as r < d), then xd − 1 does not divide
xn − 1.
Exercise 13.5.4.
The first assertion is analogous to the last exercise.
Now, Fpd is defined as the field whose pd elements are the roots of xp
d − x over Fp. Similarly is
defined Fpn . Take a = p. So, d divides n if and only if pd−1 divides pn−1, and that occurs exactly when
xp
d−1 − 1 divides xpn−1 − 1 (by last exercise). Thus, if d divides n, any root of xpd − x = x(xpd−1 − 1)
must be a root of xp
n − x = x(xpn−1 − 1), hence Fpd ⊆ Fpn . Conversely, if Fpd ⊆ Fpn , then xpd−1 − 1
divides xp
n−1 − 1, so d divides n.
13
13.5 Separable and Inseparable Extension
Exercise 13.5.5.
Let f(x) = xp−x+a. Let α be a root of f(x). First we prove f is separable. Since (α+1)p−(α+1)+a =
αp + 1 − α − 1 + a = 0, then α + 1 is also a root of f(x). This gives p distinct roots of f(x) given by
α+ k with k ∈ Fp, so f is separable.
Now we prove f is irreducible. Let f = f1f2 · · · fn where fi(x) ∈ Fp[x] is irreducible for all 1 ≤ i ≤ n.
Let 1 ≤ i < j ≤ n and let αi be a root of fi and αj be a root of fj , so that fi is the minimal polynomial
of αi and fj the minimal polynomial of αj . We prove deg fi = deg fj . Since αi is a root of fi, it is a
root of f , hence there exists k1 ∈ Fp such that αi = α + k1. Similarly, there exists k2 ∈ Fp such that
αj = α + k2. Thus, αi = αj + k1 − k2, so fi(x+ k1 − k2) is irreducible having αj as a root, so it must
be its minimal polynomial. Hence, fi(x+ k1 − k2) = fj(x), so deg fi = deg fj , as claimed. Since i and
j were arbitrary, then all fi are of the same degree, say q. Then p = deg f = nq, so n = 1 or n = p (as
p is prime). If n = p, then all roots of f are in Fp, so α ∈ Fp and thus 0 = αp − α+ a = a, contrary to
the hypothesis. Therefore, n = 1, so f is irreducible.
Exercise 13.5.6.
By definition, Fpn is the field whose pn elements are the roots of xp
n − x over Fp. Since xpn − 1 =
x(xp
n−1 − 1), clearly
xp
n−1 − 1 =
∏
α∈F×
pn
(x− α).
Set x = 0. Then
−1 =
∏
α∈F×
pn
(−α) = (−1)pn−1
∏
α∈F×
pn
α
⇒ (−1)pn−1(−1) = (−1)pn−1(−1)pn−1
∏
α∈F×
pn
α
⇒ (−1)pn =
∏
α∈F×
pn
α.
Hence, the product of the nonzero elements is +1 if p = 2 and −1 is p is odd. For p odd and n = 1, we
have
−1 =
∏
α∈F×p
α,
so taking module p we find [1][2] · · · [p− 1] = [−1], i.e., (p− 1)! ≡ −1 (mod p).
Exercise 13.5.7.
Let a ∈ K such that a 6= bp for every b ∈ K. Let f(x) = xp − a. We prove that f is irreducible and
inseparable. If α is a root of xp−a, then xp−a = (x−α)p, so α is a multiple root of f (with multiplicity
p), hence f is inseparable. Now, let g(x) be an irreducible factor of f(x). Note that α 6∈ K, otherwise
a = αp, contrary to the hypothesis. Then g(x) = (x−α)k for some k ≤ p. Using the binomial theorem,
we have
g(x) = (x− α)k = xk − kαxk−1 + · · ·+ (−α)k.
Therefore, kα∈ K. Since α 6∈ K, then k = p, so g = f . Hence, f is irreducible. We conclude that K(α)
is an inseparable finite extension of K.
Exercise 13.5.8.
Let f(x) = anxn + . . .+ a1x+ a0 ∈ Fp[x]. Since Fp has characteristic p, then (a+ b)p = ap + bp for any
a, b ∈ Fp. Easily we can generalize this to a finite number of terms, so that (x1+· · ·+xn)p = xp1+· · ·+xpn
for any x1, · · · , xn ∈ Fp. Furthermore, by Fermat’s Little Theorem, ap = a for every a ∈ Fp. So, over
Fp, we have
f(x)p = (anx
n + . . .+ a1x+ a0)
p = apnx
np + . . .+ ap1x
p + ap0 = anx
np + . . .+ a1x
p + a0 = f(x
p).
14
13.6 Cyclotomic Polynomials and Extensions
Exercise 13.5.9.
By the binomial theorem, we have
(1 + x)pn =
pn∑
i=0
(
pn
i
)
xi,
so the coefficient of xpi in the expansion of (1 + x)pn is
(
pn
pi
)
.
Since Fp has characteristic p, we have (1+x)pn = 1+xpn = (1+xp)n, so over Fp this is the coefficient
of (xp)i in (1 + xp)n. Furthermore, (1 + x)pn = (1 + xp)n implies
(1 + xp)n =
n∑
i=0
(
n
i
)
(xp)i =
pn∑
i=0
(
pn
k
)
xi = (1 + x)pn
over Fp, hence
(
pn
pi
) ≡ (ni) (mod p).
Exercise 13.5.10.
This is equivalent to prove that for any prime number p, we have f(x1, x2, . . . , xn)p = f(x
p
1, x
p
2, . . . , x
p
n)
in Fp[x1, x2, . . . , xn]. Let
f(x1, x2, . . . , xn) =
∑
γ1,...,γn=0
aγ1,...,γnx
γ1
1 . . . x
γn
n
be an element of Fp[x1, x2, . . . , xn].
Since Fp has characteristic p, then (x1 + · · · + xn)p = xp1 + · · · + xpn for any x1, · · · , xn ∈ Fp.
Furthermore, by Fermat’s Little Theorem, ap = a for every a ∈ Fp. Hence, over Fp we have
f(x1, x2, . . . , xn)
p = (
∑
aγ1,...,γnx
γ1
1 . . . x
γn
n )
p =
∑
(aγ1,...,γnx
γ1
1 . . . x
γn
n )
p
=
∑
apγ1,...,γn(x
γ1
1 . . . x
γn
n )
p =
∑
aγ1,...,γn(x
pγ1
1 . . . x
pγn
n ) = f(x
p
1, x
p
2, . . . , x
p
n).
Exercise 13.5.11.
Let f(x) ∈ F [x] with no repeated irreducible factors in F [x]. We can suppose f is monic. Then
f = f1f2 · · · fn for some monic irreducible polynomials fi(x) ∈ F [x]. Since F is perfect, f is separable,
hence all fi has distinct roots. Thus, f splits in linear factors in the closure of F , hence splits in linear
factors in the closure of K. Therefore, f(x) has no repeated irreducible factors in K[x].
13.6 Cyclotomic Polynomials and Extensions
Exercise 13.6.1.
Since (ζmζn)mn = 1, then ζmζn is an mnth root of unity. Now, let 1 ≤ k < mn. Then (ζmζn)k = ζkmζkn.
For a contradiction, suppose ζkmζkn = 1. Then, ζkm = 1 and ζkn = 1, hence m divides k and n divides k,
so mn divides k (as m 6= n). Since 1 ≤ k < mn, this is impossible. So, (ζmζn)k 6= 1 for all 1 ≤ k < mn
and thus the order of ζmζn is mn. Therefore, ζmζn generates the cyclic group of all mnth roots of unity,
that is, ζmζn is a primitive mnth root of unity.
Exercise 13.6.2.
Since (ζdn)(n/d) = ζnn = 1, then ζdn is an (n/d)th root of unity. Now let 1 ≤ k < (n/d). Then (ζdn)k = ζkdn .
Since 1 ≤ kd < n, then ζkdn 6= 1, so (ζdn)k 6= 1. Hence, the order of ζdn is (n/d), so it generates the cyclic
group of all (n/d)th roots of unity, that is, ζdn is a primitive (n/d)th root of unity.
Exercise 13.6.3.
Let F be a field that contains the nth roots of unity for n odd and let ζ be a 2nth root of unity. If
ζn = 1, then ζ ∈ F , so suppose ζn 6= 1. Since ζ2n = 1, then ζn is a root of x2 − 1. Since the roots of
this polynomial are 1 and −1, and ζn 6= 1, then ζn = −1. Hence, (−ζ)n = (−1)n(ζ)n = (−1)n+1 = 1
(since n is odd), so −ζ ∈ F . Since F is a field, then ζ ∈ F .
Exercise 13.6.4.
Let F be a field with char F = p. The roots of unity over F are the roots of xn−1 = xpkm−1 = (xm−1)pk ,
so are the roots of xm− 1. Now, since m is relatively prime to p, so is xm− 1 and its derivative mxm−1,
15
13.6 Cyclotomic Polynomials and Extensions
so xm− 1 has no multiple roots. Hence, the m different roots of xm− 1 are precisely the m distinct nth
roots of unity over F .
Exercise 13.6.5.
We use the inequality ϕ(n) ≥ √n/2 for all n ≥ 1. Let K be an extension of Q with infinitely many
roots of unity. Let N ∈ N. Then, there exits n ∈ N such that n > 4N2 and there exits some nth root
of unity ζ ∈ K. Then [K : Q] ≥ [Q(ζ) : Q] = ϕ(n) ≥ √n/2 > N . Since N was arbitrary, we conclude
that [K : Q] > N for all N ∈ N, so [K : Q] is infinite. Therefore, in any finite extension of Q there are
only a finite number of roots of unity.
Exercise 13.6.6.
Since Φ2n(x) and Φn(−x) are irreducible, they are the minimal polynomial of any of its roots. Thus, it
is sufficient to find a common root for both. Let ζ2 be the primitive 2th root of unity and let ζn be a
primitive nth root of unity. Note that ζ2 = −1, so that ζ2ζn = −ζn. Since n is odd, 2 and n are relatively
prime. So, by Exercise 1, ζ2ζn is a primitive 2nth root of unity, i.e, a root of Φ2n(x). Furthermore,
−ζn is clearly a root of Φn(−x). Thus, −ζn is a common root for both Φ2n(x) and Φn(−x), hence
Φ2n(x) = Φn(−x).
Exercise 13.6.7.
The Möbius Inversion Formula sates that if f(n) is defined for all nonnegative integers and F (n) =∑
d|n f(d), then f(n) =
∑
d|n µ(d)F (
n
d ). So lets start with the formula
xn − 1 =
∏
d|n
Φd(x).
We take natural logarithm in both sides and obtain
ln(xn − 1) = ln(
∏
d|n
Φd(x)) =
∑
d|n
ln Φd(x).
So, we use the Möbius Inversion Formula for f(n) = ln Φn(x) and F (n) = ln(xn − 1) to obtain
ln Φn(x) =
∑
d|n
µ(d) ln(xn/d − 1) =
∑
d|n
ln(xn/d − 1)µ(d).
Hence, taking exponentials we obtain
Φn(x) = exp(
∑
d|n
ln(xn/d − 1)µ(d)) =
∏
d|n
(xn/d − 1)µ(d) =
∏
d|n
(xd − 1)µ(n/d).
Exercise 13.6.8.
(a) Since p is prime, in Fp[x] we have (x− 1)p = xp − 1, so Φ`(x) = x`−1x−1 = (x−1)
`
x−1 = (x− 1)l−1.
(b) Note that ζ has order ` as being a primitive `th root of unity. Since pf ≡ 1 mod `, then pf − 1 = q`
for some integer q, hence ζp
f−1 = ζq` = 1, so ζ ∈ Fpf . Now we prove that f is the smallest integer
with that property. Suppose ζ ∈ Fpn for some n. Then ζ is a root of xpn−1 − 1, hence ` divides pn − 1
(see Exercise 13.5.3). Since f is the smallest power of p such that pf ≡ 1 mod `, is the smallest integer
such that ` divides pf − 1, so n ≥ l, as desired. This in fact proves that Fp(ζ) = Fpf , so the minimal
polynomial of ζ over Fp has degree f .
(c) Since ζa ∈ Fp(ζ), clearly Fp(ζa) ⊂ Fp(ζ). For the other direction we follow the hint. Let b the
the multiplicative inverse of a mod `, i.e ab ≡ 1 mod `. Then (ζa)b = ζ, so ζ ∈ Fp(ζa) and thus
Fp(ζ) ⊂ Fp(ζa). The equality follows.
Now, consider Φ`(x) as a polynomial over Fp[x]. Let ζi for 1 ≤ i ≤ ` be ` distinct primitive `th roots
of unity. The minimal polynomial of each ζi has degree f by part (b). Hence, the irreducible factors of
Φ`(x) have degree f . Since Φ` have degree ` − 1, then there must be `−1f factors, and all of them are
different since Φ`(x) is separable.
(d) If p = 7, then Φ7(x) = (x − 1)6 by part (a). If p ≡ 1 mod 7, then f = 1 in (b) and all roots have
degree 1, so Φ7(x) splits in distinct linear factors. If p ≡ 6 mod 7, then f = 2 is the smallest integer
16
13.6 Cyclotomic Polynomials and Extensions
such that pf = p2 ≡ 36 ≡ 1 mod 7, so we have 3 irreducible quadratics. If p ≡ 2, 4 mod 7, then f = 3 is
the smallest integer such that p3 ≡ 23, 43 ≡ 8, 64 ≡ 1 mod 7, so we have 2 irreducible cubics. Finally, if
p ≡ 3, 5 mod 7, then f = 6 is the smallest integer such that p6 ≡ 36, 56 ≡ 729, 15626 ≡ 1 mod 7, hence
we have an irreducible factor of degree 6.
Exercise 13.6.9.
Let A be an n by nmatrix over C for which Ak = I for some integer k ≥ 1. Then the minimal polynomial
of A divides xk − 1. Since we are working over C, there are k distinct roots of this polynomial, so the
minimal polynomial of A can be split in linear factors. Hence, A is diagonalizable.
Now consider A =
(
1 α
0 1
)
where α is an element of a field ofcharacteristic p.
Computing powers of A, we can prove (by induction) that An =
(
1 nα
0 1
)
for every positive integer
n. Since pα = 0, then Ap = I. Now, if A is diagonalizable, there exists some non-singular matrix P
such that A = PDP−1, where D is a diagonal matrix whose diagonal entries are the eigenvalues of A.
Since A has only one eigenvalue 1, then D = I, and thus A = PIP−1 = I. So, if A is diagonalizable, it
must be α = 0.
Exercise 13.6.10.
Let a, b ∈ Fpn . Then ϕ(a+ b) = (a+ b)p = ap + bp = ϕ(a) +ϕ(b), and ϕ(ab) = (ab)p = apbp = ϕ(a)ϕ(b),
so ϕ is and homomorphism. Moreover, if ϕ(a) = 0, then ap = 0 implies a = 0. Hence, ϕ is injective.
Since Fpn is finite, then ϕ is also surjective so it is an isomorphism. Furthermore, since every element
of Fpn is a root of xp
n − x, then ϕn(a) = apn = a for all a ∈ Fpn , so ϕn is the identity map. Now, let m
be an integer such that ϕm is the identity map. Then ap
m
= a for all a ∈ Fpn , so every element of Fpn
must be a root of xp
m − x. Hence, Fpn ⊂ Fpm and thus n divides m (Exercise 13.5.4), so n ≤ m.
Exercise 13.6.11.
Note that the minimal polynomial of ϕ is xn−1, for if ϕ satisfies some polynomial xn−1 + · · ·+a1x+a0
of degree n − 1 (or less) with coefficients in Fp, then xpn−1 + · · · + a1xp + a0 for all x ∈ Fpn , which
is impossible. Since Fpn has degree n as a vector space over Fp, then xn − 1 is also the characteristic
polynomial of ϕ, hence is the only invariant factor. Therefore, the rational canonical form of ϕ over Fp
is the companion matrix of xn − 1, which is
0 0 · · · 0 1
1 0 · · · 0 0
0 1 · · · 0 0
...
...
. . .
...
...
0 0 · · · 1 0
 .
Exercise 13.6.12.
We’ll work over the algebraic closure of Fpn , to ensure the field to contain all eigenvalues. In last exercise
we proved that the minimal and characteristic polynomial of ϕ is xn−1. Moreover, the eigenvalues of ϕ
are the nth roots of unity. We use Exercise 4 and write n = pkm for some prime p and some m relatively
prime to p, so that xn − 1 = (xm − 1)pk and we get exactly m distinct nth roots of unity, each one with
multiplicity pk. Since all the eigenvalues are zeros of both the minimal and characteristic polynomial of
multiplicity pk, we get m Jordan blocks of size pk. Now, fix a primitive mth root of unity, say ζ. Then,
each Jordan block each of the form
Ji =

ζi 1 0 · · · 0 0
0 ζi 1 · · · 0 0
0 0 ζi · · · 0 0
...
...
...
. . .
...
...
0 0 0 · · · ζi 1
0 0 0 · · · 0 ζi

17
13.6 Cyclotomic Polynomials and Extensions
for some 0 ≤ i ≤ m− 1. We already know the Jordan canonical form is given by
J0 0 · · · 0
0 J1 · · · 0
0 0 · · · 0
...
...
. . .
...
0 0 · · · Jm−1
 .
Exercise 13.6.13.
(a) Z is a division subring of D, and it is commutative by definition of the center, so Z is a field. Since
it is finite, its prime subfield is Fp for some prime p, so it is isomorphic to Fpm for some integer m. Let
q = pn, so that Z is isomorphic to Fq. Since D is a vector space over Z, then |D| = qn for some integer
n.
(b) Let x ∈ D× and let CD(x) be the set of the elements inD that commutes with x. Clearly Z ⊂ CD(x).
We prove that every element a ∈ CD(x) has an inverse in CD(x). Since a ∈ CD(x), then ax = xa.
Since D is a division ring, then a−1 ∈ D. Moreover, we have a−1ax = a−1xa and thus x = a−1xa, so
xa−1 = a−1x and a−1 ∈ CD(x). Hence, CD(x) is a division ring. Now, since Z ⊂ CD(x), then CD(x)
is a Z-vector space, so |CD(x)| = qm for some integer m. If x 6∈ Z, then CD(x) is a proper subset of D
and hence m < n.
(c) The class equation for the group D× is
|D×| = |Z(D×)|+
r∑
i=1
|D× : CD×(xi)|,
where the xi are the representatives of the distinct conjugacy class. By (a) we have |D×| = qn − 1,
|Z(D×)| = q− 1 and |CD×(xi)| = qmi − 1. Then |D× : CD×(xi)| = q
n − 1
|CD×(xi)|
=
qn − 1
qmi − 1 . Replacing in
the class equation we obtain
qn − 1 = (q − 1) +
r∑
i=1
qn − 1
|CD×(xi)|
= (q − 1) +
r∑
i=1
qn − 1
qmi − 1 .
(d) Since |D× : CD×(xi)| is an integer, then |D× : CD×(xi)| = q
n − 1
qmi − 1 is an integer. Hence, q
mi − 1
divides qn− 1, so (Exercise 13.5.4) mi divides n. Since mi < n (no xi is in Z), no mthi root of unity is a
nth root of unity. Therefore, as Φn(x) divides xn−1, it must divide (xn−1)/(xmi−1) for i = 1, 2, . . . , r..
Letting x = q we have Φn(q) divides (qn − 1)/(qmi − 1) for i = 1, 2, . . . , r.
(e) From (d), Φn(q) divides (qn − 1)/(qmi − 1) for i = 1, 2, . . . , r, so the class equation in (c) implies
Φn(q) divides q − 1. Now, let ζ 6= 1 be a nth root of unity. In the complex plane q is closer to 1 that ζ
is, so |q − ζ| > |q − 1| = q − 1. Since Φn(q) =
∏
ζ primitive(q − ζ) divides q − 1, this is impossible unless
n = 1. Hence, D = Z and D is a field.
Exercise 13.6.14.
We follow the hint. Let P (x) = xn + · · · + a1x + a0 be a monic polynomial of degree ≥ 1 over Z. For
a contradiction, suppose there are only finitely many primes dividing the values P (n), n = 1, 2, . . ., say
p1, p2, . . . , pk. Let N be an integer such that P (N) = a 6= 0. Let Q(x) = a−1P (N + ap1p2 . . . pkx).
Then, using the binomial theorem, we have
Q(x) = a−1P (N + ap1p2 . . . pkx)
= a−1((N + ap1p2 . . . pkx)n + · · ·+ a1(N + ap1p2 . . . pkx) + a0)
= a−1(Nn + an−1Nn−1 + · · ·+ a1N + a0 +R(x))
= a−1(P (N) +R(x))
= 1 + a−1R(x)
18
13.6 Cyclotomic Polynomials and Extensions
for some polynomial R(x) ∈ Z[x] divisible by ap1p2 . . . pk. Hence, Q(x) ∈ Z[x]. Moreover, for all n ∈ Z+,
P (N + ap1p2 . . . pkn) ≡ a (mod p1, p2, . . . , pk), so Q(n) = a−1P (N + ap1p2 . . . pkn) ≡ a−1a = 1 (mod
p1, p2, . . . , pk). Now let m be a positive integer such that |Q(m)| > 1, so that Q(m) ≡ 1 (mod pi) for
all i. Therefore, none of the pi’s divide Q(m). Since |Q(m)| > 1, there exists a prime q 6= pi for all i
such that q divides Q(m). Then q divides aQ(m) = P (N + ap1p2 . . . pkm), contradicting the fact that
only the primes p1, p2, . . . , pk divides the numbers P (1), P (2), . . ..
Exercise 13.6.15.
We follow the hint. Since Φm(a) ≡ 0 (mod p), then am ≡ 1 (mod p). Hence, there exist b such that
ba ≡ 1 mod p (indeed, b = am−1), so a is relatively prime to p. We prove that the order of a is m.
For a contradiction, suppose ad ≡ 1 (mod p) for some d dividing m, so that Φd(a) ≡ 0 (mod p) for
some d < m. Thus, a is a multiple root of xm − 1, so is also a root of its derivative mam−1. Hence,
mam−1 ≡ 0 mod p, impossible since p does not divide m nor a. Therefore, the order of a in (Z/pZ)× is
precisely m
Exercise 13.6.16.
Let p be an odd prime dividing Φm(a). If p does not divide m, then, by (c), a is relatively prime to p
and the order of a in F×p is m. Since |F×p | = p− 1, this implies m divides p− 1, that is, p ≡ 1 (mod m).
Exercise 13.6.17.
By Exercise 14, there are infinitely many primes dividing Φm(1),Φm(2),Φm(3), . . .. Since only finitely
of them can divide m, then, by Exercise 16, there must exists infinitely many primes p with p ≡ 1 (mod
m).
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	Field Theory
	Basic Theory and Field Extensions
	Algebraic Extensions
	Classical Straightedge and Compass Constructions
	Splitting Fields and Algebraic Closures
	Separable and Inseparable Extension
	Cyclotomic Polynomials and Extensions