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

(b) Extra Credit


2


(a) Is (√ --
  3 + √ --
  5)


Because √-
 3 + √-
 5 (√-
 3,√ -
  5) then (√ -
  3 + √ -
  5) (√ -
  3,√-
 5). Similarly, because
 (   -    -        -    -)    -
1 (√ 3+ √ 5)3 - 14(√ 3+ √ 5) = √3
4

and

- 1-( √   √- 3     √-   √- )   √-
 4  ( 3 +  5) - 18( 3 +  5)  =  5

then we also have (√ -
  3,√-
 5) (√-
 3 + √ -
  5). Putting these two results together indicates that (√ -
  3,√ -
  5) and (√ -
  3 + √-
 5) are one in the same.

(b) Extra Credit: Show that (√--
 3,√ --
  5) has a finite number of subfields)


Any subfield K (√ -
  3,√ -
  5) will be a subspace of (√ -
  3,√-
 5) over . Since (√ -
  3,√-
 5) 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.

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 = and K = (√ ---
  - 3), then the non trivial element σ of Aut(K∕Q) is defined by σ(a + b√ ---
  - 3) = a-b√---
 - 3 for all a + b√---
 - 3 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√---
 - 3)xi for nonzero b F, then (σ*g)(x) will have the term (a-b√---
 - 3)xi and therefore not be equal to g(x). Hence g(x) = (σ*h)(x).

(b) Extra Credit


(c) Extra Credit