Math 502: Abstract Algebra

Homework 9

Lawrence Tyler Rush

<me@tylerlogic.com>

January 5, 2014

http://coursework.tylerlogic.com/courses/upenn/math502/homework09

1

(a)

(b) Extra Credit

2

(a) Via Zorn’s Lemma, prove that a nonzero v ∈ V must be contained in some basis of V

Fix some nonzero v ∈ V and define

to be the family of subsets of V defined by

L = {S ⊆ V | S is linearly independent and v ∈ S}

Then is a poset regarding what it means to be a subset. Let C = {S_α} be a chain in . Because each S_α is linearly independent, then C must have a maximal element in since V is a vector space and a basis in a vector space is a maximal linearly independent set. Thus by Zorn’s Lemma, must also have a maximal element. Such an element, by definition, is a basis of V .

(b)

For later contradiction, assume that j is not injective. Then there exist two distinct v₁,v₂ ∈ V with j(v₁) = j(v₂), or in other words j(v₁)(λ) = j(v₂)(λ) for all λ ∈ V ^∨. This implies that

λ(v1) = λ(v2)

(2.1)

for all λ ∈ V ^∨.

However, let’s let be a basis containing v₁ but not containing v₂, and define γ ∈ V ^∨ to be the map

v ↦→ a

where a is the coordinate for v₁ when v is written in the basis . With this we have that γ(v₁) = 1 and γ(v₂) = a with a≠1 since v₁ and v₂ were assumed distinct. Hence γ(v₁)≠γ(v₂), which contradicts equation 2.1. Therefore j must be injective.

j is F-linear Let v,u ∈ V , a,b ∈ F and λ ∈ V ^∨. Then we have the following

j(av1 + bv2)(λ) = λ(av1 + bv2) = aλ (v1)+ bλ(v2) = aj(v1)(λ)+ bj(v2)(λ) = (aj(v1)+ bj(v2))(λ)

by the linearity of λ. Hence j is F-linear.

(c)

(d)

3

(a) Show that (x,y) in ℂ[x,y] is not principle.

For later contradiction, assume that (x,y) is principle. Then there is some element of f ∈ C[x,y] that generates the ideal (x,y). Since x ∈ (x,y) and y ∈ (x,y), then f must divide both x and y. However, this is a contradiction with the fact that there is no element of C[x,y] that divides both x and y.

(b) Extra Credit

(c) Extra Credit

4

Let T ∈ End_F(V ) for a finite dimensional vector space V over a field F. Denote the dimension of V by n.

(a) Show that if T is diagonalizable, then T is semisimple.

Assume that T is diagonalizable. Then it’s charateristic polynomial is

char(T) = (λ - a1)(λ- a2)⋅⋅⋅(λ - an)

(4.2)

where a₁,…,a_n are the diagonal entries of T as represented in the basis of its eigenvectors. Because the minimal polynomial divides the characteristic polynomial, according to the Caley-Hamilton theorem, then equation 4.2 indicates that the minimal polynomial is

(λ- b1)(λ - b2)⋅⋅⋅(λ- bm)

where m ≤ n and b₁,…,b_m are the distinct elements of {a₁,…,a_n}. Hence T is semisimple.

(b)

(c)

(d) Extra Credit

(e) Extra Credit

5

Let R be a commuitative ring.

(a) Show that R[x] is an integral domain iff R is an integral domain.

Let R[x] be an integral domain. Let a,b ∈ R be elements with ab = 0. Then a and b are also elements of R[x] as constant polynomials. Hence either a or b must be zero as R[x] has no zero divisors.