Math 503: Abstract Algebra

Homework 7

Lawrence Tyler Rush

<me@tylerlogic.com>

April 16, 2014

http://coursework.tylerlogic.com/courses/upenn/math503/homework07

1 Cyclotomic Polynomials

(a)

(b)

(c)

(d) Extra Credit

2

Let K = ℚ[T]∕(T³ - 2) and let x be the image of T in K.

(a) Show that K is an extension of degree 3 over ℚ

The polynomial T³ - 2 ∈ ℚ[T] is irreducible, and therefore K is generated over ℚ by 1,x,x² and is therefore a vector space of degree deg(T³ - 2) = 3 over ℚ.

(b)

Let α be a real number such that α³ = 2. Define τ_j : ℚ[T] → ℂ for each j ∈ ℤ∕3ℤ by

√-- τj(f(T)) = e2jπ -1∕3f(α )

Therefore T³ - 2 is contained within kerτ_j for each j. Since K = ℚ[T]∕(T³ - 2), this implies the existence of a unique ring homomorphism τ_j : K → ℂ such that τ_j(α) = τ_j(α) = e^2jπ∕3α

(c)

For γ₀, γ₁, and γ₂, let γ_n = √3-
2

e^2πn∕3. Then since these are each roots of T³ - 2 over ℚ then

~ ~ Kj = K = ℚ(γn)

for each n. Therefore L ~
= ℚ(γ₀,γ₁,γ₂), however, we have the linear relation γ₀ + γ₁ + γ₂ = 0 due to each γ_n being distinct third roots of 2. This implies that ℚ(γ₀,γ₁,γ₂) is actually isomorphic to ℚ(γ₁,γ₂), without loss of generality in selecting two elements of γ₀,γ₁,γ₂, since any two are linearly independent. Hence, so is L. Now because ℚ(γ₁) and ℚ(γ₂) are have trivial intersection and each is generated by three elements over ℚ, then L must be generated by six elements over ℚ, i.e. [L : ℚ] = 6.

(d) Show that there exists a surjective ring homomorphism K ⊗_ℚK → L

(e)

3

Let L₁,L₂ be finite extensions of a field k with degrees 3 and 2, respectively.

(a)

Since the degrees of L₁ and L₂ over k are finite, then they are both algebraic over k. Letting α ∈ L₁ -k, we have [k(α) : k] divides [L₁ : k] and [k(α) : k] > 1, but since [L₁ : k] = 3, i.e. [L₁ : k] is prime, then [k(α) : k] must also be 3. Similarly for β ∈ L₂ - k, [k(β) : k] = 2. Therefore L₁

k(α) and L₂

k(β), which in turn implies

L1 ~= k[x]∕(m α(x)) and L2 ~= k[x]∕(m β(x))

where m_α(x) is the minimal polynomial of α, and likewise for β and m_β(x). This reveals L₁ ⊗_kL₂k[x] ∕ m_α(x)⊗_kk[x] ∕ m_β(x), but because m_α(x) is irreducible over k[x] of degree [k(α) : k] = 3 and m_β(x) is irreducible over k[x] of degree [k(β) : k] = 2, then m_α(x) and m_β(x) are coprime. Therefore