## Math 503: Abstract Algebra

Homework 8 <me@tylerlogic.com>
April 16, 2014
http://coursework.tylerlogic.com/courses/upenn/math503/homework08

### 1

#### (a)

Since cos(2π∕5) = 14( - 1), then (cos(2π∕5)) is simply (), therefore being an extension of degree two since x2 - 5 is the minimal polynomial of . Also since cos(2π∕6) = cos(π∕3) = 12 then (cos(2π∕6)) is just ; and hence it is an extension of degree one.

### 2

#### (a) Is ℚ(+)

Because + (,) then ( + ) (,). Similarly, because

and

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

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

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

### 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 = and K = (), 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)xi for nonzero b F, then (σ*g)(x) will have the term (a-b)xi and therefore not be equal to g(x). Hence g(x) = (σ*h)(x).