Topics in algebra Chapter IV: Commutative rings and modules I - 1

Topics in algebra
Chapter IV: Commutative rings and modules I
Hsin-Ju Wang
Contents
4.1 Rings, ideals and homomorphisms.
4.2 Basic properties of commutative rings.
4.3 Rings of quotients and localization.
4.4 Modules, homomorphisms and exact sequences.
4.5 Projective modules and injective modules.
4.6 Hom and tensor product.
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1 Rings, ideals and homomorphisms
Definition 1.1 A ring is a nonempty set R together with two binary operations
(usually denoted as addition + and multiplication) such that :
(i) (R, +) is an additive (abelian) group;
(ii) (ab)c = a(bc) for all a, b, c ∈ R;
(iii) a(b + c) = ab + ac and (a + b)c = ac + bc for all a, b, c ∈ R.
If in addition:
(iv) ab = ba for all a, b ∈ R,
then R is said to be a commutative ring. If R contains an element 1R such
that
(v) 1Ra = a1R = a for all a ∈ R.
then R is said to be a ring with identity.
Definition 1.2 Let R be a ring.
(i) An element a ∈ R is said to be a zero divisor if there is a nonzero element
b ∈ R such that ab = 0 or ba = 0.
(ii) An element a ∈ R is said to be nilpotent if there is a positive integer n
such that a n = 0.
(iii) A nonzero element u ∈ R is said to be invertible or a unit if u has a
multiplicative inverse in R.
Definition 1.3 A commutative ring with identity and no zero divisor (except
0) is called an (integral) domain. A ring D with identity in which every
nonzero element is a unit is called a division ring. A field is a commutative
division ring.
Example 1.4 Z is an integral domain. Q, R and C are fields. 2Z is a
commutative ring without identity. Zn is a commutative ring with identity
and is a field if and only if n is prime.
Example 1.5 The polynomial ring over any field (e.g. Q or R or C) in one
variable is an integral domain.
Example 1.6 The n × n matrices (n ≥ 2) over any field F (e.g. Q or R or
C) form a noncommutative ring with identity (denoted by Mn(F )).
There is a famous division ring. The division ring of real quaternions.
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Example 1.7 Let R be the field of real numbers. Let D be the abelian group
R ⊕ R ⊕ R ⊕ R. Write the elements of D as formal sums (a0, a2, a2, a3) =
a0 + a1i + a2j + a3k. (i, j, k are symbols)
The addition in D is given by
(a0 + a1i + a2j + a3k) + (b0 + b1i + b2j + b3k)
= (a0 + b0) + (a1 + b1)i + (a2 + b2)j + (a3 + b3)k .
The multiplication in D is obtained by multiplying the formal sums term by
term subject to the following relations: (i) ri = ir, rj = jr, rk = kr for
all r ∈ R; (ii) i 2 = j 2 = k 2 = ijk = −1; ij = −ji = k; jk = −ki = i;
ki = −ik = j.
Example 1.8 Let G be an abelian group and End G be the set of all group
homomorphism from G to itself. Define addition in End G by (f + g)(a) =
f(a) + g(a) and multiplication in End G by composition of functions. Then
End G is a ring with identity.
Definition 1.9 Let R, S be rings. A function f : R −→ S is a homomorphism
of rings if for all a, b ∈ R:
f(a + b) = f(a) + f(b) and f(ab) = f(a)f(b).
f is called a monomorphism (resp. epimorphism, isomorphism) if f is injective
(resp. surjective, bijective). The kernel of f, denoted by kerf, is the set
{r ∈ R | f(r) = 0}.
We use the symbol ∼ = for ring isomorphisms.
Example 1.10 (i) The canonical map Z −→ Zn given by a −→ ā is an
epimorphism.
(ii) The canonical map Q[x] −→ Q[ √ 2] given by f(x) −→ f( √ 2) is an
epimorphism.
(iii) The identity map 2Z −→ Z given by a −→ a is a monomorphism.
Definition 1.11 Let R be a ring. If there is a least integer n such that na =
0 for all a ∈ R, then R is said to have characteristic n (write charR = n).
If no such n exist, then R is said to have characteristic 0 (write charR = 0).
Lemma 1.12 Let R be an integral domain; then charR is a prime number.
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Definition 1.13 Let (R, +, ·) be a ring and S a nonempty subset of R that
is closed under +, ·. If S itself a ring under + and ·, then S is called a
sub-ring of R. An sub-ring I of a ring R is a left ideal (denoted by I l R) if
r ∈ R and x ∈ I =⇒ rx ∈ I;
I is a right ideal (denoted by I r R) if
r ∈ R and x ∈ I =⇒ xr ∈ I;
I is an ideal (denoted by I R) if I is both a left and right ideal.
Example 1.14 Z is a sub-ring of Q, Q is a sub-ring of R and R is a sub-ring
of C.
Example 1.15 Let R be a ring then the set C(R) = {a ∈ R | ra = ar∀r ∈
R} is a sub-ring of R and is called the center of R. In general C(R) is not
an ideal of R. For example, the center of M2(R) is R and R is not an ideal
of M2(R).
Example 1.16 Any ideal of Z is of the form nZ.
Lemma 1.17 Every ring R may be embedded into a ring S with identity.
Proof. Let S be the abelian group R ⊕ Z and define multiplication in S by
(r1, n1)(r2, n2) = (r1r2 + n2r1 + n1r2, n1n2),
where ri ∈ R and ni ∈ Z. Then S is a ring with identity (0, 1) containing R.
Definition 1.18 Let X be a subset of a ring R. Let {Ii | i ∈ Λ} be the set
of all ideals of R which contain X. Then the ideal ∩i∈ΛIi is called the ideal
of R generated by the set X. This ideal is denoted (X). If X = {a}, then
(X) = (a) = {ra + as + na + n i=1 riasi | r, s, ri, si ∈ R, n ∈ Z}. In this case,
(a) is called a principal ideal of R.
We use aR for the set {ar | r ∈ R} and Ra for the set {ra | r ∈ R}. In general
a /∈ aR and a /∈ Ra. If R is commutative with identity, then aR = (a) = Ra.
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Example 1.19 Let f : R −→ S be a ring homomorphism; then ker f is an
ideal of R.
Example 1.20 Let R be a ring and I, J be ideals of R. We use IJ for the
smallest ideal containing both I and J. If R is commutative, then IJ =
{ n
i=1 aibi | ai ∈ I, bi ∈ J}.
A commutative ring with identity is called a principal ideal ring if every
ideal of the ring is principal.
Example 1.21 Z is a principal ideal domain (P.I.D. for short.).
Let R be a ring and I be an ideal of R; then the set of coset of I,
{a + I | a ∈ R}, forms a ring.
Lemma 1.22 Let R be a ring and I be an ideal of R; then the abelian group
R/I = {a + I | a ∈ R} is a ring with multiplication given by
(a + I)(b + I) = ab + I.
R/I is called the quotient ring of R with respect to I. If R is commutative,
then so is R/I.
Example 1.23 It is easy to see that the quotient ring Z/nZ is isomorphic
to Zn.
To end this section, we introduce the fundamental theorems of ring theory.
Theorem 1.24 (First isomorphism theorem) If f : R −→ S is a homomorphism
of rings, then f induces an isomorphism of rings R/kerf ∼ = Imf,
where Imf = {f(a) | a ∈ R}.
Theorem 1.25 (Second isomorphism theorem) Let I and J be ideals of a
ring R. Then there is an isomorphism of rings I/(I ∩ J) ∼ = (I + J)/J.
Theorem 1.26 (Third isomorphism theorem) Let I and J be ideals of a ring
R. If I ⊆ J, then J/I is an ideal of R/I and there is an isomorphism of
rings (R/I)/(J/I) ∼ = R/J.
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Problem set
1. Find a formula for the inverse of a nonzero element a0 +a1i+a2j +a3k
in the division ring of real quaternions.
2. Let R be a commutative ring with identity; then the set of all units of
R form a multiplicative subgroup of R.
3. Let R be a commutative ring; then the set of all nilpotent elements of
R is an ideal of R.
4. Let R be a ring and I be a left ideal of R then the set annR(I) = {r ∈
R | ra = 0∀a ∈ I} is an ideal of R.
5. Let R be a commutative ring and I be an ideal of R. Then the set
rad(I) = √ I = {r ∈ R | r n ∈ I for some n} is an ideal of R.
6. Let R be a ring and I be an ideal of R. Then the sets I : R = {r ∈
R | ar ∈ I∀a ∈ R} and 0 : I = {r ∈ R | ar ∈ 0∀a ∈ I} are ideals of R.
7. Let S be the ring of all 2 × 2 matrices over a field F . Prove the following:
a 0
(a) The center of S consists of all matrices of the form .
0 a
(b) The center of S is not an ideal of S.
What is the center of Mn(F )?
8. A ring R such that a 2 = a for all a ∈ R is called a Boolean ring. Prove
that every Boolean ring is commutative and a + a = 0 for all a ∈ R.
9. Let G = Z ⊕ Z. Prove that End G is a noncommutative ring with
identity.
10. Let G = {1, −1, i, −i, j, −j, k, −k} be the subset of the division ring
of real quaternions. Prove that G is a group.
11 Prove that a ring R with identity is a division ring if and only if R has
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no left ideal other than R and 0.
12. An element e ∈ R is said to be idempotent if e 2 = e. Find all idempotents
of M2(Q).
13. Let f : R −→ S be a ring homomorphism, I R and J S.
(a) Prove that f −1 (J) = {r ∈ R | f(r) ∈ J} is an ideal of R.
(b) If f is surjective, then f(I) is an ideal of S.
14. Find all idempotents of the ring Z17·19.
15. Let D be the real quaternions group. Let x = a0+a1i+a2j+a3k ∈ D;
then the conjugate of x is the element x ∗ = a0 − a1i − a2j − a3k. Prove that
(xy) ∗ = y ∗ x ∗ for every elements x, y ∈ D.
16. Use Exercise 15 to show the Euler’s Identity for real numbers ai, bi, ci, di,
i = 1, 2.:
(a 2 1 + b 2 1 + c 2 1 + d 2 1)(a 2 2 + b 2 2 + c 2 2 + d 2 2) = (a1a2 + b1b2 + c1c2 + d1d2) 2
+(a1b2 − b1a2 + c1d2 − d1c2) 2
+(a1c2 − c1a2 + d1b2 − b1d2) 2
+(a1d2 − d1a2 + c1b2 − b1c2) 2
17. Let R be a Boolean ring. Prove that a nonzero proper ideal P is a
prime ideal if and only if it is a maximal ideal.
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2 Basic properties of commutative rings.
From now on, every ring is commutative with identity.
Definition 2.1 Let R be a ring and P be an ideal of R. P is said to be
prime if P = R and for any ideal I, J of R, we have
IJ ⊆ P =⇒ I ⊆ P or J ⊆ P.
Prime ideals have the following properties.
Lemma 2.2 If P is an ideal in a ring R such that P = R, then P is prime
if and only if for all a, b ∈ R,
ab ∈ P =⇒ a ∈ P or b ∈ P. (1)
Proof. If P is a prime ideal and ab ∈ P for some a, b ∈ R, then (a)(b) ⊆ P ,
so that (a) ⊆ P or (b) ⊆ P , it follows that a ∈ P or b ∈ P . Conversely,
assume that P satisfies (1), and IJ ⊆ P for some ideals I, J of R. Assume
that I P , then there is an element a ∈ I such that a /∈ P but ab ∈ P for
all b ∈ J, it follows that J ⊆ P .
From the above lemma, it is easy to see the following:
Corollary 2.3 If P is an ideal in a ring R such that P = R, then P is
prime if and only if R/P is an integral domain.
Another important ideals are maximal ideals.
Definition 2.4 Let R be a ring and M be an ideal of R such that M = R.
M is said to be maximal if M is maximal with respect to the inclusion.
It is easy to show by Zorn’s lemma that R has at least one maximal ideal.
It is possible that R has more than one maximal ideals.
Lemma 2.5 If M is an ideal in a ring R such that M = R, then M is
maximal if and only if R/M is a field.
Proof. Observe first that R/M is a commutative ring with identity.
(⇒) Let M be a maximal ideal of R and a /∈ M; we need to show that ā
(= a + M) is a unit in R/M. However, (a) + M = R, there are r ∈ R and
b ∈ M such that 1 = ar + b, it follows that ā¯r = ¯1.
(⇐) Assume that R/M is a field. If N is an ideal of R such that M N,
then there is an element a ∈ N such that a /∈ M, so that there is an element
r ∈ R such that ā¯r = ¯1, it follows that R ⊆ (a) + M ⊆ N.
An immediately consequence is the following.
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Corollary 2.6 Every maximal ideal is prime.
Corollary 2.7 The following are equivalent for a ring R:
(i) R has only two ideals: R and 0.
(ii) R is a field.
(iii) 0 is a maximal ideal of R.
(iv) Every ring homomorphism from R to any ring is an injection.
Definition 2.8 Let R be an integral domain. An nonzero element a is said
to be irreducible if a is not a unit and if a = bc for some b, c ∈ R then b or c
is a unit. An nonzero element a is said to be prime if (a) is a prime ideal.
An easy consequence is that if a is prime then a is irreducible. If R is a
U.F.D. (unique factorization domain), which we will introduce later, then
the converse holds.
Example 2.9 Every prime ideal of Z is 0 or of the form pZ. Every maximal
ideal is of the form pZ.
Example 2.10 Let F [x] be the polynomial ring in one variable over a field
F . Every prime ideal of F [x] is 0 or of the form (f(x)), where f is an
irreducible polynomial of F [x]. Moreover, every maximal ideal of F [x] of the
form (f(x)), where f is an irreducible polynomial of F [x].
We next introduce a well-known theorem: Chinese Remainder Theorem.
Definition 2.11 Let {Ri | i ∈ I} be a nonempty family of rings and
i∈I Ri
the
direct product of the additive abelian groups Ri. If the multiplication of
i∈I Ri is defined by {ai}i∈I{bi}i∈I = {aibi}i∈I, then
i∈I Ri is a ring.
If I is a finite set in the above definition, say I = {1, . . . , n}, then we often
write the direct product ring as R1 × R2 × · · · × Rn.
Definition 2.12 Let R be a ring and I, J be ideals of R. I and J are said
to be co-maximal if I + J = R.
Example 2.13 If (n, m) = 1, then nZ and mZ are co-maximal.
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Lemma 2.14 Let I1, . . . , In be ideals of a ring R such that Ii and Ij are
co-maximal for i = j, Then R = Ii + ∩j=iIj for every i.
Proof. It suffice to show that R = I1 + ∩j≥2Ij. Notice that for every j ≥ 2,
R = I1+Ij. Therefore R = R n−1 =
j≥2 (I1+Ij) ⊆ I1+I2 · · · In ⊆ I1+∩j≥2Ij.
Theorem 2.15 (Chinese Remainder Theorem) Let R be a ring and I1, . . . , In
be ideals of R such that Ii and Ij are co-maximal for i = j. Then
R/(I1 ∩ · · · ∩ In) ∼ = R/I1 × R/I2 × · · · × R/In.
Proof. Let φ : R −→ R/I1 × R/I2 × · · · × R/In be the canonical map given
by φ(r) = (r + I1, . . . , r + In); then it is clear that φ is a ring homomorphism
and kerφ = I1 ∩ · · · ∩ In, therefore to finish the proof, it suffices to show that
φ is onto, or equivalently, to show that if b1, . . . , bn ∈ R, then there is an
element b ∈ R such that
b ≡ bi (mod Ii).
By Lemma 2.14, there are ai ∈ Ii, ri ∈ ∩j=iIj such that bi = ai + ri and
ri ≡ bi (mod Ii) and ri ≡ 0 (mod Ij) for j = i.
Now let b = r1 + · · · + rn; then it is easy to check that b ≡ bi (mod Ii) for
every i.
Corollary 2.16 Let m1, . . . , mn be positive integers such that (mi, mj) = 1
if i = j. If b1, . . . , bn are non-negative integers, then there is an integer b
such that
b ≡ bi (mod mi) ∀i.
The second part of this section is devoted to discussing the properties of
three kinds of rings. Namely, U.F.D., P.I.D. and Euclidean domain.
Definition 2.17 An integral domain R is a unique factorization domain
(U.F.D.) provided that:
(i) Any nonzero nonunit element a ∈ R can be written as a = c1 · · · cn with
c1, . . . , cn irreducible.
(ii) If a = c1 · · · cn and a = d1 · · · dm (ci, di irreducible) then n = m and
after a rearrangement, ci and di are associates for every i.
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In a domain R, two elements a, b are associates if there is a unit u ∈ R such
that a = bu.
Lemma 2.18 Let R be a U.F.D. and a ∈ R. If a is irreducible then a is a
prime element.
Proof. Let b, c ∈ R such that bc ∈ (a) and b /∈ (a); then bc = ad for some
d ∈ R. Since b /∈ (a), every irreducible factor of b is not associate with
a, therefore there is an irreducible factor of c that associates with a, thus
c ∈ (a).
Theorem 2.19 Let R be an integral domain; then R is a U.F.D. if and only
if the following hold:
(i) R has the ascending chain condition (a.c.c.) on principal ideals.i.e., if
(a1) ⊆ (a2) ⊆ · · ·
is an ascending chain, then there is an integer n such that (an) =
(an+1) = · · · .
(ii) Every irreducible element is a prime element.
Proof. (⇒) Suppose that R is a U.F.D.. By Lemma 2.18, (ii) is satisfied. To
prove (i), let (a1) ⊆ (a2) ⊆ · · · be an ascending chain of principal ideals. We
may assume that (ai) (ai+1) for every i. Write a1 = c1 · · · cn with c1, . . . , cn
irreducible. Then it is easy to see that (an) = (an+1) = · · · . Therefore the
ascending chain must stationary.
(⇐) We first show that every nonzero non-unit element can be written as a
finite product of irreducible elements. Suppose not. Then the set T = {a ∈
R | a = 0, a is not a unit, a is not a finite product of irreducible elements}
is not empty. Let a1 ∈ T ; then a is reducible, so that there are non-unit
b, c ∈ R such that a1 = bc, it follows that b ∈ T or c ∈ T by the definition
of T . We may assume b ∈ T and write a2 for b. Then (a1) (a2). By using
the fact that a2 ∈ T , we can construct a strictly ascending chain of principal
ideals, a contradiction. Therefore, we conclude that every nonzero nonunit
element a ∈ R can be written as a finite product of irreducible elements.
Now, suppose that a = c1 · · · cn and a = d1 · · · dm (ci, di irreducible). Since
c1 is a prime element and d1 · · · dm ∈ (c1), there is an integer i such that di ∈
(c1). However, d1 is irreducible, we see that c1 and di are associate. We may
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assume w.l.o.g. that i = 1 and d1 = c1. It follows that c2 . . . cn = d2 · · · dm.
Therefore, by induction, m = n and di is associated with ci for every i ≥ 2
after a rearrangement.
Theorem 2.20 Every principal ideal domain is a unique factorization domain.
Proof. Let R be a P.I.D.. We first show that R has the ascending chain
condition on principal ideals. Let (a1) ⊆ (a2) ⊆ · · · be an ascending chain of
principal ideals of R and let I = ∪ ∞ i=1(ai); then I is an ideal and I = aR for
some a ∈ R as R is a P.I.D.. Since a ∈ ∪ ∞ i=1(ai), there is an integer n such
that a ∈ (an), it follows that (a) = (an) = (an+1) = · · · .
To finish the proof, by Theorem 2.19, we need to show that every irreducible
element is a prime element. Let a be an irreducible element. If we can show
that (a) is a maximal ideal in R, then (a) is a prime ideal, and so a is a prime
element as desired. To show (a) is maximal, let (a) I; then I = (b) as R
is a P.I.D., so that a = bc for some nonunit c ∈ R, it follows that b is a unit
and I = R.
Definition 2.21 Let R be a ring and N be the set of all non-negative integers.
R is called a Euclidean ring if there is a map φ : R − {0} −→ N such
that
(i) if a, b ∈ R and ab = 0, then φ(a) ≤ φ(ab);
(ii) if a, b ∈ R and b = 0, then there exist q, r ∈ R such that a = bq + r with
r = 0, or r = 0 and φ(r) < φ(b).
A Euclidean ring which is an integral domain is called a Euclidean ring.
Example 2.22 The ring Z of integers with φ(x) = |x| is a Euclidean domain.
Example 2.23 If F is a field, let φ(x) = 1 for all x ∈ F . Then F is a
Euclidean domain.
Example 2.24 If F is a field, then the polynomial ring in one variable F [x]
is a Euclidean domain with φ(f) = degree(f).
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Example 2.25 Let R = Z[i] = {a + bi | a, b ∈ Z} (i = √ −1); then R is a
ring and is called the domain of Gaussian integers. Define φ(a+bi) = a 2 +b 2 .
Then it is easy to check that Z[i] is a Euclidean domain under φ.
Definition 2.26 Let R be a sub-ring of S and a ∈ S. a is said to be integral
over R if there is a monic polynomial f(x) ∈ R[x] such that f(a) = 0, or
equivalently, there are r0, . . . , rn−1 ∈ R such that a n + r1a n−1 + · · · + rn−1a +
rn = 0. S is said to be integral over R if every element of S is integral over
R.
Lemma 2.27 Let R be a sub-ring of S; then the set of elements of S integral
over R is a sub-ring of S, denoted by ¯ R S , called the integral closure of R in
S.
Definition 2.28 Let Q( √ d) be the quadratic field, where d is a square free
integer. Let Ad be the integral closure of Z in Q( √ d). Elements of Ad are
called integers of Q( √ d).
Lemma 2.29 Ad = Z[ √ d] if d ≡ 2, 3 (mod 4). Ad = Z[ 1 + √ d
] if d ≡
2
1 (mod 4).
Theorem 2.30 Any Euclidean domain is a principal ideal domain.
Proof. Let R be a Euclidean domain with φ : R − {0} −→ N and I be a
non-zero ideal of R. Let n = min{φ(a) | a ∈ I, a = 0}; then there is an
element b ∈ I such that φ(b) = n. It is easy to check that I = (b).
It is well-known that A−19 is a P.I.D. and is not a Euclidean domain.
Moreover, Ad is a P.I.D. if it is a U.F.D..
To end this section, we prove that if D is a U.F.D. then so is D[x], the
polynomial ring in one variable over D.
Definition 2.31 Let X be a non-empty subset of a ring R. An element
d ∈ R is said to be a greatest common divisor (g.c.d.) of X if
(i) d|a for all a ∈ X;
(ii) If c|a for all a ∈ X, then c|d.
Greatest common divisors are not always exist. However, any two greatest
common divisors of X are clearly associates.
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Lemma 2.32 If R is a U.F.D. then for every finite set {a1, . . . , an} of R
there exists a greatest common divisor.
Definition 2.33 If d is a greatest common divisor of a finite set {a1, . . . , an},
then we use d = (a1, . . . , an) for this fact. If (a1, . . . , an) = 1, then we say
that a1, . . . , an are relatively prime.
Definition 2.34 Let D be a U.F.D. and f =
n
aix i ∈ D[x] be a poly-
nomial. A greatest common divisor of the coefficients a0, . . . , an is called a
content of f and is denoted C(f). f is said to be primitive if C(f) is a unit.
In particular, if f is monic then f is primitive.
Lemma 2.35 (Gauss lemma) If D is a U.F.D. and f, g ∈ D[x], then C(fg) =
C(f)C(g). In particular, the product of any two primitive polynomials is
primitive.
Proof. Since f = C(f)f1 and g = C(g)g1 with f1, g1 primitive, C(fg) =
C(C(f)f1C(g)g1) = C(f)C(g)C(f1g1). Therefore, it suffices to show that
n
f1g1 is primitive. Let f1 = aix i m
and g1 = bix i m+n
; then f1g1 = ckx k ,
with ck =
i+j=k
i=0
i=0
i=0
k=0
aibj. If f1g1 is not primitive, then there is a prime element
p ∈ R such that p|ck for all k. Since C(f1) is a unit, there is an integer s
such that
p|ai for i < s and p ∤ as.
Similarly, there is an integer t such that
p|bi for i < t and p ∤ bt.
Since p is a factor of cs+t = b0as+t+b1as+t−1+· · ·+btas+· · ·+bs+t−1a1+bs+ta0,
asbt ∈ (p), a contradiction. Therefore, f1g1 is primitive.
Lemma 2.36 Let D be a U.F.D. with quotient field F and let f and g be
primitive polynomials of D[x] such that f and g are associate in F [x]; then
f and g are associate in D[x].
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Proof. By assumption, there are a, b ∈ D such that f = a/bg, or equivalently,
af = bg. We may assume that (a, b) = 1. If a is not a unit then there is a
prime factor of a that is a factor of C(g), which is impossible. Therefore a
is a unit in D. Similarly, b is a unit. Thus, we conclude that f and g are
associate in D[x].
Lemma 2.37 Let D be a U.F.D. with quotient field F and let f be primitive
polynomial of D[x] of positive degree such that f is irreducible in D[x]; then
f is irreducible in F [x].
Proof. Suppose that f is reducible in F [x]. Then there are polynomials
g, h ∈ F [x] of positive degree such that f = gh. Write g = a
b g1 and h = c
d h1
with a, b, c, d ∈ D and g1, h1 ∈ D[x]. Therefore, f = ac
bd g1h1. We may assume
that (ac, bd) = 1 and g1, h1 are primitive. From the above lemma, we see that
f and g1h1 are associate in D[x] and f is reducible in D[x], a contradiction.
Theorem 2.38 If D is a U.F.D. then so is D[x].
Proof. Let F be the quotient field of D.
Let f be a polynomial of D[x] of positive degree; then f = C(f)f1 for some
primitive polynomial of D[x]. Since F [x] is a P.I.D., f1 can be written as
g1 · · · gk for some irreducible polynomials of F [x]. Therefore, there are a, b ∈
D such that (a, b) = 1 and f1 = a
b h1 · · · hk for some primitive irreducible
polynomials of D[x]. However, f1 is primitive and (a, b) = 1, we see that
f1 = h1 · · · hk. Since C(f) is either a unit or a product of irreducible elements
of D. f can be written as a finite product of irreducible elements of D[x].
Let f be a polynomial of D[x] of positive degree. Assume that
f = c1 · · · cng1 · · · gk
= d1 · · · dmh1 . . . hl
where ci, di, gi, hi are irreducible in D[x]. Then gi and hi are primitive. Since
c1 · · · cn and d1 · · · dm are associate and D is a U.F.D., we see that m = n
and ci, di are associate after a rearrangement. W.l.o.g. we assume that
g1 · · · gk = h1 · · · hl. Since F [x] is a U.F.D., k = l and gi, hi are associate in
F [x] after a rearrangement. Therefore gi and hi are associate in D[x] as well
as gi and hi are primitive.
111
,

Theorem 2.39 (Eisenstein’s Criterion) Let D be a U.F.D. with quotient
field F . If f(x) =
n
aix i ∈ D[x], deg(f) ≥ 1 and p is an irreducible
i=0
element of D such that
p ∤ an; p|ai for i = 0, . . . , n − 1; p 2 ∤ a0,
then f is irreducible in F [x]. If f is primitive then f is irreducible in D[x].
Proof. Since C(f) is a unit in F , we may assume that f is primitive. Suppose
that f is reducible in F [x]. Then f is reducible in D[x]. Therefore there are
m
primitive polynomials g = bix i k
, h = cix i of D[x] of positive degree
i=0
such that f = gh. Since p|a0 and p 2 ∤ a0, we may assume that p|b0 and p ∤ c0.
Let t be the integer such that
i=0
p|bi for i = 0, . . . , bt; p ∤ bt+1.
Notice that t + 1 ≤ m < n as k ≥ 1. Therefore, since at+1 = bt+1c0 + btc1 +
· · · + b0ct+1 is a multiple of p, we conclude that bt+1c0 is a multiple of p, a
contradiction. Thus, f is irreducible in F [x].
Problem set
1. Determine all prime ideals and maximal ideals of Zn.
2. Let f : R −→ S be a ring homomorphism and Q is a prime ideal of S;
then f −1 (Q) is a prime ideal of R.
3. Let f : R −→ S be a ring epimorphism and Q is a maximal ideal of
S; then f −1 (Q) is a maximal ideal of R.
4. Let N be the ideal of all nilpotent elements of R; then R/N has no
nontrivial nilpotent element.
5. Find an integral domain R and an element a ∈ R such that a is irreducible
but a is not a prime element of R.
112

6. Find all solutions of the equations:
(a) x ≡ 1 (mod 3), x ≡ 1 (mod 5) and x ≡ 1 (mod 7).
(b) x ≡ 1 (mod 3), x ≡ 2 (mod 5) and x ≡ 3 (mod 7).
(c) x ≡ 1 (mod 2), x ≡ 4 (mod 5) and x ≡ 6 (mod 7).
(d) x ≡ 1 (mod 3), x ≡ 2 (mod 7) and x ≡ 3 (mod 11).
7. Prove that Z[i] is a Euclidean domain.
8. Prove that the integral closure of Z in Q is Z.
9. Find an example of a ring R so that there is a set X with no greatest
common divisor.
10. Let R be a P.I.D. and d is a greatest common divisor of {a1, . . . , an}.
Prove that d = r1a1 + · · · + rnan for some ri ∈ R.
11. Find a, b, c ∈ Z so that 6a + 10b + 15c = 1.
12 Let R = Z[ √ d] be the sub-ring of C, where d is a square free integer.
Let N : R −→ Z be the map given by N(a + b √ d) = a 2 − db 2 . Prove the
following:
(i) For any α, β ∈ R, N(αβ) = N(α)(β).
(ii) α is a unit if and only if N(α) = ±1.
13. Let R = {a + b √ 10 | a, b ∈ Z} be the sub-ring of R and N : R −→ Z
be the map given by N(a + b √ 10) = (a + b √ 10)(a − b √ 10) = a 2 − 10b 2 .
Prove the following:
(i) 2, 3, 4 + √ 10, 4 − √ 10 are irreducible elements of R.
(ii) 2, 3, 4 + √ 10, 4 − √ 10 are not prime elements of R.
(iii) R is not a U.F.D..
14. Let R = {a + b √ 5 | a, b ∈ Z} be the sub-ring of R and N : R −→ Z
be the map given by N(a + b √ 5) = (a + b √ 5)(a − b √ 5) = a 2 − 5b 2 . Prove
the following:
(i) 2, 1 + √ 5, 1 − √ 5 are irreducible elements of R.
(ii) 2, 1 + √ 5, 1 − √ 5 are not prime elements of R.
(iii) R is not a U.F.D..
113

15. Find all units of Z[i].
16. Find all units of Z[ √ −3].
17. Find all units of Z[ √ 2].
18. Let I be a proper ideal of a ring R; then there is a maximal ideal M
of R such that I ⊆ M.
19. A nonzero ideal in a principal ideal domain is maximal if and only if
it is prime.
20. If R is a U.F.D. then every nonzero prime ideal of R contains a
nonzero principal prime ideal.
21. Let R be a Euclidean ring and a ∈ R; then a is a unit if and only if
φ(a) = φ(1).
22. Let R = Z[ √ −3]; then 2 and 5 are irreducible in R but 3 and 7 are not.
23. rove the following polynomials are irreducible in [x]: (a) x 3 + x + 15.
(b) X 4 + nx + 1, where n is an odd integer.
24. Factor both x 8 − 1 and x 6 − 1 into irreducible factors in each of the
following rings: Z[x], Z2[x] and Z3[x].
25. Find the l.c.m. and g.c.d. of 11 + 3i and 8 − i in Z[i].
26. Use Eisenstein’s Criterion to show that x 2 + y 2 + 1 is irreducible in
C[x, y].
27. Find the prime factors of 5 in Z[i].
28. Let a = 3 + 8i, b = −2 + 3i ∈ Z[i]. Find c, d = x + yi ∈ Z[i] such that
a = cb + d, where either d = 0 or x 2 + y 2 < 13.
29. In Z[ √ −5], the greatest common divisor of 2 and 1 + √ −5 is 1. However,
the greatest common divisor of 6(1 − √ −5) and 3(1 + √ −5)(1 − √ −5)
114

does not exist.
30. In Z[i], find gcd(9 − 5i, −9 + 13i).
31. If R is a subring of a U.F.D., then is R a U.F.D.?
115

3 Rings of quotients and localization
Definition 3.1 Let R be a ring and S be a subset of R. S is said to be
multiplicative closed if 1 ∈ S and ab ∈ S whenever a, b ∈ S.
Example 3.2 Let R be a ring; then the following sets are all multiplicative
closed subsets of R.
(i) S = {a n | n ≥ 0}, where a is not a nilpotent element of R.
(ii) S = 1 + I = {1 + a | a ∈ I}, where I is an ideal of R.
(iii) S = R − ∪ n i=1Pi, where Pi is a prime ideal for every i.
(iv) S = {a ∈ R | a is not a zero divisor}.
Definition 3.3 Let R be a ring and S be a multiplicative subset of R. Define
a relation on the set
R × S = {(r, s) | r ∈ R, s ∈ S}
as follows: (a, s) ∼ (b, t) if and only if there is an element u ∈ S such that
u(at − bs) = 0. It is easy to show that ∼ is an equivalent relation. Let D =
(R × S)/ ∼; and we use [a, s] for the equivalent class {(b, t) | (b, t) ∼ (a, s)};
then any element in D is of the form [a, s]. Moreover, the addition on D is
given by
[a, s] + [b, t] = [at + bs, st]
and the multiplication is given by
[a, s] · [b, t] = [ab, st];
then it is easy to see that D is a commutative ring with identity [1, 1] and
0D = [0, 1]. For convenience, we replace [a, s] by a
s and D by RS. We call
RS the ring of quotient (fraction) of R with respect to S.
Remark 3.4 (a) If S consists of NZD (non-zero-divisor) then we can define
the equivalent relation by (a, s) ∼ (b, t) if and only if (at − bs) = 0.
(b) RS has the following universal property: Let R be a ring and S be a
multiplicative subset of R. Let f : R −→ R ′ be a ring homomorphism such
that f(a) is a unit in R ′ for every a ∈ S; then there is an unique ring
homomorphism ˜ f : RS −→ R ′ induced by f such that f = ˜ f ◦ iS, where
iS : R −→ RS is the canonical ring homomorphism given by iS(a) = a
1 .
116

Definition 3.5 Let R be a ring and S be the set of all NZD of R; then
the ring of quotient of R w.r.t. S is called the total quotient ring of R. In
particular, if R is a an integral domain then RS is the quotient ring of R.
Example 3.6 The quotient ring of Z is Q and the quotient ring of the polynomial
ring F [x] is F (x), where F is a field.
Example 3.7 Let a be a nonzero integer; then the set R = { m
| n ∈ N ∪
an {0}, m ∈ Z} is a ring between Z and Q as R = ZS and S = {an | n ∈ N∪{0}}.
Example 3.8 Let p be a prime number; then the set Z(p) = { m
| p ∤ n, m ∈
n
Z} is a ring between Z and Q as Z(p) = ZS and S = Z − pZ.
Let R be a ring and S be a multiplicative closed subset of R. In order
to discover the relation between ideals of R and ideals of RS, we study the
canonical ring homomorphism iS : R −→ RS. For an ideal I of R, we use Ie for the ideal of RS generated by iS(I), i.e., Ie = iS(I)RS. Let IS = { a
| a ∈
s
I, s ∈ S}; then it is easy to see that
I e = IRS = IS.
Moreover, for any ideal J of RS, we use J c for the ideal i −1
S (J) ( often written
as J ∩ R).
Proposition 3.9 Let iS : R −→ RS be the canonical ring homomorphism;
then the following hold:
(a) iS is a monomorphism if and only if S consists of non-zero-divisors.
(b) IS = RS if and only if I ∩ S = ∅.
(c) For every ideal J of RS, J ce = J. In particular, J = IS for some ideal
I of R.
(d) If P is a prime ideal of R such that P ∩ S = ∅, then P ec = P .
(e) If P is a prime ideal of R then PS = P RS is either RS or is a prime
ideal of RS.
117

Proof. (a) If S contains a non-zero zero-divisor, say s, then there is a nonzero
element a ∈ R such that as = 0, it follows that a ∈ ker(iS) as a
= 0 in RS.
1
If iS is not a monomorphism, then there is a non-zero element a ∈ R such
that a
1 = 0 in RS, or equivalently, there is an element s ∈ S such that as = 0.
It follows that s is a zero-divisor.
(b) If IS = RS, then 1 a
= for some a ∈ I and s ∈ S, so that there is an t ∈ S
1 s
such that t(a − s) = 0, it follows that st ∈ I and then I ∩ S = ∅. Conversely,
if I ∩ S = ∅, then there is an element a ∈ I ∩ S, so that 1 a
=
1 a ∈ IS, it
follows that RS = IS.
(c) It is clear that J ce ⊆ J. Let a a
∈ J; then
s 1 ∈ J so that a ∈ J c , it follows
that a 1 a
= ·
s s 1 ∈ J ce .
(d) It is clear that P ⊆ P ec . Conversely, let a ∈ P ec ; then a b
= for some
1 s
b ∈ P and s ∈ S, so that there is an t ∈ S such that t(as − b) = 0, it follows
that ast ∈ P . However, P ∩ S = ∅, we conclude that a ∈ P and P ec ⊆ P .
(e) If P ∩ S = ∅ then PS = RS by (b). So, we may assume that P ∩ S = ∅.
Let a b
·
s t ∈ PS; then ab ∈ P ec = P by (d), so that a ∈ P or b ∈ P , it follows
that a
s ∈ PS or b
∈ PS.
t
From Proposition 3.9, we see that if I is an ideal of R with I ∩ S = ∅ then
I = I ec does not hold in general. However, if P is a prime ideal of R with
P ∩ S = ∅, then P ec = P . Therefore we may ask that what kind of ideals in
R will satisfy I ec = I. The answer is primary ideals.
Definition 3.10 Let Q be an ideal of a ring R; Q is said to primary if
ab ∈ Q then either a ∈ Q or b n ∈ Q for some positive integer n.
Lemma 3.11 Let Q be a primary ideal of a ring R; then √ Q is a prime
ideal.
Proof. Let P = √ Q and ab ∈ P with b /∈ P ; then b n /∈ Q for every n, so that
by the definition of primary ideal a ∈ Q ⊆ P .
From now on, if Q is a primary ideal of a ring R with √ Q = P , then we say
that Q is a P -primary ideal.
Corollary 3.12 The following are equivalent for an ideal of a ring R:
118

(a) Q is a P -primary ideal.
(b) √ Q = P is a prime ideal and if ab ∈ Q with b /∈ P then a ∈ Q.
Lemma 3.13 Let iS : R −→ RS be the canonical ring homomorphism; then
for every P -primary ideal Q, we have Q ec = Q and √ QS = PS if P ∩ S = ∅.
Proof. It is clear that Q ⊆ Qec . Conversely, let a ∈ Qec ; then a b
=
1 s for
some b ∈ Q and s ∈ S, so that there is an t ∈ S such that t(as − b) = 0, it
follows that ast ∈ Q. However, P ∩ S = ∅, we conclude by Corollary 3.12
that a ∈ Q and Qec ⊆ Q.
To show that QS is PS-primary; let a b
·
s t ∈ QS with b
t /∈ √ QS; then ab ∈
Qec = Q with b /∈ P , so that a ∈ Q, it follows that a
s ∈ QS. Moreover, it is
easy to see that √ QS = PS.
From the above, we obtain the following:
Corollary 3.14 Let R be a ring and S be a multiplicative closed subset of
R; then there is a one to one correspondence between
and
{Q | Q is a primary ideal of R with Q ∩ S = ∅}
{Q | Q is a primary ideal of RS}.
Definition 3.15 Let R be a ring; then R is said to be local if R has only one
maximal ideal. R is said to be semi-local if R has only finitely many maximal
ideals.
Example 3.16 Let p be a prime number; then the set Z(p) = { m
n
n, m, n ∈ Z} is a local ring.
| p ∤
Proof. Let P = { m
| p ∤ n, p|m, m, n ∈ Z}; then it is easy to see that P is an
n
ideal of Z(p). Moreover, if I is an ideal of Z(p) such that I P , then there is
an a ∈ I such that a /∈ P . However, a = m
for some m, n such that p ∤ m
n
and p ∤ n, therefore a is a unit of Z(p) and I = Z(p).
Lemma 3.17 The following are equivalent for a ring R:
119

(a) R is a local ring.
(b) The set of non-units of R is contained in a proper ideal of R.
(c) The set of non-units of R forms an ideal.
Proof. Let T be the set of all non-units of R.
(a) ⇒ (b) Let P be the unique maximal ideal of R and a /∈ P . If a is not
a unit, then (a) ⊆ M for some maximal ideal of R, so that a ∈ M = P , a
contradiction. Therefore a is a unit and T ⊆ P .
(b) ⇒ (c) Suppose that T ⊆ P for some proper ideal P . As a proper ideal,
P consists of non-units, therefore P ⊆ T and then T = P is an ideal of R.
(c) ⇒ (a) By assumption, T is an ideal of R. Let I be an ideal of R such
that I T ; then there is an element a ∈ I \ T , so that a is a unit and
I = R. Therefore every proper ideal of R is contained in T and T is the
unique maximal ideal of R.
Example 3.18 Let R be a ring and S = R − ∪ n i=1Pi, where Pi is a prime
ideal of R for every i; then RS is a semi-local ring and the maximal ideals of
RS is contained in {P e 1 , . . . , P e n}. If S = R − P for some prime P of R, then
RS (often written as RP ) is a local ring with maximal ideal PS = P RP .
The ring RP is called the localization of R at P . There is a well-known
local ring: the formal power series ring.
Definition 3.19 Let F be a field; then the set
∞
F [[x]] = {
i=0
aix i | ai ∈ F }
of all formal power series is a ring under the formulas:
and
where ck =
i+j=k
∞
aix i +
i=0
∞
bix i =
i=0
∞
(ai + bi)x i
i=0
∞
( aix i ∞
) · ( bix i ) =
i=0
i=0
∞
ckx k ,
k=0
aibj = a0bk + a1bk−1 + · · · + akb0.
120

Example 3.20 Let F be a field; then the formal power series ring F [[x]] is
a local ring.
Proof. It is enough to show that (x) is the unique maximal ideal of F [[x]]. For
∞
this, let f =
i=0
aix i /∈ (x); then a0 = 0. Let b0 = a −1
0 and bi = −a −1
0 (bi−1a1+
∞
· · · + b0ai) for every i ≥ 1; then it is easy to see that f · ( bix i ) = 1 and f
is a unit in F [[x]].
Problem set
1. Let R be an integral domain and S be a multiplicative subset of R.
Define a relation on the set R × S as follows: (a, s) ∼ (b, t) if and only if
(at − bs) = 0. Prove that ∼ is an equivalent relation.
2. Find a ring R and a multiplicative closed subset S of R such that iS
is not 1-1.
3. Find a ring R and a multiplicative closed subset S of R such that there
are ideals I = J of R such that IS = JS.
4. Let p be a prime number. Prove that any (p)-primary ideal of Z is of
the form (p n ).
5. Let R be a ring and S be a multiplicative subset of R. Prove that
(I1 ∩ I2) e = I e 1 ∩ I e 2, (I1I2) e = I e 1I e 2 and (I1 + I2) e = I e 1 + I e 2.
6. Prove that if R is a P.I.D., then so is RS, where S is a multiplicative
closed subset of R.
7. Let R be a ring and S be a multiplicative subset of R. Let I be an
ideal of R; then ( √ I)S = √ IS.
8. If S = {2, 4} and R = Z6, then RS ∼ = Z3.
121
i=0

9. Let R be a ring. Then R is local if and only if for all r, s ∈ R, r +s = 1
implies r or s is a unit.
10. Let R be a ring and M be a maximal ideal of R; then R/M n is a
local ring, where n is a positive integer.
11. Let R be a ring. Prove that the following are equivalent:
(a) R has a unique prime ideal.
(b) Every nonunit is nilpotent.
(c) R has a unique minimal prime which contains all zero-divisors, and all
non-units of R are zero-divisor.
12. Let R and R ′ be rings. R ′ is said to a homomorphic image of R if
there is a epimorphism from R to R ′ . Prove that every homomorphic image
of a local ring is local.
13. Let F be a field. Prove that F [x]S is a local ring, where S =
n
{ aix i | a0 = 0, ai ∈ F }.
i=0
14. Find the inverse of 1 + x in the power series ring F [[x]].
15. Let S and T be two multiplicative closed subsets of a ring R. Prove
that RST ∼ = (RS)iS(T ), where iS is the canonical ring homomorphism from R
to RS.
16. Let R be a ring. Prove that if a, b ∈ R are NZDs then so is ab.
122

4 Modules, homomorphisms and exact sequences
Definition 4.1 Let R be a ring. A left R-module is an additive abelian group
M together with a function R × M −→ M (the image of (r, x) is denoted by
rx) such that for all r, s ∈ R and x, y ∈ M:
(a) r(x + y) = rx + ry.
(b) (r + s)x = rx + sx.
(c) r(sx) = (rs)x.
(d) 1Rx = x.
Right R-modules are defined in a similar fashion.
Definition 4.2 Let R be a ring. M is an R-module if M is both a left and
a right R-module, and rx = xr for all r ∈ R and x ∈ M.
Example 4.3 Let G be an abelian group then G is a Z-module.
Example 4.4 Let F be a field; then any F -module is a vector space over F .
Example 4.5 Let R be a ring; then R is an R-module and every ideal of R
is an R-module.
Example 4.6 Let {Mi | i ∈ Λ} be R-modules; then,
Mi, the direct prod-
uct of the abelian groups Mi is an R-module with the action of R given by
r{ai} = {rai}.
Example 4.7 Let R be a ring and R n = ⊕ n i=1R be the direct sum of n copies
of the abelian group R; then R n is an R-module with the action of R given
by r(a1, · · · , an) = (ra1, · · · , ran).
Example 4.8 Let f : R −→ S be a ring homomorphism; then S can be
viewed as an R-module by define ra = f(r)a for all r ∈ R, a ∈ S. In
particular, every quotient ring R/I (I is an ideal of R) is an R-module as
r(a + I) = ra + I.
123
i∈Λ

Example 4.9 Let M be an R-module and I be an ideal of R; then the set
n
IM = { aixi | ai ∈ I, xi ∈ M} is an R-module.
i=1
Definition 4.10 Let M and N be two R-modules. A function f : M −→ N
is a homomorphism of R-modules (R-homomorphism or R-linear map) if for
all x, y ∈ M, r ∈ R:
f(x + y) = f(x) + f(y) and f(rx) = rf(x).
f is called a monomorphism (resp. epimorphism, isomorphism) if f is injective
(resp. surjective, bijective). The kernel of f, denoted by kerf, is
the set {x ∈ M | f(x) = 0}. The image of f, denoted by Imf, is the set
{f(x) | x ∈ M}.
Notice that if f : M −→ N is an R-module homomorphism, then f is a
monomorphism if and only if kerf = 0.
In the sequel, we use the symbol ∼ = for R-module isomorphisms.
Definition 4.11 Let R be a ring and M be an R-module. A subgroup N of
M is a submodule of M if rx ∈ N whenever r ∈ R and x ∈ N.
Definition 4.12 Let N ⊆ M be R-modules; then the abelian group M/N =
{x + N | x ∈ M} is an R-module with r(x + N) = rx + N for all x ∈ M and
r ∈ R.
Example 4.13 Let f : nZ −→ mZ given by f(na) = ma is a Z-module
isomorphism. Hence nZ ∼ = mZ as Z-module.
Example 4.14 Let R be a ring; then every ideal of R is a sub-module of R.
Example 4.15 Let f : M −→ N be an R-module homomorphism; then
kerf is a submodule of M and Imf is a submodule of N.
Example 4.16 Let M be an R-module and I be an ideal of R, then IM is
a submodule of M. Moreover, if x ∈ M, then xR is a submodule of M.
Example 4.17 Let {Mi | i ∈ Λ} be R-modules; then ⊕i∈ΛMi, the direct sum
of the abelian groups Mi, is a submodule of
Mi.
124
i∈Λ

As in ring theory, we have some fundamental theorems for modules.
Theorem 4.18 If f : M −→ N is a homomorphism of R-modules, then f
induces an isomorphism of R-modules M/kerf ∼ = Imf.
Theorem 4.19 Let M and N be R-submodules of an R-modules L. Then
there is an isomorphism of R-modules M/(M ∩ N) ∼ = (M + N)/N.
We next discuss the notion: generators.
Definition 4.20 Let M an R-module and X be a non-empty subset of M;
then the submodule of M generated by X, is denoted by (X), is the intersection
of all submodules of M containing X, or equivalently, the set
n
{ rixi | ri ∈ R, xi ∈ X}. If X = {x1, . . . , xn} is a finite set, then
i=1
(X) = (x1, . . . , xn)R = x1R + · · · + xnR.
Definition 4.21 An R-module M is said to be finitely generated (f.g. for
short) if there are x1, . . . , xn ∈ M such that M is generated by these elements.
M is said to by cyclic if M is generated by a single element x, i.e., M = xR.
Lemma 4.22 Let M be an R-module. M is cyclic if and only if M ∼ = R/I
for some ideal I of R.
Proof. (⇒) By assumption, there is an element x ∈ M such that M = xR.
Let φ : R −→ xR be the map given by φ(r) = xr; then it is easy to see that
φ is an R-module epimorphism. Let I = kerφ; then R/I ∼ = M as R-module.
(⇐) If M ∼ = R/I for some ideal I of R then since R/I is generated by 1 + I,
M is cyclic.
Lemma 4.23 Let M be an R-module generated by {x1, . . . , xn}; then there
is an epimorphism from R n to M.
Proof. Let φ : R n −→ M be the map given by φ((a1, . . . , an)) =
n
aixi;
then it is easy to check that φ is an epimorphism of R-modules.
An important tool in studying modules theory is the notion: exact sequences.
125
i=1

Definition 4.24 A pair of modules homomorphisms, L f
−→ M g
−→ N is
said to be exact at M if kerg = Imf. A finite sequence of modules homo-
morphisms, M0
f1
−→ M1
f2
−→ · · · fn−1
−→ Mn−1
fn
−→ Mn is said to be exact (resp.
a complex) if kerfi+1 = Imfi (resp. Imfi ⊆ kerfi+1) for i = 1, 2, . . . , n − 1.
An infinite sequence of modules homomorphisms, · · · fi−1
−→ Mi−1
fi
−→ Mi
fi+1
−→
Mi+1
fi+2
−→ · · · is said to be exact (resp. a complex) if kerfi+1 = Imfi (resp.
Imfi ⊆ kerfi+1) for every i.
Definition 4.25 A finite sequence of modules homomorphisms, 0 −→ L f
−→
M g
−→ N −→ 0 is said to be a short exact sequence (s.e.s. for short ) if it is
exact.
Remark 4.26 If 0 −→ L f
−→ M g
−→ N −→ 0 is a s.e.s. then f is 1-1 and
g is onto.
Example 4.27 Let R be a ring and I be an ideal of R; then
0 −→ I
i
−→ R π
−→ R/I −→ 0
is a s.e.s., where i is the inclusion map and π is the canonical surjective map.
Example 4.28 Let f : M −→ N be an R-homomorphism; then f induces
an exact sequence:
0 −→ kerf
i
−→ M f
−→ N π
−→ N/Imf −→ 0,
where i is the inclusion map and π is the canonical surjective map.
Example 4.29 Let M and N be R-modules; then
0 −→ M f
−→ M ⊕ N g
−→ N −→ 0
is a s.e.s., where f is the R-homomorphism given by f(x) = (x, 0) and g is
the R-homomorphism given by g(x, y) = y, where x ∈ M and y ∈ N.
There are several important results of short exact sequences.
Lemma 4.30 Let 0 −→ L f
−→ M g
−→ N −→ 0 be a s.e.s.; then the following
hold:
126

(a) If M is f.g. then so is N.
(b) If L and N are f.g. then so is M.
Proof. (a) If M is generated by {x1, . . . , xn}, then it is clear that N is generated
by {g(x1), . . . , g(xn)}.
(b) Suppose that L is generated by {x1, . . . , xn} and N is generated by
{y1, . . . , yk}. Let {z1, . . . , zk} ⊆ M such that g(zi) = yi for every i; then
we shall see that M is generated by {z1, . . . , zk, f(x1), . . . , f(xn)}. To show
this, let x ∈ M; then g(x) ∈ N, so that there are r1 . . . , rk ∈ R such that
k
k
g(x) = riyi, it follows that x − rizi ∈ kerg = Imf. Therefore, there
i=1
are s1, . . . , sn ∈ R such that x −
i=1
k
rizi =
i=1
n
sif(xi). Thus, we conclude
that M is generated by {z1, . . . , zk, f(x1), . . . , f(xn)}.
There is a well-known result about s.e.s.: called The snake’s lemma.
Lemma 4.31 (The snake’s lemma) Let
i=1
0 −→ L
f
−→ M g
−→ N −→ 0
↓ α ↓ β ↓ γ
0 −→ L ′ f ′
−→ M ′ g ′
−→ N ′ −→ 0
be a commutative diagram of R-modules and R-homomorphisms such that
each row is a s.e.s.. Then there is a long exact sequence
0 −→ kerα f1
−→ kerβ g1
−→ kerγ d
−→ L ′ /Imα f ′ 1 ′
g
−→ M /Imβ ′ 1 ′
−→ N /Imγ −→ 0,
where f1 (resp. g1) is an R-homomorphism induced by f (resp. g) and f ′ 1
(resp. g ′ 1) is an R-homomorphism induced by f ′ (resp. g ′ ).
Proof. (Sketch) 0 −→ kerα f1
−→ kerβ is exact:
Let x ∈ kerα such that f1(x) = 0; then x = 0 as f is 1-1, so that
kerf1 = 0.
kerα f1
−→ kerβ g1
−→ kerγ is exact:
Let x ∈ kerg1; then g(x) = 0. Since kerg = Imf, there is an element
y ∈ L such that x = f(y), it follows that (f ′ ◦ α)(y) = (β ◦ f)(y) = β(x) = 0.
Since f ′ is 1-1, α(y) = 0 and then y ∈ kerα. Therefore x = f1(y).
127

M ′ /Imβ g′ 1 ′ ′ −→ N /Imγ −→ 0 is exact as g is onto.
L ′ /Imα f ′ 1 ′
g
−→ M /Imβ ′ 1 ′ −→ N /Imγ is exact:
Let x + Imβ ∈ kerg ′ 1; then g ′ (x) ∈ Imγ, so that there is an element
y ∈ N such that g ′ (x) = γ(y). Since g is onto, there is an element x ′ ∈ M
such that g(x ′ ) = y. It follows that g ′ (x − β(x ′ )) = g ′ (x) − (γ ◦ g)(x ′ ) = 0.
Therefore there is an element z ∈ L ′ such that f ′ (z) = x − β(x ′ ). Hence
f ′ 1(z + Imα) = x + Imβ.
To finish the proof, we need to know the definition of d. The map d :
kerγ −→ L ′ /Imα is given as follows: Let z ∈ kerγ. Since g is onto and
z ∈ N, there is an element y ∈ M such that g(y) = z. Furthermore,
(g ′ ◦ β)(y) = (γ ◦ g)(y) = 0, therefore β(y) ∈ kerg ′ = Imf ′ . Hence, there
is an element x ∈ L ′ such that f ′ (x) = β(y). Now, d(z) = x + Imα.
From the above, we need to show that d is well-defined, i.e., the definition is
independent of the choice of y. Suppose there is another element y ′ ∈ M such
that g(y ′ ) = z. Then β(y ′ ) = f ′ (x ′ ) for some x ′ ∈ L ′ . On the other hand,
since y−y ′ ∈ kerg = Imf, there is an element x ′′ ∈ L such that f(x ′′ ) = y−y ′ .
Since f ′ is 1-1, α(x ′′ ) = x − x ′ . This shows that x + Imα = x ′ + Imα.
kerβ g1
−→ kerγ d
−→ L ′ /Imα is exact:
Let z ∈ kerγ and z ∈ kerd. Let x ∈ L ′ and y ∈ M be such that g(y) = z
and f ′ (x) = β(y). Since d(z) = 0, x ∈ Imα, so there is an element x ′ ∈ L
such that x = α(x ′ ). It follows that β(y) = (f ′ ◦ α)(x ′ ) = (β ◦ f)(x ′ ).
Therefore y − f(x ′ ) ∈ kerβ. Since g(y − f(x ′ )) = g(y) = z, z ∈ Img1.
−→ L ′ /Imα f ′ 1
Let x ∈ L ′ such that f ′ (x) ∈ Imβ; then there is an element y ∈ M such
that β(y) = f ′ (x). Since (g ′ ◦ f ′ ) = 0, (γ ◦ g)(y) = (g ′ ◦ β)(y) = 0. Let
z = g(y); then z ∈ kerγ, so that by the definition of d, d(z) = x + Imα and
kerf ′ 1 ⊆ Im(d).
kerγ d
−→ M ′ /Imβ is exact:
Corollary 4.32 (The short five lemma) Let
0 −→ L
f
−→ M g
−→ N −→ 0
↓ α ↓ β ↓ γ
0 −→ L ′ f ′
−→ M ′ g ′
−→ N ′ −→ 0
be a commutative diagram of R-modules and R-homomorphisms such that
each row is a s.e.s.. Then the following hold:
(a) If α and γ are monomorphism then so is β.
128

(b) If α and γ are epimorphism then so is β.
(c) If α and γ are isomorphism then so is β.
Theorem 4.33 Let 0 −→ L f
−→ M g
−→ N −→ 0 be a s.e.s.; then the
following conditions are equivalent:
(a) There is an R-homomorphism g ′ : N −→ M with g ◦ g ′ = idN.
(b) There is an R-homomorphism f ′ : M −→ L with f ′ ◦ f = idL.
(c) The s.e.s. is isomorphic (with identity maps on L and N) to the s.e.s.
0 −→ L i
−→ L ⊕ N π
−→ N −→ 0. In particular, M ∼ = L ⊕ N.
Proof. (a)⇒(c) The homomorphisms f and g ′ induce a module homomorphism
φ : L ⊕ N −→ M, given by φ(a, b) = f(a) + g ′ (b). It is easy to see
that the following diagram
0 −→ L
i
−→ L ⊕ N π
−→ N −→ 0
↓ idL ↓ φ ↓ idN
0 −→ L
f
−→ M
g
−→ N −→ 0
is commutative. By Corollary 4.32, φ is an isomorphism.
(b)⇒(c) The homomorphisms f ′ and g induce a module homomorphism ψ :
M −→ L ⊕ N, given by φ(a) = (f ′ (a), g(a)). It is easy to see that the
following diagram
0 −→ L
f
−→ M
g
−→ N −→ 0
↓ idL ↓ ψ ↓ idN
0 −→ L
i
−→ L ⊕ N π
−→ N −→ 0
is commutative. By Corollary 4.32, ψ is an isomorphism.
(c) ⇒ (a) and (c) ⇒ (b) are easy.
Definition 4.34 Let M be an R-module. M is said to be simple if M has
no submodules other than 0 and M itself.
Example 4.35 If p is a prime number then Z/pZ is a simple Z-module. In
fact, if m is a maximal ideal of a ring R, then R/m is a simple R-module.
129

Lemma 4.36 Let M be a simple R-module; then there is a maximal ideal m
of R such that M ∼ = R/m.
Definition 4.37 Let M be an R-module. M is said to be a finite length
module if there is a finite chain of R-submodules of M:
0 = M0 ⊆ M1 ⊆ · · · ⊆ Mn−1 ⊆ Mn = M
such that Mi+1/Mi is simple for i = 0, . . . , n − 1. If Mi = Mi+1 for every i,
we say that the length of M is n and denoted by lR(M) = n.
We will show later that lR(M) is well-defined.
Example 4.38 Z/30Z is a finite length module as
0 ⊆ 15Z/30Z ⊆ 5Z/30Z ⊆ Z/30Z
and 15Z/30Z ∼ = Z/2Z, 5Z/30Z
15Z/30Z ∼ = Z/3Z and Z/30Z
5Z/30Z ∼ = Z/5Z. In particular,
lZ(Z/30Z) = 3.
Later we will show that if 0 −→ L f
−→ M g
−→ N −→ 0 is a s.e.s. of
R-modules, then M is a finite length R-module if and only if L and N are
both finite length R-modules. If this is the case, lR(M) = lR(L) + lR(N).
Problem set
1. Let M be an R-module; prove that annRM = {r ∈ R | rx = 0 ∀x ∈
M} is an ideal of R.
2. Let a be an element of a ring R and M be an R-module; then
(0 :M a) = {x ∈ M | ax = 0} is a submodule of M.
3. Let M be an R-module and x ∈ M; then (0 :R x) = {r ∈ R | rx = 0}
is an ideal of R.
4. Let m be a factor of n and k = n
m . Prove that mZ/nZ ∼ = Z/kZ.
130

5. Find a s.e.s. 0 −→ L f
−→ M g
−→ N −→ 0 such that M is f.g. but L
is not f.g..
6. Let M be a finite length R-module and P is a prime ideal of R but
not maximal. Prove that there is no monomorphism from R/P to M.
7. Let R be a local ring with maximal ideal m and M be a finite length
R-module. Prove that there is a module monomorphism from R/m to M.
(Hint: Find a simple R-submodule of M.)
8. Let M be an R-module and I be an ideal of R; then the following
conditions are equivalent:
(a) There is an R-monomorphism from R/I to M.
(b) There is an element x ∈ M such that I = (0 :R x) = {r ∈ R | rx = 0}.
9. Let M be an R-module and I = annRM; then M can be viewed as an
R/I-module.
10. Prove that L f
−→ M g
−→ N is a complex if and only if g ◦ f = 0.
11. If 0 −→ L f
−→ M g
−→ R −→ 0 is a s.e.s. of R-modules, then
M ∼ = L ⊕ R.
12. Find lZ(Z/120Z) and lZ(Z/200Z).
13. Let M be an R-module and x ∈ M; then the set annR(x) = {r ∈
R | rx = 0} is an ideal of R. If annR(x) = 0, then x is said to be a torsion
element of M. Prove that if R is an integral domain then the set T (M) of
all torsion elements of M is a submodule of M.
14. Let M be an R-module and I be an ideal of R. If IM = 0, then M
can be viewed as R/I module.
131

5 Projective modules and injective modules
We first introduce the notion: free modules.
Definition 5.1 Let F be an R-module. F is a free module if there is a
subset X = {ei | i ∈ Λ} (which is called a basis) of F such that F = (X),
i.e., for every x ∈ F , there are e1, . . . , en ∈ X and r1, . . . , rn ∈ R such
n
that x = riei, and that X is linear independent, i.e., for every finite set
i=1
{e1, . . . , en} of X with
F ∼ = ⊕i∈ΛR.
n
riei = 0, then ri = 0 for every i. In this case,
i=1
If F is free with a finite basis then F ∼ = R n for some n and F is called a free
R-module of rank n.
We now introduce the notion: projective modules.
Definition 5.2 Let P be an R-module. P is said to be projective if for any
exact sequence M g
−→ N −→ 0 and R-homomorphism f : P −→ N there is
an R-homomorphism h : P −→ M such that g ◦ h = f.
It is easy to see that free modules are projective.
There are several equivalent characterizations for projective modules.
Proposition 5.3 Let R be a ring. Then the following are equivalent for an
R-module P :
(a) P is projective.
(b) Any s.e.s. of the form 0 −→ L f
−→ M g
−→ P −→ 0 splits.
(c) P is a direct summand of some free R-module F .
Proof. (a) ⇒ (b) Let 0 −→ L f
−→ M g
−→ P −→ 0 be a s.e.s.. Let i : P −→ P
be the identity map; then there is a R-homomorphism h : P −→ M such
that g ◦ h = i, so that g ◦ h = idP ,i.e., the s.e.s. splits.
(b) ⇒ (c) Let X = {xi | i ∈ Λ} be a generating set of P ; then there is a free
module F = ⊕i∈ΛR and a R-epimorphism g : F −→ P . Let K be the kernel
of g; then 0 −→ K i
−→ F g
−→ P −→ 0 is a s.e.s., so that the sequence splits
132

y assumption, it follows that F ∼ = K ⊕ P and P is a direct summand of F .
(c) ⇒ (a) Let P be a direct summand of a free R-module F . Then there are
R-monomorphism i : P −→ F and R-epimorphism π : F −→ P such that
π ◦ i = idP . Let M g
−→ N −→ 0 be an exact sequence and f : P −→ N be
an R-homomorphism. Since f ◦ π : F −→ N is an R-homomorphism and F
is free, there is an R-homomorphism h : F −→ M such that g ◦ h = f ◦ π,
therefore h ◦ i is an R-homomorphism from P to M such that g ◦ (h ◦ i) =
f ◦ π ◦ i = f.
Projective modules are not free modules in general as the following example
suggests.
Example 5.4 Let R = Z6; then Z2 and Z3 are Projective R-modules as
there is a R-isomorphism Z6 ∼ = Z2 ⊕ Z3. However, Z2 is not a free R-module
as ¯1 is a torsion element.
Lemma 5.5 Let {Pi | i ∈ Λ} be R-modules; then ⊕i∈ΛPi is projective if and
only if Pi is a projective R-module for every i.
Proof. (⇒) For every j, there are R-homomorphism ij : Pj −→ ⊕i∈ΛPi and
g
R-epimorphism πj : ⊕i∈ΛPi −→ Pj such that πj◦ij = idPj . Let M −→ N −→
0 be an exact sequence and f : Pj −→ N be an R-homomorphism. Since
f ◦πj : ⊕i∈ΛPi −→ N is an R-homomorphism and ⊕i∈ΛPi is projective, there
is an R-homomorphism h : ⊕i∈ΛPi −→ M such that g ◦ h = f ◦ πj, therefore
h◦ij is an R-homomorphism from Pj to M such that g◦(h◦ij) = f ◦πj◦ij = f.
(⇐) For every j, there are R-homomorphism ij : Pj −→ ⊕i∈ΛPi and R-
g
epimorphism πj : ⊕i∈ΛPi −→ Pj such that πj ◦ij = idPj . Let M −→ N −→ 0
be an exact sequence and f : ⊕i∈ΛPi −→ N be an R-homomorphism. For
every j, since Pj is projective and f ◦ ij is an R-homomorphism from Pj to
N, there is an R-homomorphism hj : Pj −→ M such that g ◦ hj = f ◦ ij. Let
h : ⊕i∈ΛPi −→ M be the R-homomorphism induced by hj, i.e., h ◦ ij = hj;
then it is easy to see that g ◦ h = f.
The second part of this section is to introduce the notion: injective modules.
Definition 5.6 A module E over a ring R is said to be injective if for every
exact sequence 0 −→ L f
−→ M and an R-homomorphism g : L −→ E, g may
be extended to an R-homomorphism h : M −→ E. i.e., h ◦ f = g.
As projective modules, injective modules have some equivalent characterizations.
133

Lemma 5.7 Let E be an R-module. E is an injective module if and only if
for every ideal I of R, any R-homomorphism I −→ E may be extended to
an R-homomorphism R −→ E.
Proof. (⇒) Since 0 −→ I −→ R is exact, the assertion follows.
(⇐) Let 0 −→ N g
−→ M be an exact sequence and f : N −→ E be an R
homomorphism; to show that E is injective, we must find a homomorphism
h : M −→ E with h ◦ g = f. Let T be the set of all R-homomorphism
h : M ′ −→ E with h ◦ g = f, where Img ⊆ M ′ ⊆ M. Partially order T by
the natural way and verify the assumptions of Zorn’s Lemma, we conclude
that T contains a maximal element h : M ′ −→ E with h ◦ g = f. We will
show that M ′ = M.
If M ′ = M and x ∈ M − M ′ , then I = {r ∈ R | rx ∈ M ′ } is an ideal of
R. The map i : I −→ E given by i(r) = h(rx) is an R-homomorphism. By
assumption, there is an R-homomorphism j : R −→ E such that j(r) = h(rx)
for all r ∈ I. Let y = j(1R) and define a map ˜ h : M ′ + xR −→ E by
˜h(a + rx) = h(a) + ry. ˜ h is well-defined: If a1 + r1x = a2 + r2x, then
a1 − a2 = (r2 − r1)x ∈ M ′ ∩ xR, so that r2 − r1 ∈ I and
h(a1)−h(a2) = h(a1−a2) = h((r2−r1)x) = j(r2−r1) = (r2−r1)j(1R) = (r2−r1)y.
It is easy to see that ˜ h is an R-homomorphism. This contradicts the maximality
of h and hence M ′ = M. Therefore E is injective.
There is an important fact about injective modules: Every R-module
can be imbedded in an injective R-module. To show this, we need some
preparation.
Definition 5.8 An abelian group G is said to be divisible if for every a ∈ G
and 0 = n ∈ Z there exists b ∈ G such that bn = a.
Example 5.9 Q is a divisible abelian group.
Lemma 5.10 An abelian group D is divisible if and only if D is an injective
Z-module.
Proof. (⇒) Let D be a divisible abelian group and f : I −→ D be a Zmodule
homomorphism. Let I = nZ and f(n) = a; then there is an element
b ∈ D such that nb = a. Let g : Z −→ D be the map given by g(m) = mb,
then it is easy to check that g is a Z-homomorphism and is an extension of
134

f.
(⇐) Let D be an injective Z-module, a ∈ D and 0 = n ∈ Z. Let f :
nZ −→ D be the Z-homomorphism induced by f(n) = a; then there is a Zhomomorphism
g : Z −→ D such that g(n) = a. Let b = g(1); then bn = a.
Example 5.11 Q is an injective Z-module.
Lemma 5.12 Every abelian group G may be embedded in a divisible abelian
group.
Proof. Let G be a Z-module; then there is a free Z-module F and an epimorphism
F −→ G with kernel K so that F/K ∼ = G. Since F is a direct sum of
copies of Z and Z ⊆ Q, there is a monomorphism f : F −→ D where D is
a direct sum of copies of Q. Therefore A ∼ = F/K ∼ = f(F )/f(K) ⊆ D/f(K).
Since D is divisible and so is D/f(K), the assertion follows.
Let R be a ring; then R is an abelian group. Let G be an abelian group;
then the set of all group homomorphisms (Z-homomorphisms) from R to G,
HomZ(R, G), is an R-module with (rf)(a) = f(ra) for all a ∈ R, where
r ∈ R and f ∈ HomZ(R, G).
Lemma 5.13 Let D be an injective Z module; then HomZ(R, D) is an injective
R module.
Proof. Let I be an ideal of R and f : I −→ HomZ(R, D) be an Rhomomorphism;
then the map g : I −→ D given by g(a) = [f(a)](1R) is
a group homomorphism induced by f. Since D is an injective Z module,
there is a group homomorphism ˜g : R −→ D such that ˜g|I = g. Define
˜f : R −→ HomZ(R, D) by ˜ f(r)(x) = ˜g(xr) for all x ∈ R. ˜ f is well-defined
as ˜ f(r) is a group homomorphism from R to D. ˜ f is an R-homomorphism:
Let r, s, x ∈ R; then ˜ f(rs)(x) = ˜g(xrs) = ˜g((xs)r)) = ˜ f(r)(xs), so that
by the R-module structure of HomZ(R, D), ˜ f(r)(xs) = s ˜ f(r)(x), therefore
˜f(rs) = s ˜ f(r). To finish the proof, it remains to check that ˜ f is an extension
of f. Let r ∈ I and x ∈ R; then ˜ f(r)(x) = ˜g(xr) = g(xr) = [f(xr)](1R).
Since f is an R-homomorphism f(xr) = xf(r). Moreover, HomZ(R, D) is an
R-module, [xf(r)](1R) = f(r)(1Rx) = f(r)(x). We conclude that ˜ f|I = f.
Corollary 5.14 Any R-module can be imbedded in an injective R-module.
135

Proof. Let M be an R-module. Since M is an abelian group, M is embedded
in an injective Z-module D. By the previous lemma, HomZ(R, D) is an
injective R-module. Therefore, to finish the proof, it remains to show that M
is an R-submodule of HomZ(R, D). Notice that the group monomorphism
M −→ D induces the R-monomorphism HomZ(R, M) −→ HomZ(R, D).
Moreover, let i : M −→ HomZ(R, M) be the map given by i(x)(r) = xr for
x ∈ M and r ∈ R; then it is easy to check that i is an R-monomorphism.
We now conclude that M is an R-submodule of HomZ(R, D).
Proposition 5.15 Let R be a ring. Then the following are equivalent for
an R-module E:
(a) E is injective.
(b) Any s.e.s. of the form 0 −→ E f
−→ M g
−→ N −→ 0 splits.
(c) E is a direct summand of any R-module M of which it is a submodule.
Proof. (a) ⇒ (b) Let 0 −→ E f
−→ M g
−→ N −→ 0 be a s.e.s.. Let i : E −→
E be the identity map; then there is a R-homomorphism h : M −→ E such
that h ◦ f = i, so that h ◦ f = idE,i.e., the s.e.s. splits.
(b) ⇒ (c) Suppose E is a submodule of M. Then there is a s.e.s. 0 −→
E i
−→ M π
−→ M/E −→ 0. Since it splits, E is a direct summand of M.
(c) ⇒ (a) Since E can be embedded in an injective R-module E ′ by the
previous corollary and E is a direct summand of E ′ by assumption, therefore
E is an injective module.
Problem set
1. Let R be a field; then every module is free.
2. The following are equivalent for a ring:
(a) Every R-module is projective.
(b) Every s.e.s. splits.
(c) Every R-module is injective.
3. Let E be an R-module; then E is injective if and only if for every ideal
I and R-homomorphism g : I −→ E there is an element x ∈ E such that
136

g(a) = ax for all a ∈ I.
4. Q is not a projective Z module.
5. Let D be a direct sum of copies of Q; then D is an injective Z module.
6. Every homomorphic image of a divisible abelian group is a divisible
abelian group. Every direct summand of a divisible abelian group is a divisible
abelian group.
7. Let E be a direct summand of an injective R-module; then E is injective.
8. Let {Ei | i ∈ Λ} be R-modules; then
Ei is injective if and only if
Ei is injective for every i.
9. Let R be an P.I.D.; then the quotient field of R is an injective Rmodule.
10. Let p be a prime number and let Z(p ∞ ) be the subset of the additive
group Q/Z:
i∈Λ
Z(p ∞ ) = {a/b ∈ Q/Z | a, b ∈ Z, b = p i for some i ≥ 0}.
Prove that Z(p ∞ ) is a subgroup of Q/Z.
11. Prove that Z(p ∞ ) is a divisible abelian group.
12. A Z-module M is said to be torsion-free if T (M) = 0. Prove that if
M is a torsion-free divisible then M is a direct sum of copies of Q.
137

6 Hom and tensor product
The first part of this section is to study the functors HomR(−, M) and
HomR(M, −), where M is an R-module. Recall that the set of all Rhomomorphism
from M to N, denoted by HomR(M, N), is an R-module
with the action of R given by rf(x) = f(rx) for r ∈ R and x ∈ M.
Lemma 6.1 Let 0 −→ N1
φ
−→ N2
ψ
−→ N3 be an exact sequence; then
HomR(M, −) induces the exact sequence
0 −→ HomR(M, N1)
˜φ
˜ψ
−→ HomR(M, N2) −→ HomR(M, N3),
where ˜ φ(f) = φ ◦ f for f ∈ HomR(M, N1), and ˜ ψ(g) = ψ ◦ g for g ∈
HomR(M, N2).
Proof. ˜ φ is 1-1:
Let f ∈ HomR(M, N1) with ˜ φ(f) = 0; then φ(f(x)) = 0 for all x ∈ M, so
that f(x) = 0 for all x ∈ M as φ is 1-1, it follows that f = 0.
˜φ
˜ψ
HomR(M, N1) −→ HomR(M, N2) −→ HomR(M, N3) is exact:
It is easy to see that ˜ ψ ◦ ˜ φ = 0. Therefore Im ˜ φ ⊆ ker ˜ ψ. To show ker ˜ ψ ⊆
Im ˜ φ, let g ∈ ker ˜ ψ; then ψ(g(x)) = 0 for every x ∈ M. Let x ∈ M; then
there is a unique element y ∈ N1 such that φ(y) = g(x). Now, define a
map f : M −→ N1 by f(x) = y; then it is easy to check that f is an
R-homomorphism and ˜ φ(f) = g.
We say that HomR(M, −) is exact if it preserve every short exact se-
quences,i.e., if 0 −→ N1
HomR(M, N1)
φ
−→ N2
˜φ
−→ HomR(M, N2)
ψ
−→ N3 −→ 0 is a s.e.s., then 0 −→
˜ψ
−→ HomR(M, N3) −→ 0 stays exact.
Lemma 6.2 Let N1
φ
−→ N2
ψ
−→ N3 −→ 0 be an exact sequence; then
HomR(−, M) induces the exact sequence
0 −→ HomR(N3, M)
˜ψ
˜φ
−→ HomR(N2, M) −→ HomR(N1, M),
where ˜ ψ(f) = f ◦ ψ for f ∈ HomR(N3, M), and ˜ φ(g) = g ◦ φ for g ∈
HomR(N2, M).
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Proof. ˜ ψ is 1-1:
Let f ∈ HomR(N3, M) such that ˜ ψ(f) = 0; then f(ψ(x)) = 0 for every
x ∈ N2. Since ψ is onto, f(y) = 0 for all y ∈ N3, therefore f = 0.
˜ψ
˜φ
HomR(N3, M) −→ HomR(N2, M) −→ HomR(N1, M) is exact:
It is easy to see that ˜ φ◦ ˜ ψ = 0. Therefore Im ˜ ψ ⊆ ker ˜ φ. To show ker ˜ φ ⊆ Im ˜ ψ,
let g ∈ HomR(N2, M) such that ˜ φ(g) = 0; then g(φ(N1)) = 0. Let x ∈ N3;
then there is an element y ∈ N2 such that ψ(y) = x. Now, define a map
f : N3 −→ M by f(x) = g(y), then f is well-define as g(φ(N1)) = 0.
Moreover, it is easy to check that f is an R-homomorphism and ˜ ψ(f) = g.
We say that HomR(−, M) is exact if it preserve every short exact sequences,i.e.,
φ ψ
˜ψ
if 0 −→ N1 −→ N2 −→ N3 −→ 0 is a s.e.s., then 0 −→ HomR(N3, M) −→
˜φ
HomR(N2, M) −→ HomR(N1, M) −→ 0 stays exact.
Lemma 6.3 The following are equivalent for an R-module P :
(a) P is projective.
(b) HomR(P, −) is exact.
Proof. Let P be an R-module and 0 −→ N1
φ
−→ N2
ψ
−→ N3 −→ 0 be a s.e.s.;
then P is projective if and only if for every R-homomorphism f : P −→ N3
there is an R-homomorphism g : P −→ N2 such that ψ ◦ g = f. Therefore P
is projective if and only if ˜ ψ : HomR(P, N2) −→ HomR(P, N3) is surjective,
or equivalently, P is projective if and only if HomR(P, −) is exact.
Lemma 6.4 The following are equivalent for an R-module E:
(a) E is injective.
(b) HomR(−, E) is exact.
φ ψ
Proof. Let E be an R-module and 0 −→ N1 −→ N2 −→ N3 −→ 0 be a s.e.s.;
then E is injective if and only if for every R-homomorphism f : N1 −→ E
there is an R-homomorphism g : N2 −→ E such that g ◦ φ = f. Therefore E
is injective if and only if ˜ φ : HomR(N2, E) −→ HomR(N1, E) is surjective,
or equivalently, E is injective if and only if HomR(−, E) is exact.
The second part of the section is to introduce the notion: tensor product
⊗.
139

Definition 6.5 Let M, N, P be three R-modules. A mapping f : M ×N −→
P is said to be R-bilinear if for each x ∈ M the mapping y −→ f(x, y) of N
into P is R-linear, and for each y ∈ N the mapping x −→ f(x, y) of M into
P is R-linear. Or equivalently, f satisfies the conditions:
f(x1 + x2, y) = f(x1) + f(x2),
f(x, y1 + y2) = f(x, y1) + f(x, y2),
f(rx, y) = f(x, ry) = rf(x, y).
Proposition 6.6 Let M, N be R-modules. Then there exists a pair (T, i)
consisting of an R-module T and an R-bilinear mapping i : M × N −→ T ,
with the following property:
Given any R-module P and any R-bilinear mapping f : M × N −→ P
there exists a unique R-linear mapping f ′ : T −→ P such that f = f ′ ◦ i.
Proof. Let C denote the free R-module R M×N ,i.e., free abelian group on
the set M × N. The elements of C are formal combinations of the form
n
i=1 ai(xi, yi) (ai ∈ R, xi ∈ M, yi ∈ N). Let D be the submodules of C
generated by all elements of C of the following types:
(x + x ′ , y) − (x, y) − (x ′ , y)
(x, y + y ′ ) − (x, y) − (x, y ′ )
(ax, y) − a(x, y)
(x, ay) − a(x, y).
Let T = C/D. For each element (x, y) of C, let x ⊗ y denotes its image in
T . Then T is generated by the element of the form x ⊗ y and we have
(x + x ′ ) ⊗ y = x ⊗ y + x ′ ⊗ y
x ⊗ (y + y ′ ) = x ⊗ y + x ⊗ y ′
ax ⊗ y = a(x ⊗ y) = (x ⊗ ay).
It is clear that i : M × N −→ T is given by i((x, y)) = x ⊗ y. Now, if
f : M × N −→ P is a R-bilinear map then f(D) = 0, so f induces an
R-homomorphism f ′ : T −→ P such that f = f ′ ◦ i.
If h is another R-homomorphism from T to P with f = h◦i, then h(x⊗y) =
140

f((x, y)) = x ⊗ y = f ′ (x ⊗ y). Since T is generated by elements of the form
x ⊗ y, h = f ′ .
The module T constructed above is called the tensor product of M and
N, and is denoted by M ⊗R N.
Lemma 6.7 (Adjoint associativity) Let M, N, L be R-modules; then there is
a natural R-isomorphism
HomR(M ⊗ N, L) ∼ = HomR(M, HomR(N, L)).
Proof. (Sketch) Let α : HomR(M ⊗ N, L) −→ HomR(M, HomR(N, L))
be the map given by [α(f)(a)](b) = f(a ⊗ b), where f ∈ HomR(M ⊗
N, L), a ∈ M and b ∈ N. It is easy to check that α is well-defined,i.e.,
α(f)(a) ∈ HomR(N, L), α(f) ∈ HomR(M, HomR(N, L)) and α is an Rhomomorphism.
Let β : HomR(M, HomR(N, L)) −→ HomR(M ⊗ N, L)
be the map given by (βg)(a ⊗ b) = [g(a)](b), where a ∈ M, b ∈ N and
g ∈ HomR(M, HomR(N, L)). It is easy to check that β is well-defined,i.e,
βg ∈ HomR(M ⊗ N, L) and β is an R-homomorphism. Now, it is also easy
to see that every compositions α ◦ β, β ◦ α is the identity map. We conclude
that α is an isomorphism.
Lemma 6.8 Let N1
φ
−→ N2
ψ
−→ N3 −→ 0 be an exact sequence of Rmodules;
then ⊗RM induces the exact sequence
N1 ⊗R M φ⊗i
−→ N2 ⊗R M ψ⊗i
−→ N3 ⊗R M −→ 0,
where i is the identity map of M.
φ ψ
Proof. Let E. be the s.e.s. N1 −→ N2 −→ N3 −→ 0; then we need to show
that E. ⊗R M is exact. Let P be any module; then HomR(E., HomR(M, P ))
is exact as E. is exact; so by the above lemma HomR(E. ⊗R M, P ) is exact,
it follows by Exercise 6 that E. ⊗R M is exact.
Remark 6.9 From the above, one can see that if N is a submodule of L then
it is not necessary that N ⊗R M is a submodule of L ⊗R M.
Example 6.10 Consider the exact sequence of Z-modules 0 −→ 2Z i
−→ Z,
where i is the canonical injection. Then the sequence 0 −→ 2Z ⊗Z Z/2Z i
−→
Z ⊗Z Z/2Z is not exact as 2 ⊗ ¯1 is a nonzero element in 2Z ⊗Z Z/2Z but
2 ⊗ ¯1 = 1 ⊗ ¯2 = 0 in Z ⊗Z Z/2Z
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Let M be an R-module; M is said to be flat if ⊗RM preserve short
exact sequences,i.e., if 0 −→ N1
φ
−→ N2
ψ
−→ N3 −→ 0 is a s.e.s., then
0 −→ N1 ⊗R M φ⊗idM
−→ N2 ⊗R M ψ⊗idM
−→ N3 ⊗R M −→ 0 stays exact.
Lemma 6.11 Let M be an R-module; then R⊗R M and M ⊗R R are canonical
isomorphic to M.
Proof. Let i : R ⊗R M −→ M be the map given by i(r ⊗ x) = rx; then i
is well-defined and is an R-homomorphism as i is induced by the R-bilinear
map ĩ : R × M −→ M with ĩ(r, x) = rx. It is easy to see that i is an
isomorphism and R ⊗R M ∼ = M. Similarly, M ⊗R R ∼ = M.
Corollary 6.12 R is a flat R-module.
φ
Proof. Let 0 −→ N1 −→ N2 be an exact sequence of R-modules; then ⊗RR
induces the sequence 0 −→ N1 ⊗R R φ⊗idR
−→ N2 ⊗R R. Moreover,
0 −→ N1 ⊗R R φ⊗idR
−→ N2 ⊗R R
↓ ↓
0 −→ N1
φ
−→ N2
is a commutative diagram of R-modules and R-homomorphisms such that
the second row is exact and the vertical homomorphisms are isomorphisms.
Therefore we conclude that φ ⊗ idR is 1-1.
Let R be a ring and S be a multiplicative subset of R; then the construction
of RS can be carried through with an R-module M in place of the ring R.
Definition 6.13 Let R be a ring, S be a multiplicative subset of R and M be
an R-module. Define a relation on the set M × S = {(x, s) | x ∈ M, s ∈ S}
as follows: (x, s) ∼ (y, t) if and only if there is an element u ∈ S such that
u(xt − ys) = 0. It is easy to show that ∼ is an equivalent relation. As
the above (M × S)/ ∼ can be made into an RS-module. We use x
for the
s
equivalent class of the element (x, s), and MS for the module (M × S)/ ∼.
As before, if S = R−P for some prime P , then we use MP instead of MS.i.e.,
MP = { x
| x ∈ M, s /∈ P }.
s
Lemma 6.14 Let M be an R-module and S be a multiplicative subset of R;
then M ⊗R RS is canonically isomorphic to MS, so is RS ⊗R M.
142
.

Proof. Let ˜ f : M × RS −→ MS be the R-bilinear map f(x, a xa
) =
s s ;
then ˜ f induces the canonical R-homomorphism f : M ⊗R RS −→ MS with
f(x ⊗ a xa
) = . To finish the proof, it remains to show that f is 1-1 as f is
s s
n
clearly a surjection. For this, let xi ⊗ (ai/si) be any element of M ⊗R RS.
i=1
If s =
i si, ti =
j=i sj, we have
n
xi ⊗ ( ai
n
) = xi ⊗ ( aiti
n
) = aitixi ⊗
s 1
n
= ( aitixi) ⊗
s 1
s ,
i=1
si
i=1
so that every element of M ⊗R RS is of the form x ⊗ 1
. Suppose that
s
f(x ⊗ 1
x
) = 0. Then = 0, hence tx = 0 for some t ∈ S, and therefore
s s
x ⊗ 1
s
= x ⊗ t
st
i=1
= xt ⊗ 1
st
= 0 ⊗ 1
st
i=1
= 0.
Hence f is 1-1 and therefore an isomorphism.
As in the proof of the fact that R is a flat R-module, we have the following
consequence.
Corollary 6.15 Let R be a ring and S be a multiplicative subset of R; then
RS is a flat R-module.
Problem set
If G and H are abelian group then Hom(G, H) is the set of all group homomorphisms
from G to H. Moreover, G ⊗ H = G ⊗Z H.
1. Let M be an R-module and I be an ideal of R. Prove that HomR(R/I, M) ∼ =
(0 :M I) = {x ∈ M | xI = 0}. In particular, HomR(R, M) ∼ = M.
2. Let I, J be ideals of a ring R; then HomR(R/I, R/J) ∼ = (J : I)/J,
where (J : I) = {r ∈ R | rI ⊆ J}.
3. Let R be an integral domain and I be a non-zero ideal of R; then
HomR(R/I, R) = 0. In particular, Hom(Zn, Z) = 0.
143

4. Hom(Zn, Zm) ∼ = Zd, where d is g.c.d. of m and n.
5. The sequence 0 −→ N1
φ
−→ N2
R-module M, the sequence 0 −→ HomR(M, N1)
HomR(M, N3) is exact.
6. The sequence N1
φ
−→ N2
R-module M, the sequence 0 −→ HomR(N3, M)
HomR(N1, M) is exact.
ψ
ψ
−→ N3 is exact if and only if for every
˜φ
−→ HomR(M, N2)
˜ψ
−→
−→ N3 −→ 0 is exact if and only if for every
˜ψ
−→ HomR(N2, M)
7. Let M and N be R-modules; then M ⊗R N ∼ = N ⊗R M.
8. Prove that free modules are flat.
˜φ
−→
9. Let M be an R-module and I be an ideal of R; then M ⊗R R/I ∼ =
M/IM.
10. Let I and J be ideals of R; then R/I ⊗R R/J ∼ = R/(I + J).
11. Zm ⊗ Zn ∼ = Zd, where d is g.c.d. of m and n.
12. Let G be a torsion abelian group; then G ⊗ Q = 0.
13. Let M be an R-module and Mm = 0 for every maximal ideal m of R;
then M = 0.
14. Let f : N −→ M be an R-homomorphism and fm : Nm −→ Mm is an
injection for every maximal ideal m of R; then f is an injection.
15. Q ⊗ Q ∼ = Q.
16. If P is a projective module, then PS is a projective RS module, where
S is a multiplicative subset of R.
144