Dummit and Foote Soln

7/24/2019 Dummit and Foote Soln

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MAT7400 Assignment #3 1

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Dummit & Foote Text Exercise 3.4-2: Exhibit all 3 composition series for Q8 and all 7 compositionseries for D8 . List the composition factors in each case.

Solution: In Exercise 3.1-32, we showed every subgroup ofQ8 is normal. Of the normal subgroups,i,j, andk are maximal. In addition, these three only have one non-trivial subgroup. Every quotient ineach series has order two:Q8/

i

,

i

/

1

/,

1

/1,

Q8/j,j/1/,1/1,Q8/k,k/1/,1/1,So the every quotient is isomorphic to the simple abelian group Z/2Z. This gives us the following threecomposition series for Q8:

1 1 i Q8,1 1 j Q8,1 1 k Q8.

Exercise 3.1-33 informed us that there are three maximal subgroups ofD8 which are normal:s, r2,r,andrs,r2. Each have order four.s, r2 has three subgroups of order two: s,r2,r2s. Since each of these subgroups have index two,they are normal tos, r

2

.r has one subgroup of order two:r2 and since its index is two, it is normal in r.rs,r2 has three subgroups of index two (and hence normal): r2,rs,r3s.Again, every quotient is isomorphic to the simple abelian group Z/2Z. This gives us the following sevencomposition series for D8:1. 1 s s, r2 D82. 1 r2s s, r2 D83. 1 r2 s, r2 D84. 1 r2 r D85. 1 r2 rs,r2 D86. 1 rs rs,r2 D87. 1 r3s rs,r2 D8

Dummit & Foote Text Exercise 3.4-5: Prove that subgroups and quotient groups of a solvable groupare solvable.

Solution: A subnormal series of a group G is a chain of subgroups

G= Gs> G2 > G1 > G0such thatGi Gi+1 for all i.

LetG be a solvable group. So G has a subnormal series 1 =G0 G1 Gs1 Gs= G with abelianfactors: for each i = 1, . . . , s, Gi+1/Gi is an abelian group. Let H G. ConsiderH Gi andH Gi+1.Each is a group. Ifg H Gi andh H Gi+1, thenhgh1 Hsinceg, h H. Also, hgh1 Gi. SinceGi Gi+1, H Gi H Gi+1. This also implies that H Gi+1 NG(H Gi). HenceH is solvable andthe composition (subnormal) series is

1 H G1 H Gs1 H Gs= H G= H.We can also use the second (diamond) isomorphism theorem to show this as well. Again let H Gand foreach i let Hi = Gi H. Then for every i, Gi1 Gi, and we have the subnormal series

1 = H0 H1 Hs1 Hs= H.By the second (diamond) isomorphism theorem, for every i,

Hi/Hi1 = Hi/(Gi1 H) =(HiGi1)/Gi1,


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which is a subgroup of the abelian group Gi/Gi1 and therefore is abelian. Hence, Hhas a subnormalseries with abelian factors and that implies H is solvable.

(I had help with this part of problem.)LetKbe a quotient group ofG with K=G/N with N G. Define : G Kas the projectionhomomorphism. and last define Ki= (Hi). Then for each i, Ki1 Ki. We have now obtained thesubnormal series:

1 =K0 K1 Ks1 Ks = K.Again by the second (diamond) isomorphism theorem we obtain

Ki/Ki1 = (GiN)/(Gi1N) =Gi/(Gi Gi1N)which is a quotient ofGi/Gi1 by the third isomorphism theorem sinceGi1 Gi Gi1N . Thus, thequotient groups Ki/Ki1 are abelian and therefore K is solvable.//

Dummit & Foote Text Exercise 3.5-10: Find a composition series for A4. Deduce that A4 is solvable.

Solution: It is quite fortuitous for me as Prof. Okoh had a very similar problem assigned as an examproblem!

LetH= {(), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}. We know thatA4 = {(), (1, 2, 3), (1, 3, 2), (2, 3, 4), (2, 4, 3), (1, 2, 4), (1, 4, 2), (1, 3, 4), (1, 4, 3), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}.SinceHis a finite and nonempty subset ofA4 and closed under the binary operation, it is a subgroup ofA4.To show it is a normal subgroup. We need to show that aha1 Hfor all h H anda A4.Since conjugation preserves shapes (see Example 7.2.3 in Beachy and Blairs text),aH a1 preserve thecycle structure ofH. SinceHcontains all of elements ofA4 that have this cycle structure, aH a

1 =H.HenceHis normal in A4.The definition of solvable is a chain of normal subgroups G = N0 N1 N2 Nn = {e} such that

(1)Ni Ni1 fori = 1, 2, 3, . . . n(2)Ni1/Ni is abelian for i = 1, 2, 3, . . . n(3)Nn =

{e

}.

The chain to verify is A4 H {e} where G = N0, A4 = N1, H= N2 andN3 = {e}.(1)H A4 (just proven) and{e} H since{e} is the trivial normal subgroups to all subgroups.(2) By Lagrange, the factor group A4/Hwill have three cosets. Since this is a group of prime order,A4/H= Z/3Z. So it is cyclic and therefore abelian.H/{e} = {(1), (12)(34), (13)(24), (14)(23)} =H. So His the identity in the factor group and the identitycommutes with itself. So H/{e} is abelian.(3)N3 = {e}.ThereforeA4 is solvable.//

Dummit & Foote Text Exercise 3.5-11: Prove that 4 has no subgroup isomorphic to Q8.

Solution: Assume a subgroup H 4 exists such thatH=Q8.Q8 contains 6 elements of order 4 namely (i, i,j, j, k, andk). 4 also contains exactly 6 elements oforder 4: the 4-cycles. So H contains all 4-cycles in Sigma4. Since His a subgroup, it is closed under thebinary operation. That implies Hcontains elements of order three (e.g., (1234)(1342) = (143) and(1234)(1423) = (243)). Q8 has no elements of order 8 (by Lagrange). A contradiction. And # H >8 sinceit contains six elements of order four, one identity, and more than one element of order three. Anothercontradiction. Therefore, 4 has no subgroup isomorphic to Q8.//


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Dummit & Foote Text Exercise 7.1-1 Let R be a ring with 1. Show that (1)2 = 1 in R.Solution:

(1)2 + (1) = (1)(1) + (1)= (1)((1) + 1)= (1)(0)= 0 (1)2 = 1.

and(1) + (1)2 = (1) + (1)(1)

= (1 + (1))(1)= (0)(1)= 0 (1)2 = 1.

Dummit & Foote Text Exercise 7.1-2: Prove that if u is a unit in R then so isu.Solution: Sinceu is a unit, there exists some v R such that uv = vu = 1. From Exercise 7.1.1, we knowthat (1)2 = 1. So 1 =uv = u 1 v= u(1)2v= u(1)(1)v= (u)(v) and1 = vu = v 1 u= v(1)2u= v(1)(1)u= (v)(u). Henceu is a unit.//

Dummit & Foote Text Exercise 7.1-4: Prove that the intersection of any nonempty collection ofsubrings of a ring is also a subring.

Solution: Let R be a ring and let Xbe a nonempty set of subrings ofR.We know that

X R is a subgroup. To show thatXis closed under multiplication, let a, b X.Thena, b Sfor all S X, and xy Sfor all S X. HencexyX, and by definitionX R is asubring. Note that the distributive properties did not have to be checked as those are inherited ifmultiplicative closure holds.//

Dummit & Foote Text Exercise 7-1-5: Decide which of the following (a)- (f) are subrings ofQ:(a) the set of all rational numbers with odd denominators (when written in lowest terms)(b) the set of all rational numbers with even denominators (when written in lowest terms)(c)the set of nonnegative rational numbers(d) the set of squares of rational numbers(e)the set of all rational numbers with odd numerators (when written in lowest terms)(f )the set of all rational numbers with even numerators (when written in lowest terms).

Solution: (a) Let Xbe the given set.Associativtity in Xfor addition is inherited from Q.Xis closed under addition. If a

b, cd X, thenb and d are odd. Therefore bd is odd. Next a

b+ c

d= ad+bc

bd

and is equal to some other fraction pq

in lowest terms, such that q|bd. Because bd is odd, qmust also beodd. So Xis closed under addition.0

1 is the additive identity element for X since 0

1+ a

b = a

b = a

b+ 0

1 for all a

b X.

For every ab X, a

b X. Because a

b+ a

b = 0

b = 0

1 and a

b + a

b = 0

b = 0

1, every element ofXhas an

additive inverse.SoX is a group under addition. iffracab and c

dare written in lowest terms and b and d are odd, then

whenac/bd is written in lowest terms, its denominator must divide bd and therefore is odd. Hence X isclosed under multiplication and Xis a subring.//

(b) Let Xbe the given set. Ifa is odd, then 2a is even. So 12a

Xbut 12a

+ 12a

= 1a X. Since X is not

closed under addition, Xcannot be a subring ofQ.

(c)This set does not contain additive inverses for all of its elements and therefore it is not a group underaddition. Hence, it cannot be a subring ofQ.

(d) This set X, is not closed under addition. (13

)2 Xbut ( 13

)2 + ( 13

)2 = 29 X since 2 Q and

therefore cannot be in X.


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(e)This set Xis not closed under addition. 15

+ 15

= 25 X but 1

5 X.

(f )Let Y be the given set. If ab

and cd

are in lowest terms and a and c are even, then ab

+ cd

= adbcbd

. We

must have b and d be odd because when reduced to lowest terms, the numerator of adbcbd

is even since 2divides a and b.SinceYcontains 2 = 2

1, Y Q is a subgroup. Now if a

b and c

d Yand in lowest terms, then b and d are

odd. Thus when expressed in lowest terms, 2 must divide ac. Thus Yis closed under multiplication. Now

to check for the multiplicative identity. Assume Yhas an identity element cd . Then for all

ab we haveca

db= a

b, so that cab = dab. Thusc = d. Sincec must be even and d odd, we have a contradiction. So Y is a

not ring with a multiplicative identity. Yis a ring but not a unital ring. The definition of subring requiresthe identity element of the group operation (addition) and Y meets that definition. Ycan be considered asubring ofQ.

Dummit & Foote Text Exercise 7.1-7: The center of a ring R is{z R | zr = rz for all r R}(i.e., isthe set of all elements which commute with every element ofR). Prove that the center of a ring is asubring that contains the identity. Prove that the center of a division ring is a field.

Solution: Z(R) contains the zero element (additive identity). 0 Z(R) because 0 r= 0 = r 0 for allr RHence, Z(R) is nonempty. Ifa, b Z(R) andr R, then (a b)r= ar br= ra rb= r(a b). Bythe subgroup criterion, Z(R)

R. Next abr = arb = rab, so that ab

Z(R). Hence by definition of a

subring, Z(R) is a subring.IfR has a 1 (multiplicative identity), then, 1 a= a1 = a for all a R. Hence 1 Z(R).Next letR be a division ring, and consider its center Z(R). Ifa Z(R), then by the cancellation law, theinverse ofa is unique. Let the inverse ofa be a1. We know that (r1)1 =r. Since (ab)(b1a1) = 1, wehave (ab)1 =b1a1. Now letr R. a1r1 = (ra)1 = (ar)1 =r1a1. Since r1 R, a1 Z(R).SinceZ(R) is a commutative division ring, it is a field.

Dummit & Foote Text Exercise 7.1-13: An elementX in R is called nilpotent ifxm = 0 for somem Z+. (a) Show that ifn = akb for some integers a and b then ab is a nilpotent element ofZ/nZ.(b) Ifa Z is an integer, show that the element a Z/nZis nilpotent if and only if every prime divisor ofn is also a divisor ofa. In particular, determine the nilpotent elements ofZ/72Z explicitly.(c)Let R be the ring of functions from a nonempty setXto a field F. Prove that R contains no nonzero

nilpotent elements.

Solution: (a) Let n = akb, where k 1. Then (ab)k =akbk = (akb)bk1 =nbk1 0 mod n.(b) () Assume a Z/(n) is nilpotent. Then am =nk for some m and k . Ifp is a prime that divides n,thenp divides am, so that it divides a. Hence every prime that divides n divides a.() Let n = pb11 pbkk anda = pc11 pckk m, where 1 bi, ci for all i and m is an integer. Let t = max{bi}.Thenat = (pc11 pckk m)t =pcit1 pcitk mt, where cit bi for eachi. Hence at =nd for some integerd, andwe have at 0 mod n.Since 72 has prime factorization 2332 and since every prime that dividesn must dividea the only nilpotentelements in Z/nZare those a such that a is a multiple of 6 (for any representative choice ofa):{0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 50, 66}.(c) Proof by contradiction. Assume f R is nilpotent. Iff= 0, then there exists x Xsuch thatf(x) = 0. Let m be the least positive integer such that f(x)m = 0. Then f(x)f(x)m

1

= 0 andf(x) = 0andf(x)m1 = 0. We have our desired contradiction as F contains zero divisors. Therefore no nonzeroelement ofR is nilpotent.//


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Dummit & Foote Text Exercise 7.1-14: Letx be a nilpotent element of the commutative ring R (cf.the preceding exercise).(a) Prove thatx is either zero or a zero divisor.(b) Prove that rxis nilpotent for all r R.(c)Prove that 1 + x is a unit in R.(d) Deduce that the sum of a nilpotent element and a unit is a unit.

Solution: (a) Let n be the least positive integer such that xn = 0. Ifn = 1, then x = 0. Ifn >1, thenx = 0, xn1 = 0 and x xn1 = 0, then x is a zero divisor by definition of a zero divisor.//(b) Let x be defined as in part (a). Because R is commutative, (rx)n =rnxn =rn 0 = 0.//(c)Let x be defined as in part (a). Ifx = 0, then clearly 1 = 1 + x is a unit in R. So assume x R andx = 0. We know that 1 = (1 xn) = (1 + x)(1 x + x2 + (1)n1xn1). We also know from the closureproperties of a ring, (1 x + x2 + xn1) R. Thus 1 + x is a unit in R.//(d) Let x be defined as in part (a). Ifu is any unit, then u + x= u(1 + u1x) is the product of two units(using the results of part (b). Hence (u + x) is a unit.//

Dummit & Foote Text Exercise 7.1-19: LetIbe any nonempty index set and letRi be a ring for eachi I. Prove that the direct productiIRi is a ring under componentwise addition and multiplication.Solution: In Dummit & Foote Text Exercise 5.1-15 we proved that the direct product of groups is a groupand it certainly holds for abelian groups. We need to show associativity of multiplication and the left andright distributive properties.

If (

ai), (

bi), (

ci)

IRi, then(

ai)((

bi)(

ci)) = (

ai)(

(bici)) =

ai(bici) =

(aibi)ci = (

aibi)(

ci) = ((

ai)(

bi))(

ci),so multiplication is associative.

Next, (

ai)((

bi) + (

ci)) = (

ai)(

(bi+ ci)) =

ai(bi+ ci) =

(aibi+ aici) = (

aibi) + (

aici) =(

ai)(

bi) + (

ai)(

ci). So, multiplication distributes over addition on the left.

((

ai) +

bi)(

ci) = (

(ai+ bi))(

ci)) =

(ai+ bi)(ci) =

(aici+ bici) = (

aici) + (

bici) =(

ai)(

ci) + (

bi)(

ci). So, multiplication distributes over addition on the right as well.

Thus

IRi is a ring under componentwise addition and multiplication.//

Dummit & Foote Text Exercise 7.1-26: LetKbe a field. A discrete valuation on Kis a functionv: K Z satisfying(i) v(ab) = v(a) + v(b) (i.e., vis a homomorphism from the multiplicative group of nonzero elements ofKto Z,(ii) v is surjective, and(ii) v(x + y) min{v(x), v(y)} for all x, y K withx + y= 0. The setR = {x K | v(x) 0} {0}iscalled the valuation ring ofv .

(a) Prove thatR is a subring ofKwhich contains the identity. (In general, a ring R is called a discretevaluation ring if there is some field Kand some discrete valuation v onKsuch thatR is the valuation ringofv .)(b) Prove that for each nonzero element x Keither x or x1 is in R.(c)Prove that an element x is a unit ofR if and only ifv(x) = 0.Solution: v is an epimorphism (surjective homomorphism) since we are given thatv is is surjective and ithas the homomorphic property for addition. So ifx R, then v(x) = v(1 x) = v(1) + v(x). This impliesv(1) = 0 by cancellation. We can use this to show v(x) = v(x) as follows:0 = v(1) =v((1)(1)) =v(1) + v(1) implies v(1) = v(1) and that in turn implies v(1) = 0. Soifx R, then v(x) = v((1)x) = v(1) + v(x) = v(x). Hence, v(x) = v(x). This fact is necessary forpart (a).

(a) By definition, 0 R and that implies R is nonempty. Apply the subgroup criterion. Let x, y R andconsiderx y. Ifx = 0 and y = 0, thenv(x y) = v(x) 0. Ifx = 0 andy= 0, then


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v(x y) = v(y) = v(y) 0. Ifx = y = 0, then x y = 0 R. Ifx, y= 0, then either x y= 0 R orx y= 0, and then v(x y) min(v(x), v(y)) = min(v(x), v(y)) 0.Now to check for multiplicative closure. Ifx, y R and xy = 0, then xy R. Ifxy= 0, thenv(xy) = v(x) + v(y) 0, so that xy R. Thus R is a subring ofK. We showed earlier that v(1) = 0, sothat 1 R.

(b)Assumex K is nonzero. So, 0 =v(1) =v(xx1

) = v(x) + v(x

1

). Hence, v(x) = v(x1

), and eitherv(x) orv(x1) is nonnegative and therefore either x orx1 R. (c)() Ifu R is a unit then u1 R.From part (b), v(u) = v(u1), and both v(u) and v(u1) are nonnegative. Hence v(u) = v(u1) = 0.() Ifv(u) = 0, then0 = 0 = v(u) = v(u1), so that u1 R and u R is a unit.//

Dummit & Foote Text Exercise 7.1.27: A specific example of a discrete valuation ring (cf. thepreceding exercise) is obtained when p is a prime, K= Q and

vp : Q Z by vpa

b

= where

a

b =p

c

d, p |c and p |d.

Prove that the corresponding valuation ring R is the ring of all rational numbers whose denominators arerelatively prime to p. Describe the units of this valuation ring.

Solution: We are asked to prove that R = {a/b Q| a = 0 or gcd(b, p) = 1}.(R {a/b Q| a = 0 or gcd(b, p) = 1}). Assume thatvp(ab ) 0, and let a = pmcand b = pnd, where pdoes not divide c or d. By definition, vp(

ab

) = m n 0. Take the case that ab

is in lowest terms theneitherm or n is 0. Ifn = 0, then m = 0, and vp(ab )< 0, a contradiction. Hence n = 0 and gcd(b, p) = 1.({a/b Q| a = 0 or gcd(b, p) = 1} R). If a

b Qand p does not divide b, then a

b =p) c

d for some 0.

Hencevp(fracab) 0, and we have ab R.//

Dummit & Foote Text Exercise 7.2-2 Letp(x) = anxn + anlx

nl + + a1x + a0 be an element ofthe polynomial ring R[x]. Prove that p(x) is a zero divisor inR[x] if and only if there is a nonzero b Rsuch thatbp(x) = 0. [Let g(x) = bmx

m + bm1xml + + b0 be a nonzero polynomial of minimal degree

such thatg(x)p(x) = 0. Show that bman = 0 and soang(x) is a polynomial of degree less than m that alsogives 0 when multiplied by p(x). Conclude thatang(x) = 0. Apply a similar argument to show byinduction on i that anig(x) = 0 fori = 0, 1, . . . , nand show that this implies bmp(x) = 0.]

Solution:) Ifbp(x) = 0 for some nonzero b R, then letting b(x) = b states thatp(x) is a zero divisor.() Assume p(x) is a zero divisor. For some q(x) = mi=0 bixi, p(x)q(x) = 0. We may choose q(x) to haveminimal degree among the nonzero polynomials with this property.

Now for the induction argument to show that aiq(x) = 0 for all 0 i n. I obtained help for this partof the problem.

The base case: p(x)q(x) =n+m

k=0

i+j=kaibj

xk = 0. The coefficient ofxn+m in this product isanbm

on the left side and 0 on the right side. Thusanbm= 0. Now anq(x)p(x) = 0, and the coefficient ofxm inq

is anbm = 0. Thus the degree ofanq(x) is strictly less than that ofq(x). Sinceq(x) has minimal degree

among the nonzero polynomials which multiply p(x) to 0, anq(x) = 0. So, anbi= 0 for all 0 i m.For the inductive step, suppose that for some 0 s < 4, we have arq(x) = 0 for all s < r n. Now

p(x)q(x) =n+m

k=0

i+j=kaibj

xk = 0. The coefficient ofxm+s is

i+j=m+s aibj on side side of the

equation and it is 0 on the other. Thus

i+j=m+s aibj = 0. By the induction hypothesis, ifi s, thenaibj = 0. Hence all terms such that i sare zero. Ifi < s, then we must have j > m, a contradiction.Thus we have asbm = 0. As in the base case, asq(x)p(x) = 0 andasq(x) has degree strictly less than thatofq(x), so that by minimality, asq(x) = 0.

By induction, aiq(x) = 0 for all 0 i n. In particular, aibm = 0. Hence bmp(x) = 0.


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Dummit & Foote Text Exercise 7.2-3: Define the set R[[x]] of formal power series in theindeterminatex with coefficients from R to be all formal infinite sums

n=0

anxn =a0+ a1x + a2x

2 + a3x3 + .

Define addition and multiplication of power series in the same way as for power series with real or complex

coefficients i.e., extend polynomial addition and multiplication to power series as though they werepolynomials of infinite degree:

n=0

anxn +

n=0

bnxn =

n=0

(an+ bn)xn

n=0

anxn

n=0

bnxn =

n=0

nk=0

akbnk

xn.

(The term formal is used here to indicate that convergence is not considered, so that formal power seriesneed not represent functions on R.)(a) Prove thatR[[x]] is a commutative ring with 1.(b) Show that 1

x is a unit in R[[x]] with inverse 1 + x + x2+

.

(c)Prove that

n=0 anxn is a unit in R[[x]] if and only ifa0 is a unit in R.

Solution: (a) Let r =

n=0 anxn, s =

n=0 bnxn, and v =

n=0 cnxn.

Addition is Associative:

(r+ s) + v = ((

n=0 anxn) + (

n=0 bnxn)) + (

n=0 cnxn)

= (

n=0(an+ bn)xn) + (

n=0 cnxn)

=

n=0((an+ bn) + cn)xn

=

n=0(an+ (bn+ cn))xn

= (n=0

anxn) + (

n=0(bn+ cn)x

n)

= (

n=0 anxn) + ((

n=0 bnxn) + (

n=0 cnxn))

= r+ (s + v).

Addition is Commutative:

r+ s =

n=0

anxn

+

n=0

bnxn

=

n=0(an+ bn)x

n

=

n=0

(bn+ an)xn

=

n=0

bnxn

+

n=0

anxn

=s + rn=00 xn = 0 is theadditive identity.


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r+ 0 = (

n=0 anxn) + (

n=00 xn)

=

n=0(an+ 0)xn

= (

n=0 anx

n) =

=

n=0(0 + an)x

n

= (

n=00 xn) + (

n=0 anxn)

= 0 + r

R[[x]] contains left and right inverses. Definer R[[x]] as r = n=0(an)xn. Then we haver+ r = (

n=0 anxn) + (

n=0(an)xn)

=

n=0(an an)xn

= n=00 xn

= 0

=

n=00 xn

=

n=0(an+ an)xn

= (

n=0(an)xn) + (

n=0 anxn)

= r+ r.

Hence the left and right inverses coincide. r is the additive inverse for r .

I had a lot of help sorting this one out. The definition of the product is identical to that of what electrical

engineers know as convolution. Multiplication is Associative:

(rs)v =

n=0

anxn

n=0

bnxn

n=0

cnxn

=

n=0

i+j=n

aibj

xn

n=0

cnxn

=

n=0

t+k=n

i+j=t

aibj

ck

xn

=

n=0

t+k=n

i+j=t

aibjck

xn


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=

n=0

i+j+k=n

aibjck

xn

=

n=0

i+s=n

j+k=s

aibjck xn

=

n=0

i+s=n

ai

j+k=s

bjck

xn

=

n=0

anxn

n=0

j+k=n

bjck

xn

=

n=0anx

n

n=0bnx

n

n=0cnx

n

= r(sv).

Distributivity:

r(s + v) =

n=0

anxn

n=0

bnxn

+

n=0

cnxn

=

n=0

anxn

n=0

(bn+ cn)xn

=

n=0

i+j=n a

i(bj + cj)

x

n

=

n=0

i+j=n

aibj+ aicj

xn

=

n=0

i+j=n

aibj

+

i+j=n

aicj

xn

=

n=0

i+j=n

aibj

x

n

+

n=0

i+j=n

aicj

x

n

=

n=0

anxn

n=0

bnxn

+

n=0

anxn

n=0

cnxn

= rs + rv.


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(r+ s)v =

n=0

anxn

+

n=0

bnxn

n=0

cnxn

=

n=0

(an+ bn)xn

n=0

cnxn

=

n=0

i+j=n

(aj+ bj)ci

xn

=

n=0

i+j=n

aicj+ bicj

xn

=

n=0

i+j=n

aicj

+

i+j=n

bicj

xn

=

n=0

i+j=n

aicj

xn+

n=0

i+j=n

bicj

xn

=

n=0

anxn

n=0

cnxn

+

n=0

bnxn

n=0

cnxn

= rv+ sv.

Hence multiplication distributes over addition on the left and right.

IfR is commutative for multiplication, then so is R[[x]]:

rs =

n=0

anxn

n=0

bnxn

=

n=0

i+j=n

aibj

xn

=

n=0

j+i=n

bjai

xn

=

n=0 bnx

n n=0 anx

n= sr.


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IfR has a multiplicative identity, then so does R[[x]]. Define 1 =

n=0 enxn bye0 = 1 and ei+1 = 0.

That gives us

r 1 =n=0

anxn

n=0

enxn

=

n=0

i+j=n

aiej

xn

=

n=0

ane0xn

=

n=0

anxn

= r

1 r = n=0

enxn

n=0

anxn

=

n=0

i+j=n

ejai

xn

=

n=0

e0anxn

=

n=0anx

n

= r

HenceR[[x]] has the multiplicative identity 1.

(b). Let 1 x= n=0 bnxn whereb0 = 1, b1 = 1, and bi= 0 for i 2. Let r =n=0 xn. Taking theproduct gives

(1 x)r=n=0

bnxn

n=0

xn

=

n=0

i+j=n

bi

xn.

So ifn 1, theni+j=n di = 0, andd0 = 1. Hence (1 x)r = 1.(c)() Supposeu R[[x]] is a unit, with inverse u1 =

n=0 anxn. So,

1 = uu1 =

n=0

anxn

n=0

anxn

=

n=0

i+j=n

aiaj

xn.

The coefficient ofx0 in this power series is 1 on the left hand side and a0a0 on the right hand side. Thusa0a0 = 1 R. We also have

1 = u1u=

n=0

anxn

n=0

anxn

=

n=0

i+j=n

aiaj

xn.


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Again, the coefficient ofx0 in this power series is 1 on the left hand side and a0a0 on the right hand side.Hencea0a0 = 1 R and a0 is a unit in R.I had help on the following part (could not keep the indices on the sums consistent).() Supposea0 Ris a unit, with a0a0 = a0a0 = 1. Define u =

n=0 bnx

n withb0 = a0 andbk+1 = a0

i+j=k+1,jkaibj . Then

uu=

n=0

anxn

n=0

bnxn

=

n=0

i+j=n

aibj

xnIfn = 0,

i+j=n aibj =a0a0 = 1. Ifn 1, then

i+j=n

aibj = a0bn+

i+j=n,j


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r1N g1+ + rnN gn = r1N+ + rnN sinceN g= N.Thenr1N+ + rnN=r1g1N+ + rngnN since gN=N.Then by the distributive property, we have r1g1N+ + rngnN= (r1g1+ + rngn)N.SoN(r1g1+ + rngn) = (r1g1+ + rngn)N implies Nis in the center ofRG. //

Dummit & Foote Text Exercise 7.3-2: Prove that the rings Z[x] and Q[x] are not isomorphic.

Solution: Suppose : Q[x] Z[x] is a ring homomorphism. Then since (1) = 1 by definition of a ringhomomorphism,

(2) =(1 + 1) =(1) + (1) = 1 + 1 = 2.

We also have1 = (1) =(2 1/2) = 2 (1/2).

Thus (1/2) Z[x]. By by Proposition 7.2.4(2) Z[x] = Z = {+1, 1} . Since 2 1 = 1 and 2 1 = 1,we have a contradiction. Hence no such homomorphism exists.//

Dummit & Foote Text Exercise 7.3-4: Find all ring homomorphisms from Z to Z/30Z. In each casedescribe the kernel and the image.

Solution: A homomorphism : Z Z/30Z requires that(x)(x 1) = x (1). We need to determinehow many images of 1 are possible since each of them defines the map. We know that(x + y) = (x + y)(1) =x(1) + y(1) =(x) + (y). Also, (xy) = xy(1) while(x)(y) = x(1)y(1) =xy(1)2 This implies that we have a homomorphism if and only if(1) =(1)2

in Z/30Z. The eight elements ofZ/30Z which meet this requirement are{0, 1, 6, 10, 15, 16, 21, 25}.Map Kernel Image

(1) 0 Z {0}

(1) 1 30 1

(1) 6 5 6

(1) 10 3 10

(1) 15 2 15

(1) 16 15 16 = 2

(1) 21 10 21 = 3

(1) 25 6 25 = 5


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Dummit & Foote Text Exercise 7.3-5: Describe all ring homomorphisms from the ring Z Z to Z.Solution: Z Zhas two generators (0, 1) and (1, 0). So any homomorphism has to send generators togenerators. If : mathbbZZ Z is a ring homomorphism then is determined by where it sends thetwo generators, say (1, 0) = a and (0, 1) = b.So(1, 1) = (1, 0) + (0, 1) = a + band (1, 1) = ((1, 1)(1, 1)) =(1, 1)(1, 1) = (a + b)2. This impliesthata + b= 0 or a + b= 1.

Next we have for all (s, t) Z Z, (s, t) = (s 1, t 1) = sa + yb and(s, t) = ((1, 1)(s, t)) = (a + b)(sa + sb). So(s, t) = sa + tb= (a + b)(sa + tb).Ifa + b= 0, (s, t) = 0 (s, t) Z Z. this implies = 0.The remaining case isa + b= 1. Since is a ring homomorphism,((s, t)(u, v)) = (s, t)(u, v) = (as + bt)(au + bv) and ((s, t)(u, v)) = (su, tv) = asu + btv. This impliesasu + btv= a2su + (sv+ tu)ab + b2tu for all integers s, t, v andv .Ift = v = 0 and s, u = 0, then asu = a2su. Hence a2 =a. Since a is an integer, we have a {0, 1}. Thatimplies (a, b) {(1, 0), (0, 1)}.So we have three ring homomorphisms (maps):1. (1, 0) 0 and (0, 1) 0.2. (1, 0) 1 and (0, 1)0.3. (1, 0) 0 and (0, 1)1.

Dummit & Foote Text Exercise 7.3-6: Decide which of the following are ring homomorphisms fromM2(Z) toZ:

(a)

a bc d

a (projection onto the 1,1 entry.)

(b)

a bc d

a + d (the trace of the matrix)

(c)

a bc d

ad bc (the determinant of the matrix).

Solution: (a) IfA =

1 11 0

, then(AA) =

2 11 0

= 2 and(A)(A) = 1, Since they are not equal,

this mapping is not a ring homomorphism.

(b) IfA =

1 11 0

, then(AA) =

2 11 0

= 3 and(A)(A) = 1.Since they are not equal, this

mapping is not a ring homomorphism.

(c)IfA =

1 00 0

and B =

0 00 1

, then (A + B) = 1 and (A) + (B) = 0. Since they are not equal,

this mapping is not a ring homomorphism.

Dummit & Foote Text Exercise 7.3-7: LetR = {

a b0 d

| a,b,d Z} be the subring of all upper

triangular matrices. Prove that the map

: R Z Z defined by a b0 d

(a, d)is a surjective ring homomorphism and describe its kernel.

Solution: Let A, B R with A =

a1 b10 d1

and B =

a2 b20 d2

.

is a ring homomorphism since


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(A + B) =

a1 b10 d1

+

a2 b20 d2

= (a1+ a2, d1+ d2)

= (a1+ a2, d1+ d2)

=

a1 b10 d1

+

a2 b20 d2

= (A) + (B)

and

(AB) =

a1 b10 d1

a2 b20 d2

= (a1a2, d1d2)

= (a1, d1)(a2, d2)

=

a1 b10 d1

a2 b20 d2

= (A) (B).

is surjective since if (a, d) Z2 , then

a 00 d

= (a, d).

If

a b0 d

ker, then

a b0 d

= (a, d) = (0, 0) which implies a = d = 0. Since

0 b0 0

= (0, 0),

ker=

0 b0 0

b Z

.

Dummit & Foote Text Exercise 7.3-8: Decide which of the following are ideals of the ring Z Z.(a){(a, a)| a Z}(b){(2a, 2b)| a, b Z}(c){(2a, 0)| a Z}(d){(a, a)| a Z}.Solution:LetR = Z Z for all four of these subproblems. (a) Let R = Z Zand S= {(a, a)| a Z}. Sis not an ideal ofR becauseSdoes not absorb R. (1, 1) Sbut (1, 0)(1, 1) = (1, 0) / S.(b). We need to verify that S= {(2a, 2b)|a, b Z}is a subring ofR and absorbs R. Let(2a1, 2b1), (2a2, 2b2) Iand (x, y) R. (0, 0) I, so thatIis not empty.(2a1, 2b1) (2a2, 2b2) = (2(a1 a2), 2(b1 b2)) S, so thatSis an additive subgroup.(2a1, 2b1)(2a2, 2b2) = (4a1a2, 4b1b2) S shows Sis closed under multiplication, so that Sis a subring.(2a1, 2b1)(x, y) = (2a1x, 2b1y) S, so thatSabsorbs R. Since R is a commutative ring, we dont need tocalculate absorption on the right. Hence S is an ideal ofR.(c). We need to show that S= {(2a, 0)|a Z} is a subring and absorbs R on one side (since R iscommutative. . (Absorption on one side is sufficient since R is commutative.) Let (2a1, 0), (2a2, 0) I and(x, y) R. (0, 0) S, so thatSis not empty. (2a1, 0) (2a2, 0) = (2(a1 a2), 0) S, so that Sis anadditive subgroup. (2a1, 0)(2a2, 0) = (4a1a2, 0) S shows Iis closed under multiplication, so that S is asubring. (2a1, 0)(x, y) = (2a1x, 0) S, so that IabsorbsR. Hence S is an ideal ofR.(d) The setS= {(a, a)| a Z}is not closed under the multiplication operation: (1, 1) Sbut(1, 1)(1, 1) = (1, 1) / S. SoSis not a subring and hence cannot be an ideal.


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Dummit & Foote Text Exercise 7.3-10: Decide which of the following are ideals of the ring Z[x]:

(a) The set of all polynomials whose constant term is a multiple of 3(b) The set of all polynomials whose coefficient ofx2 is a multiple of 3(c)The set of all polynomials whose constant term, coefficient ofx, and coefficient ofx2 are all zero(d) Z[x2] (the set of all polynomials in which only even powers ofx appear)(e)The set of all polynomials whose coefficients sum to zero

(f )The set of polynomials p(x) such that p(0) = 0, wherep(x) is the usual first derivative of p withrespect to x.

Solution: (a) Let Sbe the given set and = 3a + xp(x),= 3b + xq(x) Sand = r+ xt(x) Z[x].We know that 0 Sand = 3(a b) + x(p(x) q(x)) S. This implies Sis a subgroup. S is asubring since= 9ab + x(3bp(x) + 3aq(x) + xp(x)q(x)) S. Lastly, S is closed under multiplication since= 3ar+ x(3at(x) + rp(x) + xp(x)t(x)) S. HenceSis an ideal.(b) This set is not closed under multiplication since x Sbutxx = x2 / S. Hence Sis not a subring andtherefore cannnot be an ideal.

(c)Let Sbe the given set and = x3p(x), = x3q(x) Sand c(x) Z[x]. We know that 0 S, and = x3(p(x) q(x)) S. SoSis a subgroup. = x6p(x)q(x) Sshows that Sis closed undermultiplication. Hence t Sis a subring. Sabsorbs Z[x]. since c(x) = x3p(x)c(x) S. Hence, Sis an ideal.(d) Sinc3x2 is in this set, but x2x= x3 is not. Since this subset does not absorb Z[x] on the left, it is notan ideal.

(e)A polynomial p(x) is in this set Swhenp(1) = 0. Suppose p, q Sandr Z[x]. We know that 0 S.Since (p q)(1) =p(1) q(1) = 0, implies p q S, S is subgroup. Also, since (pq)(1) = (1), Sis closedunder multiplication and hence S is a subring. Since (pr)(1) =p(1)r(1) = 0 r(1) = 0, Sabsorbs Z[x].HenceSis an ideal.

(f )Let Sbe the given subset and let p(x) = x2 a and q(x) = x. Sincep(x) = 2x, p S. But(qp) =qp +pq, so that (qp)(0) =q(0)p(0) +p(0)q(0) =p(0) = a. Since Sdoes not absorb Z[x] on theleft, it is not an ideal.

Dummit & Foote Text Exercise 7.3-11: LetR be the ring of all continuous real valued functions on

the closed interval [0, 1]. Prove that the map : R R given by (f) = 10 f(x)dx is a homomorphism ofadditive groups but not a ring homomorphism.

Solution: We do remember from calculus that integration is a linear operation. That is,

(f+ g) =

10

(f+ g)(x)dx=

10

f(x) + g(x)dx=

10

f(x)dx +

10

g(x)dx= (f) + (g)

. So, is an additive group homomorphism. does not meet the homomorphic property for multiplication. For example, (x3) = 1/4 while(x)3 = 1/8. //

Dummit & Foote Text Exercise 7-3.26: The characteristic char(R) of a ring R is the smallest positiveintegern such that 1 + 1 + + 1 = 0 (n times) in R; if no such n exists, then the characteristic ofR issaid to 0. For example Z/(nZ) is a ring of characteristic n for each positive integer n and Zis a ring ofcharacteristic 0.(a) Prove that the map Z R defined by

k

1 + 1 + + 1(k times) if k >00 ifk = 01 1 1 (k times) ifk


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(b) Determine the characteristics of the rings Q, Z[x], Z/nZ[x].

(c)Prove that ifp is a prime and ifR is a commutative ring of characteristic p, then (a + b)p =ap + bp forall a, b R.Solution: (a). I had help with this part. My approach to the induction argument wasnt as clean as thisone. We will show that (a + b) = (a) + (b) for nonnegativeb by induction. For the base case b = 0, we

have (a + 0) =(a) = (a) + 0 =(a) + (0). Now assume that for some b 0, for all a,(a + b) = (a) + (b). Ifa + b > 0, then(a + (b + 1)) =((a + b) + 1) =(a + b) + 1 =(a) + (b) + 1 =(a) + (b + 1). Hence inductionhypothesis holds for all positiveb. Note the case where a + b < 0 is handled by the next case. Assumeb < 0. Ifa 0, then (a + b) = (b + a) = (b) + (a) = (a) + (b). Ifa


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Dummit & Foote Text Exercise 7.3-29: LetR be a commutative ring. Recall (cf. Exercise 13, Section1) that an element x R is called nilpotent ifxn = 0 for some n Z+. Prove that the set of nilpotentelements form an ideal called the nilradical ofR denoted by N(R) . [Hint: Use the Binomial Theorem toshow N(R) is closed under addition.]

Solution: Ifx, y N(R), then for some positive integers n and m, xn =ym = 0. Since we are given thehint to use the Binomial Theorem, consider (x + y)n+m. The Binomial Theorem tells us

(x + y)n+m =

n+mk=0

n + m

k

xkyn+mk.

Ifk n,xk = 0. Ifk < n, thenyn+mk = 0. Hence (x + y)n+m = 0 andx + y N(R). So this set is closed.The set has inverses for its elements since we have (x)n = (1)nxn = 0, so thatx N(R). The set hasan identity since 0 = 01 N(R), N(R) is an additive subgroup ofR.SinceR is commutative, ifr Rthen (rx)n =rnxn = 0 and (xr)n =xnrn = 0. So, N(R) absorbs R on theleft and the right. Hence N(R) an ideal.//

Dummit & Foote Text Exercise 7.3-30: Prove that ifR is a commutative ring and N(R) is its

nilradical (cf. the preceding exercise), then zero is the only nilpotent element ofR/N

(R) i.e., prove thatN(R/N(R)) = 0.

Solution: From Exercise 7.3-29, we know that N(R) is an ideal ofR. Ifx + N(R) N(R/N(R)), then forsome positive integer n, (x + N(R))n =xn + N(R) = N(R). Hence xn N(R). Then for some positiveintegerm, (xn)m =xnm = 0. Hence x N(R). This implies that x + N(R) = N(R). That is x = 0. Inother words, the additive subgroup N(R/N(R)) contains only one element, the additive identity. //

Dummit & Foote Text Exercise 7.3-34: LetI , Jbe ideals ofR.(a) Prove thatI+ J is the ideal ofR containing bothIand J.(b) Prove that I Jis an ideal contained in I J.(c)Give an example where I J=I J.(d) Prove that ifR is commutative and ifI+ J=R, then I J=I J.Solution: I had help with this problem. (a) Let a1+ b1, a2+ b2 I+ Jfor someai Iandbi J. Then(a1+ b1) (a2+ b2) = (a1 a2) + (b1 b2) I+ J since I andJare closed under subtraction. Let r R.Then, r(a + b) = ra + rb I+ J since I andJabsorbR on the right. (a + b)r I+ J since I andJabsorbR on the left. Thus I+ Jis an ideal ofR. Now, letKbe an ideal ofR that contains IandJ. Thatis, I , J K. Ifa + b I+ J, thena + b K sinceKis closed under addition. So I+ J K.(b) Let x, y IJ, with x = aibi andy = cidi, whereai, ci Iandbi, di J. We have x + y IJ.Letr R. Since I is an ideal ofR, rx=(rai)bi IJ. Similarly, xr IJ. HenceI J is an ideal ofR.Again let x =

aibi. Sinceai IandIis an ideal, aibi I, and thus x I. Similarly, x J. ThusIJ I J.(c)IfR = Z, I= 2Z and J= 4Z, then (2Z)(4Z) = 8Z = 2Z 4Z = 4Z.(d) Assume I+ J=R and R is commutative. Part (c) of this exercise tells us that I J

I

J. We have(I J)(I+ J) = (I J)R= I J. Ifa I J, d = c(a + b) for some a I andb J andc I J. Thend= ca + cb= ac + cb IJ. HenceI J=I J.//

Dummit & Foote Text Exercise 7.4-4: Assume R is commutative. Prove thatR is a field if and onlyif 0 is a maximal ideal.

Solution: () AssumeR is a field and letIbe an ideal ofR which properly contains zero. This impliesthat there exists an element x = 0 in I. Then a is a unit since R is field. Proposition 9 tells us that I= R.Hence,R is the only ideal ofR that properly contains zero. This implies that 0 is a maximal ideal in R.


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() Assume that zero is a maximal ideal in R. Let x R and x = 0. Since zero is maximal and since R isgiven to by commutative, (a) = R. Note thatR contains the multiplicative identity, 1. Therefore, existselements y and Z inRsuch thatxy = za= 1. This implies thatx has a two-sided inverse. Therefore, everynonzero element in R is invertible and that implies R is a field. //

Dummit & Foote Text Exercise 7.4-7: LetR be a commutative ring with 1. Prove that the principalideal generated by x in the polynomial ring R[x] is a prime ideal if and only ifR is an integral domain.Prove that (x) is a maximal ideal if and only ifR is a field.

Solution: IfR is a commutative ring then : R[x] R defined by (p(x)) = p(0) is an surjective ringhomomorphism and ker = (x). It is clear that is surjective. Let p(x) =

aix

i andq(x) =

bixi. We

then have(p + q) =

(ai+ bi)xi

=a0+ b0 = (p) + (q)

and

(pq) =

k

(

i+j=k

aibj)xk

=a0b0 = (p)(q)

which show that is a ring homomorphism.

To show that ker= (x), assume that p ker. We can write p(x) = a0+ xq(x), and p(0) =a0 = 0. Thusp(x) = xq(x) (x). So ker (x). Conversely, ifp(x) = xq(x), then p(0) = 0 q(x) = 0, and(p) = 0.(x) ker. Hence ker = (x).By the First Isomorphism Theorem for rings, we have R[x]/ker = R[x]/(x) =R. Proposition 13 tells usthat ifR[x] is commutative, then the ideal (x) is a prime ideal in R[x] if and only if the quotient ringR[x]/(x) is an integral domain. Since R[x]/(x) =R, R is an integral domain if and only ifR[x]/(x) =R isan integral domain.

Proposition 12 tells us that ifR[x] is commutative, then the ideal (x) is maximal if and only if the quotientring R[x]/(x) is a field. We have shown by the first isomorphism theorem that R[x]/(x) =R. SoR is afield if and only ifR[x]/(x) is a field.//

Dummit & Foote Text Exercise 7.4-8: LetR be an integral domain. Prove that (a) = (b) for someelements a, b R, if and only ifa = ub for some unit uofR.Solution: () If (a) = (b), thena (b). This implies that a = ub for some element u R. We also haveb (a) and therefore b = va for some element v R. Soa = (uv)aand this implies va= (vu)va. Sincesince R is an integral domain we have uv = vu = 1. Henceu is a unit.

() Leta = ub and u R be a unit. Then by Proposition 9 in the text we obtain(a) = aR = ubR = buR = bR = (b).//

Dummit & Foote Text Exercise 7.4-9: LetR be the ring of all continuous functions on [0, 1] and letIbe the collection of functions f(x) in R with f(l/3) = f(l/2) = 0. Prove that I is an ideal ofR but is not aprime ideal.

Solution: Iis an ideal. Iff , g I, then (f g)(1/2) = f(1/2) g(1/2) = 0 0 = 0 and(f g)(1/3) = f(1/3) g(1/3) = 0 0 = 0. So, f g I. Since 0 I, Iis an additive subgroup ofR bythe subgroup criterion. Ifh R, then (f h)(1/2) = f(1/2)h(1/2) = 0 h(1/2) = 0 and(f h)(1/2) = (hf)(1/2) = (f h)(1/3) = (hf)(1/3) = 0. So we have shown that I is an ideal ofR.Iis not a prime ideal. Iff(x) = x2 1/4 and g(x) = x2 1/9, (f g)(1/2) = (f g)(1/3) = 0 then f g I.However, f / Iand g / I sincef(1/3) = 0 and g (1/2) = 0. //

Dummit & Foote Text Exercise 7.4-10: Assume R is commutative. Prove that ifPis a prime ideal ofR and Pcontains no zero divisors then R is an integral domain.


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Solution: Let x, y R such that xy = 0. Since P is a prime ideal and xy P, x P. The followingequally valid ify is chosen instead ofx. Ifx = 0 then since Pcontains no zero divisors in R, y = 0, //

Dummit & Foote Text Exercise 7.4-15: Letx2 + x + 1 be an element of the polynomial ringE= F2[x] and use the bar notation to denote passage to the quotient ring F2[x]/(x

2 + x + 1).(a) Prove that E has 4 elements: 0, 1, x, and x + 1.

(b) Write out the 4 4 addition table for Eand deduce that the additive group Eis isomorphic to theKlein 4-group.

(c)Write out the 4 4 multiplication table for Eand prove that E is isomorphic to the cyclic group oforder 3. Deduce that E is a field.

Solution: Note: This problem is identical to Example 4.3.4 in Beachy & Blairs text, pages 208-209.(a) The congruence classes modulo x2 + x + 1 are polynomials that that have degree less than 2 over F2.The congruence classes are 0, 1, x, and 1 + x.(b)

+ 0 1 x 1 + x

0 0 1 x 1 + x

1 1 0 1 + x x

x x 1 + x 0 1

1 + x 1 + x x 1 0

Since the group has order four and each element except the identity has order 2, this group is isomorphicto the Klein four group.

(c) 0 1 x 1 + x

0 0 0 0 0

1 0 1 x 1 + x

x 0 1 + x 0 1

1 + x 0 1 + x 1 x

E

= {1, x, 1 + x}. Since the order ofE is three, its isomorphic to a cyclic group of order three.(d) x2 + x + 1 is irreducible over the field F2 since it has no roots in F2. Therefore F2[x]/(x2 + x + 1) is afield.//

Dummit & Foote Text Exercise 7.3-16: Letf(x) = x4 16 be an element ofE= Z[x] and use the barnotation to to denote passage to the quotient ring row Z[x]/(x4 16). (a) Find a polynomial of degree 3

which is congruent to g(x) = 7x13

11x9

+ 5x5

2x3

+ 3 modulo x4

16.(b) Prove that x + 2 andx 2 are zero divisors in E.Solution: (a). Using the division algorithm (and Matlab), we obtain the following:

g(x) = (7 x9 + 101 x5 + 1621 x) (x4 16) + (2 x3 + 25936 x + 3)and thus,2 x3 + 25936 x + 3 is the desired polynomial.(b)x + 2 and x 2 are zero divisors since they are factors ofx4 16. That is,x4 16 = (x 2)(x + 2)(x2 + 4) implies 0 (x 2)(x + 2)(x2 + 4) modx4 16.


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Dummit & Foote Text Exercise 7.4-18: Prove that ifR is an integral domain and R[[x]] is the ring offormal power series in the indeterminate x then the principal ideal generated by x is a prime ideal (cf.Exercise 3, Section 2). Prove that the principal ideal generated byx is a maximal ideal if and only ifR is afield.

SolutionWe approach this problem in the same manner as Exercise 7.4-7.

IfR is a commutative ring then : R[[x]]

R defined by rixi =r0 is an surjective ringhomomorphism and ker = (x). It is clear that is surjective. Let p(x) = aixi andq(x) = bixi. Wethen have

(p + q) =

(ai+ bi)xi

=a0+ b0 = (p) + (q)

and

(pq) =

k

(

i+j=k

aibj)xk

=a0b0 = (p)(q)

which show that is a ring homomorphism. Not concering ourselves regarding convergence makes me a bitworried that the above is not entirely true.

Now to show ker= (x). (ker (x)) Assume rixi ker. Then r0 = 0 (using the definition that00 = 1), and rix

i =x ri+1xi (x). ((x) ker) Ify = x aix

i, then y (x) and that implies(y) = 0, so thaty ker.By the First Isomorphism Theorem for rings, we have R[[x]]/ker = R[[x]]/(x) =R. Proposition 13 tellsus that ifR[[x]] is commutative, then the ideal (x) is a prime ideal in R[[x][ if and only if the quotient ringR[[x]]/(x) is an integral domain. Since R[[x]]/(x) =R, R is an integral domain if and only ifR[[x]]/(x) =R is an integral domain.Proposition 12 tells us that ifR[[x]] is commutative, then the ideal (x) is maximal if and only if thequotient ringR[[x]]/(x) is a field. We have shown by the first isomorphism theorem that R[[x]]/(x) =R.SoR is a field if and only ifR][x]]/(x) is a field.//

Dummit & Foote Text Exercise 7.4-19: LetR be a finite commutative ring with identity. Prove thatevery prime ideal ofR is a maximal ideal.

Solution

: IfP is a prime ideal of a ring R that is commutative and contains 1, then R/Pis an integraldomain. Suppose that P is prime, and leta + P andb + Pbe elements ofR/P. Suppose that theirproduct ab + P= 0 + P, the zero element ofR/P. This implies ab0 Por equivalently ab P. SinceP isprime, eithera P or b P. Ifa P, then a + P= 0 + Pand ifb P, thenb + P = 0 + P. Either way,the only way a product equals zero is when one of the factors equals zero, which along with R/P iscommutative implies R/Pis an integral domain.

IfR is a finite integral domain, then R is a field. Prof. Robert Ash of UIUC had a very simple proof this.Ifa R and a = 0,the map x ax, x R,is injective because R is an integral domain (no zero divisors).IfR is finite,the map is surjective as well, so that ax = 1 for some x. HenceR is a field.

SinceR/Pis a finite integral domain, it is a field and therefore P R is maximal.//

Dummit & Foote Text Exercise 7.4-26: Prove that a prime ideal in a commutative ring R contains

every nilpotent element (cf. Exercise 13, Section 1). Deduce that the nilradical ofR (cf. Exercise 29,Section 3) is contained in the intersection of all the prime ideals ofR. (It is shown in Section 15.2 that thenilradical of R is equal to the intersection of all prime ideals ofR.)

Solution: Let P R be a prime ideal and let r R be nilpotent with rn = 0. Let 1 m n be minimalsuch thatrm P. R/Pis an integral domain and xm + P = 0 R/P. Ifm 2, we have(r+ P)(rm1 + P) = 0, so that r + P is a zero divisor in R/P, a contradiction sinceR/P is an integraldomain. Therefore m = 1, and r P. Hence N(R) P. IfP(R) is the collection of all prime ideals ofR,we have N(R) P(R).


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Dummit & Foote Text Exercise 7.4-28: Prove that if R is a commutative ring andN= (a1, a2, . . . , an) where eachai is a nilpotent element, then Nis a nilpotent ideal (cf. Exercise 37,Section 3). Deduce that if the nilradical ofR is finitely generated then it is a nilpotent ideal.

Solution: First we complete Exercise 7.4-12 as we will need the result from it. Assume R is commutative

and supposeI= (a1, a2, . . . , an) and J= (b1, b2 . . . bm) are two finitely generated ideals inR. Prove thatthe product ideal I J is finitely generated by the elements aibj fori = 1, 2, . . . , nand j = 1.2, . . . , m.

Proof: Let Hbe the set containing the elements aibj fori = 1, 2, . . . , nand j = 1.2, . . . , m.

(IJ H) Letx =i risi IJ, whereri = jti,jaj andsi = kui,kbk. Then since R is commutative,x=

i(

jti,jaj)(

kui,kbk) =

i

j

kti,jui,kajbk (H).

(IJ H) Letx = ri,jaibj (H). Since I is an ideal, x IJ.Back to the given problem. Assume amii = 0 for allai A for some positive integer mi. M=

mi. From

Exercise 7.4-12 (which we just completed), we have NM =XwhereX= { akii | ki = M}. So for eachelement

akii X, someki is at least mi and therefore X= 0 and that implies NM = 0. Thus N is anilpotent ideal.

IfN(R) is finitely generated, then each generator is (by definition) nilpotent. ThusN(R) is a nilpotentideal.//

Dummit & Foote Text Exercise 7.4-30: LetIbe an ideal of the commutative ring R and define

radI= {r R | rn I for some n Z+}called the radical ofI. Prove that radIis an ideal containing Iand that (radI)/I= N(R/I) (cf. Exercise29, Section 3).

Solution: For all a I, a1 I, so I radI. Hence radI is nonempty. Let a, b radI withan, bm I. Thebinomial theorem tells us that

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

m + n

k

akbm+nk.

SinceI is an ideal, ifk

n, then ak

I, and ifk < n, thenbm+nk

I. Since R is commutative, this

implies every term of (a + b)n+

m I, and consequently (a + b)m+n I since Iis a group under addition.Next, for all r R, (ra)n =rnan I, and (a)n R. Hence radIis an ideal ofR. //x + I radI/I x radI xn I for some n 1 xn + I= 0 R/Ifor some n 1 (x + I)n = 0 R/I for some n 1 x + I N(R/I). Hence radI/I= N(R/I). //

Dummit & Foote Text Exercise 7.4-31: An ideal Iof the commutative ring R is called a radical idealifradI= I .

(a) Prove that every prime ideal ofR is a radical ideal.

(b) Let n >1 be an integer. Prove that 0 is a radical ideal in Z/nZ is radical if and only ifn is a productof distince primes to the first power (i.e., n is squarefree). Deduce that (n) is a radical ideal in Z if andonly ifn the product of distinct primes in Z.

Solution: (a) Let P R be a prime ideal. Assume x radP. Then for some n 1, xn P. Assume thatn is minimal with this property. Ifn 2, then xxn1 P. SincePis prime, then either x P orxn1 Pwhich contradicts our assumption that n is minimal. Therefore n = 1 andx P.(b) () Let the prime factorization ofn ben = pkii . Assume that 0 is a radical ideal in Z/nZ, andassume someki 2. Then

pi is nonzero in Z/nZ, and (

pi)max ki = 0, a contradiction. This implies nis squarefree.

() Assume n is squarefree and let a Z/nZ with ak = 0 for some positive integer k . If some prime pdivides n but nota, then no power ofa can be divisible by n, a contradiction sinceak 0 mod n and pdivides n. Hencen divides a, and rad0 = 0.


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I had help with this part of the problem. Let R be a commutative ring , let I R be an ideal, and letJ R be an ideal containing I. Then J/Iis radical ideal in R/I if and only ifJ is a radical ideal in R.Proof: ( Assume J/Iis a radical ideal in R/I, and let x rad J. Then xm Jfor somem 1. Next(x + I)m =xm + I J/I, so x + I radJ/I= J/I. Thus x J, and J is a radical ideal in R.() Assume Jis radical in R and let x + I radJ/I. Then (x + I)m =xm + I J/Ifor some m 1, sothatxm J. Thus x radJ=J, andx + I J/IHence J/Iis a radical ideal.If (n) is an ideal ofZ, then (n) is a radical ideal in Zif and only if 0 is a radical ideal in Z/(n) if and onlyifn is square free. //

Dummit & Foote Text Exercise 7.4-33: LetR be the ring of all continuous functions from the closedinterval [0, 1] to R and for each c [0, 1] let Mc = {f R | f(c) = 0} (recall thatMc was shown to be amaximal ideal ofR).(a) Prove that ifMis any maximal ideal ofR then there is a real numberc [0, 1] such that M=Mc.(b) Prove that ifb and c are distinct points in [0, 1] then Mb=Mc.(c)Prove thatMc is not equal to the principal ideal generated by x c.(d) Prove that Mc is not a finitely generated ideal.

Solution: I had a lot of help on this problem. It was good to apply material from MAT5600.

(a) Proof by contradiction. LetM

R be a maximal ideal and assume M=Mc

c

[0, 1]. Then for all

such c, there exists a function fc M such that fc(c) = 0. Assume thatfc(c)> 0. Sincefc is continuous,there exists a positive real number c > 0 such that fc[(c c, c + c)] = 0. The set{(c c, c + c) [0, 1]|c [0, 1]}covers[0, 1]. Since [0, 1] is compact (closed and bounded subset ofR, thiscover has a finite subcover by the definition of compact. Let K [0, 1] be the finite subcover Kis finiteand{(c c, c + c) [0, 1]| c K} covers [0, 1]. For each c K, define uc(x) = 1 + (x c)/c ifx (c c, c) [0, 1], uc(x) = 1 + (c x)/c ifx [c, c + c) [0, 1], anduc(x) = 0 otherwise. Then uc Randuc is zero outside of (c c, c + c). Next considerg =

ucfc. Since fc M, we have g M.

However, for all x [0, 1], g(x) is positive since each uc(x)fc(x) is nonnegative and someuc(x)fc(x) ispositive. Thus g(x)> 0 for all x, and therefore 1/g R. This implies that Mcontains a unit, acontradiction. Thus M=Mc for some c [0, 1].(b) Assume b =c. Thenx b Mb but (x b)(c) = c b = 0. This implies x b / Mc. Therefore,Mb=Mc.(c)AssumeMc = (x c). Then|x c| =f(x)(x c) for some f(x) R. Ifx =c, then f(x) =|x c|

x c . Asx approaches c from the right, f(x) approaches, while f(x) approachesas x approachesc from theleft. So, no extension off is in R. Therefore, Mc is not generated by x c.(d) Proof by contradiction. AssumeMc = (A) is finitely generated, with A = {ai(x)| 1 i n}. Letf= |ai|. Then fis continuous on [0, 1] and f Mc. Thus f= riai for some continuous

functionsri R. If we let r = |ri|, we have f(x) = ri(x)ai(x) |ri(x)||ai(x)| r(x)f(x). For

each b =c, there must exist a function ai such thatai(b) = 0, otherwise h(b) = 0 for all h Mc andx c Mc. Thusc is the only zero off. From

f(x) r(x)f(x), forx =c we haver(x) 1/

f(x). Asx

approachesc, f(x) approaches 0, so that 1/

f(x) is unbounded and that implies r(c) does not exist whichis a contradiction sincer(x) R. Thus Mc is not finitely generated.

Dummit & Foote Text Exercise 7.4-37: A commutative ring R is called a local ring if it has a uniquemaximal ideal. Prove that ifR is a local ring with maximal ideal Mthen every element ofR M is a unit.Prove conversely that ifR is a commutative ring with 1 in which the set of nonunits forms an ideal M,thenR is a local ring with unique maximal ideal M.

Solution: Let x R M. If the ideal (x) is proper then it must be contained in some maximal ideal, andM is the only maximal ideal. Thus (x) M, a contradiction sincex / M. Thus (x) = R, and for somey R we have xy = yx = 1. This implies x is a unit.Assume the set Mof nonunits in R form an ideal. To show that Mis maximal, assume M I for someideal I. Then Icontains a unit, so that I=R. To show that Mis the unique maximal ideal, assume there


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is an ideal N R such that N M. Then Ncontains some element x not in M, which is a unit. ThusN=R. In particular, every proper ideal ofR is contained inM. Thus Mis the unique proper ideal ofR,andR is a local ring.

Dummit & Foote Text Exercise 7.4-38: Prove that the ring of all rational numbers whosedenominators is odd is a local ring whose unique maximal ideal is the principal ideal generated by 2.

Solution: R is a commutative ring with 1 = 0. Note that 1 = 11

. Let M Rbe the set of nonunits. SinceR Q, every element a/b R has an inverse in Q, b/a. Ifa/b is invertible in R then its inverse is b/a.However, a/b is not invertible in R whena is even (including zero). Thus M= {2a/b|a, b Z, b = 0}. So,M (2). Since 2 itself is a nonunit in R, M= (2), that is, Mis an ideal. By the previous exercise(7.4-37),R is local with maximal ideal M= (2).

Dummit & Foote Text Exercise 7.4-39: Following the notation of Exercise 26 in Section 1, let Kbe afield, letv be a discrete valuation on Kand let R be the valuation ring ofv . For each integer k 0 defineAk = {r R | v(r) k} {0}.(a) Prove thatAk is a principal ideal and that A0 A1 A2 . (b) Prove that ifI is any nonzeroideal ofR, then I=Ak for some k

0. Deduce thatR is a local ring with unique maximal ideal A1.

Solution: (a) It is clear that Ak+1 Ak for all k . Claim: Ak is an ideal. Since 0 Ak, Ak is nonempty.Leta, b Ak. If one or both ofa and b is 0, then a + b Ak. Ifa + b= 0, then a + b Ak. Ifa + b = 0, wehave v(a + b) min(v(a), v(b)) k, so that a + b Ak. Next, v(a) = v(a) and that impliesa Ak. Ifr R thenv(r) 0 and v(ra) = v(r) + v(a) k, so that ra Ak. SinceKis commutative, Ak is an idealofR.

To show that Ak is principal, choose x Ak such that v(x) = k. We can choose one because that elementexists sincev is surjective. Note that x1 exists inK, and thatv(x1) = v(x). Letr Ak. Thenv(r) kandv(x1r) = v(x1) + v(r) = v(r) v(x) 0. That implies x1r R and r = xx1r. Thus r (x), andwe have Ak (x), and thus Ak = (x). In particular, Ak is generated by any element of valuation k .(b) Let I Ak be a nonzero ideal ofR. Letk be minimal among v(r) forr I. Leta Isuch thatv(a) = k. In particular, we have I Ak. Moreover, since Ak = (a), we have Ak I. HenceI= Ak. Frompart (a), every proper ideal ofR is contained in A1 and therefore A1 is the unique maximal ideal ofR.

Dummit & Foote Text Exercise 7.4-40: Assume R is commutative. Prove that the following areequivalent: (see also Exercises 13 and 14 in Section 1)(i) R has exactly one prime ideal(ii) every element ofR is either nilpotent or a unit(iii) R/(R) is a field (cf. Exercise 29, Section 3).

Solution: (i) (ii) AssumeR has exactly one prime ideal. Since every maximal ideal ofR is prime, R islocal. By Exercise 7.4-26, N(R) is the unique maximal ideal. By Exercise 7.4-37, every element ofR N(R) is a unit, and (by definition) the remaining elements are nilpotent.(ii) (iii) Assume x + N(R) is nonzero. Then x / N(R). Sincex is not nilpotent in R, x is a unit. Thenx1 exists in R, and we have (x + N(R))(x1 + N(R)) = 1. Thus every nonzero element ofR/N(R) is a

unit. Since R is commutative, R/N(R) is a field.(iii) (i) Assume R/N(R) is a field. By Exercise 7.4-26, N(R) is contained in every prime ideal ofR. Bythe Lattice Isomorphism Theorem for rings, the only possible proper prime ideal is N(R). Also, N(R) isprime since it is maximal in R. Thus N(R) is the unique prime ideal ofR.

Dummit & Foote Text Exercise 7.5-3: LetFbe a field. Prove that F contains a unique smallestsubfieldF0 and thatF0 is isomorphic to either Q or Z/pZ for some prime p (F0 is called the prime subfieldofF). [See Exercise 26, Section 3.]


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Solution: We need to make use of the following fact and we will prove it:

LetFbe a field and letXbe a set of subfields ofF. ThenX is a subfield ofF. Proof:X is a subringofF, contains 1 since 1 Efor all E X. Let r F. Then r E for eachE X, and so r1 E. Thenr1 F, and henceF is a field. //LetFbe a field, and letFdenote the set of all subfields ofF. ThenF0 =

F is a subfield ofFand is

contained in every other subfield.

Consider the ring homomorphism : Z F0 with (1) = 1. The induced map : Z/(n) F0 is aninjective ring homomorphism, Exercises 7.4-26 and 7.4-28 informs us that n is a prime or 0. For all nonzeroa Z/(n), (a) is a unit inF0. Thus we have an monomorphism (injectivehomomorphism) :F(Z/(n)) F0, whereF(Z/(n)) denotes the field of fractions. Note that im is asubring ofFwhich is a field, and thus im =F0. Thus is an isomorphism. Ifn = 0, then F0= Q, and ifn= p is prime, then F0= Z/(p).//

Dummit & Foote Text Exercise 7.5-4: Prove that any sub field ofR must contain Q.

Solution: Exercise 7.5-3 showed us that , R contains a unique inclusion-smallest subfield which isisomorphic either to Z/(p) for a prime p or to Q.

Assume that the unique smallest subfield ofR is isomorphic to Z/(p), and let a in this subfield be nonzero.

Thenpa = 0 R, and since p R is a unit implies a = 0, which is a contradiction. Hence the uniquesmallest subfield ofRis isomorphic to Q. Any subfield ofR contains a subfield isomorphic to Q.

In Section 10.1s exercises R is a ring with 1 and M is a left R-module.

Dummit & Foote Text Exercise 10.1-2: Prove thatRx and Msatisfy the two axioms in Section 1.7for a group action of the multiplicative group Rx on the set M.

Solution: R is a group under the multiplication. The restriction of the group action to R M is amappingR M M. SinceR is a multiplicative group, it contains the multiplicative identity, 1. For allm M, the second axiom is satisfied, 1 m= m. For all r1, r2 R, (r1r2) m= r1 (r2 m) and therebysatisfies the first axiom.//

Dummit & Foote Text Exercise 10.1-5: For any left ideal IofR define

IM= {finite

ai mi| ai I, mi M

to be the collection of all finite elements of the form am where a I andm M. Prove thatI M is asubmodule ofM.

Solution: Apply the definition of the submodule criterion! Since 0 0 = 0 IM , I Mis not empty. Leti ai mi,

kbk nk IM and letr R. Then(

i ai mi) + r (

kbk nk) =

i ai mi+

k(rbi) nk IM. Hence I M Mis a submodule. //

Dummit & Foote Text Exercise 10.1-15: IfMis a finite abelian group then Mis naturally aZ-module. Can this action be extended to make Minto a Q-module?.

Solution: Proof by contradiction. LetMbe a (multiplicative) finite abelian group of order k . AssumethatM is a Q-module and that the action ofQ on Mextends the natural action ofZ given by k m= mk.Letx M be any nonidentity (and non-zero) element. Then 1

k x= y for some y M. That leads to

k ( 1k x) = k y. Sox = yk = 1, a contradiction. Therefore no such module structure exists.

Dummit & Foote Text Exercise 10.1-18: LetF = R, let V = R2 and letTbe the lineartransformation from V toVwhich is rotation clockwise about the origin by /2 radians. Show that V and0 are the only F[x]-submodules for thisT.


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Solution: I had help on this problem as I didnt quite understand fully F[x] modules. Note thatT :V V is given by (x, y) (y, x). Let Nbe an F[x]-submodule ofV . Assume that there exists anonzero element (a, b) N. This implies either a = 0 or b = 0 and therefore a2 + b2 >0. Let (x, y) V,r1 =

ax + by

a2 + b2, andr2 =

bx aya2 + b2

. Then

(r1+ r2x)

(a, b) = r1

(a, b) + r2

(x

(a, b))

= (r1a, r1b) + (r2b, r2a)= (r1a + r2b, r1b r2a)= (r1a + r2b, r1b r2a)

=

axa + bya

a2 + b2 +

bxb ayba2 + b2

,axb + byb

a2 + b2 bxa aya

a2 + b2

=

a2x + b2x

a2 + b2 ,

b2y+ a2y

a2 + b2

=

x

a2 + b2

a2 + b2, y

b2 + a2

a2 + b2

= (x, y).

Hence (x, y) N andN=V sinceN V. So, every nonzero F[x]-submodule ofV is V . Therefore, 0andV arethe only F[x]-submodules ofV .

Dummit & Foote Text Exercise 10.1-19: LetF = R, let V = R2 and letTbe the lineartransformation from V toVwhich is projection onto the y-axis. Show thatV , 0, thex-axis, and the y-axisare the only F[x]-submodules for this T.

Solution: Let N V be anF[x] submodule. Let thex-axis be represented by X= R 0 and they axis be represented as Y = 0 R.Assume there exists an element (a, b)

Nwhere a, b

= 0 and let (x, y)

V. Then

(xa + ( yb xa )x) (a, b) = (x, y). Thus V N, and since N V, that implies N=V.Next assume either a = 0 or b = 0 for every element (a, b) Nand assume that (a, 0) N witha = 0 andthere exists (0, b) N withb = 0. Then (a, b) N with a, b = 0, a contradiction. Thus N X. Now let(x, 0) X. Since x

a (a, 0) = (x, 0),X N. Hence N=X.

Next assume (0, b) N with b = 0. As before, N Y. If (0, y) Y, then yb (0, b) = (0, y), and so Y N.

Thus N=Y.

Last, if for all (a, b) N a= b = 0, then N= 0.So, the only possible submodules ofV are 0, X, Y, andV . 0 and Vare trivial submodules. So, we need toverify that Xand Y are submodules by the submodule criterion.

Sincee (0, 0) X, Xis nonempty. Let (a, 0), (b, 0) X, p(x) F[x], and p(x) = p0+ xp(x). Thenx (b, 0) = (0, 0) and

(a, 0) +p(x) (b, 0) == (a, 0) +p0 (b, 0) +p(x) (x (b, 0))= (a, 0) + (p0b, 0) + (0, 0)= (a +p0b, 0) X

By the submodule criterion, X V is a submodule.We repeat the process for Y . Since (0, 0) Y, Y is nonempty. Let (0, a), (0, b) X , p(x) F[x] and

p(x) =

pixi. Then T2(v) = T(v). In particular, xk v= x v for all k 1. Ifv Y, thenT(v) = v. We

obtain the following:


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(0, a) +p(x) (0, b) = (0, a) + (pixi) (0, b)= (0, a) +

pi (xi (0, b))= (0, a) +

pi (0, b)= (0, a +

pib) Y.So, Y V is a submodule.

Dummit & Foote Text Exercise 10.2-1 : Use the submodule criterion to show that kernels and imagesofR-module homomorphisms are submodules.

Solution: LetR be a ring with 1 and letM andNbe leftR-modules, and let : M Nbe an R-modulehomomorphism. (0) = 0 implies that ker and im are both nonempty.

Leta, b ker and let r R. Then (a + r b) = (a) + r (b) = 0, so that a + r b ker. Hence keris anR-submodule ofM by the submodule criterion.

Letx, y Mand letr R. Then(x) + r (y) = (x + r y). Since (x) and (y) are arbitrary in im,im is an R-submodule ofN by the Submodule criterion.

Dummit & Foote Text Exercise 10.2-2 : Show that the relation is R-module isomorphic to is anequivalence relation on any set ofR-modules.

Solution: M=Nif there exists an R-module isomorphism : M N. To prove an equivalence relation,we need to show that the relation is reflexive, symmetric, and transitive. Let Sbe a set of left R-modules.

(Reflexive). LetM S. The identity mapping id: M M is an R-module isomorphism sinceid(x + r y) = x + r y= id(x) + r id(y) for all x, y M andr R. HeneeM=M, and so the relation=is reflexive.(Symmetric). Assume that M , N Sand that M=N. This implies that there exists an Rmodulehomomorphism from : M N. Since is a bijection, has in inverse that is well defined. Let r Randx, y N. Let x = (a) andy = (b) for somea, b M. Then

1(x + r y) = 1((a) + r (b))= 1((a + r

b))

= a + r b= 1(x) + r 1(y).

So1 :N M is an R-module isomorphism. Hence N=Mand the relation is symmetric.(Transitive). LetM , N , P SandM=N andN=P. Then there exist R-module isomorphisms: M N and : N P. We know that : M Tis a bijection (since function composition ofbijections is a bijection). To show that it is also an R-module homomorphism, let x, y M andr R.Then

( )(x + r y) = ((x + r y))= ((x) + r (y))= ((x)) + r ((y))= (

)(x) + r

(

)(y)

So : M P is anR-module isomorphism, and we have M=P. Hence the relation is transitive.Since all three requirement of an equivalence relation have been met, is R-module isomorphic to is anequivalence relation.

Dummit & Foote Text Exercise 10.2-3: Give an explicit example of a map from one R-module toanother which is a group homomorphism but not an R-module homomorphism.

Solution: Let R be a non-commutative ring with 1. Choose an element a not in the center ofR and thenconsiderR as a left module over itself via multiplication. That is, M=R. If(x) = a x defines the map


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: M M, we have a group homomorphism (an endomorphism) since(x + y) = a (x + y) = a x + a y= (x) + (y). However, ifb R does not commute with a, then(b) = a b= ab but b (1) =b (a 1) = b a= ba =ab. Hence, although a group homomorphism, is notanR-module homorphism.

Dummit & Foote Text Exercise 10.2-13: LetIbe a nilpotent ideal in a commutative ring R (cf.Exercise 37, Section 7.3), let M andN be R-modules and let : M Nbe anR-module homomorphism.Show that if the induced map : M/IM N/IN is surjective, then is surjective.Solution: The induced map is given by (m + IM) = (m) + IN.Since is surjective, N/IN=[M/IM] = ([M] + IN)/IN. By the Lattice Isomorphism theorem formodules, we have N=[M] + IN.

We prove the statement N=[M] + ItNfor all t 1 by induction. The base case t = 1 has been shown tobe true. Assume for the induction step the equation holds for some t >1. Then

N = [M] + ItN= [M] + It([M] + IN)= [M] + It[M] + It+1N= [M] + It+1N sinceIt[M]

[M].

The induction hypothesis is proved. Since Ik = 0, N=[M] + IkN=[M] and implies is surjective.

Dummit & Foote Text Exercise 10.2-14: LetR = Z[x] be the ring of polynomials in x and letA= Z[t1, t2, . . . , ] be the ring of polynomials in the independent indeterminates t1, t2, . . . . Define an actionofR on A as follows: 1) let 1 R act on A as the identity, 2) for n 1, let xn 1 = tn, letxn ti = tn+ifori = 1, 2, . . . and let xn act as 0 on monomials in A of (total) degree at least two, and 3) extendZ-linearly, i.e., so that the module axioms 2(a) and 2(c) are satisfied.

(a) Show thatxp+q ti = xp (xq ti) = tp+q+i and use this to show that under this action the ring A is a(unital)R-module.

(b)Show that the map : R Adefined by (r) = r 1A is anR-module homomorphism of the ring Rinto the ring A mapping 1R to 1A, but is not a ring homomorphism from R to A.

Solution: (a)xp+q ti = tp+q+i = xp (tq+i) = xp (xq ti). HencemakesA into a unital left R-module.(b) Let a, b, r R. Then

(a + r b) = (a + r b) 1= a 1 + r (b 1)= (a) + r (b)

so that is an R-module homomorphism.

To show its not a ring homomorphism, we use contradiction. Assume that is a ring homomorphism.Then(x2) = x2 1 = t2 but(x2) = (x)(x) = (x 1)(x 1) = t12 contradiction. Hence is not a ringhomomorphism.

Dummit & Foote Text Exercise 10.3-1: Prove that ifA and B are sets of the same cardinality, thenthe free modules F(A) and F(B) are isomorphic.

Solution: I had help on this problem. Prof. R.D. Maddux of Iowa State University had the most lucidexplanation for this problem: http://orion.math.iastate.edu/maddux/505-Spring-2010/hw05.3.pdf

SinceA and B have the same cardinality, there exists a bijection f :A B. Sincefis a bijection, itsinverse is also a bijection, that is, f1 :B A. A. By Theorem 10.6 the free modules F(A) andF(B)have the universal mapping property. We now apply this property to f. Note that the image off is Bsincef is surjective, and B is a subset of the free module F(B). Thus we have another function that mapsA into F(B) and is equal tofon all elements ofA, that is, g : A F(B) and g(a) = f(a) for all a A.


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By the universal mapping property there is a uniqueR-module homomorphism : F(A) F(B) whichextendsg (andf). That is, f(a) = g(a) = (a) for all a = inA. Using the same reasoning, with A and Binterchanged and f replaced by f1, we obtain anotherR-module homomorphism : F(B) F(A)extendingf1. The composition ofR-module homomorphisms is again an R-module homomorphism: : F(A) F(A). For every a A, ( )(a) = ((a)) = (f(a). Since f(a) B and extends f1,(f(a)) = f1(f(a)) = a. This shows that is an extension of the identity map : A F(A) whichsends every element ofA to itself (i.e, (a) = a for everya A. By the universal mapping property, has aunique extension to an R-module homomorphism from F(A) to itself. The identity map from F(A) toF(A) is an R-module homomorphism ofF(A) onto itself, and by the universal mapping property ofF(A)it is the onlyR-module homomorphism ofF(A) onto itself. However, is a homomorphism ofF(A)onto itself that extends the identiy map on A so must be the identity map from F(A) toF(A).Similarly, is the identity map from F(B) toF(B). Thus and areR-module homomorphismsbetween F(A) and F(B), and they are inverses of each other, so they are both injective and surjective, andtherefore are isomorphisms between F(A) and F(B). Thus, F(A) =F(B) whenever|A| = |B|. //

Dummit & Foote Text Exercise 10.3-2: Assume R is commutative. Prove that Rn=Rm if and onlyifn = m, i.e., two freeR-modules of finite rank are isomorphic if and only if they have the same rank.[Apply Exercise 12 of Section 2 withIa maximal ideal ofR. You may assume that ifF is a field, thenFn

=Fm if and only ifn = m, i.e., two finite dimensional vector spaces over F are isomorphic if and only

if they have the same dimension - this will be proved later in Section 11.1.]

Solution: There are two solutions. One easy and one a bit more involved.

Simpler way. Assume that Rn=Rm. LetIbe a maximal ideal ofR. Then Rn/IRn=Rm/IRm. Exercise10.2-12 (not assigned) tells that Rn/IRn=R (R/IR) (R/IR) (R/IR) (n times). This implies(R/I)n=(R/I)m. Since these are vector spaces over the field R/I, we haven = m. //The more complicated way.It is clear that ifn = m, then the result holds. Assume hat Rm=Rn, where R : Rm Rn is anR-module isomorphism. LetI R be a maximal ideal. Next : Rm Rn/IRn, where denotes thenatural projection.

Claim: ker = I Rm.(IRm ker ): Ifx = ai (ri,j) IRm, then (ai (ri,j)) = ai ((ri,j)) IRn. Thus( )(x) = 0, and we have I R

m

ker .(ker IRm): Assumex = (ri) ker . Then 0 = ( )(x) = (x) + IRn. Thus (x) IRn. Let(x) =

ai (ri,j), where (ri,j) Rn. Since is surjective, we have (ri,j) = ((si,j)) for some (si,j) Rm.Then(x) = eo

ai((si,j)) = (

ai(si,j)). Since is injective, we have x =

ai (si,j), and thusx IRm as desired. Certainly is also surjective.By the first isomorphism theorem, the induced map : Rm/IRm Rn/IRn is anR-module isomorphism.The result of Exercise 10.2-12 implies (as with the easier method) that (R/I)m=R (R/I)n. SinceI ismaximal,R/Iis a field, and therefore m = n.//

Dummit & Foote Text Exercise 10.3-4: An R-module Mis called a torsion module if for each m Mthere is a nonzero element r Rsuch that rm= 0, where r may depend onm (i.e., M=Tor(M) in thenotation of Exercise 8 of Section 1). Prove that every finite abelian group is a torsion Zmodule. Give anexample of an infinite abelian group that is a torsion Zmodule.Solution: Let Mbe a finite abelian group. This implies it is also a Z-module. Letn be the order ofM.Then for every a M, by Lagrange, the order ofa divides the order ofn of the the abelian group M.Hence,na = 0. Sincen Z+ andna is the result of the action ofn on a in the module M, we havea Tor(M), so M Tor(M). Tor(M) Mis obviously true. Therefore M=Tor(M). //LetN=

iN Z/2Zand this implies that Nis a direct product of countably many copies of the 2-elementcyclic group Z/2Z. But Nis an infinite abelian group whose cardinality is the same as the set of realnumbers (by Cantors Diagonal Argument). Every element ofNhas order 2 so 2 n= 0 for every n N.Hence Tor(N) = N.


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Dummit & Foote Text Exercise 10.3-6 Prove that ifMis a finitely generated R-module that isgenerated byn elements then every quotient ofMmay be generated by n (or fewer) elements. Deduce thatquotients of cyclic modules are cyclic.

Solution: We need to show M /N= (ai+ N| i {1, 2, 3 , n 1, n}).((ai+ N| i {1, 2, 3 , n 1, n}) M/N) holds by the definition of a quotient.(M/N)

(ai+ N

|i

{1, 2, 3

, n

1, n

}). Assume x + N

M/N where x =i ri ai. Thenx + n= ri (ai+ N) as required. Hence{ai+ N}ni=1 is an R-module generating set for M /N.//

IfMis cyclic, then it has a generating set consisting of a single element. Moreover, the quotient ofM isthen generated by at most one element. Since every module is generated by at least one element, everyquotient of M is also cyclic.

Dummit & Foote Text Exercise 10.3-7: LetNbe a submodule ofM. Prove that if both M /N andNare finitely generated then so is M.

Solution: IfM /N= (ai+ N| 1 i m) andN= (bj| 1 j n), thenM= (ai, bj| 1 i m, 1 j n).M (ai, bj| 1 i m, 1 j n)) Assumem M. Thenm + N=

i ri (ai+ N) = (

i ri ai) + N.Thenm

i ri ai Nand that implies m i ri ai= jsjbj . This leads tom=i ri ai+jsj bj (ai, bj| 1 i m, 1 j n). Hence M is finitely generated.//(ai, bj| 1 i m, 1 j n) M) holds.

Dummit & Foote Text Exercise 10.3-9: An R-module Mis called irreducible ifM= 0 and if 0 andMare the only submodules ofM. Show thatMis irreducible if and only ifM= 0Mis a cyclic modulewith any nonzero element as generator. Determine all the irreducible Z-modules.

Solution: () IfM is irreducible, then M= 0 by definition. Let x Mbe nonzero. Then Rx M is anonzero submodule since 1 x= x Rx. Since M is irreducible, Rx = M. Hence any nonzero element ofMgenerates M.

() Assume M= 0 and that ifx = 0, then Rx = M. Let N Mbe a nonzero submodule. Then thereexists some nonzero x

N, and M=Rx

N. Thus N=M. Since the only submodules ofMare 0 and

M, M is irreducible.//

We know that Z-modules are abelian groups, and that Z-submodules are the subgroups of abelian groups.IfMis an irreducible Z-module, it is a abelian group whose subgroups that has no nontrivial propersubgroups). Hence Mis cyclic and therefore it is of prime order.

Dummit & Foote Text Exercise 10.3-10: Assume R is commutative. Show that an R-module M isirreducible if and only ifMis isomorphic (as an R-module) toR/IwhereIis a maximal ideal ofR. [Bythe previous exercise, ifMis irreducible there is a natural map R Mdefined byr rm, wherem is anyfixed nonzero element ofM.]

Solution: () AssumeMis irreducible, let m Ma fixed element and be nonzero, and definem : R Mbym(r) = r m. Then for all x, y R and r R,m(x + r y) = (x + r y) m= x m + r (y m) = m(x) + r m(y). So m is anR-modulehomomorphism. Sincem = 0 and M is irreducible, by Exercise 10.3.9, M=Rm. Ifb M, then thereexistsa R such that b = a m= m(a). Hencem is surjective.We need to show that kerm is a maximal ideal. Let x +ker m be nonzero. Then x m = 0. Since M isirreducible,M=R(x m). This implies m = y (x m) = (yx) mfor some y R. Then1 m (yx) m= 0, so that 1 yx kerm. That is, (y+ker m)(x +ker m) = 1 +ker m. So everynonzero element ofR/kerm has a left inverse. Because R/kerm is commutative, R/kerm is a field, sothat kerm is a maximal ideal ofR.

By the First Isomorphism Theorem,=R R/kerm, where kerm is a maximal ideal.


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() Assume I R is a maximal ideal. Then R/I is a field. Letx + Ibe nonzero. Then there exists y Rsuch that (y+ I)(x + I) = (1 + I). Ifr + I R/I, thenr + I= (ry+ I)(x + I) which impliesR/I=R(x + I). That is, R/Iis generated (as an R-module) by any nonzero element. By Exercise 10.3-9,R/Iis an irreducibleR-module.

Dummit & Foote Text Exercise 10.3-11: Show that ifM1 andM2 are irreducible R-modules, thenany nonzero R-module homomorphism from M1 toM2 is an isomorphism. Deduce that ifM is irreduciblethenEndR(M) is a division ring (this result is called Schurs Lemma). [Consider the kernel and the image.]

Solution: Assume : M1 M2 is an R-module homomorphism with M1 andM2 irreducible. In Exercise10.2-1, we proved that ker and im are submodules ofM1 andM2 respectively. Since is not the zerohomomorphism, its kernel is not all ofM1, and since M1 is irreducible ker = 0. Thus is injective.Similarly, im is not the zero submodule, and consequently M2 is not the zero submodule. Hence issurjective. Therefore is an R-module isomorphism.

LetMbe an irreducible unital left R-module. We know that EndR(M) is a ring under pointwise additionand composition. If EndR(M) is nonzero, then by the first part of this exercise, it is an isomorphismand so has an inverse 1 which is also in EndR(M). So EndR(M) is a division ring. We are not given anyadditional information that would lead us to conclude that EndR(M) is commutative and therefore a field.

Dummit & Foote Text Exercise 10.3-12: LetR be a commutative ring and let A, B , and M beR-modules. Prove following isomorphisms ofR-modules:

(a) HomR(A B, M) =HomR(A, M) HomR(B, M)(b) HomR(M, A B) =HomR(M, A) HomR(M, B).Solution: I had A LOT of help with this problem.(a) Let f :A M andg : B M , be R-module homomorphisms defined by (f, g) : A B M by(f, g)(a, b) = f(a) + g(b). Then for all a1, a2 A, b1, b2 B, and r R, we have

(f, g)((a1, b1) + r (a2, b2)) = (f, g)((a1+ r a2, b1+ r b2))= f(a1+ r a2) + g(b1+ r b2)= f(a1) + r f(a2) + g(b1) + r g(b2)= (f, g)(a1, b1) + r (f, g)(a2, b2).

So(f, g) : A B M is anR-module homomorphism.Now consider the map : HomR(A, M) HomR(B, M) HomR(A B, M). We will show that is anR-module homomorphism. Letf1, f2 : A M andg1, g2 : B M be R-module homomorphisms and letr R. Then

((a1, b1) + r (a2, b2))(a, b) = (f1+ r f2, g1+ r g2)(a, b)= (f1+ r f2)(a) + (g1+ r g2)(b)= f1(a) + r f2(a) + g1(b) + r g2(b)= (f1, g1)(a, b) + r (f2, g2)(a, b)= ((f1, g1) + r (f2, g2))(a, b).

So is an R-module homomorphism. Next we show is aninjection. Assume (f, g) ker. Then0 = (f, g)(a, 0) = f(a) for all a

A, so thatf= 0. Similarly, g = 0. Thus (f, g) = 0 and ker = 0. Hence

is an injection.

Next we show is an surjection. Let :A B Mbe an R-module homomorphism, and definef :A M andg :B M byf(a) = (a, 0) andg(b) = (0, b). Then

f(a1+ r a2) = (a1+ r a2, 0= (a1, 0) + r (a2, 0)= f(a1) + r f(a2).


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Sof is an R-module homomorphism. Similarly, g is an R-module homomorphism. Since

(f, g)(a, b) = f(a) + g(b)= (a, 0) + (0, b)= (a, b),

we have (f, g) = . Hence is a surjection.

Thus, we have HomR(A B, M) =R HomR(A, M) HomR(B, M). //(b) Now let A : A B A and B :A B B be the left and right coordinate projections respectivelyand these are R-module homomorphisms. Define : HomR(M, A B) HomR(M, A) HomR(M, B) by() = (A , B ). If, HomR(M, A B) and r R, then

( + r ) = (A ( + r ), B ( + r ))= (A + r (A ), B + r (B ))= (A , A ) + r (B , B )= () + r ().

Thus is an R-module homomorphism. We now show that is an injection. Assume ker. Ifm M, then

0 = ()(m, m)

= ((A )(m), (b )(m))= (A((m)), B((m)) = (m).

So = 0, hence ker = 0, and thus is an injection.

Next we show thatP si is a surjection. Assume (f, g) HomR(M, A) HomR(M, B). Definef,g : M A B byf,g(m) = (f(m), g(m). f,g is an R-module homomorphism. Moreover,(f,g)(m) = ((A f,g)(m), (B f,g)(m)) = (f(m), g(m)). Thus (f,g) = (f, g), and so is asurjection . Hence HomR(M, A B) =R HomR(M, A) HomR(M, B).//

Dummit & Foote Text Exercise 10.3-13: LetR be a commutative ring and let Fbe a free R-moduleof finite rank. Prove the following isomorphism ofR-modules: HomR(F, R) =F.Solution: Let Fbe a free R-module of finite rank. That is, let Fbe free on the set

{ai

}ni=1. Then every

element ofFcan be written uniquely as

ri ai for someri R. Next, define : HomR(F, R) F by() =

(ai) ai. is an R-module homomorphism. If, HomR(F, R) and r R, then

( + r ) = ( + r )(ai) ai= (

(ai) ai) + r (

(ai) ai)= () + r ()

shows that is an R-module homomorphism.

is injective. Assume ker. Then 0 = () = (ai) ai. Since F is free on the ai, we have(ai) = 0 for all ai, and thus = 0. So ker = 0, and thus is injective.

is surjective. Let

ti ai F. Defineg : F R by g (

ri ai) =

riti . So , g is an R-modulehomomorphism. Moreover, g(ai) = ti, so that (g) = ti ai. Hence is surjective.Thus is an R-module isomorphism, so that we have HomR(F, R) =F.//

Dummit & Foote Text Exercise 10.3-20: LetIbe a no empty index set and for each i I let Mi beanR-module. The direct productof the modulesMi is defined to be their direct product as abelian groups(cf. Exercise 15 in Section 5.1) with the action ofR componentwise multiplication. The direct sum of themodules Mi is defined to be the restricted direct product of the abelian groups Mi (cf. Exercise 17 inSection 5.1) with the action of R componentwise multiplication. In other words, the direct sum of the Misis the subset of the direct product,

iIMi, which consists of all elements

iImi such that only finitely

many of the components mi are nonzero; the action ofR on the direct product or direct sum is given by


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r

iImi=

iIrmi (cf. Appendix I for the definition of Cartesian products of infinitely many sets). Thedirect sum will be denoted by

iIMi.

(a) Prove that the direct product of the Mis is anR-module and the direct sum of the Mis is asubmodule of their direct product.

(b)Show that ifR = Z, I= Z+ andMi is the cyclic group of order i for eachi, then the direct sum of theMis is not isomorphic to their direct product. [Look at torsion.]

Solution: (a) We know from Exercise 5.1-15, that

iIMi is an abelian group under point-wise additionand multiplication. So we just need to verify the three left-module axioms.Letr, s R and let (mi), (ni)

iIMi. Then

(r+ s) (mi) = ((r+ s) mi)= (r mi+ s mi)= (r mi) + (s mi)= r (mi) + s (mi),

(rs) (mi) = ((rs) mi)= (r (s mi))= r (s mi)= r (s (mi)),

r ((mi) + (ni))

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