Lecture Notes on Rings and Modules

8/18/2019 Lecture Notes on Rings and Modules

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Lecture Notes on Rings and Modules

Thanasin V. NampaisarnJacobs University [email protected]

Spring Semester, 2015

mailto:[email protected]

mailto:[email protected]



Algebra (Spring 2015) Thane V. Nampaisarn

0 Factorization in Polynomial Rings

Lemma 0.1. Let D be a unique factorization domain with field of fractions F and x an indetermi-nate. Two primitive polynomials f (x) and g(x) in D[x] are associates in D[x] if and only if they areassociates in F [x].

Proof of Lemma: The direct implication is trivial, so we shall verify the converse. Supposethat f (x) and g(x) are associates in the integral domain F [x]. We have that f = g u for someunit u(x) of F [x]. However, we then have that u(x) is an element of F , whence u(x) = u0 for

some u0 ∈ F . Now, u0 = b

c for some b, c ∈ D with c 0D. That is, c f = b g. Since the contents

C (f ) and C (g) are units in D,

c ≈ c C (f ) ≈ C (c f ) = C (b g) ≈ b C (g) ≈ b , (1)

where, for p, q ∈ D, p ≈ q means that p and q are associates in D. Hence, b = cv for some unitv ∈

D, and so c f = b g = cv g, or f = v g. The result follows immediately. #

Theorem 0.2 (Gauss’s Lemma). Let D be a unique factorization domain with field of fractions F and f (x) a primitive polynomial of positive degree in D[x], where x is an indeterminate. Then, f isirreducible over D if and only if f is irreducible over F .

Proof: The converse is trivial. We shall prove the other direction by contrapositivity. Suppose thata nonconstant primitive polynomial f (x) ∈ D[x] is reducible over F . Then, there are nonconstant

polynomials g(x) and h(x) in F [x] such that f = gh. Write g(x) =m

i=0

ai

bi

xi and h(x) =n

j=0

c j

d j

x j

for some ai, bi, c j, d j ∈ D with bi 0D and d j 0D for every i = 0, 1, 2, . . . , m and j = 0, 1, 2, . . . , n.Consider the polynomial g1(x) := b g(x)

∈ D[x], where b := b0b1b2

· · ·bn. Then, g1(x) = a g2(x),

where a is the content of g1, and g2(x) ∈ D[x] is primitive. Therefore, g = ab

g2 with a, b ∈ D,

b 0D, and g2(x) ∈ D[x] is primitive such that deg(g) = deg (g2). Similarly, we can write h = c

d h2

with c, d ∈ D, d 0D, and h2(x) ∈ D[x] is primitive such that deg(h) = deg(h2). Since f isprimitive and bd f = bdgh = (b g) (d h) = (a g2) (c h2) = ac g2h2, we have

bd ≈ bd C (f ) ≈ C (bd f ) = C (ac g2h2) ≈ ac C (g2h2) ≈ ac C (g2) C (h2) ≈ ac . (2)

Consequently, ac and bd are associates in D, whence f and g2h2 are associates in D. This meansf is reducible over D. Q.E.D.

Example 0.3. Let x be an indeterminate. The polynomial f (x) := x3

− 3x − 1 is irreducible inQ[x], since it is irreducible in Z[x]. To show that f is irreducible over Z, we note that, if it werereducible, it would have a linear factor, and since f is monic, it would then have an integer root.Clearly, if f had an integer root, then the root could only be +1 or −1. Since both numbers are notroots of f , the claim follows.

Exercise 0.4. Let a1, a2, . . ., an, and A be integers with A > 0. Prove that the polynomial

P (x) := An

i=1

(x − ai) − 1 is irreducible over Q, where x is an indeterminate.

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Theorem 0.5. If D is a unique factorization domain, then so is the poylnomial ring D [x1, x2, . . . , xn],where n ∈ N and x1, x2, . . . , xn are indeterminates.

Proof: Clearly, we only need to show that D[x] is also a unique factorization domain, where

x is an indeterminate. The rest follows by induction on the number of indeterminates (usingD [x1, x2, . . . , xn] =

D [x1, x2, . . . , xn−1]

[xn]).

Let f (x) ∈ D[x] be of a positive degree. Then, f = C (f ) f 1 for some primitive polynomialf 1(x) ∈ D[x] of the same positive degree as that of f (x). Let F be the field of fractions of D,so that F [x] is a unique factorization domain. Therefore, f 1 = p1 p2 · · · pd for some d ∈ N andirreducible polynomials p1(x), p2(x), . . . , pd(x) in F [x]. We can show that, for i = 1, 2, . . . , d, there

exists a primitive polynomial q i(x) ∈ D[x] and ai, bi ∈ D such that bi 0D and pi = ai

bi

q i. That

is, f 1 =

a1a2 · · · ad

b1b2 · · · bd

q 1q 2 · · · q d. Since f 1, q 1, q 2, . . ., q d are primitive over D, we conclude that f

and q 1q 2· · ·

q d are associates over D. Since D is a unique factorization domain, C (f ) is associateto a product of irreducible elements of D. Thus, f = C (f ) f 1 is a product of irreducible elementsin D[x].

The uniqueness of factorization in D[x] follows immediately from the facts that D is a uniquefactorization domain, that a factorization in D[x] is also a factorization in F [x], that irreduciblepolynomials in D[x] are irreducible in F [x] as well, and that F [x] is also a unique factorizationdomain. The proof is now complete. Q.E.D.

Corollary 0.6. Let D be a unique factorization domain and X a set of indeterminates. Then,the polynomial ring D[X ] is a unique factorization domain.

Theorem 0.7 (Extended Eisenstein’s Criterion). Let D be a unique factorization domain. Letf (x) =

ni=0

aixi, where n ∈ N, a0, a1, a2, . . . , an ∈ D, and x is an indeterminate. If, for some k ∈ N

with k < n and for some irreducible p ∈ D, we have p | a0, p | a1, . . ., p | ak−1, p ak, and p2 a0,then f (x) has an irreducible factor in D[x] of a degree at least k .

Proof: Write f = C (f ) f 1, where f 1(x) ∈ D[x] is primitive and C (f ) ∈ D is not divisible by p. Itsuffices to show that f 1(x) has an irreducible factor of degree at least k in D[x]. Since p dividesthe constant term of f 1(x), there exists an irreducible factor g(x) of f 1(x) whose constant term is

divisible by p. Write g(x) =l

j=0

b j x j for some l ∈ N and b0, b1, . . . , bl ∈ D such that bl 0. Suppose

that f 1(x) = g(x) h(x) for some h(x) =r

j=0

c j x j, where r ∈ N0 and c0, c1, . . . , cr ∈ D such that

cr 0D.

Because p2 a0, we conclude that p c0. The first-degree term of f 1 has coefficient b1c0 + b0c1,which is divisible by p. Since p c0 and p | b0, we conclude that p | b1. By induction, we can showthat p divides b0, b1, b2, . . . , bk−1. Since p ak, we can also verify that p bk. Therefore, l ≥ k andthe result follows immediately. Q.E.D.

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Corollary 0.8 (Eisenstein’s Criterion). Let D be a unique factorization domain with field of

fractions F . Let f (x) =n

i=0

aixi, where n ∈ N, a0, a1, a2, . . . , an ∈ D, and x is an indeterminate. If,

for some irreducible p ∈ D, p | a0, p | a1, . . ., p | an−1, p an, and p2 a0, then f is irreducible over

F . If f is primitve, it is irreducible over D.

Example 0.9. Let p ∈ N be a prime number and x an indeterminate. The p-th cyclotomic

polynomial Φ p(x) def == x p−1 + x p−2 + · · · + x + 1 is irreducible, since Φ p(x) =

x p − 1

x − 1 , whence

Φ p(x + 1) = (x + 1) p − 1

x = x p−1 +

p−1r=1

p

r

xr−1 , (3)

where Eisenstein’s Criterion applies.

Exercise 0.10. Let x be an indeterminate. Find all positive integers n such that the polynomial

P n(x) := xn − 2 xn−1 − 15 is reducible over Z.

Definition 0.11 (Derivatives of Polymomials). Let R be a ring and X a set of indeterminates. For

y ∈ X , define ∂

∂y

c

x∈X

xkx

to be kyc yky−1

x∈X \{y}

xkx for every (kx)x∈X ∈ NX 0 such that, for only

finitely many x ∈ X , kx 0, and for any c ∈ R. The operator ∂

∂y is defined on R [X ] via term-by-

term application. Write x for (x)x∈X . For f (x) ∈ R [x] = R[X ] and y ∈ X , ∂

∂yf (x) is called the

(first) formal derivative of f (x) with respect to y.

If x is an indeterminate, then

d

dx is the same as

∂

∂x on R[x]. We usually write f

(x) for

d

dxf (x),

where f (x) ∈ R[x]. For f (x) ∈ R[x], define f (0)(x) def == f (x) and, for r ∈ N, f (r)(x)

def ==

f (r−1)

(x).

For every f (x) ∈ R[x] and r ∈ N0, the polynomial f (r)(x) is called the r-th formal derivative of f .

Proposition 0.12. Let D be an integral domain, X a set of indeterminates, and x := (x)x∈X . Forf (x) , g (x) ∈ D [x] and y ∈ X , we have

(a) ∂

∂y

c f (x)

= c

∂

∂yf (x) for every c ∈ R,

(b)

∂

∂y

f (x) + g (x)

= ∂

∂y f (x)

+ ∂

∂y g (x)

,

(c) ∂

∂y

f (x) g (x)

= f (x)

∂

∂yg(x)

+

∂

∂yf (x)

g(x), and

(d) ∂

∂y

f (x)

k= k

f (x)k−1

∂

∂yf (x)

for every k ∈ N0.

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Definition 0.13 (Multiple Roots). Let D be an integral domain and x an indeterminate. For c ∈ Dand f (x) ∈ D[x], we say that c is a root of f with multiplicity m ∈ N if f (x) = (x − c)m g(x) forsome g(x) ∈ D[x] not divisible by x − c. We say that c is a simple root of f if it has multiplicity 1.If c is a root of multiplicity m > 1, then we say that c is a multiple root .

Theorem 0.14. Let D be an integral domain which is a subring of an integral domain E , and x aninderminate. A polynomial f (x) ∈ D[x] and c ∈ E are given.

(1) The element c is a multiple root of f if and only if f (c) = 0E and f (c) = 0E .

(2) If D is a field and f is relatively prime to f , then f has no multiple roots in E .

(3) If D is a field, f is irreducible over D, and E contains a root of f , then f has no multiple rootsin E if and only if f (x) 0D.

(4) If D is of characteristic 0, then c is a root of f of multiplicity m ∈ N in E if and only if f (r)(c) = 0E

for all r = 0, 1, 2, . . . , m − 1 and f

(m)

(c)

0E .Example 0.15. Let F be a field of characteristic p 2. The polynomial P (x,y,z ) := x2+y2+z 2 is

irreducible in F [x,y,z ], where x, y,z are indeterminates. To prove this, we start with the observationthat F [x,y,z ] = D[x], where D := F [y, z ] is a unique factorization domain. We want to use

Eisenstein’s Criterion with P (x,y,z ) = x2 +

y2 + z 2

, and therefore, we need to show that there is

no irreducible element q (y, z ) ∈ F [y, z ] = D such that

q (y, z )2

divides Q(y, z ) := y2 + z 2. Suppose

contrary that such q (y, z ) ∈ D exists.Since D = F [y, z ] = E [y], where E := F [z ] is an integral domain, and Q(y, z ) is a monic

polynomial of degree 2 in the variable y , we conclude that q (y, z ) ∈ F [y, z ] = E [y] is linear in y . LetK be the field of fraction of E (i.e., K = F (z )). This means the equationq (y, z ) = 0F in the variable

y has a solution in K and, therefore, Q(y, z ) as an element of K [y] has a multiple root c(z ) ∈ K (i.e.,

Q

c(z ), z

= 0F ). By the previous theorem, we conclude that Q(y, z ) and ∂

∂yQ(y, z ) as elements of

K [y] must have a common factor. However,

Q(y, z ) −

y

2

∂

∂yQ(y, z )

=

y2 + z 2

−

y

2

(2y) = z 2 (4)

is a unit of K , where we have used to hypothesis that the characteristic of F is p 2. This means

Q(y, z ) and ∂

∂yQ(y, z ) are relatively prime in K [y], and we have a contradiction.

Note that, if char(F ) = 2, P (x,y,z ) is reducible. A factorization is P (x,y,z ) = (x + y + z )2.

Exercise 0.16. Let D and E be integral domains with D ⊆ E , and x an indeterminate. Supposethat f (x) ∈ D[x] is irreducible over D. If D is of characteristic 0, then show that f has no multipleroots in E . If p is a positive prime integer, give an example where f has a multiple root in E withD having characteristic p.

4




1 Definitions and Basic Properties of Ring Modules

Definition 1.1 (Left Modules). Let (R, +, ·) be a ring. A left R-module is an abelian group (M, +)together with a scalar left-multiplication ∗ : R × M → M such that

(i) r ∗ (m + n) = (r ∗ m) + (r ∗ n),

(ii) (r + s) ∗ m = (r ∗ m) + (s ∗ m), and

(iii) (r · s) ∗ m = r ∗ (s ∗ m),

for every r ∈ R, s ∈ S , and m, n ∈ M . If R is unital with identity 1R, then we say that M is unitary if 1R ∗ m = m for all m ∈ M . If R is a division ring, then a left unitary R-module is called a left R-vector space .

Definition 1.2 (Right Modules). Let (R, +, ·) be a ring. A right R-module is an abelian group(M, +) together with a scalar right-multiplication

∗ : M

×R

→ M such that

(i) (m + n) ∗ r = (m ∗ r) + (n ∗ r),

(ii) m ∗ (r + s) = (m ∗ r) + (m ∗ s), and

(iii) m ∗ (r · s) = (m ∗ r) ∗ s,

for every r ∈ R, s ∈ S , and m, n ∈ M . If R is unital with identity 1R, then we say that M is unitary if m ∗ 1R = m for all m ∈ M . If R is a division ring, then a right unitary R-module is called a right R-vector space .

Definition 1.3 (Bimodules). Let (R, +,

·) and (S, +,

·) be rings. An (R, S )-bimodule M is a left

R-module M equipped with a right S -module structure in such a way that (r ∗ m) ∗ s = r ∗ (m ∗ s)for each r ∈ R, m ∈ M , and s ∈ S .

Remark 1.4.

(1) Without specification, a module or a vector space is taken to be a left module or a left vectorspace, respectively.

(2) If R is commutative, then a left R-module M is also a right R-module via the identification

m ∗ r def == r ∗ m for all r ∈ R and m ∈ M , and vice versa. Therefore, every (left or right) R-

module is an (R, R)-bimodule. Without specification, an R-module is given the aforementioned

(R, R)-bimodule structure if R is commutative.

(3) We shall ignore right modules for the most part of our discussion since results for left moduleshold equivalently for right modules.

Example 1.5.

(1) Every abelian group is a unitary Z-module.

(2) If R is a subring of a ring S , then an S -module is also an R-module.

5




(3) If I is a left ideal of a ring R, then R/I is a left R-module via the definition r ∗ (s + I ) def == r ·s + I

for all r, s ∈ R. Similarly, if I is a right ideal of R, then R/I is a right R-module.

(4) If R and S are rings and φ : R → S is a ring homomorphism, then any S -module M can be made

into an R-module via the pullback along φ: for every r ∈ R and m ∈ M , r ∗ m

def

== φ(r) ∗ m.

(5) For a ring R, any abelian group A can be given a trivial module structure by defining r ∗ a def == 0A

for all r ∈ R and a ∈ A. Here, 0A is the additive identity of A.

Exercise 1.6. Let R and S be rings. The set of all ring homomorphisms from R to S is given byHom(R, S ), which is clearly an abelian group under (f + g)(x) = f (x) + g(x) for all f, g ∈ Hom(R, S )and x ∈ R. For f ∈ Hom(R, S ), we define r ∗ f and f ∗ s for r ∈ R and s ∈ S by

(a) (r ∗ f )(x) = f (x · r) and

(b) (f

∗s)(x) = f (x) ·

s,

for all x ∈ R. Verify that this gives an (R, S )-bimodule structure on Hom(R, S ).Alternatively, if we define f ∗ r and s ∗ f , for r ∈ R, s ∈ S , and f ∈ Hom(R, S ), by

(a) (f ∗ r)(x) = f (r · x) and

(b) (s ∗ f )(x) = s ·

f (x)

,

for all x ∈ R. Justify that this gives an (S, R)-bimodule structure on Hom(R, S ). In fact, we canalso give Hom(R, S ) an (R, R)-bimodule structure, as well as an (S, S )-bimodule structure, using thesame scalar multiplications as above.

Definition 1.7 (Submodules). Let R be a ring and M an R-module. An R-submodule N of M isa subset of M which is also an R-module. We say that N is proper if N M . If M is a nontrivialR-module without a nonzero proper R-submodule, then we say that M is simple . If R is a divisionring, then an R-submodule of an R-vector space M is called an R-vector subspace .

Definition 1.8 (Module Generators). If X is a subset of a module M over a ring R, then theintersection X of all R-submodules of A containing X is called the R-submodule generated by X ,or the R-submodule spanned by X . If X is finite and X generates the module B, then B is said tobe finitely generated . If X is a singleton {x}, then the module x it generates is called cyclic .

Example 1.9.

(1) The zero module 0 is generated by the empty set.

(2) Every simple module is cyclic. In fact, a module is simple if and only if any nonzero elementgenerates the whole module.

(3) The Z-module Z is cyclic. However, neither is it simple nor does it have a simple Z-submodule.

6




Exercise 1.10. Let G be a group and R a ring. The group ring R[G] is an R-module of the form

g∈G

tg g ||| tg ∈ G for all g ∈ G and there are only finitely many g ∈ G such that tg 0R

, (5)

equipped with scalar multiplication

r ∗

g∈G

tg g

def

==g∈G

(r · tg) g , (6)

for every r ∈ R andg∈G

tgg ∈ R[G], and with ring multiplication in accordance with the multiplication

rule in G. That is, if rg and sg are elements of R for g ∈ G such that there are only finitely many

g ∈ G such that rg 0R and sg 0R, then

g∈G

rg g

·g∈G

sg g

def == g∈G

h∈G

rhsh−1g

g .

Now, consider the ring S := CS 3, whereS k is the k-th symmetric group for every k ∈ N0.

Show that S is unital and all unitary S -modules are vector spaces over C. Prove that every simpleS -module is a finite-dimensional C-vector space. Describe all simple S -modules.

Theorem 1.11. Let R be a ring and M an R-module.

(a) For m ∈ M , R ∗ m def == {r ∗ m ||| r ∈ R} is an R-submodule of M .

(b) For m ∈ M , m = {r ∗ m + k m ||| r ∈ R and k ∈ Z}. If R is unital and 1R ∗ m = m, then m isunitary and equals R ∗ m.

(c) If X is a subset of M , then

X =

si=1

ri ∗ xi +t

j=1

k j y j ||| s, t ∈ N0, i ∈ [s], j ∈ [t], xi, y j ∈ X, ri ∈ R , and k j ∈ Z

, (7)

where [l] def == {1, 2, . . . , l} for all l ∈ N and [0]

def == ∅. If R is unital and 1R ∗x = x for every x ∈ X ,

then X is unitary and

X =

si=1

ri ∗ xi ||| s ∈ N0, xi ∈ X, ri ∈ R , and k j ∈ Z for all i = 1, 2, . . . , s

. (8)

Definition 1.12 (Sum of Submodules). Let M be a module over a ring R. For an index set J along with a collection {N j ||| j ∈ J } of R-submodules of M , we write

j∈J

N j for the R-submodule

j∈J

N j

, which is called the sum of the modules N j’s. If J = {1, 2, . . . , n} for some n ∈ N0, we also

write N 1 + N 2 + · · · + N n in place of

j∈J

N j.

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Theorem 1.13. Let J be an index set. The sum of the family {N j ||| j ∈ J } of R-submodules of amodule M over a ring R consists of all finite sums ni1 + ni2 + · · · + nil, where l ∈ N0, i1, i2, . . . , il ∈ J ,and nik ∈ N ik for k = 1, 2, . . . , l.

Theorem 1.14 (Direct Sum and Direct Product of Modules). Let J be an index set and{

M j||| j ∈

J }a family of modules over a ring R.

(a) The direct product

j∈J

M j of abelian groups M j’s is indeed an R-module with the action of R

given by r ∗ (a j ) j∈J

def == (r ∗ a j )

j∈J . We call

j∈J

M j the (external) direct product of the family

{M j ||| j ∈ J } of R-modules.

(b) The direct sum j∈J

M j of abelian groups M j’s is indeed an R-submodule of

j∈J

M j . We call

j∈J

M j the (external) direct sum of the family {M j ||| j ∈ J } of R-modules.

(c) If J is finite, then the direct product and the direct sum coincide. If J = {1, 2, . . . , l} for somel ∈ N0, we may write M 1 × M 2 × · · · × M l for

j∈J

M j and M 1 ⊕ M 2 ⊕ · · · ⊕ M l for j∈J

M j; of

course,

j∈J

M j = M 1 × M 2 × · · · × M l = M 1 ⊕ M 2 ⊕ · · · ⊕ M l = j∈J

M j.

Theorem 1.15 (Quotient Modules). Let R be a ring. Suppose that N is an R-submodule of anR-module M . Then the quotient abelian group M/N has an R-module structure given by

r ∗ (m + N ) def == (r ∗ m) + N (9)

for all r ∈ R and m ∈ M . The quotient M/N with this R-module structure is called the quotient module of M by N .

Exercise 1.16. Let R be a ring and M an R-module. Suppose that M has a maximal properR-submodule N . Prove that, if R ∗M is not contained in N (which automatically holds if R is unitaland M is a unitary R-module), then M/N is a simple R-module. In particular, if J is a maximal(left) ideal of R such that R · R is not contained in J (which holds automatically if R is unital), thenshow that R/J is a simple R-module.

8




2 Homomorphisms of Ring Modules

Definition 2.1. Let M and N be modules over a ring R. A homomorphism f : M → N of abeliangroups is called an R-module homomorphism if it satisfies the property that f (r ∗ m) = r ∗

f (m)

for all r ∈

R and m ∈

M . The set of R-module homomorphisms from M to N is denoted by

HomR(M, N ). The set of R-module endomorphisms of M is EndR(M ) def == HomR(M, M ). Note that,

if R is a division ring, then an R-module homomorphism is called an R-linear transformation .We say that f is monic , or that f is a monomorphism , if it is injective. We say that f is epic , or

that f is an epimorphism , if it is surjective. We say that f is iso, or that it is an isomorphism , if itis bijective. If there exists an isomorphism between R-modules M and N , then we say that M andN are isomorphic and write M N . An R-module endomorphism which is an isomorphism is calledan R-module automorphism . The set of R-module automorphisms on M is denoted by AutR(M ).

For a homomorphism f : M → N of R-modules M and N , the kernel of f is defined to be the

set ker(f ) def ==

m ∈ M ||| f (m) = 0M

. The image of f is the set im(f )

def ==

f (m) ||| m ∈ M

. The

cokernel of f is coker(f )

def

== N/ im(f ) and the coimage of f is coim(f )

def

== M/ ker(f ).If S is a ring, then an (R, S )-bimodule homomorphism is a left R-module homomorphism whichis also a right S -module homomorphism. Other definitions such as Hom(R,S ), End(R,S ), Aut(R,S ), and(R, S )-bimodule monomorphisms remain similar.

Remark 2.2.

(1) A module homomorphism is a monomorphism if and only if its kernel is trivial.

(2) For a ring R, the class Mod(R) of all R-modules together with R-module homomorphisms formsa concrete category. If R is unital, then the class UMod(R) of unitary R-modules along withR-module homomorphisms is a full subcategory of Mod(R). The zero module 0 over R is both

initial (cofinal) and final (coinitial) in Mod(R) (as well as in UMod(R) if R is unital). In eitherof these categories, the categorical definitions of monomorphisms, epimorphisms, isomorphisms,kernel, cokernel, image, and coimage coincide with our definition above.

(3) Let R be a ring. Define the abelian group S = Z⊕R to be a ring with the following multiplicationrule:

(k1, r1) · (k2, r2) def == (k1k2, k1 r2 + k2 r1 + r1 · r2) (10)

for every k1, k2 ∈ Z and r1, r2 ∈ R. The multiplicative identity of S is clearly 1S = (1, 0R),whence S is unital. We call S the unital closure of R. Observe that Mod(R) is the same as

UMod(S ). Hence, it is sufficient to study only unitary modules over an arbitrary unital ring.Example 2.3.

(1) The zero map 0 between two modules over the same ring is a module homomorphism.

(2) Every homomorphism of abelian groups is a Z-module homomorphism.

(3) Let R be a ring and M an R-module. For m ∈ M , the map R → R ∗ m sending r → r ∗ m forevery r ∈ R is an R-module epimorphism.

9




(4) Let M and N be a simple module over a ring R. Any R-module homomorphism from M to N is either the zero homomorphism or an R-module isomorphism.

(5) If R is a ring and x is an indeterminate, the map µ : R[x] → R[x] sending f (x) ∈ R[x] to x f (x)is an R-module homomorphism, but not a ring homomorphism.

(6) Let M and N be modules over a ring R such that M ⊇ N . The canonical epimorphism (orthe canonical projection ) π : M → M/N sending m → m + N for all m ∈ M is an R-moduleepimorphism with kernel N .

(7) If f : M → N is a module homomorphism of modules M and N over a ring R, then ker(f )is an R-submodule of M and im(f ) is an R-submodule of N . More generally, if M and N

are R-submodules of M and N , respectively, then the image f

M

=

f (m) ||| m ∈ M

is an

R-submodule of N and the preimage f −1

N

=

m ∈ M ||| f (m) ∈ N

is an R-submodule of N .

Exercise 2.4. Consider the ring R = C S

3. In M :=

C3

, write e1 := (1, 0, 0), e2 := (0, 1, 0),and e3 := (0, 0, 1) for the standard basis vectors. There is a natural R-module structure on M in

which σ ∗ ei = eσ(i) for every σ ∈S 3 and i = 1, 2, 3. Describe all R-module endomorphisms of M .

Theorem 2.5. Let M and N be modules over a ring R. Suppose that L is an R-submodule of ker(f ). Then, there is a unique R-module homomorphism f : M/L → N such that f (m + L) = f (m)

for all m ∈ M . Furthermore, im

f

= im(f ) and ker

f

= ker(f )/L. The map f is an R-module

isomorphism if and only if f is an R-module epimorphism and L = ker(f ).

Corollary 2.6. Let M and N be modules over a ring R. Suppose that M and N are R-submodulesof M and N , respectively. For an R-module homomorphism f : M → N such that f

M ⊆ N ,

f induces an R-module homomorphism f : M/ M → N/ N given by m + M → f (m) + N for allm ∈ M . The map f is an R-module isomorphism if and only if im(f ) + N = N and f −1

N ⊆ M .

In particular, f is an R-module isomorphism, if f is an R-module epimorphism such that f

M

= N

and ker(f ) ⊆ M .

Theorem 2.7 (Isomorphism Theorems). Let R be a ring. Suppose that M , N , and L are R-modules.

(a) Let f : M → N be an R-module homomorphism. Then, we have im(f ) M/ ker(f ).

(b) If N and L are R-submodules of M , then N/(N ∩ L) (N + L)/L.

(c) If M ⊇ N ⊇ L, then N/L is an R-submodule of M/L, and (M/L)/(N/L) M/N .

Theorem 2.8. Let M and N be modules over a ring R such that M ⊇ N . There is a one-to-one correspondence between the set of all R-submodules of M containing N and the set of allR-submodules over M/N , given by the map L → L/N , where L is an R-submodule of M containingN .

10




Exercise 2.9. Let R, S , and T be rings. Suppose that M is an (S, R)-bimodule and N is an (S, T )-

bimodule. Then, justify that HomS (M, N ) is an (R, T )-bimodule via (f + g)(x) def == f (x) + g(x),

(r ∗ f )(x) def == f (x ∗ r), and (f ∗ t)(x)

def == f (x) ∗ t for every r ∈ R, t ∈ T , x ∈ M , and maps

f, g

∈ HomS (M, N ). In addition, show that the set End(S,R)(M ) = Hom(S,R)(M, M ) of (S, R)-

bimodule endomorphisms is an (R, R)-submodule of EndS (M ) and also a unital associative (R, R)-algebra under the composition; i.e., show that the composition has an identity, is associative, and is(R, R)-bilinear (that is, f ◦(g+h) = f ◦g+f ◦h, (f +g)◦h = f ◦h+g◦h, f ◦(r∗g) = r∗(f ◦g) = (r∗f )◦g,and f ◦ (g ∗ r) = (f ◦ g) ∗ r = (f ∗ r) ◦ g, for all f ,g, h ∈ End(S,R)(M ) and r ∈ R).

Theorem 2.10 (Categorical Product of Modules). Let J be an index set and {M j ||| j ∈ J } a familyof modules over a ring R. For k ∈ J , the canonical projection prk :

j∈J

M j → M k in the category of

abelian groups is indeed an R-module epimorphism. Furthermore,

j∈J

M j together with

pr j

j∈J

is a

categorical product of {M j ||| j ∈ J } in Mod(R), which is unique up to isomorphisms. In other words,

for any R-module N along with a family {ϕ j : N → M j ||| j ∈ J } of R-module homomorphisms, thereis a unique R-module homomorphism ϕ : N → j∈J

M j such that pr j ◦ ϕ = ϕ j for all j ∈ J .

Theorem 2.11 (Categorical Coproduct of Modules). Let J be an index set and {M j ||| j ∈ J } a familyof modules over a ring R. For k ∈ J , the canonical injection ink : M k →

j∈J

M j in the category of

abelian groups is indeed an R-module monomorphism. Furthermore, j∈J

M j together with (in j ) j∈J

is a categorical coproduct of {M j ||| j ∈ J } in Mod(R), which is unique up to isomorphisms. In otherwords, for any R-module N along with a family {ψ j : M j → N ||| j ∈ J } of R-module homomorphisms,there is a unique R-module homomorphism ψ :

j∈J

M j

→ N such that ψ

◦in j = ψ j for all j

∈ J .

Theorem 2.12. Let R be a ring and n ∈ N0. For R-modules A, A1, A2, . . ., An, we have thatA A1 ⊕ A2 ⊕ ·· · ⊕ An if and only if, for i = 1, 2, . . . , n, there are R-module homomorphismsπi : A → Ai and ιi : Ai → A such that

(i) πi ◦ ιi = idAi for i = 1, 2, . . . , n,

(ii) πi ◦ ι j = 0 for i, j = 1, 2, . . . , n such that i j , and

(iii) ι1 ◦ π1 + ι2 ◦ π2 + · · · + ιn ◦ πn = idA.

Theorem 2.13 (Internal Direct Sums). Let R be a ring and M an R-module. Suppose that J is anindex set and {M j ||| j ∈ J } is a family of R-submodules of M such that

(i) M =

j∈J

M j, and

(ii) for each k ∈ J , M k ∩

j∈J \{k}

M j = 0,

11




where 0 is the zero R-module. Then, there is an isomorphism M j∈J

M j and we say that M is the

(internal) direct sum of a family of R-submodules {M j ||| j ∈ J }. Furthermore, each M j is said to bea direct summand of M .

Definition 2.14 (Semisimple Modules). A module M over a ring R is said to be semisimple if andonly if

(i) for any nonzero m ∈ M , R ∗ m is nontrivial, and

(ii) every R-submodule of M is a direct summand of M .

It is known that an R-module M is semisimple if and only if it is the sum of a family of the simpleR-submodules, or equivalently, it is the (internal) direct sum of a family of its simple R-submodules.Another equivalent notion for M to be semisimple is that M is isomorphic to a direct sum of simpleR-modules.

Definition 2.15 (Indecomposable Modules). A module M over a ring R is said to be indecomposable if M cannot be written as an (internal) direct sum N ⊕L of nonzero R-submodules N and L; otherwise,we say that M is decomposable .

Example 2.16.

(1) Every simple module is indecomposable and semisimple.

(2) If R is a unital ring and M is a semisimple R-module, then M is unitary.

(3) The Z-module Z is indecomposable, but not semisimple.

Exercise 2.17. Let F be a field (in fact, F can be any division ring) and V a finite-dimensionalvector space over F . Take R to be the ring EndF (V ) of F -linear transformations on V (obviously, Ris unital). An R-module M is provided.

(a) Prove that there are R-submodules N and L of M satisfying the following conditions:

(i) N is a trivial R-module,

(ii) L is a unitary R-module, and

(iii) M = N ⊕ L.

(The statement above holds for any unital ring R and any R-module M .)

(b) Show that L is a vector space over F .

(c) If n := dimF (V ), then verify that a simple R-module is a vector space over F of dimension nand it is unique up to isomorphisms.

(d) Justify that L is a semisimple R-module.

(e) If l := dimF (L) is finite, then show that n divides l.

12




3 Exact Sequences of Ring Modules

Definition 3.1. Let A, B, and C be modules over a ring R. We say that a pair of R-module

homomorphisms A f −→ B

g−→ C is exact if im(f ) = ker(g). Let J be an interval of Z. For

every j ∈ J , an R-module A j is provided. For an arbitrary sequence of module homomorphisms. . .

f j−2−→ A j−1f j−1−→ A j

f j−→ A j+1f j+1−→ . . . is said to be exact if any pair of adjacent homomorphisms is

exact. A short exact sequence of R-modules is an exact sequence of the form 0 → A → B → C → 0,where A, B, and C are R-modules

Example 3.2. Let R be a ring.

(1) For R-modules A, B , and C , if A f −→ B

g−→ C is exact, then g ◦ f = 0.

(2) For R-modules A and B, 0 −→ A f −→ B is exact if and only if f is an R-module monomorphism.

On the other hand, A f

−→ B

−→ 0 is exact if and only if f is an R-module epimorphism.

(3) For R-modules A and B, 0 −→ A ι−→ A ⊕ B

π−→ B −→ 0 is a short exact sequence, where ιand π are the canonical injection and the canonical projection, respectively.

(4) For R-modules A and B such that A ⊆ B, 0 −→ A ι−→ B

π−→ B/A −→ 0 is a short exactsequence, where ι is the inclusion map and π is the quotient map.

(5) Let f : A → B be a homomorphism of R-modules A and B. Then, we have three exact

sequences 0 −→ ker(f ) ⊆−→ A −− coim(f ) −→ 0, 0 −→ im(f )

⊆−→ B −− coker(f ) −→ 0, and

0 −→ ker(f ) ⊆−→ A

f −→ B −− coker(f ) −→ 0. The double-headed arrows denote the quotientmaps.

Exercise 3.3. Let n ∈ N0 and F a field. Suppose that 0 → V 0 → V 1 → V 2 → . . . → V n → 0 is an

exact sequence of finite-dimensional F -vector spaces. Prove thatn

i=0

(−1)i dimF (V i) = 0.

Definition 3.4 (Commutative Diagrams). Let R be a ring. A commutative diagrams of R-modulesis a collection of maps {φi,j : Ai → A j ||| (i, j) ∈ S }, where Ai is an R-module for every i in an indexset J and S ⊆ J × J , such that, whenever one can form two sequences

Ai0

φi0,i1−−−→ Ai1

φi1,i2−−−→ Ai2

φi2,i3−−−→ . . .φim−2,im−1−−−−−−−→ Aim−1

φim−1,im−−−−−→ Aim (11)

and

A j0

φj0,j1−−−→ A j1

φj1,j2−−−→ A j2

φj2,j3−−−→ . . .φjn−2,jn−1−−−−−−→ A jn−1

φjn−1,jn−−−−−→ A jn , (12)

where m, n ∈ N0, i0 = j0, and im = jn, the following equality holds:

φim−1,im ◦ φim−2,im−1 ◦ · · · ◦ φi1,i2 ◦ φi0,i1 = φ jn−1,jn ◦ φ jn−2,jn−1 ◦ · · · ◦ φ j1,j2 ◦ φ j0,j1 . (13)

13




Exercise 3.5 (The Four Lemma). Let R be a ring. The following commutative diagram of R-modules is provided:

A B C D

A B C D ,

α β γ δ (14)

where the rows are exact.

(a) If α is epic, while β and δ are monic, then show that γ is monic.

(b) If δ is monic, while α and γ are epic, then show that β is epic.

Theorem 3.6 (The Five Lemma). Let R be a ring. The following commutative diagram of R-modules is provided:

A B C D E

A B C D E ,

α β γ δ (15)

where the rows are exact.

(a) If α is epic, whilst β and δ are monic, then γ is monic.

(b) If is monic, whilst β and δ are epic, then γ is epic.

(c) If α is epic, β and δ are iso, and is monic, then γ is iso.

Corollary 3.7 (The Short Five Lemma). Let R be a ring. The following commutative diagramof R-modules is provided:

0 A B C 0

0 A B C 0 ,

α β γ (16)

where the rows are exact. Such a triplet (α,β,γ ) of R-module homomorphisms is called a homomor-phism of short exact sequences of R-modules .

(a) If α and γ are monic, then β is monic. In this case, (α,β,γ ) is a monomorphism of short exact sequences of R-modules .

(b) If α and γ are epic, then β is epic. In this case, (α,β,γ ) is an epimorphism of short exact sequences of R-modules .

(c) If α and γ are iso, then β is iso. In this case, (α,β,γ ) is an isomorphism of short exact sequences of R-modules .

14




Definition 3.8 (Category of Short Exact Sequences). Let R be a ring. The class SES(R) of all shortexact sequences of R-modules together with homomorphisms of short exact sequence of R-modulesform a category. Note that SES(R) is a nonconcrete category. The definitions of monomorphisms ,epimorphisms , and isomorphisms coincide with the definitions given in the corollary above. If there

is an isomorphism between two short exact sequences, then we say that the two exact sequences areisomorphic .

Theorem 3.9 (The Splitting Lemma). Let R be a ring and 0 → A f −→ B

g−→ C → 0 a short exactsequence of R-modules. The following conditions are equivalent:

(i) There is a retract r : B → A of f , namely, an R-module homomorphism r : B → A such thatr ◦ f = idA;

(ii) There is a section s : C → B of g, namely, an R-module homomorphism s : C → B such thatg ◦ s = idC ; and

(iii) The short exact sequence 0 → A

f

−→ B

g

−→ C → 0 is isomorphic to the direct-sum short exactsequence 0 → A

inA−−→ A ⊕ C prC −−→ C → 0.

If the short exact sequence 0 → A → B → C → 0 satisfies any of the conditions above, we say that0 → A → B → C → 0 splits .

Proof:

[(i)⇒(iii)] Let ψ : B → A ⊕ C be the map ψ(b) :=

r(b), g(b)

for every b ∈ B . Then, clearly, ψ is anR-module homomorphism. We have the commutative diagram

0 A B C 0

0 A A ⊕ C C 0 ,

f

idA

r

g

ψ idC

⊆ prC

(17)

where we then use the Short Five Lemma to verify that ψ is an isomorphism.

[(ii)⇒(iii)] Let φ : A ⊕ C → B be the map φ(a, c) := f (a) + s(c) for every a ∈ A and c ∈ C . Then,clearly, ψ is an R-module homomorphism. We have the commutative diagram

0 A A ⊕ C C 0

0 A B C 0 ,

⊆

idA

prC

φ idC

f g

s

(18)

where we then use the Short Five Lemma to verify that φ is an isomorphism.

[(iii)⇒(i)] Suppose that we have the commutative diagram

0 A A ⊕ C C 0

0 A B C 0 ,

inA

αprA

prC

β γ

f g

(19)

15




where α, β , and γ are isomorphisms of R-modules. The map r := α ◦ prA ◦ β −1 is a requiredretract of f .

[(iii)⇒(ii)] Suppose that we have the commutative diagram

0 A A ⊕ C C 0

0 A B C 0 ,

inA

α

prC

β γ inC

f g

(20)

where α, β , and γ are isomorphisms of R-modules. The map s := β ◦ inC ◦ γ −1 is a requiredsection of g .

Q.E.D.

Corollary 3.10. Let R be a ring. Suppose 0 → A → B → C → 0 is a splitting short exact

sequence of R-modules. Then, B A ⊕ C .

Corollary 3.11. Let R be a ring. An R-module M is decomposable if and only if there exists a

nonzero proper submodule N of M such that the short exact sequence 0 → N ⊆−→ M M/N → 0,

where the two-headed arrow indicates the quotient map, is nonsplitting.

Example 3.12.

(1) Let n ∈ N with n > 1. The short exact sequence 0 → Z ×n−→ Z Z/nZ → 0 of abelian groups

does not split, whence Z is an indecomposable Z-module.

(2) Let F be a field. Any short exact sequence 0

→ U

→ V

→ W

→ 0 of F -vector spaces is

splitting.

(3) The short exact sequence 0 → Z i−→ Z ⊕ (Z/2Z)N p−→ (Z/2Z)N → 0 of abelian groups does not

split if i(x) :=

2 x, (2Z, 2Z, 2Z, . . .)

and p

x, (y1, y2, y3, . . .)

:= (x + 2Z, y1, y2, y3, . . .) for every

x ∈ Z and y1, y2, y3, . . . ∈ Z/2Z.

Exercise 3.13. Let m, n ∈ N. If m and n are relatively prime, then prove that any short exactsequence 0 → Z/mZ → A → Z/nZ → 0 of abelian groups always splits. If m and n are not relativelyprime, then show that there exists a nonsplitting exact sequence 0 → Z/mZ → Z/mnZ → Z/nZ → 0

of abelian groups.

16




4 Free Modules over a Ring

Definition 4.1 (Linear Independence). Let R be a ring. For a subset X of an R-module M , an

R-linear combination is a finite sum of the formn

i=1

ri

∗ xi, where n

∈ N, r1, r2, . . . , rn

∈ R, and

x1, x2, . . . , xn ∈ X . A subset X of an R-module M is said to be R-linearly independent providedthat, for mutually distinct elements x1, x2, . . . , xn ∈ X and for r1, r2, . . . , rn ∈ R, where n ∈ N0, the

equalityn

j=1

r j ∗ x j = 0M implies that r j = 0M holds for every j = 1, 2, . . . , n. If M is generated as

an R-module by a subset Y ⊆ M , then we say that Y spans M over R. An R-linearly independentsubset of M that spans M over R is called an R-basis of M .

Remark 4.2.

(1) If R is a unital ring and M is a unitary R-module, then a subset Y of M spans M over R if and

only if every element of M is an R-linear combination

n

i=1 ri∗yi for some n ∈

N0, r1, r2, . . . , rn ∈ R,

and y1, y2, . . . , yn ∈ Y .

(2) For any ring R, the empty set is R-linearly independent and it is the sole basis of the zeroR-module.

(3) A subset X of a module M over a ring R is said to be R-linearly dependent if there exist n ∈ N,pairwise distinct elements x1, x2, . . . , xn ∈ X , and r1, r2, . . . , rn ∈ R such that ri 0 for some

i ∈ {1, 2, . . . , n} and thatn

i=1

ri ∗ xi = 0M .

Definition 4.3 (Free Modules). Let R be a ring. For a set X , we say that an R-module F is free on X if it is a free object on X in Mod(R). That is, F is a free R-module on a set X if and only if there exists a function ι : X → F such that, for any R-module M and for each function f : X → M ,there is a unique R-module homomorphism f : F → M with f ◦ ι = f . An R-module F is a free R-module if and only if it is a free object in Mod(R).

Let R be a unital ring. For a set X , we say that a unitary R-module F is free on X if it is afree object on X in UMod(R). That is, F is a unitarily free R-module on a set X if and only if there exists a function ι : X → F such that, for any unitary R-module M and for each functionf : X → M , there is a unique R-module homomorphism f : F → M with f ◦ ι = f . An R-moduleF is a unitarily free R-module if and only if it is a free object in UMod(R).

Exercise 4.4. For a ring R, show that R is not a free object in Mod(R).

Theorem 4.5 (Existence of Free Modules). Let R be a ring. The following results hold whether C

is Mod(R) or, if R is unital, C is UMod(R).

(a) The zero object of C is free on ∅.

(b) If J is an index set and, for each j ∈ J , F j is a free objects in C on a set X j, then j∈J

F j is a

free object on the disjoint union

j∈J

X j .

17




(c) Let F be a free object on a singleton of C. If R is unital and C is UMod(R), then F R asa unitary R-module. If R is unital and C is Mod(R), then F Z ⊕ R, where Z is taken to bea trivial R-module. If R is nonunital and C is Mod(R), then F S , where the abelian groupS := Z ⊕ R is the unital closure of R, and S has an R-module structure via the embedding

R → S sending r → (0, r) for every r ∈ R (that is, r ∗ (k, r) def

== (0, r) · (k, r) for every k ∈ Z andr, r ∈ R).

(d) For any set X , there is a free object on X in C, which is unique up to isomorphisms.

(e) Let F be a free object on a singleton of C. For an arbitrary set X , the free object on X in C

is (isomorphic to) F ⊕|X | def ==x∈X

F . Hence, a free object on a set X in C depends only on the

cardinality of X .

Proof:

(a) This part is obvious.

(b) Suppose that ι j : X j → F j satisfies the free-object condition. Let X :=

j∈J

X j and

ι :=

j∈J

ι j . We claim that F together with ι forms a free object on X . Let M be

an arbitrary object with a function f : X → M . Write ini for the canonical injectionini : X i →

j∈J

X j, for each i ∈ J . We have a function f ◦ ini : X i → M , for every

i ∈ J . For each i ∈ J , because F i together with ιi is free on X i, we conclude that thereexists uniquely an R-module homomorphism f i : F i → M . Ergo, we have an R-modulehomomorphism f :=

i∈J

f i :i∈J

F i → M which satisfies f ◦ ι = f . It is clear that f is

unique.(c) Suppose that X = {t}. If R is unital and C is UMod(R), we can take F := R and

ι : X → R to be the map sending t to 1R. For any unitary R-module M with a functionf : X → M , we let f : R → M to be the map f (r) := r ∗ f (t) for every r ∈ R. Obviously,f is the unique R-module homomorphism such that f ◦ ι = f .If R is unital and C is Mod(R), we take F := Z ⊕ R, where Z is taken to be a trivialR-module, and ι : X → F to be the map ι(t) := (1, 1R). If M is an R-module, we canwrite M = N ⊕ L, where N ⊆ M is a trivial R-module and L ⊆ M is a unitary R-module.For a given function f : X → M , we write f (t) = n + l, where n ∈ N and l ∈ L. Define

an R-module homomorphism f : F → M to be f

(k, r)

:= k n + r ∗ l for all k ∈ Z and

r ∈ R. Ergo, ˜f is the unique required map.Finally, if R is nonunital and C is Mod(R), we take F := S and ι : X → S to be the map

ι(t) := 1S = (1, 0R). For an R-module M with a function f : X → M , define f : F → M

via f

(k, r)

:= k f (t) + r ∗ f (t) for all k ∈ Z and r ∈ R. Hence, f is the only map withthe desired property.

(d) This part follows from Part (b) and Part (c) immediately.

(e) This part follows from Part (c) and Part (d) easily.

Q.E.D.

18




Corollary 4.6. Let R be a ring and S its unital closure. Then, an R-module is a free object inMod(R) if and only if it is a free object in UMod(S ).

Exercise 4.7. Let R be an arbitrary ring. If R is either of characteristic p ∈ N or with multi-plicative identity, then show that, for any cardinal number c, there exist a set X with cardinality c

and an R-module M such that X is an R-basis of M , but M is not a free R-module. Now, if R isa nonunital ring of characteristic 0, then prove that R-module with a basis is free if and only if Rcontains no element s for which there exists n ∈ N such that r · s = n r for every r ∈ R. Finally, if nis an integer with n > 1, give an example of an (nZ)-module which has a basis but is not free.

Theorem 4.8. Let R be a unital ring and F a unitary R-module. The following conditions areequivalent:

(i) F is a free object in UMod(R);

(ii) There exists a set X and a function ι : X → F with the property that, given any unitary

R-module M and a function f : X → M , there exists a unique R-module homomorphismf : F → M such that f ◦ ι = f .

(iii) F has a basis;

(iv) F is the internal direct sum of a family of cyclic R-submodules, each of which is isomorphic toR as a unitary R-module; and

(v) F is isomorphic as an R-module to a direct sum of copies of the unitary R-module R.

Theorem 4.9. Let R be a ring. The category C is either Mod(R) or UMod(R) (if R is unital).Then, every object M in C is a homomorphic image of a free object F in C. If M is finitely generated,

then we may take F to be finitely generated as well.Remark 4.10.

(1) Let R be a ring. The category C is either Mod(R) or UMod(R) (if R is unital). Let F , togetherwith a function ι : X → F , be a free object in C on a set X . Then, ι(X ) is an R-basis of F .

(2) A submodule of a free module over an arbitrary ring need not be free. Likewise, a submodule of a unitarily free module over an arbitrary unital ring need not be free. For example, if R = Z/6Z,then R is a unitarily free R-module, but its R-submodule 2R is not free.

(3) For an arbitrary unital R-module, a unitarily free module F may have bases of different cardi-

nalities. Take for example, R := HomK (F, F ), where K is a unital ring and F is a unitarily freeK -module with an infinitely countable basis {e1, e2, e3, . . .}. It turns out that as, for any n ∈ N,R is isomorphic to R⊕n as a unitary R-module.

Exercise 4.11. Let R be a ring and M an R-module. Let C be a subcategory of Mod(R).A left resolution , or a resolution , of M in C is an exact sequence of objects in C of the form. . . → E 2 → E 1 → E 0 → M → 0. Prove that, if C is Mod(R) itself, then there exists a free resolution of M , i.e., a resolution . . . → F 2 → F 1 → F 0 → M → 0, where F k is a free R-module forall k = 0, 1, 2, . . .. Prove also that, if R is unital, M is unitary, and C is UMod(R) itself, then there

19




exists a unitarily free resolution of M , i.e., a resolution . . . → F 2 → F 1 → F 0 → M → 0, where F k isa unitarily free R-module for all k = 0, 1, 2, . . .. For example, there exists a unitarily free resolutionin UMod(Z) of Z/nZ, where n ∈ N, of the form . . . → 0 → 0 → Z → Z → Z/nZ → 0.

Proposition 4.12. Let R be a ring and 0

→ M

→ N

→ F

→ 0 a short exact sequence of R-

modules. If F is free, then the sequence splits. If R is unital, F is unitarily free, and M and N areunitary R-modules, then the sequence splits.

Proof: We shall work both cases simultaneously. Let φ be the given map N → F . Suppose F is(unitarily) free on a set X , with ι : X → F . Define f : X → N by assigning each x ∈ X withan element ax ∈ N such that φ (ax) = ι(x). By the universal property of free objects, there exists(uniquely) an R-module homomorphism f : F → N such that f ◦ ι = f .

Now, for x ∈ X ,

φ ◦ f

ι(x)

= φ

f ◦ ι

(x)

= φ

f (x)

= φ (ax) = ι(x) = (idF ◦ ι) (x). Thus,

φ ◦ f ◦ ι = idF ◦ ι By the uniqueness of such a map, φ ◦ f = idF , so we have a section of φ.

Thence, the given short exact sequence splits. Q.E.D.

20




5 The Invariant Dimension Property

Theorem 5.1. Let R be a unital ring and F a unitarily free R-module with an infinite basis X .Then, every basis of F has the same cardinality as X .

Proof: Suppose that Y is another basis of F . Firstly, we claim that Y is infinite. If Y were finite,there would exist a finite subset Z of X such that every element of Y is a linear combination of elements of Z , whence F would be generated by a proper subset Z of X , which would contradictthe hypothesis that X is a basis of F .

Let K (Y ) be the set of all finite subsets of Y . Define a map f : X → K (Y ) by x → {y1, y2, . . . , yn},

where x ∈ X can be written asn

i=1

ri ∗ yi for some n ∈ N, where r1, r2, . . . , rn ∈ R \ {0R} and

y1, y2, . . . , yn ∈ Y . Since Y is a basis, f is well defined. Furthermore, im(f ) = f (X ) is infinitebecause

V ∈im(f )

V is a subset of Y that generates X , whence also F .

Next, we shall verify that the preimage f −1(T ) is a finite subset of X for each T ∈ im(f ) ⊆ K (Y ).Let F T be the R-submodule of F generated by T . Since T is finite, there is a finite subset S of X such that F T is contained in the R-submodule F S of F generated by S . If x ∈ f −1(T ), then x iscontained in F T . That is, x ∈ f −1(T ) implies that x ∈ F S , whence x is an R-linear combination of elements of S . Because X is R-linearly independent, x ∈ X , S ⊆ X , and x ∈ F S , it follows thatx ∈ S . Therefore, f −1(T ) ⊆ S , making f −1(T ) finite.

For each T ∈ im(f ), order the elements of f −1(T ), say x1, x2, . . . , xn, for some n ∈ N, and define aninjective map gT : f −1(T ) → im(f ) × N by xk → (T, k) for k = 1, 2, . . . , n. Since sets of the formf −1(T ), where T ∈ im(f ), form a partition of X , we conclude that the map g : X → im(f ) × N

defined by x

→ gT (x) if x

∈ f −1(T ) is a well defined injection. Therefore,

|X | ≤ im(f ) × N

= im(f ) |N| =

im(f ) ≤ K (Y )

. (21)

We now need to verify thatK (Y )

= |Y |. However, K (Y ) =

k∈N

Y

k

, where

V

c

is the set of

all subsets of cardinality c of a set V . Because

Y

k

clearly has the same cardinality as Y for all

k ∈ N, we conclude that there is a bijection from K (Y ) to

k∈N

Y , which has the same cardinality

as Y × N. We then have |X | ≤K (Y )

= |Y × N| = |Y | |N| = |Y |.

Using the same argument, we can show that |Y | ≤ |X |. The Schroeder-Bernstein Theorem guar-antees that |Y | = |X |. Q.E.D.

Corollary 5.2. Let R be a unital ring and F a free R-module with an infinite basis X . Then,every basis of F has the same cardinality as X .

Definition 5.3 (Invariant Dimension Property). Let R be a ring such that, for every free R-moduleF , any two bases of F have the same cardinality. Then, R is said to have the invariant dimension property for left R-modules and the cardinal number of any basis of a free R-module F is called the

21




dimension or rank of F over R, which is denoted by dimR(F ) or rkR(F ). We can similarly define theinvariant dimension property for right R-modules .

Let R be a unital ring such that, for every unitarily free R-module F , any two bases of F havethe same cardinality. Then, R is said to have the invariant dimension property for unitary left R-

modules and the cardinal number of any basis of a unitarily free R-module F is called the dimension or rank of F over R, which is denoted by dimR(F ) or rkR(F ). We can similarly define the invariant dimension property for unitary right R-modules .

Theorem 5.4. Let R be a ring and S its unital closure. Then, R has the invariant dimensionproperty for its left modules if and only if S has the invariant dimension property for its unitary leftmodules. Similarly, R has the invariant dimension property for its right modules if and only if S hasthe invariant dimension property for its unitary right modules.

Theorem 5.5. A unital ring has the invariant dimension property for its unitary left modules if andonly if it has the invariant dimension property for its unitary right modules. Hence, it is safe just tosay that a unital ring has the invariant dimension property for unitary modules , or has the invariant basis number , without specifying whether the property is with repect to left or right modules.

Proof: We can ignore unitarily free modules on infinite bases due to the previous theorem. Then,we can rephrase the definition of the invariant dimension property in both cases in the followingreformulation:

A unital ring R has the invariant dimension property if and only if whenever A is anm-by-n matrix over R and B is an n-by-m matrix over R, where m, n ∈ N0, such thatA · B = Im×m and B · A = In×n, where Ik×k is the k-by-k identity matrix over R forevery k ∈ N0, we must have m = n.

Since this definition is left-right symmetric, the claim follows. Q.E.D.

Corollary 5.6. A ring has the invariant dimension property for its left modules if and only if ithas the invariant dimension property for its right modules. Hence, it is safe just to say that a unitalring has the invariant dimension property without specifying whether the property is with repect toleft or right modules.

Proposition 5.7. Let R be a unital ring. The following conditions are equivalent:

(i) The ring R has the invariant basis number;

(ii) If F is a unitarily free R-module with two finite bases X and Y , then |X | = |Y |; and(iii) For m, n ∈ N0, Rm

Rn as R-modules if and only if m = n.

Example 5.8. Every finite unital ring has the invariant basis number.

Exercise 5.9. Let R be a unital ring. Prove that R has the invariant basis number if either of the following conditions hold:

(i) All finitely generated unital subrings of R have the invariant basis number; or

22




(ii) There is a unital subring S of R with the invariant basis number such that R is a finitelygenerated unitarily free S -module.

Proposition 5.10. Let F 1 and F 2 be unitarily free modules over a unital ring R with the invariantbasis number. Then, F 1 F 2 as unitary R-modules if and only if dimR (F 1) = dimR (F 2).

Lemma 5.11. Let R be a unital ring, I a proper two-sided ideal of R, F a unitarily free R-modulewith basis X , and π : F → F/(I ∗ F ) the canonical projection (where I ∗ F is the R-submodule

ni=1

ri ∗ ai ||| n ∈ N0 , r1, r2, . . . , rn ∈ I , and a1, a2, . . . , an ∈ F

of F ). Then, F /(I ∗F ) is a unitarily

free (R/I )-module with basis π(X ) andπ(X )

= |X |, where the (R/I )-action on F /(I ∗ F ) is given

by (r + I ) ∗ (a + I ∗ F ) def == r ∗ a + I ∗ F for every r ∈ R and a ∈ F .

Proof of Lemma: For each u ∈ F , u =n

i=1

ri ∗ xi for some n ∈ N0, r1, r2, . . . , rn ∈ R, and

x1, x2, . . . , xn ∈ X . That is,

u + I ∗ F =

ni=1

ri ∗ xi

+ I ∗ F =

ni=1

(ri ∗ xi + I ∗ F )

=n

i=1

(ri + I ) ∗ (xi + I ∗ F ) =n

i=1

(ri + I ) ∗ π (xi) . (22)

Ergo, π(X ) generates F/(I ∗ F ) as a unitary (R/I )-module.

We now need to show that π(X ) is an (R/I )-linearly independent set. Suppose that we havem

j=1

(s j + I )∗

π (y j) = I ∗

F for some m ∈

N, elements s1, s2, . . . , sm

∈ R, and pairwise distinct

y1, y2, . . . , ym ∈ X . We must then have

m

j=1

s j ∗ y j

+ I ∗ F =

m j=1

(s j + I ) ∗ (y j + I ∗ F ) =m

j=1

(s j + I ) ∗ π (y j) = I ∗ F . (23)

Consequently,m

j=1

s j ∗ y j ∈ I ∗ F , whencem

j=1

s j ∗ y j =l

k=1

tk ∗ z k for some l ∈ N0, elements

t1, t2, . . . , tl ∈ I , and pairwise distinct z 1, z 2, . . . , z l ∈ X . The R-linear independence of X impliesthat m = l, and, up to reordering, s j = t j and y j = z j for j = 1, 2, . . . , m. Thus, s j + I = I for

every j = 1, 2, . . . , m, so π(X ) is indeed an (R/I )-linearly independent set.Now, it is easily seen that π|X : X → F/(I ∗ F ) is an injective mapping. Because it is alsosurjective onto its image π(X ), we conclude that π|X : X → π(X ) is bijective, implying thatπ(X )

= |X |. #

Exercise 5.12. Let R be a unital ring satisfying the invariant basis number. For a unitarily freeR-module F of infinite rank α and for a cardinal number β , 0 < β ≤ α, show that F has infinitelymany proper unitarily free R-submodules of rank β .

23




Proposition 5.13. Let R and S be unital rings and f : R → S a nonzero ring epimorphism. If S has the invariant basis number, then so does R.

Proof: Let I := ker(f ); then, I is a proper two-sided ideal of R and S R/I . Suppose that F is

a unitarily free R-module with bases X and Y . Let π : F → (I ∗ F ) be the quotient map. By theprevious lemma, F/(I ∗ F ) is a unitarily free (R/I )-module (hence, a free S -module) with bases

π(X ) and π(Y ). Since |X | =π(X )

, |Y | =π(Y )

, and R/I S has the invariant dimension

property, we have thatπ(X )

= π(Y ). That is, |X | = |Y |. Thence, R also satisfies the invariant

dimension property. Q.E.D.

Corollary 5.14. If R is a unital ring that has a homomorphic image which is a division ring, thenR has the invariant basis number.

Proof of Corollary: This follows immediately from the fact (to be given in the next section)that every division ring has the invariant dimension property. #

Corollary 5.15. Every commutative unital ring has the invariant basis number.

Proof of Corollary: Let R be a commutative unital ring. Then, R has a maximal ideal I .Since R/I is a field, whence a division ring, this cororally follows immediately from the previousone. #

Exercise 5.16. Let R be a ring without zero divisors (i.e., R has neither left zero divisors nor rightzero divisors) such that, for all r, s ∈ R, there exist a, b ∈ R, not both zero, such that a ·r + b ·s = 0R.Prove that R, as an R-module, is indecomposable. If R is unital, prove that R has the invariantbasis number.

Theorem 5.17. Let F 1 and F 2 be unitarily free modules over a unital ring R with the invariantbasis number. Then, rkR (F 1 ⊕ F 2) = rk (F 1) + rkR (F 2).

Corollary 5.18. Let F 1 and F 2 be unitarily free modules over a unital ring R with the invariantbasis number. If there exists a short exact sequence 0 → F 1 → E → F 2 → 0, where E is anR-module, then E is a unitarily free R-module, E F 1 ⊕ F 2, and rkR(E ) = rkR (F 1) + rkR (F 2).

24




6 Vector Spaces over Division Rings

Theorem 6.1. Let D be a division ring and V a D-vector space. For any D-linearly independentsubset X of V and for any spanning subset Y of V over D such that X ⊆ Y , there exists a D-basisB of V with X

⊆ B

⊆ Y .

Proof: Let I be the set of D-linearly independent subsets of Y that contains X . Note that X ∈ I ,whence I is nonempty. Order elements of I by inclusion. Let C be a nonempty chain in I . Weclaim that C :=

S ∈C

S is an element of I (so that every nonempty chain in I has an upper bound in

I ). However, it is clear that C is D-linearly independent and X ⊆ C ⊆ Y . Hence, the hypothesisof Zorn’s Lemma is satisfied for the partially ordered set ( I , ⊆). That is, I has a maximal elementB, which is what we desire. Q.E.D.

Corollary 6.2. Let D be a division ring and V a D-vector space. Then, any D-linearly indepen-dent subset X of V can be extended to a D-basis B of V .

Corollary 6.3. Let D be a division ring and V a D-vector space. Then, any D-spanning subsetY of V contains a D-basis B of V .

Corollary 6.4. Let D be a division ring. Then, every vector space over D has a D-basis.

Corollary 6.5. Every vector space over a division ring D is a unitarily free D-module.

Corollary 6.6. Let D be a division ring. Any D-submodule of a D-vector space is a D-vectorspace, whence a unitarily free D-module.

Exercise 6.7. Let G be a nontrivial group that is not cyclic of order 2. Prove that G has a

nonidentity group automorphism.

Theorem 6.8. Let D be a unital ring. Then, D is a division ring if and only if every unitaryD-module is unitarily free.

Proof: The direct implication is given in a corollary above. We shall prove the converse. Supposethat every unitary D-module is unitarily free, but D is not a division ring. We claim that D hasa maximal left ideal L which is nontrivial. As D is not a division ring, there is a nonzero elementu ∈ D which is not invertible. That is, D · u is a nontrivial proper left ideal of D. Hence, the setL of all nontrivial proper left ideals of D is nonempty. Order L by inclusion. For any nonemptychain

C in

L, we see that C :=

I ∈C

I is a nontrivial left ideal of D. Furthermore, C is a proper left

ideal since 1D I forall I ∈ C. Therefore, every nonempty chain in L has an upper bound, so Lhas a maximal element L, which is what we claim.

Consider the short exact sequence of D-modules: 0 → L ι−→ D

π−→ D/L → 0, where ι is the inclusionmap and π is the quotient map. Because D/L is a unitarily free module, the short exact sequencesplits, therefore D is isomorphic to L ⊕ (D/L) as D-modules.

Now, as D/L is free, it has a D-basis. Take x to be an element of a D-basis of D/L. Observe thatD/L is a simple D-module, hence it is generated by any nonzero element. This means D/L = D∗x.

25




However, because x is a basis element, D∗x D as a unitary D-module. Thence, D/L is isomorphicto D as a D-module. That is, as a D-module, D is simple. However, as D L ⊕ (D/L), we seethat D has a proper nonzero D-submodule, which is a contradiction. Hence, D must be a divisionring. Q.E.D.

Theorem 6.9. Every division ring has the invariant basis number.

Proof: Let D be a division ring and V a D-vector space. Take two bases X and Y of V . We canassume that both X and Y are finite, say X = {x1, x2, . . . , xm} and Y = {y1, y2, . . . , yn}, wherem, n ∈ N0, x1, x2, . . . , xm ∈ V , and y1, y2, . . . , yn ∈ D. Without loss of generality, we assume that

m ≤ n. Since X and Y are bases, we get 0V y1 =m

i=1

di ∗ xi for some d1, d2, . . . , dm ∈ D. If k is

the smallest positive integer such that dk 0D, then xk = d−1k ∗ y1 −

mi=k+1

d−1

k · di

∗ xi. That is,

the set X 1 :=

{y1

} ∪ {xi

|||i

∈ S 1

}, with S 1 :=

{1, 2, . . . , m

} \ {k

} is a D-spanning set of V .

Now, suppose the D-spanning set X j = {y1, y2, . . . , y j} ∪ {xi ||| i ∈ S j} and S j ⊆ {1, 2, . . . , m} are

given, where j ∈ N, j < m, |S j| = m − j . We must have y j+1 = j

µ=1

sµ ∗ yµ +

ν ∈S j

tν ∗ xν for

some sµ, tν ∈ D, where µ = 1, 2, . . . , j and ν ∈ S j. Let λ be the smallest element of S j suchthat tλ 0D (which must exist, otherwise y j+1 would be a D-linear combination of y1, y2, . . . , y j,contradicting the assumption that Y is a D-basis). Define S j+1 := S j \ {λ} ⊆ {1, 2, . . . , m} andX j+1 := {y1, y2, . . . , y j+1}∪{xi ||| i ∈ S j+1}. Clearly, |S j+1| = m − ( j + 1) and X j+1 spans V over D.

Finally, the process above yields the set X m = {y1, y2, . . . , ym} (as S m = ∅), which spans V . As Y is a base and X m ⊆ Y , we must have X m = Y , whence m = n, as desired. Q.E.D.

Definition 6.10. The dimension of a vector space V over a division ring D is denoted by dimD(V ),which is the cardinality of a D-basis of V . If dimD(V ) is finite, then we say that V is finite-dimensional ; otherwise, V is infinite-dimensional .

Exercise 6.11 (Odd-Even Town Problem). A town has n residents, where n ∈ N0. A club is aset of residents in this town. Clubs in this town must satisfy the following conditions:

(i) Every club must have an odd number of members; and

(ii) Any two different clubs must share an even number of members.

Prove that there can be at most n clubs in this town. Give an example where there are exactly nclubs.

Theorem 6.12. Let D be a division ring, and V and W D-vector spaces with V ⊇ W .

(a) We have the inequality dimD(V ) ≥ dimD(W ).

(b) If dimD(V ) = dimD(W ) and V is finite-dimensional over D, then V = W .

(c) We have the equality dimD(V ) = dimD(W ) + dimD(V /W ).

26




Proof: Parts (a) and (b) are easy. We shall prove Part (c). Let X and Y be D-bases of V andW , respectively, such that X ⊇ Y . Let x1, x2, . . . , xk be mutually distinct elements of X \ Y . Weshall prove that x1 + W, x2 + W , . . . , xk + W are D-linearly independent elements of V /W . Suppose

thatk

i=1

di

∗(xi + W ) = W for some d1, d2, . . . , dk

∈ D. Then,

k

i=1

di

∗xi + W = W , making

ki=1

di ∗ xi ∈ W . Because the xi’s are not in Y , we then have d1 = d2 = . . . = dk = 0D.

Let π be the quotient map π : V → V /W . From the result above, π|X \Y : X \Y → V /W is injective.

If we show that π(X \Y ) spans V /W , then we see that dimD(V /W ) =π(X \Y )

= |X \Y |. However,

this is clear since X spans V . Thence, dimD(V ) = |X | = |Y | + |X \ Y | = dimD(W )+dimD(V /W ),as desired. Q.E.D.

Corollary 6.13. Let D be a division ring, and V and W vector spaces over D. If f : V → W isa D-linear transformation, then there exists a D-basis X of V such that X

∩ker(f ) is a D-basis of

ker(f ) and

f (x) ||| f (x) 0 and x ∈ X

is a basis of im(f ). In particular,

dimD(V ) = dimD

ker(f )

+ dimD

im(f )

. (24)

Proof of Corollary: Let W := ker(f ). Let Y be a D-basis of W and X a D-basis of V ⊇ W containing Y . Then, X has the desired property. Also, as V /W = V / ker(f ) im(f ), we havethe desired equality relating dimensions of V , ker(f ), and im(f ). #

Theorem 6.14. Let D be a division ring. If V and W are D-vector subspaces of a D-vector space,then

dimD(V ) + dimD(W ) = dimD(V ∩ W ) + dimD(V + W ) . (25)

Proof: Let X be a D-basis of V ∩W , Y a basis of V containing X , and Z a basis of W containingX . Then, X ∪ (Y \ X ) ∪ (Z \ X ) is a D-basis of V + W , where X , Y \ X , and Z \ X are pairwisedisjoint. Consequently,

dimD(V + W ) + dimD(V ∩ W ) =|X | + |Y \ X | + |Z \ X |

+ |X |

=|X | + |Y \ X |

+|Z \ X | + |X |

= |Y | + |Z | = dimD(V ) + dimD(W ) . (26)

Q.E.D.

Theorem 6.15. Let R, S , and T be division rings such that R ⊆ S ⊆ T . Then,

dimR(T ) = (dimS (T )) (dimR(S )) . (27)

Furthermore, dimR(T ) is finite if and only if both dimS (T ) and dimR(S ) are finite.

27




Proof: Let U be a basis of T over S , and let V be a basis of S over R. Then, the cardinality of the set W := {v · u ||| u ∈ U and v ∈ V } is |U | |V |, as the elements of the form v · u, where u ∈ U and v ∈ V are pairwise distinct, due to the S -linear independence of U . We need to verify that W is indeed an R-basis of T .

Let t ∈ T . Then, there exist m ∈ N0, u1, u2, . . . , um ∈ U , and s1, s2, . . . , sm ∈ S such that

t =m

i=1

si ∗ ui. The si’s can be written as R-linear combinations of v1, v2, . . . , vn ∈ V for some

n ∈ N0. That is, for i = 1, 2, . . . , m, there are ri,j ∈ R for j = 1, 2, . . . , n such thatn

j=1

ri,j ∗ v j = si.

Hence, t =m

i=1

n

j=1

ri,j ∗ v j

∗ ui =

mi=1

n j=1

ri,j ∗ (v j · ui). Hence, W spans T over R.

For m, n ∈ N, take pairwise distinct u1, u2, . . . , um ∈ U and pairwise distinct v1, v2, . . . , vn ∈V . Suppose that there exists ri,j ∈ R, where i = 1, 2, . . . , m and j = 1, 2, . . . , n, such that

mi=1

n j=1

ri,j ∗ (v j · ui) = 0T . Then, 0T =m

i=1

n j=1

ri,j · v j

∗ ui. By S -linear independence of U ,

we haven

j=1

ri,j ∗ v j =n

j=1

ri,j · v j = 0S for every i = 1, 2, . . . , m. By R-linear independence of V ,

ri,j = 0R for every i = 1, 2, . . . , m and j = 1, 2, . . . , n. Ergo, W is R-linearly independent. Q.E.D.

Exercise 6.16. What are all possible fields K such that Q K Q 4√

2?

28




7 Projective Modules

Definition 7.1 (Projective Modules). Let R be a ring, and C either Mod(R) or, if R is unital,UMod(R). An object P in C is said to be projective if, for any epimorphism f : A → B of objects A and B in C and for any R-module homomorphism g : P

→ B , there exists an R-module

homomorphism h : P → A such that f ◦ h = g. In other words, P is projective if and only if, for anydiagram

P

A B 0

g

f

(28)

of R-module homomorphisms in C, there is an R-module homomorphism h : P → A such that thediagram

P

A B 0

gh

f

(29)

is commutative. We shall use the term projective R-modules for projective objects in Mod(R), andthe term unitarily projective R-modules for projective objects in UMod(R), if R is unital.

Theorem 7.2. Let R be a ring and S := Z ⊕ R the unital closure of R. Then, an R-module P isprojective in Mod(R) if and only if it is projective in UMod(S ).

Theorem 7.3. Every free module F over a ring R is projective. If R is unital, every unitarily freeR-module is unitarily projective.

Proof: Suppose that F is (unitarily) free on a set X , with ι : X

→ F . A diagram of homomor-

phisms of (unitary) R-module

F

A B 0

g

f

(30)

is given. For each x ∈ X , define φ(x) to be an element ax ∈ A such that f (ax) = (g ◦ ι)(x) (whichalways exists as f is surjective). Thus, we have a map φ : X → A. As F is (unitarily) free, thereexists a homomorphism h : F → A of (unitary) R-modules such that h ◦ ι = φ.

Now, consider f ◦h : F → B. Note that (f ◦h) ◦ ι = f ◦ (h ◦ ι) = f ◦ φ = g ◦ ι. Thus, f ◦h = g bythe uniqueness of such a map. Q.E.D.

Corollary 7.4. Let R be a ring. Every R-module is a homomorphic image of a projective R-module. If R is unital, every unitary R-module is a homomorphic image of a unitarily projectiveR-module.

Exercise 7.5. Let R be a ring, and C either Mod(R) or, if R is unital, UMod(R). For an objectM in C, we say that a resolution . . . → P 2 → P 1 → P 0 → M → 0, where P 0, P 1, P 2, . . . are objectsin C, is a projective resolution if each of the P i’s is projective in C. Prove that every object M in C

has a projective resolution (hence, C has enough projectives ).

29




Theorem 7.6. Let R be a ring, and C either Mod(R) or, if R is unital, UMod(R). The followingconditions on an object P of C are equivalent:

(i) The object P is projective in C;

(ii) Every short exact sequence 0 → A → B → P → 0 of objects in C splits; and

(iii) There is a free object F and an object K in C such that F K ⊕ P .

Proof:

[(i)⇒(ii)] Suppose P is projective and 0 → A f −→ B

g−→ P → 0 is provided. Consider the followingdiagram

P

B P 0

idP

g

(31)

where the bottom row is exact. By projectivity of P , there exists s : P → B such thatg ◦ s = idP . Thus, s is a section, and so the short exact sequence splits.

[(ii)⇒(iii)] Suppose that (ii) holds. Since P is a homomorphic image of a free object F , there is an

epimorphism ϕ : F → P . Ergo, we have the short exact sequence 0 → K ⊆−→ F

ϕ−→ P → 0, if K := ker(ϕ). Because this short exact sequence must split, F K ⊕ P .

[(iii)⇒(i)] Let π : F → P be the composition of an isomorphism F −→ K ⊕P and the canonical projection

K ⊕ P → P . Also, let ι : P → F be the composition of the canonical injetion P → K ⊕ P with the inverse isomorphism K ⊕ P

−→ F of the previous isomorphism. A diagram

P

A B 0

f

g

(32)

of homomorphisms of objects in C, where the bottom row is exact, is provided. Consider thediagram

F

P

A B 0 .

πι

f

g

(33)

Since F is projective, there is an R-module homomorphism ψ : F → A such that g ◦ψ = f ◦π.Define h := ψ ◦ ι. Then, g ◦ h = g ◦ (ψ ◦ ι) = (g ◦ ψ) ◦ ι = (f ◦ π) ◦ ι = f ◦ (π ◦ ι) = f ◦ idP = f .That is, P is projective.

Q.E.D.

Corollary 7.7. Let R be a ring, and C either Mod(R) or, if R is unital, UMod(R). For a shortexact sequence 0 → A → B → P → 0 of objects in C, B A ⊕ P .

30




Exercise 7.8. Prove that Q is not a unitarily projective Z-module.

Theorem 7.9 (Dual Basis Lemma). Let R be a unital ring. Then, a unitary R-module P is unitarilyprojective if and only if there are an index set J , a family ( p j)

j∈J of elements of P , and a family

(f j) j∈J

of R-module homomorphisms in HomR(P, R) such that

(i) for every x ∈ P , the set { j ∈ J ||| f j(x) 0R} is finite, and

(ii) for every x ∈ P , x =

j∈J

f j(x) ∗ p j.

Proof: Suppose that P is a unitarily projective R-module. Then, there exists a unitarily freemodule F =

j∈J

R for some index set J such that P is a direct summand of F , with π : F → P

and ι : P → F being the canonical projection and the canonical injection, respectively. Let p j := π (a j) for all j ∈ J . Now, F =

i∈J

R with the projection pr j : F → R being the map giving

the component indexed by j for every j ∈

J . Define f j to be pr j ◦

ι, and we are done.

Conversely, suppose that the two conditions are satisfied by a unitary R-module P . It is clearthat, if F :=

j∈J

R is the unitarily free R-module on J , then the map (r j) j∈J

→ j∈J

r j ∗ p j, where

(r j) j∈J

∈ F is an epimorphism of F to P . However, the map x → (f j (x)) j∈J

, for every x ∈ P ,provides a section of this epimorphism, whence F = K ⊕P , for some unitary R-module K . Q.E.D.

Corollary 7.10. Let R be a ring and S := Z ⊕ R its unital closure. Then, an R-module P isprojective if and only if there are an index set J , a family ( p j)

j∈J of elements of P , and a family

(f j) j∈J

of R-module homomorphisms in HomR(P, S ) such that

(i) for every x

∈ P , the set

{ j

∈ J

|||f j(x) 0S

} is finite, and

(ii) for every x ∈ P , x =

j∈J

f j(x) ∗ p j, where (k, r) ∗ u def == k u + r ∗ u for every k ∈ Z, r ∈ R, and

u ∈ P .

Example 7.11. If R is the unitary ring Z/6Z, then both Z/2Z and Z/3Z are unitary R-modules.As there is an R-module isomorphism R (Z/2Z) ⊕ (Z/3Z), we conclude that both Z/2Z and Z/3Z

are projective, although both of them are not unitarily free.

Theorem 7.12. Let R be a ring, and C either Mod(R) or, if R is unital, UMod(R). For an indexset J and j ∈ J , P j is an object in C. The direct sum

j∈J

P j of R-modules is projective in C if and

only if, for every j ∈ J , P j is a projective object of C.

Proof:

(⇐) Suppose that P j is projective in C for every j ∈ J . Then, there exists, for every j ∈ J , an

object K j of C such that K j ⊕ P j is free in C. That is,

j∈J

K j

⊕

j∈J

P j

j∈J

(K j ⊕ P j)

is free in C. That is, j∈J

P j is a direct summand of a free object in C, whence it is projective.

31




(⇒) Suppose that j∈J

P j is projective. For i ∈ J , let ιi and πi be the canonical injection P i → j∈J

P j

and the canonical projection j∈J

P j → P i, respectively. For a fixed k ∈ J , a diagram

P k

A B 0

f k

g

(34)

be given, where the bottom row is exact. For j ∈ J \ {k}, define f j : P j → B to be thezero map. Hence, we have f :

j∈J

P j → B, where f := j∈J

f j is the unique R-module

homomorphism such that f ◦ ι j = f j for every j ∈ J . Consequently, we have the diagram

j∈J

P j

A B 0 .

f

g

(35)

Since j∈J

P j is projective, there exists an R-module homomorphism h : j∈J

P j → A such that

g ◦ h = f . Now, consider the extended diagram

P k

j∈J

P j

A B 0 .

πkιk

f h

g

(36)

Hence, hk := h ◦ πk : P k → A satisfies the condition that

g ◦ hk = g ◦ (h ◦ πk) = (g ◦ h) ◦ πk = f ◦ πk = f k . (37)

That is, P k is projective. Since k is arbitrary, we are done.

Q.E.D.

Exercise 7.13. Let a1, a2, . . . , an, where n ∈ N, be elements of a unital ring R such that theleft ideal they generate is R itself. Define M to be the R-submodule of the unitary R-module Rn

consisting of all n-tuples (x1, x2, . . . , xn) ∈ Rn such thatn

i=1

xi · ai = 0R. Prove that M is a unitarily

projective R-module.

32




8 Injective Modules

Definition 8.1 (Injective Modules). Let R be a ring, and C either Mod(R) or, if R is unital,UMod(R). An object I in C is said to be injective if, for any monomorphism f : A → B of objects A and B in C and for any R-module homomorphism g : A

→ I , there exists an R-module

homomorphism h : B → I such that h ◦ f = g. In other words, I is injective in C if and only if, forany diagram

0 A B

I

f

g (38)

of R-module homomorphisms in C in which the top row is exact, there is an R-module homomorphismh : B → I such that the diagram

0 A B

I

f

g

h

(39)

is commutative. We shall use the term injective R-modules for injective objects in Mo d(R), and theterm unitarily injective R-modules for injective objects in UMod(R), if R is unital.

Example 8.2. The zero module over any ring is injective. If the ring is unital, the zero module isalso unitarily injective.

Exercise 8.3. Properties of projective or unitarily projective modules do not usually have dualforms for injective or unitarily injective modules. The problem therein is the lack of the duals of free objects in the related categories. We shall show that such duals, i.e., cofree objects , do not existexcept for a single circumstance. Let now R be a ring, and C either Mod(R) or, if R is unital,UMod(R).

An object K in C is said to be cofree on a set X if there is a function σ : K → X such that, forany R-module M and for each function f : M → X , there exists a unique R-module homomorphismf : M → K such that σ ◦ f = f . Prove that, for a set X , there exists a cofree object in C on X if and only if |X | = 1. Show that, if X is a singleton, then the only cofree object in C on X is the zeromodule over R.

Theorem 8.4. Let R be a ring, and C either Mod(R) or, if R is unital, UMod(R). For an indexset J and j ∈ J , I j is an object in C. The direct product

j∈J

I j of R-modules is injective in C if and

only if, for every j ∈ J , I j is an injective object of C.

Proof:

(⇐) Suppose that I j is injective in C for every j ∈ J . Suppose we have a diagram

0 A B

j∈J

I j

f

g (40)

33




of R-module homomorphisms of objects in C in which the top row is exact. Let πi :

j∈J

I j → I i

be the canonical projection, where i ∈ J . Hence, we have the diagram

0 A B

I i

f

πi◦g (41)

and, by the injectivity condition, there exists, for each i ∈ J , an R-module homomorphismhi : B → I i such that hi ◦ f = πi ◦ g. By the universal property of products, there existsuniquely an R-module homomorphism h :=

j∈J

h j such that πi ◦ h = hi for every i ∈ J . That

is, h ◦ f : B → j∈J

I j is a map that satisfies πi ◦ (h ◦ f ) = (πi ◦ h) ◦ f = hi ◦ f = gi = πi ◦ g

for every i ∈ J . Again, by the universality of categorical products regarding the uniqueness

of such maps, we conclude that h ◦ f = g.(⇒) We can easily prove this direction by dualizing our proof for the dual statement for projective

modules.

Q.E.D.

Theorem 8.5. Let R be a ring and S := Z ⊕ R the unital closure of R. Then, an R-module I isinjective in Mod(R) if and only if it is injective in UMod(S ).

Theorem 8.6 (Baer’s Criterion). Let R be a unital ring. A unitary R-module I is unitarily injectiveif and only if, for every left ideal L of R, any R-module homomorphism L → I may be extended to

an R-module homomorphism R → I .

Proof: Let ϕ : L → I be an R-module homomorphism. Note that L ⊆−→ R is a monomorphism of

R-modules. If I is injective, then there must exist an R-module homomorphism ϕ : R → I whichmakes the diagram below commutative:

0 L R .

I

⊆

ϕϕ

(42)

That is, ϕ extends ϕ.

Conversely, suppose that, for any left ideal L of R, any R-module homomorphism ϕ : L → I canbe extended to ϕ : R → I . Let A and B be unitary R-modules with an R-module monomorphismf : A → B and an R-module homomorphism g : A → I .

Take S to be the set of all R-module homomorphism η : M → I , where M is a unitary R-modulesuch that im(f ) ⊆ M ⊆ B and that η ◦ f = g. Note that S is nonempty since it contains

g ◦

f |im(f )

−1: im(f ) → I . Partially order S by extension, i.e., for h1, h2 ∈ S , we say that

h1 h2 if and only if the domain Dom (h1) of h1 is contained in the domain Dom (h2) of h2, and

34




h2|Dom(h1) = h1. We can easily that the partially ordered set (S , ) satisfies the hypothesis of Zorn’s Lemma. Hence, there exists a maximal element h : H → I of S with h ◦ f = g . We shallcomplete the proof by showing that H = B.

If H B and b

∈ B

\ H , then L :=

{r

∈ R

|||r

∗b

∈ H

} is a left ideal of R. Now, ϕ : L

→ I

sending r ∈ L to h(r ∗ b) is a well defined R-module homomorphism. This map must then extendsto ϕ : R → I , by the hypothesis on I . We shall now look at the map h : H + R ∗ b → I which sendsx + r ∗ b → h(x) + r ∗ ϕ (1R), where x ∈ H and r ∈ R. It is easily seen that h is a well definedR-module homomorphism that is an element of S . This, however, contradicts the maximality of hbecause b H and so H H + R ∗ b. Therefore, H = B, whence I is injective. Q.E.D.

Exercise 8.7. Let R be a unital ring. Show that a unitary R-module I is injective if and only if,for every left ideal L of R and for any R-module homomorphism ϕ : L → I , there exists m ∈ I suchthat ϕ(r) = r ∗ m for every r ∈ L.

Theorem 8.8. Let I be a unitarily injective Z-module and R a unital ring. Then, HomZ(R, I ) is aunitarily injective R-module.

Proof: Recall that, for each f ∈ HomZ(R, I ) and for r, x ∈ R, (r ∗ f )(x) = f (x · r). Let L bea left ideal of R with an R-module homomorphism ϕ : L → HomZ(R, I ). Define f : L → I

by f (x) =

ϕ(x)

(1R) for every x ∈ L. Note that f is a homomorphism of unitary Z-modules.Therefore, we have the diagram

0 L R

I

⊆

f (43)

of unitary Z-modules, where the top row is exact. As I is a unitarily injective Z-module, there

exists a Z-module homomorphism ˜f : R → I such that

˜f |L = f .

Now, we take ϕ : R → HomZ(R, I ) to be the map given by

ϕ(r)

(x) := f (x · r) for every x, r ∈ R.Observe that ϕ is a well-defined R-module homomorphism. Furthermore, ϕ clearly extends ϕ. ByBaer’s Criterion, HomZ(R, I ) is injective. Q.E.D.

Theorem 8.9. Every unitary module M over a unital ring R may be embedded in a unitarilyinjective R-module.

Proof: We shall firstly assume the fact (to be proven in the next section) that every abelian group(i.e., every unitary Z-module) can be embedded in an injective abelian group (i.e., a unitarilyZ-module). Now, since M is an abelian group, there is an injective abelian group I along with

a monomorphism f : M → I of abelian groups. Define ˜f : HomZ(R, M ) → HomZ(R, I ) byf (g) := f ◦ g. Clearly, f is a monomorphism of R-modules.

Next, HomR(R, M ) is an R-submodule of HomZ(R, M ). Define the map δ : M → HomR(R, M )

by

δ (m)

(r) := r ∗ m. It is easily seen that this is an R-module isomorphism with inverse

: HomR(R, M ) → M being the map (ψ) := ψ (1R) for every ψ ∈ HomR(R, M ). That is, we havethe sequence of R-module monomorphisms

M −→ HomR(R, M )

⊆−→ HomZ(R, M )f −→ HomZ(R, I ) . (44)

35




The composition of these maps gives an embedding M → HomZ(R, I ), where HomZ(R, I ) is injec-tive. The proof is now complete. Q.E.D.

Corollary 8.10. Every module M over a ring R may be embedded in an injective R-module.

Exercise 8.11. Let R be a ring and M an R-module. Let C be a subcategory of Mod(R).A right resolution , or a coresolution , of M in C is an exact sequence of objects in C of the form0 → M → E 0 → E 1 → E 2 → . . ..

Suppose now that C is either Mod(R) or, if R is unital, UMod(R). For an object M in C, wesay that a coresolution 0 → M → I 0 → I 1 → I 2 → . . ., where I 0, I 1, I 2, . . . are objects in C, is aninjective resolution (or, to be precise, an injective coresolution ) if each of the I i’s is injective in C.Prove that every object M in C has an injective resolution (hence, C has enough injectives ).

Theorem 8.12. Let R be a ring, and C either Mod(R) or, if R is unital, UMod(R). The followingconditions on an object I in C are equivalent:

(i) The object I is injective;

(ii) Every short exact sequence 0 → I → B → C → 0 splits; and

(iii) If M is a object in C containing I as an R-submodule, then I is a direct summand of M .

Proof:

[(i)⇒(ii)] This part follows easily by dualizing our proof for a similar statement for projective modules.

[(ii)⇒(iii)] Suppose (ii) holds. Then, if we have an embedding I ⊆−→ M for some object M in C, then

0

→ I

⊆

−→ M

→ M/I

→ 0 is a short exact sequence of objects in C, which splits. Therefore,

M I ⊕ (M/I ), as required.

[(iii)⇒(i)] Suppose (iii) holds. Then I is an R-submodule of an injective object Q of C. Then, by thehypothesis, Q = I ⊕ J I × J for some R-submodule J of Q. Since the direct productof objects in C is injective if and only if each term is injective, we conclude that I itself isinjective.

Q.E.D.

36




9 Divisible Modules

Definition 9.1 (Divisible Modules). A left module M over a ring R is said to be divisible if andonly if, for every nonzero r ∈ R which is not a right zero divisor and for every m ∈ M , there existsm

∈ M such that m = r

∗ m. A right module M over a ring R is said to be divisible if and only if,

for every nonzero r ∈ R which is not a left zero divisor and for every m ∈ M , there exists m ∈ M such that m = m ∗ r.

Exercise 9.2. Let R be a ring and M a divisible R-module. Prove that any homomorphic imageof M is also a divisible R-module. In particular, if N is an R-submodule of M , then justify thatM/N is divisible.

Theorem 9.3. Let R be a ring. A family (M j) j∈J

of R-modules is given, where J is an index set.

(a) The direct product of

j∈J

M j is divisible if and only if each of the M j’s is divisible.

(b) The direct sum of j∈J

M j is divisible if and only if each of the M j ’s is divisible. In particular,

every direct summand of a divisible R-module is divisible.

(c) If each of the M j’s is a divisible R-submodule of an R-module M , then the sum

j∈J

M j is a

divisible R-submodule of M .

Theorem 9.4. Let R be a unital ring. Then, every unitarily injective R-module is divisible.

Proof: Let r be a nonzero element of R which is not a right zero divisor. For a unitarily injective

R-module I and m ∈ I , consider the R-module homomorphism f : R · r → I sending s · r → s ∗ mfor every s ∈ R. This is a well defined R-module homomorphism. Now, note that we have thediagram

0 R · r R

I

⊆

f (45)

in which the top row is exact. Hence, there exists f : R → I that extends f . Now, left m := f (1R).We must then have r ∗ m = r ∗ f (1R) = f (r · 1R) = f (r) = f (r) = m. Q.E.D.

Definition 9.5. A principal left ideal ring is a ring in which every left ideal is principal, i.e., generated

as a left ideal by a single element. A principal right ideal ring is a ring in which every right idealis principal, i.e., generated as a right ideal by a single element. A principal left-and-right ideal ring is a ring all left ideals and all right ideals are principal. A principal ideal ring is a ring in whichany two-sided ideal is principal, i.e., generated as a two-sided ideal by a single element. A principal ideal domain is a principal ideal ring which is an integral domain. (Recall that an integral domainis a commutative unital ring without a zero divisor. Therefore, a principal ideal domain, being anintegral domain, is unital.)

37




Theorem 9.6. Let R be a principal left ideal unital ring. A unitary R-module is unitarily injectiveif and only if it is divisible.

Proof: We only need to prove the converse. Suppose that a unitary R-module I is divisible. LetL be a left ideal of R. As R is a principal ideal ring, L is generated by a single element s

∈ R,

or L = R · s. Now, for an arbitrary R-module homomorphism ϕ : L → I , let m := ϕ(s). As I isdivisible, there exists m ∈ I for which s∗ m = m. Now, we extends ϕ to ϕ : R → I via ϕ(r) := r ∗ mfor every r ∈ R. By Baer’s Criterion, I must be a unitarily injective R-module. Q.E.D.

Corollary 9.7. Let D be a principal ideal domain. A unitary D-module is unitarily injective if and only if it is divisible. In particular, every divisible abelian group (i.e., every divisible unitaryZ-module) is an injective abelian group.

Corollary 9.8. Let D be a division ring. Then, every D-vector space is an injective D-vectorspace (i.e., a unitarily injective D-module). In fact, every D-vector space is free, projective, andinjective as a unitary D-module.

Example 9.9. The abelian groups Q, Q/Z, and R are divisible, whence injective.

Exercise 9.10. Let D be a principal ideal domain and M a unitary D-module. Show that M isthe internal direct sum N ⊕ L, where N is a divisible D-submodule and L is reduced (i.e., L has nonontrivial divisible D-submodule).

Theorem 9.11. Let D be a principal ideal domain. Every unitary D-module can be embedded ina divisible unitary D-module.

Proof: Let F be the field of quotient of D. For any unitary D-module M , there is a unitarily freeD-module E on a set J , so that E

j∈J

D, and a D-module epimorphism E →

M with kernel K so

that E /K M . Now, embed E in j∈J

F via the D-module monomorphism ι : E → j∈J

F . Hence,

ι induces a D-module isomorphism E/K ι(E )/ι(K ). Since

j∈J

F

/ι(K ) is a homomorphic

image of a divisible D-module, it must be divisible, Then, we have the embedding of M into thedivisible D-module D/f (K ) via the composition

M −→ E/K

−→ ι(E )/ι(K ) ⊆−→

j∈J

F

/ι(K ) . (46)

Q.E.D.

Corollary 9.12. Every abelian group may be embedded in a divisible abelian group.

Definition 9.13 (Torsion Elements). Let R be a ring and M an R-module. An element m ∈ M issaid to be a torsion element if r ∗ m = 0M for some r ∈ R with r 0R. If M has no torsion elementsother than 0M , then we say that M is torsion-free . The set of all torsion elements of M is denotedby Tor(M ).

38




Theorem 9.14. Let D be an integral domain and M a unitary D-module.

(a) The set Tor(M ) is a D-submodule of M , whence we call Tor(M ) the torsion D-submodule of M .

(b) The unitary D-module M/Tor(M ) is torsion-free.

(c) If M is divisible, then Tor(M ) is also divisible.

(d) If D is a principal ideal domain and M is divisible, then M = Tor(M ) ⊕ N for some torsion-freeD-submodule N of M .

Proof: Parts (a), (b), and (c) are easy exercises. Part (d) follows from Parts (b), along with thefact that Tor(M ) is a unitarily injective D-module due to Part (c). Q.E.D.

Theorem 9.15. Let D be a principal ideal domain and M a divisible unitary D-module. If M istorsion-free, then M is a direct sum of (possibly zero) copies of F , where F is the field of fractions

of D.

Proof: We only need to show that M is an F -vector space. For m ∈ M and d ∈ D with d 0D,

there is a unique element m ∈ M such that d ∗ m = m. Hence, we shall denote m by 1

dm. Now,

for a, b ∈ D, with b 0D, and x ∈ M , we define

a

b

∗ x := a ∗

1

bx

. It can easily be seen that

this definition makes M a unitary F -module, and the proof is complete. Q.E.D.

Theorem 9.16. Let D be a principal ideal domain, p an irreducible element of D, and F the field of

fractions of D. Define D

1

p

to be a D-submodule of F consisting of elements of the form

a

b, where

a, b ∈ D with b = pk for some k ∈ N0. Then, the Prüfer D-module D ( p∞) def == D

1

p

/D associated

to p is a divisible unitary D-module.

Proof: Let a ∈ D and k ∈ N0. For d ∈ D with d 0D, we need to show that there exists an

element x of D

1

p

such that d ∗ (x + D) =

a

pk + D. Suppose that d = pr ·s with r ∈ N0 and s ∈ D

such that p s. That is, gcd pk, s

is a unit, whence there exist u, v ∈ D for which pk ·u+s·v = 1D.

Take x := a · v

pk+r. Then,

d · x − a

pk =

s · (a · v)

pk − a

pk =

a · (s · v − 1D)

pk = − pk · u

pk = −u ∈ D . (47)

Hence, d ∗ (x + D) = a

pk + D, and so D ( p∞) is divisible. Q.E.D.

39




Corollary 9.17. Let p ∈ N be a prime integer. The Prüfer p-group Z ( p∞) is the quotient Z

1

p

/Z

of abelian groups, where Z

1

p

denotes the abelian subgroup of Q containing all rational numbers

that can be written in the form a

b , where a, b ∈ Z with b = pk for some k ∈ N0. Then, Z ( p∞) is aninjective abelian group.

Theorem 9.18. Let D be a principal ideal domain and F its field of fractions. For a given divisibleunitary D-module M , M is a direct sum of (possibly zero) copies of F and Tor(M ). The torsionsubmodules Tor(M ) is a direct sum of (possibly zero) copies of D-modules of the form D ( p∞) forvarious irreducible elements p of D.

Proof: We only need to be concerned about the structure of T := Tor(M ). Two elements a and bof D are said to be associates in D if there is a unit u ∈ D such that b = a · u, in which case weshall write a

≈ b. Note that

≈ is an equivalence relation on D. Let P be a set of representatives

of the partition (D/ ≈) which are prime in D.

Define, for each p ∈ P , T p to be the D-submodule of T consisting of elements x ∈ T such that pk ∗ x = 0M for some k ∈ N0. Clearly, T =

p∈P

T p. Now, the problem is reduced to studying each

T p individually, for p ∈ P .

Fix a prime p ∈ P such that T p is nonzero. A sequence (m0, m1, m2, . . .) of elements of T p is saidto be good , if p ∗ mi = mi−1 for every i = 1, 2, . . .. For a good sequence m := (m0, m1, m2, . . .), wedefine N m to be the D-submodule

i∈N0

(D ∗ mi), which is isomorphic as a D-module to D ( p∞). A

set S of good sequences is said to be excellent if the sum m∈S

N m is direct. Let now S be the set

of all excellent sets. Clearly, S is nonempty. Order S by inclusion. Observe that (S , ⊆) satisfiesthe hypothesis of Zorn’s Lemma. Hence, it must have a maximal element Z . We claim that thesum N :=

m∈Z

N m, which is direct, equals T p itself. If not, then there exists x ∈ T p \ N such that

p ∗ x = 0M . Create a good sequence x := (x0, x1, x2, . . .) with x = x0. Then, we can easily see thatthe sum N + N x is direct, making Z ∪ {x} an element of S , contradicting the maximality of Z .

Hence, T p =m∈Z

N m =m∈Z

N m

D ( p∞)⊕|Z |

. Q.E.D.

Corollary 9.19. Every divisible abelian group is a direct sum of (possibly zero) copies of Q and

(possibly zero) copies of Prüfer groups Z

( p

∞

) for various primes p ∈ N

.Exercise 9.20. Let D be a principal ideal domain and F its field of fractions. Prove that the

unitary D-module F/D is divisible and isomorphic to the D-module

p∈P

D ( p∞), where P is a set

containing representatives of equivalence classes of irreducible elements that are associates. Provealso that Q/Z is isomorphic as an abelian group to

p∈P+

Z ( p∞), where P+ is the set of positive prime

integers.

40




10 Final Exam Preparatory Problems with Solutions

Problem 10.1. Let (G, ·) be a group such that the automorphism group Aut(G) is cyclic. Provethat G is abelian.

Solution: We claim that, for an arbitrary group H , H/C(H ) is isomorphic to the subgroup Inn(H )of inner automorphisms of H . To show this, we consider the map ϕ : H → Aut(H ), where ϕ(h) isthe conjugation by g for every h ∈ H . Hence, im(ϕ) = Inn(H ) and ker(ϕ) = C(H ). By the FirstIsomorphism Theorem for Groups, H/C(H ) Inn(H ).

We now claim that, for a group H , if H/C(H ) is cyclic, then H is abelian. Suppose that H is a group such that H/C(H ) is cyclic, say with generator h · C(H ). Then, for elements a andb in H , we can write a = hi · x and b = h j · y , for some i, j ∈ Z and x, y ∈ C(H ). Thus,

a·b =

hi · x·

h j · y

=

hi · h j·(x·y) = hi+ j ·(x·y), since x, y ∈ C(G). Similarly, b·a = hi+ j ·(x·y).

Therefore, a · b = b · a for every a, b ∈ H , whence H is abelian.

Finally, suppose that G is a group such that Aut(G) is cyclic. Then, the subgroup Inn(G) of Aut(G) is also cyclic. However, this means G/C(G) Inn(G) is cyclic. By the result above, G isabelian.

Problem 10.2. Let (G, ·) be a group containing a subgroup H of finite index. Show that G containsa normal subgroup N of finite index such that N ⊆ H .

Solution: Suppose H is a subgroup of G of index n ∈ N. Then, G acts on the n left cosets g · H ,where g ∈ G, by left translation. This gives a group homomorphism ϕ : G →S n. The kernel N of ϕ is a normal subgroup of G. Furthermore, as G/N im(ϕ), by the First Isomorphism Theorem

of Groups, we conclude that [G : N ] = |G/N | = im(ϕ) ≤ S n < ∞. We now prove that N ⊆ H .

For g ∈ N , ϕ(g) must be the identity element of S n, whence g ·H = H . That is, g ∈ H . Therefore,N ⊆ H , as desired.

Problem 10.3. Let p and q be positive prime integers. Show that any group G of order p2q is notsimple.

Solution: Let G be a group of order p2q . If p > q , then a Sylow p-subgroup H of G is of index q .Since [G : H ] = q is the smallest prime divisor of |G|, H must be normal, whence G is not simple.

If p = q , then G is a p-group. Hence, the center C(G) is nontrivial. If C(G) G, then C(G) isa normal subgroup of G, whence G is not simple. If C(G) = G, then G is commutative and any

subgroup of G is normal. Take an arbitrary element x ∈ G of order p, which exists by Cauchy’sTheorem. The subgroup Z generated by x is a cyclic group of order p, which is a normal subgroupof G, and so G is once again nonsimple.

If p < q , then let n be the number Sylow q -subgroups of G. By the Third Sylow Theorem, n divides|G| = p2q as well as n ≡ 1 (mod q ). That is, n ∈

1,p ,p2

. Clearly, n p since p < q , making

p ≡ 1 (mod q ) impossible. If n = p2, then there are p2 subgroups of order q , and the intersectionbetween any two different such subgroups is trivial. Hence, there are p2(q − 1) elements of G withorder q , leaving p2 elements in the group. The Sylow p-subgroup P of G must then consist of all

41




of these elements. However, this means P is unique, and so P is normal in G. Therefore, G isnonsimple. Finally, if n = 1, then there exists a single Sylow q -subgroup Q of G, which must benormal. Hence, G is nonsimple.

Indeed, we can show that G is a solvable group. From our work earlier, we conclude that G

has a normal subgroup N such that either |N | = p2 or |N | = q . Hence, the subnormal seriesG N {1G} is a solvable series since the factor groups G/N and N are abelian (since one of them must have order p2 and the other is of order q ). That is, G is solvable.

Problem 10.4. Let F be a field and V an F -vector space. The general linear group GL(V ) is thegroup of invertible F -linear transformations on V under composition. The affine group on V is thegroup Aff (V ) := GL(V ) V =

(A, v) ||| A ∈ GL(V ) and v ∈ V

with the following multiplication

rule: (A1, v1) · (A2, v2) :=

A1 ◦ A2, A1 (v2) + v1

for every A1, A2 ∈ GL(V ) and v1, v2 ∈ V . Now,

let V be F itself. Show that, if F has at least three elements, then the group Aff (F ) is solvable, butnot nilpotent.

Solution: Note that G := Aff (F ) is F × F , where F × is the multiplicative group of F consistingof nonzero elements, with the multiplication rule (a, b) · (c, d) = (a · c, a ·d + b), where a, c ∈ F × andb, d ∈ F . Note that 1G := (1F , 0F ) is the identity element of G. Firstly, we shall find the center of G. Suppose that (a, b) ∈ C(G). Then, for any (x, y) ∈ G, we must have (a, b) · (x, y) = (x, y) · (a, b).Thus, (a · x, a · y + b) = (x · a, x · b + y) for every (x, y) ∈ G. This means (a − 1F ) · y = b · (x − 1F )for every (x, y) ∈ G, whence a = 1F and b = 0F . Hence, the center of G is trivial. Now, as Ck(G)is the preimage of C (G/Ck−1(G)) under the canonical projection for k ∈ N, with C0(G) = {1G},we see that Ck(G) = {1G} for every k ∈ N0. Thus, G is not nilpotent.

We then proceed to show that G is solvable. For (c, d) ∈ G, we have (c, d)−1 =

c−1, −c−1 · d

. Now,

for (x, y), (z, w) ∈ G, (x, y)·(z, w)·(x, y)−1

·(z, w)−1

= (x·z, x·w+y)·x−1

, −x−1

· y·z

−1

, −z −1

· w

.Consequently, (x, y) · (z, w) · (x, y)−1 · (z, w)−1 = (z, x · w − y · z + y) ·

z −1, −z −1 · w

, which equals

(1F , x · w − w − y · z + y). Obviously, this means the commutator [G, G] of G contains {1F } × F ,which is indeed a subgroup of G. Thus, [G, G] = {1F } × F . Now, recall that the k-th derived

subgroup G(k) of G isG(k−1), G(k−1)

for each k ∈ N and G(0) is G itself. The derived series of G

then consists of G(0) = G = F × F , G(1) = {1F } × F , and G(2) = G(3) = . . . = {1G}. This meansG is solvable.

Problem 10.5. Let A be an abelian group. Show that A is finite if and only if it has a compositionseries of finite length. (In Hungerford’s Textbook, a composition series for a group is assumed to be

of finite length.)

Solution: Firstly, we claim that every simple abelian group is finite with prime order. Let (S, +)be a simple abelian group with identity 0S . Take x 0S in S (which exists as S is nontrivial).The subgroup x generated by x is a nontrivial normal subgroup of S . Hence, S = x. If xhas an infinite order, then 2 x is clearly a proper nontrivial normal subgroup of S , which is acontradiction. Therefore, x has a finite order p ∈ N with p > 1. If p is not prime, there exists aproper divisor d 1, with d ∈ N, of p. Then, d x is a proper nontrivial normal subgroup of S ,again a contradiction. This mean p must be prime, whence |S | = p is prime.

42




Since every finite group has a composition series of finite length, we only need to prove the converse:if an abelian group has a composition series of finite length, then the group is finite. Assume that(A, +) is an abelian group with a composition series A = A0 A1 A2 . . . An = {0A},where n ∈ N0, 0A is the identity of A, and A0, A1, A2, . . . , An are abelian subgroups of A. Because

each composition factor Ai−1/Ai, where i = 1, 2, . . . , n, is a simple abelian group, we must have[Ai−1 : Ai] = |Ai−1/Ai| = pi for some prime pi ∈ N, for each i = 1, 2, . . . , n. Note from Lagrange’sTheorem that |A| = |A0| = [A0 : A1] |A1|, and so we have |A| = [A0 : A1] [A1 : A2] |A2|. Thus,by repetitively using Lagrange’s Theorem, we get |A| = [A0 : A1] [A1 : A2] · · · [An−1 : An] |An|,as |An| = 1, we obtain |A| =

n j=1

[A j−1 : A j ]. Therefore, the order of A is given by the product

|A| =n

i=1

[Ai−1 : Ai] =n

i=1

pi < ∞. Thus, A is a finite abelian group, and the proof is complete.

Problem 10.6. Let R be a nonzero ring such that, for each nonzero a ∈ R, there exists a uniqueb

∈ R for which a

·b

·a = a. Prove that R is a division ring.

Solution: For a ∈ R with a 0R, we shall write a−1 for the unique b ∈ R such that a · b · a = a.Firstly, we proceed by showing that R has no zero divisors. Suppose z ∈ R with z 0R is a leftzero divisor or a right zero divisor. Then, there exists y 0R such that z · y = 0R or y · z = 0R.However, this means z ·

y + z −1

· z = z , which is a contradiction to the uniqueness of z −1. Hence,

R has no zero divisors.

We now claim that a−1 · a · a−1 = a−1 for every nonzero a ∈ R. Let a ∈ R be such that a 0R.From a · a−1 · a = a, we then see that

a−1 · a · a−1

· a = a−1 ·

a · a−1 · a

= a−1 · a. Because R

has no zero divisors, a−1 · a · a−1 = a−1.

Next, for any r, a ∈

R with a 0R, there exists a unique s ∈

R such that r = s·

a. Let r, a ∈

R besuch that a 0R. The uniqueness of s such that r = s · a is trivial (as R has no zero divisors). We

note that, if s := r · a−1, then (s · a) · a−1 =

r · a−1· a

· a−1 = r ·

a−1 · a · a−1

= r · a−1. This

means r = s · a, as claimed. Similarly, for any r, s ∈ R with a 0R, there exists a unique t ∈ Rsuch that r = a−1 · t (i.e., when t is equal to a · r).

Now, for a fixed b ∈ R with b 0R. Define 1R := b−1 · b. We claim that 1R is the uniquemultiplicative identity of R. To show this, take an arbitrary r ∈ R. We see that there exists anelement s ∈ R such that r = s·b. Hence, r·1R = (s·b)·

b−1 · b

= s·

b · b−1 · b

= s·b = r. Similarly,

let t ∈ R be such that r = b−1 · t, then 1R · r =

b−1 · b

·

b−1 · t

=

b−1 · b · b−1

· r = b−1 · t = r.Ergo, 1R is indeed a multiplicative identity of R, which must also be unique.

Finally, for any nonzero a ∈ R, a · 1R = a = a · a−1 · a = a · a−1 · a

. That is, a−1 · a = 1R.

Likewise, 1R · a = a = a · a−1 · a =

a · a−1· a, implying a · a−1 = 1R. Therefore, a−1 is the unique

multiplicative inverse of a. Hence, R is a division ring.

Problem 10.7. Let R be a unital ring and S a ring without zero divisors. If a nonzero ringhomomorphism f : R → S exists, show that S is unital with multiplicative identity f (1R).

43




Solution: Let e := f (1R). Clearly, e 0S and e2 = e. For an arbitrary s ∈ S , we note that(s · e) · e = s · e2 = s · e, whence s · e = s, as S lacks zero divisors. Similarly, e · s = s for everys ∈ S . That is, S is unital and e = 1S .

Problem 10.8. For n ∈

N0, let Mat

n×n(R) be the ring of n-by-n matrices over a ring R. For a

division ring D , show that, for every n ∈ N, Matn×n(D) has only two two-sided ideals.

Solution: Let D be a division ring and n ∈ N. The ring Matn×n(D) is denoted by M . We wantto show that the only two-sided ideals of M are the zero ideal {0M } and the whole ring M itself.Suppose J is a nonzero two-sided ideal of M . Then, J contains a nonzero element A.

Let [n] := {1, 2, . . . , n}. For i, j ∈ [n], write E ji for the n-by-n matrix with 1D at the (i, j)-entry

and 0D everywhere else. Note that E ji · El

k = δ jk El

i for every i, j, k,l ∈ [n] (here, δ is the Kroneckedelta—for j, k ∈ [n], δk

j = 1R if j = k , and δk j = 0R if j k). Hence, A =

i,j∈[n]

Ai j · E

ji for some

Ai j

∈ F with i, j

∈ [n]. Suppose that Aµ

ν 0D, for some µ, ν

∈ [n].

For p ∈ [n], observe that Eµ p · A =

i,j∈[n]

Ai j ·

Eµ p · E

ji

=

i,j∈[n]

Ai j ·δ

µi E j

p

=

j∈[n]

Aµ j · E j

p. For

p, q ∈ [n], we then have Eµ p · A · Eq

ν =

j∈[n]

Aµ j ·

E j p · Eq

ν

=

j∈[n]

Aµ j ·δ

jν E

q p

= Aµ

ν · Eq p. Since

Aµν 0D, we conclude that α · Eq

p ∈ J for every p, q ∈ [n] and α ∈ D. Since the matrices of theform α · Eq

p, with p, q ∈ [n] and α ∈ D, generate the whole ring M , we conclude that J = M .

Indeed, for any unital ring R and n ∈ N0, the set T n of all two-sided ideals of Matn×n(M ) is in abijective correspondence with the set I n of all two-sided ideals of R. We associate each two-sidedideal J of Matn×n(M ) with the two-sided ideal f n(J ) of R generated by the entries of all matricesin J . Conversely, for a two-sided ideal K of R, we take gn(K ) to be the set of all matrices in

Matn×n(R) with entries in K (i.e., gn(K ) = Matn×n(K )). Clearly, the function f n : T n → I n is theinverse of the function gn : I n → T n, and vice versa.

Problem 10.9. Let R be the ring Z[x]/

x2 + 5

. Prove that R is not a principal ideal ring.

Solution: Note that R is isomorphic to A := Z[α], with α :=√ −5. Elements of A take the form

x + y α with x, y ∈ Z. Consider the ideal I := (2, 1 + α). We shall prove that I is not principal.Firstly, define N : A → Z via N (x + y α) := x2 + 5 y2 for every x, y ∈ Z. Observe that N ismultiplicative (i.e., N (u · v) = N (u) · N (v) for every u, v ∈ A).

Firstly, we shall prove that I is a proper ideal of A. To this end, we show that 1 I . Suppose

contrary that 1 = 2 ·u +(1+ α) ·v for some u, v ∈ A; then, 1 −2 u = (1 + α) ·v. However, N (1−2 u)is odd, while N

(1 + α) · v

is divisible by N (1 + α) = 6 is even. This is a contradiction, whence

1 I , so I is proper.

Now, for the sake of contradiction, assume that I is principal and generated by β ∈ A. This meansβ must divide 2 and 1 + α. Because N (2) = 4 and N (1 + α) = 6, we conclude that N (β ) ∈ {1, 2}.Since x2 + 5 y2 = 2 has no solution (x, y) ∈ Z2 and the only integral solutions to x2 + 5 y2 = 1 are(+1, 0) and (−1, 0), we see that the only possible values of β are +1 and −1. This contradicts ourearlier conclusion that I is not a proper ideal of R. Hence, I cannot be principal.

44




Problem 10.10. The ring Z[i] of Gaussian integers is composed by elements of the form a + b i,where a, b ∈ Z. The multiplication rule on Z[i] is given by (a + b i) · (c + d i) = (ac − bd) + (ad + bc) ifor a, b, c, d ∈ Z. This ring is a Euclidean domain with the Euclidean function N (a + b i) := a2 + b2

for every a, b ∈ Z. For a nonzero ideal I of Z[i], show that the quotient ring Z[i]/I is finite. What

are all possible fields that come from the quotient Z[i]/I of Z[i] by an ideal I ?

Solution: Since R := Z[i] is a Euclidean domain, it is a principal ideal domain. Therefore, anonzero ideal I is generated by a single element α 0R. For every β ∈ R, we can perform theEuclidean division on β and write β = α q + r for some q, r ∈ R with N (r) < N (α). Hence, eachelement β + I of R/I can be represented by a single element r ∈ R such that N (r) < N (α). Sincer = u + v i for some x, y ∈ Z, the condition N (r) < N (α) is equivalent to u2 + v2 < N (α), whichhas only finitely many solutions (u, v) ∈ Z. Therefore, R/I is a finite commutative unital ring.

To describe fields of the form R/I , where I is an ideal of R, we firstly determine when I = (α),with α ∈ R, is a maximal ideal. Since R is a principal ideal domain, I is a prime ideal, whence αis a prime element of R. For β

∈ R, if β = a + b i with x, y

∈ Z, then the conjugate of β is given

by β := a − b i, so that N (β ) = β · β . Now, note that α · α = N (α) must be either a positive primeinteger or the square of a positive prime integer.

For a positive prime integer p, if p ≡ −1 (mod 4), then the solution to t2 ≡ −1 (mod p) has nosolution t ∈ Z, which means there are no α ∈ R with N (α) = p. Furthermore, p must then be primein R. Hence, if I := ( p), then R/I is the field F p[i] F p2. Elements of F p[i] takes the form x+y i, withx, y ∈ F p. The multiplication rule on F p[i] is given by (x+y i)·(z +w i) = (x·z −y ·w)+(x·w+y ·z ) i,where x,y,z, w ∈ F p.

Now, if p is a positive prime integer such that p = 2 or p ≡ +1 (mod 4), then t2 ≡ −1 (mod p)has a solution t ∈ Z. Therefore, p divides (t + i) · (t − i) for some t ∈ Z. Since p divides neithert + i nor t

−i, we conclude that it is not prime in R, so there exists a prime π of R such that π

divides p. Clearly, this means p = N (π) = π · π. We want to show that, if I := (π), then R/I isthe field F p. Recall that π must divide t + i or t − i, for some t ∈ Z. Without loss of generality,suppose that π divides t − i. For a, b ∈ Z, identify (a + b i) + I ∈ R/I with a + b t ∈ F p. Evidently,this identification is a ring isomorphism, and so R/I F p.

In summary, every possible field in the form R/I , where I is an ideal of R, takes either the formF p2 if p ≡ −1 (mod 4) is a positive prime integer, or the form F p if p = 2 or p ≡ +1 (mod 4) is apositive prime integer. Our work is now complete.

Problem 10.11. Let D be a unique factorization domain and S ⊆ D a multiplicative subset notcontaining 0D. Prove that the localization S −1D is also a unique factorization domain. Furthermore,

if D is a principal ideal domain, then show that S −1D is also a principal ideal domain.

Solution: Let T be the set of irreducible elements of D that divide an element of S , and U the setof all irreducible elements of D not in T . For an irreducible element p of D, we claim that p ∈ T if and only if p is a unit in S −1D. To show the direct implication, assume p ∈ T . Then, there

exists s ∈ S such that s = p · x for some x ∈ D. Therefore,

ps

s

·

x

s

=

p · x · s

s · s =

s2

s2 = 1S −1D.

Hence, p = ps

s is a unit in S −1D. Conversely, assume that an irreducible p ∈ D is a unit in

45




S −1D. Therefore, there exists s, t ∈ S and x ∈ D such that

p · s

s

·

x

t

= 1S −1D. Hence,

p · x · s

s · t = 1S −1D =

s

s. Consequently, p · x · s2 = s2 · t. Therefore, p · x = t, or p ∈ T .

Now, assume that p

∈ U . We shall prove that p is irreducible in S −1D. Suppose contrary that p is

reducible in S −1D. Then, for s ∈ S , there exist x, y ∈ D and t, u ∈ S such that p · ss

=

xt

·yu

.

Therefore, p · s · t · u = s · x · y, whence p · t · u = x · y. Now, p must divide either x or y, but notboth. Say, p divides x and x = p · v for some v ∈ D. Then, we have v · y = t · u. This implies thatall irreducible factors of v and y must lie in T , as t, u ∈ S . By our earlier work, y , being a product

of irreducible elements in T , must be a unit in S −1D. Therefore, y

u is a unit in S −1D. Hence, p is

irreducible in S −1D.

Next, we shall verify that every irreducible element of S −1D is associate (in S −1D) to an element

in U . Suppose that a

s is an irreducible element of S −1D. Then, a must be divisible by some

p ∈

U . Write a = p ·

b with b ∈

D. If b has a factor q ∈

U , say b = q ·

c for some c ∈

D,

then as

=

p · ss

·

q · ss

·

cs

, whence a

s is reducible, which is a contradiction. Therefore, all

irreducible factors in b can only come from T . Hence, b

s is a unit. Thus,

a

s =

p · b

s =

p · s

s

·

b

s

is associate to p in S −1D.

For a ∈ D and s ∈ S , we want to find a factorization of a

s into irreducible factors in S −1D. We

start with the factorization a = υ ·

mi=1

pkii

· n

j=1

q lj

j

in D, where m, n ∈ N0, p1, p2, . . . , pm ∈

T , k1, k2, . . . , km ∈ N, q 1, q 2, . . . , q n ∈ U , l1, l2, . . . , ln ∈ N, and υ is a unit in D. Hence, we

have a factorization as

= Y · n

j=1

q j · s

s

lj into irreducible factors in S −1D, where Y denotes

υ

s

·

mi=1

pi · s

s

ki

, which is a unit in S −1D. The uniqueness of such factorizations of a

s can be

easily justified using the unique factorization property in D.

Finally, assume that D is a principal ideal domain and J is an ideal of S −1D. Since J must takethe form S −1I for some ideal I in D and I is generated as an ideal of D by a single element d ∈ D,we conclude that J is generated by d as an ideal of S −1D. Since this is true for any ideal J of S −1D, S −1D must then be a principal ideal domain.

Problem 10.12. Let R be a commutative unital ring. For an ideal I of R, define I ·0 := R andI ·k := I · I ·(k−1) for every k ∈ N. Let M be a maximal ideal of R. For any positive integer n, provethat R/M ·n has exactly one prime ideal. Prove also that R/M ·n is a local ring.

Solution: Firstly, let S be any commutative ring. An element x of S is said to be nilpotent if xk = 0S for some k ∈ N. It is easily seen that the nilpotent elements of S form an ideal called thenilradical of S , denoted by Nil(S ). We shall prove that, if S is unital, then Nil(S ) is the intersectionof all prime ideals of S . Firstly, we shall show that N := Nil(S ) is contained in any prime ideal I of

46




S . For a prime ideal I and u ∈ N , let k be the smallest positive integer k such that uk ∈ I (whichmuch exists as u is nilpotent). If k > 1, then we see that either u ∈ I or uk−1 ∈ I , since u · uk−1 ∈ I and I is prime. This is a contradiction to the minimality assumption on k. Hence, k = 1, andu ∈ I . Therefore, N ⊆ I . Now, let N be the intersection of all prime ideals of S . Pick an element

a ∈ S such that a is not nilpotent. Let T be the set of all ideals of R not containing any elementof the form ar with r ∈ N. Since the zero ideal is in T , T is nonempty. Order T by set inclusion.Clearly, T satisfies the hypothesis of Zorn’s Lemma. Therefore, T has a maximal element J . Ourtask is to show that J is a prime ideal. Suppose otherwise; then, there exist b, c ∈ S \ J such thatb · c ∈ J . Then, the set Y of elements y ∈ S such that b · y ∈ J is evidently an ideal of R properlycontaining J . However, c ∈ Y \ J ; therefore, by maximality of J , Y contains aµ for some µ ∈ N.Similarly, the set Z of elements z ∈ S such that aµ · z ∈ J is an ideal of R properly containing J (as b ∈ Z \ J ). Thus, Z contains an element of the form aν for some ν ∈ N. However, this meansaµ+ν = aµ · aν ∈ J , a contradiction. Thus, J is a prime ideal not containing a. This means N = N .

Now, we see that the ideal M/M ·n is in the nilradical of R/M ·n. Hence, any prime ideal P of R/M ·n

must contain M/M ·n. However, P must arise as a quotient P /M ·n for some ideal P containing M ·n

of R. As P contains M/M ·n, we must have M ⊆ P . Because M is a maximal ideal of R, we eitherhave P = M or P = R. The latter case does not happen, since otherwise P = R/M ·n, contradictingthe assumption that P is a prime ideal. Thus, P = M and P = M/M ·n. Consequently, the ringR/M ·n has exactly one prime ideal.

Finally, we shall verify that R/M ·n is a local ring. Suppose that J is a maximal ideal of R/M ·n.However, every maximal ideal is a prime ideal. Therefore, J = P , and so R/M ·n is local.

Problem 10.13. Let F be a field and x an indeterminate. Show that the localization of F [x] at theprime ideal (x) is isomorphic to a subring of the ring of power series F x.

Solution: Let P denote the ideal (x) and R the localization of F [x] at P . For g(x) ∈ F [x] \ P ,there exists an inverse g(x) in F x. We define ψ : R → F x sending each

f (x)

g(x) → f (x) · g(x)

for every f (x) ∈ F [x] and g(x) ∈ F [x] \ P . It is easy to show that this map is a well defined ringmonomorphism. Therefore, R is isomorphic to im(ψ) which is a subring of F x.

Problem 10.14. Prove that the polynomial x p−1 + 2 x p−2 + 3 x p−3 + · · · + ( p − 1) x + p is irreducibleover Q, where p is a positive prime integer and x is an indeterminate.

Solution: If p = 2, then the said polynomial is x+2, which is irreducible. Now, assume that p > 2.Let P (x) denote the polynomial x p−1 + 2 x p−2 + 3 x p−3 + · · · + ( p − 1) x + p. By Gauss’s Lemma, we

can show that P (x) is irreducible over Z. Note that (x − 1) · P (x) = x p

+ x p−1

+ · · · + x − p, and so(x−1)·P (x) =

x p − 1

x − 1

·x− p. Define Q(x) := P (x+1). Then, x·Q(x) =

(x + 1) p − 1

x

·(x+1)− p.

That is, x · Q(x) = (x + 1) ·

pr=1

p

r

xr−1

− p, which means Q(x) = x p−1 +

p−2r=0

p + 1

r + 2

xr.

The coefficients of 1, x1, x2, . . . , x p−3 in Q(x) are all divisible by p, whilst the coefficient of x p−2 isnot divisible by p and the constant term is not divisible by p2. Hence, by the Extended Eisenstein’sCriterion, Q(x) has an irreducible factor R(x) of degree at least p − 2, which may be taken to be

47




monic. If R(x) Q(x), then Q(x) must have a linear factor, whence Q(x) has an integer root.That is, P (x) has an integer root. If r is an integer root of P (x), then r must divides p (since

P (r) = 0 means − p = r

r p−2 + 2 r p−3 + · · · + ( p − 2) r + ( p − 1)

. Therefore, r ∈ {±1, ± p}.

Clearly, P (+1) and P (+ p) are positive, so r ∈ {−

1,−

p}

. Now, P (−

1) = p + 1

2 > 0. Finally,

observe that P (− p) ≡ − p( p − 1) + p ≡ 2 p 0 (mod p2). Hence, P (x) has no integer root, whichis a contradiction. Therefore, R(x) = Q(x), or Q(x) is irreducible over Z, which then implies thatP (x) is also irreducible over Z, whence over Q as well.

Problem 10.15. Let x be an indeterminate. Determine all positive prime integers p such that thepolynomial x3 + 5 x2 − 2 x + 1 has a multiple root in F p.

Solution: Let P (x) := x3 + 5 x2 − 2 x + 1. The first derivative of P (x) is given by the polynomialP (x) = 3 x2 + 10 x − 2. If p = 3, then we have P (x) = x − 2 in F3. That is, 2 is the only possibleroot of P (x) in F3. However, P (2) 0 in F3. Therefore, P (x) has no multiple roots over F3.

Now, assume that p 3. Therefore, it is possible to define Q(x) := x2

+ 10

3 x2

− 2

3 . We observe

that P (x) −

x + 5

3

· Q(x) = −62

9 x +

19

9 . That is, P (x) and Q(x) has a common factor if

only if p {2, 3, 31} and x = 19

62 is a common root for both P (x) and P (x) = 3 Q(x). Now,

P

19

62

=

211025

238328 and P

19

62

=

5175

3844. Since gcd(211025, 5175) = 575 = 52 × 23, the only

possible values of p for which P (x) has a multiple root are p = 5 and p = 23. To verify the result,P (x) = (x − 2)2 · (x − 1) over F5 and P (x) = (x − 17)2 · (x − 7) over F23.

Problem 10.16. Let R be a ring. For R-modules A and B, suppose that f : A → B and g : B → Aare R-module homomorphisms such that g

◦f = idA. Prove that B = im(f )

⊕ker(g).

Solution: For each b ∈ B, we see that (f ◦ g)(b) ∈ im(f ). Furthermore, for b ∈ B, we have

g

b − (f ◦ g)(b)

= g(b) −

g ◦ (f ◦ g)

(b) = g(b) −

(g ◦ f ) ◦ g

(b), and, as g ◦ f = idA, we obtain

g

b − (f ◦ g)(b)

= g(b) − (idA ◦ g) (b) = g(b) − g(b) = 0B. Therefore, B = im(f ) + ker(g).

We now need to verify that the sum is direct. Suppose β ∈ im(f ) ∩ ker(g). Then, β = f (x) for

some x ∈ A and g(β ) = 0A. However, this means x = idA(x) = (g ◦ f )(x) = g

f (x)

= g(β ) = 0A.

Consequently, β = f (0A) = 0B. Hence, im(f ) ∩ ker(g) = {0B}, or B = im(f ) ⊕ ker(g).

Problem 10.17. Let R be a ring. Consider the following commutative diagram of R-module homo-

morphismsA1 B1 C 1 0

0 A2 B2 C 2

f 1

α

g1

β γ

f 2 g2

(48)

in which the rows are exact. Prove that there exists an exact sequence

ker(α)f 1−→ ker(β )

g1−→ ker(γ ) ∂ −→ coker(α)

f 2−→ coker(β ) g2−→ coker(γ ) (49)

48




of R-module homomorphisms, where f 1 := f 1|ker(α), g1 := g1|ker(β), f 2 is induced by f 2, and g2 is

induced by g2. Prove also that, if f 1 is monic, then f 1 is monic as well, and that, if g2 is epic, theng2 is epic as well. (This result is known as the Snake Lemma .)

Solution: Firstly, the induced maps f 2 and g2 are given by f 2 (a + im(α)) := f 2 (a) + im(β ) forevery a ∈ A2 and g2 (b + im(β )) := g2 (b) + im(γ ) for every b ∈ B2. We firstly need to show thatf 1, g1, f 2, and g2 are well defined.

Secondly, we must show that im

f 1 ⊆ ker(β ). Let a ∈ ker(α). Then, β

f 1(a)

= β (f 1(a)),

which yields β

f 1(a)

= (β ◦ f 1) (a) = (f 2 ◦ α) (a) = f 2

α(a)

= f 2 (0A2) = 0B2

. Hence, we have

im

f 1 ⊆ ker(β ), as desired. Similarly, im (g1) ⊆ ker(γ ).

Now, we want to show that f 2 is well defined. Suppose that a1 + im(α) = a2 + im(α), wherea1, a2 ∈ A2, then a1 − a2 ∈ im(α). That is, a1 − a2 = α(x) for some x ∈ A1. Now, therefore,

we have f 2 (a1 − a2) = f 2

α(x)

= (f 2 ◦ α) (x) = (β ◦ f 1) (x) = β (f 1(x)) ∈ im(β ). Consequently,˜

f 2 (a1 + im(α)) = f 2 (a1) + im(β ) = f 2 (a2) + im(β ) = ˜f 2 (a2 + im(α)). Hence,

˜f 2 is well defined,and similarly, so is g2.

Next, we define ∂ : ker(γ ) → coker(α) as follows. Take an arbitrary function s : C 1 → B1 such that

g1◦s = idC 1 . Define ∂ := π◦τ ◦β ◦s, where τ : im (f 2) → A2 is the map

f 2|im(f 2)

−1, whereas π is the

canonical projection A2 → coker(α). We claim that ∂ is an R-module homomorphism that does notdepend on the choice of s. Fix c ∈ ker(γ ). Suppose b1, b2 ∈ B1 are such that g1 (b1) = c = g1 (b2).We then have g1 (b1 − b2) = 0C 1 . That is, b1 − b2 ∈ ker(g1) = im (f 1). Suppose that a ∈ A1 issuch that b1 − b2 = f 1 (a). Then, (f 2 ◦ α) (a) = (β ◦ f 1) (a) = β (f 1(a)) = β (b1 − b2). That is,

β (b1 − b2) = f 2

α(a) ∈ im (f 2). Furthermore, (τ ◦ β ) (b1 − b2) = α(a) ∈ im(α). This means

(π ◦ τ ◦ β ) (b1 − b2) = im(α) = 0coker(α) and so (π ◦ τ ◦ β ) (b1) = (π ◦ τ ◦ β ) (b2), i.e., ∂ does not

depend on the section s. We have to now check that im(β ◦ s) ⊆ im (f 2). For c ∈ ker(γ ), wehave c = g1(b) for some b ∈ B1. Now, 0C 2 = γ (c) = (γ ◦ g1) (b) = (g2 ◦ β ) (b) = g2

β (b)

. Hence,

β (b) ∈ ker (g2) ∈ im (f 2). This means im(β ◦ s) ⊆ im (f 2), as required. Moreover, it is obvious that∂ is an R-module homomorphism.

For the next step, we shall verify the exactness of the derived sequence. For the exactness atker(β ), notice that ker (g1) = ker(g1) ∩ ker(β ) = i m (f 1) ∩ ker(β ) ⊇ im

f 1

. If b ∈ ker (g1),

then b ∈ ker(g1) = im (f 1), so b = f 1(a) with a ∈ A1. Note that (f 2 ◦ α) (a) = (β ◦ f 1) (a),

leading to f 2

α(a)

= β (b) = 0B2. Because f 2 is monic, α(a) = 0A1

, so a ∈ ker(α). That is,

b = f 1(a) = f 1(a) ∈ im

f 1

. This means im

f 1

= ker (g1).

Next, for b ∈ ker(β ), choose any section s : C 1 → B1 such that s (g1(b)) = b. Then, we have∂ (g1(b)) = (π◦τ ◦β ◦s) (g1(b)) = (π◦τ )

β (b)

, whence ∂ (g1(b)) = (π ◦τ ) (0B2) = im(α) = 0coker(α).

Hence, we get im(g1) ⊆ ker(∂ ). On the other hand, suppose that ∂

c

= 0coker(α) for some

c ∈ ker(γ ). Then, (τ ◦ β ◦ s)(c) ∈ im(α). Suppose that (τ ◦ β ◦ s)(c) = α(a) for some element

a ∈ A1; i.e., (β ◦ s)(c) = f 2

(τ ◦ β ◦ s)(c)

= (f 2 ◦ α) (a) = (β ◦ f 1) (a). Consequently, we

obtain β (s(c) − f 1(a)) = 0B2, implying that s(c) − f 1(a) lies within ker(β ). Now, we observe that

g1 (s(c) − f 1(a)) = (g1 ◦ s) (c) − (g1 ◦ f 1) (a) = c, since g1 ◦ s = idC 1 and g1 ◦ f 1 = 0. Therefore,c ∈ im (g1), so im (g1) = ker(∂ ).

49




Then, we verify the exactness at coker(α). Firstly, observe that f 2◦π : A2 → coker(β ) sends a ∈ A2

to f 2 (a) + im(β ). Therefore, if b ∈ im (f 2), we have (f 2 ◦ τ ) (b) = b. It follows immediately that

f 2

(π◦τ )(b)

= b+im(β ) for every b ∈ im (f 2). In particular, for c ∈ ker(γ ), (β ◦s)(c) ∈ im(β ). That

is, we have f 2◦

∂ (c) = f 2(π

◦τ )(β

◦s)(c)+ im(β ) = (β

◦s)(c) + im(β ) = im(β ) = 0coker(β).

Thus, im(∂ ) ⊆ ker

f 2

. Also, if a ∈ A2 is such that f 2

a + im(α)

= 0coker(β) = im(β ), then

f 2(a) ∈ im(β ), whence β (b) = f 2(a) for some b ∈ B1. Then, by the commutativity of the givendiagram, γ (g1(b)) = (γ ◦ g1) (b) = (g2 ◦ β ) (b) = (g2 ◦ f 2) (a) = 0C 2 , making g1(b) ∈ ker(γ ). Clearly,

∂ (g1(b)) = a + im(α), yielding a + im(α) ∈ im(∂ ). Hence, im(∂ ) = ker

f 2

.

For the last exactness, we have the following equalities: ker(g2) =

g−12

im(γ )

+ im(β )

/ im(β )

and im

f 2

= (im(f 2) + im(β )) / im(β ). Evidently, as im (f 2) = ker(g2), im

f 2 ⊆ ker (g2).

Suppose now that b ∈ B2, b + im(β ) ∈ ker(g2). Then, g2(b) ∈ im(γ ), whence g2(b) = γ (c)for some c ∈ C 1. As g1 is epic, there exists an element b ∈ B1 such that g1 (b) = c. Now,we have (g

2 ◦β ) (b) = (γ

◦g1) (b) = γ (c) = g

2(b). This means b

− β (b)

∈ ker(g

2), whence

b + im(β ) =

b − β (b)

+ im(β ) ∈

ker(g2) + im(β )

/ im(β ). Ergo, b + im(β ) ∈ im

f 2

, where

we use the fact that ker (g2) = im (f 2).

Finally, suppose that f 1 is monic. Then, f 1 = f 1|ker(α) : ker(α) → ker(β ) is also monic. Likewise, if g2 is epic, then g2 : coker(β ) → coker(γ ) must be epic as well.

Problem 10.18. Let R be the ring Z[x], where x is an indeterminate. Consider the sequence of

R-module homomorphisms 0 → R f −→ R

g−→ Z → 0, where f p(x)

:= x · p(x) and g

p(x)

:= p(0)

for all p(x) ∈ R, and where Z is given an R-module structure via x ∗ k := 0 for every k ∈ Z. Provethat this is a short exact sequence of R-modules. Does it split as a short exact sequence of abelian

groups? Does it split as a short exact sequence of R-modules?

Solution: Firstly, im(f ) = R · x. For p(x) ∈ R, p(0) = 0R means that the constant term of p(x)is zero, whence p(x) = x · q (x) for some q (x) ∈ R. Hence, p(x) ∈ im(f ). Obviously, this meansim(f ) = ker(g), and the sequence is exact.

This short exact sequence splits as abelian groups because Z is a free abelian group. On the otherhand, if the sequence splits as R-modules, then there must exist an R-module homomorphisms : Z → R such that g ◦ s = idZ. Note that, for every a ∈ Z, s(a) = f ( pa(x)) + a for some

pa(x) ∈ R, since a = (g ◦ s)(a). Now, for a, b ∈ Z, we note that s(a · b) = a · s(b) as well as

s(a ·b) = ss(a)∗b = s(a)∗s(b) = s(a) ·s(b). Ergo, from the expansions a ·s(b) = a ·f ( pb(x))+a ·b

and s(a) · s(b) = f ( pa(x)) · f ( pb(x)) + a · f ( pb(x)) + b · f ( pa(x)) + a · b, we get the equalityf ( pa(x)) ·

f ( pb(x)) + b

= 0R. For a fixed a ∈ Z, this equality holds for every b ∈ B. Clearly, if

b 0, f ( pb(x)) + b 0R. Ergo, f ( pa(x)) = 0R for every a ∈ Z. That is, s(a) = a for every a ∈ Z.However, if a 0, we have x ∗s(a) = a ·x 0R, whilst s(x∗a) = s(0) = 0R. This is a contradiction,and so the short exact sequence of R-modules does not split.

Problem 10.19. Let d ∈ N and F a field. For n ∈ N, suppose there are pairwise distincts1, s2, . . . , sn ∈ F and polynomials f 1(x), f 2(x), . . . , f n(x) ∈ F [x] with x being an indeterminate. If,

50




for every i = 1, 2, . . . , n, we have deg (f i) < d and f i (si) 0F , but f i (s j) = 0F for all j ∈ {1, 2, . . . , n}with i j , then show that n ≤ d.

Solution: The set of polynomials in F [x] of degree less than d form a d-dimensional vector space

over F with basis

1, x , x2

, . . . , xd−1

. If we can show that f 1(x), f 2(x), . . . , f n(x) are F -linearlyindependent, then we are done.

Suppose thatn

i=1

ai·f i(x) = 0F for some a1, a2, . . . , an ∈ F . Plugging in x ⇔ s j, for j ∈ {1, 2, . . . , n},

we get a j · f j (s j) = 0F , whence a j = 0F . Since this is true for every j = 1, 2, . . . , n, we geta1, a2, . . . , an = 0F . That is, the polynomials f 1(x), f 2(x), . . . , f n(x) form an F -linearly independentset.

Problem 10.20. Let R be a ring, and C is either Mod(R) or, if R is unital, UMod(R). Provethat the following conditions on R are equivalent:

(i) Every object in C is projective;

(ii) Every short exact sequence of objects in C splits; and

(iii) Ever object in C is injective.

Solution: To show that (i) implies (ii) or that (iii) implies (ii) is easy. To show that (ii) implies(i), let M be an object in C. Then, M is a homomorphic image of a free module F under an

R-module epimorphism π. Therefore, we have a short exact sequence 0 → ker(π) ⊆−→ F

π−→ M → 0.If (ii) holds, then the short exact sequence must split and F ker(π) ⊕ M . Consequently, M isprojective.

To show that (ii) implies (iii), we suppose that (ii) holds and let M be an object in C. Since M can be embedded into an injective object I , say by an R-module monomorphism ι, we obtain theshort exact sequence 0 → M

ι−→ I coker(ι) → 0, where the two-headed arrow represents thecanonical projection from I to I/ im(ι) = coker(ι). By our hypothesis, the short exact sequencesplits, whence I = M ⊕ coker(ι). Now, as M is a direct summand of an injective object, M itself must then be injective.

51




11 Solutions to the Preliminary Final Examination

Problem 11.1. Describe (up to isomorphisms) all groups of order 153.

Solution: Let G be a group of order 153 = 32

× 17. Write m and n for the number of Sylow

3-subgroups and Sylow 17-subgroups of G. Using the Third Sylow Theorem, m ≡ 1 (mod 3) andm divides |G|, whence m = 1. Similarly, n = 1. Therefore, there are a unique Sylow 3-subgroup H and a unique Sylow 17-subgroup K of G, both of which are normal in G. Note that |H | = 32 = 9and |K | = 17. Clearly, both H and K are abelian. By the Classification Theorem of Finitely

Generated Abelian Groups, we have H C 3 ×C 3 or H C 9, and K C 17, whereC n is thecyclic group of order n ∈ N0.

Now, G must then contain a subgroup N := H ·K isomorphic to H ×K (since H and K are normalsubgroups in G such that H ∩ K = {1G}). However, as |N | = |H | × |K | = |G|, we conclude that

G = N H × K . That is, G is abelian, and we have G C 3 ×C 3 ×C 17 or G C 9 ×C 17.

Problem 11.2. Prove that a solvable group is finite if and only if it has a composition series of finitelength. Give an example (with proof) of an infinite group which has a composition series of finitelength.

Solution: Let G be a solvable group. If it is finite, then it automatically has a composition seriesof finite length (since all finite groups do). If it has a composition series

G =: G0 G1 G2 . . . Gl−1 Gl := {1G} (50)

for some l ∈ N0, then we can assume by Schreier’s Theorem that this is also a solvable series (i.e.,each factor Gi−1/Gi for i = 1, 2, . . . , l is abelian); otherwise, we can find a common refinement of

this composition series and a solvable series of G. Thus, for i = 1, 2, . . . , l, Gi−1/Gi is the trivialgroup or a simple abelian group, which has a prime order. Let pi := |Gi−1/Gi| for i = 1, 2, . . . , l.It follows from Lagrange’s Theorem that

|G| = |G0| = [G0 : G1] |G1| = [G0 : G1] [G1 : G2] |G2| = . . .

= [G0 : G1] [G1 : G2] · · · [Gl−1 : Gl] |Gl| =l

j=1

[G j−1 : G j] , (51)

because |Gl| = 1. That is, |G| =l

j=1

[G j−1 : G j] =l

j=1

|G j−1/G j | =l

j=1

p j < ∞.

Now, for k ∈ N, writeA k for the k-th alternating group. For m, n ∈ N with m ≤ n, consider thegroup monomorphism ιm,n :A m →A n sending each permutation σ ∈A m to the permutation

σ ∈A n such that σ(i) := σ(i) for i = 1, 2, . . . , m, and σ(i) = i for i = m + 1, m + 2, . . . , n.

The direct limitA ∞def == lim

−→k∈N

A k under the embeddings {ιm,n ||| m, n ∈ N}, being the direct limit

of simple groups, is a simple group. (To show this, suppose that N is a normal subgroup of A ∞.

Then, for k ∈ N, we considerA k as a subgroup of A ∞, so we see that N k := N ∩A k is a normal

52




subgroup of A k. For all integers k ≥ 5,A k is simple, so N k =A k or N k =

1A ∞

. If N t =A t

for some t ∈ N with t ≥ 5, then it follows that N k =A k for all k ∈ N with k ≥ 5, so N =A ∞.Otherwise, N k =

1A ∞

for every k = 5, 6, . . ., whence N =

1A ∞

. Therefore,A ∞ is simple.)

Hence, a composition series of A ∞ isA ∞ =: A0 A1 := 1A ∞, which is of finite length.

Problem 11.3. Describe (with proof) all left ideals of the ring of matrices Mat3×3(C).

Solution: For a ring R and n ∈ N0, write Matn×n(R) for the ring of n-by-n matrices with entriesin R. For a fixed n ∈ N0 and for a unital ring R, we claim that the set Ln of all left ideals of thering M n := Matn×n(R) is in a one-to-one correspondence with the set S n of all left R-submodules

of the unitary left R-module R⊕n :=n

r=1

R and this bijection is given by f n : S n → Ln sending

each V ∈ S n to the left ideal

f n(V ) :=

[vi,j]i∈[n],j∈[n] ||| for i, j ∈ [n], vi,j ∈ R, and, for i ∈ [n], (vi,1, vi,2, . . . , vi,n) ∈ V

, (52)

where [n] denotes {1, 2, . . . , n}.

Firstly, it is obvious that f n is an injective function. Secondly, for V ∈ S n and X ∈ f n(V ), then

we can write X =

x1

x2

.

.

.

xn

, where xi =

xi,1 xi,2 · · · xi,n

with (xi,1, xi,2, . . . , xi,n) ∈ V for every

i ∈ [n]. Thus, if A ∈ M n, then A = [ai,j]i∈[n],j∈[n], so that

A · X =

n

j=1

a1,j · xj

nj=1

a2,j · xj

.

.

.nj=1

an,j · xj

, (53)

whence A · X ∈ f n(V ). Clearly, f n(V ) is closed under addition. Therefore, f n(V ) is indeed a leftideal of M n.

Now, we must prove that all left ideals of M n arise this way. Define gn : Ln → S n as follows: for

I ∈ Ln, let gn(I ) be the left R-submodules of R⊕n

spanned over R by all elements of the form(v1, v2, . . . , vn), where

v1 v2 · · · vn

is a row of a matrix in I . We claim that gn is the inverse

of f n. Obviously, gn ◦ f n = idS n. Furthermore, it is clear that, for every I ∈ Ln, I ⊆ (f n ◦ gn) (I ),so we need to verify that this is indeed an equality.

Let I ∈ Ln and Y ∈ I . For i, j ∈ [n], write E ji for the matrix with 1R at the (i, j)-entry and 0R

everywhere else. Observe that E ji · El

k = δ jk · El

i for i, j, k,l ∈ [n] (here, δ is the Kronecke delta—for j, k ∈ [n], δk

j = 1R if j = k, and δk j = 0R if j k). Hence, Y =

i,j∈[n]

Y i j · E

ji for some Y i

j ∈ R with

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i, j ∈ [n]. Clearly, this means

d · Eν µ · Y =

i,j∈[n]

d · Y i j ·

Eν µ · E

ji

=

i,j∈[n]

d · Y i j ·δ

ν i · E j

µ

=

j∈[n]

d · Y ν j · E j

µ , (54)

for every µ, ν ∈ [n]. Thus,

d · Eν µ

· Y is the matrix whose µ-th row is the scalar multiplication onthe left by d of the ν -th row of Y, and whose other rows are all zero. What this means is that I contains the left ideal J generated by matrices of the form α · Z, where α ∈ R and Z is a matrixone of whose rows is identical to a row of a matrix in I and whose other rows are all zero. However,J ⊇ (f n ◦ gn) (I ) by the definition. Therefore, I ⊇ J ⊇ (f n ◦ gn) (I ), and so we have the equalityI = (f n ◦ gn) (I ). Thus, f n ◦ gn = idLn, as desired. Hence, we have found all left ideals of M n.

Alternative Solution: For a division ring D and n ∈ N0, we consider elements of the matrixring M n := Matn×n(D) as left D-linear transformation in EndD (W ), where W := D⊕n. (Thisproof requires the assumption that D is a division ring.) As before, let Ln be the set of all leftideals of M n and S n the set of all left D-vector subspaces of the left D-vector space W . Construct

F n : Ln → S n via I → T∈I

ker(T) for every I ∈ Ln. We shall show that F n is a bijection with the

inverse map Gn : S n → Ln sending V → {T ∈ M n ||| V ⊆ ker(T)} for all V ∈ S n.

Firstly, it is trivial that V ⊆ (F n ◦ Gn) (V ) for every V ∈ S n. We claim that there exists a D-linear transformation τ N ∈ M n such that ker(τ V ) = V . Firstly, observe that the short exact

sequence 0 → V ⊆−→ W W/V → 0 of D-vector spaces, where the two headed arrow represents

the quotient map. Since W/V is a D-vector space, which is a free object in the category of D-vector spaces, the short exact sequence splits, whence W is the internal direct sum V ⊕ W V , whereW V is a D-vector subspace of W isomorphic to W/V . Define τ V ∈ M n to be the composition

W = V

⊕W V

πW V

−−→ W V

ιW V

−−→ V

⊕W V = W , where πW V and ιW V are the canonical projection and

the canonical injection, respectively. Then, V = ker (τ V ) ⊇ T∈Gn(V )

ker(T) = (F n ◦ Gn) (V ) ⊇ V .

Hence, we have V =

T∈Gn(V )

ker(T) = (F n ◦ Gn) (V ).

Now, we want to show that I = (Gn ◦ F n) (I ) for every I ∈ Ln. Obviously, we have the inclusionI ⊆ (Gn ◦ F n) (I ). From our work before, if ρ(I ) is the set of all possible rows that occur in thematrices in I , then I contains any matrix whose rows are in ρ(I ). Furthermore, if t1, t2, . . . , tn are

the rows of T ∈ M n, then ker(T) =n

i=1

ker (ti), where the ti are treated as D-linear transformations

from W = D⊕n to D. Ergo, F n(I ) =

T∈I

ker(T) =

r∈ρ(I )

ker(r). It follows easily that ρ(I ) is a left

D-vector space. If we can show that ρ(I ) contains any row matrix r such that ker(r) ⊇ F n(I ), thenwe are done, since (Gn ◦ F n) (I ) is precisely the set of all matrices in M n whose rows annihilateF n(I ).

For I ∈ Ln, suppose m is the dimension of ρ(I ) over D. Then, dimD (F n(I )) = dimD

r∈ρ(I )

ker(r)

is equal to n − m. The set Y (I ) of row matrices r that annihilate F n(I ) is a left D-vector spacecontaining ρ(I ). Note that each r ∈ Y (I ) must factor through W/F n(I ), i.e., there is a unique

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homomorphism ϕr : W/F n(I ) → D of left D-vector spaces such that ϕr ◦ κF n(I ) = r, whereκF n(I ) : W → W/F n(I ) is the quotient map. On the other hand, any row in Y (I ) must arise thisway. Therefore, Y (I ) HomD (W/F n(I ), D) W/F n(I ) (as W/F n(I ) is finite-dimensional over

D), or dimD Y (I ) = dimD (W/F n(I )) = dimD(W ) − dimD (F n(I )) = n − (n − m) = m. Thus,

dimD

Y (I )

= dimD

ρ(I )

. As ρ(I ) ⊆ Y (I ), we conclude that ρ(I ) = Y (I ), and the proof iscomplete.

Problem 11.4. Prove or disprove that C[x, y]/

(x + y)2

is a principal ideal ring, where x and yare indeterminates.

Solution: We claim that the ring C[x, y]/

(x+y)2

is not a principal ideal ring. Firstly, we observe

that C[x, y]/

(x+y)2 C[u, v]/

u2

, where u and v are indeterminates, via the ring isomorphism

sending z +

(x+y)2 → z +

u2

, x+

(x+y)2 → u + v

2 +

u2

and y+

(x+y)2 → u − v

2 +

u2

,

where z

∈ C.

Next, observe that C[u, v]/

u2

is isomorphic to the polynomial ring R[v] via the ring isomorphism

sending z +

u2 → z · 1R, u +

u2 → u, and v +

u2 → v, where R := C[u]/

u2 · C[u]

, z ∈ C,

1R := 1+u2·C[u], and u := u+u2 ·C[u]. Note that R[v] is a commutative ring which is a

C[v], C[v]

-bimodule with generators 1R and u, with the multiplication rule 1R ·1R = 1R, 1R ·u = u = u·1R, andu · u = 0R, where 0R := u2 ·C[u]. Notice that elements of R[v] take the form p(v) ·1R + u ·q (v), where

p(v), q (v) ∈ C[v]. If p1(v), p2(v), q 1(v), q 2(v) ∈ C[v] satisfy p1(v) ·1R + u ·q 1(v) = p2(v) ·1R + u ·q 2(v),then it follows immediately that p1(v) = p2(v) and q 1(v) = q 2(v). Our task now is to show thatR[v] is not a principal ideal ring.

Consider the ideal J := u

·v, v2

·1R of R[v]. If J is principal, then there are f (v), g(v)

∈ C[v] such

that J = (((f (v) ·1R + u ·g(v)))). Consequently, u ·v = (α(v) · 1R + u · β (v)) · (f (v) · 1R + u · g(v)) andv2 · 1R = (γ (v) · 1R + u · δ (v)) · (f (v) · 1R + u · g(v)), for some α(v), β (v), γ (v), δ (v) ∈ C[v]. That is,

u · v =

α(v) · f (v)

· 1R + u ·

β (v) · f (v) + α(v) · g(v)

(55)

and

v2 · 1R =

γ (v) · f (v)

· 1R + u ·

δ (v) · f (v) + γ (v) · g(v)

. (56)

The first equation implies that α(v) · f (v) = 0, while the second equation implies that f (v) 0.Thus, α(v) = 0. That is, β (v) · f (v) = v , from the first equation. Without loss of generality, we

can assume that f (v) is monic. Hence, f (v) = 1 or f (v) = v.If f (v) = 1, then we have 1R + u · g(v) = f (v) · 1R + u · g(v) ∈ J =

u · v, v2

. Then, for some

λ(v), µ(v), ν (v), ρ(v) ∈ C[v], we must have

1R + u · g(v) =

λ(v) · 1R + u · µ(v)

· (u · v) +

ν (v) · 1R + u · ρ(v)

·

v2 · 1R

=

v2 · ν (v)

· 1R + u ·

v · λ(v) + v2 · ρ(v)

. (57)

Thus, 1 = v2 · ν (v), which clearly is absurd. Consequently, f (v) = v.

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Now, J = (((v · 1R + u · g(v)))). Then, for some λ(v), µ(v), ν (v), ρ(v) ∈ C[v], we must have

v · 1R + u · g(v) =

λ(v) · 1R + u · µ(v)

· (u · v) +

ν (v) · 1R + u · ρ(v)

·

v2 · 1R

= v2

· ν (v) ·

1R + u

· v

·λ(v) + v2

· ρ(v) . (58)

Hence, v = v2 · ν (v), a blatant contradiction. Therefore, J cannot be principal, so R[v] is not aprincipal ideal ring.

According to this proof, we see that C[x, y]/

(x + y)2

is not a principal ideal ring. The ideal

I :=

x2, x · y, y2

/

(x + y)2

is an ideal of C[x, y]/

(x + y)2

which is not principal.

Alternative Solution: We start with C[x, y]/

(x + y)2 C[u, v]/

u2

, where u and v are

indeterminates, as before. We claim that the ideal (u, v)/

u2

is not principal in C[u, v]/

u2

.

Suppose contrary that (u, v)/

u2

is principal in C[u, v]/

u2

. Therefore, there exists a polynomial

q (u, v) ∈ C[u, v] such that (u, v)/

u2

is generated by q (u, v) +

u2

, or (u, v) =

u2, q (u, v)

inC[u, v]. Consequently,

u = u2 · s(u, v) + q (u, v) · t(u, v) and v = u2 · σ(u, v) + q (u, v) · τ (u, v) (59)

for some s(u, v), t(u, v), σ(u, v), τ (u, v) ∈ C[u, v]. Without loss of generality, we may assume thatq (u, v) = q 1(v) + u · q 2(v), t(u, v) = t1(v) + u · t2(v), and τ (u, v) = τ 1(v) + u · τ 2(v) for someq 1(v), q 2(v), t1(v), t2(v), τ 1(v), τ 2(v) ∈ C[u, v]. That is,

u2 · (s(u, v) + q 2(v) · t2(v)) + u · (q 2(v) · t1(v) + q 1(v) · t2(v) − 1) + (q 1(v) · t1(v)) = 0 (60)

and

u2 · (σ(u, v) + q 2(v) · τ 2(v)) + u · (q 2(v) · τ 1(v) + q 1(v) · τ 2(v)) + (q 1(v) · τ 1(v) − v) = 0 (61)

The second equation suggests that q 1(v) ·τ 1(v) = v; hence, we can assume without loss of generalitythat q 1(v) = 1 or q 1(v) = v. Because the ideal (u, v) of C[u, v] is mapped to the zero ideal of C

under the evaluation homomorphism u → 0 and v → 0, we conclude that q 1(v) = v. The firstequation suggests that t1(v) = 0 and q 2(v) · t1(v) + q 1(v) · t2(v)−1 = 0, so q 2(v) ·0 + v · t2(v)−1 = 0,or v · t2(v) = 1, a contradiction.

That is, C[u, v]/

u2

, whence also C[x, y]/

(x + y)2

, is not a principal ideal ring. This proof

suggests that the ideal (x, y)/

(x + y)

2 of C[x, y]/

(x + y)

2 is not principal.

Problem 11.5. Let R be a commutative unital ring. Prove that R is local if and only if, for everyr, s ∈ R, the condition r + s = 1R implies that either r or s is a unit in R.

Solution: If R is a local ring, then the set of all nonunits of R is the unique maximal ideal M of R. Therefore, if r, s ∈ R such that r and s are nonunits, then we have r, s ∈ M . This meansr + s ∈ M . As 1R M , we get that r + s 1R. Hence, the direct implication is verified viacontrapositivity.

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Now, to show the converse, assume that, for r, s ∈ R, if r + s = 1R, then r or s is a unit of R.Let M be a maximal ideal of R. We want to show that 1R − M = R×, where R× is the set of allunits of R. By the hypothesis, 1R − M ⊆ R×. If there exists u ∈ R× such that u 1R − M , then1R −u is a nonunit that is not in M . The ideal M + (1R − u) ·R then properly contains M , whence

M + (1R − u) · R = R. That is, 1R = m + (1R − u) · ρ for some m ∈ M and ρ ∈ R. This means(1R − u) · ρ = 1R − m ∈ R×. Consequently, both 1R − u and ρ must be in R×. This contradictsthe fact that 1R − u is not a unit. That is, 1R − M = R×. Ergo, M = 1R − R× is the set of allnonunits of R, and so, R is local.

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Lecture Notes on Rings and Modules

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