algebra-13-dummit-1

Solutions to
Abstract Algebra
Chapter 13 - Field Theory
David S. Dummit & Richard M. Foote
Solutions by positrón0802
https://positron0802.wordpress.com
1 January 2021
Contents
13 Field Theory
13.1 Basic Theory and Field Extensions . . . . . . . . . .
13.2 Algebraic Extensions . . . . . . . . . . . . . . . . .
13.3 Classical Straightedge and Compass Constructions
13.4 Splitting Fields and Algebraic Closures . . . . . . .
13.5 Separable and Inseparable Extension . . . . . . . .
13.6 Cyclotomic Polynomials and Extensions . . . . . .
13
13.1
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1
1
5
12
13
15
19
Field Theory
Basic Theory and Field Extensions
Exercise 13.1.1. The polynomial 𝑝 (𝑥) = 𝑥 3 + 9𝑥 + 6 is irreducible in Z[𝑥] by Eisenstein Criterion
with 𝑝 = 3. By Gauss Lemma, it is thien rreducible in Q[𝑥]. To find (1 + 𝜃 ) −1, we apply the
Euclidean algorithm to 𝑝 (𝑥) and 1 + 𝑥 to find
𝑥 3 + 9𝑥 + 6 = (1 + 𝑥) (𝑥 2 − 𝑥 + 10) − 4.
Evaluating at 𝜃, we have (1 + 𝜃 ) (𝜃 2 − 𝜃 + 10) = 4. Therefore
(1 + 𝜃 ) −1 =
𝜃 2 − 𝜃 + 10
.
4
1
13.1
Basic Theory and Field Extensions
Exercise 13.1.2. Let 𝑓 (𝑥) = 𝑥 3 − 2𝑥 − 2. The polynomial 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 𝜃 3 = 2𝜃 + 2, so
that
(1 + 𝜃 ) (1 + 𝜃 + 𝜃 2 ) = 1 + 2𝜃 + 2𝜃 2 + 𝜃 3 = 3 + 4𝜃 + 2𝜃 2 .
1+𝜃
For computing
, first we compute (1 + 𝜃 + 𝜃 2 ) −1 . Applying the Euclidean algorithm, we
1 + 𝜃 + 𝜃2
obtain
𝑥 3 − 2𝑥 − 2 = (𝑥 2 + 𝑥 + 1) (𝑥 − 1) − 2𝑥 − 1
and
𝑥2 𝑥 7
9
𝑥 − 2𝑥 − 2 = (2𝑥 + 1)
− −
− .
2
4 8
8
3
Evaluating at 𝜃, from these equalities it follows that
2
(𝜃 + 𝜃 + 1) (𝜃 − 1) = 2𝜃 + 1
so that
and
(2𝜃 + 1)
−1
8 𝜃2 𝜃 7
=
,
− −
9 2
4 8
𝜃2 𝜃 7
8 2
(𝜃 + 𝜃 + 1) (𝜃 − 1) ( − − ) = 1.
9
2
4 8
Then
2
(𝜃 + 𝜃 + 1)
−1
2
8
2𝜃 2 𝜃 5
𝜃
𝜃 7
= (𝜃 − 1)
=−
− −
+ + ,
9
2
4 8
3
3 3
where we used 𝜃 3 = 2𝜃 + 2 again. It follows that
2𝜃 2 𝜃 5
𝜃 2 2𝜃 1
1+𝜃
=
(1
+
𝜃
)
−
=− +
+
+
+ .
2
1+𝜃 +𝜃
3
3 3
3
3 3
Exercise 13.1.3. Since 03 + 0 + 1 = 1 and 11 + 1 + 1 = 1 in F2, it follows that 𝑥 3 + 𝑥 + 1 is irreducible
over F2 . Since 𝜃 is root of 𝑥 3 + 𝑥 + 1, we have 𝜃 3 = −𝜃 − 1 = 𝜃 + 1. Hence, the powers of 𝜃 in F2 (𝜃 )
are
𝜃, 𝜃 2, 𝜃 3 = 𝜃 + 1, 𝜃 4 = 𝜃 2 + 𝜃, 𝜃 5 = 𝜃 2 + 𝜃 + 1, 𝜃 6 = 𝜃 2 + 1, and 𝜃 7 = 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),
2
Solutions by positrón0802
13.1
Basic Theory and Field Extensions
√
√
√
√
√
so 𝜑 is an homomorphism. If 𝜑 (𝑎 + 𝑏 2) = 𝜑 (𝑐 + 𝑑 2), then 𝑎 − 𝑏 2 =√𝑐 − 𝑑 2;√ as 2 ∉ Q,
this implies
∈ Q( 2), we have
√ 𝑐 = 𝑑. Thus 𝜑 is injective. Furthermore, given 𝑎 + 𝑏 2 √
√ 𝑎 = 𝑏 and
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) = 𝑥 𝑛 + · · · + 𝑎 1𝑥 + 𝑎 0 over Z,
with gcd(𝑝, 𝑞) = 1. Then
𝑛−1
𝑛
𝑝
𝑝
𝑝
+ 𝑎𝑛−1
+ · · · + 𝑎 1 + 𝑎 0 = 0,
𝑞
𝑞
𝑞
so that
𝑞(𝑎𝑛−1𝑝 𝑛−1 + · · · + 𝑎 1𝑝𝑞𝑛−2 + 𝑎 0𝑞𝑛−1 ) = −𝑝 𝑛 .
Thus, every prime that divides 𝑞 must divide 𝑝 𝑛 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
𝑎𝑛 𝛼 𝑛 + 𝑎𝑛−1𝛼 𝑛−1 + · · · + 𝑎 1𝛼 + 𝑎 0 = 0,
then
𝑛−1
(𝑎𝑛 𝛼)𝑛 + 𝑎𝑛−1 (𝑎𝑛 𝛼)𝑛−1 + 𝑎𝑛 𝑎𝑛−2 (𝑎𝑛 𝛼)𝑛−2 + · · · + 𝑎𝑛−2
𝑛 𝑎 1 (𝑎𝑛 𝛼) + 𝑎𝑛 𝑎 0
𝑛−1
𝑛−2
𝑛−1
= 𝑎𝑛𝑛 𝛼 𝑛 + 𝑎𝑛−1
+ 𝑎𝑛−1
+ · · · + 𝑎𝑛−1
𝑛 𝑎𝑛−1𝛼
𝑛 𝑎𝑛−2𝛼
𝑛 𝑎 1 𝛼 + 𝑎𝑛 𝑎 0
= 𝑎𝑛𝑛−1 (𝑎𝑛 𝛼 𝑛 + 𝑎𝑛−1𝛼 𝑛−1 + 𝑎𝑛−2𝛼 𝑛−2 + · · · + 𝑎 1𝛼 + 𝑎 0 ) = 0.
Exercise 13.1.7. Suppose 𝑥 3 − 𝑛𝑥 + 2 is reducible; then it must have a linear factor, hence a root.
By the Rational Root Theorem, if 𝛼 is a root of 𝑥 3 −𝑛𝑥 + 2, then 𝛼 must divide 2, so that 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 𝑥 3 − 𝑛𝑥 + 2 is
irreducible for 𝑛 ≠ −1, 3, 5.
Exercise 13.1.8. We subdivide this exercise in cases and subcases.
If 𝑥 5 − 𝑎𝑥 − 1 is reducible then it has a root (linear factor) or is a product of two irreducible
polynomials of degrees 2 and 3 respectively.
Case 1. If 𝑥 5 − 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, assume that there exist 𝑓 (𝑥) and 𝑔(𝑥) irreducible monic polynomials over Z of
degrees 2 and 3 respectively, such that 𝑥 5 − 𝑎𝑥 − 1 = 𝑓 (𝑥)𝑔(𝑥). Write 𝑓 (𝑥) = 𝑥 2 + 𝑏𝑥 + 𝑐 and
𝑔(𝑥) = 𝑥 3 + 𝑟𝑥 2 + 𝑠𝑥 + 𝑡, where 𝑏, 𝑐, 𝑟, 𝑠, 𝑡 ∈ Z. Then
𝑥 5 − 𝑎𝑥 − 1 = (𝑥 2 + 𝑏𝑥 + 𝑐) (𝑥 3 + 𝑟𝑥 2 + 𝑠𝑥 + 𝑡)
= 𝑥 5 + (𝑏 + 𝑟 )𝑥 4 + (𝑏𝑟 + 𝑐 + 𝑠)𝑥 3 + (𝑏𝑠 + 𝑐𝑟 + 𝑡)𝑥 2 + (𝑏𝑡 + 𝑐𝑠) + 𝑡𝑐.
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13.1
Basic Theory and Field Extensions
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) or (𝑐, 𝑡) = (1, −1), which gives 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.
Put 𝑏 = −𝑟 into the second and third equations to obtain −𝑟 2 − 1 + 𝑠 = 0 and −𝑟𝑠 − 𝑟 + 1 = 0, that
is, 𝑟 2 + 1 − 𝑠 = 0 and 𝑟𝑠 + 𝑟 − 1 = 0. Adding these last two equations we obtain 𝑟 2 + 𝑟𝑠 + 𝑟 − 𝑠 = 0.
Thus 𝑟 2 + 𝑟𝑠 + 𝑟 + 𝑠 = 2𝑠, so that (𝑟 + 1) (𝑟 + 𝑠) = 2𝑠. Now, from 𝑟 2 + 1 − 𝑠 = 0 we have 𝑟 2 = 𝑠 − 1, so
𝑟 2 + 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.
We deduce that (𝑐, 𝑡) = (−1, 1) is impossible.
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 𝑟 2 + 𝑟 − 1 = 0. By the Rational Root Theorem,
this equation has no roots in Z. Since 𝑟 ∈ Z, we have a contradiction. Now suppose 𝑏 = −1. From
𝑏 = −𝑟 we obtain 𝑟 = 1; thus, from 𝑏𝑟 + 1 + 𝑠 = 0 we obtain 𝑠 = 0. Finally, from −𝑏 + 𝑠 = −𝑎 it
follows that 𝑎 = −1. Therefore, we find the consistent solution (𝑏, 𝑐, 𝑟, 𝑠, 𝑡) = (−1, 1, 1, 0, −1) and
the factorisation
𝑥 5 − 𝑎𝑥 − 1 = (𝑥 2 + 𝑏𝑥 + 𝑐) (𝑥 3 + 𝑟𝑥 2 + 𝑠𝑥 + 𝑡) = (𝑥 2 − 𝑥 + 1) (𝑥 3 + 𝑥 2 − 1).
4
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13.2
13.2
Algebraic Extensions
Algebraic Extensions
Exercise 13.2.1. Since the characteristic of F is 𝑝, its prime subfield is (isomorphic to) F𝑝 = Z/𝑝Z.
Then F is a vector space over F𝑝 . Since F is finite, we have [F : F𝑝 ] = 𝑛 for some 𝑛 ∈ Z+ . It follows
that
|F| = |F𝑝 | [F:F𝑝 ] = 𝑝 𝑛 .
Exercise 13.2.2. Note that 𝑔 and ℎ are irreducible over both F2 and F3 . If 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 𝜃 2 if a
root of ℎ, then F2 (𝜃 2 ) F2 /(ℎ(𝑥)) has 8 elements and F3 (𝜃 2 ) 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 +1
0
𝑥 +1
𝑥
𝑥
𝑥
0
𝑥
𝑥 +1
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 +1
0
𝑥 +1
2𝑥+2
1
𝑥 +2
2𝑥
2
𝑥
2𝑥+1
𝑥
0
𝑥
2𝑥
2𝑥+1
1
𝑥 +1
𝑥 +2
2𝑥+2
2
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 non-zero elements.
Exercise 13.2.3. Since 1+𝑖 ∉ Q, its minimal polynomial is of degree at least 2. We try conjugation,
and obtain
(𝑥 − (1 + 𝑖)) (𝑥 − (1 − 𝑖)) = 𝑥 2 − 2𝑥 + 2,
which is irreducible by Eisenstein with 𝑝 = 2. Therefore, the minimal polynomial of 1 + 𝑖 over Q
is 𝑥 2 − 2𝑥 + 2.
√
√
√
√
Exercise 13.2.4.
First, note
that (2 + 3) 2 = 4 + 4 3 + 3 = 7 + 4 3. Let 𝜃 = 2 + 3. Then
√
√
2
𝜃 2 − 4𝜃 = 7 + 4 3 − 8 − 4 3 = −1, so 𝜃 is a root of 𝑥 2 − 4𝑥 + 1. Moreover,
√ 𝑥 − 4𝑥 + 1√is irreducible
2
over Q (because 𝜃 ∉ Q), so 𝑥 − 4𝑥 + 1 is the minimal polynomial of 2 + 3. Thus 2 + 3 has degree
2 over Q.
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Solutions by positrón0802
13.2
Algebraic Extensions
√
Now let 𝛼 = 3 2 and 𝛽 = 1 + 𝛼 + 𝛼 2 . 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
𝑥 3 −2 over Q, so [Q(𝛽) : Q] is either 1 or 3. For the sake of a contradiction suppose [Q(𝛽) : Q] = 1,
so that 𝛽 ∈ Q. Then
𝛽 2 = (1 + 𝛼 + 𝛼 2 ) 2 = 1 + 2𝛼 + 3𝛼 2 + 2𝛼 3 + 𝛼 4 = 5 + 4𝛼 + 3𝛼 2,
where we used 𝛼 3 = 2, and therefore
𝛽 2 − 3𝛽 = 5 + 4𝛼 + 3𝛼 2 − 3(1 + 𝛼 + 𝛼 2 ) = 2 + 𝛼 .
But then 𝛼 = −𝛽 2 + 3𝛽 − 2 ∈ Q(𝛽) = Q, a contradiction. It follows that [Q(𝛽) : Q] = 3.
Exercise 13.2.5. Since the polynomials have degree 3, if they were reducible they would have a
linear factor, hence a root√in 𝐹√. Note that√every element of 𝐹 has the form 𝑎 + 𝑏𝑖, where 𝑎, 𝑏 ∈ Q.
The roots of 𝑥 3 − 2 are 3 2, 𝜁 3 2 and 𝜁 2 3 2, where
𝜁 is the primitive 3rd root of unity, i.e., 𝜁 =
√
√
3
1
exp(2𝜋𝑖/3) = cos(2𝜋/3) + 𝑖 sin(2𝜋/3) = − 2 + 2 . Since 3 ∉ Q, none of these elements belong to
√ √
√
𝐹, so 𝑥 3 − 2 is irreducible over 𝐹 . Similarly, the roots of 𝑥 3 − 3 are 3 3, 𝜁 3 3 and 𝜁 2 3 3, and by the
same argument none of these elements belong to 𝐹 . Hence 𝑥 3 − 3 is irreducible over 𝐹 .
Exercise 13.2.6. We have to prove that 𝐹 (𝛼 1, . . . , 𝛼𝑛 ) is the smallest field containing 𝐹 (𝛼 1 ), . . . , 𝐹 (𝛼𝑛 ).
Clearly 𝐹 (𝛼𝑖 ) ⊂ 𝐹 (𝛼 1, . . . , 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 (𝛼 1, . . . , 𝛼𝑛 ), it has the form 𝜃 = 𝑎 1𝛼 1 + · · · + 𝑎𝑛 𝛼𝑛 , where 𝑎 1, . . . , 𝑎𝑛 ∈ 𝐹 .
As every 𝑎𝑖 𝛼𝑖 belongs to 𝐾, we have 𝜃 ∈ 𝐾 . Thus 𝐹 (𝛼 1, . . . , 𝛼𝑛 ) ⊂ 𝐾 . It follows that 𝐹 (𝛼 1, . . . , 𝛼𝑛 )
contains all of the 𝐹 (𝛼𝑖 ) and is contained in every field containing all of the 𝐹 (𝛼𝑖 ), so 𝐹 (𝛼 1, . . . , 𝛼𝑛 )
is the composite of the fields 𝐹 (𝛼 1 ), 𝐹 (𝛼 2 ), . . . , 𝐹 (𝛼𝑛 ).
√ √
√ √
√ √
√ √
Exercise 13.2.7. Since 2 + √3 is an
element
of
Q(
2,
3),
clearly
Q(
2
+
3)
⊂
Q(
2, 3). On
√
√
√
√
2
3
the other hand, consider 𝜃 = 2 + 3. Then 𝜃 = 5 + 2 6, and 𝜃 = 11 2 + 9 3, so
√
√
1
1
2 = (𝜃 3 − 9𝜃 ) and
3 = (11𝜃 − 𝜃 3 ).
2
2
√
√ √
√ √
√ √
√
Therefore
2 ∈ Q(𝜃 ) and
√ √
√ 3√∈ Q(𝜃 ), so Q(√ 2,√ 3) ⊂ Q( 2 + 3). It follows that Q( 2 + 3) =
Q( 2, 3), so that [Q( 2 + 3) : Q] = [Q( 2, 3) : Q] = 4.
We also have
√
√
𝜃 4 − 10𝜃 2 = (49 + 20 6) − 10(5 + 2 6) = −1, so 𝜃 4 − 10𝜃 2 + 1 = 0.
√ √
Since
[Q(
2 + 3) : Q] = 4, the polynomial 𝑥 4 − 10𝑥 2 + 1 is irreducible over Q, and is satisfied by
√ √
2 + 3.
√ √
Exercise 13.2.8. The elements of 𝐹 ( 𝐷 1, 𝐷 2 ) can be written as
p
p
p
𝑎 + 𝑏 𝐷 1 + 𝑐 𝐷 2 + 𝑑 𝐷 1 𝐷 2, where 𝑎, 𝑏, 𝑐, 𝑑 ∈ 𝐹 .
6
Solutions by positrón0802
13.2
Algebraic Extensions
We have
p p
p p
p
p
[𝐹 ( 𝐷 1, 𝐷 2 ) : 𝐹 ] = [𝐹 ( 𝐷 1, 𝐷 2 ) : 𝐹 ( 𝐷 1 )] [𝐹 ( 𝐷 1 ) : 𝐹 ].
√
√ √
√ √
Since [𝐹 ( 𝐷 1 )√: 𝐹 ] √
= 2, [𝐹 ( √𝐷 1, 𝐷 2 ) : 𝐹 ] can be either 2 or 4. Now [𝐹 ( 𝐷 1, 𝐷 2 ) : 𝐹 ] =√ 2 if
𝐷 1, 𝐷 2 ) : 𝐹 ( 𝐷 1 )] = 1, and that occurs exactly if 𝑥 2 − 𝐷 2 is reducible in 𝐹 ( 𝐷 1 )
and only
√ if [𝐹 ( √
(i.e., if 𝐷 2 ∈ 𝐹 ( 𝐷 1 )), that is, if there exists 𝑎, 𝑏 ∈ 𝐹 such that
p
p
(𝑎 + 𝑏 𝐷 1 ) 2 = 𝐷 2, so that 𝑎 2 + 2𝑎𝑏 𝐷 1 + 𝑏 2 𝐷 12 = 𝐷 2 .
√
Note that 𝑎𝑏 = 0, as 𝑎𝑏 ≠ 0 implies 𝐷 1 ∈ 𝐹, contrary to the assumption. Then 𝑎 = 0 or 𝑏 = 0.
If 𝑏 = 0, then 𝐷 2 is a square in 𝐹, contrary to the assumption. If 𝑎 = 0, then
𝑏 2 𝐷 1 = 𝐷 2, and thus
√
𝐷2 2
𝐷 1 𝐷 2 = ( 𝑏 ) , so 𝐷 1 𝐷 2 is a square in 𝐹 . Thus 𝑥 2 − 𝐷 2 is reducible in 𝐹 ( 𝐷 1 ) if and only if 𝐷 1 𝐷 2
is a square in 𝐹 . The result follows.
p √
√
√
√
√
Exercise 13.2.9. Suppose 𝑎 + 𝑏 = 𝑚 + 𝑛 for some 𝑚, 𝑛 ∈ 𝐹, so
that 𝑎 + 𝑏 = 𝑚 + 𝑛 + 2 𝑚𝑛.
p
√
√
√
√
√
Since 𝑏 is not a square in 𝐹, this means 𝑏 = 2 𝑚𝑛. We also have 𝑎 + 𝑏 − 𝑛 = 𝑚, so
q
√
√
√
√
𝑏 = 2 𝑛( 𝑎 + 𝑏 − 𝑛).
Hence,
q
√
√
𝑏 = 2 𝑛(𝑎 + 𝑏) − 2𝑛
√
√
⇒
( 𝑏 + 2𝑛) 2 = 4𝑛(𝑎 + 𝑏)
√
√
⇒ 𝑏 + 4𝑛 𝑏 + 4𝑛 2 = 4𝑛(𝑎 + 𝑏)
⇒
𝑏 + 4𝑛 2 − 4𝑛𝑎 = 0
√
4𝑎 ± 16𝑎 2 − 16𝑏
⇒
𝑛=
8
√
2𝑛
𝑎2 − 𝑏 = ± .
⇒
𝑎
√
2
Therefore, since 𝑎 and 𝑛 belong to 𝐹, so does 𝑎 − 𝑏.
√
2 − 𝑏 is a square in 𝐹, so that 𝑎 2 − 𝑏 ∈ 𝐹 . We prove that there exist
Conversely, assume
that
𝑎
p √
√
√
𝑚, 𝑛 ∈ 𝐹 such that 𝑎 + 𝑏 = 𝑚 + 𝑛. Consider
√
√
𝑎 + 𝑎2 − 𝑏
𝑎 − 𝑎2 − 𝑏
𝑚=
and 𝑛 =
.
2
2
p √
√
√
Note that 𝑚 and 𝑛 belong to 𝐹 as char(𝐹 ) ≠ 2. We claim 𝑎 + 𝑏 = 𝑚 + 𝑛. Indeed, we have
p √ p
√
√
√ !2
√
(𝑎 + 𝑏) + 2 𝑎 2 − 𝑏 + (𝑎 − 𝑏)
𝑎+ 𝑏+ 𝑎− 𝑏
𝑚=
=
,
4
2
7
Solutions by positrón0802
13.2
and
Thus
Algebraic Extensions
p √
p
√
√
√
√ !2
(𝑎 + 𝑏) − 2 𝑎 2 − 𝑏 + (𝑎 − 𝑏)
𝑎+ 𝑏− 𝑎− 𝑏
𝑛=
=
.
4
2
p √ p
p √
p
√
√
√
𝑎+ 𝑏+ 𝑎− 𝑏
𝑎+ 𝑏− 𝑎− 𝑏
and
𝑛=
,
2
2
p
p √ p
p √
√
√
q
√
√
√
𝑎+ 𝑏+ 𝑎− 𝑏
𝑎+ 𝑏− 𝑎− 𝑏
𝑚+ 𝑛=
+
= 𝑎 + 𝑏,
2
2
√
𝑚=
so
as claimed.
p √
Finally, we use this to determine when is the field
Q(
𝑎, 𝑏 ∈ Q, biquadratic over Q.
p √ 𝑎 + 𝑏),
√
√
√ √
2
If 𝑎 − 𝑏 is a squarepin Q and 𝑏 is not, we have Q( 𝑎 + 𝑏) = Q( 𝑚 + 𝑛) = Q( 𝑚, 𝑛), so by
√
Exercise 13.2.8, Q( 𝑎 + 𝑏) is biquadratic over Q when 𝑎 2 − 𝑏 is a square in Q and none of 𝑏, 𝑚,
𝑛 or 𝑚𝑛 is a square in Q. Since
√
√
𝑎 + 𝑎2 − 𝑏 𝑎 − 𝑎2 − 𝑏 𝑏
𝑚𝑛 =
= ,
2
2
4
p √
𝑚𝑛 is never a square when 𝑏 is not. Thus, Q( 𝑎 + 𝑏) is biquadratic over Q exactly when 𝑎 2 − 𝑏
is a square in Q and none of 𝑏, 𝑚 or 𝑛 is a square in Q.
p
p √
√
Exercise 13.2.10. Note that 3 + 2 2 = 3 + 8. Recalling Exercise 13.2.9 with 𝑎 = 3 and 𝑏 =
8,we have that 𝑎 2 − 𝑏 = 9 −p8 = 1 is a square in Q and 𝑏 = 8pis not. Hence, we find (𝑚 = 2 and
√
√
√
√
𝑛 = 1 from Exercisep13.2.9) 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 automorphism of
C, so it takes squares roots√to square roots. Furthermore, it maps the first quadrant onto 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 p3 + 4𝑖 and 3 − 4𝑖 are
√
√
conjugates to each other. Now we use Exercise 13.2.9. Note that 3 + 4𝑖 = 3 + −16. With 𝑎 = 3
2
and 𝑏 = −16, we have
√ is a square in Q and 𝑏 = −16 is not.
√ 𝑎 − 𝑏 = 25
√ 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.
p √
p
√
(b) Let 𝜃 = 1 + −3 + 1 − −3. Then
q
q
√
√
√
√
√
2
𝜃 = ( 1 + −3 + 1 − −3) 2 = (1 + −3) + (2 1 + 3) + (1 − −3) = 6.
Since 𝑥 2 − 6 is irreducible over Q (Eisenstein with 𝑝 = 2), it follows that 𝜃 has degree 2 over Q.
8
Solutions by positrón0802
13.2
Algebraic Extensions
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, √[Q(𝛼 1, . . . , 𝛼𝑘 ) : Q(𝛼 1, .√. . , 𝛼𝑘−1 )] is either
√3 1 or 2.
3
3
𝑚
Then [𝐹 : Q] = 2 for some 𝑚 ∈ N. Suppose 2 ∈ 𝐹 . Then
√ Q ⊂ Q( 2) ⊂ 𝐹, so [Q( 2) : Q]
divides [𝐹 : Q], that is, 3 divides 2𝑚 , a contradiction. Thus 3 2 ∉ 𝐹 .
Exercise 13.2.14. Since 𝛼 2 ∈ 𝐹 (𝛼), clearly 𝐹 (𝛼 2 ) ⊂ 𝐹 (𝛼). Thus we have to prove 𝛼 ∈ 𝐹 (𝛼 2 ). For
this purpose, consider the polynomial 𝑝 (𝑥) = 𝑥 2 − 𝛼 2, so that 𝑝 (𝛼) = 0. Note that 𝛼 ∈ 𝐹 (𝛼 2 ) if
and only if 𝑝 (𝑥) is reducible in 𝐹 (𝛼 2 ). For the sake of a contradiction, suppose 𝑝 (𝑥) is irreducible
in 𝐹 (𝛼 2 ), so that [𝐹 (𝛼) : 𝐹 (𝛼 2 )] = 2. Then
[𝐹 (𝛼) : 𝐹 ] = [𝐹 (𝛼) : 𝐹 (𝛼 2 )] [𝐹 (𝛼 2 ) : 𝐹 ] = 2[𝐹 (𝛼 2 ) : 𝐹 ],
so [𝐹 (𝛼) : 𝐹 ] is even, a contradiction. Therefore, 𝑝 (𝑥) is reducible in 𝐹 (𝛼 2 ) and 𝛼 ∈ 𝐹 (𝛼 2 ).
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, −1 can be expressed as a sum of
squares in 𝐹 (𝛼) 𝐹 [𝑥]/(𝑓 (𝑥)). Then, there exist polynomials 𝑝 1 (𝑥), . . . , 𝑝𝑚 (𝑥), 𝑔(𝑥) such that
−1 + 𝑓 (𝑥)𝑔(𝑥) = (𝑝 1 (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 side of the equation is less
than 4𝑘 +1, so deg 𝑔 < 2𝑘 +1 as well. From the equation −1+ 𝑓 (𝑥)𝑔(𝑥) = (𝑝 1 (𝑥)) 2 + · · · + (𝑝𝑚 (𝑥)) 2,
for proving that the degree of 𝑔 is odd it suffices to prove that degree of (𝑝 1 (𝑥)) 2 + · · · + (𝑝𝑚 (𝑥)) 2
is even. Let 𝑑 be the maximal degree over all 𝑝𝑖 , we prove that 𝑥 2𝑑 is the leading term of (𝑝 1 (𝑥)) 2 +
· · · + (𝑝𝑚 (𝑥)) 2 . Note that 𝑥 2𝑑 is a sum of squares (of the leading coefficients of the 𝑝𝑖 ’s of maximal
degree). Now, since 𝐹 is formally real, 0 cannot be expressed as a sum of squares in 𝐹 . (Indeed, if
Í𝑙
Í𝑙−1
2
2
2𝑑 ≠ 0, so the degree of (𝑝 (𝑥)) 2 + · · · + (𝑝 (𝑥)) 2 is
1
𝑚
𝑖=1 𝑎𝑖 = 0, then 𝑖=1 (𝑎𝑖 /𝑎𝑙 ) = −1.) Thus 𝑥
2𝑑, and therefore the degree of 𝑔 is odd. Then 𝑔 must contain an irreducible factor of odd degree,
say ℎ(𝑥). Since deg 𝑔 < deg 𝑓 , we have deg ℎ < deg 𝑓 as well. Let 𝛽 be a root of ℎ(𝑥), hence a
root of 𝑔(𝑥). Then
𝑓 (𝑥)𝑔(𝑥)
−1 + ℎ(𝑥)
= (𝑝 1 (𝑥)) 2 + · · · + (𝑝𝑚 (𝑥)) 2,
ℎ(𝑥)
so −1 is a square in 𝐹 [𝑥]/((ℎ(𝑥)) 𝐹 (𝛽), which means that 𝐹 (𝛽) is not formally real. It follows
that 𝛽 is a root of the odd degree polynomial ℎ such that 𝐹 (𝛽) is not formally real. Since deg ℎ <
deg 𝑓 , this contradicts the minimality of 𝛼 . The result follows.
9
Solutions by positrón0802
13.2
Algebraic Extensions
Exercise 13.2.16. Let 𝑟 ∈ 𝑅 be non-zero. Since 𝑟 is algebraic over 𝐹, there exists an irreducible
polynomial 𝑝 (𝑥) = 𝑎 0 + 𝑎 1𝑥 + · · · + 𝑥 𝑛 ∈ 𝐹 [𝑥] such that 𝑝 (𝑟 ) = 0. Note that 𝑎 0 ≠ 0 since 𝑝 is
𝑛−1 + · · · + 𝑎 ). Since 𝑎 ∈ 𝐹 ⊂ 𝑅 and 𝑟 ∈ 𝑅, we have 𝑟 −1 ∈ 𝑅.
irreducible. Then 𝑟 −1 = −𝑎 −1
1
𝑖
0 (𝑟
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, it follows that [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). Then
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 unique factorisation domain and 𝑘 (𝑡)
is its field of fractions, it follows from the Gauss Lemma that 𝑃 (𝑋 ) − 𝑡𝑄 (𝑋 ) is irreducible in
𝑘 ((𝑡)) [𝑋 ] if 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], it 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). Finally, 𝑥 is clearly a root of 𝑃 (𝑋 ) −𝑡𝑄 (𝑋 )
𝑃 (𝑥)
since 𝑃 (𝑥) − 𝑡𝑄 (𝑥) = 𝑃 (𝑥) −
𝑄 (𝑥) = 𝑃 (𝑥) − 𝑃 (𝑥) = 0.
𝑄 (𝑥)
(b) Let 𝑛 = max{deg 𝑃 (𝑥), deg 𝑄 (𝑥)}. Write
𝑃 (𝑥) = 𝑎𝑛 𝑥 𝑛 + · · · + 𝑎 1𝑥 + 𝑎 0
and 𝑄 (𝑥) = 𝑏𝑛 𝑥 𝑛 + · · · + 𝑏 1𝑥 + 𝑏 0,
where 𝑎𝑖 , 𝑏𝑖 ∈ 𝑘 for all 𝑖, so at least one of 𝑎𝑛 or 𝑏𝑛 is non-zero. The degree of 𝑃 (𝑋 )−𝑡𝑄 (𝑋 ) is clearly
≤ 𝑛, we shall prove it is precisely 𝑛. If either 𝑎𝑛 or 𝑏𝑛 is zero then clearly deg (𝑃 (𝑋 )−𝑡𝑄 (𝑋 )) = 𝑛, so
assume 𝑎𝑛 , 𝑏𝑛 ≠ 0. Then 𝑎𝑛 , 𝑏𝑛 ∈ 𝑘, but 𝑡 ∉ 𝑘, so it cannot be that 𝑎𝑛 = 𝑡𝑏𝑛 . Thus (𝑎𝑛 − 𝑡𝑏𝑛 )𝑋 𝑛 ≠ 0
and the degree of 𝑃 (𝑋 ) − 𝑡𝑄 (𝑋 ) is precisely 𝑛.
(c) Since 𝑃 (𝑋 ) − 𝑡𝑄 (𝑋 ) is irreducible over 𝑘 (𝑡) and 𝑥 is a root by part (a), it follows that
[𝑘 (𝑥) : 𝑘 (𝑡)] = deg 𝑃 (𝑋 ) − 𝑡𝑄 (𝑋 ), and this degree equals max{deg 𝑃 (𝑥), deg 𝑄 (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
an 𝐹 -linear transformation 𝑇𝛼 of 𝐾 . Denote by 𝑇𝛼 the matrix
of 𝑇𝛼 with respect to the basis
previously fixed. Then define 𝜑 : 𝐾 → 𝑀𝑛 (𝐹 ) by 𝜑 (𝛼) = 𝑇𝛼 . We claim that 𝜑 is an isomorphism
onto its image. Indeed, if 𝛼, 𝛽 ∈ 𝐾, then 𝑇 (𝛼+𝛽) (𝑘) = (𝛼 + 𝛽) (𝑘) = 𝛼𝑘 + 𝛽𝑘 = 𝑇𝛼 (𝑘) + 𝑇𝛽 (𝑘) for
every 𝑘 ∈ 𝐾, so 𝑇 (𝛼+𝛽) = 𝑇𝛼 + 𝑇𝛽 . We also have 𝑇 (𝛼 𝛽) (𝑘) = (𝛼𝛽) (𝑘) = 𝛼 (𝛽𝑘) = 𝑇𝛼 𝑇𝛽 (𝑘) for every
𝑘 ∈ 𝐾, so 𝑇 (𝛼 𝛽) = 𝑇𝛼 𝑇𝛽 . 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.
10
Solutions by positrón0802
13.2
Algebraic Extensions
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
𝐴) = 0 in 𝐾 . Therefore 𝑝 (𝛼) = 0.
√3 − 𝛼𝑘 = 0, so det(𝐼𝛼
√3 √−
3
Now, consider
the
field
Q(
2)
with
basis
{1,
2,
4}
Q. Denote the elements of this basis
√3
√3
√3
√over
√
3
by 𝑒 1 = 1, 𝑒 2 = 2 and 𝑒 3 = 4. Let 𝛼 = 2 and 𝛽 = 1 + 2 + 3 4. Then 𝛼 (𝑒 1 ) = 𝑒 2, 𝛼 (𝑒 2 ) = 𝑒 3 and
𝛼 (𝑒 3 ) = 2𝑒 1 . We also have 𝛽 (𝑒 1 ) = 𝑒 1 + 𝑒 2 + 𝑒 3, 𝛽 (𝑒 2 ) = 2𝑒 1 + 𝑒 2 + 𝑒 3 and 𝛽 (𝑒 3 ) = 2𝑒 1 + 2𝑒 2 + 𝑒 3 . Thus,
the associated matrices of the their linear transformations are, respectively,
0 0 2
©
ª
𝐴𝛼 = ­1 0 0®
«0 1 0¬
and
1 2 2
©
ª
𝐴 𝛽 = ­1 1 2 ® .
«1 1 1 ¬
The characteristic
polynomial of 𝐴𝛼 is 𝑥 3 − 2, hence is the monic polynomial of degree 3 satisfied
√3
by 𝛼 = 2. Furthermore, the characteristic polynomial of 𝐴𝛽 is 𝑥 3 − 3𝑥 2 − 3𝑥 − 1, hence is the
√ √
monic polynomial of degree 3 satisfied by 𝛽 = 1 + 3 2 + 3 4.
Exercise 13.2.21. The matrix
“multiplication
√ of the linear transformation √
√ 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
√
𝑎 𝑏𝐷
𝑎 𝑏𝐷
. Now let 𝜑 : 𝐾 → 𝑀2 (Q) be defined by 𝜑 (𝑎 + 𝑏 𝐷) =
.
the matrix is
𝑏 𝑎
𝑏 𝑎
We have
√
√
√
√
𝑐 𝑑𝐷
𝑎 𝑏𝐷
𝑎 + 𝑐 (𝑏 + 𝑑)𝐷
= 𝜑 (𝑎 + 𝑏 𝐷) + 𝜑 (𝑐 + 𝑑 𝐷),
+
=
𝜑 (𝑎 + 𝑏 𝐷 + 𝑐 + 𝑑 𝐷) =
𝑑 𝑐
𝑏 𝑎
𝑏 +𝑑
𝑎 +𝑐
and
√
√
𝑎𝑐 + 𝑏𝑑𝐷
𝜑 ((𝑎 + 𝑏 𝐷) · (𝑐 + 𝑑 𝐷)) =
𝑎𝑑 + 𝑏𝑐
(𝑎𝑑 + 𝑏𝑐)𝐷
𝑎 𝑏𝐷
𝑐 𝑑𝐷
=
·
𝑎𝑐 + 𝑏𝑑𝐷
𝑏 𝑎
𝑑 𝑐
√
√
= 𝜑 (𝑎 + 𝑏 𝐷)𝜑 (𝑐 + 𝑑 𝐷),
so 𝜑 is an homomorphism. Since 𝐾 is a field, its only ideals are {0} and 𝐾, so ker(𝜑) is either
trivial or all of 𝐾 . But 𝜑 (𝐾) is clearly non-zero, so ker(𝜑) ≠ 𝐾 and thus ker(𝜑) = {0}. Hence 𝜑 is
injective. It follows that 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, 𝑎 1, 𝑎 2 ∈ 𝐾 and 𝑏, 𝑏 1, 𝑏 2 ∈ 𝐾2 . Then
𝜑 ((𝑎 1, 𝑏) + (𝑎 2, 𝑏)) = 𝜑 (𝑎 1 + 𝑎 2, 𝑏) = (𝑎 1 + 𝑎 2 )𝑏 = 𝑎 1𝑏 + 𝑎 2𝑏 = 𝜑 (𝑎 1, 𝑏) + 𝜑 (𝑎 2, 𝑏),
and
𝜑 ((𝑎,1 𝑏) + (𝑎, 𝑏 2 )) = 𝜑 (𝑎, 𝑏 1 + 𝑏 2 ) = 𝑎(𝑏 1 + 𝑏 2 ) = 𝑎𝑏 1 + 𝑎𝑏 1 = 𝜑 (𝑎, 𝑏 1 ) + 𝜑 (𝑎, 𝑏 2 ).
We also have, for 𝑟 ∈ 𝐹, 𝜑 (𝑎𝑟, 𝑏) = (𝑎𝑟 )𝑏 = 𝑎(𝑟𝑏) = 𝜑 (𝑟𝑏). Thus 𝜑 is a 𝐹 -bilinear map, so it induces
an 𝐹 -algebra homomorphism Φ : 𝐾1 ⊗𝐹 𝐾2 → 𝐾1𝐾2 . We shall use Φ to prove both directions. Note
that 𝐾1 ⊗𝐹 𝐾2 have dimension [𝐾1 : 𝐹 ] [𝐾2 : 𝐹 ] as a vector space over 𝐹 .
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Solutions by positrón0802
13.3
Classical Straightedge and Compass Constructions
First, we assume [𝐾1𝐾2 : 𝐹 ] = [𝐾1 : 𝐹 ] [𝐾2 : 𝐹 ] and shall prove that 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 ), it follows that 𝐿 is a field (Exercise 13.2.16). Hence 𝐿 is a field containing both 𝐾1 and 𝐾2,
and since 𝐾1𝐾2 is the smallest such field (by definition), we have 𝐿 = 𝐾1𝐾2 . So Φ is surjective, as
claimed. It follows that 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 assume that 𝐾1 ⊗𝐹 𝐾2 is a field. In this case Φ is a field homomorphism, so it either
injective or trivial. It is clearly non-trivial since Φ(1 ⊗ 1) = 1, so it must be injective. Hence,
[𝐾1 : 𝐹 ] [𝐾2 : 𝐹 ] ≤ [𝐾1𝐾2 : 𝐹 ]. As we already have [𝐾1𝐾2 : 𝐹 ] ≤ [𝐾1 : 𝐹 ] [𝐾2 : 𝐹 ] (by Proposition
21 in the book), the equality follows.
13.3
Classical Straightedge and Compass Constructions
Exercise 13.3.1. Suppose that 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.
Then 𝛼 = ∠𝑄𝑃𝑂, 𝛽 = ∠𝑅𝑄𝑂, 𝛾 = ∠𝑄𝑅𝑂, and 𝜃 is an exterior angle of 4𝑃𝑅𝑂. Since 4𝑃𝑄𝑂
is isosceles, 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 . These two angles equal 𝛼, so 𝛽 = 2𝛼 . Now,
4𝑄𝑅𝑂 is isosceles, so 𝛽 = 𝛾 . Finally, since 𝜃 is an exterior angle of 4𝑃𝑅𝑂, it equals the sum of the
two remote interior angles, which are 𝛼 and 𝛾 . It follows that 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.
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Solutions by positrón0802
13.4
Splitting Fields and Algebraic Closures
From the figure, using similar triangles for (a), (b) and (c), and Pythagoras Theorem for (d),
the 4 relations are clear, that is,
√
√
𝑦
1 − 𝑘2
𝑏 +𝑘
1 − 𝑘2
𝑦=
, 𝑥 =𝑎
,
=
and (1 − 𝑘 2 ) + (𝑏 + 𝑘) 2 = (1 + 𝑎) 2 .
1+𝑎
1+𝑎
𝑥 −𝑘
3𝑘
√
3𝑘𝑦
Thus 𝑦 (1 + 𝑎) = 1 − 𝑘 2 = 𝑥−𝑘 , which implies 3𝑘 = (𝑥 − 𝑘) (1 + 𝑎). From the equation for 𝑥 above,
4𝑘+𝑘𝑎
we find 3𝑘 = ( 𝑎 (𝑏+𝑎)
1+𝑎 − 𝑘) (1 + 𝑎) = 𝑎(𝑏 + 𝑘) − 𝑘 (1 + 𝑎), so 𝑏 + 𝑘 =
𝑎 . Using this in the last
equation and reducing, we obtain
⇒
⇒
(1 − 𝑘 2 ) + (𝑏 + 𝑘) 2 = (1 + 𝑎) 2
2
4𝑘 + 𝑘𝑎
2
(1 − 𝑘 ) +
= (1 + 𝑎) 2
𝑎
𝑎 2 (1 − 𝑘 2 ) + (4𝑘 + 𝑘𝑎) 2 = 𝑎 2 (1 + 𝑎) 2
⇒ 𝑎 2 − (𝑘𝑎) 2 + (4𝑘) 2 + 8𝑘 2𝑎 + (𝑘𝑎) 2 = 𝑎 2 + 2𝑎 3 + 𝑎 4
⇒
𝑎 4 + 2𝑎 3 − 8𝑘 2𝑎 − 16𝑘 2 = 0.
We let 𝑎 = 2ℎ to obtain
ℎ 4 + ℎ 3 − 𝑘 2ℎ − 𝑘 2 = 0.
We find ℎ = 𝑘 2/3, so that 𝑎 = 2𝑘 2/3 . Finally, from 𝑏 = 4𝑘+𝑘𝑎
− 𝑘 we find 𝑏 = 2𝑘 1/3 . It follows that
𝑎
1/3
2/3
we can construct 2𝑘 and 2𝑘 using Conway’s construction.
Exercise 13.3.4. Let 𝑝 (𝑥) = 𝑥 3 + 𝑥 2 − 2𝑥 − 1 and 𝛼 = 2 cos(2𝜋/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, so [Q(𝛼) : 𝑄] cannot be
a power of 2. Since we cannot construct 𝛼, it follows that the regular 7-gon is not constructible
by straightedge and compass.
Exercise 13.3.5. Let 𝑝 (𝑥) = 𝑥 2 + 𝑥 − 1 = 0 and 𝛼 = 2 cos(2𝜋/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, we deduce that 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(2𝜋/5) is also constructible. Finally, as sin(2𝜋/5) =
p
1 − cos2 (2𝜋/5), sin(2𝜋/5) is also constructible. We conclude that the regular 5-gon is constructible by straightedge and compass.
13.4
Splitting Fields and Algebraic Closures
√ √ √
√
Exercise 13.4.1.√Let 𝑓 (𝑥) = 𝑥 4 −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). This field 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 𝑥 4 − √
2 over Q, we have [Q( 4 2) : Q] √= 4.
Furthermore,
𝑖 ∉ Q( 4 2), so 𝑥 2 √
+ 1 is irreducible over Q( 4 2) having 𝑖 as a root. So [Q(𝑖, 4 2) :
√4
Q( 2)] = 2 and therefore [Q(𝑖, 4 2) : Q] = 8.
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Solutions by positrón0802
13.4
Splitting Fields and Algebraic Closures
Exercise 13.4.2. Let 𝑓 (𝑥)√= 𝑥 4 + 2. Let 𝐾 be the splitting field of 𝑓 and let 𝐿 be the splitting field
of 𝑥 4 − 2, that is, 𝐿 = Q(𝑖, 4 2) (Exercise 13.4.1). We claim 𝐾 = 𝐿, so that [𝐾 : Q] = 8 by Exercise
13.4.1.
√
√
√
Let 𝜁 = 22 + 𝑖 22 . First we prove 𝜁 ∈ 𝐿 and 𝜁 ∈ 𝐾 . That 𝜁 ∈ 𝐿 is easy: let 𝜃 = 4 2. Since
√
√
𝜃 ∈ 𝐿, we have 𝜃 2 = 2 ∈ 𝐿. We
√ also have 𝑖 ∈ 𝐿, so 2, 𝑖 4 ∈ 𝐿 implies 𝜁 ∈ 𝐿. We 4now prove
𝜁 ∈ 𝐾 . We shall prove 𝑖 ∈ 𝐾 and 2 ∈ 𝐾 . Let 𝛼 be a root of 𝑥 + 2 and 𝛽 be a root of 𝑥 − 1. Then
(𝛼𝛽) 4 = 𝛼 4 𝛽 4 = −2, so 𝛼𝛽 is also a root of 𝑥 4 + 2. Since the roots of 𝑥 4 − 1 are ±1, ±𝑖, the roots of
𝑥 4 + 2 are ±𝛼 and ±𝑖𝛼 . Since 𝐾 is generated over Q by these roots, it follows that 𝑖𝛼/𝛼 = 𝑖 ∈ 𝐾 .
2
Now√let 𝛾 = 𝛼 2√∈ 𝐾 . As 𝛾 2 = 𝛼 4 = −2, we have that 𝛾 is a root of 𝑥 2 + 2. Since the roots
√ of 𝑥 + 2
are 𝑖 2 and −𝑖 2, 𝛾 must be one of this roots. In either case 𝛾/𝑖 ∈ 𝐾, which implies 2 ∈ 𝐾, and
therefore 𝜁 ∈ 𝐾 .
Now we prove 𝐿 = 𝐾 . On the one hand, let 𝛼 be a root of 𝑥 4 + 2 and 𝜃 be a root of 𝑥 4 − 2. Then
4
𝛼 = −2 and 𝜃 4 = 2. Note that 𝜁 2 = 𝑖, so 𝜁 4 = −1. Hence (𝜁 𝜃 ) 4 = 𝜁 4𝜃 4 = −2, so 𝜁 𝜃 is a root of 𝑥 4 +2.
Then, as we proved earlier, the roots of 𝑥 4 + 2 are ±𝜁 𝜃 and ±𝑖𝜁 𝜃 . We also have (𝜁 𝛼) 4 = 𝜁 4𝛼 4 = 2,
so 𝜁 𝛼 is a root of 𝑥 4 − 2. Then, by Exercise 13.4.1, the roots of 𝑥 4 − 2 are ±𝜁 𝛼 and ±𝑖𝜁 𝛼 . Now, since
𝜁 and 𝛼 are in 𝐾, we have 𝜁 𝛼 ∈ 𝐾 . We also have 𝑖 ∈ 𝐾, so all the roots of 𝑥 4 − 2 are in 𝐾 . Since 𝐿
is generated by these roots, it follows that 𝐿 ⊂ 𝐾 . Similarly, 𝜁 and 𝜃 belong to 𝐿, so 𝜁 𝜃 ∈ 𝐿; since
𝑖 ∈ 𝐿, all the roots of 𝑥 4 + 2 are in 𝐿. Since 𝐾 is generated by these roots, it follows that 𝐾 ⊂ 𝐿.
We conclude that 𝐾 = 𝐿; in particular, [𝐾 : Q] = 8.
Exercise 13.4.3.
Let 𝑓 (𝑥) = 𝑥√4 + 𝑥 2 + 1. Note that 𝑓 (𝑥) = (𝑥 2 + 𝑥 + 1) (𝑥 2 − 𝑥 + 1), so the roots of 𝑓
√
are ± 12 ± 𝑖 23 . Write 𝑤 = 21 − 𝑖 23 , so that these roots are precisely 𝑤, −𝑤, 𝑤, −𝑤, where 𝑤 denotes
the complex conjugate of 𝑤 . Hence, the splitting field of 𝑓 is Q(𝑤, 𝑤). Since 𝑤 + 𝑤 = 1, we have
Q(𝑤, 𝑤) = Q(𝑤). Furthermore, 𝑤 is a root of 𝑥 2 − 𝑥 + 1, which is irreducible over Q since 𝑤 ∉ Q.
Therefore, the degree of the splitting field of 𝑓 is [Q(𝑤) : Q] = 2.
√
Exercise
13.4.4.
Let 𝑓 (𝑥) = 𝑥 6 − 4. Note that 𝑓 (𝑥) = (𝑥 3 − 2) (𝑥 3 + 2). The roots of 𝑥 3 − 2 are 3 2,
√3
√
𝜁 2 and 𝜁 2 3 2, where
𝜁 denotes the primitive 3rd root of unity, i.e., 𝜁 = exp(2𝜋𝑖/3) = cos(2𝜋/3) +
√
√
√
√
3
1
𝑖 sin(2𝜋/3) = − 2 + 2 . Furthermore, the roots of 𝑥 3 + 2 are − 3 2, −𝜁 3 2 and −𝜁 2 3 2. Therefore, the
√
√
√
√
√
√
splitting field of 𝑓 is Q(𝜁 , 3 2). We have [Q(𝜁 , 3 2) : Q]
= [Q(𝜁 , 3 2) : Q( 3 2)] [Q( 3 2) : Q]. As 3 2
√
Furthermore, 𝜁 is a
is a root of the irreducible polynomial 𝑥 3 − 2 over
Q, 3 2 has degree 3 over Q. √
√
3
3
2
root of 𝑥 + 𝑥 + 1, which is irreducible over Q(
√3 2), so 𝜁 has degree 2 over Q( 2). It follows that
the degree of the splitting field of 𝑓 is [Q(𝜁 , 2) : Q] = 6.
Exercise 13.4.5. We follow the hint. First assume that 𝐾 is a splitting field over 𝐹 . Then 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 of 𝑓 over 𝐹 (𝑎), and 𝐾 (𝛽) is the splitting field of 𝑓 over 𝐹 (𝛽). Therefore,
∼
by Theorem 28, 𝜑 extends to an isomorphism 𝜎 : 𝐾 (𝛼) −
→ 𝐾 (𝛽). Since 𝐾 = 𝐾 (𝛼), we have
[𝐾 : 𝐹 ] = [𝐾 (𝛼) : 𝐹 ] = [𝐾 (𝛽) : 𝐹 ], and therefore 𝐾 = 𝐾 (𝛽). Thus 𝛽 ∈ 𝐾 .
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Separable and Inseparable Extension
Conversely, assume that every irreducible polynomial in 𝐹 [𝑥] that has a root in 𝐾 splits completely in 𝐾 [𝑥]. Since [𝐾 : 𝐹 ] is finite, we have 𝐾 = 𝐹 (𝛼 1, . . . , 𝛼𝑛 ) for some 𝛼 1, . . . , 𝛼𝑛 . For every
1 ≤ 𝑖 ≤ 𝑛, let 𝑝𝑖 be the minimal polynomial of 𝛼𝑖 over 𝐹, and let 𝑓 = 𝑝 1𝑝 2 · · · 𝑝𝑛 . Since every 𝛼𝑖
is in 𝐾, every 𝑝𝑖 has a 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 be 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 . Then 𝑓1 𝑓2 splits completely in 𝐾1𝐾2 and 𝐾1𝐾2 is generated over 𝐹 by its
roots, hence is the splitting field of 𝑓1 𝑓2 (𝑥) ∈ 𝐹 [𝑥].
(b) We follow the hint. By Exercise 13.4.5, we shall prove that every irreducible polynomial in
𝐹 [𝑥] that has a root in 𝐾1 ∩ 𝐾2 splits completely in (𝐾1 ∩ 𝐾2 ) [𝑥]. Thus, let 𝑓 (𝑥) be an irreducible
polynomial in 𝐹 [𝑥] that has a root, say 𝛼, in 𝐾1 ∩ 𝐾2 . By Exercise 13.4.5, 𝑓 splits completely in
𝐾1 and splits completely in 𝐾2 . Since 𝐾1 and 𝐾2 are contained in 𝐾, by the uniqueness of the
factorisation of 𝑓 in 𝐾, the roots of 𝑓 in 𝐾1 must coincide with its roots in 𝐾2 . It follows that 𝑓
splits completely in (𝐾1 ∩ 𝐾2 ) [𝑥].
13.5
Separable and Inseparable Extension
Exercise 13.5.1. Let 𝑓 (𝑥) = 𝑎𝑛 𝑥 𝑛 + · · · + 𝑎 1𝑥 + 𝑎 0 and 𝑔(𝑥) = 𝑏𝑚 𝑥 𝑚 + · · · + 𝑏 1𝑥 + 𝑏 0 be two
polynomials. We may assume, without any loss of generality, that 𝑛 ≥ 𝑚. Thus, we can write
𝑔(𝑥) = 𝑏𝑛 𝑥 𝑛 + · · · + 𝑏 1𝑥 + 𝑏 0, where some of the last coefficients 𝑏𝑖 could be zero. We have
𝑓 (𝑥) + 𝑔(𝑥) = (𝑎𝑛 + 𝑏𝑛 )𝑥 𝑛 + · · · + (𝑎 1 + 𝑏 1 )𝑥 + (𝑎 0 + 𝑏 0 ), so
𝐷𝑥 (𝑓 (𝑥) + 𝑔(𝑥)) = 𝑛(𝑎𝑛 + 𝑏𝑛 )𝑥 𝑛−1 + · · · + 2(𝑎 2 + 𝑏 2 )𝑥 + (𝑎 1 + 𝑏 1 ) = 𝐷𝑥 (𝑓 (𝑥)) + 𝐷𝑥 (𝑔(𝑥)).
Í
Now, for ℓ = 1, . . . , 2𝑛, set 𝑐𝑛 = 𝑛𝑘=0 𝑎𝑘 𝑏𝑛−𝑘 , where we also set 𝑎𝑖 = 0 and 𝑏𝑖 = 0 if 𝑖 > 𝑛. Then
!
! 𝑛
2𝑛 Õ
ℓ
2𝑛
𝑛
Õ
Õ
Õ
Õ
𝑗
𝑖
𝑏𝑗𝑥 =
( 𝑎𝑘 𝑏 ℓ−𝑘 )𝑥 ℓ =
𝑐ℓ 𝑥 ℓ ,
𝑎𝑖 𝑥
𝑓 (𝑥)𝑔(𝑥) =
𝑗=0
𝑖=0
so that
𝐷𝑥 (𝑓 (𝑥)𝑔(𝑥)) = 𝐷𝑥
ℓ=0 𝑘=0
2𝑛
Õ
!
𝑐ℓ 𝑥
ℓ
=
ℓ=0
2𝑛−1
Õ
ℓ=0
(ℓ + 1)𝑐 ℓ+1𝑥 ℓ .
ℓ=0
Thus, the coefficient of 𝑥 ℓ in 𝐷𝑥 (𝑓 (𝑥)𝑔(𝑥)) is (ℓ + 1)𝑐 ℓ+1 . On the other hand, we have
! 𝑛
! 2𝑛−1 ℓ
!
𝑛−1
Õ
Õ Õ
Õ
𝑘
𝑘
(𝑘 + 1)𝑎𝑘+1𝑏 ℓ−𝑘 𝑥 ℓ
(𝑘 + 1)𝑎𝑘+1𝑥
𝑏𝑘 𝑥 =
𝐷𝑥 (𝑓 (𝑥)) 𝑔(𝑥) =
𝑘=0
ℓ=0
𝑘=0
𝑘=0
and
𝑓 (𝑥)𝐷𝑥 (𝑔(𝑥)) =
𝑛
Õ
𝑘=0
!
𝑎𝑘 𝑥 𝑘
𝑛−1
Õ
!
(𝑘 + 1)𝑏𝑘+1𝑥 𝑘 =
2𝑛−1
Õ
(
ℓ
Õ
𝑎𝑘 (ℓ − 𝑘 + 1)𝑏 ℓ−𝑘+1 )𝑥 ℓ ,
ℓ=0 𝑘=0
𝑘=0
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13.5
Separable and Inseparable Extension
so the coefficient of 𝑥 ℓ in 𝐷𝑥 (𝑓 (𝑥))𝑔(𝑥) + 𝐷𝑥 (𝑔(𝑥)) 𝑓 (𝑥) is
!
!
ℓ
ℓ
Õ
Õ
(𝑘 + 1)𝑎𝑘+1𝑏 ℓ−𝑘 +
(ℓ − 𝑘 + 1)𝑎𝑘 𝑏 ℓ−𝑘+1
𝑘=0
𝑘=0
= (ℓ + 1)𝑎 ℓ+1𝑏 0 +
= (ℓ + 1)𝑎 ℓ+1𝑏 0 +
= (ℓ + 1)𝑎 ℓ+1𝑏 0 +
ℓ−1
Õ
𝑘=0
ℓ
Õ
𝑘=1
ℓ
Õ
!
(𝑘 + 1)𝑎𝑘+1𝑏 ℓ−𝑘 +
ℓ
Õ
!
(ℓ − 𝑘 + 1)𝑎𝑘 𝑏 ℓ−𝑘+1 + (ℓ + 1)𝑎 0𝑏 ℓ+1
𝑘=1
!
𝑘𝑎𝑘 𝑏 ℓ−𝑘+1 +
ℓ
Õ
!
(ℓ − 𝑘 + 1)𝑎𝑘 𝑏 ℓ−𝑘+1 + (ℓ + 1)𝑎 0𝑏 ℓ+1
𝑘=1
!
(ℓ + 1)𝑎𝑘 𝑏 ℓ−𝑘+1 + (ℓ + 1)𝑎 0𝑏 ℓ+1
𝑘=1
= (ℓ + 1)
ℓ+1
Õ
!
𝑎𝑘 𝑏 ℓ−𝑘+1 = (ℓ + 1)𝑐 ℓ+1 .
𝑘=0
We deduce that 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. ≠ 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 if 𝑓 (0) = 𝑓 (1) = 1.
Hence, the only irreducible polynomial of degree 2 over F2 is 𝑥 2 + 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 (otherwise it would be 0), and must have constant term 1
(otherwise 𝑥 would be a factor). We are left with
𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1,
𝑥 4 + 𝑥 2 + 1,
𝑥 4 + 𝑥 3 + 1,
𝑥 4 + 𝑥 + 1.
For any of these polynomials to be irreducible, it cannot be factorised as a product of two quadratic
irreducible factors. Since 𝑥 2 + 𝑥 + 1 is the only irreducible polynomial of degree 2 over F2, only
(𝑥 2 + 𝑥 + 1) 2 = 𝑥 4 + 𝑥 2 + 1 of these four is not irreducible. Hence, the irreducible polynomials of
degree 4 over F2 are precisely 𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1, 𝑥 4 + 𝑥 3 + 1 and 𝑥 4 + 𝑥 + 1.
Now, since 𝑥 + 1 = 𝑥 − 1 in F2, we have (𝑥 + 1) (𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1) = 𝑥 5 − 1. We also have
2
(𝑥 + 𝑥 + 1) (𝑥 4 + 𝑥 + 1) (𝑥 4 + 𝑥 3 + 1) = 𝑥 10 + 𝑥 5 + 1. It follows that the product of all these irreducible
polynomials is
𝑥 (𝑥 + 1) (𝑥 2 + 𝑥 + 1) (𝑥 4 + 𝑥 + 1) (𝑥 4 + 𝑥 3 + 1) (𝑥 4 + 𝑥 3 + 𝑥 2 + 𝑥 + 1)
= 𝑥 (𝑥 5 − 1) (𝑥 10 + 𝑥 5 + 1) = 𝑥 16 − 𝑥 .
Exercise 13.5.3. We follow the hint. First assume that 𝑑 divides 𝑛, so that 𝑛 = 𝑞𝑑 for some 𝑞 ∈ Z+ .
Then 𝑥 𝑛 − 1 = 𝑥 𝑞𝑑 − 1 = (𝑥 𝑑 − 1) (𝑥 𝑞𝑑−𝑑 + 𝑥 𝑞𝑑−2𝑑 + . . . + 𝑥 𝑑 + 1), so 𝑥 𝑑 − 1 divides 𝑥 𝑛 − 1.
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Separable and Inseparable Extension
Conversely, assume that 𝑑 does not divide 𝑛. Then 𝑛 = 𝑞𝑑 + 𝑟 for some 𝑞 ∈ Z ≥0 and 0 < 𝑟 < 𝑑,
so that
𝑥 𝑛 −1 = (𝑥 𝑞𝑑+𝑟 −𝑥 𝑟 )+(𝑥 𝑟 −1) = 𝑥 𝑟 (𝑥 𝑞𝑑 −1)+(𝑥 𝑟 −1) = 𝑥 𝑟 (𝑥 𝑑 −1) (𝑥 𝑞𝑑−𝑑 +𝑥 𝑞𝑑−2𝑑 +· · ·+𝑥 𝑑 +1)+(𝑥 𝑟 −1).
Since 𝑥 𝑑 − 1 divides the first term, but does nt divide 𝑥 𝑟 − 1 (as 𝑟 < 𝑑), it follows that 𝑥 𝑑 − 1 does
not divide 𝑥 𝑛 − 1.
Exercise 13.5.4. The first assertion follows analogously as in Exercise 13.5.3. Now, F𝑝𝑑 is defined
𝑑
as the field whose 𝑝 𝑑 elements are the roots of 𝑥 𝑝 − 𝑥 over F𝑝 , and similarly F𝑝 𝑛 . Take 𝑎 = 𝑝.
𝑑
Thus, 𝑑 divides 𝑛 if and only if 𝑝 𝑑 − 1 divides 𝑝 𝑛 − 1, and that occurs exactly when 𝑥 𝑝 −1 − 1 divides
𝑑
𝑛
𝑑
𝑥 𝑝 −1 − 1 (by Exercise 13.5.3). Thus, if 𝑑 divides 𝑛, any root of 𝑥 𝑝 − 𝑥 = 𝑥 (𝑥 𝑝 −1 − 1) must be a
𝑛
𝑑
𝑛
root of 𝑥 𝑝 − 𝑥 = 𝑥 (𝑥 𝑝 −1 − 1). Hence F𝑝𝑑 ⊆ F𝑝 𝑛 . Conversely, if F𝑝𝑑 ⊆ F𝑝 𝑛 , then 𝑥 𝑝 −1 − 1 divides
𝑛
𝑥 𝑝 −1 − 1, so 𝑑 divides 𝑛.
Exercise 13.5.5. Let 𝑓 (𝑥) = 𝑥 𝑝 − 𝑥 + 𝑎. Let 𝛼 be a root of 𝑓 (𝑥). First we prove 𝑓 is separable.
Since (𝛼 + 1) 𝑝 − (𝛼 + 1) + 𝑎 = 𝛼 𝑝 + 1 − 𝛼 − 1 + 𝑎 = 0, it follows that 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 = 𝑓1 𝑓2 · · · 𝑓𝑛 where 𝑓𝑖 (𝑥) ∈ F𝑝 [𝑥] is irreducible for all
1 ≤ 𝑖 ≤ 𝑛. Let 1 ≤ 𝑖 < 𝑗 ≤ 𝑛, let 𝛼𝑖 be a root of 𝑓𝑖 and 𝛼 𝑗 be a root of 𝑓 𝑗 , so that 𝑓𝑖 is the minimal
polynomial of 𝛼𝑖 and 𝑓 𝑗 the minimal polynomial of 𝛼 𝑗 . We prove deg 𝑓𝑖 = deg 𝑓 𝑗 . Since 𝛼𝑖 is a
root of 𝑓𝑖 , it is a root of 𝑓 , hence there exists 𝑘 1 ∈ F𝑝 such that 𝛼𝑖 = 𝛼 + 𝑘 1 . Similarly, there exists
𝑘 2 ∈ F𝑝 such that 𝛼 𝑗 = 𝛼 + 𝑘 2 . Thus 𝛼𝑖 = 𝛼 𝑗 + 𝑘 1 − 𝑘 2, so 𝑓𝑖 (𝑥 + 𝑘 1 − 𝑘 2 ) is irreducible having
𝛼 𝑗 as a root, so it must be its minimal polynomial. It follows that 𝑓𝑖 (𝑥 + 𝑘 1 − 𝑘 2 ) = 𝑓 𝑗 (𝑥), so
deg 𝑓𝑖 = deg 𝑓 𝑗 , as claimed. Since 𝑖 and 𝑗 were arbitrary, all the 𝑓𝑖 are of the same degree, say 𝑞.
Then 𝑝 = deg 𝑓 = 𝑛𝑞, so we must have either 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 = 𝛼 𝑝 − 𝛼 + 𝑎 = 𝑎, contrary to the assumption. Therefore
𝑛 = 1, so 𝑓 is irreducible.
𝑛
Exercise 13.5.6. By definition, F𝑝 𝑛 is the field whose 𝑝 𝑛 elements are the roots of 𝑥 𝑝 − 𝑥 over
𝑛
𝑛
F𝑝 . Since 𝑥 𝑝 − 1 = 𝑥 (𝑥 𝑝 −1 − 1), clearly
Ö
𝑛
𝑥 𝑝 −1 − 1 =
(𝑥 − 𝛼).
𝛼 ∈F𝑝×𝑛
Setting 𝑥 = 0, we have
−1 =
Ö
(−𝛼) = (−1) 𝑝
𝑛 −1
𝛼 ∈F𝑝×𝑛
⇒
(−1) 𝑝
𝑛 −1
Ö
𝛼
𝛼 ∈F𝑝×𝑛
(−1) = (−1) 𝑝
𝑛 −1
(−1) 𝑝
𝑛 −1
Ö
𝛼
𝛼 ∈F𝑝×𝑛
⇒
𝑛
(−1) 𝑝 =
Ö
𝛼.
𝛼 ∈F𝑝×𝑛
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Separable and Inseparable Extension
It follows that the product of the non-zero elements of F𝑝 𝑛 is +1 if 𝑝 = 2 and −1 is 𝑝 is odd.
For 𝑝 odd and 𝑛 = 1 we have
Ö
𝛼,
−1 =
𝛼 ∈F𝑝×
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 ≠ 𝑏 𝑝 for every 𝑏 ∈ 𝐾 . Let 𝑓 (𝑥) = 𝑥 𝑝 − 𝑎. We prove that 𝑓
is irreducible and inseparable. If 𝛼 is a root of 𝑥 𝑝 − 𝑎, then 𝑥 𝑝 − 𝑎 = (𝑥 − 𝛼) 𝑝 , so 𝛼 is a multiple
root of 𝑓 (with multiplicity 𝑝) and hence 𝑓 is inseparable. Now let 𝑔(𝑥) be an irreducible factor
of 𝑓 (𝑥). Note that 𝛼 ∉ 𝐾, otherwise 𝑎 = 𝛼 𝑝 , contrary to the assumption. Then 𝑔(𝑥) = (𝑥 − 𝛼)𝑘 for
some 𝑘 ≤ 𝑝. Using the binomial theorem, we have
𝑔(𝑥) = (𝑥 − 𝛼)𝑘 = 𝑥 𝑘 − 𝑘𝛼𝑥 𝑘−1 + · · · + (−𝛼)𝑘 ,
so that 𝑘𝛼 ∈ 𝐾 . Since 𝛼 ∉ 𝐾, it follows that 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) = 𝑎𝑛 𝑥 𝑛 + . . . + 𝑎 1𝑥 + 𝑎 0 ∈ F𝑝 [𝑥]. Since F𝑝 has characteristic 𝑝, we have
(𝑎 + 𝑏) 𝑝 = 𝑎𝑝 + 𝑏 𝑝 for any 𝑎, 𝑏 ∈ F𝑝 . We can easily generalise this to a finite number of terms,
𝑝
𝑝
so that (𝑐 1 + · · · + 𝑐𝑛 ) 𝑝 = 𝑐 1 + · · · + 𝑐𝑛 for any 𝑐 1, . . . , 𝑐𝑛 ∈ F𝑝 . Furthermore, by Fermat’s Little
𝑝
Theorem, 𝑎 = 𝑎 for every 𝑎 ∈ F𝑝 . Thus, over F𝑝 , we have
𝑝
𝑝
𝑝
𝑓 (𝑥) 𝑝 = (𝑎𝑛 𝑥 𝑛 + · · · + 𝑎 1𝑥 + 𝑎 0 ) 𝑝 = 𝑎𝑛 𝑥 𝑛𝑝 + · · · + 𝑎 1 𝑥 𝑝 + 𝑎 0 = 𝑎𝑛 𝑥 𝑛𝑝 + · · · + 𝑎 1𝑥 𝑝 + 𝑎 0 = 𝑓 (𝑥 𝑝 ).
Exercise 13.5.9. From the binomial theorem we have
𝑝𝑛 Õ
𝑝𝑛 𝑖
(1 + 𝑥) 𝑝𝑛 =
𝑥,
𝑖
𝑖=0
so the coefficient of 𝑥 𝑝𝑖 in the expansion of (1 +𝑥) 𝑝𝑛 is 𝑝𝑛
𝑝𝑖 . Since F𝑝 has characteristic 𝑝, we have
𝑝 𝑖
𝑝 𝑛
(1 + 𝑥) 𝑝𝑛 = 1 + 𝑥 𝑝𝑛 = (1 + 𝑥 𝑝 )𝑛 , so over F𝑝 we have that 𝑝𝑛
𝑝𝑖 is the coefficient of (𝑥 ) in (1 + 𝑥 ) .
𝑝𝑛
𝑝
𝑛
Furthermore, (1 + 𝑥) = (1 + 𝑥 ) implies
𝑝 𝑛
(1 + 𝑥 ) =
𝑛 Õ
𝑛
𝑝 𝑖
(𝑥 ) =
𝑝𝑛 Õ
𝑝𝑛
𝑖
𝑖=0
𝑖=0
over F𝑝 , so that
𝑝𝑛 𝑝𝑖
≡
𝑛
𝑖
𝑥 𝑖 = (1 + 𝑥) 𝑝𝑛
𝑘
(mod 𝑝).
Exercise 13.5.10. This is equivalent to proving that for any prime number 𝑝, we have 𝑓 (𝑥 1, 𝑥 2, . . . , 𝑥𝑛 ) 𝑝 =
𝑝 𝑝
𝑝
𝑓 (𝑥 1 , 𝑥 2 , . . . , 𝑥𝑛 ) in F𝑝 [𝑥 1, 𝑥 2, . . . , 𝑥𝑛 ]. Let
Õ
𝛾
𝛾
𝑓 (𝑥 1, 𝑥 2, . . . , 𝑥𝑛 ) =
𝑎𝛾1,...,𝛾𝑛 𝑥 1 1 . . . 𝑥𝑛𝑛
𝛾 1 ,...,𝛾𝑛 =0
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Cyclotomic Polynomials and Extensions
be an arbitrary element of F𝑝 [𝑥 1, 𝑥 2, . . . , 𝑥𝑛 ]. Since F𝑝 has characteristic 𝑝, we have (𝑐 1 +· · ·+𝑐𝑛 ) 𝑝 =
𝑝
𝑝
𝑐 1 + · · · + 𝑐𝑛 for any 𝑐 1, · · · , 𝑐𝑛 ∈ F𝑝 . Furthermore, by Fermat’s Little Theorem, 𝑎𝑝 = 𝑎 for every
𝑎 ∈ F𝑝 . Hence, over F𝑝 ,
Õ
Õ
Õ
𝛾
𝛾 𝑝
𝛾
𝛾
𝑝𝛾
𝑝𝛾
𝑓 (𝑥 1, 𝑥 2, . . . , 𝑥𝑛 ) 𝑝 =
𝑎𝛾1,...,𝛾𝑛 𝑥 1 1 . . . 𝑥𝑛𝑛 =
(𝑎𝛾1,...,𝛾𝑛 𝑥 1 1 . . . 𝑥𝑛𝑛 ) 𝑝 =
𝑎𝛾1,...,𝛾𝑛 (𝑥 1 1 . . . 𝑥𝑛 𝑛 )
𝑝
𝑝
𝑝
= 𝑓 (𝑥 1 , 𝑥 2 , . . . , 𝑥𝑛 ).
Exercise 13.5.11. Let 𝑓 (𝑥) ∈ 𝐹 [𝑥] have no repeated irreducible factors in 𝐹 [𝑥]. We may assume
that 𝑓 is monic. Then 𝑓 = 𝑓1 𝑓2 · · · 𝑓𝑛 for some monic irreducible polynomials 𝑓𝑖 (𝑥) ∈ 𝐹 [𝑥]. Since 𝐹
is perfect, 𝑓 is separable, hence all the 𝑓𝑖 have distinct roots. Thus, 𝑓 splits in linear factors in the
closure of 𝐹, hence splits in linear factors in the closure of 𝐾 . It follows that 𝑓 (𝑥) has no repeated
irreducible factors in 𝐾 [𝑥].
13.6
Cyclotomic Polynomials and Extensions
Exercise 13.6.1. Since (𝜁𝑚 𝜁𝑛 )𝑚𝑛 = 1, it follows that 𝜁𝑚 𝜁𝑛 is an 𝑚𝑛 th root of unity. Now assume
that for some positive integer 𝑘 we have (𝜁𝑚 𝜁𝑛 )𝑘 = 1. Then (𝜁𝑚 )𝑘𝑛 = (𝜁𝑚 )𝑘𝑛 (𝜁𝑛 )𝑘𝑛 = 𝜁 𝑘𝑛 = 1,
so that 𝑚 divides 𝑘𝑛. Since 𝑚 and 𝑛 are relatively prime, it follows that 𝑚 divides 𝑘. Similarly, 𝑛
divides 𝑘. Thus 𝑘 is a common multiple of 𝑚 and 𝑛; since they are coprime, it follows that 𝑘 is a
multiple of 𝑚𝑛. We conclude that 𝜁𝑚 𝜁𝑛 is a primitive 𝑚𝑛 th root of unity.
Exercise 13.6.2. Since (𝜁𝑛𝑑 ) (𝑛/𝑑) = 𝜁𝑛𝑛 = 1, it follows that 𝜁𝑛𝑑 is an (𝑛/𝑑) th root of unity. Now let
1 ≤ 𝑘 < (𝑛/𝑑). Then (𝜁𝑛𝑑 )𝑘 = 𝜁𝑛𝑘𝑑 . Since 1 ≤ 𝑘𝑑 < 𝑛, we have 𝜁𝑛𝑘𝑑 ≠ 1, so (𝜁𝑛𝑑 )𝑘 ≠ 1. Hence, the
order of 𝜁𝑛𝑑 is precisely (𝑛/𝑑), so it generates the cyclic group of all (𝑛/𝑑) th roots of unity, that is,
𝜁𝑛𝑑 is a primitive (𝑛/𝑑) th root of unity.
Exercise 13.6.3. Let 𝐹 be a field containing the 𝑛 th roots of unity for 𝑛 odd and let 𝜁 be a 2𝑛 th
root of unity. If 𝜁 𝑛 = 1, then 𝜁 ∈ 𝐹, so assume 𝜁 𝑛 ≠ 1. Since 𝜁 2𝑛 = 1, we have that 𝜁 𝑛 is
a root of 𝑥 2 − 1. The roots of this polynomial are 1 and −1, and 𝜁 𝑛 ≠ 1, so 𝜁 𝑛 = −1. Hence,
(−𝜁 )𝑛 = (−1)𝑛 (𝜁 )𝑛 = (−1)𝑛+1 = 1 (since 𝑛 is odd), so −𝜁 ∈ 𝐹 . Since 𝐹 is a field, we deduce that
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
𝑘
𝑘
𝑥 𝑛 − 1 = 𝑥 𝑝 𝑚 − 1 = (𝑥 𝑚 − 1) 𝑝 , so are the roots of 𝑥 𝑚 − 1. Now, since 𝑚 is relatively prime to
𝑝, so is 𝑥 𝑚 − 1 and its derivative 𝑚𝑥 𝑚−1, so 𝑥 𝑚 − 1 has no multiple roots. Hence, the 𝑚 different
roots of 𝑥 𝑚 − 1 are precisely the 𝑚 distinct 𝑛 th roots of unity over 𝐹 .
√
Exercise 13.6.5. We use the inequality 𝜑 (𝑛) ≥ 𝑛/2 for all 𝑛 ≥ 1, where 𝜑 denotes the Euler’s
phi-function. Let 𝐾 be an extension of Q with infinitely many roots of unity. Let 𝑁 ∈ N. Then
there exists 𝑛 ∈ N such that 𝑛 > 4𝑁 2 and there exists some 𝑛 th root of unity 𝜁 ∈ 𝐾 . Thus
√
[𝐾 : Q] ≥ [Q(𝜁 ) : Q] = 𝜑 (𝑛) ≥ 𝑛/2 > 𝑁 . Since 𝑁 was arbitrary, we deduce that [𝐾 : Q] > 𝑁
for all 𝑁 ∈ N, so [𝐾 : Q] is infinite. It follows that in any finite extension of Q there are only a
finite number of roots of unity.
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Cyclotomic Polynomials and Extensions
Exercise 13.6.6. Since Φ2𝑛 (𝑥) and Φ𝑛 (−𝑥) are irreducible, they are the minimal polynomial of
any of its roots. Thus, it suffices to find a common root of both. Let 𝜁 2 = −1 be the primitive 2th
root of unity and let 𝜁𝑛 be a primitive 𝑛 th root of unity, so that 𝜁 2𝜁𝑛 = −𝜁𝑛 . Since 𝑛 is odd, 2 and
𝑛 are relatively prime. Thus, by Exercise 13.6.1, 𝜁 2𝜁𝑛 is a primitive 2𝑛 th 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 of both Φ2𝑛 (𝑥)
and Φ𝑛 (−𝑥), so Φ2𝑛 (𝑥) = Φ𝑛 (−𝑥).
Exercise 13.6.7. The Möbius Inversion Formula states that if 𝑓 (𝑛) is defined for all non-negative
Í
Í
integers and 𝐹 (𝑛) = 𝑑 |𝑛 𝑓 (𝑑), then 𝑓 (𝑛) = 𝑑 |𝑛 𝜇 (𝑑)𝐹 ( 𝑑𝑛 ). So let us start with the formula
Ö
𝑥𝑛 − 1 =
Φ𝑑 (𝑥).
𝑑 |𝑛
We take natural logarithm in both sides and obtain
©Ö
ª Õ
ln Φ𝑑 (𝑥).
Φ𝑑 (𝑥) ® =
ln(𝑥 𝑛 − 1) = ln ­
𝑑
|𝑛
𝑑
|𝑛
«
¬
Thus, we use the Möbius Inversion Formula for 𝑓 (𝑛) = ln Φ𝑛 (𝑥) and 𝐹 (𝑛) = ln(𝑥 𝑛 − 1) to obtain
Õ
Õ
ln Φ𝑛 (𝑥) =
𝜇 (𝑑) ln(𝑥 𝑛/𝑑 − 1) =
ln(𝑥 𝑛/𝑑 − 1) 𝜇 (𝑑) .
𝑑 |𝑛
𝑑 |𝑛
Taking exponentials we deduce
Ö
©Õ
ª Ö 𝑛/𝑑
Φ𝑛 (𝑥) = exp ­ ln(𝑥 𝑛/𝑑 − 1) 𝜇 (𝑑) ® =
(𝑥
− 1) 𝜇 (𝑑) =
(𝑥 𝑑 − 1) 𝜇 (𝑛/𝑑) .
𝑑 |𝑛
« 𝑑 |𝑛
¬ 𝑑 |𝑛
Exercise 13.6.8. (a) Since 𝑝 is prime, in F𝑝 [𝑥] we have (𝑥 − 1) 𝑝 = 𝑥 𝑝 − 1, so
Φℓ (𝑥) =
𝑥 ℓ−1
(𝑥 − 1) ℓ
=
= (𝑥 − 1) ℓ−1 .
𝑥 −1
𝑥 −1
(b) Note that 𝜁 has order ℓ, being a primitive ℓ th root of unity. Since 𝑝 𝑓 ≡ 1 mod ℓ, we have
𝑓
− 1 = 𝑞ℓ for some integer 𝑞, so that 𝜁 𝑝 −1 = 𝜁 𝑞ℓ = 1 and hence 𝜁 ∈ F𝑝 𝑓 . Now we prove that 𝑓 is
𝑛
the smallest integer with this property. Suppose 𝜁 ∈ F𝑝 𝑛 for some 𝑛. Then 𝜁 is a root of 𝑥 𝑝 −1 − 1,
so ℓ divides 𝑝 𝑛 − 1 (see Exercise 13.5.3). Since 𝑓 is the smallest power of 𝑝 such that 𝑝 𝑓 ≡ 1 mod
ℓ, it is the smallest integer such that ℓ divides 𝑝 𝑓 − 1, so 𝑛 ≥ 𝑙, as desired. This in fact proves that
F𝑝 (𝜁 ) = F𝑝 𝑓 , so the minimal polynomial of 𝜁 over F𝑝 has degree 𝑓 .
𝑝𝑓
(c) Since 𝜁 𝑎 ∈ F𝑝 (𝜁 ), clearly F𝑝 (𝜁 𝑎 ) ⊂ F𝑝 (𝜁 ). For the other direction we follow the hint. Let
𝑏 be the multiplicative inverse of 𝑎 mod ℓ, i.e 𝑎𝑏 ≡ 1 mod ℓ. Then (𝜁 𝑎 )𝑏 = 𝜁 , so 𝜁 ∈ F𝑝 (𝜁 𝑎 ) and
thus F𝑝 (𝜁 ) ⊂ F𝑝 (𝜁 𝑎 ). It follows that F𝑝 (𝜁 𝑎 ) = F𝑝 (𝜁 ).
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Cyclotomic Polynomials and Extensions
Now, consider Φℓ (𝑥) as a polynomial over F𝑝 [𝑥]. Let 𝜁𝑖 , for 1 ≤ 𝑖 ≤ ℓ, be ℓ distinct primitive ℓ th
roots of unity. The minimal polynomial of each 𝜁𝑖 has degree 𝑓 by part (b). Hence, the irreducible
factors of Φℓ (𝑥) have degree 𝑓 . Since Φℓ have degree ℓ − 1, there must be ℓ−1
𝑓 factors, and all of
them are different since Φℓ (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 such that 𝑝 𝑓 = 𝑝 2 ≡ 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 𝑝 3 ≡ 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
𝑝 6 ≡ 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 × 𝑛 matrix over C for which 𝐴𝑘 = 𝐼 for some integer 𝑘 ≥ 1. Then
the minimal polynomial of 𝐴 divides 𝑥 𝑘 − 1. Since we are working over C, there are 𝑘 distinct
roots of this polynomial, so the minimal polynomial of 𝐴 splits into linear factors. Thus 𝐴 is
diagonalisable.
1 𝛼
, where 𝛼 is an element of a field of characteristic 𝑝.
Now consider 𝐴 =
0 1
1 𝑛𝛼
𝑛
for every positive integer
Computing powers of 𝐴, inductively it follows that 𝐴 =
0 1
𝑛. Since 𝑝𝛼 = 0, we have 𝐴𝑝 = 𝐼 . Now, if 𝐴 is diagonalizable, then 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 1 as its only, 𝐷 must be the identity and therefore also 𝐴. That is, if
𝐴 is diagonalisable, we must have 𝛼 = 0.
Exercise 13.6.10. For 𝑎, 𝑏 ∈ F𝑝 𝑛 we have 𝜑 (𝑎 + 𝑏) = (𝑎 + 𝑏) 𝑝 = 𝑎𝑝 + 𝑏 𝑝 = 𝜑 (𝑎) + 𝜑 (𝑏), and
𝜑 (𝑎𝑏) = (𝑎𝑏) 𝑝 = 𝑎𝑝 𝑏 𝑝 = 𝜑 (𝑎)𝜑 (𝑏), so 𝜑 is a homomorphism. Moreover, if 𝜑 (𝑎) = 0, then 𝑎𝑝 = 0
implies 𝑎 = 0, so 𝜑 is injective. Since F𝑝 𝑛 is finite, 𝜑 is also surjective and hence an isomorphism.
𝑛
𝑛
Furthermore, since every element of F𝑝 𝑛 is a root of 𝑥 𝑝 − 𝑥, we have 𝜑 𝑛 (𝑎) = 𝑎𝑝 = 𝑎 for all
𝑎 ∈ F𝑝 𝑛 , so 𝜑 𝑛 is the identity map. Now, let 𝑚 be a positive integer such that 𝜑 𝑚 is the identity
𝑚
𝑚
map. Then 𝑎𝑝 = 𝑎 for all 𝑎 ∈ F𝑝 𝑛 , so every element of F𝑝 𝑛 must be a root of 𝑥 𝑝 − 𝑥 . Hence,
F𝑝 𝑛 ⊂ 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 𝑥 𝑛 − 1, for if 𝜑 satisfies some polyno𝑛−1
mial 𝑥 𝑛−1 + · · · + 𝑎 1𝑥 + 𝑎 0 of degree 𝑛 − 1 (or less) with coefficients in F𝑝 , then 𝑥 𝑝 + · · · + 𝑎 1𝑥 𝑝 + 𝑎 0
for all 𝑥 ∈ F𝑝 𝑛 , which is impossible. Since F𝑝 𝑛 has degree 𝑛 as a vector space over F𝑝 , it follows
that 𝑥 𝑛 − 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 𝑥 𝑛 − 1, which is
0
©
­1
­
­0
­.
­.
­.
«0
0 ···
0 ···
1 ···
.. . .
.
.
0 ···
21
0 1
ª
0 0®
®
0 0® .
.. .. ®®
. .®
1 0¬
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13.6
Cyclotomic Polynomials and Extensions
Exercise 13.6.12. We will work over the algebraic closure of F𝑝 𝑛 , to ensure the field contains all
eigenvalues. In Exercise 13.6.11 we proved that the minimal and characteristic polynomial of 𝜑 is
𝑥 𝑛 − 1. Moreover, the eigenvalues of 𝜑 are the 𝑛 th roots of unity. We use Exercise 13.6.4 and write
𝑘
𝑛 = 𝑝 𝑘 𝑚 for some prime 𝑝 and some 𝑚 relatively prime to 𝑝, so that 𝑥 𝑛 − 1 = (𝑥 𝑚 − 1) 𝑝 and
we get exactly 𝑚 distinct 𝑛 th roots of unity, each one of multiplicity 𝑝 𝑘 . Since all the eigenvalues
are zeros of both the minimal and characteristic polynomial of multiplicity 𝑝 𝑘 , we get 𝑚 Jordan
blocks of size 𝑝 𝑘 . Now, fix a primitive 𝑚 th root of unity, say 𝜁 . Then each Jordan block has the
form
𝜁𝑖 1 0 · · · 0 0
©
ª
­ 0 𝜁𝑖 1 · · · 0 0 ®
­
®
­ 0 0 𝜁𝑖 · · · 0 0 ®
­
𝐽𝑖 = ­ .. .. .. . .
.. .. ®®
.
. .®
­. . .
­
®
­ 0 0 0 · · · 𝜁𝑖 1 ®
𝑖
«0 0 0 ··· 0 𝜁 ¬
for some 0 ≤ 𝑖 ≤ 𝑚 − 1. Finally, we already know the Jordan canonical form is given by
𝐽0
©
­0
­
­0
­.
­.
­.
«0
0 ···
0
ª
𝐽1 · · ·
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 centre,
so 𝑍 is a field. Since it is finite, its prime subfield is F𝑝 for some prime 𝑝, so 𝑍 is isomorphic to
F𝑝 𝑛 for some integer 𝑛. Let 𝑞 = 𝑝 𝑛 . Since 𝐷 is a vector space over 𝑍, we have |𝐷 | = 𝑞𝑛 for some
integer 𝑛.
(b) Let 𝑥 ∈ 𝐷 × and let 𝐶𝐷 (𝑥) be the set of the elements in 𝐷 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),
we have 𝑎𝑥 = 𝑥𝑎, and since 𝐷 is a division ring, we also have 𝑎 −1 ∈ 𝐷. Moreover, 𝑎 −1𝑎𝑥 = 𝑎 −1𝑥𝑎
and thus 𝑥 = 𝑎 −1𝑥𝑎, so 𝑥𝑎 −1 = 𝑎 −1𝑥 and 𝑎 −1 ∈ 𝐶𝐷 (𝑥). Therefore 𝐶𝐷 (𝑥) is a division ring. As
𝑍 ⊂ 𝐶𝐷 (𝑥), it follows that 𝐶𝐷 (𝑥) is a 𝑍 -vector space, so |𝐶𝐷 (𝑥)| = 𝑞𝑚 for some integer 𝑚. If
𝑥 ∉ 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
Õ
|𝐷 × : 𝐶𝐷 × (𝑥𝑖 )|,
𝑖=1
where the 𝑥𝑖 are representatives of the distinct conjugacy classes in 𝐷 × not contained in the
centre of 𝐷 × . By (a) we have |𝐷 × | = 𝑞𝑛 − 1, |𝑍 (𝐷 × )| = 𝑞 − 1 and |𝐶𝐷 × (𝑥𝑖 )| = 𝑞𝑚𝑖 − 1. Then
22
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Cyclotomic Polynomials and Extensions
|𝐷 × : 𝐶𝐷 × (𝑥𝑖 )| =
𝑞𝑛 − 1
𝑞𝑛 − 1
. Plugging these values in the class equation we obtain
= 𝑚
|𝐶𝐷 × (𝑥𝑖 )| 𝑞 𝑖 − 1
𝑞𝑛 − 1 = (𝑞 − 1) +
𝑟
𝑟
Õ
Õ
𝑞𝑛 − 1
𝑞𝑛 − 1
= (𝑞 − 1) +
.
|𝐶𝐷 × (𝑥𝑖 )|
𝑞𝑚𝑖 − 1
𝑖=1
𝑖=1
𝑞𝑛 − 1
is also an integer. Hence
𝑞𝑚𝑖 − 1
𝑞𝑚𝑖 − 1 divides 𝑞𝑛 − 1, so (Exercise 13.5.4) 𝑚𝑖 divides 𝑛. Since 𝑚𝑖 < 𝑛 (no 𝑥𝑖 is in 𝑍 ), no 𝑚𝑖th root of
unity is a 𝑛 th root of unity. Therefore, as Φ𝑛 (𝑥) divides 𝑥 𝑛 − 1, it must divide (𝑥 𝑛 − 1)/(𝑥 𝑚𝑖 − 1)
for 𝑖 = 1, 2, . . . , 𝑟 .. Letting 𝑥 = 𝑞 we deduce that Φ𝑛 (𝑞) divides (𝑞𝑛 − 1)/(𝑞𝑚𝑖 − 1) for 𝑖 = 1, 2, . . . , 𝑟 .
(d) Since |𝐷 × : 𝐶𝐷 × (𝑥𝑖 )| is an integer, |𝐷 × : 𝐶𝐷 × (𝑥𝑖 )| =
(e) From (d), Φ𝑛 (𝑞) divides (𝑞𝑛 − 1)/(𝑞𝑚𝑖 − 1) for 𝑖 = 1, 2, . . . , 𝑟, so the class equation in (c)
implies that Φ𝑛 (𝑞) divides 𝑞 − 1. Now, let 𝜁 ≠ 1 be a 𝑛 th root of unity. In the complex plane 𝑞 is
Î
closer to 1 than 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) = 𝑥 𝑛 + · · · + 𝑎 1𝑥 + 𝑎 0 be a monic polynomial
over Z of degree 𝑛 ≥ 1. For the sake of a contradiction, suppose there are only finitely many
primes dividing the values 𝑃 (𝑛), 𝑛 = 1, 2, . . . , say 𝑝 1, 𝑝 2, . . . , 𝑝𝑘 . Let 𝑁 be an integer such that
𝑃 (𝑁 ) = 𝑎 ≠ 0. Let 𝑄 (𝑥) = 𝑎 −1 𝑃 (𝑁 + 𝑎𝑝 1𝑝 2 . . . 𝑝𝑘 𝑥). Then, using the binomial theorem, we have
𝑄 (𝑥) = 𝑎 −1 𝑃 (𝑁 + 𝑎𝑝 1𝑝 2 . . . 𝑝𝑘 𝑥)
= 𝑎 −1 ((𝑁 + 𝑎𝑝 1𝑝 2 . . . 𝑝𝑘 𝑥)𝑛 + · · · + 𝑎 1 (𝑁 + 𝑎𝑝 1𝑝 2 . . . 𝑝𝑘 𝑥) + 𝑎 0 )
= 𝑎 −1 (𝑁 𝑛 + 𝑎𝑛−1 𝑁 𝑛−1 + · · · + 𝑎 1 𝑁 + 𝑎 0 + 𝑅(𝑥))
= 𝑎 −1 (𝑃 (𝑁 ) + 𝑅(𝑥))
= 1 + 𝑎 −1𝑅(𝑥)
for some polynomial 𝑅(𝑥) ∈ Z[𝑥] divisible by 𝑎𝑝 1𝑝 2 . . . 𝑝𝑘 . Thus 𝑄 (𝑥) ∈ Z[𝑥]. Moreover, for all
𝑛 ∈ Z+ we have 𝑃 (𝑁 +𝑎𝑝 1𝑝 2 . . . 𝑝𝑘 𝑛) ≡ 𝑎 (mod 𝑝 1, 𝑝 2, . . . , 𝑝𝑘 ), so 𝑄 (𝑛) = 𝑎 −1 𝑃 (𝑁 +𝑎𝑝 1𝑝 2 . . . 𝑝𝑘 𝑛) ≡
𝑎 −1𝑎 = 1 (mod 𝑝 1, 𝑝 2, . . . , 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 such that 𝑞 ≠ 𝑝𝑖 for all 𝑖 and such that 𝑞 divides 𝑄 (𝑚). Then 𝑞 divides 𝑎𝑄 (𝑚) =
𝑃 (𝑁 +𝑎𝑝 1𝑝 2 . . . 𝑝𝑘 𝑚), contradicting the fact that only the primes 𝑝 1, 𝑝 2, . . . , 𝑝𝑘 divide the numbers
𝑃 (1), 𝑃 (2), . . . .
Exercise 13.6.15. We follow the hint. Since Φ𝑚 (𝑎) ≡ 0 (mod 𝑝), we have 𝑎𝑚 ≡ 1 (mod 𝑝). Then
there exists 𝑏 such that 𝑏𝑎 ≡ 1 mod 𝑝 (indeed, 𝑏 = 𝑎𝑚−1 works), so 𝑎 is relatively prime to 𝑝. We
prove that the order of 𝑎 is precisely 𝑚. For the sake of a contradiction, suppose 𝑎𝑑 ≡ 1 (mod 𝑝)
for some 𝑑 dividing 𝑚, so that Φ𝑑 (𝑎) ≡ 0 (mod 𝑝) for some 𝑑 < 𝑚. Then 𝑎 is a multiple root of
𝑥 𝑚 − 1, so it is also a root of its derivative 𝑚𝑎𝑚−1 . But then 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.
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13.6
Cyclotomic Polynomials and Extensions
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 F𝑝× is 𝑚. Since |F𝑝× | = 𝑝 − 1, this implies that 𝑚 divides
𝑝 − 1, that is, 𝑝 ≡ 1 (mod 𝑚).
Exercise 13.6.17. By Exercise 13.6.14, there are infinitely many primes dividing Φ𝑚 (1), Φ𝑚 (2), Φ𝑚 (3), . . . .
Since only finitely many of them can divide 𝑚, it follows from by Exercise 13.6.16 that there must
exist infinitely many primes 𝑝 with 𝑝 ≡ 1 (mod 𝑚).
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