Abstract Algebra Homework.pdf

8/9/2019 Abstract Algebra Homework.pdf

1/37

Abstract Algebra Homework

Variously Kenneth Brown, Bryan Hanson, Jing Liu,Austin Mohr, Ranil Wanigasiri, Pam Welch

April 26, 2012

Problem 0Classify up to similarity all the square matrices over the complex numbers with mini-

mal polynomialm(x) = (x1)2(x2)2 and characteristic polynomialc(x) = (x1)6(x2)5.

LetA be a matrix of the desired form. Observe first that, since we are working over C,A can be expressed in Jordan form. Now, the roots ofc(x) are precisely the eigenvaluesofA. Furthermore, a roots multiplicity in m(x) gives the size of the largest elementaryJordan block in A. Since, up to similarity, permutations of the blocks are irrelevant, wehave the following possibilities for A:

1 10 1

1 10 1

1 10 1

2 10 2

2 10 2

2

1


2/37

1 10 1

1 10 1

1 10 1

2 10 2

22

2

1 10 1

1 10 1

11

2 10 2

2 10 22

1 10 1

1 10 1

1

1 2 10 2

22

2

2


3/37

1 10 1

11

11

2 10 2

2 1

0 22

1 10 1

11

11

2 10 2

22

2

3


4/37

Problem 1

LetT :V V be a linear transformation of rank 1 on a finite dimensional vector spaceVover any field. Prove that either T is nilpotent or Vhas a basis of eigenvectors ofT.

Proof. Let dim(V) = n and let { v0} be a basis for R(T) (since rank(T) = 1). Observe

that v0 = T( v0) for some v0 V. By the dimension theorem, the nullity ofT isn 1, and

so let{ w0, , wn2}be a basis for N(T).Observe that, since T = 0, it cannot be that T is nilpotent and V has a basis of

eigenvectors ofT.SupposeT is not nilpotent. We claim that{ v0, w0, , w1} is a basis of eigenvectors

for V. Since T is not nilpotent, we have that T( v0) =T2( v0)= 0. Since Thas rank 1, it

must be thatT( v0) = v0 for some = 0. In other words, v0 is an eigenvector for T. Now,T( wi) = 0 wi for alli, and so each wi is an eigenvector for Twith eigenvalue 0.

Now, the set { w0, , wn2} is independent, since it is a basis for N(T). Suppose v0can be expressed as a linear combination of the wi. Then

vi=n2i=0

ai wi

T(vi) =T(n2i=0

ai wi)

T(vi) =n2i=0

aiT( wi)

T(vi) = 0

which is a contradiction. Hence,{ v0, w0, , wn2}is an independent set ofn eigenvectorsofT, and so a basis for V.

4


5/37

Problem 2

Let Vbe a vector space over a field K.

a. Prove that ifU0 and U1 are subspaces ofV such that U0 U1 and U1 U0, thenV =U0 U1.

Proof. Since U0 U1, there exists a basis vector U0 u0 U1. Similarly, since U1 U0,there exists a basis vectorU0 u1 /U1. Now, consider the linear combination u0 +u1 V.

u0+ u1 /U0 (since u0 /U0)u0+ u1 /U1 (since u1 /U1)

Hence, u0+ u1 /U0 U1, and so U0 U1=V.

b. Prove that ifU0, U1, and U2 are subspaces ofV such that Ui Uj when i=j andKhas at least 3 elements, then V =U0 U1 U2.

Proof. Since no subspace is contained in any other, we can find basis vectors u0 / U0,u1 /U1, and u2 /U2, ui V. Observe that if all the ui are identical, then we have founda vector in Vwhich is in none of the Ui, and soU0 U1 U2=V. Let it be, instead, thatat least one of the ui differs from the rest. Without loss of generality, let u0 differ from therest of the ui. Now, consider the linear combination u0+ u1+ u2 V. It is possible thatu1 = u2, but since K is of characteristic 3, we are assured that u1 +u2 = 0. It follows that,for alli u0+ u1+ u2 /Ui (since ui /Ui)

Hence, u0+ u1+ u2 /U0 U1 U2, and soU0 U1 U2=V.

c. State a prove a generalization of (b) for nsubspaces.

Proposition 1. LetVbe a finite-dimensional vector space over a fieldKof characteristicat leastn. Let{Ui |0 i n 1}be a set of subspaces such thatUiUj fori =j . Then,

n1i=0

Ui =V

Proof. Since no subspace is contained in any other, we can find basis vectors V ui /Uifor each 0 i n 1. Observe that if all the ui are identical, then we have found a vector

in Vwhich is in none of the Ui, and son1i=0

Ui =V. Let it be, instead, that at least one of

the ui differs from the rest. Without loss of generality, let u0 differ from the rest of the ui.

Now, consider the linear combinationn1i=0

ui V. It is possible that u1= u2= = un1,

5


6/37

but since K is of characteristic n, we are assured thatn1

i=1

ui = 0. It follows that, for all

0 j n 1

n1i=0

ui /Uj (since uj /Uj)

Hence,

n1i=0

ui /n1i=0

Ui

and so

n1

i=0

Ui=V.

For the following problems, let

A= A, iB= B,i

be algebras with r-ary operations for each i in some indexed set I. Then, we have thealgebra

A B= A B, i

where

i((a0, b0), . . . , (ar1, br1)) = (i(a0, . . . , ar1),i(b0, . . . , br1))

withaj A and bj B for 0 j r 1.

Problem 3Prove that the congruence relations ofA are exactly those subuniverses ofA Awhich

happen to be equivalence relations on A.

Proof. () Let be an equivalence relation on A. Define the algebraT= T, i for alli Iwhere

T ={(a, a)| a, a A, aa}

It is clear that T A A and defines an equivalence relation (since is an equivalencerelation). To see that it preserves i, let (a0, a

0), . . . , (ar1, a

r1) T. Then, for each i I

i((a0, a0), . . . , (ar1, a

r1)) = (i(a0, . . . , ar1), i(a

0, . . . , a

r1))

6


7/37

Now, since is a congruence relation, we have that

i(a0, . . . , ar1) i(a0, . . . , ar1)

and so

(i(a0, . . . , ar1), i(a0, . . . , a

r1)) T

() LetE= E, ifor all i Ibe an algebra where

E={(a, a)| a, a A, ais equivalent to a }

and soEis a subuniverse ofA A. Define the relation

aa if and only if (a, a) E.

It is clear that is an equivalence relation (since Eis an equivalence relation). To see that preservesi, supposea0a

0, . . . , ar1a

r1. Then, (a0, a

0), . . . , (ar1, a

r1) E. Hence,

for alli I

i((a0, a0), . . . , (ar1, a

r1)) = (i(a0, . . . , ar1), i(a

0, . . . , a

r1))

SinceEis closed under i, we have that

(i(a0, . . . , ar1), i(a0, . . . , a

r1)) E

and so

i(a0, . . . , ar1) i(a0, . . . , a

r1)

Problem 4Prove that the homomorphisms from A toB are exactly those subuniverses ofA B

which are functions from A toB .

Proof. () Let h : A B be a homomorphism. Define the algebra H = H, i for alli Iwhere

H={(a, h(a))| a A}

It is clear that HA B and defines a function from A toB (namelyh). To see that itpreserves i, let a0, . . . , ar1 A. Then, for each i I

i((a0, h(a0)), . . . , (ar1, h(ar1))) = (i(a0, . . . , ar1),i(h(a0), . . . , h(ar1)))

= (i(a0, . . . , ar1), h(i(a0, . . . , ar1)))

H

() LetF= F, i for alli Ibe an algebra where

7


8/37

F ={(a, f(a))| a A}

for some functionf :A B (and soFis a subuniverse ofA B). SinceF is an algebra,we have for a0, . . . , ar1 A and for each i I

i((a0, f(a0)), . . . , (ar1, f(ar1))) = (i(a0, . . . , ar1),i(f(a0), . . . , f (ar1))) F

and so

f(i(a0, . . . , ar1)) = i(f(a0), . . . , f (ar1))

In other words, fis a homomorphism.

Problem 5Prove that the projection functions associated with A Bare homomorphisms.

Proof. Define

0: A B A0(a, b) =a for alla A and b B

Now, let a0, . . . , ar1 A and b0, . . . , br1 B. Then, for each i I,

0(i((a0, b0), . . . , (ar1, br1))) =0((i(a0, . . . , ar1),i(b0, . . . , br1)))

=i(a0, . . . , ar1)

=i(0((a0, b0)), . . . , 0((ar1, br1)))

Similarly, define1: A B B

1(a, b) =b for all a A andb B

Then, for all i I

1(i((a0, b0), . . . , (ar1, br1))) =1((i(a0, . . . , ar1),i(b0, . . . , br1)))

= i(b0, . . . , br1)

= i(1((a0, b0)), . . . , 1((ar1, br1)))

Problem 6Derive a list of equations that follow from the equations axiomatizing the theory of

groups.

Claim 1. The distinguished element 1 in a group is the unique identity.

8


9/37

Proof. Suppose that there exists another identity element 1. Then

1 = 1 1 = 1

Claim 2. The elementx1 is the unique inverse of the elementx.

Proof. Suppose that x has another inverse x. Then

x =x 1 =x (xx1) = (xx)x1 = 1 x1 =x1

Claim 3. (xy)1 =y1x1

Proof. (xy)(y1x1) =x(yy1)x1 =x 1 x1 =xx1 = 1

Claim 4. xnxk =xn+k

Proof. xnxk = (x x) n times

(x x) k times

= x x n + k times

=xn+k

Claim 5. (xn)1 = (x1)n

Proof.

x

n

(x

1

)

n

= (x x) n times

(x

1

x

1

) n times

= (x x) n-1 times

(xx1) (x1 x1) n-1 times

= (x x) n-1 times

1 (x1 x1) n-1 times

= (x x) n-1 times

(x1 x1) n-1 times

=

=xx1

= 1

9


10/37

Problem 7

The five equations used to axiomatize groups are not all needed. Find a simpler set ofequations that will serve.

We take only the following equations for our axiomatization of groups. For all elementsx,y,zin the algebra A= A, , 1, 1

x(yz) = (xy)z

xx1 = 1

x 1 =x

From these, we derive the following.Claim 6. x1x= 1

Proof.

x1x= (x1 1)x

= (x1(xx1))x

= (x1x)(x1x)

(x1x)(x1x)1 = (x1x)(x1x)(x1x)1

1 =x1x

Claim 7. 1 x= x

Proof.

1 x= (xx1)x

=x(x1x)

=x 1

=x

Problem 8State and prove a version of the isomorphism theorem from class that holds for algebraic

systems generally, not just for groups.

10


11/37

Theorem 2. (The Second Isomorphism Theorem) LetA be an algebra and let and be

congruence relations onA such that . Then is a congruence relation onA

and

A= A

Proof. Define h : AAby

h(a) =afor all a A

Claim 8. h is well-defined.

Proof. Suppose a=b. Then (a, b) , and so

a=b.

Claim 9. h is onto A.

Proof. LetaA. We see that h(

a) =a.

Claim 10. h is a homomorphism.

Proof.

h(a b) =h(

ab)

=ab

=a b

=h(a) h(b)

Claim 11. Ker(h) =

Proof.

(a,b) K er(h)h(

a) =h(b)

a=

b

(a, b)

11


12/37

Hence

Ker(h) ={(a, b)| (a, b) }

=

Now, by the Homomorphism Theorem, there is an isomorphism betweenA

and

A, that isA

=A.

Problem 10

Let G be a group. Prove thatG cannot have four distinct proper normal subgroupsN0, N1, N2, N3 so that N0 N1 N2 G and so that N1N3= G andN2 N3 = N0.

Proof. Suppose we have normal subgroups of the desired form. Letn2 N2 G. Then,n2= n1n3for somen1 N1andn3 N3(sinceN1N3 = G). We have thatn3= n

11 n2 N2

(since n11 N1 N2). Hence,n3 N2 N3 = N0 N1. Since n3 N1, we have thatn2= n1n3 N1. Therefore,N2 N1, which implies that N1 = N2. This is a contradictionwith the fact that N1 andN2 are distint.

Problem 11LetH and K be subgroups of the groupG each of finite index inG. Prove thatH K

is also a subgroup of finite index in G.

Claim 12. H Kis a subset ofG.

Proof. Leta H K. Then,a H. SinceH is a subgroup ofG, we have thata G.

Claim 13. HK contains the identity.

Proof. Hand K are both groups, so we have 1 Hand 1 K. Therefore, 1 H K.

Claim 14. HK is closed under the taking of inverses.

Proof. Leta H K. Then,a Handa K. SinceH and K are both groups, we have

that a1 Hand a1 K. Therefore, a1 H K.

Claim 15. HK is closed under addition.

Proof. Let a, b H K. Then, a, b belong to both H and K. Since H and K are bothgroups, we have that a +b belongs to both H and K. Therefore,a +b H K.

12


13/37

By the previous claims, we see that H K is a subgroup ofG.

Claim 16. HK has finite index inG.

Proof. Define the function

f :GH KGH

GKf(a(H K)) = (aH,aK) for all a G

To see that this function is well-defined, suppose a(H K) = b(H K). Then, a and b

belong to the same equivalence class in GH K. We see that

a(H K) aH

b(H K) aHaH=bH

and similarly

a(H K) aK

b(H K) aK

aK=bK

Hence, (aH,aK) = (bH,bK), and so f is well-defined.

We claim that this function is one-to-one. Suppose (aH,aK) = (bH,bK). ThenaH=bHandaK=bK, which implies that a and b belong to the same equivalence class in GH as

well as the same equivalence class in GK. Therefore,aand b belong to the same equiva-

lence class inGH K. That is,a(H K) =b(H K).

Now, f is a one-to-one map sending each coset of GH K into a finite range (sinceeach ofH andKhave finite index in G), so it must be that the domain is finite. That is,H Khas finite index in G.

Problem 12

Prove that there is no groupG such thatGZ(G)

=Z, where Z denotes the group of

integers under addition.

Proof. Suppose GZ(G)= Z for some group G. Since Z is cyclic, GZ(G) is cyclic. Let

aZ(G) generateGZ(G). Then, every coset is of the formanZ(G) for some integern. Since

13


14/37

the cosets partitionG, we have that every element ofG is of the form ancfor some integer

n and some element c Z(G). Let g1, g2 be arbitrary elements ofG with g1 =an1

c1 andg2 = an2c2. It follows that

g1g2= an1c1a

n2c2

=an2c2an1c1 (since c1, c2 Z(G) and powers ofa commute)

=g2g1

Hence, any two elements ofG commute. That is, Z(G) =G, and soGZ(G)= {1}, which

contradicts our assumption that GZ(G)= Z.

Problem 13Let p be the smallest prime that divides the cardinality of the finite group G. Provethat any subgroup ofG of index p must be normal.

Proof. Let H be a subgroup of G having index p in G. Let H be the action of leftmultiplication by G on the cosets of H in G. Since piH is a homomorphism, ker(H)is normal in G and [G : K] = [G : H][H : ker(H)]. Now, since H has p left cosets,Gker(H)

is isomorphic to some subgroup of ofS Y Mp. By Lagranges theorem, we have

that|Gker(H)|= p[H :ker(H)] dividesp!. That is, [H :ker(H)] divides

p!p

= (p 1)!.

The minimality ofp implies that [H :ker(H)] = 1, and so ker(H) =H. Therefore,H isnormal in G.

Problem 14How many elements of order 7 are there in a simple group of order 168?

Proof. Note first that 168 = 23 3 7. By Sylows theorem, we have that |Syl7(G)| 1(mod 7) and |Syl7(G)| | 24. So, |Syl7(G)| is equal to 1 or 8. SinceG is simple, it cannotbe that |Syl7(G)| = 1, as this would imply that G has a unique (and so normal) Sylow7-subgroup. Now, all elements of order 7 appear in a Sylow 7-subgroup. Furthermore,the 8 Sylow 7-subgroups are disjoint except for the identity element. Therefore, there are8 6 = 48 elements of order 7.

Problem 15LetNbe a normal subgroup of the finite group G and let Kbe a Sylow p-subgroup ofNfor some prime p. Prove that G = NG(K)N.

14


15/37

Proof. Let g be an element ofG. Since K is a Sylow p-subgroup of the normal subgroup

N, we have that gK g1

is also a Sylow p-subgroup ofN. By Sylows theorem, all Sylowp-subgroups are conjugate, so we can find an element n of N so that ngKg1n1 = K.The lefthand side can be rewritten as (ng)K(ng)1, and so we see that ng is an element ofNG(K). That is, g = n

1(ng) N NG(K). Since N is normal in G, N NG(K) =NG(K)N,and so we have shown that g NG(K)N.

The reverse inclusion is easy, since NG(K) G and NG. Therefore,G = NG(K)N.

Problem 16Prove that there is no simple group G of order 56.

Proof. Note first that 56 = 23 7. By Sylows theorem, we have that|Syl7(G)| 1 (mod7) and|Syl7(G)| |8. So,|Syl7(G)|is equal to 1 or 8. If|Syl7(G)|= 1, then G has a unique(and so normal) Sylow 7-subgroup. If |Syl7(G)| = 8, then there are 8 6 = 48 elementsof order 7, leaving 8 elements which comprise a unique (and so normal) Sylow 2-subgroup.In either case, we see that G contains a normal subgroup, and so we see that G cannot benormal.

Problem 17Prove that ifG,H, andK are finite Abelian groups andG H=G K, thenH =K.

Proof. Since G, H, and K are finite Abelian groups, they are each isomorphic to somedirect product of cyclic groups of prime power order.

Observe that, sinceG H=G Kand all the groups are finite,

|G||H|= |G H|= |G K|= |G||K|

Hence,|H|= |K|.Now, suppose for the sake of contradiction thatHis not isomorphic toK. Since they are

of the same finite order, it must be that one of the cyclic groups comprising Happears withdifferent multiplicity than inK (note one of these multiplicites may be 0). Let the order ofthis cyclic factor be pk for some prime p. We see that H and Knecessarily have differingnumbers of elements of order pk. This implies that G H and G K do not have thesame number of elements of order pk, and so are not isomorphic, which is a contradiction.Therefore, it must be thatH =K.

Problem 18Prove that every group of order 35 is cyclic.

Proof. Observe first that 35 = 5 7. Consider the Sylowp-subgroups ofG. Sylows theoremgives

15


16/37

n5 1 (mod 5) andn5| 7, so n5 = 1

n7 1 (mod 7) andn7| 5, so n7 = 1

Hence, G has a unique Sylow 5-subgroup N5 and a unique Sylow 7-subgroup N7. Theuniqueness of each of these groups implies that they are normal in G.

Observe next that N5 N7 is trivial, since only the identity element can have orderdividing both|N5| and |N7|. By the Third Isomorphism theorem, we have

N5N7N7=N5N5 N7= N5

Applying Lagranges theorem, we have

N5N7N7 =|N5||N5N7|

|N7| =|N5|

|N5N7|= |N5||N7|= |G|

Hence,G =N5 N7.Now, both ofN5 andN7 are cyclic (and so Abelian), since they are of prime order. We

claim thatG is also Abelian. Let (a1, b1) and (a2, b2) be elements ofG. It follows that

(a1, b1) (a2, b2) = (a1a2, b1b2)

= (a2a

1, b

2b

1)

= (a2, b2)(a1, b1)

Hence,G is Abelian. We see thatG is a finite Abelian group withk2 not dividing its orderfor allk >1. By problem 20, we conclude that G is cyclic.

16


17/37

Problem 19

Describe, up to isomorphism, all groups of order 1225.Proof. Observe first that 1225 = 5272. Consider the Sylow p-subgroups of G. Sylowstheorem gives

n5 1 (mod 5) andn5| 49, so n5 = 1

n7 1 (mod 7) andn7| 25, so n7 = 1

Hence, G has a unique Sylow 5-subgroup N5 and a unique Sylow 7-subgroup N7. Theuniqueness of each of these groups implies that they are normal in G.

Observe next that N5 N7 is trivial, since only the identity element can have order

dividing both|N5| and |N7|. By the Third Isomorphism theorem, we haveN5N7N7

=N5N5 N7= N5

Applying Lagranges theorem, we haveN5N7N7 =|N5|

|N5N7|

|N7| =|N5|

|N5N7|= |N5||N7|= |G|

Hence,G =N5 N7.We proceed by showing N5 is Abelian. Since N5 is of prime power order, it has a

nontrivial center Z(N5). Furthermore, Z(N5) is normal in N5, soN5Z(N5)is a group of

size 1 or 5. If it is of size 5, then it is cyclic. We claim that this is impossible in general.

Claim 17. If a group G properly contains its center, thenGZ(G) is not cyclic.

Proof. Suppose, to the contrary, that GZ(G) is cyclic generated by aZ(G). We argue

that G is Abelian. Let band c be elements ofG. We can find integers m and n so that

bZ(G) =amZ(G)

cZ(G) =a

n

Z(G)This further implies that we can find elements d ande in Z(G) so that

b= amd

c= ane

17


18/37

Observe that d and e commute freely with any element since they are in the center. Fur-

thermore, powers ofa commute with each other. It follows that

bc= (amd)(ane)

= (ane)(amd)

=cb

Hence, G is Abelian, so Z(G) = G, which contradicts our assumption that G properly

contains its center. Therefore, we conclude thatGZ(G)is not cyclic.

Citing the claim above, we conclude that N5Z(N5)is of size 1. In other words, N5 is

Abelian.Similarly, we can show that N7 is Abelian (replace every occurrence of 5 with 7 inthe argument forN5).

18


19/37

Applying the Fundamental Theorem of Finite Abelian Groups, we conclude that G is

isomorphic to one ofZ49 Z25

Z49 Z5 Z5Z25 Z7 Z7

Z7 Z7 Z5 Z5

Problem 20LetGbe a finite Abelian group. Prove that if|G| is not divisible by k 2 for anyk >1,

thenG is cyclic.

Proof. By the Fundamental Theorem of Finite Abelian Groups, G has a unique decompo-sition as a direct product of cyclic groups of prime power order. More precisely,

G= Zp1n1 Zpknk

where each pi is a prime number and each ni 1. Since k2 does not divide the order of

G for any k >1, it must be that ni = 1 for all i (otherwise, p2i divides the order ofG for

some i). We have now

G= Zp1 Zpk

We see that |G|= p1 pk. We claim that G has an element of order p1 pk, and henceis cyclic.

Consider the element (1, . . . , 1) k times

ofG. We see that (1, , 1)m = (0, , 0) if and only

ifm is the least common multiple ofp1, , pk. Since the pi are prime (and so relativelyprime),m = p1 pk. That is, (1, , 1) has order p1 pk, and so G is cyclic.

Problem 21.Prove that Aut(Sn)=Sn for every natural number n.

Proof. Let Sn and define f() = 1 for all Sn. We claim that f is an

automorphism of Sn: for all , Sn we have f( ) = 1 = ( 1)(1) =

f()f(). So f is a homomorphism. Now, f() = f() 1 = 1 = so

f is one-to-one. Also, for all Sn, there exists 1 Sn, such that f(1 ) =(1 )1 = . So, f is onto. Hence, f is an isomorphism from Sn to Sn; i.e. f Aut(Sn)

19


20/37

Now, let : Sn Aut(Sn). By using = f to denote () = f, we have () =

1

for all Sn. Also, is a homomorphism: Let Sn. Then for all , Sn, wehave () = ()()1 =11 =()

1 = (()) = ().

Claim: is one-to-one.

ker

= IdSn () = Id() Sn

1 = Sn

= Sn

SinceSnis the group of permutations, the identity is the only element inSnthat commutes

with every element in Sn. So, ker = {1}. Therefore, is one-to-one. Since both Snand Aut(Sn) are finite, it follows from begin one-to-one that is onto. Thus is anisomorphism. It follows that Aut(Sn)=Sn.

Problem 22.Let p be a prime number. Prove that ifa and b are elements of the symmetric group Sp,wherea has order p andb is a transposition, then {a, b} generates Sp.

Proof. Let a, b Sp and let bbe a transposition. Without loss of generality, let b= (0 1).As a has order p and p is prime p, we have that a is p-cycle. Therefore, ak = (0 1 . . .)for some k. We can re-index the other elements so that we have ak = (0 1 . . . p 1). Let

c = ak

. Then cbc1

= (0 1 . . . p 1)(0 1)(p 1 . . . 0 1) = (0)(1 2)(3)...(p 1) = (1 2).By induction, we have ckbck = c(ck1bck+1)c1 = (k+ 1 k + 2). Therefore, we havethat (0 1), (1 2),..., (p 2 p 1) are generated by {a, b}. Let (xy) be a transposition.Then (x x +1)(x + 1 x + 2) (y 1 y) = (x y) and (x y) is also generated by{a, b}. Asevery permutation can be decomposed into transpositions we conclude that {a, b} generatesSp.

Problem 23.LetHG. Prove that NG(H)/CG(H) is embeddable into Aut(H).

Proof. We begin by recalling the definitions ofNG(H) andCG(H).

NG(H) = {g G | gH=H g}= {g G | H=gH g1}

CG(H) = {g G | gh = hg for allh H}

= {g G | h = ghg1 for all h H}

20


21/37

Now, define a maph : NG(H) Aut(H) byh(n) =nfor eachn NG(H), where we have

n(h) =nhn1

forh H. Since n acts on Hby conjugation, we see that n Aut(H).Claim: h: NG(H) Aut(H) is a homomorphism. Let a, b NG(H). We have

h(ab) = ab

= ab

= h(a)h(b)

Claim: ker h= CG(H). We have

ker h = {n| n(h) =h for allh H}

= {n| nhn

1 =h for all h H}= CG(H)

By the isomorphism theorems, we now have NG(H)/CG(H) = imh Aut(H) which iswhat we wanted to establish.

Problem 24.Let G be a group of order n. Define : G G by(a) = an

2+3n+1 for all a G. Provethat is an automorphism ofG.

Proof. First, we show that an = 1, for all a G. Let a G and|a|= m. By Lagranges

Theorem, |a| divides |G|, i.e. m | n. Let n = mq for some q N. Thenan

= amq

=(am)q = 1. Now, (a) = an

2+3n+1 = an2+3n a = (an)n+3 a = 1 a = a, for all a G.

Therefore,is the identity map fromG to G. So,is an automorphism ofG.

Problem 25.(1) LetIandJbe ideals of a commutative ring R withI+ J=R. Prove that I J=I J.(2) Let I, J, and K be ideals of a principal ideal domain. Prove that I(J+K) =I J+I K

Proof. 1) Let x I J. Then x= xi+xj where xi I and xj J. Therefore, sinceI andJ are ideals, by definition,xixj Iand xixj J. Thus,IJI J.Let x

I

J. Since I + J = R, then 1 = i+ j where i

I and j

J. Then

x= 1x= (i +j)x= ix+jx and therefore,x I J. So, I JI J andIJ=I J.

Problem 26.Let R be a commutative ring and Ibe a proper prime ideal ofR such that R/I satisfiesthe descending chain condition on ideals. Prove that R/Iis a field.

21


22/37

Proof. SinceIis a proper prime ideal ofR,R/Iis an integral domain. Now, pick a R/I

witha = 0, and consider the ideal generated by a. SinceR/Isatisfies the descending chaincondition, we have

(a) (a2) (a3) (ak) = (ak+1) =

So we have ak = bak+1 for some b. Now, since R/I is an integral domain, we see thatbak+1 ak = 0 and so ba = 1. Thus b = a1, and soR/Iis a field.

Problem 27.Let R be a commutative ring and I be an ideal which is contained in a prime ideal P.Prove that the collection of prime ideals contained in P and containing Ihas a minimal

member.

Proof. LetF be the set of all prime ideals that containIand are contained in P. LetCbechain of prime ideals in Fordered by . Observe that Cis an upper bound. Furthermore,I CP. We claim that C is indeed a prime ideal ofR.

We see that 0 is an element of every prime ideal in the chain, so 0 is an element ofC.Let a and b be elements ofC. We have thata and b belong to every prime ideal in

the chain, so a +b andabelong to every prime ideal in the chain. Hence, a +b andabelong to C.

Let a be an element of C and r an element of the ring R. We see that a belongsto every prime ideal in the chain, so ra and ar belong to every prime ideal in the chain.

Hence,raand ar belong toC.Let ab be an element ofC. We see that ab belongs to every prime ideal in the chain.

Hence, either a or b (or both) belongs to every prime ideal in the chain. Since the chainis ordered by , every prime ideal contains a or every prime ideal contains b (or both).In other words, we cannot have prime ideals Pi and Pj with Pi Pj, a Pi b, andaPj b. Hence, either a or b (or both) belongs toC.

The preceeding observations imply that C is indeed a prime ideal ofR. By ZornsLemma, we conclude that Fhas a minimal element (i.e. a maximal element with respectto).

Problem 28.

Let Xbe a finite set and let R be the ring of functions from X into the field R of realnumbers. Prove that an ideal M ofR is maximal if and only if there is an element a Xsuch that M={f |fR andf(a) = 0}.

22


23/37

Proof. To prove the first direction, let X = {x1, x2, . . . , xn}, R = {f | f : X R}.

Without of loss of generality, let a=x1 X, let M = {f | f R and f(x1) = 0}. Now,we show that M is a maximal ideal. Clearly, Mis a proper ideal ofR. LetIbe an ideal ofR such that M I and M =I. Then there existsf1 I, f1 /M. So, f1(x1)= 0. Now,definefi for 2 i n, 1 j n by

fi(xj) =

1 ifj = i0 ifj =i

Then,fi(x1) = 0 for i = 2, . . . , n. So, fi(x) MI for i = 2, . . . , n. Now, let

h(x) := f1(x)

f1(x1)+f2(x) + +fn(x) I

Notice that h(x) = 1 for allxi X. Therefore I=R. This means that M is an maximalideal ofR.To prove the other direction, Let Mbe a maximal ideal ofR. Assume that for all xi X,there existsgi M, such thatgi(xi)= 0. Now, definehi(x) fori= 1, . . . , nandj = 1, . . . , nby

hi(xj) =

1gi(xi)

ifj = i

0 ifj =i

Then,hi(x)gi(x) M for i = 1, . . . , n. Let

(x) :=h1(x)g1(x) +h2(x)g2(x) + +hn(x)gn(x) M.

Also, notice that (x) 1. This means that M = R. This is a contradiction. Thereforeour assumption must be false. SoM {f |fR andf(a) = 0} for somea X. Since wealready showed that the right set is a maximal ideal. So, M={f |fR andf(a) = 0}.

Problem 29.LetR be a commutative ring and letnbe a positive integer. LetJ, I0, I1, . . . , I n1be idealsofR so thatIk is a prime ideal for everyk < n and so that JI0 In1. Prove thatJIk for some k < n.

Proof. When n = 1, the claim is trivial. Suppose it holds for n = k. That is, if J iscontained in a the union ofk prime ideals, then J is contained in one of them. Now, forn= k + 1, J I0 Ik. Suppose that each Ii is necessary. That is, for eachi, we can

find ai Jwith the property that ai Ii but ai /j=i

Ij. If this were not the case (that

23


24/37

is, no such element exists for some Ii) thenJI0 Ii1 Ii+1 Ik, and we are

done after an appeal to the inductive hypothesis. Now, consider the sum j

i=j

ai. This

sum belongs to J (since J is an ideal), so it must belong to the union of the prime ideals.Hence, it belongs to at least one of the prime ideals. Without loss of generality, suppose itisI0. Sincea0 I0 andI0 is an ideal, all terms containing a0 belong toI0. By subtraction,we conclude that a1 ak I0. SinceI0 is prime, we have that one of the ai belongs to I0,which is a contradiction. Therefore, it must be that one of theIi is not necessary, and soJis contained in the union ofk prime ideals. By the inductive hypothesis, we conclude thatJis contained in one of these ideas.

Problem 30.

Let R be a nontrivial commutative ring and let J be the intersection of all the maximalproper ideals ofR. Prove that 1 + ais a unit ofRfor all a J.

Proof. Let a J. Then a Mfor each M. Since each Mis a proper maximal ideal, we have1 /M, and in particular, 1+a /M. As each Mis maximal,R/Mis a field. Since 1+a /M,we see that 1 + a + M= 0 + M. Now,R/Mis a field, so there exists bM+ MR/M suchthat (1+a +M)(bM+M) = 1+M. Then (1+a)bM+M= 1+M(1 +a)bM1 0 +M;in other words, (1 +a) is a unit ofR.

Problem 31.Let F be a field and let p(x) F[x] be a polynomial of degree n. Prove thatp(x) has at

most n distinct roots in F.

Proof. We proceed by induction on the degreen. Base step: ifn= 1, then letp(x) =ax +bfor somea, b F,a= 0. Letp(x) = 0, we have ax = b. SinceFis a field, we can multiplyboth sides of the equation by the inverse ofa. It follows thatx = b

aF is a root ofp(x).

Induction step: Assumep(x) has at most n roots for n m. Now, consider n = m+ 1,let c F such that p(c) = 0. Then, by the factor theorem,p(x) = q(x)(x c) for someq(x) F[x] with deg q(x) m. By the induction hypothesis, q(x) has at most m distinctroots. Therefore,p(x) has at most m+ 1 =n distinct roots.

Problem32. LetF be a field and let F be its (multiplicative) group of nonzero elements.

LetGbe any finite subgroup ofF

. Prove that G must be cyclic.

Proof. As G is a finite abelian group, the fundamental theorem of finite abelian groupsgives us

G=Cd1 Cd2 Cdr

24


25/37

where each di is a prime power. Let m= lcm[d1, d2, . . . , dr] and note that m d1d1 dr.

For each ai Cdi we have am

i = 1 since adi

i = 1 in Cdi and di |m. Thus for alla G wehave am = 1, and so every element ofG is a zero of the polynomial xm 1. By a previoushomework problem, we know that xm 1 has at most m zeros in F. Since am 1 = 0for all elements ofG we must have m d1d2 dr. Hence m= d1d2 dr. Therefore theprimes in the prime powers di are distinct, and we see that G =Cm. ThusG is cyclic.

Problem 33.Suppose that D is a commutative ring such that D[x] is a principal ideal domain. ProvethatD is a field.

Proof. Let d D, d = 0. We want to show that d has a multiplicative inverse. Let

I= (d, x), the ideal generated by dandx. ThenI={u(x) d+v(x) x| u(x), v(x) D[x]}.Since D[x] is a principal ideal domain, we may write I = (a(x)) for some a(x) D[x].Notice thatd I, soa(x)| d. This shows that deg a(x) = 0. On the other hand,x I, soa(x)| x. Since x is irreducible in D[x], in other words, the only possible divisors ofx areunits and associates ofx, we see that a(x) is a unit. Therefore 1 I. So, I=D[x]. Also,notice that Iis the set of polynomials whose constant term is a multiple ofd. Since 1 I,we conclude that 1 is a multiple ofd. This means that d has a multiplicative inverse. So,D is a field.

Problem 34.Is the polynomial y 3 x2y2 +x3y+x+x4 irreducible inZ[x, y]?

Proof. Suppose that P = y3 x2y2 +x3y+ x+x4 = g(x, y)h(x, y) where g, h Z[x, y].Without loss of generality, assume that there exists a 1 term in g and anx term inh. Thenthere can not be an x, x2, or x3 term in g as multiplying any of these terms gives a termnot inP. Similarly, there cannot be 1, y ,y 2, or y 3 term inh. However, this contradictsy 3

being a term inp. Thus,Pcannot not be factored. As the multiplicative inverse ofP is 1P

and 1P

is not inZ[x, y] we have P is not a unit. Thus,P is irreducible.

Alternatively, one can observe that Z[x, y] = (Z[x])[y]. Since Z is a unique factorizationdomain, Z[x] is a unique factorization domain. We can view the given polynomial as apolynomial in the indeterminate y having coefficients in the ring Z[x]. Now, let P = (x),

the ideal ofZ[x] generated by the element x. We see that P is a prime ideal and thatx2 and x3 are elements ofP, while 1 is not an element ofP. Furthermore, observe that

P2 ={i,j

xnixnj | ni 1, nj 1}. Hence,x + x4 is not an element ofP2. By Eisensteins

Criterion, we conclude that the specified polynomial is irreducible in Z[x, y].

25


26/37


27/37

Similarly, rJ

= r

J. With these facts, we see

( rI J

) = ( rI

, rJ

)

= (r

I,

r

J)

=( r

I J)

Claim is a homomorphism.

Proof of Claim: We see that respects addition

( r

I J +

s

I J) =(

r+s

I J)

= (r+s

I ,

r+s

J )

= (r

I +

s

I,r

J +

s

J)

= (r

I,r

J) + (

s

I,s

J)

=( r

I J) +(

s

I J)

and multiplication

( r

I J

s

I J) =(

r s

I J)

= (r s

I ,

r s

J )

= (r

I

s

I,r

J

s

J)

= (r

I,r

J) (

s

I,s

J)

=( r

I J) (

s

I J)

Claimker() =I J

Proof of Claim:r

I J ker() (

r

I J) = (

0

I,0

J)

r I andr J

r I J

27


28/37

Claimim() = RI

RJ

Proof of Claim: Let ( rI

, rJ

) RI

RJ. We have ( r

IJ) = ( r

I, rJ

), so ( rI

, rJ

) im().

Finally, invoking the Homomorphism Theorem, we have

RIJ

= RI RJ

Theorem 3 (First Isomorphism Theorem).Letf : G H be a homomorphism with kernel K. ThenK is a normal subgroup of GandG/K=imf.

Theorem 4 (Second Isomorphism Theorem).IfH andNare subgroups ofG withNnormal inG, thenHN is a subgroup ofG and

HN/N=H/(H N).

Theorem 5 (Third Isomorphism Theorem).LetN andKbe normal subgroups ofG such thatN is a subgroup ofK. ThenK/N is anormal subgroup ofG/N and

(G/N)/(K/N)=G/K

Theorem 6 (Lagranges Theorem).LetG be a group and letHbe a subgroup ofG. Then

|G|= [G: H]|H|

In particular, ifG is finite, then|H| divides|G|.

Problem 1.Let G be a finite group, let Hbe a subgroup ofG, and let Nbe a normal subgroup ofG.Suppose further that|N| and [G: N] are relatively prime and that |H|divides|N|. Provethat HN.

28


29/37

Proof. First, observe that HNH=H N[H :H N] = 1. So we wish to show

that [H :H N] = 1. The Second Isomorphism Theorem tell us that

HN/N=H/(H N)

So we have [HN :N] = [H :H N]. Recall that |N| and [G: N] are relatively prime. Soevery factor of |N| is relatively prime to every factor of [G : N]. In particular, since |H|divides |N|,|H| is relatively prime to every factor of [G: N]. But [G: N] = [G: H N][HN :N]. So |H| is relatively prime to [HN :N]. Thus|H| is relatively prime to [H :H N].However, Lagrange tells us that [H :H N] divides |H|. Therefore [H :H N] = 1, andso HN.

Problem 2.Let G be a finite group, let Hbe a subgroup ofG, and let Nbe a normal subgroup ofG.Suppose further that|N| and [G: N] are relatively prime and that |H|= |N|. Prove thatH=N.

Proof. The Second Isomorphism Theorem tell us that

HN/N=H/(H N)

so that [HN :N] = [H :H N]. Now, HNG, so Lagrange tells us that [G: N] = [G:HN][HN : N]. Since [G : N] and |N| are relatively prime, it follows that [HN : N] and

|N| are relatively prime. Now, [HN :N] = [H :H N], so we see that [H :H N] and|N|are relatively prime. Since|N|= |H|, it follows that [H :H N] and|H|are relativelyprime. However, Lagrange tells us that [H : H N] divides |H|. Thus [H : H N] = 1.SoH=H N, and so HN. Since|H|= |N| and H, Nare finite, we have H=N.

Problem 3.Let G be a finite group, let Hbe a subgroup ofG, and let Nbe a normal subgroup ofG.Suppose further that |N| and [G: H] are relatively prime. Prove that NH.

Proof. By the Second Isomorphism Theorem, we have

HN/N=H/(H N)

so that|HN|

|N| =

|H|

|H N|.

29


30/37

Rearranging the terms, we have

|HN|

|H| =

|N|

|H N|.

Thus [HN : H] = [N : H N]. By Lagrange, [HN : H] divides [G : H] since HN G.Similarly, we have [N :H N] divides |N|. Since |N| and [G: H] are relatively prime, itfollows that [HN :H] = [N :H N] = 1, and thus N=H N. Hence NH.

Ifa G, then we define thecentralizerofa in G, denoted by CG(a), to be the set ofall g G that commute with a.

CG(a) ={g| g G and gag1

=a}

It is immediate that CG(a) is a subgroup ofG.IfHG, then we define the normalizerofH in G, denoted byNG(H), to be

NG(H) ={a| a G andaH a1 =H}

It is immediate thatNG(H) is a subgroup ofG. Also, notice thatHNG(H). By definition,NG(H) is the largest subgroup ofG in which His normal.

Theorem 7 (Sylow I).LetG be a finite group, letp be a prime number, and letk N. Ifpk | |G|, thenG has asubgroup of orderpk.

Theorem 8 (Sylow II).LetG be a finite group and letp be a prime.

a) All Sylowp-subgroups ofG are conjugate. In particular, ifP is a Sylowp-subgroupofG, then all Sylowp-subgroups ofG are conjugate to P.

b) Letnp denote the number of Sylowp-subgroups ofG. Thennp 1 modp.

c) np| [G: P] for any Sylowp-subgroup P ofG.

d) IfHG and|H| is a power ofp, thenHP for some Sylowp-subgroup ofG.

30


31/37

Problem 4. Letp, q Zbe primes withq p and p 1 (mod q). Prove that any group

of order pqis abelian.Proof. Let P be a Sylow p-subgroup ofG and let Q be a Sylow q-subgroup ofG. Since|P| = p and |Q| = q, we have P = Zp and Q = Zq. From the Sylow theorems, we havenp 1 (mod q) and np| q. Sincep1 (mod q) and p, qare primes, we have nq = 1 sincenq |p. (Otherwise nq =kq+ 1 1 (mod q)).) So nq = 1 gives us that Q is a unique, andhence normal Sylow q-subgroup ofG. Similarly, we havenp = 1 since q p. ThusP isalso a unique normal Sylow p-subgroup ofG. Since gcd(p, q) = 1, we have P Q= {e}.From the isomorphism theorems, we get PQ/Q = Q/(P Q) and since |P Q| = 1,|P Q|= |P||Q|. ThusG= P G, and soG=P Q

Problem 5.LetG be a finite group and letHGso that|H|= pk for some primep and some positiveinteger k . Prove that H is contained in every Sylow p-subgroup ofG.

Proof. Let G be a group with p| |G|. Since H is a p-subgroup ofG, the Sylow theoremstell us that H is contained in some Sylow p-subgroup P ofG. Now, for g G we havegHg1 gP g1 =P, whereP is a Sylow p-subgroup ofG. Thus every conjugate ofH iscontained in the corresponding conjugate ofP. Now,His a normal subgroup ofG, so thismeans that H is contained in every conjugate ofP. ThusH is contained in every Sylow

p-subgroup ofG.

Problem 6.Let G be a finite group and let P be a Sylow p-subgroup of G. Suppose that H is a

p-subgroup ofNG(P). Prove that HP.

Proof. RecallNG(P) ={g G| gP g

1 =P}

We havePNG(P). SinceH is ap-group, the Sylow theorems tell us that H is containedin some Sylow p-subgroupP. Also by the Sylow theorems, we know thatP and P mustbe conjugates in NG(P). However,P is a normal subgroup ofNG(P), so P is the unique

Sylow p-subgroup ofG. Thus P =P

, and henceHP.

Problem 7.LetG be a finite group and let Pbe a Sylow p-subgroup ofG, wherep is a prime number.Prove that p and [NG(P) :P] are relatively prime.

31


32/37

Proof. We need to show that p[NG(P) :P]. By the definition of Sylowp-subgroups, we

know that |P| = pk

, where pk

| |G| but pk+1

|G|. Now, by Lagrange, we have |G| = [G:P]|P|. Thus p[G: P]. Also by Lagrange, we have [G: P] = [G: NG(P)][NG(P) :P], sowe see that p [NG(P) :P].

Problem 8. (Frattini Argument)LetNbe a normal subgroup of the finite groupG, and let Pbe a Sylowp-subgroup ofN.IfNG(P) is the normalizer ofP inG, show that G= NG(P)N.

Proof. Let g NG(P)N. As NG(P) and N are subgroups ofG, we clearly have g G.Thus NG(P)N G. For the other direction, suppose a G. By normality ofN, we havea1P a a1Na= N, soP anda1P aare conjugate Sylowp-subgroups ofN. Thus there

existsn Nso that

P =n(a1P a)n1

= (an1)1P(an1)

From this, we conclude that an1 NG(P), and so a NG(P)N. ThusG NG(P)N.HenceG = NG(P)N.

Problem 9.LetGbe a nonabelian group. Prove that G/Z(G) is not cyclic.

Proof. Recall that we define the centerofGby

Z(G) ={a G| ag = ga for everyg G]}

We prove the contrapositive. Suppose that G/Z(G) is cyclic. We need to show thatG is abelian. Suppose aZ(G) is a generator for G/Z(G). Then if g1, g2 G, we haveg1Z(G) = a

iZ(G) for some i. Thus we have g1(ai)1 = zi Z(G). Similarly, we have

g2Z(G) = ajZ(G) for some j. So g2(aj)1 = z2 Z(G). Since cyclic groups are abelian,

we have

g1g2= aiz1a

jz2

=z1z2ai+j

=z2a1aj+i

=z2ajz1a

i

=g2g1

We have shown that g1g2= g2g1 forg1, g2 G. Thus G is an abelian group.

32


33/37

Problem 10.

Prove that every finite nontrivial p-group has a nontrivial center.Proof. Let Pbe a finite nontrivial p-group. Let Pact on itself by conjugation. From theconjugacy class equation, we have

|P|= |Z(P)| +x

[P :C(x)]

where the sum runs over conjugacy class representatives with |orbitx| > 1. Now, |P| is aprime power, and each [P :C(x)] is divisible by p, so we must have |Z(P)| divisible by p.Thus|Z(P)| = 1, and so Z(P) is nontrivial.

Problem 11.Prove that the polynomialy 3 x2y2 +x3y+x+x4 is irreducible inZ[x, y]

Proof. Let

f=y3 x2y2 +x3y+x+x4

=a3y3 +a2y

2 +a1y+a0

Notice that x is an irreducible element ofZ[x]. So considerfas an element of (Z[x])[y].We have x | a2, x | a1, x | a0, and x a3. Also, we havex

2 a0, so Eisensteins criteriontells us that f is irreducible in (Z[x])[y] = Z[x, y].

Problem 12.Prove that the polynomialx3y+x2y xy2 +x2 +y is irreducible in Z[x, y].

Proof. Let

f=x3y+x2y xy2 +x3 +y

= (y+ 1)x3 +yx2 + (y2)x+y

Now,y is an irreducible element ofZ[y], so consider fas an element of (Z[y])[z]. Observethat y | y , y | (y2), and y | y , but y (y+ 1). Also, y2 y , so Eisensteins criterion tellsus that f is irreducible in (Z[y])[x] = Z[x, y].

33


34/37

Problem 13.

LetRbe a commutative ring, and letN={r| r R and rn = 0 for some positive integer n}.

a) Prove thatNis an ideal ofR.

b) Prove thatNP for every prime ideal P ofR.

Proof. a)Clearly 0 N since 01 = 0 and we can take n= 1. Now, let a N and r R.Picknso thatan = 0. Since we are in a commutative ring, we have (ra)n =rnan =rn0 = 0.Thus ra N. Now, suppose a, b N. We must show that a+ b N. Pick positiveintegers m, n such that am = bn = 0. We know that the Binomial Theorem holds in any

commutative ring, so we have

(a +b)m+n =m+nk=0

m+n

k

akbn+mk

In each term in the sum above, either ak = 0 or bm+nk = 0. This means (a+b)m+n = 0,so a +b N as desired.

b)LetPbe a prime ideal and leta N. Thenan Pfor somensincean = 0 P. Weargue by induction on n to show that a P. In the base step, we clearly have a1 =a P.So suppose ak P for all k n. Then we have ak+1 = aak P since either a P orak Pby the induction hypothesis. Thus NP.

Problem 14.LetRbe a commutative ring, let Ibe an ideal ofR, and let

N={r| r R andrn I for some positive integer n}.

a) Prove thatNis an ideal ofR.

b) Prove thatNP for every prime ideal P ofR such that IP.

Proof. a)First we check that N is an ideal. We see that 0 N since 01 = 0 I becauseIis an ideal. Next, ifr, s Nwe can pick positive integers m, nso that rn, sm I. Sincethe Binomial Theorem holds in commutative rings, we have

(r+s)m+n =m+nk=0

m+n

k

rksm+nk

34


35/37


36/37

Let C be a chain of ideals belonging to F. We know that the union of any chain of

ideals is again an ideal. So all we need to see is that an

/ Cfor every positive integer n.If it were the case that an C, then we could pickI Cso that an I. However,I Fso we know that an /I. Thus

C F. So

Ccan serve as the desired upper bound in

F. Then Zorn tells us that Fhas a maximal member.

Problem 16.Let R be a ring, let Ibe an ideal ofR, and let a R so that an / I for every positiveinteger n. Prove that there is an ideal P with the following properties:

i) an /P for all positive integers n

ii) IfJis an ideal that properly includes P, then am J for some positive integer m.

Proof. Let F = {K | K is an ideal with I Kand an / Kfor all positive integers n}.Notice that F is not empty since I F. Also notice thatF is ordered by set inclusion.Let Cbe any chain in F. We know that

C is an ideal, since the union of any chain of

ideals is an ideal. Plainly,

C is an upper bound ofCand also

C includes I. So to seethat

C F, we only need that an /

C for each positive integer n. But since an /K

for all K C, this follows immediately. So by Zorns Lemma, there is Pwith is maximalinF. Now, supposeJ is an ideal that properly includesP. ThenJ / Fby the maximalityofP. This means that there must be a positive integer m so that am J.

Problem 17.Provide an example of a unique factorization domain that is not a principal ideal domain.Prove that your example has the desired properties.

Proof. Let F be any field. Then F[x, y] is a UFD. Let Ibe ideal generated by {x, y}.So Iconsists of those polynomials with constant term 0. Suppose, for contradiction, thatthere exists a polynomial p(x, y) that generates I. Then p(x, y)| x andp(x, y)| y and theconstant term ofp(x, y) is 0. This means that there are polynomials f(x, y) and g(x, y)such that x = f(x, y)p(x, y) and y = g(x, y)p(x, y). Let the x-degree of a polynomial bethe largest n such that xn occurs in the polynomial. Similarly define they-degree. Now,multiplying polynomials never decreases these degrees. So we find thatp(x, y) must havex-degree 0 sincey has x-degree 0. Similarly,p(x, y) must havey-degree 0 sincex hasy-degree0. This means that neither x nor y occur inp(x, y). Since the constant term ofp(x, y) is 0,we can only conclude that p(x, y) is the zero polynomial. But the zero polynomial cannotgenerateI. So we have the desired contradiction.

36


37/37

Alternatively, one can show that Z[x] is a UFD but not a PID. Since Z is a UFD, Z[x]

is also a UFD. Let Ibe the ideal generated by {2, x}. Suppose, for contradiction, that I isgenerated by a single element, say f. Then f |2 and f |x, so there exist g, h Z[x] suchthatf g= 2 andf h= x. Now,f g= 2 shows thatf Z. Alsof h= x shows thatf=1.Thus we must have I=Z[x]. However, the constant term of every polynomial in (x, 2) isa multiple of 2, so we see that 1 /(2, x) =I. ThusIis a proper ideal ofZ[x], and we havethe desired contradiction.

37

Abstract Algebra Homework.pdf

Documents

Transcript of Abstract Algebra Homework.pdf