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MTH 513 Homework April 3, 2019
1. In your own words, state (explain? Prove?) why
(a) if R is a commutative ring with 1, then R[x], the polynomial ring, is also a commutative ring.
(b) if R is an integral domain, then R[x] is also an integral domain.
2. Show that for the numbers 50 and 20, 50 = 2 × 20 + 10 is the unique representation of the
form 50 = × 20 + where are integers. (See the uniqueness part of the proof of
Theorem 3.2 on pages 31-32.)
3. = 60, = 74.
(a) Find integers such that gcd( , ) = + .
(b) Are these integers unique? If so, why? If not, find another pair of integers
′ ′ ℎ ℎ gcd( , ) = ′ + ′ .
3. Can a fifth-degree polynomial over the real numbers (i.e. with coefficients in R) be
irreducible? If so, give an example. If not, explain why not. Graph your polynomial.
4. Can a sixth-degree polynomial over the real numbers (i.e. with coefficients in R) be
irreducible? If so, give an example. If not, explain why not. Graph your polynomial function.
5. Do the following exercises that appear on the attached sheets.
Page 93/ 1, 2, 3, 5, 6(i), (ii), 7, 8
An undergraduate course in
Abstract Algebra
Course notes for MATH3002 Rings and Fields
Robert Howlett
An undergraduate course in
Abstract Algebra
Robert Howlett
typesetting by TEX
Chapter 0:
The integers
Quotients of the ring of integers
Equivalence relations
Congruence relations on the integers
The ring of integers modulo n
Properties of the ring of integers modulo n
Chapter 5:
Introduction to rings
Two basic properties of the integers
The greatest common divisor of two integers
Factorization into primes
Chapter 4:
Ruler and compass constructions
Operations on sets
The basic definitions
Two ways of forming rings
Trivial properties of rings
Chapter 3:
Three problems
Some examples of constructions
Constructible numbers
Chapter 2:
Concerning notation
Concerning functions
Concerning vector spaces
Some very obvious things about proofs
Chapter 1:
Some Ring Theory
Subrings and subfields
The characteristic of a ring
Chapter 6:
More Ring Theory
More on homomorphisms
More on ideals
Congruence modulo an ideal
Quotient rings
The Fundamental Homomorphism Theorem
Chapter 8:
Addition and multiplication of polynomials
Constant polynomials
Polynomial functions
Evaluation homomorphisms
The division algorithm for polynomials over a field
The Euclidean Algorithm
Irreducible polynomials
Some examples
Factorization of polynomials
Irreducibility over the rationals
Chapter 7:
Field Extensions
Ideals in polynomial rings
Quotient rings of polynomial rings
Fields as quotient rings of polynomial rings
Field extensions and vector spaces
Extensions of extensions
Algebraic and transcendental elements
Ruler and compass constructions revisited
Finite fields
Index of notation
Index of examples
Foreword. . .
The purpose of this book is to complement the lectures and thereby decrease,
but not eliminate, the necessity of taking lecture notes. Reading the appropriate sections of the book before each lecture should enable you to understand
the lecture as it is being given, provided you concentrate! This is particularly
important in this course because, as theoretical machinery is developed, the
lectures depend more and more heavily upon previous lectures, and students
who fail to thoroughly learn the new concepts as they are introduced soon
become completely lost.
Proofs of the theorems are an important part of this
course. You cannot expect to do third year Pure Mathematics
without coming to grips with proofs. Mathematics is about proving
theorems. You will be required to know proofs of theorems for the
It is the material dealt with in the lectures, not this book, which defines the
syllabus of the course. The book is only intended to assist, and how much
overlap there is with the course depends on the whim of the lecturer. There
will certainly be things which are in the lectures and not in the book, and
vice versa. The lecturer will probably dwell upon topics which are giving
students trouble, and omit other topics. However, the book will still provide
a reasonable guide to the course.
Students will be assumed to be familiar with the material mentioned in this
preliminary chapter. Anyone who is not should inform the lecturer forthwith.
Concerning notation
When reading or writing mathematics you should always remember that the
mathematical symbols which are used are simply abbreviations for words.
Mechanically replacing the symbols by the words they represent should result
in grammatically correct and complete sentences. The meanings of a few
commonly used symbols are given in the following table.
{ … | … }

To be read as
the set of all . . . such that . . .
in or is in
greater than or is greater than
Thus for example the following sequence of symbols
{x ∈ X | x > a} =
6 ∅
is an abbreviated way of writing the sentence
The set of all x in X such that x is greater than a is not the empty set.
When reading mathematics you should mentally translate all symbols in this
fashion. If you cannot do this and obtain meaningful sentences, seek help
from your tutor. And make certain that, when you use mathematical symbols
yourself, what you write can be translated into meaningful sentences.
Chapter Zero: Prerequisites
Concerning functions
The terminology we use in connection with functions could conceivably differ
from that to which you are accustomed; so a list of definitions of the terms
we use is provided here.
• The notation ‘f : A → B’ (read ‘f , from A to B’) means that f is a
function with domain A and codomain B. In other words, f is a rule which
assigns to every element a of the set A an element in the set B denoted
by ‘f (a)’.
• A map is the same thing as a function. The term mapping is also used.
• A function f : A → B is said to be injective (or one-to-one) if and only
if no two distinct elements of A yield the same element of B. In other words,
f is injective if and only if for all a1 , a2 ∈ A, if f (a1 ) = f (a2 ) then a1 = a2 .
• A function f : A → B is said to be surjective (or onto) if and only if for
every element b of B there is an a in A such that f (a) = b.
• If a function is both injective and surjective we say that it is bijective
(or a one-to-one correspondence).
• The image of a function f : A → B is the subset of B consisting of all
elements obtained by applying f to elements of A. That is,
im f = { f (a) | a ∈ A }.
An alternative notation is ‘f (A)’ instead of ‘im f ’. Clearly, f is surjective if
and only if im f = B.
• The notation ‘a 7→ b’ means ‘a maps to b’; in other words, the function
involved assigns the element b to the element a. Thus ‘a 7→ b under f ’ means
exactly the same as ‘f (a) = b’.
• If f : A → B is a function and C a subset of B then the inverse image
or preimage of C is the subset of A
f −1 (C) = { a ∈ A | f (a) ∈ C }.
(The above line reads ‘f inverse of C, which is the set of all a in A such that
f of a is in C.’ Alternatively, one could say ‘The inverse image of C under
f ’ instead of ‘f inverse of C’.)
Chapter Zero: Prerequisites
Concerning vector spaces
Vector spaces enter into this course only briefly; the facts we use are set out
in this section.
Associated with each vector space is a set of scalars. In the common
and familiar examples this is R, the set of all real numbers, but in general it
can be any field. (Fields are defined in Chapter 2.)
Let V be a vector space over F . (That is, F is the associated field of
scalars.) Elements of V can be added and multiplied by scalars:
If v, w ∈ V and λ ∈ F then v + w, λv ∈ V .
These operations of addition and multiplication by scalars satisfy the following properties:
(i) (u + v) + w = u + (v + w)
for all u, v, w ∈ V .
(ii) u + v = v + u
for all u, v ∈ V .
(iii) There exists an element 0 ∈ V such that v + 0 = v for all v ∈ V .
(iv) For each v ∈ V there exists an element −v ∈ V such that v + (−v) = 0.
(v) λ(µv) = (λµ)v
for all λ, µ ∈ F and all v ∈ V .
(vi) 1v = v
for all v ∈ V .
(vii) λ(v + w) = λv + λw
for all λ ∈ F and all v, w ∈ V .
(viii) (λ + µ)v = λv + µv
for all λ, µ ∈ F and all v ∈ V .
The properties listed above are in fact the vector space axioms; thus in
order to prove that a set V is a vector space over a field F one has only to
check that (∗) and (i)–(viii) are satisfied.
Let V be a vector space over F and let v1 , v2 , . . . vn ∈ V . The elements
v1 , v2 , . . . vn are said to be linearly independent if the following statement is
If λ1 , λ2 , . . . , λn ∈ F and λ1 v1 + λ2 v2 + · · · + λn vn = 0
then λ1 = 0, λ2 = 0, . . . , λn = 0.
The elements v1 , v2 , . . . vn are said to span the space V if the following
statement is true:
For every v ∈ V there exist λ1 , λ2 , . . . , λn ∈ F
such that v = λ1 v1 + λ2 v2 + · · · + λn vn .
A basis of a vector space V is a finite subset of V whose elements are linearly
independent and span V .
Chapter Zero: Prerequisites
We can now state the only theorem of vector space theory which is used
in this course.
0.1 Theorem If a vector space V has a basis then any two bases of V will
have the same number of elements.
Comment …
If V has a basis then the dimension of V is by definition the
number of elements in a basis.

Some very obvious things about proofs
When trying to prove something, the logical structure of what you are trying to prove determines the logical structure of the proof. The following
observations seem trivial, yet they are often ignored.
• To prove a statement of the form
If p then q
your first line should be
Assume that p is true
and your last line
Therefore q is true.
• The statement
p if and only if q
is logically equivalent to
If p then q and if q then p,
and so the proof of such a statement involves first assuming p and proving q,
then assuming q and proving p.
• To prove a statement of the form
All xxxx’s are yyyy’s,
the first line of your proof should be
Let a be an xxxx
and the last line should be
Therefore a is a yyyy.
(The second line could very well involve invoking the definition of ‘xxxx’
or some theorem about xxxx’s to determine things about a; similarly the
second to last line might correspond to the definition of ‘yyyy’.)
Chapter Zero: Prerequisites
When trying to construct a proof it is sometimes useful to assume
the opposite of the thing you are trying to prove, with a view to obtaining
a contradiction. This technique is known as “indirect proof” (or “proof
by contradiction”). The idea is that the conclusion c is a consequence of
the hypotheses h1 , h2 , . . . , if and only if the negation of c is incompatible
with h1 , h2 , . . . . Hence we may assume the negation of c as an extra
hypothesis, along with h1 , h2 etc., and the task is then to show that the
hypotheses contradict each other. Note, however, that although indirect
proof is a legitimate method of proof in all situations, it is not a good policy
to always use indirect proof as a matter of course. Most proofs are naturally
expressed as direct proofs, and to recast them as indirect proofs may make
them more complicated than necessary.
#1 Suppose that you wish to prove that a function λ: X → Y is injective.
Consult the definition of injective. You are trying to prove the following
For all x1 , x2 ∈ X, if λ(x1 ) = λ(x2 ) then x1 = x2 .
So the first two lines of your proof should be as follows:
Let x1 , x2 ∈ X.
Assume that λ(x1 ) = λ(x2 ).
Then you will presumably consult the definition of the function λ to derive
consequences of λ(x1 ) = λ(x2 ), and eventually you will reach the final line
Therefore x1 = x2 .
#2 Suppose you wish to prove that λ: X → Y is surjective. That is, you
wish to prove
For every y ∈ Y there exists x ∈ X with λ(x) = y.
Your first line must be
Let y be an arbitrary element of Y .
Somewhere in the middle of the proof you will have to somehow define an
element x of the set X (the definition of x is bound to involve y in some
way), and the last line of your proof has to be
Therefore λ(x) = y.
Chapter Zero: Prerequisites
#3 Suppose that A and B are sets, and you wish to prove that A ⊆ B.
(That is, A is a subset of or equal to B.) By definition the statement ‘A ⊆ B’
is logically equivalent to
All elements of A are elements of B.
So your first line should be
Let x ∈ A
and your last line should be
Therefore x ∈ B.
#4 Suppose that you wish to prove that A = B, where A and B are sets.
The following statements are all logically equivalent to ‘A = B’:
(i) For all x, x ∈ A if and only if x ∈ B.

(ii) (For all x) (if x ∈ A then x ∈ B) and (if x ∈ B then x ∈ A) .
(iii) All elements of A are elements of B and all elements of B are elements
of A.
(iv) A ⊆ B and B ⊆ A.
You must do two proofs of the general form given in #3 above.
Ruler and compass constructions
Abstract algebra is essentially a tool for other branches of mathematics.
Many problems can be clarified and solved by identifying underlying structure and focussing attention on it to the exclusion of peripheral information
which may only serve to confuse. Moreover, common underlying structures
sometimes occur in widely varying contexts, and are more easily identifiable
for having been previously studied in their own right. In this course we shall
illustrate this idea by taking three classical geometrical problems, translating them into algebraic problems, and then using the techniques of modern
abstract algebra to investigate them.
Three problems
Geometrical problems arose very early in the history of civilization, presumably because of their relevance to architecture and surveying. The most
basic and readily available geometrical tools are ruler and compass, for constructing straight lines and circles; thus it is natural to ask what geometrical
problems can be solved with these tools.†
It is said that the citizens of Delos in ancient Greece, when in the
grips of a plague, consulted an oracle for advice. They were told that a god
was displeased with their cubical altar stone, which should be immediately
replaced by one double the size. The Delians doubled the length, breadth
and depth of their altar; however, this increased its volume eightfold, and
the enraged god worsened the plague.
Although some historians dispute the authenticity of this story, the
so-called “Delian problem”
† Note that the ruler is assumed to be unmarked; that is, it is not a measuring
device but simply an instrument for ruling lines.
Chapter One: Ruler and compass constructions
(1) Given a cube, construct another cube with double the volume
is one of the most celebrated problems of ancient mathematics. There are
two other classical problems of similar stature:
(2) Construct a square with the same area as a given circle
(3) Trisect a given angle.
In this course we will investigate whether problems (1), (2) and (3) can be
solved by ruler and compass constructions. It turns out that they cannot.
We should comment, however, that although the ancient mathematicians were unable to prove that these problems were insoluble by ruler and
compass, they did solve them by using curves other than circles and straight
Some examples of constructions
Before trying to prove that some things cannot be done with ruler and compass, we need to investigate what can be done with those tools. Much of
what follows may be familiar to you already.
#1 Given straight lines AB and AC intersecting at A the angle BAC can
be bisected, as follows. Draw a circle centred at A, and let X, Y be the
points where this circle meets AB, AC. Draw circles of equal radii centred
at X and Y , and let T be a point of intersection of these circles. (The radius
must be chosen large enough so that the circles intersect.) Then AT bisects
the given angle BAC.
#2 Given lines AB and AC intersecting at A and a line P Q, the angle
BAC can be copied at P , as follows. Draw congruent circles CA , CP centred
at A and P . Let CA intersect AB at X and AC at Y , and let CP intersect
P Q at V . Draw a circle with centre V and radius equal to XY , and let T
be a point of intersection of this circle and CP . Then the angle T P Q equals
the angle BAC.
#3 Given a point A and a line P Q, one can draw a line through A parallel
to P Q. Simply draw any line through A intersecting P Q at some point X,
and then copy the angle AXQ at the point A.
#4 Given a line AB one can construct a point T such that the angle T AB
equals π3 radians (60 degrees). Simply choose T to be a point of intersection
of the circle centred at A and passing through B and the circle centred at B
and passing through A.
Chapter One: Ruler and compass constructions
#5 Given line segments of lengths r, s and t one can construct a line
segment of length rt/s, as follows. Draw distinct lines AP , AQ intersecting
at A and draw circles Cr , Cs and Ct of radii r, s and t centred at A. Let Cr
intersect AP at B and let Cs , Ct intersect AQ at X, Y . Draw a line through
Y parallel to XB, and let C be the point at which it intersects AP . Then
AC has the required length.
#6 There are simple constructions for angles equal to the sum and difference of two given angles, lengths equal to the sum and difference of two
given lengths, and for a/n and na, where a is a given length and n a given
positive integer. See the exercises at the end of the chapter.
#7 Given line segments of lengths √
a and b, where a ≥ b, it is possible
to construct a line segment of length ab, as follows. First, construct line
segments of lengths r1 = 21 (a + b) and r2 = 21 (a − b), and draw circles of
radii r1 and r2 with the same centre O. Draw a line through O intersecting
the smaller circle at P , and draw a line through P perpendicular to OP . (A
right-angle can be constructed, for instance, by constructing an angle of π3 ,
bisecting it, and adding on another angle of π3 .) Let this perpendicular meet
the large circle at Q. Then P Q has the required length.
Further ruler and compass constructions are dealt with in the exercises.
Constructible numbers
Consider the Delian Problem once more: we are given a cube and wish to
double its volume. We may as well choose our units of length so that the
given cube has sides
√ of length one. Then our problem is to construct a line
segment of length 3 2. The other problems can be stated similarly. A circle of
unit radius has area π; to√construct a square of this area one must construct
a line segment of length π. A right-angled triangle with unit hypotenuse
and an
angle θ has other sides cos θ and sin θ; to trisect θ one must construct
cos 3 . So the problems become:

(1) Given a unit line segment, construct one of length π.

(2) Given a unit line segment, construct one of length 3 2.
(3) Given line segments of lengths 1 and cos θ, construct one of length
cos θ3 .
Chapter One: Ruler and compass constructions
To show that Problem 3 cannot be solved by ruler and compass, it will
be sufficient to sh …
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