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 L to be the family of subsets of V defined by
L = {S ⊆ V | S is linearly independent and v ∈ S}

Then L is a poset regarding what it means to be a subset. Let C = {Sα} be a chain in L. Because each Sα is linearly independent, then C must have a maximal element in L since V is a vector space and a basis in a vector space is a maximal linearly independent set. Thus by Zorn’s Lemma, L 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 B 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 B. With this we have that γ(v1) = 1 and γ(v2) = a with a1 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 [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 EndF(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,,an 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,,bm are the distinct elements of {a1,,an}. 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.

Conversely assume that R is an integral domain. ????

(b)


(c)


(d)


(e) Extra Credit


(f) Extra Credit


(g) Extra Credit