## Math 503: Abstract Algebra

Homework 7 <me@tylerlogic.com>
April 16, 2014
http://coursework.tylerlogic.com/courses/upenn/math503/homework07

### 2

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

#### (b)

Let α be a real number such that α3 = 2. Define τj : [T] for each j 3 by

Therefore T3 - 2 is contained within kerτj for each j. Since K = [T](T3 - 2), this implies the existence of a unique ring homomorphism τj : K such that τj(α) = τj(α) = e23α

#### (c)

For γ0, γ1, and γ2, let γn = e2πn3. Then since these are each roots of T3 - 2 over then

for each n. Therefore L(γ012), however, we have the linear relation γ0 + γ1 + γ2 = 0 due to each γn being distinct third roots of 2. This implies that (γ012) is actually isomorphic to (γ12), without loss of generality in selecting two elements of γ012, since any two are linearly independent. Hence, so is L. Now because (γ1) and (γ2) are have trivial intersection and each is generated by three elements over , then L must be generated by six elements over , i.e. [L : ] = 6.

### 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 L1k(α) and L2k(β), which in turn implies

where mα(x) is the minimal polynomial of α, and likewise for β and mβ(x). This reveals L1 kL2k[x] mα(x)kk[x] mβ(x), but because mα(x) is irreducible over k[x] of degree [k(α) : k] = 3 and mβ(x) is irreducible over k[x] of degree [k(β) : k] = 2, then mα(x) and mβ(x) are coprime. Therefore

but furthermore that mα(x),mβ(x) is maximal, implying k[x] mα(x),mβ(x) is a field and thus L1 kL2 is as well.