SERGE LANG’S ALGEBRA CHAPTER 3 EXERCISE SOLUTIONS …people.math.sc.edu/kellerlv/Lang_Ch3 (1).pdf · SERGE LANG’S ALGEBRA CHAPTER 3 EXERCISE SOLUTIONS 3 Since fis injective, M0˘=Imf

SERGE LANG’S ALGEBRA CHAPTER 3 EXERCISESOLUTIONS

KELLER VANDEBOGERT

1. Problem 1

Let {u1, . . . , un} and {w1, . . . , wm} be bases for U and W , respec-

tively. Without loss of generality, we may assume that {u1, . . . , uk}

and {v1, . . . , vk} form bases for U ∩W . This implies

Span{u1, . . . , uk} = Span{w1, . . . , wk}

so that {u1, . . . , un, wk+1, . . . , wm} forms a basis for U +W . Counting

cardinalities,

dim(U +W ) = dim(U) + dim(W )− dim(U ∩W )

=⇒ dim(U +W ) + dim(U ∩W ) = dim(U) + dim(W )

2. Problem 2

As S−1M ∼= S−1R⊗RM , we may assume without loss of generality

that R is local. Let m denote the maximal ideal; R/m is a field, so that

R/m⊗RM is a vector space.

Choosing a basis yields a generating set for the preimage, and con-

versely, every generating set can be refined to a basis in R/m ⊗R M .

Since vector spaces have the invariant basis property, we deduce that

M does as well.

Date: March 7, 2018.1

2 KELLER VANDEBOGERT

3. Problem 3

Since R is an integral domain, the homothety map s 7→ rs is injec-

tive. Extending by linearity over k, we have an injective map over a

finite dimensional vector space. But this means that we have an iso-

morphism, thus there exists s ∈ R such that rs = 1, whence R is a

field.

4. Problem 4

(a). Assume first that

0 // M ′ f// M

g// M ′′ // 0

splits, so that M ∼= M ′ ⊕M ′′. Then, take h : M ⊕M ′′ → M ′ as the

natural projection onto M ′. By definition, hf ≡ id.

Similarly, we may take i : M ′′ ↪→ M ′ ⊕ M ′′ to be the standard

inclusion. Again by definition, we have that ig ≡ id.

Assume no that f is left invertible, with left inverse h. Observe first

that for any m ∈M ,

h(m− f ◦ h(m)

)= h(m)− h(m) = 0

So that m − f(h(m)) ∈ Ker k. This immediately gives that M =

Kerh + Im f , since m = (m − f(h(m))) + f(h(m)). Indeed, we can

say even more, since if m ∈ Kerh ∩ Im f , then m = f(m′) for some

m′ ∈M ′, and

0 = h(m) = h(f(m′)) = m′

So m′ = 0 =⇒ f(m′) = 0. Hence

M = Kerh⊕ Im f

SERGE LANG’S ALGEBRA CHAPTER 3 EXERCISE SOLUTIONS 3

Since f is injective, M ′ ∼= Im f . It remains to show that

Kerh ∼= M ′′

Since g is surjective, every m′′ ∈ M ′′ is of the form g(m) for some

m ∈M . As M = Kerh⊕ Im f and Ker g = Im f by exactness,

g(M) = g(Kerh) = M ′′

And, as g|Kerh is injective by exactness,

Kerh ∼= M ′′

so that M ∼= M ′ ⊕M ′′.

Suppose now that g is right invertible with right inverse i. Consider

m− i(g(m))

We again see that

g(m− i(g(m))

)= g(m)− g(m) = 0

So that m − i(g(m)) ∈ Ker g = Im f (by exactness). Also, if m ∈

Ker g ∩ Im i, then m = i(m′′) for some m′′ ∈M ′′ and

0 = g(m) = g(i(m′′)) = m′′

So that m = 0 as well. Hence

M ∼= Ker g ⊕ Im i

Since Ker g = Im f and f is injective, Im f = M ′. Similarly, we deduce

that

g(M) = g(Im i) = M ′′

and by the first isomorphism theorem, this must be an isomorphism.

Hence,

M ∼= M ′ ⊕M ′′


as desired.

(b). Let

i : E →⊕i

Ei

π :⊕i

Ei → E

be the given maps. These are trivially R-module homomorphisms. We

also see

π ◦ i(x) = π(ψ1x, . . . , ψmx)

= φ1 ◦ ψ1x+ · · ·+ φm ◦ ψmx

=(∑

i

φiψi

)(x)

= x

i ◦ π(x1, . . . , xm) = i(φ1x1 + · · ·+ φmxm)

= ψ1φ1x1, . . . , ψmφmxm)

= (x1, . . . , xm)

Whence i and π are inverses of each other, so they are isomorphisms.

If each φi is the natural inclusion Ei ↪→⊕

iEi and ψi the natural

projection⊕

iEi → ei, we see

ψiφi = id, ψj ◦ φi ≡ 0 (i 6= j)

We also see for (e1, . . . , em) ∈⊕

iEi,

φ1ψ1(e1, . . . , em) + · · ·+ φmψm(e1, . . . , em)

=φ1(e1) + · · ·+ φm(em)

=(e1, 0, . . . , 0) + · · ·+ (0, . . . , 0, em)

=(e1, . . . , em)

So that∑

i φiψi = id, as required.


5. Problem 5

Proceed by induction on the maximal amount of linearly independent

element of A over R. For the base case n = 1, this is by definition.

Let {v1, . . . , vm} be a maximal set of such elements. Consider any

subgroup A0 contained in the space generated by {v1, . . . , vm−1}. By

the inductive hypothesis, these can all be generated by integral linear

combinations.

Now, denote by S the subset of A such that

v = a1v1 + · · ·+ amvm, ai ∈ R

0 6 ai < 1

0 6 am 6 m

Choose v′m such that the coefficient am is minimal and nonzero in S.

Note that such an element exists since S is finite by assumption and if

every am = 0, {v1, . . . , vm−1} generates S and by scaling, {v1, . . . , vm−1}

generates A. Employing the inductive hypothesis would yield the re-

sult, whence we may find am > 0.

We want to now show that

{v1, . . . , vm−1, v′m}

is a basis for A over Z. Let v ∈ A, so that v = a1v1 + · · · + amvm.

Then we may find a sufficiently large N ∈ N such that v/N ∈ S; by

definition, am/N > a′m, where a′m is the mth coefficient of v′m.

Let k be the smallest positive integer such that ka′m > am/N . If

ka′m 6= am/N , then by minimality of k,

amN− ka′m < a′m


But this may not happen, so in fact

ka′m =amN

and if

v′m = a′1v1 + · · ·+ a′mvm

for some coefficients a′i, we may multiply by the above by Nk and use

that Nka′m = am and see

amvm = −Nka′1v1 − · · · −Nka′m−1 +Nkv′m

And, substituting this for the expression of v,

v = (a1 −Nka′1)v1 + · · ·+ (am−1 −Nka′m−1)vm−1 +Nkv′m

Subtracting we see that v−Nkv′m ∈ A0, so by the inductive hypothesis

we may find ji ∈ Z such that

v −Nkv′m = j1v1 + . . . jm−1vj−1

Whence we finally see v ∈ SpanZ{v1, . . . , v′m}, as desired.

6. Problem 6

Confer Lang’s Algebraic topology book for the correct statement.

The statement given here is not true.

7. Problem 7

(a). Let u, v ∈ W . Then,

|u− v| 6 |u|+ |v| = 0

=⇒ |u− v| = 0

So this is a subgroup.


(b). For convenience, we may assume M0 = {0}. Let M1 = (v1, . . . , vr)

and let d ∈ Ann(M/M1). Then, dM ⊂M1, and we may choose nj,j to

be the smallest integer such that there exist

nj,1, . . . , nj,j−1 ∈ Z

such that

nj,1v1 + · · ·+ nj,jvj = dwj

for some wj ∈ M . Without loss of generality, we may assume 0 6

nj,k 6 d − 1. It remains to show our elements {w1, . . . , wr} forms a

basis.

By selection

Span{w1, . . . , wr} = Span{v1, . . . , vr}

And, since the cardinality of the wi matches that of the vi, linear

independence is a triviality. Finally, since 0 6 nj,k 6 d− 1,

|wi| =∣∣∣ r∑j=1

nj,kdvj

∣∣∣6

r∑j=1

|vj|

As desired.

8. Problem 8

(a). We certainly have that the kernel is ±1. Let (a, b) = (a′, b′) = 1.

If x = a/b, y = a′/b′,

h(xy) = log max(|a||a′|, |b||b′|

)6 log

(max

(|a|, |b|

)max

(|a′|, |b′|

))= h(x) + h(y)


(b). M is certainly finitely generated, since if not, we could find an

infinite irredundant generating set for M1, which is a contradiction.

Using Problem 7, after completing the xi to a basis for M , we may

bound any generating set appropriately.

9. Problem 9

(a). Define our localization as equivalence classes

m

s=m′

s′⇐⇒ ∃ r ∈ A such that r(s′m− sm′) = 0

This is given the trivial S−1A-module structure

a

b

(ms

):=

am

bs

Well definedness/distributeivity follow immediately from the fact that

M is itself an A-module.

(b). Let

0 // M ′ φ// M

ψ// M ′′ // 0

We exact. Then, φ and ψ extend to localized maps by defining

φ(m′s

):=

φ(m′)

s

ψ(ms

):=

ψ(m)

s

And then extending by linearity. Suppose then that φ(m′)/s = 0, so

that some t ∈ A must annihilate φ(m′), implying that φ(tm′) = 0.

Since φ is injective, tm′ = 0, whence m′/s = 0, so the localized φ is

also injective.

Now let us check exactness at S−1M . We have Imφ ⊂ Kerψ by

definition. Suppose that ψ(m/s) = 0, so there exists t ∈ A such that


ψ(tm) = 0. By exactness, tm ∈ Imφ, so φ(m′) = tm for some m′ ∈M ′.

Then,

m

s=tm

ts

=φ(m′)

ts

= φ(mts

)∈ Imφ

So our sequence is exact at M . It remains to show that ψ is surjective.

Observe that ψ on M is surjective, so given m′′ ∈ M ′′, there exists

m ∈M such that ψ(m) = m′′. Then,

m′′

s=ψ(m)

s= ψ

(ms

)∈ Imψ

So that

0 // S−1M ′ φ// S−1M

ψ// S−1M ′′ // 0

is exact.

10. Problem 10

(a). Our map is

M →∏

p∈m-Spec(A)

Mp

m 7→(m

1

)p∈m-Spec(A)

Now, suppose m 7→ (0). Then for each p ∈ m-Spec(A), there exists

ap /∈ p such that apm = 0. But then Ann(m) is not contained in any

maximal ideal, whence Ann(m) = A, so m = 0, yielding surjectivity.

(b). We already know the forward direction from part (b) of the pre-

vious problem. Let φ, ψ be our maps φ : M ′ →M , ψ : M →M ′′, and

consider the converse.


Firstly, suppose φ(m′) = 0 for some m′ ∈ M ′. Then for all p ∈

m-Spec(A), there exists ap /∈ p such that apm′ = 0. To see this, note

that

φ(m′) = 0 =⇒ φ(m′

1

)=⇒ m′

1= 0 for all p ∈ m-Spec(A)

By identical reasoning as in part (a), Ann(m′) = A, so that m′ = 0,

and φ is injective.

We know that Kerψ ⊂ Imφ. For the reverse inclusion, note that

ψ ◦ φ(m′

1

)= 0

=⇒ ψ ◦ φ(m′)

1= 0 for all p ∈ m-Spec(A)

=⇒ Ann(ψ ◦ φ(m′)) = A

whence Imφ ⊂ Kerψ, giving exactness at M . Finally, surjectivity is a

tautology, so that

0 // M ′ φ// M

ψ// M ′′ // 0

is exact.

(c). Suppose that M → Mp and m 7→ m/1 = 0. Then, there exists

ap /∈ p such that apm = 0. Since M is torsion free, m 6= 0 implies that

ap = 0. But then ap ∈ p, which cannot happen, so we deduce that

m = 0 and our natural inclusion is thus injective, as desired.

11. Problem 11

Let p ∈ m-Spec(o). Then Mp is still finitely generated and torsion

free over op. By problems of the previous chapter, op is a PID, and

hence Mp is projective (since it is free). Let

Ff// M // 0


be exact. We want to show that f is right invertible. We know that

the induced map fp is right invertible by freeness of Mp.

Hence for all p ∈ m-Spec(o), there exists gp such that fpgp = idMp .

We can then find cp ∈ o with cp /∈ p and

cpgp(M) ⊂ F

(this is merely by definition), since gp(Mp) ⊂ Fp. Then, we want to

show that {cp}p∈Spec(o) generate all of o; this follows since {cm}m∈m-Spec(o)

is not contained in any maximal ideal, hence generates all of o. Thus

there exist xi, cpi ∈ o such that∑i

xicpi = 1

Set g :=∑

i xicpigpi . Then for all a/b ∈ o, m ∈ m-Spec(o):

(f ◦ g)m

(ab

)=

1

b

∑i

fm ◦ (xicpigpi(a))

=1

b

∑i

xicpifpigpi(a)

=a

b

∑i

xicpi =a

b

Since the maximal ideal m was arbitrary, we deduce that f ◦ g ≡ idM .

12. Problem 12

(a). We have the following short exact sequence

0 // a ∩ b // a⊕ b // a + b // 0

Assume then that a and b are coprime, so that a ∩ b = ab, a + b = o.

As o is projective, the sequence above splits, so

a⊕ b = o⊕ ab


Now, employing exercise 19 of the previous chapter, choose x, y ∈ o

such that xa and yb are coprime. Then,

a⊕ b = xa⊕ yb

= o⊕ xyab

= o⊕ ab

Whence the general case. Indeed, by induction, one easily sees

a1 ⊕ · · · ⊕ an = on−1 ⊕ a1a2 . . . an

(b). Let f : a→ b be our isomorphism. Then, fk|a = f , and

fk(a) = fk(1) · a = ca

for each a ∈ a. Hence, f is merely the homothety mc : x 7→ cx, where

c := fk(1).

(c). Let f ∈ Hom(a, o). Certainly 1 /∈ a, and we may extend f to all

of k by linearity as in (b). Then the association

f 7→ fk(1) ∈ a−1

is an isomorphism. Note that well definedness follows since if f(a) ∈ o

for a ∈ a, then fk(1) · a ∈ o, so that fk(1) ∈ a−1 by definition.

Injectivity is easy since if fk(1) = gk(1), then for all a ∈ a, fk(1) ·a =

gk(1) · a =⇒ f(a) = g(a), whence f ≡ g. Surjectivity follows from

part (b), so this is indeed an isomorphism.

In particular,

Hom(a, o) = a−1

and

a∨∨ = (a−1)−1 = a =⇒ a∨∨ = a


13. Problem 13

M is projective, hence a direct summand of a free module F . This

immediately gives that M is torsion free, so the free module F ′ gen-

erated by the non torsion elements is contained in M . By definition

(since we have only removed torsion elements) the rank of F ad F ′

must coincide, since else F ′ would have nontrivial torsion. Thus, there

exists F and F ′ free such that

F ′ ⊂M ⊂ F, rankF = rankF ′

(b). Proceed by induction on the rank of M . When n = 1, there is

nothing to prove.

Assume now that M has rank n. Choose generators e1, . . . , en−1

linearly independent with span denoted by N . We have the short exact

sequence

0 // N // M // M/N // 0

By the inductive hypothesis, N = a1⊕· · ·⊕an−1. Also, counting ranks

yields that M/N has rank 1, whence we may choose a generator of

M/N . Its preimage will by linearly independent with N since it has

nonzero class in M/N , in which case we see that M ∼=⊕

i ai.

As o is Noetherian, an is finitely generated and M/N is torsion free/

Thus M/N is projective, so our sequence splits

=⇒ M = a1 ⊕ · · · ⊕ an

As desired.

(c). We may assume without loss of generality that the ai are pairwise

coprime. By Problem 12 part (a),

M = on−1 ⊕ a


where a = a1 . . . an. Suppose now that for any two fractional ideals a,

b ∈ o, that on−1⊕ a = om−1⊕ b. We want to show that this is possible

if and only if a = b and n = m.

If on−1 ⊕ a = om−1 ⊕ b, taking the rank of both sides immediately

yields m = n. If we take the (n+ 1)th exterior power, we find that

D1on−1 ⊗ a = D2o

n−1 ⊗ b, Di ∈ o

=⇒ a = b

where we have used the fact that the exterior power converts our direct

sum to a tensor product (the Di are our determinants). Whence the

map M 7→ [a] is an isomorphism, and we are done.

14. Problem 14

We have the following commutative diagram with exact rows, which

will referenced each part of this problem:

M ′

f��

φ// M

g

��

ψ// M ′′

h��

// 0

0 // N ′φ′// N

ψ′// N ′′

(a). Suppose that f and h are monomorphisms. Let m ∈ Ker g. By

commutativity, there exists m′ ∈ M ′ with φ(m′) = m. By commuta-

tivity,

φ′(f(m′)) = 0

Since φ′ is injective by exactness, f(m′) = 0.

But f is also injective, so that m′ = 0 and φ(0) = 0 = m, and g is

also a monomorphism.

(b). Suppose that f and h are surjective. Let n ∈ N . Then, ψ′(n) Im Imh,

since h is surjective, so there exists m′′ ∈M ′′ such that h(m′′) = ψ′(n).


By exactness, ψ is surjective, so there exists m ∈ M such that

ψ(m) = m′′. By commutativity of the diagram, ψ′(g(m)) = ψ(n),

so that g(m) − n ∈ Kerψ′ = Imφ′, so there exists n′ ∈ N ′ such that

φ′(n′) = g(m)− n, and since f is surjective, there exists m′ ∈M ′ such

that f(m′) = n′. By commutativity of the diagram,

g ◦ φ(m′) = g(m)− n

=⇒ n = g(m− φ(m′)) ∈ Im g

So that g is surjective.

(c). Assume 0→M ′ →M and N → N ′′ → 0 are exact. By the Snake

Lemma,

0 // Ker f // Ker g // Kerh

δ// Coker f // Coker g // Cokerh // 0

is also exact. However, we observe that if any of the above two kernels

and cokernels vanish, so must the other. Hence the statement is a

triviality.

15. Problem 15

The diagram that will be referenced in each part of this question is

the following:

M1

f1��

a1// M2

f2��

a2// M3

f3��

a3// M4

f4��

a4// M5

f5��

N1b1// N2

b2// N3

b3// N4

b4// N5

Note the above is commutative with exact rows. The format of the

solutions will be a string of implications so as to make it very easy

for the reader to follow along the diagram. Also, any element of its


corresponding set will be denoted with the lower case letter with the

same index (ie, m3 ∈M3 always).

(a). We have:

m3 ∈ Ker f3

=⇒ f4(a3(m3)) = 0 (commutativity)

=⇒ a3(m3) = 0 (injectivity of f4)

=⇒ m3 ∈ Im a2 (exactness)

=⇒ a2(m2) = m3 (by definition)

=⇒ b2(f2(m2)) = 0 (commutativity)

=⇒ b1(n1) = f2(m2) (exactness)

=⇒ f1(m1) = n1 (surjectivity of f1)

=⇒ f2(a1(m1)) = f2(m2) (commutativity)

=⇒ f2(a1(m1)−m2) = 0

=⇒ a1(m1) = m2 (injectivity of f2)

=⇒ m2 ∈ Im a1 = Ker a2 (exactness)

=⇒ m3 = a2(m2) = 0

=⇒ f3 is injective


(b). Employing the same convention as in part (a), we see

n3 ImN3

=⇒ f4(m4) = b3(n3) (surjectivity of f4)

=⇒ f5(a4(m4)) (exactness and commutativity)

=⇒ a4(m4) = 0 (injectivity of f5)

=⇒ a3(m3) = m4 (commutativity)

=⇒ b3(f3(m3)− n3) = 0

=⇒ b2(n2) = f3(m3)− n3 (exactness)

=⇒ f2(m2) = n2 (surjectivity of f2)

=⇒ f3(a2(m2)) = f3(m3)− n3 (commutativity)

=⇒ n3 = f3(m3 − a2(m2))

So that f3 is surjective.

16. Problem 16

Let {Si, (fji}i∈I denote our inverse system, where each fji : Sj → Si

are all surjective. By simplicity, this implies that each fji is either

trivial or an isomorphism.

If every Si = 1, then we are done. Hence, suppose not. Given Si,

Sj, there exists k such that k > i, j. Then

Sk ∼= Si, Sk ∼= Sj

=⇒ Si ∼= Sj

Then any two nontrivial groups in our inverse system are necessarily

isomorphic. Let S denote the common isomorphism. By assumption

S is simple, it remains only to show that

lim←−Si = S


The isomorphism is not so difficult to specify. Choose i such that Si is

nontrivial. The inclusion

I :S ↪→ lim←−Si

x 7→ (x)i∈I

is injective, since any nonzero element must represent a nonzero class

in some Si. It remains to show surjectivity. Let (xi) ∈ lim←−Si. For

every j 6 i, fji(xj) = xi, and for every k > i, fik(xi) = xk. Inverting

yields fki(xk) = xi, so that every element is completely determined by

the ith spot; whence i(xi) = (xi), and we have an isomorphism.

17. Problem 17

(a). We have the inverse system

(Z/pn, πnm)

with

πnm(a+ (pn)) = a+ (pm)

By definition, πnn ≡ id. Now, set Zp := lim←−Z/pn. Let

ρn :Zp → Z/pn

(a+ (pm))m∈N 7→ a+ (pn)

This is certainly surjective as any m+ (pn) has preimage

(n+ (pn))n∈N

There are no zero divisors, since if k is a zero divisor, then pn|k for all

n ∈ N, which is possible only if k = 0.

The maximal ideal is merely pZp, since one immediately sees that

Zp/pZp = Z/pZ


which is, in particular, a field. This is also the unique maximal ideal

since any other

x = (xi + (pn))n∈N

is pointwise invertible by merely noting that Zp is isomorphic to Z/p[[p]].

An element in the ring of formal power series is invertible if and only

if the first term is a unit, which corresponds to elements x /∈ pZp, so

that every element not contained in pZp is a unit, so this is maximal.

This also gives that p is the only prime in this ring, since any other

prime would generate an idea contained in pZp, and hence divide p.

Finally, for the UFD property, we can actually do better, since Zp

is a PID. To see this, merely note that every ideal must be an ideal

in each entry, and ideals in every entry are principal. Hence, Zp is

principal, hence a UFD.

(b). By the Chinese Remainder theorem,

Z/(a) ∼=⊕i

Z/(pαii )

where a = pα11 . . . pαk

k is the prime factorization of a. Using this and

the fact that inverse limits preserve direct products,

lim←−(a)

Z/(a) =∏

p prime

lim←−n

Z/(pn)

=∏

p prime

Zp

As asserted.


18. Problem 18

(a). The diagram

An+1 ×Mn+1

��

// Mn+1

��

An ×Mn// Mn

commutes, so that

gn+1,n(an+1mn+1) = fn+1,n(an+1) ·mn

Then, let lim←−An act on lim←−Mn by

(an)n∈N(mn)n∈N := (anmn)n∈N

By the above commutative diagram, this action is well defined and

preserves the structure of the inverse limit.

(b). Using part (a), we consider our maps. Observe that Mn = A for

each n, and, An = /bbz/(pn). Then, we have

ψ :A→ A

a 7→ pa

φ :Z/(pn)→ Z/(pn+1)

m+ (pn) 7→ m+ (pn+1)

with trivial action

Z/(pn)× A→ A

(m+ (pn), a) 7→ ma

We then see that

(m+ (pn), a) 7→ ma 7→ pma

and

(m+ (pn), a) 7→ (m+ (pn+1), pa) 7→ pma


Whence the diagram in part (a) commutes, and using the result of (a),

lim←−A := Tp(A) is a module over lim←−Z/(pn) = Zp.

(c). We see that

(m+ (pn), a, b) 7→ m(a, b) 7→ pm(a, b)

and,

(m+ (pn), a, b) 7→ (m+ (pn+1), pa, pb) 7→ m(pa, pb)

And, since pm(a, b) = m(pa, pb), the diagram in (a) commutes so the

result follows immediately.

19. Problem 19

By definition, if a 7→ 0 ∈ lim−→An, then fik(a) = 0 for some k > i.

20. Problem 20

Note that

lim−→i

lim−→j

Aij and lim−→j

lim−→i

Aij

both satisfy the following universal property: for all (i, j) 6 (r, s), there

exist maps f(i,j) and f(r,s) for every

f(i,j),(r,s) : Aij → Ars

making the following commute:

lim−→ilim−→j

Aij

Aij

f(i,j)99

f(i,j),(r,s)

// Ars

f(r,s)ee


lim−→jlim−→i

Aij

Aij

f(i,j)99

f(i,j),(r,s)

// Ars

f(r,s)ee

And any other object satisfying the above must factor through the

direct limits. Whence they factor through each other, and we have a

natural isomorphism

lim−→i

lim−→j

Aij ∼= lim−→j

lim−→i

Aij

Similarly, if we merely reverse the directions of the arrows in the above

diagram, lim←−i lim←−j Aij and lim←−j lim←−iAij also satisfy the same universal

property, and are hence naturally isomorphic.

21. Problem 21

First, we need some notation. Elements of our direct limit can be

written as classes [Mi,mi] with mi ∈Mi and group operation

[Mi,mi] + [Mj,mj] := [Mk, φik(mi) + φjk(xj)]

with k > i, j. By definition of direct limits, this is well defined. We

also have induced maps u, v such that

u[M ′i ,m

′i] = [Mi, ui(m

′i)]

v[Mi,mi] = [M ′′i , vi(mi)]

where

0 // M ′i

ui// Mi

vi// M ′′

i// 0

is exact for every i. Now we may prove exactness. Suppose first that

u([Mi,mi]) = 0. Then,

[Mi, ui(m′i)] = 0 =⇒ fij(ui(m

′i)) = 0


for j > i. But,

fij(ui(m′i)) = uj(fij(m

′i)) = 0

and since each uj is a monomorphism, fij(m′i) = 0 for every > i, and

we see that

[M ′i ,m

′i] = [M ′

i , 0]

so that u is a monomorphism. We also see:

vu([M ′i ,m

′i]) = v([Mi, ui(m

′i)])

= [M ′′i , viui(m

′i)]

= [M ′′i , 0]

where the last equality follows from the fact that Imui ⊂ Ker vi for

every i, and we thus deduce that Imu ⊂ Ker v. Let us consider the

reverse inclusion now; suppose

v([Mi,mi]) = [M ′′i , vi(mi)]

= [M ′′i , 0]

Then, for all j > i, gij(vi(mi)) = vj(gij(ui)) = 0, so that given gij(mi) ∈

Ker vj, there exists m′j ∈M ′j such that uj(m

′j) = gij(mi), in which case

[Mi,mi] = [Mj, uj(m′j)]

= u([M ′i ,m

′i]) ∈ Imu

So that Ker v = Imu. Finally, let [M ′′i ,m

′′i ] ∈ lim−→Mi. Then for each i,

ui(mi) = m′′i for some mi ∈Mi, so that

[M ′′i ,m

′′i ] = M ′′

i , ui(mi)]

= u([Mi,mi]) ∈ Imu

And we conclude that

0 // lim−→M ′i

v// lim−→Mi

u// lim−→M ′′

i// 0

is also exact.


22. Problem 22

(a). Consider the universal property of the direct sum. If we apply

the contravariant functor Hom(−, N), we reverse the direction of the

inclusion maps in our universa property. We then get an induced map

u : Hom(⊕i

Mi, N)→∏i

Hom(Mi, N)

We also get an inverse map∏i

Hom(Mi, N)→ Hom(⊕i

Mi, N)

(fi) 7→ f

Where f(mi) =∑

i fi(mi), (mi) ∈⊕

iMi. Whence,

Hom(⊕i

Mi, N) ∼=∏i

Hom(Mi, N)

(b). We have a similar universal property. When we apply the covari-

ant functor Hom(N,−), we preserve the direction of our arrows and

get an induced map

u : Hom(N,∏i

Mi)→∏i

Hom(N,Mi)

And we have an inverse map∏i

Hom(N,Mi)→ Hom(N,∏i

Mi)

(fi) 7→ f

where f is such that

f(n) = (fi(n)) ∈∏i

Mi

and we conclude ∏i

Hom(N,Mi) ∼= Hom(N,∏i

Mi)


23. Problem 23

We have the diagram

lim←−iMi

{{ ##

Mj// Mi

for i 6 j. Applying Hom(N,−), we have the induced diagram

Hom(N, lim←−iMi)

vv ((

Hom(N,Mj) // Hom(N,Mi)

So by the universal property, there exists a map

u : Hom(N, lim←−i

Mi)→ lim←−i

Hom(N,Mi)

And we construct an inverse

lim←−i

Hom(N,Mi)→ Hom(N, lim←−i

Mi)

[Hom(N,Mi), fi] 7→ f

where f is such that

f(n) = [Mi, fi(n)]

whence

Hom(N, lim←−i

Mi) ∼= lim←−i

Hom(N,Mi)

24. Problem 24

Let M be an R-module. Consider the set of finitely generated sub-

modules of M , ordered by inclusion. We have the direct system

{Mi, iij}


where iij is the natural inclusion Mi ↪→ Mj. We want to show that

M = lim−→Mi. The map is naturally defined as

m 7→ [Mi,mi]

Now, if m 7→ [Mi, 0], then m is 0 is some finitely generated submodule

of M , hence m = 0.

Now, given [Mi,mi] ∈ lim−→Mi, note that M =⋃iMi, and we may

take the preimage as mi ∈ M for any i. This is well defined, since if

i 6 j, iij(mi) = mj, but iij(mi) = mi, merely viewed as an element of

Mj. Hence,

M = lim−→Mi

25. Problem 25

We have an exact sequence

0 // K // F // M // 0

Consider the poset

S := {(N, I) | |I| <∞, N ⊂ K ∩RI , N f.g}

Under the partial order

(N, I) 6 (N ′, I ′)

⇐⇒ N 6 N ′ and I ⊂ I ′

Now consider lim−→RI/N . We want to show that this is isomorphic to

F/K, as F/K ∼= M . The map is trivial, we merey send

f +K 7→ [RI/N, f +N ]

and by identical steps as in the previous problem, this is an isomor-

phism. For each RI/N , we have the exact sequence

N // RI // RI/N // 0


As N ⊂ K ∩RI , this is a finite presentation.

26. Problem 26

We first show this is a monomorphism. Consider

lim−→Hom(E,Mi)→ Hom(E, lim−→Mi)

[Hom(E,Mi), fi] 7→ f

where f(n) = [Mi, fi(n)]. Suppose [Hom(E,Mi), fi] 6= 0, so fi 6= 0 for

some i. Then there exists n ∈ E such that fi(n) 6= 0, and injectivity

follows by taking the contrapositive.

Suppose now that E is finitely generated and free, so that E = Rn

for some n ∈ N. We then see

lim−→Hom(E,Mi) = lim−→Hom(Rn,Mi)

= lim−→(

Hom(E,Mi))n

= lim−→Mni

=(

lim−→Mi

)nand

Hom(E, lim−→Mi) = Hom(Rn, lim−→Mi)

=(

Hom(R, lim−→Mi))n

=(

lim−→Mi

)nSo that these are indeed isomorphic in the free and finitely generated

case. Suppose E is finitely presented and choose a presentation

F0// F1

// E // 0


Apply the left exact contravariant functor Hom(−,Mi) and then the

exact functor (by Problem 21) lim−→ to get the commutative diagram

0 // lim−→Hom(E,Mi)

��

// lim−→Hom(F0,Mi)

��

// lim−→Hom(F1,Mi)

��

0 // Hom(E, lim−→Mi) // Hom(F0, lim−→Mi) // Hom(F1, lim−→Mi)

Where the vertical arrows are the natural maps. Then, using exactness

we easily deduce that first vertical arrow must be a surjection, since

we have already shown that the second 2 are. Whence we have an

isomorphism

lim−→Hom(E,Mi) ∼= Hom(E, lim−→Mi)

27. Problem 27

Define the product

(x+ An−1)(y + Am−1) := xy + An+m−1

This is well defined and preserves the graded structure as xy ∈ An+m,

so

AnAn−1

· AmAm−1

⊂ An+mAn+m−1

And this is the multiplication rule for the associated graded module

gr(A).

28. Problem 28

(a). We have the natural definition

gri(L) :gri → gri(B)

a+ Ai−1 7→ L(a) +Bi−1


Let us show this is well defined. Suppose that a + Ai−1 = a′ + Ai−1.

Then a− a′ ∈ Ai−1, so that

L(a− a′) ∈ Bi−1 =⇒ L(a) +Bi−1 = L(a′) +Bi−1

So this is well defined.

(b). Let b ∈ Bi, and without loss of generality assume that b /∈ Bi−1.

Since gri(L) is an isomorphism, there exists a0 ∈ Ai such that b −

L(a0) ∈ Bi−1. Similarly, we may find a1 ∈ Ai−1 such that

(b− L(a0))− L(a1) ∈ Bi−2

Iterating this, after i+ 1 times we have found ak ∈ Ai−k such that

b−i∑

k=0

L(ak) ∈ B−1 = {0}

Whence b−∑i

k=0 L(ak) = 0, implying that

b = L( i∑k=0

ak

)so that L is surjective.

Suppose now that L(a) = 0 for a ∈ A. Then, a ∈ Ai for some i, and

since gri(L) is an isomorphism, a ∈ Ai−1. Iterating this, we see that

a ∈ Aj for all j 6 i, and in particular, a ∈ A−1 = {0}, so that a = 0,

and L is an isomorphism.

29. Problem 29

(a). These are algebras just by definition, and indeed we see that

det(N − λI) = λn for N ∈ ni, whence Nn = 0.


(b). Closure follows from

(I +X)(I + Y ) = I +X + Y +XY

Since n is an algebra, this remains in our set. Associativity follows

from associativity of matrix multiplication. Lastly, I = I + 0 is the

identity element.

Finally, suppose that X is nilpotent of degree i; we have

(I +X)(I −X +X2 − · · ·+ (−1)i−1X i−1) = I −X i = I

So all elements are invertible, and we have a group.

(c). Note that exp is a polynomial function, where

exp(X) =∞∑n=0

Xn

n!

The sum is not actually infinite since X is nilpotent, so we have a

polynmial function. To show this is a bijection, we only need show

that log is the inverse. We see:

log(exp(X)) =∞∑n=1

(−1)n+1

n

(exp(X)− I

)n=∞∑n=1

(−1)n+1

n

n∑k=0

(−1)n−k(n

k

)· exp(kX)

=∞∑n=1

(−1)n+1

n

n∑k=0

(−1)n−k(n

k

)·∞∑m=0

Xm

m!· km

= −∞∑m=0

Xm

m!

( m∑n=1

1

n

n∑k=0

(−1)k(n

k

)km

+∞∑

n=m+1

1

n

n∑k=0

(−1)k(n

k

)km)

Now, considern∑k=0

(n

k

)(−1)kkmxk


This is the resulting of applying the operator(x ddx

)mso (1−x)n. When

n < m, we see that the end result will still have a factor of 1 − x, so

that, setting x = 1, whenever m < n we haven∑k=0

(n

k

)(−1)kkm = 0

And the above sum becomes

−∞∑m=0

Xm

m!

m∑n=1

1

n

n∑k=0

(−1)k(n

k

)km

=−∞∑m=0

Xm

m!

m∑n=1

n∑k=0

(−1)k1

k

(n− 1

k − 1

)km

=−∞∑m=0

Xm

m!

m∑k=1

(−1)k · km−1m∑n=k

(n− 1

k − 1

)Now consider

m∑n=k

(n− 1

k − 1

)This is the coefficient of xk−1 in the sum

(1 + x)k−1 + (1 + x)k + · · ·+ (1 + x)m−1

which, by the geometric sum formula, is just equal to

(1 + x)m−k+1

x− (1 + x)k−1

x

This has no degree k− 1 terms, but for the form before that, xk−1 has

coefficient (m− kk

)=

(m

k

)So we find

m∑k=n

(n− 1

k − 1

)=

(m

k

)and our sum becomes

−∞∑m=0

Xm

m!

m∑k=1

(−1)k · km−1(m

k

)


Now, when m > 1, we have already shown above thatm∑k=1

(−1)k · km−1(m

k

)=

m∑k=0

(−1)k · km−1(m

k

)= 0

and, when m = 1,1∑

k=1

(−1)k(

1

k

)= −1

So that all terms of order m > 1 vanish, and we are merely left with

X. Hence, log(exp(X)) = X. Also by rearranging the terms in our

series, we also see that

log(exp(X)) = exp(log(X)) = X

so this log is a left and right inverse, giving that exp is indeed a bijec-

tion, as desired.

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SERGE LANG’S ALGEBRA CHAPTER 3 EXERCISE SOLUTIONS …people.math.sc.edu/kellerlv/Lang_Ch3 (1).pdf · SERGE LANG’S ALGEBRA CHAPTER 3 EXERCISE SOLUTIONS 3 Since fis injective, M0˘=Imf