Abstract Algebra
Muhammad Atif Zaheer
Syed Babar Ali School of Science and Engineering,
Lahore University of Management Sciences (LUMS)
Contents
Preface
ii
1. Introduction to Groups
1
2. Introduction to Rings
2.1. Basic Definition and Examples
2.2. Polynomial Rings and Matrix Rings
2
2
5
i
Preface
This book is based on various sources.
ii
Chapter 1
Introduction to Groups
1
Chapter 2
Introduction to Rings
2.1 Basic Definition and Examples
Definition 2.1.1. A ring R is a set together with two binary operations called addition
(denoted by a + b) and multiplication (denoted by ab) satisfying the following axioms:
(a) (R, +) is an abelian group,
(b) multiplication is associative: (ab)c = a(bc) for all a, b, c ∈ R,
(c) the distributive laws hold: a(b+c) = ab+ac and (a+b)c = ac+bc for all a, b, c ∈ R.
The ring R is called commutative if multiplication is commutative. The ring R is said to
have an identity if there is an element 1 ∈ R with 1a = a1 = a for all a ∈ R.
Remarks. Note that there is a unique ring having 1 = 0, namely the trivial ring {0}.
Definition 2.1.2. A ring R with identity 1 ̸= 0 is called a division ring (or skew field )
if every nonzero element a ∈ R has a multiplicative inverse, i.e., there exists b ∈ R such
that ab = ba = 1. A commutative division ring is called a field.
Examples 2.1.3.
(a) Every abelian group is a ring.
(b) The integers, Z, form a commutative ring with identity under usual addition and
multiplication. Note that under multiplication Z\{0} is not a group (in fact, very
few elements admit inverses).
(c) Q, R, C are all examples of commutative rings with identity (in fact, they are
fields).
(d) For any positive integer, the set of integers modulo n, Z/nZ, form a commutative
ring with identity.
(e) Let X be a nonemepty set and let A be a ring. The collection, R, of all (set)
functions f : X → A is a ring under the usual definition of pointwise addition and
multiplication of functions. It is easily verified that the ring R is commutative if
and only if A is commutative and R has an identity if and only if A has an identity.
If X and A have more structure, we can restrict our attention to the functions that
respect those structures and get a new ring. For instance, if X = [0, 1] and A = R,
then the set of all continuous functions from [0, 1] to R also form a ring.
2
(f) The ring of even integers, 2Z, doesn’t contain the identity element.
The set of all functions f : R → R having compact support is a commutative ring
without identity.
Proposition 2.1.4. Let R be a ring. Then
(a) 0a = a0 = 0 for all a ∈ R,
(b) (−a)b = a(−b) = −(ab) for all a, b ∈ R.
Definition 2.1.5. Let R be a ring.
(a) A nonzero element a ∈ R is called a zero divisor if there is a nonzero element b ∈ R
such that either ab = 0 or ba = 0.
(b) Assume R has an identity 1. An element u ∈ R is called a unit in R if there is
some v ∈ R such that uv = vu = 1. The set of units in R is denoted R× .
It is easy to see that the units in a ring R form a group under multiplication. Moreover,
it is easily observed that a zero divisor cannot be a unit.
Definition 2.1.6. A commutative ring with identity 1 ̸= 0 is called an integral domain
if it has no zero divisors.
Proposition 2.1.7. Assume a, b and c are elements of any ring with a not a zero divisor.
If ab = ac, then either a = 0 or b = c. In particular, if a, b, c are any elements in an
integral domain and ab = ac, then either a = 0 or b = c.
Proposition 2.1.8. In a finite ring with identity, every nonzero element is either a zero
divisor or a unit.
Proof. Let R be a finite ring and let a be a nonzero element that is not a zero divisor.
Then the map x 7→ ax is an injective function. Since R is finite this map is also surjective.
Thus, there is an element b ∈ R such that ab = 1. Using a symmetric argument, we get
that there is an element c ∈ R such that ca = 1. Now note that
c = c1 = c(ab) = (ca)b = 1b = b.
Hence, a is a unit.
Corollary 2.1.9. Any finite integral domain is a field.
A similar but comparatively difficult result is that any finite division ring is a field.
This result is known as Wedderburn’s Theorem.
Examples 2.1.10.
(a) The ring of integers Z has no zero divisors and Z× = {±1}.
(b) It is easily verified that if n > 1 then a nonzero element [a] ∈ Z/nZ is either a unit
or a zero divisor. Furthermore, [a] ∈ Z/nZ is a unit if and only if (a, n) = 1.
3
(c) If R is the ring of all functions from R to R then the units of R are the functions
that are not zero at any point. If f is not a unit and not zero then f is a zero
divisor because if we define
(
0, if f (x) ̸= 0
g(x) =
1, if f (x) = 0
then g is not the zero function but f (x)g(x) = 0 for all x. Observe that this is the
kind of behaviour that the finite rings with identity have.
(d) If R is the ring of all continuous functions from R to R then the units of R are still
the functions that are not zero at any point, but now there are functions that are
neither units nor zero divisors. For instance, f (x) = x has a zero at x = 0 so f is
not a unit. On the other hand if f g = 0 then g(x) = 0 for all x ̸= 0, and the only
continuous function with this property is the zero function. Hence f is neither a
unit nor a zero divisor.
Example 2.1.11 (Hamilton Quaternions). Let H = R4 be the abelian group with componentwise addition. We then define multiplication on H as follows:
(a, b, c, d)·(a′ , b′ , c′ , d) = (aa′ −bb′ −cc′ −dd′ , ab′ +ba′ +cd′ −dc′ , ac′ −bd′ +ca′ +db′ , ad′ +bc′ −cb′ +da′ )
We use the following notations:
a = (a, 0, 0, 0),
i = (0, 1, 0, 0),
j = (0, 0, 1, 0),
k = (0, 0, 0, 1).
The following relations are then easily verified:
i2 = j 2 = k 2 = −1,
ij = −ji = k,
jk = −kj = i,
ki = −ik = j.
(2.1)
It can be easily verified that H forms a ring (the associativity is a bit tedious to prove).
The inverse of the nonzero element a + bi + cj + dk is given by
(a + bi + cj + dk)−1 =
a − bi − cj − dk
.
a2 + b2 + c2 + d2
It is easily observed that H is a noncommutative ring due to relations (2.1). Thus, H is
a division ring.
Definition 2.1.12. A subring of a ring R is a subset S of R that forms a ring under the
addition and multiplication operations inherited from R.
Proposition 2.1.13 (Subring criterion). Let R be a ring. Then a subset S of R is a
subring if and only if the following conditions hold:
(a) S is a subgroup of R,
(b) S is closed under multiplication.
Example 2.1.14. Let D be a rational number that is not a perfect square in Q and
define
√
√
Q( D) = {a + b D : a, b ∈ Q}.
4
√
√
It is easily verified that Q( D) is a subring of C. In fact, Q( D) is a commutative
ring with identity. It is easy to show
√ that that the assumption that D is not a square
√
implies that every element of Q( D) may√be written in the √
unique form a + b D.
This assumption √
also implies
D ̸= 0 and therefore
√ that if a + b D ̸= 0 then a − b √
a2 − Db2 = (a + b D)(a − b D) ̸= 0. It thus follows that if a + b D ̸= 0 then
√
√ −1 a − b D
(a + b D) = 2
.
a − Db2
√
√
This shows that every nonzero element in Q( D) is a unit; that is, Q( D) is a field. It
is referred to as a quadratic field.
The rational number D can be written as D = f 2 D′ for some rational number f and
a unique integer D′ is not divisible by the square of any integer greater than 1; that is, D′
is either −1 or ±1 times the product
primes. √
The unique√integer D′ is called
√ of distinct
√
the square free part of D. Then D = f D′ , and so Q( D) = Q( D′ ). So there is no
loss in generality in assuming D to be a square free integer.
√
√
D]
=
{a
+
b
D : a, b ∈ Z}.
Example 2.1.15. Let D be √
a square free integer and
let
Z[
√
It then easily follows that Z[ D] is a subring of Q( D). If D ≡ 1 (mod 4) then
"
√ #
√
1+ D
1+ D
: a, b ∈ Z}
Z
= {a + b
2
2
√
is also a subring of Q( D).
√
The ring of integers in the quadratic field Q( D) is defined to be
O = OQ(√D) = Z[ω] = {a + bω : a, b ∈ Z}
where
√

 D
√
ω = 1+ D


2
if D ≡ 2, 3
if D ≡ 1
(mod 4)
(mod 4).
:
Note that OQ(i) = Z[i] = {a
√ + bi a, b ∈ Z}.
The field norm N : Q( D) → Q is defined by
√
√
√
N (a + b D) = (a + b D)(a − b D) = a2 − Db2 .
√
√
Note that N (a + b √D) = 0 if and only if a + b D = 0. This norm gives a measure of the
size in the field Q( D). It is quite easy
√ to show that this norm is multiplicative; that is,
N (αβ) = N (α)N (β) for all α, β ∈ Q( D).
√
√
√
√
Observe that if we define a + b D = a−b D for any a+b D ∈ Q( D) then N (α) =
αα. In particular, if a + bω ∈ Z[ω] then N (a + bω) = (a + bω)(a + bω) = (a + bω)(a + bω).
2.2 Polynomial Rings and Matrix Rings
Definition 2.2.1. Let R be a ring. Then a polynomial with coefficients in R is an
infinite sequence f = (a0 , a1 , a2 , . . . ) in R with finite support; that is, only finitely
5
many entries ai are nonzero. The element ai is called the ith coefficient of f .
f = (a0 , a1 , . . . , an , 0, 0, . . . ) then we write
f (x) = a0 + a1 x + · · · + an x
n
or f (x) =
n
X
If
ai x i .
i=0
The set of all polynomials with coefficients in R is denoted by R[x].
If f = (a0 , a1 , a2 , . . . ) and g = (b0 , b1 , b2 , . . . ) are polynomials in R[x] then the addition
and multiplication of f and g are defined by
f + g = (a0 + b0 , a1 + b1 , a2 + b2 , . . . ),
f g = (c0 , c1 , c2 , . . . ) with ck =
k
X
X
ai bk−i =
i=0
ai b j .
i+j=k
Proposition 2.2.2. If R is a ring then R[x] is also a ring.
Proof. It is quite trivial to show that R[x] is closed under addition and forms an abelian
group. Now suppose that f = (a0 , a1 , . . . , an , 0, . . . ) and g = (b0 , b1 , . . . , bm , 0, . . . ). If
k > n + m then
X
X
X
ai bj = 0 + 0 = 0.
ai b j +
ai b j =
ck =
i+j=k
i+j=k
i≤n
i+j=k
i>n
Thus f g = (c0 , c1 , c2 , . . . ) has a finite support and so is a polynomial. Hence R[x] is
closed under multiplication.
We now prove that the multiplication is associative. Let h = (d0 , d1 , d2 , . . . ) be a
polynomial and
(f g)h = (e0 , e1 , e2 , . . . )
then
ek =
X
i+j=k
ci d j =
X X
i+j=k
u+v=i
au b v d j =
X
au b v d j .
u+v+j=k
If we now compute f (gh) in a similar manner, we find exactly the same coefficients for
f (gh) as for (f g)h, thereby proving associativity. Finally, the multiplicative laws are
quite straightforward to prove.
Observe that the ring R[x] contains a copy of R, namely the set of constant polynomials.
Proposition 2.2.3. Let R be a ring. Then
(a) R is commutative if and only if R[x] is commutative;
(b) R has an identity if and only if R[x] has an identity.
Definition 2.2.4. Let R be a ring and f = (a0 , a1 , a2 , . . . ) ∈ R[x] be a polynomial with
coefficients in R. Then the degree of f is defined to be the largest integer n such that
an ̸= 0. In this case, we write deg f = n and the coefficient an is called the leading
coefficient of f . If no such integer exists, that is, if f = 0 then the degree of f is defined
to be −∞.
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Proposition 2.2.5. Let f (x) and g(x) be polynomials in R[x], where R is an integral
domain. Then
deg f (x)g(x) = deg f (x) + deg g(x).
Furthermore, R[x] is an integral domain.
Proof. If f (x) = 0 or g(x) = 0 then the identity clearly holds as both sides become −∞.
Now suppose that
f (x) = an xn + · · · + a1 x + a0
and
g(x) = bm xm + · · · + b1 x + b0
where an ̸= 0 and bm ̸= 0. The degrees of f (x) and g(x) are n and m respectively. Then
f (x)g(x) = cn+m xn+m + · · · + c1 x + c0
and cn+m = an bm . Since R is an integral domain and both an and bm are nonzero therefore
cn+m is nonzero. Thus
deg f (x)g(x) = n + m = deg f (x) + deg g(x).
Since f (x) ̸= 0 and g(x) ̸= 0 imply that f (x)g(x) ̸= 0, we know that R[x] must be an
integral domain.
Corollary 2.2.6. Suppose R is an integral domain. Then
R[x]× = R× .
Proof. Since R[x] has a copy of R as the set of constant polynomials and R[x] and R
have the same identity, it follows that R× ⊂ R[x]× .
To prove the other inclusion, suppose that f (x) ∈ R[x]× . This means there is g(x) ∈
R[x] such that f (x)g(x) = 1. By Proposition, 2.2.5, we have
deg f (x) + deg g(x) = deg f (x)g(x) = deg 1 = 0.
This, in particular, implies that f (x) and g(x) are both nonzero, and so their degrees are
at least 0. Therefore deg f (x) and deg g(x) are two nonnegative integers that add up to
0. Hence both of them are zeros. That means f (x) = a and g(x) = b for some a, b ∈ R,
and f (x)g(x) = ab = 1. This implies that a ∈ R× , which finishes the proof.
Remarks. Note that the above corollary does not hold for general rings. For instance,
the nonconstant polynomial 1 + 2x is a unit in Z4 [x] as 1 is a unit and 2x is a nilpotent
element. (See Exercises.)
7
Solutions to Exercises
8