## 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]∕(T^{3} - 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 T^{3} - 2 ∈ ℚ[T] is irreducible, and therefore K is generated over ℚ
by 1,x,x^{2} and is therefore a vector space of degree deg(T^{3} - 2) = 3 over ℚ.

#### (b)

Let α be a real number such that α^{3} = 2. Define τ_{j} : ℚ[T] → ℂ for each j ∈ ℤ∕3ℤ
by
Therefore T^{3} - 2 is contained within kerτ_{j} for each j. Since K = ℚ[T]∕(T^{3} - 2), this implies the existence of a unique ring
homomorphism τ_{j} : K → ℂ such that τ_{j}(α) = τ_{j}(α) = e^{2jπ∕3}α

#### (c)

For γ_{0}, γ_{1}, and γ_{2}, let γ_{n} = e^{2πn∕3}. Then since these are each roots of
T^{3} - 2 over ℚ then
for each n. Therefore Lℚ(γ_{0},γ_{1},γ_{2}), however, we have the linear relation γ_{0} + γ_{1} + γ_{2} = 0 due to each γ_{n} being distinct
third roots of 2. This implies that ℚ(γ_{0},γ_{1},γ_{2}) is actually isomorphic to ℚ(γ_{1},γ_{2}), without loss of generality in selecting
two elements of γ_{0},γ_{1},γ_{2}, since any two are linearly independent. Hence, so is L. Now because ℚ(γ_{1}) and ℚ(γ_{2}) 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 ⊗_{ℚ}K → L

#### (e)

### 3

Let L_{1},L_{2} be finite extensions of a field k with degrees 3 and 2, respectively.

#### (a)

Since the degrees of L_{1} and L_{2} over k are finite, then they are both algebraic over
k. Letting α ∈ L_{1} -k, we have [k(α) : k] divides [L_{1} : k] and [k(α) : k] > 1, but since [L_{1} : k] = 3, i.e. [L_{1} : k] is prime, then
[k(α) : k] must also be 3. Similarly for β ∈ L_{2} - k, [k(β) : k] = 2. Therefore L_{1}k(α) and L_{2}k(β), which in turn
implies
where m_{α}(x) is the minimal polynomial of α, and likewise for β and m_{β}(x). This reveals L_{1} ⊗_{k}L_{2}k[x] ∕ m_{α}(x)⊗_{k}k[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

but furthermore that m_{α}(x),m_{β}(x) is maximal, implying k[x] ∕ m_{α}(x),m_{β}(x) is a field and thus L_{1} ⊗_{k}L_{2} is as
well.

#### (b)

#### (c)

### 4

#### (a)

#### (b)

#### (c)