## Math 503: Abstract Algebra

Homework 8
Lawrence Tyler Rush

<me@tylerlogic.com>

April 16, 2014

http://coursework.tylerlogic.com/courses/upenn/math503/homework08
### 1

#### (a)

Since cos(2π∕5) = 1∕4( - 1), then ℚ(cos(2π∕5)) is simply ℚ(), therefore
being an extension of degree two since x^{2} - 5 is the minimal polynomial of . Also since cos(2π∕6) = cos(π∕3) = 1∕2 then
ℚ(cos(2π∕6)) is just ℚ; and hence it is an extension of degree one.

#### (b) Extra Credit

### 2

#### (a) Is ℚ( + )

Because + ∈ ℚ(,) then ℚ( + ) ⊆ ℚ(,). Similarly,
because
and

then we also have ℚ(,) ⊆ ℚ( + ). Putting these two results together indicates that ℚ(,) and ℚ( + )
are one in the same.

#### (b) Extra Credit: Show that ℚ(,) has a finite number of subfields)

Any subfield K ⊆ ℚ(,) will be a subspace of ℚ(,) over ℚ. Since
ℚ(,) is a finite extension, it is therefore a finite vector space over ℚ, and hence only has a finite number of vector
subspaces. Thus there are only a finite number of subfields, too.

### 3

#### (a)

#### (b) Extra Credit

### 4

Let f(x) be an irreducible polynomial over a field F. Let K be a finite extension field
of F. Let g(x) and h(x) are two irreducible factors of f(x) in K[x].

#### (a)

Let F = ℚ and K = ℚ(), then the non trivial element σ of Aut(K∕Q) is
defined by σ(a + b) = a-b for all a + b ∈ K. Since σ_{*} : K[x] → K[x] is defined by σ_{*}(x) = x (where x is the
indeterminate) and σ_{*}|_{K} = σ.
Since f(x) is irreducible over F[x] but reducible as f(x) = g(x)h(x) over K[x] where g(x),h(x) are irreducible in K[x],
then it must be the case that either g(x) or h(x) has some term with a coefficient in K - F, otherwise f(x) would be
reducible in F[x]. Without loss of generality, let g(x) have a term with coefficient in K - F. Since f(x) ∈ F[x], then
σ_{*}(f(x)) = σ_{*}(g(x)h(x)) = (σ_{*}g)(x)(σ_{*}h)(x) = f(x). Because g(x) divides f(x), then it divides (σ_{*}g)(x)(σ_{*}h)(x), but
because g(x) has a term (a + b)x^{i} for nonzero b ∈ F, then (σ_{*}g)(x) will have the term (a-b)x^{i} and therefore not
be equal to g(x). Hence g(x) = (σ_{*}h)(x).

#### (b) Extra Credit

#### (c) Extra Credit