Exam III Review Sheet
Math 4201
The third exam will be on Monday, April 29, 2013. The syllabus for Exam III is sections 1 – 3 of
Chapter 10. Some of the main examples and facts from this material are listed below.
∙ If 𝐹 is an extension field of a field 𝐾, then the Galois group of 𝐹 over 𝐾, denoted gal(𝐹 : 𝐾)
is the group, under composition, of field automorphism of 𝐹 which fix 𝐾, i.e. 𝜎 ∈ gal(𝐹 : 𝐾)
if 𝜎 : 𝐹 → 𝐹 is a field isomorphism and 𝜎(𝑎) = 𝑎 for all 𝑎 ∈ 𝐾.
∙ The Galois group of a polynomial 𝑓 (𝑥) ∈ 𝐾[𝑥] is the Galois group of 𝐸 over 𝐾 where 𝐸 is a
splitting field for 𝑓 (𝑥) over 𝐾.
∙ If 𝐹 is an extension field of 𝐾, 𝑓 (𝑥) ∈ 𝐾[𝑥], 𝑢 ∈ 𝐹 , and 𝜎 ∈ gal(𝐹 : 𝐾), then 𝜎(𝑓 (𝑢)) =
𝑓 (𝜎(𝑢)). In particular, if 𝑢 is a root of 𝑓 (𝑥), then 𝜎(𝑢) is also a root.
∙ As a consequence of the previous item, if 𝐹 is the splitting field of 𝑓 (𝑥) ∈ 𝐾[𝑥], then the
Galois group of 𝐹 over 𝐾 is isomorphic to a subgroup of the group of permutations of the
roots of 𝑓 (𝑥). In particular, if deg 𝑓 (𝑥) = 𝑛 then gal(𝐹 : 𝐾) is isomorphic to a subgroup of
the symmetric group 𝑆𝑛 . (Theorem 3, Page 417.)
∙ Let 𝐾 be a field, let 𝑓 (𝑥) be a polynomial of positive degree in 𝐾[𝑥], and let 𝐹 be a splitting
field for 𝑓 (𝑥) over 𝐾. If no irreducible factor of 𝑓 (𝑥) has repeated roots in 𝐹 , then
∣gal(𝐹 : 𝐾)∣ = [𝐹 : 𝐾].
(Corollary, Page 419.)
∙ If 𝐹 is a finite field with char(𝐹 ) = 𝑝, the map 𝜙 : 𝐹 → 𝐹 defined by 𝜙(𝑥) = 𝑥𝑝 , for all 𝑥 ∈ 𝐹 ,
is an automorphism of 𝐹 called the Frobenius automorphism of 𝐹 .
∙ Let 𝐾 be a finite field with ∣𝐾∣ = 𝑝𝑟 , where 𝑝 = char(𝐾). Let 𝐹 be an extension field of 𝐾
with [𝐹 : 𝐾] = 𝑚, and let 𝜙 be the Frobenius automorphism of 𝐹 . The gal(𝐹 : 𝐾) is a cyclic
group of order 𝑚, generated by 𝜙𝑟 .
∙ A polynomial 𝑓 (𝑥) ∈ 𝐾[𝑥] has no multiple roots if and only if gcd(𝑓 (𝑥), 𝑓 ′ (𝑥)) = 1.
∙ A polynomial 𝑓 (𝑥) ∈ 𝐾[𝑥] is separable if its irreducible factors have only simple roots.
∙ An algebraic extension 𝐹 of 𝐾 is called separable if the minimal polynomial of each element
of 𝐹 is separable.
∙ If 𝐹 is a finite separable extension of 𝐾, then 𝐹 is a simple extension, i.e. 𝐹 = 𝐾(𝑢) for some
𝑢 ∈ 𝐹.
∙ If 𝐾 is an extension field of a field 𝐹 , know the Galois correspondence between fields between
𝐹 and 𝐾 and subgroups of Gal𝐹 (𝐾). If 𝐸 is a field between 𝐾 and 𝐹 (i.e., 𝐹 ⊆ 𝐸 ⊆ 𝐾) and
if 𝐻 ⊆ gal(𝐾 : 𝐹 ), then the correspondence is given by:
𝐸 7→ gal(𝐾 : 𝐸),
𝐻 7→ 𝐾𝐻 = {𝑎 ∈ 𝐾 : 𝜎(𝑎) = 𝑎 for all 𝜎 ∈ 𝐻} .
∙ 𝐾 is said to be an Galois extension of 𝐹 (or 𝐾 is Galois over 𝐹 ) is 𝐾 is finite-dimensional
over 𝐹 and
𝐾gal(𝐾:𝐹 ) = 𝐹.
That is, the only elements of 𝐾 fixed by all 𝜎 ∈ gal(𝐾 : 𝐹 ) are the elements of 𝐹 .
1
Exam III Review Sheet
Math 4201
∙ An extension 𝐾 of 𝐹 is a Galois extension if and only if 𝐾 is the splitting field of a separable
polynomial over 𝐹 .
∙ If the characteristic of 𝐹 is 0, then 𝐾 is a Galois extension of 𝐹 if and only if 𝐾 is a splitting
field of a polynomial 𝑓 (𝑥) ∈ 𝐹 [𝑥].
∙ Know the Fundamental Theorem of Galois Theory, which states briefly that if 𝐹 is a splitting
field of a separable polynomial over the field 𝐾, then the Galois correspondence is bijective
and preserves normality (of subgroups and subfields). More precisely:
Fundamental Theorem of Galois Theory Let 𝐹 be the splitting field of a separable
polynomial over the field 𝐾, and let 𝐺 = gal(𝐹 : 𝐾).
1. For an intermediate field 𝐾 ⊆ 𝐸 ⊆ 𝐹 , the Galois group gal(𝐹 : 𝐸) is a subgroup of
gal(𝐹 : 𝐾) and it has fixed field
𝐹gal(𝐹 :𝐸) = 𝐸.
Furthermore,
∣gal(𝐹 : 𝐸)∣ = [𝐹 : 𝐸] and
[𝐸 : 𝐾] = [gal(𝐹 : 𝐾) : gal(𝐹 : 𝐸)].
2. For a subgroup 𝐻 ⊆ gal(𝐹 : 𝐾), its fixed field 𝐾 ⊆ 𝐹𝐻 ⊆ 𝐹 has Galois group
gal(𝐹 : 𝐹𝐻 ) = 𝐻.
Furthermore,
[𝐹 : 𝐹𝐻 ] = ∣𝐻∣
and
[𝐹𝐻 : 𝐾] = [gal(𝐹 : 𝐾) : 𝐻].
3. An intermediate field 𝐾 ⊆ 𝐸 ⊆ 𝐹 is a splitting field of a separable polynomial over 𝐾
if and only if the corresponding group gal(𝐸 : 𝐾) is a normal subgroup of gal(𝐹 : 𝐾),
and in this case
gal(𝐸 : 𝐾) ∼
= gal(𝐹 : 𝐾)/ gal(𝐹 : 𝐸).
∙ If 𝐸𝑛 denotes the splitting field of 𝑥𝑛 − 1 over a field 𝐾 of characteristic 0, then gal(𝐸𝑛 : 𝐾)
is isomorphic to a subgroup of ℤ∗𝑛 and hence is an abelian group. (Theorem 1, Page 438.)
∙ Know what are roots of unity and primitive roots of unity.
∙ If 𝐾 is a field of characteristic 0 that contains a primitive 𝑛𝑡ℎ root of unity, and if 𝑢 is a root
of 𝑥𝑛 − 𝑎 ∈ 𝐾[𝑥] in some extension field of 𝐾, then 𝐹 = 𝐾(𝑢) is a splitting field of 𝑥𝑛 − 1
over 𝐾 and gal(𝐹 : 𝐾) is abelian. (Lemma 4, Page 438.)
∙ Know what it means for an equation 𝑓 (𝑥) = 0 to be solvable by radicals.
∙ Know what it means for a group to be solvable: 𝐺 is solvable if there is a chain of subgroups
𝐺 = 𝐺0 ⊇ 𝐺1 ⊇ 𝐺2 ⊇ ⋅ ⋅ ⋅ ⊇ 𝐺𝑛−1 ⊇ 𝐺𝑛 = {𝑒}
such that (1) 𝐺1 is a normal subgroup of 𝐺𝑖−1 for 1 ≤ 𝑖 ≤ 𝑛, and (2) 𝐺𝑖−1 /𝐺𝑖 is abelian for
all 𝑖.
∙ 𝑆𝑛 is not solvable if 𝑛 ≥ 5.
∙ Know Galois’s criterion for solvability by radicals: 𝑓 (𝑥) = 0 is solvable by radicals if and only
if the Galois group of 𝑓 (𝑥) is a solvable group.
2
Exam III Review Sheet
Math 4201
∙ Know how to use Galois’ criterion to show that there are rational polynomials of degree 5
that are not solvable by radicals.
∙ The 𝑛𝑡ℎ cyclotomic polynomial is the polynomial
∏
Φ𝑛 (𝑥) =
(𝑥 − 𝛼𝑘 )
(𝑘, 𝑛)=1, 1≤𝑘<𝑛
where 𝛼 = 𝑒2𝜋𝑖/𝑛 is a complex primitive 𝑛𝑡ℎ root of unity. Thus, Φ𝑛 (𝑥) is the complex
polynomial which has all primitive 𝑛𝑡ℎ roots of unity as roots. Properties of Φ𝑛 (𝑥) are:
1. deg(Φ𝑛 (𝑥)) = 𝜑(𝑛);
∏
2. 𝑥𝑛 − 1 = 𝑑∣𝑛 Φ𝑑 (𝑥);
3. Φ𝑛 (𝑥) is monic, with integer coefficients;
4. Φ𝑛 (𝑥) is irreducible over ℚ for every positive integer 𝑛.
∙ Know how to use the above properties to compute Φ𝑛 (𝑥).
∙ For every positive integer 𝑛, the Galois group of the 𝑛𝑡ℎ cyclotomic polynomial Φ𝑛 (𝑥) over ℚ
is isomorphic to ℤ∗𝑛 , the group of units of the ring ℤ𝑛 .
Review Exercises
Here are some sample exercises such as might be given on the exam.
1. If 𝐸 = ℚ(𝑒2𝜋𝑖/8 ), compute gal(𝐸 : ℚ).
▶ Solution. Since 𝑒2𝜋𝑖/8 is a primitive 8-th root of unity it follows that 𝐸 is the splitting
field of 𝑥8 − 1 and gal(𝐸 : ℚ) ∼
◀
= ℤ∗8 ∼
= ℤ2 × ℤ2 .
2. If 𝐸 = ℚ(𝑒2𝜋𝑖/6 ), compute gal(𝐸 : ℚ).
▶ Solution. Since 𝑒2𝜋𝑖/6 is a primitive 6-th root of unity it follows that 𝐸 is the splitting
field of 𝑥6 − 1 and gal(𝐸 : ℚ) ∼
◀
= ℤ∗6 ∼
= ℤ2 .
3. If 𝐸 = ℚ(𝑖,
√
7), compute gal(𝐸 : ℚ).
▶ Solution. This is the splitting field of (𝑥2 + 1)(𝑥2 − 7) and gal(𝐸 : ℚ) ∼
= ℤ2 × ℤ2 . A similar
example was done in class.
◀
√
4. If 𝐸 = ℚ( 4 2), show that gal(𝐸 : ℚ) ∼
= ℤ2 .
√
▶ Solution. Since√the minimal polynomial of 4 2 is 𝑥4 − 2 it follows
√ that any
√ automorphism
𝜎 of 𝐸 must take 4 2 to another root of 𝑥4 − 2 in 𝐸. That is, 𝜎( 4 2) = ± 4 2. Thus, there
are at most 2√automorphisms
of 𝐸 over ℚ.
But each
of √
these is realized. Namely, 𝜎1 (𝑥) = 𝑥
√
√
√
√
4
4
4
4
4
and 𝜎2 (𝑎 + 𝑏 2 + 𝑐 4 + 𝑑 8) = 𝑎 − 𝑏 2 + 𝑐 4 − 𝑑 4 8 are two automorphisms. Hence
gal(𝐸 : ℚ) ∼
◀
= ℤ2 .
√ √ √
5. Let 𝐾 = ℚ( 2, 3, 5). Determine [𝐾 : ℚ], prove that 𝐾 is a splitting field of a polynomial
over ℚ, and determine gal(𝐾 : ℚ)
3
Exam III Review Sheet
Math 4201
▶ Solution. There is a chain of field extensions:
√
√ √
√ √ √
ℚ ⊂ ℚ( 2) ⊂ ℚ( 2, 3) ⊂ ℚ( 2, 3, 5) = 𝐾.
Since each successive extension is obtained by adjoining a square root, it follows from the
tower theorem that [𝐾 : ℚ] = 2 × 2 × 2 = 8. 𝐾 is the splitting field of the polynomial
(𝑥2 −2)(𝑥2 −3)(𝑥2 −5). Analogous to adjoining 2 square roots, gal(𝐾 : ℚ) ∼
= ℤ2 ×ℤ2 ×ℤ2 . ◀
6. Let 𝐾 be the splitting field over ℚ of the polynomial 𝑓 (𝑥) = (𝑥2 − 2𝑥 − 1)(𝑥2 − 2𝑥 − 7).
Determine gal(𝐾 : ℚ), and find all intermediate fields explicitly.
√
√
▶ Solution. The roots√of 𝑓 (𝑥) are 1 ± 2 and 1 ± 4√ 2. Since all of these are included in the
quadratic extension ℚ( 2) it follows that 𝐾 = ℚ( 2) is the splitting field of 𝑓 (𝑥) over ℚ.
Since [𝐾 : ℚ] = 2 it follows that gal(𝐾 : ℚ) ∼
= ℤ2 . Since [𝐾 : ℚ] = 2 there are no intermediate
fields other than 𝐾 and ℚ.
◀
7. Let 𝐹 be a splitting field of a separable polynomial over a field 𝐾 such that gal(𝐹 : 𝐾) ∼
= ℤ12 .
How many intermediate fields 𝐿 are there such that (a) [𝐿 : 𝐾] = 4, (b) [𝐿 : 𝐾] = 9, (c)
gal(𝐹 : 𝐿) ∼
= ℤ6 ?
▶ Solution. Let 𝐺 = gal(𝐹 : 𝐾). Then, by the Fundamental Theorem of Galois Theory,
any intermediate field 𝐿 has the form 𝐿 = 𝐹𝐻 for some subgroup 𝐻 of 𝐺 and [𝐹 : 𝐿] = ∣𝐻∣
while [𝐿 : 𝐾] = [𝐺 : 𝐻]. Hence:
(a) [𝐿 : 𝐾] = 4 if and only if [𝐺 : 𝐻] = 4 if and only if ∣𝐻∣ = 12/4 = 3. There is exactly one
subgroup of ℤ12 of order 3 and hence there is exactly one subfield 𝐿 of 𝐹 with [𝐿 : 𝐾] = 4.
(b) Since 9 does not divide 12, there is no subfield 𝐿 with [𝐿 : 𝐾] = 9.
(c) There is exactly one subgroup of ℤ12 isomorphic to ℤ6 . Hence there is exactly on subfield
𝐿 of 𝐹 containing 𝐾 such that gal(𝐹 : 𝐿) ∼
◀
= ℤ6 .
8. Let 𝑓 (𝑥) = 𝑥3 + 5 ∈ ℚ[𝑥].
√
√
(a) Show that the splitting field of 𝑓 (𝑥) is 𝐾 = ℚ( −3, 3 5), and compute [𝐾 : ℚ].
√
√
√
▶ Solution.
The roots of 𝑓 (𝑥) are − 3 5, − 3 5𝜔, and − 3 5𝜔 2 , where 𝜔 = 𝑒2𝜋𝑖/3 =
√
−1 + 3𝑖
is a primitive cube root of unity in ℂ. Thus, the splitting field of 𝑓 (𝑥)
2
√
√
√
√
√
−1 + 3𝑖
−1 + −3
3
3
3
2
is 𝐾1 = ℚ(− 5, − 5𝜔, − 5𝜔 ). Note that 𝜔 =
=
∈ 𝐾 =
2
2
√
√
√
√
√
ℚ( −3, 3 √
5), and√clearly, − 3 √
5 ∈ 𝐾 = ℚ( −3, 3 5). Thus, each of the roots of 𝑓 (𝑥),
3
3
namely, − 5, − 5𝜔, and − 3 5𝜔 2 are in 𝐾. Hence 𝐾1 ⊆ 𝐾. To show
the reverse
√
√
3
inclusion, it is sufficient
to show that each of
the generators of 𝐾, namely, 5 and −3,
√
√
3
is in 𝐾1√. That √
5 is in 𝐾1 is clear since − 3 5 is one of the√generators of 𝐾1 . Moreover,
3
3
𝜔 = − 5𝜔/(− 5) ∈ 𝐾1 , and similarly 𝜔 2 ∈ 𝐾1 . Hence −3 = 𝜔 − 𝜔 2 ∈ 𝐾1 . Thus,
𝐾 ⊆ 𝐾1 and hence 𝐾 = 𝐾1 . The degree of the extension is calculated by the tower
formula:
√
√
3
3
[𝐾 : ℚ] = [𝐾 : ℚ( 5)][ℚ( 5) : ℚ] = 3 × 2 = 6.
◀
4
Exam III Review Sheet
Math 4201
(b) Find the Galois group gal(𝐾 : ℚ).
▶ Solution. Since gal(𝐾 : ℚ) is isomorphic to a subgroup of 𝑆3 of order [𝐾 : ℚ] = 6,
it follows that gal(𝐾 : ℚ) ∼
◀
= 𝑆3 .
(c) How many different fields 𝐿 (other than ℚ and 𝐾) satisfy ℚ ⊂ 𝐿 ⊂ 𝐾? That is, how
many intermediate fields are there between 𝐾 and ℚ?
▶ Solution. By the Fundamental Theorem of Galois Theory, there is a one-to-one
correspondence between intermediate fields of 𝐾 and subgroups of gal(𝐾 : ℚ) ∼
= 𝑆3 .
But the subgroups of 𝑆3 are {(1)}, ⟨(1 2)⟩, ⟨(1 3)⟩, ⟨(2 3)⟩, ⟨(1 2 3)⟩, and 𝑆3 . Since
ℚ corresponds to 𝑆3 , and {(1)} corresponds to 𝐾, it follows that there are exactly 4
additional intermediate fields.
◀
(d) For each intermediate field 𝐿, compute the degree [𝐿 : ℚ] and write 𝐿 in the form
𝐿 = ℚ(𝛼) for an appropriate 𝛼 ∈ 𝐾.
▶ Solution. It is only necessary to find 4 distinct intermediate fields, since we know that
that is the number of intermediate fields, according to the fundamental theorem. Each
of the roots of the polynomial 𝑓 (𝑥) generates a subfield√of 𝐾 of degree
deg 𝑓 (𝑥) =
3 over
√
√
3
3
3
ℚ. √
Hence, 3 of the 4 intermediate
fields
are:
𝐿
=
ℚ(−
5)
=
ℚ(
5),
𝐿
=
ℚ(−
5𝜔)
=
1
2
√
√
3
3
3
2
2
field is obtained
ℚ( 5𝜔), and 𝐿3 = ℚ(− 5𝜔 ) = ℚ( 5𝜔
√ ). The fourth intermediate
√
from the other generator of 𝐾, namely −3, so that 𝐿4 = ℚ( −3). The degrees are:
[𝐿1 : ℚ] = [𝐿2 : ℚ] = [𝐿3 : ℚ] = 3
and
[𝐿4 : ℚ] = 2.
◀
9. Let 𝐹 ⊆ 𝐿 ⊆ 𝐾 be fields. Prove or disprove:
(a) If 𝐾 is Galois over 𝐹 , then 𝐾 is Galois over 𝐿.
▶ Solution. This is item (4) in the Main Theorem of Galois Theory. Page 429
◀
(b) If 𝐾 is Galois over 𝐹 , then 𝐿 is Galois over 𝐹 .
√
▶ Solution. This is false. ℚ( 3 2, 𝜔) where 𝜔 is a primitive cube root of unity is the
splitting field of 𝑥3√− 2 over ℚ and hence it is a Galois extension. √However, the intermediate field 𝐿 = ℚ( 3 2) is not a Galois extension
of ℚ since gal(ℚ( 3 2) : ℚ) is the identity
√
3
group, and hence it has fixed field ℚ( 2) ∕= ℚ.
◀
(c) If 𝐿 is Galois over 𝐹 , and 𝐾 is Galois over 𝐿, then 𝐾 is Galois over 𝐹 .
√
√
▶ Solution.
This is also√false. Consider the chain of subfields ℚ ⊂ ℚ( 2) ⊂ ℚ(√4 2).
√
4
2
ℚ(√
2) is Galois over
√ ℚ( 2) since it is the splitting field of the polynomial 𝑥 − 2 ∈
ℚ( 2)[𝑥], while ℚ( 2) is Galois
over ℚ since it is the splitting field of the polynomial
√
4
𝑥2 − 2 ∈ ℚ[𝑥].
However,
ℚ(
2)
is
not Galois over ℚ since the polynomial
𝑥4 − 2 has a
√
√
4
4
root in ℚ( 2) but does not split in this field since the roots ±𝑖 2 are not in this field.
(See Theorem 3, part (b), Page 427.)
◀
5
Exam III Review Sheet
Math 4201
√
10. Let 𝑤 = 4 2 and 𝐾 = ℚ(𝑤, 𝑖). Then 𝐾 is a splitting field over ℚ of the polynomial 𝑓 (𝑥) =
𝑥4 − 2. Label the roots of 𝑓 (𝑥) as 𝑟1 = 𝑤, 𝑟2 = −𝑤, 𝑟3 = 𝑤𝑖, and 𝑟4 = −𝑤𝑖. Give an example
(with justification) of a permutation of these roots which does not correspond to an element
of the Galois group gal(𝐾 : ℚ).
▶ Solution. The 3-cycle (𝑟2 , 𝑟3 , 𝑟4 ) cannot come from an automorphism 𝜎 ∈ gal(𝐾 : ℚ).
Suppose there is such a 𝜎. Since 𝜎 must permute the roots of 𝑥4 − 2, we must have 𝜎(𝑤) = 𝑤.
But if this is true, then 𝜎(−𝑤) = −𝜎(𝑤) = −𝑤. Thus, 𝜎(−𝑤) cannot be 𝑤𝑖 for any 𝜎 that
fixes 𝑤.
◀
6