MATH 201, Algebra II, Final (March 12, 1999)

(Winter 1999, R. Boltje)

1. Find an element α ∈ ℂ such that ℚ(α)/ℚ is a Galois extension of degree 3.

2. Find an element α ∈ ℂ such that ℚ(\sqrt{2}, \sqrt{3}) = ℚ(α). Is ℚ(α)/ℚ a Galois extension? Compute all subfields of ℚ(α).

3. Let ℚ̄ be the set of all complex numbers which are algebraic over ℚ.

(a) Show that ℚ̄ is a subfield of ℂ containing ℚ.

(b) Show that ℚ̄ is algebraically closed.

(ℚ̄ is called the field of algebraic numbers.)

4. Construct a finite field with 27 elements and find a nonzero element of maximal multiplicative order.

5. Let p be a prime and let F be a field with p^{25} elements. Does F contain a subfield with p^{10} elements?

6. Is the Galois group of f(X) = X^{4} − 2 over ℚ abelian?

7. Let ζ := e^{2πi/7} ∈ ℂ. Is \sqrt{7}i ∈ ℚ(ζ)?

Solutions

1. and 7. It is not easy to write down a Galois extension of degree 3 over ℚ directly. But we can easily find a Galois extension E/ℚ with abelian Galois group of order 6, namely E = ℚ(ζ) with ζ := e^{2πi/7}: We have G := Gal(E/ℚ) ≅ (ℤ/7ℤ)^{×} which is a cyclic group of order 6. The subgroup {1 + 7ℤ, −1 + 7ℤ} of (ℤ/7ℤ)^{×} corresponds to a Galois extension F/ℚ of degree 6/2 = 3, since every subgroup of G is normal in G. The element 1 + 7ℤ corresponds to the identity of E and −1 + 7ℤ corresponds to the automorphism τ: E → E with τ(ζ) = ζ^{−1}. Obviously, α := ζ + ζ^{−1} is fixed by τ. Therefore, α ∈ F. If we can show that α ∉ ℚ, we have F = ℚ(α), since [F : ℚ] = 3 is a prime. But if ζ + ζ^{6} = α ∈ ℚ, we can replace ζ^{6} using 1 + ζ + ζ^{2} + ⋯ + ζ^{6} = 0 (remember Φ_{7} = 1 + X + ⋯ + X^{6}) and obtain (a + 1) + ζ^{2} + ζ^{3} + ζ^{4} + ζ^{5} = 0 which is impossible, since 1, ζ, ζ^{2}, ζ^{3}, ζ^{4}, ζ^{5} are a ℚ-basis of E.

Since Gal(E/ℚ) is a cyclic group of order 6, it has exactly one subgroup of order 3, and therefore, there exists exactly one intermediate field M in E/ℚ with [M : ℚ] = 2. Therefore, \sqrt{7}i ∈ E if and only if ℚ(\sqrt{7}i) = M. We will compute M. The subgroup of order three in Gal(E/ℚ) is obviously given by {σ_{1}, σ_{2}, σ_{4}}, where σ_{i}(ζ) = ζ^{i} for i = 1, …, 6. One can easily find an element which is fixed under these three automorphisms. For example β := ζ + ζ^{2} + ζ^{4} has this property.

Moreover, since 1, ζ, ζ^{2}, ζ^{4} are ℚ-linearly independent, β ∉ ℚ. Next we determine the minimal polynomial of β over ℚ, which has to be a polynomial of degree 2. From this we should be able to decide, if ℚ(β) = ℚ(\sqrt{7}i). Using the relation ζ^{6} = −(1 + ζ + ⋯ + ζ^{5}), we obtain β^{2} = −2 − ζ − ζ^{2} − ζ^{4} = −2 − β. Therefore, β is a root of the polynomial X^{2} + X + 2. But we can easily determine its roots by the well-known formula as (−1 ± \sqrt{1 − 4 · 2})/2 = (−1 ± \sqrt{7}i)/2. Therefore, 2β = −1 + \sqrt{7}i or 2β = −1 − \sqrt{7}i. This implies ℚ(β) = ℚ(\sqrt{7}i) and \sqrt{7}i ∈ E.

Remark: One can actually show in general, that if p is an odd prime, then the unique quadratic subfield in the cyclotomic extension ℚ_{p}/ℚ is given by ℚ(\sqrt{p}) if p ≡ 1 mod 4 and by ℚ(\sqrt{p}i) if p ≡ 3 mod 4.

2. Obviously, E := ℚ(\sqrt{2}, \sqrt{3}) is a splitting field of the polynomial f(X) := (X^{2} − 2)(X^{2} − 3) ∈ ℚ[X] over ℚ. Since ℚ has characteristic 0, f(X) is separable over ℚ. This implies that E/ℚ is a Galois extension. Since \sqrt{2} ∉ ℚ, and since \sqrt{2} is a root of the polynomial X^{2} −2 of degree 2, F_{1} := ℚ(\sqrt{2})/ℚ has degree 2. We claim that \sqrt{3} ∉ ℚ(\sqrt{2}). Assume that \sqrt{3} = a + b\sqrt{2} with a, b ∈ ℚ. Then \sqrt{3} − a = b\sqrt{2} and squaring both sides gives 3 + a^{2} − 2a\sqrt{3} = 2b^{2}. But since \sqrt{3} ∉ ℚ, this implies a = 0. Therefore, \sqrt{3} = b\sqrt{2}. Multiplication by \sqrt{2} gives \sqrt{6} = 2b ∈ ℚ which is a contradiction. Therefore \sqrt{3} ∉ ℚ(\sqrt{2}) and [E : F_{1}] = 2. Altogether we have [E : ℚ] = [E : F_{1}][F_{1} : ℚ] = 2 · 2 = 4. Therefore G := Gal(E/ℚ) has order 4. All groups of order 4 are abelian. There are two possibilities: Either G ≅ ℤ/4ℤ or G ≅ ℤ/2ℤ×ℤ/2ℤ. If G is cyclic of order 4 then G would have only one subgroup of order 2 and therefore, E/ℚ would have just one intermediate field of degree 2 over ℚ. But besides F_{1} = ℚ(\sqrt{2}) we immediately see the intermediate field F_{2} := ℚ(\sqrt{3}) of degree 2 over ℚ. Therefore, G ≅ ℤ/2ℤ × ℤ/2ℤ. This group has 3 subgroups of order 2 (anyone of the non-identity elements together with the identity). Therefore, E/F has a third intermediate field F_{3} with [F_{3} : ℚ] = 2, and no more.

Besides the identity map there are three more elements in G. Since E = F_{1}(\sqrt{3}), H_{1} := Gal(E/F_{1}) = \{1, σ_{1}\} with σ_{1}(\sqrt{2}) = \sqrt{2} and σ_{1}(\sqrt{3}) = −\sqrt{3}. Similarly, since E = F_{2}(\sqrt{2}), H_{2} := Gal(E/F_{2}) = \{1, σ_{2}\} with σ_{2}(\sqrt{2}) = −\sqrt{2} and σ_{2}(\sqrt{3}) = \sqrt{3}. Since H_{1} and H_{2} are subgroups of G, σ_{1}, σ_{2} ∈ G. Moreover, σ_{3} := σ_{1} ∘ σ_{2} has the properties σ_{3}(\sqrt{2}) = −\sqrt{2} and \sqrt{3} = −\sqrt{3}, which shows that the four elements 1, σ_{1}, σ_{2}, σ_{3} are pairwise distinct and form the group G. Obviously \sqrt{6} = \sqrt{2}\sqrt{3} ∈ E and σ_{3}(\sqrt{6}) = \sqrt{6}. Therefore, F_{3} := ℚ(\sqrt{6}) is the missing intermediate field

In order to find an element α with E = ℚ(α), we only have to find an element α ∈ E which is not contained in any of the three subfields F_{1}, F_{2}, F_{3}. Therefore, it suffices to find an element α ∈ E such that σ_{i}(α) ≠ α for all i = 1, 2, 3. The element α := \sqrt{2} + \sqrt{3} obviously satisfies these conditions.

3. (a) If a ∈ ℚ, then a is a root of the polynomial X − a ∈ ℚ[X] and therefore is algebraic over ℚ. This shows, that ℚ̄ contains ℚ. If a and b are in ℚ̄, then ℚ(a)/ℚ is a finite field extension, and also ℚ(a, b)/ℚ(a) is a finite field extension, because b is also algebraic over ℚ(a), if it is over ℚ. Therefore ℚ(a, b)/ℚ is finite. This implies that ℚ(a, b)/ℚ is an algebraic field extension. In particular, a+b, a−b, and ab are in ℚ(a, b) and therefore algebraic over ℚ. Also, if a ≠ 0, a^{−1} ∈ ℚ(a) and therefore is algebraic over ℚ. This shows that a ± b, ab, a^{−1} ∈ ℚ̄ and that ℚ̄ is a subfield of ℚ.

(b) Let f(X) = a_{n}X^{n} + ⋯ + a_{1}X + a_{0} ∈ ℚ̄[X] be a polynomial which is not constant. We will show that f has a root in ℚ̄. Since ℂ is algebraically closed, f has a root α in ℂ. Consider the field F := ℚ(a_{1}, …, a_{n}). Since a_{1}, …, a_{n} are algebraic over ℚ, F/ℚ is a finite field extension and f ∈ F [X]. Since α is a root of f, F (α)/F is algebraic and finite. Therefore also F (α)/ℚ is finite. This implies that α is algebraic over ℚ and therefore already in ℚ̄.

4. Let F := ℤ/3ℤ. Then F is a field with 3 elements. We want to find an irreducible polynomial f(X) ∈ F [X] of degree 3. To check that such a polynomial is irreducible, it suffices to check that it has no root in F. We try f(X) = ¯1 − X + X^{3} and find f(¯0) = ¯1, f(¯1) = ¯1, and f(¯2) = ¯1. Therefore, f(X) is irreducible in F [X] and E := F [X]/(f(X)) is a field with 3^{3} = 27 elements. We set a := X̄ = X + (f(X)) ∈ E. Then ¯1, a, a^{2} is an F -basis of E and we have a^{3} = a − ¯1. E^{×} is a cyclic group with 26 elements. Every element in E^{×} has order 1, 2, 13, or 26. We consider for example a. We have a ≠ 1, a^{2} ≠ 1, and a^{13} = a^{8}a^{4}a with a^{4} = a(a − ¯1) = a^{2} − a and a^{8} = (a^{2} − a)^{2} = a^{4} − ¯2a^{3} + a^{2} = (a^{2} − a) − ¯2(a − ¯1) + a^{2} = ¯2a^{2} + ¯2. This gives a^{12} = a^{8}a^{4} = (¯2a^{2} +¯2)(a^{2} −a) = ¯2a^{4} +a^{3} +¯2a^{2} +a = ¯2(a^{2} −a)+(a−¯1)+¯2a^{2} +a = a^{2} + ¯2, and a^{13} = a^{12}a = (a^{2} + ¯2)a = a^{3} − a = −¯1. Since a has neither order 1, nor 2, nor 13, it has to have order 26 in E^{×}.

5. If F had a subfield F_{0} of order p^{10}, then F would be a vector space of finite dimension k over F_{0}. But then F would have (p^{10})^{k} = p^{10k} elements which implies 10k = 25 and k = 5/2 ∉ ℕ. This is a contradiction.

6. E := ℚ(\sqrt[4]{2}, i) is a splitting field of X^{4} − 2 over ℚ. Assume that the Galois group G := Gal(E/F) is abelian. Then every subgroup of G is normal in G. This implies that every intermediate field of E/ℚ would be a Galois extension of ℚ. But the intermediate field F := ℚ(\sqrt[4]{2}) is not a Galois extension of ℚ, since it is not normal over F. In fact, X^{4} − 2 has a root in F but does not split into linear factors in F [X], because X^{4} − 2 = (X^{2} − \sqrt{2})(X^{2} + \sqrt{2}) and the last factor has no root in the subfield F which consists of real numbers. This is a contradiction, and therefore, G is not abelian.