Concrete Algebra and Equations

Lecture 1
Concrete Algebra and Equations
Abstract — Withdrawn or seperated from matter, from material embodiment, from practice,
or from particular examples. Opposed to concrete.
Concrete — Hence, generally, Combined with, or embodied in matter, actual practice, or a
particular example; existing in a material form or as an actual reality, or pertaining to that
which so exists. Opposed to abstract.
Oxford English Dictionary, Second Edition on CD-ROM,
Oxford University Press, 2009 (v4.0.0.3)
Although it may not seem like it at the time, most of this unit will be targeted at
solving polynomial equations. As the Unit Title suggests, much of the content of the unit
will involve quite abstract ideas. The reason for studying those ideas is that they will help
us solve some quite practical problems. As you will find if you study more algebra, we must
first agree on some terms. Rather than have several definitions all in a row I’ll collect some
of terms we’ll need all in one list. After this list we will use familiar algebra to solve some
equations. After this lecture our algebra will move from the concrete to the abstract for a
while.
1.1 Definitions In the following the a∗ are constants, be they integers, rationals, reals
or complex numbers, while i and n are non-negative integers.
(1) A polynomial is any expression of the form
anxn + an-1xn-1 + · · · + a2x2 + a1x + a0.
Note: This isn’t the real definition, but it will do for this unit. For example, a
polynomial might have several things like x, for example 2×2 + 3xy + y2 is also
a polynomial.
(2) Given the polynomial above aixi is the ith term of the polynomial and ai is
the ith coefficient.
(3) If n is the highest power of x with a non-zero coefficient, then we say that
the polynomial has degree n, anxn is the leading term and an is the leading
coefficient.
anxn + an-1xn-1 + · · · + a2x2 + a1x + a0
leading coefficient
leading term
degree (if an is non-zero)
2nd term
coefficient constant term
(4) Two polynomials are equal, if and only if all corresponding coefficients are equal.
(5) A polynomial with leading coefficient equal to 1 is called a monic polynomial.
(6) A polynomial of degree 1 is called a linear polynomial, a polynomial of degree 2
is called a quadratic, a polynomial of degree 3 is called a cubic, a polynomial of
degree 4 is called a quartic and a polynomial of degree 5 is called a quintic. We
will not name higher degree polynomials, although we may encounter them.
(7) The term without a positive power of x, namely a0, is called the constant term.
(8) A polynomial equation is any equation of the form
anxn + an-1xn-1 + · · · + a2x2 + a1x + a0 = 0.
(9) A solution to a polynomial equation is called a root of the (polynomial) equation.
MATH216: W. N. Franzsen (28/2/2018). 1
2 Solving Equations
(10) If p(x) = anxn + an-1xn-1 + · · · + a2x2 + a1x + a0 and p(α) = 0, then we say
that α is a zero of the (polynomial) function. Thus, strictly speaking, equations
have roots, while functions have zeros. This is not a distinction that will be
examined, indeed it seems to be falling out of use.

(11)	If all the coefficients of a polynomial are integers, then we say that we have a polynomial over Z. Similarly, for polynomials over Q, R and C.

Note You should observe that I have not told you what ‘x’ is, in any of the definitions
above. This is because I don’t want to get into a philosophical discussion. Treat the x, which
is called an indeterminate if you like words, in the same way that you treat the word ‘line’
in Euclidean geometry: Everyone knows what a line is, but nobody every really defines it.
§1.1 Solving Polynomial Equations
Before dealing with new solutions we will review the ones we have known for a long time.
Degree 1 A polynomial equation of degree 1, that is a linear equation, looks like the following.
ax + b = 0
As we have been told that this polynomial equation has degree 1 we know that a 6= 0. Since
the time we were first taught any algebra, we have known that the solution to this equation,
that is the set of all values that will satisfy this equation, consists of the single number
x = –
b a
The solution to such equations, as long as everything was positive, was known to the Babylonians before 1600bc. In fact the Babylonians could only solve equations of the form ax = b
with both a and b positive, as they did not deal with negative numbers.
Degree 2 A polynomial equation of degree 2, that is a quadratic equation, looks like the
following.
ax2 + bx + c = 0
Again, we know that a 6= 0 because we have been told that this is degree 2. Before embarking
on the solution, it is worth noting that the solution to such equations, when everything was
positive, was also known to the Babylonians.
We have seen in MATH107 Foundations of Mathematics or earlier that, by completing
the square, we may show that the solutions to the above quadratic equation can by found by
using the quadratic formula.
x =
-b ± √b2 – 4ac
2a
You should observe that a typical quadratic equation has 2 solutions.
Note There are quadratic equations that have only one number that is a solution. In those
cases, where the formula gives the same number twice, we either say that the equation has a
repeated root, or that the root has multiplicity 2.
You may recall that if we call the roots α and β, then we get some relationships between
the roots and the coefficients.
α + β = – b
a
αβ = c
a
The expressions on the left of the two equations above are known as symmetric polynomials.
This is because no matter which solution you call α or β you get the same answer. For
example, swapping the solutions in the first equation gives β + α which is clearly equal
to α + β. Similarly βα = αβ.
Lecture 1 Abstract Algebra and Equations (2012) 3
Exercise It turns out that all symmetric polynomials with two letters can be written as
expressions in terms of the ones given above. For example, α3+β3 is a symmetric polynomial.
By noting that
(α + β)3 = α3 + β3 + 3αβ(α + β),
find an expression for α3 + β3 as an expression in α + β and αβ. You should actually do the
expansion to check the above equation, as we’ll be using this fact later.
Degree 3 A polynomial equation of degree 3, that is a cubic equation, looks like the following.
ax3 + bx2 + cx + d = 0
Before looking at the solution of a cubic, we will note the symmetric polynomials we get. Let
α, β and γ be the solutions of the cubic.
α + β + γ = – b
a
αβ + αγ + βγ = c
a
αβγ = -d
a
This time we get 3 symmetric polynomials α+ β + γ, αβ + αγ + βγ and αβγ. No matter how
we rearrange the roots, these always give the same answer. You should recall that there are
3! = 6 different arrangements of three things, so there are 6 rearrangements to check here.
Exercise Given the roots from above we have
ax3 + bx2 + cx + d = a(x – α)(x – β)(x – γ),
expand the right hand side and equate coefficients to verify the above equations.
We have seen that by 1600bc the Babylonians could solve quadratic equations. It took
humans a further 3000 years to find the solution to cubic equations. Just before 1535ad a
solution was found. (1535 is important because in that year Niccolo Fontana independently
rediscovered the solution, so it was known just before then.)
Major Example Following Ian Stewart, the steps in solving a general cubic equation are
given below. This is not a real example, as the numbers get too messy for a lecture, but you
can use this process to solve a cubic equation if you ever have the need (and the patience).
1. We first divide by the leading coefficient to get a monic polynomial. Indeed we usually
just assume that the polynomial is monic to start with. So we are given the following cubic
equation.
x3 + ax2 + bx + c = 0
2. We now use a sneaky trick that is know as the Tschirnhaus transformation (after
Ehrenfried Walter von Tschirnhaus 1651–1708). We let x = y – a3 .
y – a3 3 + a y – a3 2 + b y – a3 + c = 0
y3 – 3a
3
y2 + · · · + ay2 + · · · = 0
4 Solving Equations
This is messy but, as can be seen, we lose the term in y2. (If you happen to be a computer,
you don’t actually need this step as longer methods will work. But for us humans this is a
short-cut.) This simplifies the cubic to
y3 + py + q = 0,
where
p =
-a2 + 3b
3
q =
2a3 – 9ab + 27c
27
3. Because we’re at the start of this unit and not the end, we need to accept the next
step as just a very sneaky trick. Let y = √3 u + √3 v, then the exercise above shows that
y3 = u + v + 3√3 u√3 v(√3 u + √3 v).
Using these in the polynomial equation we find:
0 = y3 + py + q
0 = (√3 u + √3 v)3 + p(√3 u + √3 v) + q
0 = u + v + 3√3 u√3 v(√3 u + √3 v) + p(√3 u + √3 v) + q
0 = (u + v + q) + (√3 u + √3 v)(3√3 u√3 v + p)
This looks like a mess, and it is, but we now want to choose u and v so that both terms are
equal to 0. That is, we want to find u and v such that
u + v + q = 0
3√3 u√3 v + p = 0
3√3 u√3 v = -p
27uv = -p3
uv = –
p3
27
Thus, we need to find u and v such that
u + v = -q
uv = –
p3
27
Checking back a few pages, this means u and v are solutions to the quadratic equation
t2 + qt – p3
27
= 0.
We can solve this for u and v and track back to find the solutions to the original equation.
Summary Using some sneaky algebra we can reduce a question about a cubic equation to
a question about a quadratic equation.
Lecture 1 Abstract Algebra and Equations (2012) 5
Numerical Example We follow through the steps to solve a particular cubic. To solve
3×3 + 6×2 – 3x – 6 = 0,
we first divide through by 3.
x3 + 2×2 – x – 2 = 0.
Applying the Tschirnhaus transformation we get
y3 – 7
3
y –
20
27
= 0.
Finding the equation for u and v we get
t2 – 20
27
t +
343
729
= 0.
We can now solve for t to find u and v.
For the record
t =
10 ± i9√3
27
leading to the solutions -2, -1 and 1.
Note If you try to track back from u and v, in the above, to get to x you’ll see that there is
a bit more to do than I’ve let on. You can get to the solutions, but be prepared for a lot of
calculation. In particular, if you just take cube roots on your calculator you’ll only find one
possible solution. To get all three you need to remember some stuff from complex numbers.
We’ll see where they all come from later.
Degree 4 A polynomial equation of degree 4, that is a quartic equation, looks like the
following.
ax4 + bx3 + cx2 + dx + ε = 0,
using ε rather than e, as the symbol e is the constant 2.7182 . . .
Within 10 years of the cubic being solved, certainly by 1545, Ludovicco Ferrari had used
similar tricks to reduce the solution of a quartic to solving a cubic. It is then possible to find
the solutions to that cubic and hence solve a quartic.
It is worth noting here that for degree 4 polynomials the Tschirnhaus transformation uses
x = y – α4 . Indeed, if we have a polynomial of degree n, then the Tschirnhaus transformation
uses x = y – αn. Where α is the coefficient (of xn-1) of the monic polynomial related to the
one you start with. In the example given we would have α = ab .
Degree 5 A polynomial equation of degree 5, that is a quintic equation, looks like the
following.
ax5 + bx4 + cx3 + dx2 + εx + f = 0
With the rapid solution of the quartic using methods similar to the cubic, people tried to
attack the quintic in the same way. However, it was quickly found that no matter what was
tried the solution to a quintic equation reduced to solving an equation of degree 6 or more!
That is, all the tricks lead to a more complicated equation.
By 1800, over 200 years later, people began to feel that there was no solution. In 1824 Niels
Abel managed to prove that there is no general solution to the quintic. That is, he proved
that there is no single method that will solve all quintics. Obviously some can be solved, if
you can factorise the quintic for example, but no one method will work for all. By 1832, in
one of mathematics more romantic episodes, Evariste Galois had done something very special
(and was then killed in a duel!). Much of this unit will be an introduction to some of his
work.
6 Symmetric Polynomials
§1.2 Symmetric Polynomials
We’ve seen that the solutions and the coefficients of a polynomial equation are related via
certain expressions that happen to be symmetric polynomials. You should also notice that
the u and v used in the solution of the cubic also appear in symmetric polynomials. In
mathematics a common and powerful technique is to look for ‘invariants’ — things that are
unaffected by changes to other parts of the problem. Much of abstract algebra is concerned
with finding and using such invariants.
1.2 Definition A polynomial in x1, x2, . . . is called a symmetric polynomial if it is
unchanged by any permutation of the xi. That is, no matter how we mix up the x’s the
polynomial is unaffected.
Examples We have seen that the solutions to the cubic equation
ax3 + bx2 + cx + d = 0
satisfy conditions involving symmetric polynomials.
α + β + γ = – b
a
αβ + αγ + βγ = c
a
αβγ = -d
a
We have also met some other symmetric polynomials, plus there are many others.
α3 + β3
αβ(α + β)
1 α
+
1 β
+
1 γ
No matter how we mix up the letters, these expressions keep the same value — are invariant.
For example, if we swap α and β but leave γ fixed, in the equation for ab we find
αβ + αγ + βγ 7→ βα + βγ + αγ = αβ + αγ + βγ
This idea of symmetric polynomials leads to two questions: How many possible rearrangements are there? And is there a nice way of writing the individual rearrangements down?
The first question is easy to answer, indeed we’ve met it before. Given n things there are n!
ways of rearranging them.
We’ll soon see that the best answer to the second question will be to think of the rearrangements as functions. That is, we might have the function, f, such that
f(α) = β
f(β) = α
f(γ) = γ
So, for 3 things, we would need to find 3! = 6 such functions.
We will see in the next lecture that these 6 functions, together with composition of functions,
form what is known as a group. The idea of a group is a very powerful and useful concept.
The problem is that they are an abstract concept.
We started this lecture by looking at how to solve polynomial equations, but we ended up
looking at a set of functions! We must now justify this move to the abstract by showing that
this is a useful thing to do.
Lecture 2
Permutations and Groups
You will recall that we wanted a way of listing all the rearrangements of the roots of an
equation. A cubic, for example, has 3 roots, α, β and γ, and we know that there are 6
rearrangments. What are they, and how can we conveniently list them? Because we want to
generalise to higher degree polynomials, and because we don’t all know the Greek alphabet,
we’ll change from letters to numbers.
Question How can we conveniently write down all the rearrangements of the numbers 1, 2
and 3?
§2.1 Permutations
The answer is that we use permutation notation. In fact we will soon see that there are at
least two ways to answer this question, but the easier one first. Consider the notation given
below.
1 2 3 2 1 3
This is our new notation, taking one of the numbers in the top row, say 1, the number underneath it, 2 here, is the number it is changed to. So this is the function that sends 1 7→ 2, 2 7→ 1
and 3 7→ 3. Or, if we called that whole symbol f, then f(1) = 2, f(2) = 1 and f(3) = 3.
Using this notation we can quickly find all 6 rearrangements, or permutations. The
first row will always read 1 2 3 while we put the 6 rearrangements of those numbers in as the
second row.
1 2 3 1 2 3 1 2 3 1 3 2 1 2 3 2 1 3
1 2 3 2 3 1 1 2 3 3 1 2 1 2 3 3 2 1
Exercise Write down all 4! = 24 permutations of 1, 2, 3 and 4.
Exercise Write down all the permutations of α, β and γ.
It is important to remember that these are really just functions, so we could write
1 2 3 2 1 3 (1) = 2
if we wanted to. Most of the time we won’t need to do that, but it will come in handy for
the next bit.
You should recall that if we have two functions, f and g say, then we can find the
composite of these two functions. (I am ignoring a few important things here, what are
they?) That is, we can attempt to find f ◦ g. You should also remember that this function
is defined as follows.
f ◦ g(x) = fg(x)
Example Given the two permutations (that is, functions)
1 2 3 3 2 1 and 1 2 3 2 1 3
What is
1 2 3 3 2 1 ◦ 1 2 3 2 1 3 ?
MATH216: W. N. Franzsen (28/2/2018). 7
8 Introduction to Groups
Note that we usually just leave out the ◦ to save writing.
As with much of mathematics there is a long way and a short way. The long way is
important to see as it tells us why we do things in the order we do. To find out which function
this composite is equal to, we check what happens to each of the numbers 1, 2 and 3.
1 2 3 3 2 1 ◦ 1 2 3 2 1 3 (1) = 1 2 3 3 2 1 (2) = 2
1 2 3 3 2 1 ◦ 1 2 3 2 1 3 (2) = 1 2 3 3 2 1 (1) = 3
1 2 3 3 2 1 ◦ 1 2 3 2 1 3 (3) = 1 2 3 3 2 1 (3) = 1
That is, the final function is just 1 2 3 2 3 1 and so we can say
1 2 3 3 2 1 1 2 3 2 1 3 = 1 2 3 2 3 1
In the lecture and/or tutorial you will see the short way, and realise why it is very hard to
write down.
This procedure shows why, when multiplying permutations, we start with the one on
the right and work to the left.
Exercise Find each of the following composites/products.
1 2 3 3 1 2 1 2 3 1 2 3
1 2 3 2 3 1 1 2 3 1 3 2
It should be clear at this point that the product of two permutations is another permutation. If you think in terms of rearrangements it is even more obvious. If you rearrange
the numbers 1, 2 and 3, and then perform another rearrangement then we get a third!
§2.2 Sets and Operations
We will now find some properties of permutations and their composition/products.
First of all we should note that point we just made. The product of any two permutations is another permutation.
The next bit is very important but is also a bit abstract. For that reason we will
simplify the notation by using basic function notation, rather than permutations.
2.1 Lemma Given three functions f, g and h that we can compose, we have
f ◦ (g ◦ h) = (f ◦ g) ◦ h
That is, composition of functions is associative.
Proof To show that the two sides of this equation are equal, we need to show that they do
the same thing to all possible values.
LHS
f ◦ (g ◦ h)(x) = f(g ◦ h)(x)
= fgh(x)
Lecture 2 Abstract Algebra and Equations (2012) 9
RHS
(f ◦ g) ◦ h(x) = f ◦ gh(x)
= fgh(x)
Thus f ◦ (g ◦ h)(x) = (f ◦ g) ◦ h(x) and hence
f ◦ (g ◦ h) = (f ◦ g) ◦ h
.
So our operation, which is really just composition of functions, is associative.
Now consider the permutation 1 2 3 1 2 3 . This is the rearrangement where we don’t
actually move any of the numbers (they are fixed). If α is any other permutation it is fairly
easy to see that the following are true.
1 2 3 1 2 3 α = α
α 1 2 3 1 2 3 = α
Exercise Check that this is true for a few specific permutations.
So we can find an element, e, in our set of permutations that acts like 1 does for
multiplication and 0 does for addition. That is,
eα = α and αe = α
An element with this property is call an identity.
Final observation. Given any permutation, for example take 1 2 3 3 1 2 , we can find
a permutation that undoes this one. The easiest way to find that permutation is to turn the
given one upside down and then rearrange the columns.
3 1 2 1 2 3 = 1 2 3 2 3 1
Exercise Are we really allowed to rearrange columns like this? If these were matrices, which
is what they look like, then this would be illegal. What is it about permutation notation that
lets us get away with this?
We should now check that the permutation, that we have just found, does the job we
want it to do. You should evaluate the left hand sides to check that these do really give the
right hand side.
1 2 3 2 3 1 1 2 3 3 1 2 = 1 2 3 1 2 3 = e
1 2 3 3 1 2 1 2 3 2 3 1 = 1 2 3 1 2 3 = e
Like with matrices, these two permutations are said to be the inverse of each other. So,
everything in our set of permutations has an inverse.
Rather than summarise, we will note that the permutations of the numbers 1, 2 and 3
together with composition are our first example that fits the following definition. We will
give the formal (abstract) definition, but keep in mind the discussion above.
10 Introduction to Groups
2.2 Definition Given a set, G, and an operation, ∗, on that set, the pair (G, ∗) is called
a group if the following 4 axioms are satisfied.
(1) (closed) For all x, y ∈ G, x ∗ y ∈ G.
(2) (associative) For all x, y and z ∈ G
x ∗ (y ∗ z) = (x ∗ y) ∗ z
(3) (identity) There is an element e ∈ G such that, for all x ∈ G
x ∗ e = x
e ∗ x = x
(4) (inverses) For all x ∈ G there is a y ∈ G such that
x ∗ y = e
y ∗ x = e
Where e is the identity from axiom 3.
Examples We have seen that the permutations of the numbers 1, 2 and 3 with composition
form a group. We will now consider some more concrete examples.
Consider the integers with addition, that is, (Z, +). To decide whether this is a group
we must check all 4 axioms.
(1) We have known for many years that if x, y ∈ Z then x + y ∈ Z (the sum of two
whole numbers is a whole number). Thus (Z, +) is closed under addition.
(2) We also know that m+(n+p) = (m+n)+p, that is addition of whole numbers
is associative.
(3) The identity for addition is almost always 0. So we note that 0 ∈ Z and that
for m ∈ Z
m + 0 = m
0 + m = m
Thus, 0 is the identity for (Z, +).
(4) If m ∈ Z then we know that -m ∈ Z and
m + (-m) = 0
(-m) + m = 0
Thus, each element of Z has an inverse.
As all 4 axioms are satisfied, we know that (Z, +) is a group.
For the next example we will consider the rational numbers. It is worth remembering
the definition of the rational numbers.
Q = pq p, q ∈ Z, q 6= 0
That is, a number is rational if it can be written as the ratio of two whole numbers, with the
denominator non-zero.
Consider the set of rationals under multiplication. That is, (Q, ×).
(1) Given two rational numbers, p
q
and r
s, with q 6= 0 and s 6= 0, we have
p q
×
r s
=
pr
qs
Lecture 2 Abstract Algebra and Equations (2012) 11
Now both pq and rs are whole numbers and, as q 6= 0 and s 6= 0 we know that
qs 6= 0. Thus p
q
× r
s
∈ Q, and so (Q, ×) is closed.
(2) We have known for a while that multiplication of rational numbers is associative.
(3) If we note that 1 = 1 1 ∈ Q then we can easily see that we have our identity.
p q
× 1 =
p q
1 ×
p q
=
p q
(4) If we are given p
q
∈ Q then we want to say that it’s inverse is q
p
, but we can’t.
We know that p and q are whole numbers but we don’t know that p 6= 0. This
gives us our clue that something bad is going on. Noting that 0 = 01 ∈ Q, if x
was the inverse for 0, then
1 = 0 × x = 0
But we know that 1 6= 0 and so 0 cannot have an inverse.
Because we have only satisfied 3 out of the 4 axioms, we can conclude that (Q, ×) is not a
group.
Exercise Show that (Q∗, ×) is a group, where Q∗ means the set of all non-zero rational
numbers.
§2.3 Symmetries of Plane Figures
Before starting the main work in this section, it is worth remembering that there are 24
permutations of the numbers 1, 2, 3 and 4.
Consider a square with the vertices labelled 1, 2, 3 and 4.
2 1
3 4
What can we do to this square that leaves it looking the same, but possibly moves the
vertices? For a square, we will find that there are several different things we can do, but first
we need a name for those things.
2.3 Definition Given a subset of the plane, S ⊆ R2, if α : R2 → R2 is a function and
α(S) = S, then we say that α is a symmetry of S.
Here we are looking for symmetries of a square. It is worth noticing that one symmetry
that all plane figures have is the identity transformation, that is, the one where we leave the
figure alone. Looking for less trivial ones we can see that there are two types: Reflections
and rotations.
Reflection in a vertical line:
2 1
3 4
-→
1 2
4 3
Rotation through π2 :
2 1
3 4

-→

4 3

2

1 Looking carefully you will find that there are 8 symmetries of a square. You should also note
that each of them can be written as a permutation — this was why we numbered our vertices.
12 Symmetries and Subgroups
So the reflection in a vertical line becomes the permuation 1 2 3 4 2 1 4 3 , because
vertex 1 changed label to vertex 2, and so on. Similarly, the rotation through π2 becomes
1 2 3 4 4 1 2 3 . You should check that the following list contains all 8 symmetries of a square,
and their equivalent as permutations.
Identity 1 2 3 4 1 2 3 4
Rotation through π2 1 2 3 4 4 1 2 3
Rotation through π 1 2 3 4 3 4 1 2
Rotation through 32π 1 2 3 4 2 3 4 1
Reflection in vertical line 1 2 3 4 2 1 4 3
Reflection in horizontal line 1 2 3 4 4 3 2 1

Reflection in line joining 1 and 3

1 2 3 1 4 3 2 4

Reflection in line joining 2 and 4 3 2 1 4
You should check that you can see that each of these is the transformation we claim, and
also that the permutation is the right one.
Now we can see an interesting thing. We have a set of permutations that is also the
set of symmetries of a square. Obviously, if we perform one symmetry and then another,
the square hasn’t moved. Therefore, the composite of two symmetries of a square is another
symmetry. These are all functions and so our operation is associative. The identity is in our
set and it isn’t too hard to check that every symmetry has an inverse that is also a symmetry
of the square. Thus, these 8 permutations, all by themselves, also form a group. So we have
a little group, the 8 symmetries, sitting inside a bigger group, all 24 permutations. This idea
is so important it has a name.
2.4 Definition Given a group (G, ∗) and a subset H ⊆ G. If (H, ∗) is also a group, then
we say that (H, ∗) is a subgroup of (G, ∗), and write
(H, ∗) ≤ (G, ∗)
or (H, ∗) < (G, ∗) if we know that H 6= G. If the operation is obvious, then we often just
write H ≤ G.
Examples The set of symmetries of a regular n-gon will form a subgroup of the group of
permutations of the numbers 1, . . ., n.
The group (Z, +) is a subgroup of (Q, +), which is in turn a subgroup of (R, +).
Exercise Given the group (G, ∗), show that (G, ∗) and (e, ∗) are subgroups of this group.
Lecture 3
Cycle Notation and Some Examples
We met permutations and permutation notation in the last lecture. That notation is perfectly
fine, but can get a bit long winded. For example, when working out the value of the following
product you’d spend more time writing out the question than actually evaluating the answer.
1 2 3 4 5 6 7 8 2 7 3 1 6 4 8 5 1 2 3 4 5 6 7 8 8 7 2 6 1 4 3 5
Mathematicians like compact notation, so we’ll be introducing cycle notation. Some find
cycle notation confusing to start with, so it is important to realise that is it just permutation
notation in disguise.
§3.1 Cycle Notation
Consider one of the permutations given above.
α = 1 2 3 4 5 6 7 8 8 7 2 6 1 4 3 5
Cycle notation comes from following what happens to the symbols if we apply the permutation
repeatedly. For example, look what happens to 1.
α(1) = 8
α2(1) = α(8) = 5
α3(1) = α(5) = 1
If we continue then we will just keep going around the cycle 1 7→ 8 7→ 5 7→ 1 · · · (which is
where the name ‘cycle notation’ comes from). The shorthand for this cycle is (185) (or (851)
or (518) as long as the cycle is the same). Here we understand that the symbol (185) tells
us that 1 7→ 8, 8 7→ 5 but also that the number at the end is mapped to the number at the
start, so 5 7→ 1.
We now pick a symbol that hasn’t already appeared, say 2.
2 7→ 7
7 7→ 3
3 7→ 2
And so (273) is another cycle in our permutation. We now pick another number that we
haven’t seen before, say 4. We find that the last cycle is (46). Thus we have shown that
α = 1 2 3 4 5 6 7 8 8 7 2 6 1 4 3 5 = (185)(273)(46)
As always, it is easy to see that cycle notation is much more compact, but has the usual
downside of being harder to interpret and to multiply.
MATH216: W. N. Franzsen (28/2/2018). 13
14 Cycle Notation
Example Convert 1 2 3 4 5 6 2 1 6 3 4 5 to cycle notation, and convert (291)(38654) to permutation notation.
Following the cycles we can see that 1 7→ 2 7→ 1 and 3 7→ 6 7→ 5 7→ 4 7→ 3, and so
1 2 3 4 5 6 2 1 6 3 4 5 = (12)(3654)
To convert (291)(38654) to permutation notation we need to first decide what the
biggest number is. Here it is 9 and so this is a permutation of the symbols from 1 to 9.
We now go through the numbers in order. As 1 is at the end of a cycle it gets moved to
the number at the beginning, so 1 7→ 2, while 2 clearly gets moved to 9. Following the rest
through we see that the only awkward one is 7, there is no 7. This means that 7 isn’t moved.
Summarising:
1 7→ 2
2 7→ 9
3 7→ 8
4 7→ 3
5 7→ 4
6 7→ 5
7 7→ 7
8 7→ 6
9 7→ 1
Thus we have
(291)(38654) = 1 2 3 4 5 6 7 8 9 2 9 8 3 4 5 7 6 1
Notes Two things to realise here. The second example shows that every permutation belongs
to many groups. The last one above can be considered to be a permutation of the symbols 1
to 10, just by leaving 10 fixed:
(291)(38654) = 1 2 3 4 5 6 7 8 9 2 9 8 3 4 5 7 6 1 = 1 2 3 4 5 6 7 8 9 10 2 9 8 3 4 5 7 6 1 10
This brings up the important question of how to write down a permutation that includes an
integer with two or more digits, for example the permutation that interchanges 2 and 10.
The answer is simple, use commas: (2, 10). But notice that we now need to be careful to
remember which is a cycle and which is a point in the plane.
Multiplying permutations given in cycle notation is also slightly more complicated, but
this only takes a bit of getting used to.
Example Given α = (132)(457) and β = (13)(47)(89) evaluate αβ.
The problem becomes apparent if we write down what αβ looks like:
αβ = (132)(457)(13)(47)(89)
This looks like it is already in cycle notation, but in fact it isn’t. Looking closely, it is not
obvious what this does to the symbol 1 as it appears in two places. We need to clean this up
so that each symbol only appears once. The way to do this is to consider what this whole
product does to each symbol. Remember to start from the right. Let us see what happens
to 1. We start from the right and apply each cycle to the symbol we have at that time:
(89) : 1 7→ 1

apply (47) to 1 apply (13) to 1 apply (457) to 3 apply (132) to 3

(47) : 1 7→ 1 (13) : 1 7→ 3 (457) : 3 7→ 3 (132) : 3 7→ 2

Lecture 3 Abstract Algebra and Equations (2012) 15
We started with 1 and finished with 2 and so αβ moves 1 to 2. You now have a choice, we
can check what happens to each of the symbols to get to permutation notation, or we can
follow the symbols we find to get to cycle notation. This time we’ll aim for cycle notation.
So we follow what happens to 2.
(89) : 2 7→ 2

apply (47) to 2 apply (13) to 2 apply (457) to 2 apply (132) to 2

(47) : 2 7→ 2 (13) : 2 7→ 2 (457) : 2 7→ 2 (132) : 2 7→ 1

Thus 2 returns to 1 and we have our first cycle. Repeating this for all the symbols we find
that
αβ = (12)(75)(3)(4)(6)(89) = (12)(75)(89)
Notice that we can include cycles with a single number, such as (3) above, or not, as takes
our fancy.
Repeating this for βα we find that
βα = (13)(47)(89)(132)(457) = (23)(45)(89)
Notes
(1) It is very important to realise that αβ is not always equal to βα.
(2) If a cycle fixes everything, then we don’t write (1)(2)(3) . . . we just realise that
it is the identity and write e.
§3.2 Cycle Notation and Symmetries
There is another advantage to cycle notation that we won’t really make use of in this unit.
Consider the 6 permutations of the symbols 1 to 3. If you check you can see that they are
also the 6 symmetries of an equilateral triangle.
1
2 3
The following lists each permutation in permutation notation, cycle notation and the corresponding symmetry of the triangle.
1 2 3 1 2 3 e e
1 2 3 1 3 2 (23) reflection
1 2 3 2 1 3 (12) reflection
1 2 3 2 3 1 (123) rotation
1 2 3 3 1 2 (132) rotation
1 2 3 3 2 1 (13) reflection
Notice how the shape of the permutation in cycle notation indicates what type of symmetry
we have.
16 Groups
§3.3 Some Examples of Groups
We have already seen two important examples of groups.
3.1 Definition The group of all permutations of n symbols, with composition of functions
as the operation, is called the symmetric group on n symbols, and is denoted by Sn.
Example S3 = e, (12), (13), (23), (123), (132) .
3.2 Lemma The number of elements of the symmetric group Sn is n!.
3.3 Definition The group of symmetries of a regular n-gon, with composition of functions
as the operation, is called a dihedral group and is denoted by Dn.
Example The group of symmetries of a square is
D4 = e, (12)(34), (14)(23), (13), (24), (1234), (13)(24), (1432)
3.4 Lemma The number of elements of the dihedral group Dn is 2n. (Made up of n
reflections and n rotations, the identity being considered a rotation through the angle 0.)
Facts
(1) Dn ≤ Sn
(2) Every element of Sn is a permutation of the symbols 1, . . . , n. We can consider
any of those elements to be a permutation of the symbols 1, . . . , n + 1 by
assuming that it just fixes n + 1. In this way we have
Sn ≤ Sn+1
Exercise We have just seen that Sn ≤ Sn+1, is it ever true that Dn ≤ Dn+1? Is it ever true
that Dn ≤ Dm?
There is one more special group we will mention here that involves symmetries. Consider the group of symmetries of a non-square rectangle.
2 1
3 4
You should see that there are 4 symmetries: the identity, the rotation through π, a reflection
in a horizontal line and a reflection in a vertical line. This group, not surprisingly, is called
the four-group and is denoted by V , from the German word vier — meaning ‘four’.
V = e, (14)(23), (13)(24), (12)(34)
Before moving on to the last example, it is worth learning a new word. We have seen
that Sn has n! elements, there is a name for this.
3.5 Definition Given the group (G, ∗) there are two possibilities.
(1) If G has an infinite number of elements, then we say that the group has infinite
order, and that G is an infinite group.
(2) If G has a finite number of elements, say |G| = n, then we say that the group
has order n, and that G is a finite group.
In either case we denote the order of (G, ∗) by |(G, ∗)| or just |G|.
Examples Thus the order of Sn is |Sn| = n!, while |Dn| = 2n and (Z, +) has infinite order.
The last group example we will meet requires a new idea. We can split all permutations
into two types — odd and even permutations. There are two ways of deciding whether a
permutation is odd or even, depending upon whether we are using permutation or cycle
notation. An important theorem, that we won’t prove, shows that the two ways always give
you the same answer.
Lecture 3 Abstract Algebra and Equations (2012) 17
Permutation Notation Given a permutation we count the number of inversions: That
is, for each number in the second row we count how many numbers to the left of it are
greater than it. The total we seek is the sum of all these numbers. For example, consider the
following permutation.

α =	1 2 3 4 5 6 7 8 2 7 6 5 3 1 8 4
count	0 0 1 2 3 5 0 4

The total count is 1 + 2 + 3 + 5 + 4 = 15, as this is odd α is an odd permutation. Had the
total count been even we would have an even permutation.
Cycle Notation This is not much harder to calculate but, as you’ll see, it can be tricky
because of one awkward bit. If we convert α from above into cycle notation we get
α = (12784536)
a cycle of length 8. As this cycle has even length it is an odd permutation. (That is the
tricky part!) Thus α is an odd permutation, which agrees with our earlier calculation.
If your permutation is a product of cycles, then you just need to remember the following:
(a) odd permutation × odd permutation = even permutation
(b) odd permutation × even permutation = odd permutation
(c) even permutation × odd permutation = odd permutation
(d) even permutation × even permutation = even permutation
So, given the permutation (123)(4567) we have odd length and even length and so we have
even permutation times odd permutation, giving an odd permutation. We can check this by
converting to permutation notation.
(123)(4567) = 1 2 3 4 5 6 7 2 3 1 5 6 7 4
With a total of 5 inversions, and hence an odd permutation.
Facts
(1) Every permutation is either even or odd, the identity is even.
(2) Exactly half of the permutations of n symbols are even with the other half odd.
(3) For symmetries of plane figures, reflections are all odd and rotations are all even.
3.6 Definition The set of even permutations in Sn forms a subgroup of Sn called the
alternating group. The alternating group is denoted by An.
Example The alternating group on 3 symbols is
A3 = e, (123), (132)
Exercise Find the elements of the alternating group on 4 symbols, A4.
3.7 Lemma For all n
|An| = 1
2
|Sn|
§3.4 Why are we doing this?
We started this unit by claiming we would be solving polynomial equations, but have ended
up looking at apparently pointless permutations of symbols. Why? How do groups come into
the solution of polynomial equations? The following motivation is adapted from Stewart’s
Galois Theory, which is referenced in the Unit Outline.
18 Motivation
Example Consider the following polynomial equation.
x4 – 4×2 – 5 = 0
It should be clear that we can factorise this (it is a quadratic in x2).
0 = x4 – 4×2 – 5 = (x2 – 5)(x2 + 1) = (x – √5)(x + √5)(x – i)(x + i)
So we have 4 solutions: α = √5, β = -√5, γ = i and δ = -i.
We have seen before, that these solutions satisfy the normal symmetric polynomial
identities involving the coefficients. So, for example, we know that
α + β + γ + δ = 0
αβγδ = -5
and so on. All permutations of the symbols α, . . . leave these identities unchanged. But this
is true of all symmetric polynomials. For this specific polynomial there are other identities.
For example, α + β = 0. If we interchange α and β then this one is unaffected, but if we
interchange α and γ then this no longer valid. Similarly we have the identity γ + δ = 0, and
others. Consider the following list of valid identities.
α2 – 5 = 0, α + β = 0, γ2 + 1 = 0, γ + δ = 0, αγ – βδ = 0
We can check that if we take any of these and swap α and β then we get another valid
equation. Similarly if we swap γ and δ, or indeed both α and β and γ and δ.
That is, for this polynomial equation, there are 4 important permutations
α β γ δ α β γ δ , α β γ δ β α γ δ , α β γ δ α β δ γ and α β γ δ β α δ γ
These form a (sub-)group associated with the polynomial equation. It turns out that if we
know the group then we can decide
(1) Whether we can find a solution, or not.
(2) If we can actually find a solution, then how to find it.
Note In the above discussion you will see that we only found the group by using the solutions.
If we already know the solutions then we don’t need the group at all. The question is: Can
we find the group associated with the polynomial equation just from the equation itself?
Lecture 4
Normal Subgroups and Group Homomorphisms
Last week we met the idea of a subgroup. It turns out that, for us, most subgroups are not
particularly interesting. For example, S4 has a total of 30 subgroups but we will only be
interested in e , V , A4 and S4. These particular subgroups of S4 are important because
they are examples of what are known as normal subgroups. Because normal subgroups are
extremely important for a variety of reasons we will spend a bit of time setting them up
properly. As a consequence this lecture covers a lot of the ‘abstract’ part of the unit title.

§4.1	Left and Right Cosets
Before meeting normal subgroups we need to know about cosets. As the title suggests, there are two sorts of coset, left and right.
4.1	Definition Given a group (G, ∗), a subgroup H ≤ G and an element g ∈ G, the left

coset of H generated by g is the set
g ∗ H = g ∗ h
Similarly the right coset of H generated by g is the set
H ∗ g = h ∈ H
Note As the name suggests, most cosets are only sets. For most g ∈ G, g ∗ H will not be a
subgroup.
Example Given the group S3 = e, (12), (13), (23), (123), (132) , consider the subgroup
H = e, (12)
(As an exercise, you should check that this set does form a subgroup of S3.) If we look
at (13) ∈ S3 we can find the left coset generated by this element.
(13)H = (13) e, (12)
= (13)e, (13)(12)
= (13), (123)
We can also find the right coset.
H(13) = e, (12) (13)
= e(13), (12)(13)
= (13), (132)
You should notice that, even with the same subgroup and generator, left and right
cosets need not be the same. The example above shows that they can be different.
Example With S3, as above, we consider the subgroup A3 = e, (123), (132) with the
element (13) ∈ S3.
(13)A3 = (12), (13), (23)
A3(13) = (12), (13), (23)
And so, sometimes, left and right cosets do coincide.
Subgroups with the property that left and right cosets are always the same are the
normal subgroups mentioned earlier.
MATH216: W. N. Franzsen (28/2/2018). 19
20 Cosets and Homomorphisms
4.2 Definition Given the group (G, ∗), the subgroup H ≤ G is called a normal subgroup
of G if every left coset of H is also a right coset of H. If H is a normal subgroup of G then
we write
H ⊳ G
Example With S3, as above, we consider the subgroup A3 = e, (123), (132) . If we
complete the calculation we find that there are exactly 2 cosets of A3 in S3.
Left:
eA3 = (123)A3 = (132)A3 = e, (123), (132)
(12)A3 = (13)A3 = (23)A3 = (12), (13), (23)
Right:
A3e = A3(123) = A3(132) = e, (123), (132)
A3(12) = A3(13) = A3(23) = (12), (13), (23)
Notice that every left coset is also equal to a right coset. For example
(123)A3 = A3 = A3(132).
Thus A3 ⊳ S3.
Example Now consider the subgroup K = e, (12) ≤ S3. This has 3 left and 3 right
cosets.
Left:
eK = (12)K = e, (12)
(13)K = (123)K = (13), (123)
(23)K = (132)K = (23), (132)
Right:
Ke = K(12) = e, (12)
K(13) = K(132) = (13), (132)
K(23) = K(123) = (23), (123)
Notice that the left coset (13)K = (13), (123) is not equal to a right coset. So K is not a
normal subgroup of S3.
There are some things we can observe from these examples that are worth listing
together.
(1) Each coset may have many different names. For example, (13)K = (123)K,
from above.
(2) The subgroup itself is also a (left and right) coset, being e ∗ H = H = H ∗ e.
(3) All cosets have the same number of elements as the subgroup.
(4) Different left cosets never overlap. Similarly, different right cosets never overlap.
§4.2 Group Homomorphisms
Up to now we have been cheating a bit. For example, we have already seen the four-group
written in two ways:
V = e, (12)(34), (13)(24), (14)(23)
V = e, σv, σh, ρπ
Lecture 4 Abstract Algebra and Equations (2012) 21
That is, we have seen it as a set of permutations and as a set of symmetries of a rectangle.
Most of us are happy with this now. But if we look carefully, those two statements can’t both
be true! Equals means equals. What we really mean is that both of these groups behave the
same way. What we have here is called an isomorphism, that is, we can match the elements
in each group so that they behave in exactly the same way. I have written my sets so that
the corresponding elements match up. We can best see this with the Cayley tables. We write
down the tables for both sets making sure that they keep the order we’ve chosen.
e (12)(34) (13)(24) (14)(23)
e e (12)(34) (13)(24) (14)(23)
(12)(34) (12)(34) e (14)(23) (13)(24)
(13)(24) (13)(24) (14)(23) e (12)(34)
(14)(23) (14)(23) (13)(24) (12)(34) e
e σv σh ρπ
e e σv σh ρπ
σv σv e ρπ σh
σh σh ρπ e σv
ρπ ρπ σh σv e
What you should notice is that the pairs that match up, σv and (12)(34) for example, appear
in the tables in exactly the same places. In the tables below we’ve coloured that pair in red
to emphasise this. Similarly for the other pairs.
e (12)(34) (13)(24) (14)(23)
e e (12)(34) (13)(24) (14)(23)
(12)(34) (12)(34) e (14)(23) (13)(24)
(13)(24) (13)(24) (14)(23) e (12)(34)
(14)(23) (14)(23) (13)(24) (12)(34) e
e σv σh ρπ
e e σv σh ρπ
σv σv e ρπ σh
σh σh ρπ e σv
ρπ ρπ σh σv e
That is, the two groups behave in exactly the same way. An isomorphism is a way of matching
two groups in this way, so that every element of each group is matched with exactly one
element from the other group. We also require that the multiplications behave in the same
way.
If you have a Cayley table for the groups then it is easy to check if you have an
isomorphism. You take one table, replace each element with its matching partner and then
check to see if the new table is correct. That is, the products match the entries in the table.
Most people quickly become happy with the idea of an isomorphism. All we have really
done is change the labels on our elements. Much more important are cases where things don’t
match up exactly. In that case we get a homomorphism. That is, a way of relabelling the
elements of one group with elements from another group so that the operations still work.
This is easier to explain with a definition and examples.
22 Cosets and Homomorphisms
4.3 Definition Suppose we have two groups, (G, ∗) and (Γ, ⊗), and a function f : G → Γ.
The function f is a group homomorphism if it satisfies the following axiom.
For all x and y ∈ G we have
f(x ∗ y) = f(x) ⊗ f(y)
Note on language The bits of the words isomorphism and homomorphism come from
Greek. The -morph part means ‘form’ or ‘shape’, iso- means equal and homo- means same.
There is a subtle distinction here. Isomorphism (equal-shape) compared to homomorphism
(same-shape) are related in the same way as congruent and similar are in geometry. Isomorphic groups (as we call them) are exactly the same in all respects, while a homomorphism
just relates groups that look roughly the same in some ways but differ in others.
Example 1 Consider the two groups (Z, +), the group of 2 × 2 invertible matrices under
multiplication, M2×2, ×) and the function
f(n) = h 1 0 1 n i
We should first check that this function does what its supposed to, that is, that the answer
is a 2 × 2 matrix, which it is, and that the matrix is invertible. This time that is also easily
checked, the determinant is equal to 1 6= 0. We now have to check that
f(m + n) = f(m) × f(n)
We evaluate both sides (separately!):
LHS = f(m + n) = h 1 0 1 m+n i
RHS = f(m) × f(n)
= h 1 0 1 m i × h 1 0 1 n i
= h 1+0 0+0 0+1 n+m i
= h 1 0 1 m+n i
= LHS
Therefore, this is a group homomorphism.
For us, the last example is about as odd as things might get. All of our groups just
use multiplication, so most of our homomorphisms look pretty normal. However, before we
simplify things there are a few facts that are always true and are worth noting.
Example 2 Define the function f : V → Q, where V is the four group and Q the quaternions
that we’ve seen in tutorials. As we have finite groups here we’ll just list the function values.
f(e) = I
f(12)(34) = -I
f(13)(24) = -I
f(14)(23) = I
Decide whether this function is a homomorphism.
For finite groups, as with isomorphism, the simplest way to check is to take a Cayley
table of the domain and apply the function to each of the elements. We then check that
every product in the new table is correct. Yes, every product needs to be checked to show
Lecture 4 Abstract Algebra and Equations (2012) 23
that it is a homomorphism. However, if one fails, then you can stop as you don’t have a
homomorphism. We use a table from above.
e (12)(34) (13)(24) (14)(23)
e e (12)(34) (13)(24) (14)(23)
(12)(34) (12)(34) e (14)(23) (13)(24)
(13)(24) (13)(24) (14)(23) e (12)(34)
(14)(23) (14)(23) (13)(24) (12)(34) e
Now apply the function to each element of this table.
I -I -I I
I I -I -I I
-I -I I I -I
-I -I I I -I
I I -I -I I
A check of every product shows that all work and so we have a homomorphism.
The idea of a group homomorphism is a function that behaves well with the group
operations. This would seem to suggest that it behaves well with the other axioms of a
group. This is true, as the next lemma shows.
4.4 Lemma Suppose that (G, ∗) and (Γ, ⊗) are groups and that f : G → Γ is a group
homomorphism. We will write eG for the identity in G and eΓ for the identity in Γ. Then
(1) f(eG) = eΓ
(2) For all g ∈ G we have fg-1 = f(g)-1
Proof Being very careful, and only using the group axioms:
We know that eG ∗ eG = eG, simply because it is the identity in G.
f(eG ∗ eG) = f(eG)

f(eG) ⊗ f(eG) = f(eG)	f is a homomorphism
f(eG) ⊗ f(eG) ⊗ f(eG)-= f(eG) ⊗ f(eG)-1	Γ is a group axiom 4 axiom 3
f(eG) ⊗ eΓ = eΓ f(eG) = eΓ

1 You should now see how the second proof is going to be done. As G is a group, if g ∈ G
then it has an inverse and so
g ∗ g-1 = eG
fg ∗ g-1 = f(eG) did same to both sides

f(g) ⊗ fg-1 = eΓ f(g)-1 ⊗ f(g) ⊗ fg-1 = f(g)- 1

f is a homomorphism

1 ⊗ eΓ did same to both sides
eΓ ⊗ fg-1 = f(g)-fg-1 = f(g)-1
There is one more important idea that is useful to know about. The identity in a
group is a special element. In Example 2, you should notice that the function sends both e
and (14)(23) to the identity of Q. The lemma tells us that e had to be mapped to I, but
the example shows that other elements can be mapped to the identity. Surprisingly, this idea
turns out to be vital for all of group theory.
24 Cosets and Homomorphisms
4.5 Definition If (G, ∗) and (Γ, ⊗) are groups and f : (G, ∗) → (Γ, ⊗) a group homomorphism, then the kernel of f is the set of elements in G whose function image is eΓ, the
identity in Γ. We write ker(f) for this set.
ker(f) = g ∈ G
Note that ker(f) ⊆ G.
One of the main reasons for the importance of kernels is that they are all normal
subgroups, as we’ll see. Even more importantly, every normal subgroup can be obtained as
the kernel of a homomorphism. Indeed, if you asked me to show that a given subgroup was
a normal subgroup I would tend to find a suitable homomorphism, rather than bother with
the coset mess.
4.6 Lemma If (G, ∗) and (Γ, ⊗) are groups and f : (G, ∗) → (Γ, ⊗) a group homomorphism, then
ker(f) ⊳ G
Proof We need to show that every left coset is also a right coset. In fact, we will show a
little more by directly showing that g ∗ ker(f) = ker(f) ∗ g for any g ∈ G.
Suppose g ∈ G and consider the left coset g ∗ ker(f). If we take any x ∈ g ∗ ker(f) then
we can find k ∈ ker(f) such that
x = g ∗ k
and so we have
f(x) = f(g ∗ k)
= f(g) ⊗ f(k)

= f(g) ⊗ eΓ = f(g)

k ∈ ker(f)

That is, the image of every element of g ∗ ker(f) is equal to f(g). Now suppose that we have
an element y ∈ G such that f(y) = f(g). Let z = g-1 ∗ y.
z = g-1 ∗ y
f(z) = fg-1 ∗ y
= fg-1 ⊗ f(y)
= f(g))-1 ⊗ f(y) by Lemma 4.4
= f(g)-1 ⊗ f(g) we assumed f(y) = f(g)
= eΓ
Which tells us that z ∈ ker(f), and so
y = eG ∗ y = g ∗ g-1 ∗ y = g ∗ g-1 ∗ y = g ∗ z ∈ g ∗ ker(f)
We have therefore shown that
g ∗ ker(f) = f(x) = f(g)
We can swap the order of things in the above to show that
ker(f) ∗ g = f(x) = f(g) = g ∗ ker(f)
Thus ker(f) is a normal subgroup of G.
Lecture 5
Lagrange’s Theorem

§5.1	Lagrange’s Theorem
In the last lecture we observed some properties of cosets.
(1)	Each coset may have many different names. For example, (13)K = (123)K, from above.

(2) The subgroup itself is also a (left and right) coset, being e ∗ H = H = H ∗ e.
(3) All cosets have the same number of elements as the subgroup.
(4) Different left cosets never overlap. Similarly, different right cosets never overlap.
We will now prove all of these, except the first for which we have already provided examples.
The following lemma will talk about left cosets, similar results can be established for right
cosets.
5.1 Lemma Given a group (G, ∗) with subset H ≤ G.
(1) If x and y ∈ G then either x ∗ H = y ∗ H or (x ∗ H) ∩ (y ∗ H) = ∅.
(2) If H is a finite subgroup of G and x ∈ G then |x ∗ H| = |H|.
Proof We have a group (G, ∗), a subgroup H ≤ G and elements x, y ∈ G. To prove the first
part we will show that if x ∗ H and y ∗ H overlap then they must be equal. As then either
the two cosets do not overlap, and so their intersection is empty, or they coincide. To do this
we will show that x ∗ H ⊆ y ∗ H and that y ∗ H ⊆ x ∗ H.
1 Suppose that x ∗ H and y ∗ H overlap, that is (x ∗ H) ∩ (y ∗ H) 6= ∅, then there must be
an element in both cosets, say
z ∈ (x ∗ H) ∩ (y ∗ H)
Now x ∗ H is the set of all elements of the form x ∗ h for some h ∈ H. Thus, there is an
element r ∈ H such that z = x ∗ r. Similarly, there is an element s ∈ H such that z = y ∗ s.
Thus we have the following.
x ∗ r = y ∗ s
y-1 ∗ (x ∗ r) = y-1 ∗ (y ∗ s)
(y-1 ∗ x) ∗ r = (y-1 ∗ y) ∗ s
= e ∗ s
= s
(y-1 ∗ x) ∗ r = s
(y-1 ∗ x) ∗ r ∗ r-1 = s ∗ r-1
y-1 ∗ x = s ∗ r-1
Now H is a subgroup of G and so is a group. As r ∈ H we must have r-1 ∈ H because H
is a group and so it contains the inverse of each of its elements. But H is closed under the
operation ∗ and so, given s ∈ H and r-1 ∈ H we have
y-1 ∗ x = s ∗ r-1 ∈ H
We will now show that x ∗ H ⊆ y ∗ H. To that end suppose that w ∈ x ∗ H is an
element of x ∗ H. As before we know that we must be able to find an element t ∈ H such
MATH216: W. N. Franzsen (28/2/2018). 25
26 Lagrange’s Theorem
that w = x ∗ t. Then
w = x ∗ t
= e ∗ (x ∗ t)
= (y ∗ y-1) ∗ (x ∗ t)
= y ∗ (y-1 ∗ x) ∗ t
Now t ∈ H, and we have just shown that y-1 ∗ x ∈ H and so u = (y-1 ∗ x) ∗ t ∈ H and hence
w = y ∗ u ∈ y ∗ H.
As every element of x ∗ H is also an element of y ∗ H we must have
x ∗ H ⊆ y ∗ H
A similar proof will show that y ∗ H ⊆ x ∗ H and hence
x ∗ H = y ∗ H
So, either (x ∗ H) ∩ (y ∗ H) = ∅ or the two cosets are equal.
2 We are told that H is finite, so suppose that |H| = n, that is, that H has n elements:
H = h1, h2, . . . , hn
Consider the coset
x ∗ H = x ∗ h1, x ∗ h2, . . . , x ∗ hn
Just looking at the list of elements of x ∗ H we can see that |x ∗ H| ≤ n. The only way we
can have fewer than n elements in x ∗ H is if two of the elements on our list coincide. That
is, if we can find hi and hj with hi 6= hj but x ∗hi = x ∗hj. But, if we can find such elements
then
x ∗ hi = x ∗ hj
x-1 ∗ (x ∗ hi) = x-1 ∗ (x ∗ hj)
(x-1 ∗ x) ∗ hi = (x-1 ∗ x) ∗ hj
e ∗ hi = e ∗ hj
hi = hj
but we know that hi 6= hj and so we have a contradiction. Therefore that possibility cannot
arise and so x ∗ H cannot have fewer than n elements. Thus we have
|x ∗ H| = n = |H|.
We can now prove one of the most important results in group theory. For the record
Joseph Louis Lagrange (1736–1813) has been described as the greatest mathematician of the
eighteenth century, despite the name he was born in Turin.
5.2 Lagrange’s Theorem Given a finite group (G, ∗) and a subgroup H ≤ G, |H| is a

factor of |G|. That is,

|H| |G|.

Lecture 5 Abstract Algebra and Equations (2012) 27
Proof We first note that if g ∈ G then g is an element of g ∗ H: As H is a subgroup of G it
is a group, and so contains the identity. Thus
g = g ∗ e ∈ g ∗ H.
This tells us that every element of G lies in a coset of H. We have already seen, in Lemma 5.1,
that cosets don’t overlap, and so each element of G lies in exactly one coset of H. As G is
finite, there can only be a finite number of cosets of H. Say
G = (g1 ∗ H) ∪ (g2 ∗ H) ∪ · · · ∪ (gn ∗ H)
Where none of these cosets have anything in common. Counting elements we can see that
|G| = |g1 ∗ H| + |g2 ∗ H| + · · · + |gn ∗ H|
But, by Lemma 5.1, all cosets of H have the same number of elements as H, thus |gi∗H| = |H|
and hence
|G| = |g1 ∗ H| + |g2 ∗ H| + · · · + |gn ∗ H|
= |H| + |H| + · · · + |H|

| H
Thus |H| is a factor of |G|.
Example We have seen that e , V , S3, D4 and A4 are all subgroups of S4.
| e | = 1,	|V | = 4,	|S3| = 6,	|D4| = 8 and |A4| = 12

and these numbers are all factors of |S4| = 4! = 24.
Note A very natural question arises at this point. Given a finite group, the order of every
subgroup is a factor of the order of the group. Is the reverse the case? That is, if m is a
factor of |G| is there a subgroup of G with order m? The answer is a definite ‘maybe’ —
both cases can occur. We know that |S4| = 24 has factors
1, 2, 3, 4, 6, 8, 12 and 24
We have seen examples for all except 2, 3 and 24. You can easily check that the following
are subgroups of S4 and obviously have the appropriate orders to fill in the gaps.
e, (12) , e, (123), (132) and S4
An example of the other side of the coin is a bit harder but it can be shown (that is code
for — ‘not in this unit’) that A4 does not have a subgroup of order 6, even though |A4| = 12
and 6 is a factor of 12.
§5.2 Not Examinable
In this section we prove that A4 does not have a subgroup of order 6. This proof is not
examinable, and you shouldn’t even read it unless you really want to know.
First we list the elements of A4:
A4 = e, (12)(34), (13)(24), (14)(23),
(123), (132), (124), (142),
(134), (143), (234), (243)
28 Lagrange’s Theorem
You should note that there are a total of eight 3-cycles in S4 and so in A4.
Suppose that H ≤ A4 and |H| = 6. Notice that A4 only contains 4 elements that
aren’t 3-cycles, thus H must contain at least two 3-cycles. Relabelling my symbols if I have
to, we may assume that (123) ∈ H. As H is a group it contains the inverse of all of its
elements, thus we also know that (132) ∈ H. As all subgroups must contain the identity, we
know that
H = e, (123), (132), a, b, c
As we’ve just seen, the 3-cycles come in pairs, if you have a 3-cycle you have its inverse. We
have 3 more slots and so can only put in at most one more pair of 3-cycles. This means
that H must have at least one element that isn’t a 3-cycle or the identity — we already have
the identity.
Suppose that element was (12)(34) then, because H is a group it must contain both
(12)(34)(123) and (12)(34)(132).
(12)(34)(123) = (243)
(12)(34)(132) = (143)
But this would add 2 more pairs of 3-cycles, which is too many as we only have room for one
more pair. So the extra element can’t be (12)(34).
Checking the other possible elements.
If (13)(24)
(13)(24)(123) = (142)
(13)(24)(132) = (234)
If (14)(23)
(14)(23)(123) = (134)
(14)(23)(132) = (124)
In both cases we also add 2 more pairs of 3-cycles when we can only add one more pair, and
we have proved the following.
5.3 Lemma No subgroup of A4 has order 6.
Lecture 6
Polynomials and Irreducibility
We’ve seen how the solutions of a polynomial equation can lead to a discussion of subgroups.
Before continuing we need some more facts about polynomials. In particular, we need to
decide which polynomials we need to concentrate on. For instance, at first sight, the quartic
equation x4 + 7×2 + 12 = 0 looks awful to solve, until we remember a trick. Letting X = x2
our equation reduces to
0 = x4 + 7×2 + 12 = X2 + 7X + 12
= (X + 3)(X + 4)
= (x2 + 3)(x2 + 4)
And our awful equation reduces to two quadratic equations x2 + 3 = 0 and x2 + 4 = 0. There
is no point concentrating on a big polynomial if it factorises.
There is another consideration. Look at the quadratic equation x2 + 4 = 0. In high
school you would have said that this doesn’t factorise, but we now know that
x2 + 4 = x2 – (-4) = x2 – (2i)2 = (x – 2i)(x + 2i)
You can even meet this without the need for complex numbers. The quadratic x2 – 2x – 1
doesn’t have any nice factors, but it does factorise

x2 – 2x – 1 = (x – 1 +	√2)
These are the two considerations that we will deal with in this lecture.
(1)	How do we know if a polynomial factorises?

√2)(x – 1 – (2) What numbers can we use when finding factors?
The two are closely connected. If we use complex numbers then every polynomial factorises
into degree 1 factors.
6.1 Fundamental Theorem of Algebra If g(x) is a polynomial of degree n then there
are complex numbers α, r1, . . . , rn such that
g(x) = α(x – r1)(x – r2) · · · (x – rn)
That is, using complex numbers all polynomial factorise completely. The problem, as
we have seen, is finding the factors.

§6.1	Irreducible Polynomials
If we wish to consider the factors of polynomials we need to know exactly what we mean by the word factor.
6.2	Definitions Given polynomials f(x), g(x) and h(x).

(1) If f(x) = g(x)h(x), then we say that g(x) and h(x) are factors of f(x).
(2) If, in addition, we know that the degree of g(x) and h(x) are both greater than 0
then we say that g(x) and h(x) are proper factors of f(x).
MATH216: W. N. Franzsen (28/2/2018). 29
30 Irreducible Polynomials
Exercise Convince yourselves that, if f(x) = g(x)h(x), g(x) has degree m and h(x) has
degree k then f(x) has degree m+k. (Harder question: What does this say about the degree
of the zero polynomial, f(x) = 0, that is, the polynomial all of whose coefficients are 0?)
We now need to become very specific. We’ve seen above that the polynomial x2-2x-1
does not factorise if you want to use rational numbers, but it does factorise if you use reals.
Rather than give several separate definitions, let A represent one of the following sets of
numbers
Z, Q, R, C
and others that we will meet soon.
6.3 Definitions
(1) The polynomial anxn + an-1xn-1 + · · · + a1x + a0 is called a polynomial over A
if all the coefficients are elements of A, that is, ai ∈ A for all i.
(2) The polynomial, f(x), factorises over A if we can find polynomials g(x) and h(x)
over A that are proper factors of f(x), that is, such that f(x) = g(x)h(x) and
the degrees of g(x) and h(x) are at least 1.
(3) If a polynomial does not factorise over A then we say it is irreducible over A.
Exercises
(1) Show that any polynomial of degree 1 is irreducible.
(2) Show that if f(x) = g(x)h(x) is reducible, then the degrees of both g(x) and h(x)
are strictly less than the degree of f(x).
Examples
x2 – 5x + 18 is a polynomial over Z, and Q and . . .
x2 – 5
3x + 18 is a polynomial over Q, and R and . . ., but not Z
x2 – √2 x + 18 is a polynomial over R, and C, but not Q

x2 – ix + 18 x2 – 2x – 1 x2 + 4

is a polynomial over C, but not R is irreducible over Q but not over R is irreducible over R but not over C

Our Strategy We will consider irreducible polynomials. Obviously all degree 1 polynomials are irreducible, but their zeros are very easy to find. So we concentrate on irreducible
polynomials of degree greater than 1. As all irreducible polynomials over C are degree 1 we
are looking at irreducible polynomials over Z, Q or R with degree greater than 1. These are
the polynomials that we will mainly be looking at.
There is one more thing we can say, again the proof of this comment is not examinable:
Over R irreducible polynomials have degree 1 or 2. That is, all polynomials over R can be
factorised into linear and quadratic terms.
In fact we will concentrate, initially, on Q and Z. Here we also have one lucky fact
that reduces our work further. Given a polynomial over Q we can multiply by the lowest
common denominator of the coefficients to obtain a related polynomial over Z. But there is
one possible problem, it might be that a polynomial over Z only factorises into polynomials
over Q. For example, it might be that
x5 + 6x + 2 = 1 2×2 + 3 4x – 78 2×3 + · · ·
This is a rubbish example because, in fact, this can’t happen.
6.4 Gauss’s Lemma Given a polynomial, f(x), over Z. If f(x) is irreducible over Z
then it is irreducible over Q.
That is, if a polynomial with integer coefficients factorises over Q then it factorises
over Z.
The proof of Gauss’s Lemma is not examinable.
Lecture 6 Abstract Algebra and Equations (2012) 31
Note Carl Friedrich Gauss (1777–1855) is generally considered to be one of the greatest, if
not the greatest, mathematician.
This means that we can mostly worry about polynomials with integer coefficients. This
is a big advance, but still leaves us in the dark. For example, is the following polynomial
irreducible over Q?
x3 + 3×2 + 3x + 3
If it is irreducible, then we know that all solutions (which we can find) will be somewhat
nasty. If it is not irreducible then we should be able to find the factors. As you’ll know
by now, finding factors is hard. Like integration in calculus, there are a few tricks that
sometimes work, but not always. In fact, the polynomial just mentioned can easily be shown
to be irreducible over Q because of one nice test. There are a few such tests, if one works we
have an answer. If none work then who knows, might be irreducible, might not be.
6.5 Eisenstein’s Irreducibility Criterion (EIC) Given a polynomial over Z
f(x) = anxn + an-1xn-1 + · · · + a1x + a0
If there is a prime, p, such that
(1) p is not a factor of an
(2) p is a factor of the remaining coefficients, an-1, . . ., a1, a0

(3)	p2 is not a factor of a0
Then f(x) is irreducible over Z (and hence over Q). Proof Not examinable??

Note Ferdinand Gotthold Eisenstein (1823–1852) was a German mathematician who produced many important results before, like Galois and Abel, dying tragically young.
Examples Given f(x) = x3 + 3×2 + 3x + 3. Clearly 3 is prime, is not a factor of 1, the
leading coefficient, is a factor of all other coefficients and 32 = 9 is not a factor of the constant
term. Thus, by EIC, f(x) is irreducible over Z.
Given x19 + 7×11 – 49×8 + 28, we can see that this polynomial is irreducible by EIC
using p = 7. It is important to realise that if a term is missing then the coefficient is 0, and
every number is a factor of 0 because 0 = 0 × p. Thus 7 is a factor of all the coefficients
except the leading coefficient, and 49 is not a factor of 28.
It is important to realise that if EIC works, then the polynomial is irreducible. However,
if EIC doesn’t work then we know nothing about the irreducibility of the polynomial. An
example below will show that x4 + 1 is irreducible, even though EIC does not apply.
There are some tricks that can be used for some polynomials. One trick is to change
our polynomial into one that is easier to work with.
Example Consider the polynomial g(x) = x3 – 3×2 + 3x + 1. As no primes are factors
of the constant term, we can’t use EIC here. But we can use the following observation.
If g(x) = h(x)k(x) then clearly g(x + b) = h(x + b)k(x + b) and vice versa. Thus, g(x) is
reducible if and only if g(x+b) is reducible. Therefore g(x) is irreducible if and only if g(x+b)
is irreducible. We now note the following.
g(x + 1) = (x + 1)3 – 3(x + 1)2 + 3(x + 1) + 1
= x3 + 3×2 + 3x + 1 – 3×2 – 6x – 3 + 3x + 3 + 1
= x3 + 2
Now EIC with p = 2 shows that g(x + 1) is irreducible over Z and hence so is g(x).
32 Some Revision
Example Consider the polynomial h(x) = x4 + 1. Again EIC does not apply, but
h(x + 1) = (x + 1)4 + 1 = x4 + 4×3 + 6×2 + 4x + 2
is irreducible by EIC with p = 2. Thus x4 + 1 is Irreducible over Z and hence over Q.
From an earlier note we know that x4 + 1 is not irreducible over R as the only irreducibles in that case have degree 2. But, is it possible to find the factors? If we remember
our complex numbers we can.
x4 + 1 = x4 – (-1)
= x4 – i2
= (x2 – i)(x2 + i)
If we knew the square roots of i and -i then we could complete the factorisation. In fact
±
√2
2
(1 + i)!2 = 1 2(1 + i)2
=
1 2
(1 + 2i – 1)
= i
±
√2
2
(1 – i)!2 = 1 2(1 – i)2
= -i
Thus we have
x4 + 1 = (x2 – i)(x2 + i)
= x –
√2
2
(1 + i)! x + √22(1 + i)!
× x –
√2
2
(1 – i)! x + √22(1 – i)!
= x –
√2
2
(1 + i)! x – √22(1 – i)!
× x +
√2
2
(1 + i)! x + √22(1 – i)!
= (x2 – √2 x + 1)(x2 + √2 x + 1)
You can, of course, check that this is correct by expanding the final product. It is important
to remember that this can only be done this way if you can find all the zeros of the polynomial.

§6.2	Some Revision
While talking about finding factors of polynomials it is worth remembering those facts we learnt prior to this unit. In particular the following two results.
6.6	Remainder Theorem	Given the polynomial f(x), the remainder when f(x) is

divided by x – a is f(a). That is,
f(x) = q(x)(x – a) + f(a).
Lecture 6 Abstract Algebra and Equations (2012) 33
6.7 Factor Theorem Given the polynomial f(x), if f(a) = 0 then x – a is a factor
of f(x).
Example The polynomial f(x) = 17×5 – 12×4 + 3×3 + √2×2 – (√2 + 1)x – 7 looks to be a
very complicated polynomial until you realise that
f(1) = 17 – 12 + 3 + √2 – √2 – 1 – 7 = 0
and so x – 1 is a factor. In fact
f(x) = (x – 1)17×4 + 5×3 + 8×2 + (√2 + 8)x + 7
Example Find all the degree one factors of the polynomial
g(x) = x4 + x3 + x2 + 3x – 6
If this polynomial factorises then, by Gauss’s Lemma 6.4, it factorises over Z. As the leading
coefficient is 1, we know that any zeros must be integers. Indeed we know that they must be
factors of the constant term, namely -6.
The factors of -6 are

±1,	±2,	±3	and	± 6
Using the Factor Theorem 6.7, we look for factors by evaluating g(x) at each of these values.
g(-1) = 8 g(-2) = 0 g(-3) = 48	g(1) = 0 g(2) = 28 g(3) = 120 g(6) = 1560
g(-6) = 1092

Thus the linear factors of g(x) are x – 1 and x + 2. We were not asked to do it, but the
factorisation is
g(x) = (x – 1)(x + 2)(x2 + 3)
§6.3 A Proof of EIC
Remember that this proof is not examinable, but it is worth being available for those who
want to see it.
Proof (of EIC 6.5) Given the polynomial
f(x) = anxn + an-1xn-1 + · · · + a1x + a0
Suppose that p is a prime number such that the conditions of EIC are met, that is
(1) p is not a factor of an
(2) p is a factor of an-1, . . ., a0
(3) p2 is not a factor of a0
Suppose, however, that f(x) is not irreducible over Z, then we can find proper factors g(x)
and h(x) such that f(x) = g(x)h(x).
g(x) = bmxm + bm-1xm-1 + · · · + b0
h(x) = ckxk + ck-1xk-1 + · · · + c0
34 A Proof
Looking at the constant terms we can see that a0 = b0c0. Now p is a factor of a0 and so,
given that p is prime, p must be a factor of b0 or c0. It can’t be a factor of both, otherwise p2
would be a factor of a0. We may assume that p is a factor of b0. (If it isn’t, the argument
below can be edited to work for the other case.) We will show that in this case p is a factor
of all of the coefficients of g(x).
To prove that p is a factor of all the bi we will assume that it isn’t, and get to a
contradiction. As p is a factor of b0, we let bj be the first coefficient that isn’t a multiple
of p. So p is not a factor of bj, but is a factor of b0, b1, . . ., bj-1. We now consider aj, the
coefficient of xj in f(x).
f(x) = (bmxm + · · · + b0)(ckxk + · · · + c0)
Expanding the right hand side we can see that
aj = b0cj + b1cj-1 + · · · + bjc0
bjc0 = aj – b0cj – b1cj-1 – · · ·
Now g(x) and h(x) are proper factors of f(x) and so their degrees are less than the degree
of f(x), thus j ≤ m < n. This tells us that p is a factor of aj. We also know that p is a
factor of b0, b1, . . ., bj-1, and so p is a factor of the right hand side of . Thus p is a factor
of bjc0. But this is a contradiction as the prime p is not a factor of bj or c0 and so cannot be
a factor of bjc0. Thus all coefficients of g(x) are divisible by p.
In particular, p is a factor of bm and so is a factor of an = bmck. But this contradicts
the fact that p is not a factor of an. Hence, f(x) must be irreducible over Z.
The proof of Gauss’s Lemma follows roughly similar lines to this. That is, you consider
the primes dividing the coefficients of the product. Sadly, that proof is too long to even appear
in a section like this.
Lecture 7
Polynomial Equations and An Introduction To Fields
Suppose we have an irreducible polynomial over Q, for example
x2 – 2x – 1
We know that this is irreducible over Q but not over C.
Exercise Using EIC show that this polynomial is irreducible over Z and hence over Q.
This tells us that the zeros of x2 – 2x – 1 are not rational numbers. But, in a sense,
we don’t need all the reals. With the zeros being 1 + √2 and 1 – √2 we really only need to
‘include’ √2 and we can then factorise
x2 – 2x – 1 = (x – 1 + √2)(x – 1 – √2)
§7.1 Solving Equations Using Radicals
Radical — from latin Radix — meaning root. Radicals are square roots, or cube roots or
fourth roots etc. So examples of radical expressions are
√2, √7 11, q1 + √3 3, 23 s 11 √p313 + 2 + √17 5√719
7.1 Definition A polynomial has a solution in radicals over A if it has a zero that can be
written as an expression involving numbers in A, +, -, ×, ÷ or the extraction of nth roots.
Example All quadratics are solvable in radicals of the set of numbers that contain their
coefficients. Given a quadratic polynomial over A
ax2 + bx + c = 0
with a, b and c ∈ A, the solutions are
x =
-b ± √b2 – 4ac
2a
Which are obviously solutions in radicals over A.
Using this definition we can now frame our question properly.
Main Question Does every polynomial over A have solutions in radicals over A?
We know that the answer is yes for quadratics no matter what set of numbers we have
for A. We will also see that the same is true for cubics. Is it true for all polynomials?
§7.2 Minimal Polynomials
The discussion in this section applies to all solutions of polynomial equations, but we will
focus on radical expressions. (Remember that we do not yet know if all solutions are radical
expressions or not.)
Given a radical expression, it is possible to work from that expression to find a polynomial over Z that has that expression as a zero. You should note that proving this is
surprisingly hard. It is obvious for simple expressions, but more complicated ones show
where the problems come in.
MATH216: W. N. Franzsen (28/2/2018). 35
36 Minimal Polynomials
Examples
(1) Given √17, we let
x = √17
x2 = 17
x2 – 17 = 0
Thus we have a polynomial over Z with √17 as a zero.
Notice that, yet again, the numbers you are allowed to use as coefficients determine the polynomial. If we could use reals, then x – √17 would do!
(2) Given √2 + √3.
x = √2 + √3
x – √2 = √3
(x – √2)2 = 3
x2 – 2√2 x + 2 = 3
x2 – 1 = 2√2x
(x2 – 1)2 = 8×2
x4 – 2×2 + 1 = 8×2
x4 – 10×2 + 1 = 0
You should now see where the problem comes in. What could we do with √2 + √3 + √5? If
you check, you’ll find that this number is a zero of
x8 – 40×6 + 352×4 – 960×2 + 576
Exercise Try to confirm that the given polynomial has √2 + √3 + √5 as a zero. You’ll
quickly find the problems so don’t spend too much time looking when you get stuck. (Notice
that we said ‘when’ not ‘if’.)
Exercise We have seen that √2 + √3 + √5 is a zero of
x8 – 40×6 + 352×4 – 960×2 + 576
This is a polynomial of degree 8 and so has 8 solutions. Can you guess the 8 solutions?
Things get even worse if you consider radical expressions like √ √2+ 5+√ √3 7. In this case the
relevant polynomial is
16×8 – 960×6 + 968×4 – 240×2 + 1
Exercise Can you write down the 8 solutions in this case?
We will ignore the problem of finding these polynomials (and we hope you realise that
we are using machines to do this for us, even though we could find the polynomials ourselves
if you really wanted us to). Suppose we have a number α and a polynomial, f(x), over A
such that f(α) = 0. If f(x) factorises over A then
f(x) = g(x)h(x)
for some polynomials over A with the degrees of g(x) and h(x) less than the degree of f(x).
But
0 = f(α) = g(α)h(α)
and so either g(α) = 0 or h(α) = 0 or possibly both are true. If you keep doing this, you will
eventually find a polynomial over A of smallest degree such that α is a zero of the polynomial.
This polynomial will be irreducible.
Lecture 7 Abstract Algebra and Equations (2012) 37
7.2 Definition Given a number α, the minimal polynomial over A is the monic polynomial
over A that has α as a zero and such that no polynomial over A with smaller degree has α
as a root.
Remember ‘Monic’ means that the leading coefficient is 1 (Definition 1.1.)
7.3 Lemma Given the set of numbers A.
(1) The minimal polynomial over A for a number α is irreducible over A.
(2) The minimal polynomial over A for a number α is unique.
Proof The discussion before the definition shows that the minimal polynomial is irreducible.
If it wasn’t, then we could find a polynomial of smaller degree that had α as a zero.
Now suppose that we had two monic polynomials over A of the same smallest degree
with α as zeros. Say f(α) = 0 and g(α) = 0, where
f(x) = xm + · · ·
g(x) = xm + · · ·
If f(x) 6= g(x) then h(x) = f(x) – g(x) 6= 0 and h(x) has smaller degree than f(x) and g(x)
because the leading terms cancel. But
h(α) = f(α) – g(α) = 0 – 0 = 0
This is a contradiction as f(x) and g(x) are supposed to have the smallest degree for which
this can happen. Thus f(x) = g(x) and the minimal polynomial is unique.
Examples
(1) Find the minimal polynomial of 1 + √3 over Q.
x = 1 + √3
x – 1 = √3
x2 – 2x + 1 = 3
x2 – 2x – 2 = 0
Now x2 – 2x – 2 is a monic polynomial over Q that is irreducible by EIC with
p = 2.
(2) Find the minimal polynomial of 2 + 3i over R.
x = 2 + 3i
x – 2 = 3i
x2 – 4x + 4 = -9
x2 – 4x + 13 = 0
This is a minimal polynomial that is irreducible over R because the discriminant
is negative.
b2 – 4ac = 16 – 52 = -46 < 0
(3) Find the minimal polynomial of 1 + √3 2 over Q.
x – 1 = √3 2
x3 – 3×2 + 3x – 1 = 2
x3 – 3×2 + 3x – 3 = 0
Using EIC with p = 3 shows that the monic polynomial x3 -3×2 +3x-3 is irreducible over Z and hence over Q. Thus we have found the minimal polynomial.
38 Counter-Examples
§7.3 The Sets of Numbers
We now need to look closely at some properties of the numbers that we need to have. Keeping
the taking of radicals apart for the moment, we want to be able to add, subtract, multiply
and divide (except for division by 0). Of our number sets, we can do all of these except when
using Z. But we have seen that we can work with Z and get answers for Q, and we can divide
in Q.
7.4 Definition A set of numbers in which we can add, subtract, multiply and divide is
called a field.
More formally, using the same ideas as for groups: (A, +, ×) is called a field if the
following axioms hold.
(1) If x, y ∈ A then x + y ∈ A.
(2) If x, y, z ∈ A then x + (y + z) = (x + y) + z.
(3) There is a 0 ∈ A such that for all x ∈ A
x + 0 = x = 0 + x
(4) If x ∈ A there is a y ∈ A such that x + y = 0 = y + x.
(5) If x, y ∈ A then x + y = y + x.
(6) If x, y ∈ A then x × y ∈ A.
(7) If x, y, z ∈ A then x × (y × z) = (x × y) × z.
(8) There is a 1 ∈ A such that for all x ∈ A
x × 1 = x = 1 × x
(9) If x ∈ A and x 6= 0 there is a y ∈ A such that x × y = 1 = y × x.
(10) If x, y ∈ A then x × y = y × x.
(11) If x, y, z ∈ A then x × (y + z) = (x × y) + (x × z).
(12) If x, y, z ∈ A then (y + z) × x = (y × x) + (z × x).
You can immediately see that the definition of a field is much more complicated than
that of a group. But a closer look shows that these are just the properties we are used to
with numbers. Addition and multiplication are associative a(bc) = (ab)c, and commutative x + y = y + x, we have a zero for addition and a 1 for multiplication. We also have the
distributive laws x(y + z) = xy + xz.
As we will only be working with numbers, it is worth focussing on those properties
that aren’t always true. To do that let’s list those of the above that will always be true for
any set of numbers. Consider axioms 2, 5, 7, 10, 11 and 12. Each of these are always true
for any numbers. Take axiom 2, for example. No matter what set of numbers we have, it is
always true that
x + (y + z) = (x + y) + z
That is just the way that addition works with numbers. On the other hand, axiom 1 is not
always true for all sets of numbers. If our set of numbers was
A = 1, 2, 3, 4, 5, 6
then it is clear that axiom 1 doesn’t always hold. We have 5, 6 ∈ A but 5 + 6 = 11 is not an
element of A.
Therefore, to establish that a given set of numbers is a field, we only need to check
axioms 1, 3, 4, 6, 8 and 9.
Examples It is easy to see that Q, R and C are all fields. It is also clear that Z is not, as
we can’t divide, that is (Z, +, ×) does not satisfy axiom 9.
Lecture 7 Abstract Algebra and Equations (2012) 39
Exercise Show that A = -1, 0, 1 satisfies all axioms except for axiom 1.
§7.4 Counter-Examples
We have seen some examples of fields and some examples of sets of numbers that aren’t fields.
We have also seen that if set of numbers fails a single axiom then it isn’t a field. As we showed
above all sets of numbers satisfy some of the axioms, indeed for these sets of numbers we
only need to check axioms 1, 3, 4, 6, 8 and 9. The question arises, are there sets of numbers
that satisfy all except one of these axioms. The answer is ‘almost’. If you don’t have a 0
then you cannot have additive inverses, so if the set does not satisfy axiom 3 then it cannot
satisfy axiom 4. Similarly for axioms 8 and 9. Other than those pairs and axiom 6 there are
sets of numbers that satisfy all except one of the axioms.
Examples
(1) Consider the set X1 = -1, 0, 1. This set is clearly not closed under addition
as 1 + 1 = 2 ∈/ X1, and so axiom 1 does not hold. However, it clearly contains 0,
all elements have a negative, it is closed under multiplication, it has a 1 and
multiplicative inverses. Thus X1 satisfies all axioms except 1.
(3) (Remembering we will not have axiom 4 either.) Consider the set X3 = Q+, that
is, the set of positive rational numbers. The sum and product of two positive
rationals is a positive rational. We don’t have 0, or negatives (as advertised),
but 1 is a positive rational, and given a positive rational, rs, it is true that s r is
also a positive rational. Thus X3 satisfies all the axioms except 3 and 4.
(4) Let X4 = X3 ∪ 0 . Obviously we now have 0 and the arguments for (3) show
that X4 satisfies all the axioms except 4.
(6) In fact, it can’t happen that a set satisfies all the axioms except axiom 6. See
Bob’s Proof below.
(8) (Remembering we will not have axiom 9 either.) Let X8 = 2Z, that is, the set
of even integers. This clearly satisfies all the axioms except 8 and 9.
(9) Let X9 = Z. Again this satisfies all the axioms except 9.
Bob’s Proof If X6 satisfies all except possibly axiom 6 then, in fact, X6 satisfies all axioms.
Now 1 ∈ X6 and as X6 is closed under addition we have n ∈ X6 for all integers n. A similar
argument shows that for all x ∈ X6, nx ∈ X6. Therefore X6 contains nx-1-1 = nx -1 = nx
for all integers n. Thus mn x ∈ X6 for all integers m and n and x ∈ X6.
Now if x, y ∈ X6, then x1 + y1 -1 ∈ X6, thus
xy
x + y
∈ X6
Define x ∗ y = xxy +y , then if x, y ∈ X6 we have x ∗ y ∈ X6. So if x ∈ X6, then it also contains
y = 1 + x + x1 ∗ 1 – x – x1
=
1 + x + x1 1 – x – x1
1 + x + x1 + 1 – x – x1
=
1 – x + x1 2
2
=
1 – x2 – 2 – 1
x2
2
=
-1 – x2 – 1
x2
2
40 Counter-Examples
and hence -1 – 2y = x2 + x12 is in X6.
Applying the same reasoning to 2x we find that 4×2 + 4×12 ∈ X6 and so X6 also contains
4 4×2 + 4×12 – x2 + x12 = 15×2
And so, as X6 contains all multiples of elements by rationals, x2 ∈ X6. Thus, X6 is closed
under squaring.
Now, if x and y ∈ X6 then so is
1 4
(x + y)2 – (x – y)2 = xy
and X6 is closed under multiplication. That is, axiom 6 holds.
This proof shamelessly stolen from Dr Bob Howlett.
Lecture 8
Field Extensions and Irreducible Polynomials
We can now show the connection between irreducible polynomials and fields. What will then
remain to do will be to connect groups with them both. (And, of course, to solve the things!)
Returning to an old friend, look at the polynomial x2 – 2x – 1. This is irreducible
over Q but not over R. But we don’t need all of R. In fact, if we just add √2 this polynomial
factorises.
§8.1 Field Extensions
8.1 Definition Given a field A and a number α, the extension of A by α is the set of all
complex numbers made up from elements of A, α, addition, subtraction, multiplication and
division. The extension of A by α is denoted by A(α). (We read that symbol as “A with α”
not “A of α”.)
Example Given the field Q and √2, the extension Q(√2) includes, for example
-3,
1 2
, 1 + √2, 7 + 3√2
2 – √2
Facts
(1) The full definition of A(α) is that it is the set of all numbers of the form f(α)
g(α) ,
where f(x) and g(x) are polynomials over A and g(α) 6= 0.
(2) In fact, for radical expressions and some other numbers you only need a polynomial expression h(α). For example,
7 + 3√2
2 – √2 =
(7 + 3√2)(2 + √2)
(2 – √2)(2 + √2) = 10 +
13
2
√2
(3) You can add more than one number at a time. For example, Q(√2, √3) includes
such numbers as √2, 1 – √3 and √3 + 4√6.
Example Over Q the polynomial x2 – 2x – 1 is irreducible, but over Q(√2) it is not.
In fact Q(√2) is the smallest field in which x2 -2x -1 factorises. Finding the smallest
field in which a polynomial factorises fully is one of the tasks we need to complete.
§8.2 Polynomials and Extensions
Consider the polynomial x4 – 10×2 + 1 over Q, we’ve seen this one before, √2 + √3 is a zero.
This means that x4 -10×2 + 1 is not irreducible over Q(√2+√3). How does it factorise over
this extension? Time for some old fashioned algebra.
a = (√2 + √3)0 = 1
b = (√2 + √3)1 = √2 + √3
c = (√2 + √3)2 = 5 + 2√6
d = (√2 + √3)3 = 11√2 + 9√3
MATH216: W. N. Franzsen (28/2/2018). 41
42 Irreducible Polynomials
Because it is a field, Q(√2 + √3) must contain a, b, c and d, and so the following are also
in Q(√2 + √3).
1 2
(d – 9b) = 1
2
d – 9
2
b = √2
11
2
b – 1
2
d = √3
Thus √2 ∈ Q(√2 + √3) and √3 ∈ Q(√2 + √3). In fact it can now be easily shown that
Q(√2 + √3) = Q(√2, √3)
This means that the numbers
-√2 + √3, √2 – √3 and – √2 – √3
are also all in Q(√2 + √3) = Q(√2, √3). If you check, these are the other 3 zeros of our
polynomial.
Exercise Show that x4 – 2×2 + 7 is irreducible over Q.
In this example, we added one zero of an irreducible polynomial and got the rest for
free. The next examples show that this does not always happen.
Example A zero of x4 – 5×2 + 6 is √2, so we know that this polynomial is not irreducible
over Q(√2). We can easily check that -√2 is another zero and so
x4 – 5×2 + 6 = (x – √2)(x + √2) × g(x)
But this time we do not get all the zeros in Q(√2).
x4 – 5×2 + 6 = (x – √2)(x + √2) × g(x)
= (x2 – 2)g(x)
= (x2 – 2)(x2 – 3)
as you can check by completing the division. The remaining zeros are ±√3 which do not lie
in Q(√2). Our polynomial was not irreducible to start with. As this shows, if the polynomial
is not irreducible there need not be any connection between the zeros of the factors. We will
therefore concentrate on irreducible polynomials from now on.
It would be nice if adding one zero of an irreducible polynomial always gave you all of
the zeros. Sadly this is not true. Consider the polynomial x3 – 2 which has zeros

√3 2,	√3 2 cos 2π + i sin 23π	and	√3 2	cos	4π	+ i sin	43π
3	3

Notice that one root is real and the other two are not. If we add the real root, √3 2, to
get Q(√3 2) then all of the numbers in that field are obviously real numbers. We can’t possibly
get the remaining two zeros, which are not real. So, this time, we can add one zero and not
get the remaining ones.
So, some extensions are nice and some aren’t. The nice extensions are called normal
extensions, and are related to normal subgroups.
Lecture 8 Abstract Algebra and Equations (2012) 43
§8.3 Irreducible Polynomials and Extensions
We will consider irreducible polynomials to show how we may simplify our work with extensions. Let f(x) be a monic irreducible polynomial, and let α be a zero of f(x). Remember
that monic means that the leading coefficient of f(x) is 1, suppose
f(x) = xn + an-1xn-1 + · · · + a1x + a0
We will consider the extension Q(α).
Suppose β ∈ Q(α), then we know that we can find polynomials, g(x) and h(x), over Q
such that
β = g(α)
h(α)
Now suppose that g(x) has degree m ≥ n, say
g(x) = bmxm + · · · + b0
We know that f(α) = 0 and so
0 = αn + an-1αn-1 + · · · + a0
αn = -an-1αn-1 – · · · – a0
αm = αm-n × αn
= αm-n(-an-1αn-1 – · · · – a0)
= -an-1αm-1 – · · · – a0αm-n
Using this expression for αm in g(α) we find
g(α) = bmαm + bm-1αm-1 + · · · + b0
= bm – an-1αm-1 – · · · – a0αm-n + bm-1αm-1 + · · · + b0
And we have found a polynomial of degree m – 1 whose value at α is the same as g(α). That
is, if the degree of g(x) is at least n, then we can find a polynomial of smaller degree that
will work just as well. Thus we may assume at the top that the degree of g(x) is less than n.
Similarly, we can assume that the degree of h(x) is less than n.
Example Suppose α = √3 2, so that α is a root of the polynomial x3 – 2. Then α3 = 2 and
so we have
α4 – 3α3 + 2α2 + α – 5 = α × α3 – 3 × 2 + 2α2 + α – 5
= 2α – 6 + 2α2 + α – 5
= 2α2 + 3α – 11
And the final expression has degree 2, which is less than the degree of x3 – 2.
This tells us that if β ∈ Q(α) then we can find polynomials, g(x) and h(x), of degree
less than n over Q such that
β = g(α)
h(α)
In fact we can do much better than that.
44 Irreducible Polynomials
Example Given the polynomial, x2 -2, which is irreducible over Q, a zero is α = √2. Thus,
if β ∈ Q(√2) then we can find polynomials of degree less than 2 such that
β = g(x)
h(x)
But, a polynomial of degree less than 2 is linear. Thus we can find rationals a, b, c and d
such that
β = a + b√2
c + d√2
But you should be able to see that we can always simplify this further.
β = a + b√2
c + d√2
=
a + b√2
c + d√2 ×
c – d√2
c – d√2
=
(ac – 2bd) + (bc – ad)√2
c2 – 2d2
= r + s√2
That is, β = k(α) for some polynomial over Q with degree less than 2.
This is always true, although sadly we won’t be able to prove it. In any extension by
the zero of an irreducible polynomial we need only consider expressions of the form g(α), for
polynomials g(x).
8.2 Lemma Given a field A and a zero, α, of a polynomial of degree n over A that is
irreducible over A we have
A(α) = a0 + a1α + · · · + an-1αn-1
Proof Not examinable, as the proof needs too much information about vector spaces and
we haven’t even met them!
Example The polynomial f(x) = x3 + x + 1 is irreducible over Q, let α be a zero of f(x)
and consider the extension Q(α).
If β ∈ Q(α) then
β = aα2 + bα + c
dα2 + εα + f
We now multiply top and bottom by ξα2 + µα + ν, where
ξ = -d2 + df – ε2
µ = d2 + fε
ν = -d2 + 2df – f2 – dε – ε2
As then the denominator becomes
(dα2 + εα + f)(ξα2 + µα + ν) = -d3 – d2f + 2df2 – f3 + d2ε – 3dfε – f2ε + ε3
which is an element of A (as α has been eliminated). To verify this you need to remember
that α3 + α + 1 = 0 and so
α3 = -α – 1
α4 = -α2 – α
Exercise
(1) Check the above calculations.
(2) How did I find ξ, µ and ν?
Lecture 9
Solving Cubic and Quartic Equations
Given the cubic
f(x) = x3 + ax2 + bx + c
with zeros z1, z2 and z3. We know that f(x) factorises in A = Q(z1, z2, z3). We will now
roughly describe the Galois Correspondence. (Remember that we are leaving out many
details.)
§9.1 The Galois Group
Roughly speaking, if we consider permutations of the zeros of f(x), so for example, given the
permutation (12), we have (12) : z1 7→ z2 and vice versa. We can see (sort of, details omitted
here) that all of S3 can act on A = Q(z1, z2, z3). In this case S3 is called the Galois Group
of A over Q.
Note For specific polynomials the Galois Group might be a subgroup of S3. This can happen
if some permutations don’t make sense. For example, if we had Q(√2, √3) then we aren’t
allowed to swap √2 and √3. Why? You can only swap things that behave in the same way
algebraically and √2 and √3 don’t. For example, √2 has minimal polynomial x2 – 2 over Q
and that is not the minimal polynomial of √3. The real story here is quite complicated,
which is why we are avoiding it.
Once we have associated a group to the extension we get a correspondence (a matching)
between subgroups of the Galois Group, S3 here, and fields inside A.
Example We know that H = e, (12) ≤ S3. We look for the elements of A that are
unaffected by H. We can easily convince ourselves that this set is
Q(z1 + z2, z1z2, z3)
Notice how the symmetric polynomials crop up again. We can also see that this field is
smaller than A but larger than Q, for example the number z1 is not in Q(z1 + z2, z1z2, z3).
In fact, for every subgroup of the Galois Group we get a field and vice versa. Also
note that bigger subgroups give smaller fields. The biggest and smallest being
S3 ←→ Q and e ←→ A
As mentioned before, normal subgroups are important in all of this. What we really
need is a chain of normal subgroups starting at e and finishing with the Galois Group.
But not just any chain will do. For example, you can’t just have the chain e ⊳ G, as the
gap between them usually turns out to be too big. For the group S3 we have a nice chain
e ⊳ A3 ⊳ S3
remembering that A3 = e, (123), (132) . We will use this chain to find the solution of the
general cubic. Most of the details will make sense but some won’t as we haven’t filled in all
the gaps.
MATH216: W. N. Franzsen (28/2/2018). 45
46 Solving Cubics
§9.2 Solving Cubics
Given the cubic
f(x) = x3 + ax2 + bx + c
with zeros z1, z2 and z3. We consider the extension Q(z1, z2, z3) and look for elements fixed
by A3 but that aren’t fixed by all of S3. So we avoid the symmetric polynomials. To simplify
the algebra let ω be a complex number such that ω3 = 1. From our study of complex numbers
we know that there are three possible values for ω

cos + i sin	or	cos + i sin	or	1
2π	2π	4π	4π
3	3	3	3

If we let ω be the first of these, then the list above is ω, ω2 and ω3 = 1. (Notice the similarity
to A3, which has elements (123), (123)2 and (123)3 = e.)
Now consider
y = z1 + ωz2 + ω2z3
Applying (123) to y we get
y′ = z2 + ωz3 + ω2z1
But
ω2y = ω2z1 + ω3z2 + ω4z3
= ω2z1 + z2 + ωz3
= y′
Thus (123) : y 7→ ω2y. A similar argument shows that
(132) : y 7→ ωy
This tells us that y is not fixed by A3, but we can see that y3 is fixed.
e : y3 7→ y3
(123) : y3 7→ (ω2y)3 = (ω2)3y3 = (ω3)2y3 = y3
(132) : y3 7→ (ωy)3 = ω3y3 = y3
Thus, y3 is fixed by every element of A3. We can apply the same argument to
z = z1 + ω2z2 + ωz3
to show that z3 is also fixed by A3.
This is the first step in our chain. Now consider an element of S3 that is not an element
of A3, say (12).
(12) : y 7→ z2 + ωz1 + ω2z3 = y′′
ωz = ωz1 + ω3z + 2 + ω2z3 = z2 + ωz1 + ω2z3 = y′′
So (12) : y 7→ ωz and hence y3 7→ (ωz)3 = z3. You can check that every element of S3 that
isn’t in A3 swaps y3 and z3. While the elements of A3 fix both y3 and z3. Thus, S3 fixes
y3 + z3 and y3z3
We now recall what this means for our fields. We saw, above, that the field associated with S3
was Q, that is, anything that is fixed by S3 must be a rational number. Hence, both y3 + z3
and y3z3 are rational.
Lecture 9 Abstract Algebra and Equations (2012) 47
Group theory has told us that y3+z3 and y3z3 are the important numbers to consider.
We now look at those numbers more closely.
y3 + z3 = (z1 + ωz2 + ω2z3)3 + (z1 + ω2z2 + ωz3)3
= 2(z13 + z23 + z33)
– 3(z12z2 + z12z3 + z1z22 + z22z3 + z1z32 + z2z32)
+ 12z1z2z3
And we get 3 symmetric polynomials in z1, z2 and z3. We know that if we write these in
terms of the basic ones, then we have expressions involving the coefficients of the original
cubic.
z1 + z2 + z3 = -a
z1z2 + z1z3 + z2z3 = b
z1z2z3 = -c
Now
(z1 + z2 + z3)3 = z13 + z23 + z33
+ 3(z12z2 + z12z3 + z1z22 + z22z3 + z1z32 + z2z32)
+ 6z1z2z3
This gives us an expression for z13 + z23 + z33 that we can use in the above to show that
y3 + z3 = 2(z1 + z2 + z3)3
– 9(z12z2 + z12z3 + z1z22 + z22z3 + z1z32 + z2z32)
You can also check that
z2
1z2 + z12z3 + z1z22 + z22z3 + z1z32 + z2z32 = (z1 + z2 + z3)(z1z2 + z1z3 + z2z3)
– 3z1z2z3
Hence
y3 + z3 = 2(z1 + z2 + z3)3
– 9(z1 + z2 + z3)(z1z2 + z1z3 + z2z3)
+ 27z1z2z3
= 2(-a)3 – 9(-a)b2 + 27(-c)
y3 + z3 = -2a3 + 9ab – 27c
The calculations you’ve already done can be used to show that
y3z3 = a6 – 9a4b + 27a2b2 – 27b3
Thus we can find the values of y3 + z3 and y3z3 from the coefficients of the original cubic.
Then y3 and z3 are the solutions of
X2 – (y3 + z3)X + (y3z3) = 0
Which is a quadratic equation with rational coefficients.
48 Solving a Cubic
§9.3 Solving a Cubic
Consider the cubic equation
x3 – 6×2 + 12x – 11 = 0
From above we know that
y3 + z3 = -2(-6)3 + 9(-6)12 – 27(-11) = 81
y3z3 = (-6)6 – 9(-6)412 + 27(-6)2122 – 27(12)3
= 0
Thus y3 and z3 are the solutions of
X2 – 81X = 0
and so y3, z3 = 0, 81 . Say we take y3 = 81 and z3 = 0 and so y = 3√3 3 and z = 0. This
is all very well, but we really want z1, z2 and z3. There is also the problem that we (usually)
get a choice of cube roots. Here z is obviously equal to 0, but y could be 3√3 3 or ω3√3 3 or
even ω23√3 3. We’ll worry about that problem in a minute.
Now we know
z1 + ωz2 + ω2z3 = y
z1 + ω2z2 + ωz3 = z
z1 + z2 + z3 = -a
This is a system of simultaneous linear equations. Solving (using matrices!) we find
z1 =
1 3
(y + z – a)
z2 =
1 3
(ω2y + ωz – a)
z3 =
1 3
(ωy + ω2z – a)
Here y = 3√3 3, z = 0 and a = -6 and so
z1 = 2 + √3 3
z2 = 2 + ω√3 3
z3 = 2 + ω2√3 3
Note You have (or will have) seen in the tutorials, that if we choose y and z so that
yz = a2 – 3b
then z1, z2 and z3 given above will be the roots of the cubic.
Lecture 9 Abstract Algebra and Equations (2012) 49
Summary Given the cubic x3 + ax2 + bx + c = 0. We solve the quadratic equation
X2 – αX + β = 0
where
α = -2a3 + 9ab – 27c
β = a6 – 9a4b + 27a2b2 – 27b3
The solutions to this quadratic are y3 and z3. Choose the cube roots y and z so that
yz = a2 – 3b
The solutions to the original cubic are
z1 =
1 3
(y + z – a)
z2 =
1 3
(ω2y + ωz – a)
z3 =
1 3
(ωy + ω2z – a)
where ω = cos 2π
3 + i sin 23π. For later reference we write this as a matrix equation

z1
z2
z3

=
1 3

1 1 1
ω2 ω 1
ω ω2 1


y z -a

Note Applying the Tschirnhaus transformation first ensures that a = 0. That simplifies
much of the work, but at the cost of two extra steps.
Obviously this process is more complicated than the nice quadratic formula, you would
expect this. But, nonetheless, it appears to be relatively straight-forward. Sadly that is not
always the case. The tricky part is finding nice formulae for the cube roots of y3 and z3.
Example Consider the polynomial equation (x – 1)(x – 2)(x + 3) = x3 – 7x + 6 = 0.
This is already set up nicely, we don’t even need to think about using the Tschirnhaus
transformation. We can find the polynomial for y3 and z3
X2 + 162X + 9261 = 0
which has solutions
X = -81 ± 30i√3
Now, to find y and z we need the cube root of these numbers, but not in the obvious form
q3 -81 – 30i√3
We know that the roots are 1, 2 and -3, so we need a nice form for these cube roots. It is
possible that you might find that
(3 – 2i√3)3 = -81 – 30i√3
but, then again, you might not. With patience you might just be able to find the (obvious)
solutions to this cubic.
50 Solving Quartics
Example Consider the polynomial x(x + 1)(x – 2) = x3 – x2 – 2x = 0, which again has
obvious solutions. This time we obtain the quadratic
X2 – 16X + 343 = 0
with solutions
X = 8 ± 3i√31
Finding the cube roots this time is much harder.
Of course, the obvious solution to this is to check for nice factors first. (Which is what
any well written app does.) But in other cases, you need to keep in mind that there might
be nice solutions even though yours look nasty.
Example Consider the cubic equation x3 + x2 + x + 1 = 0. The equation for y3 and z3 is
X2 + 20X – 8 = 0
and again we find awkward numbers. But we are wise now
x3 + x2 + x + 1 = (x + 1)(x2 + 1)
and so the solutions are -1 and ±i.
§9.4 Quartics and Higher
To repeat this for quartics we need a suitable tower of normal subgroups in S4. There is such
a tower.
e ⊳ V ⊳ A4 ⊳ S4
Where V = (12)(34), (13)(24), (14)(23) , is the 4-group we have met before. We’ll see how
to solve quartics next time.
Continuing, for quintics we need a tower of normal subgroups, but we can’t find one.
The main reason is that, although we can start the chain with A5 ⊳ S5, that is where it
stops. The only normal subgroups of A5 are A5 itself and e , and the jump from A5 to e
is too big. This means that the general quintic cannot be solved. If we have a particular
quintic for which the Galois Group is a proper subgroup of S5, we might be lucky and have
a group that does possess a suitable tower. But there can be no process that will work for
all quintics. Moreover, there are quintics for which no process at all will work.
Similarly, for higher order polynomials we get an equivalent block. None of the alternating groups A6, A7, . . . have any proper normal subgroups (and there is no other way to
start the tower). Thus, no higher order polynomials have a general solution (using radicals).
§9.5 Solving Quartics
Given the quartic equation
x4 + ax3 + bx2 + cx + d = 0
Let the solutions be z1, z2, z3 and z4. We start at the bottom of the chain
V = e, (12)(34), (13)(24), (14)(23)
We want to find some expressions involving the zi that are fixed by S4, as we will then know
that those numbers are rational. To this end, we will consider V and set
y1 = (z1 + z2)(z3 + z4)
y2 = (z1 + z3)(z2 + z4)
y3 = (z1 + z4)(z2 + z3)
Lecture 9 Abstract Algebra and Equations (2012) 51
It is clear how we found these. Notice that each element of S4 changes one yi into another yj,
for example
(123) : y1 = (z1 + z2)(z3 + z4) 7→ (z2 + z3)(z1 + z4) = y3
This tells us that S4 permutes the yi and so fixes the following numbers.
α = y1 + y2 + y3
β = y1y2 + y1y3 + y2y3
γ = y1y2y3
It’s now time to state explicitly a fact we’ve hinted at before.

9.1	Lemma If the field F is a (normal) extension of the field A, so A ⊆ F , and G is the
Galois group of F over A, then the set of numbers fixed by every element of G is A.
Proof Not examinable.

Thus, the numbers α, β and γ are in the field Q(α, β, γ) and are fixed by every element
of S4, the galois group. Thus these three numbers are rational. In fact we can write them in
terms of the coefficients of the polynomial
α = 2b
β = ac + b2 – 4d
γ = abc – a2d – c2
Then, y1, y2 and y3 are solutions of the cubic
y3 – αy2 + βy – γ = 0
Which we can solve using the methods from before.
It is worth digressing for a moment to show why we should have expected to get a
cubic at this point. We haven’t really done enough group theory to do this properly but you
should get the idea. We started with the chain of normal subgroups
e ⊳ V ⊳ A4 ⊳ S4
Because of what is coming up it is worth listing the orders of these groups
1 4 12 24
We started with the group V and found an equation that the yi satisfied. Solving that
equation will solve the quartic. In group theory terms what we have done is found some
quotient groups. Don’t worry what they are, suffice it to say that they behave a little bit like
quotients of numbers. We found the quotient with respect to V and so we get the new chain
V/V ⊳ A4/V ⊳ S4/V
which have orders
4 4

= 1

12 = 3

24 = 6

4
4
The same orders as the chain for a cubic: e ⊳ A3 ⊳ S3. If we did the group theory
properly we would find that these quotients are really the same as A3 and S3. So, by using V
we should end up with an equation of degree 3.
Back to the action. We can solve the cubic equation to find values for the yi.
(z1 + z2)(z3 + z4) = y1
(z1 + z3)(z2 + z4) = y2
(z1 + z4)(z2 + z3) = y3
52 Examples
That is, we have 3 equations involving 4 unknowns (remember, we know the values of the yi).
Normally you need as many equations as unknowns, but luckily we have a fourth equation,
namely
z1 + z2 + z3 + z4 = -a
z3 + z4 = -a – (z1 + z2)
(z1 + z2)(z3 + z4) = y1
(z1 + z2) – a – (z1 + z2) = y1
(z1 + z2)2 + a(z1 + z2) + y1 = 0
z1 + z2 =
-a ± pa2 – 4y1
2
Using the quadratic formula to solve the quadratic in z1 + z2. And similarly for z1 + z2
and z1 + z4. Notice how we are cascading down the degrees: We start with a quartic, apply
some algebra to obtain a cubic, solving that we use a quadratic to extract the solutions to
the original equations. In fact, as we’ll see, we are about to solve some degree 1 equations as
well.
Note If we had applied the Tschirnhaus transformation first, then a = 0 and these reduce
to the much simpler z1 + z2 = ±√-y1. Those are much easier for humans to deal with.
Notice that we have a choice of solutions, we’ll see in a moment how to make that
choice, but for the moment let’s just assume that the choices we make are r1, r2 and r3.
z1 + z2 = r1
z1 + z3 = r2
z1 + z4 = r3
z1 + z2 + z3 + z4 = -a
A system of simultaneous linear equations. Writing this in matrix form

1 1 0 0
1 0 1 0
1 0 0 1
1 1 1 1


z1
z2
z3
z4

=

r1
r2
r3
-a

Multiplying by the inverse of the matrix we find

z1
z2
z3
z4

=
1 2

1 1 1 -1
1 -1 -1 1
-1 1 -1 1
-1 -1 1 1


r1
r2
r3
-a

To solve the quartic we only need to choose the correct ri.
ri =
-a ± pa2 – 4yi
2
If we are doing this with computer assistance then we choose the ri so that
z1z2z3 + z1z2z4 + z1z3z4 + z2z3z4 = -c
If we are doing it by hand, then we use Tschirnhaus to ensure that a = 0, then ri = ±√-yi
and the choice of sign reduces to requiring that
r1r2r3 = -c
Lecture 9 Abstract Algebra and Equations (2012) 53
§9.6 Examples
Example 1 Consider the equation
4×4 + 8×3 + 8×2 – 4x + 1 = 0
Normally we would now quickly check to see if this has any nice roots, using the remainder
theorem. Obviously I’ve chosen this so that it doesn’t. We divide by the leading coefficient
to get a monic polynomial
x4 + 2×3 + 2×2 – x + 1
4
= 0.
The coefficents are a = 2, b = 2, c = -1 and d = 14. We can calculate α, β and γ
α = 2b = 4
β = ac + b2 – 4d = 1
γ = abc – a2d – c2 = -6
So the equation for the yi is
y3 – 4y2 + y + 6 = 0
Solving
0 = y3 – 4y2 + y + 6 = (y + 1)(y – 2)(y – 3)
Thus y = -1, 2 or 3. (So we now know that the original equation wasn’t chosen at random,
normally you’d have a lot of work to do here to solve this cubic.)
We find the possible values for the ri.
r1 =
-2 ± p22 – 4y1
2
=
-2 ± √4 + 4
2
= -1 ± √2
r2 =
-2 ± p22 – 4y2

2 r3 = -2 ± p22 – 4y3

= -1 ± i

2
= -1 ± i√2
We must now check to see which choice of the sign gives us the solution. A lot of work will
show that any choice with an odd number of negatives will work. So we let
r1 = -1 – √2
r2 = -1 – i
r3 = -1 – i√2
We can now use the matrix equation to find the solutions to the original quartic.

z1
z2
z3
z4

=
1 2

1 1 1 -1
1 -1 -1 1
-1 1 -1 1
-1 -1 1 1


-1 – √2
-1 – i
-1 – i√2
-2

=
1 2

(-1 + √2)(1 + i)
(-1 + √2)(1 – i)
(-1 – √2)(1 + i)
(-1 – √2)(1 – i)

Thus the solutions to 4×4 + 8×3 + 8×2 – 4x + 1 = 0 are
–
1 2
(1 ± √2)(1 ± i)
For the record, this also gives us the real factors of our quartic.
4×4 + 8×3 + 8×2 – 4x + 1 = 2×2 + 2(1 – √2)x + 3 – 2√2
× 2×2 + 2(1 + √2)x + 3 + 2√2
54 Examples
Example 2 Consider the equation
x4 + 10×3 + 35×2 + 50x + 24 = 0
This quartic actually factorises
x4 + 10×3 + 35×2 + 50x + 24 = (x + 1)(x + 2)(x + 3)(x + 4)
But we will continue with the process just to see how nasty it can get, even when you have
a nice solution. This is already monic so the coefficients we seek are
a = 10, b = 35, c = 50 and d = 24
α = 2b = 70
β = ac + b2 – 4d = 1629
γ = abc – a2d – c2 = 12 600
So the equation for the yi is
y3 – 70y2 + 1629y – 12 600 = 0
Somewhat startlingly, this factorises. Often a nice quartic will require the use of nasty
complex numbers with surds to find the solutions.
0 = y3 – 70y2 + 1629y – 12 600 = (y – 21)(y – 24)(y – 25)
Thus y = 21, 24 or 25. Solving for the ri
r1 =
-10 ± p102 – 4y1
2
= -5 ± 2 = -7 or – 3
r2 =
-10 ± p102 – 4y2
2
= -5 ± 1 = -6 or – 4
r3 =
-10 ± p102 – 4y3
2
= -5
It can be checked that this time all choices give the solutions. So we let r1 = -7, r2 = -6
and r3 = -5.
Finally we use the matrix to find the solutions to the original quartic.

z1
z2
z3
z4

=
1 2

1 1 1 -1
1 -1 -1 1
-1 1 -1 1
-1 -1 1 1


-7
-6
-5
-10

=

-4
-3
-2
-1

And the solutions are -1, -2, -3 and -4.
Lecture 9 Abstract Algebra and Equations (2012) 55
Example 3 For the final example I asked my computer to give me a random quartic with
small integer coefficients.
x4 + 2×3 + 3×2 + 5x – 4 = 0
Calculating α, β and γ.
α = 6
β = 35
γ = 21
So the cubic equation for the yi is
y3 – 6y2 + 35y – 21 = 0
This does not factorise so we are forced to use the general solution for a cubic. Here the
coefficients are
a′ = -6, b′ = 35, c′ = -21
and the quadratic we obtain, for y3 and z3 as you recall, is
X2 + 891X – 328 509 = 0
which has solutions
3 2
– 297 ± √234 213
Now we need to find the cube roots of these numbers . . .
Clearly, most of the time, solving quartic equations involves an awful lot of hard work.
Even when the quartic has relatively nice solutions, there can be a large amount of work.
§9.7 Summary
Let z1, z2, z3 and z4 be the roots of the quartic
x4 + ax3 + bx2 + cx + d
(0) Apply the Tschirnhaus transformation, if you wish, to ensure that a = 0. Remember that this is a good thing to do if you’re human, it simplifies some of
the later tasks.
(1) Let
α = 2b
β = ac + b2 – 4d
γ = abc – a2d – c2
(2) Let y1, y2 and y3 be the roots of the cubic
y3 – αy2 + βy – γ
(This, of course, is the hard bit and may leave you with some very complicated
numbers to work with.)
(3) Let
ri =
-a ± pa2 – 4yi
2
56 Summary
(If we have applied Tschirnhaus then ri = ±√-yi and we choose the signs so
that r1r2r3 = -c.)
(4) Obtain the roots of the quartic via

z1
z2
z3
z4

=
1 2

1 1 1 -1
1 -1 -1 1
-1 1 -1 1
-1 -1 1 1


r1
r2
r3
-a


(5?)	If you applied Tschirnhaus then you need to transform your answers to get the solutions of the original quartic.

At the end of this process you should check your solutions. It is not reasonable to expect that
you will have done all the steps perfectly at the first attempt. It would be nice, if possible,
to simplify the expressions you have, but that can be a very, very hard step.
Lecture 10
Near Enough Is Good(?) Enough
We now know that we can’t solve a general polynomial of degree 5 or greater. But the
design of a critical part of our new bridge (or whatever) might rely on the solution of such
a polynomial equation. Are we stuck? Luckily, no, in most cases we don’t need the exact
solution, approximations will do as long as we can make the approximation as accurate as
we need. We also need to be able to find all the solutions, not just one of them. The three
features we need are
(1) The ability to find a solution to arbitrary accuracy.
(2) The ability to find all solutions, including complex solutions.
(3) The ability to deal with multiple roots.
As we will see, there are several methods, each with their own good and bad points.
Note Most of these methods work for any function, not just polynomials. If a method is
specific to polynomials that fact will be mentioned.
§10.1 Bisection
This is the simplest method of all, you might already have met it.
Good points Easy to use, accuracy is known at all times.
Bad points Slow!, can’t find complex solutions, or solutions where the curve does not cross
the axis (for example, can’t solve x2 = 0).
Example Find a zero of f(x) = x5 + x + 1 accurate to 1 decimal place. To start the process
we need to find two values a and b such that one of f(a) and f(b) is positive and the other
negative. Notice that
f(-1) = -1 < 0
f(0) = 1 > 0
If the function is positive at 0 but negative at -1 then it must be 0 at some point in between.
As the name suggests, we now bisect the interval [-1, 0]. The middle of that interval is -0.5,
we calculate
f(-0.5) = 0.468 . . . > 0
The function is negative at -1 and positive at -0.5 and so we concentrate on the interval [-1, -0.5]. The middle is -0.75 and
f(-0.75) = 0.012 . . . > 0
We consider the interval [-1, -0.75], the middle of which is -0.875
f(-0.875) = -0.387 . . . < 0
So we are negative at -0.875 but positive at -0.75 so the interval of interest is [-0.875, -0.75].
The middle is -0.8125
f(-0.8125) = -0.16 . . . < 0
The interval is [-0.8125, -0.75].
That is enough! We know that the solution lies in the interval [-0.8125, -0.75] and
all the numbers in that interval round to -0.8, so our solution correct to 1 decimal place
is -0.8. In fact the solution is -0.7548 . . .
MATH216: W. N. Franzsen (28/2/2018). 57
58 Newton’s Method
Exercise Continue using the bisection algorithm to find the solution correct to 2 decimal
places. (You can check your answer against the value given above.)
Note If you do this in binary (as computers do) then you get 1 extra bit of accuracy with
each step.
§10.2 Newton’s Method
You might already have met this one as well. It is also known as the Newton-Raphson
method.
Good point Faster than the bisection method — if this method works then the number of
accurate digits doubles at each step.
Bad points Doesn’t always work, can’t find complex solutions.
Suppose we have a polynomial (or, indeed, any function that we can differentiate), f(x),
and we know that a0 is close to a solution. Then we have a situation like the following.
r
a0
a0, f(a0)
We find the tangent to the curve at a0, f(a0) and see where that crosses the axis.
r
a0
a0, f(a0)
a1
This new value, a1, is (almost always) closer to the solution r. From basic calculus, we know
that the equation of the tangent is
y – f(a0) = f′(a0)(x – a0)
This crosses the axis at (a1, 0) and so
0 – f(a0) = f′(a0)(a1 – a0)
–
f(a0)
f′(a0) = a1 – a0
a1 = a0 –
f(a0)
f′(a0)
10.1 Newton’s Method If an is an approximate solution to f(x) = 0 then
an+1 = an –
f(an)
f′(an)
is usually a better approximation.
Lecture 10 Abstract Algebra and Equations (2012) 59
Example Let f(x) = x5 + x+ 1 and a0 = -1, find a better approximation to a zero of f(x).
We first note that f′(x) = 5×4 + 1.
a1 = -1 –
f(-1)
f′(-1)
= -1 =
-1
6
= –
5 6
= -0.833 . . .
a2 = a1 –
f(a1)
f′(a1)
= –
5 6
–
f -5 6
f′ -65
= –
10 138
13 263
= -0.764 . . .
For the record, a3 = -0.7550 . . . and the actual solution is -0.75487766 . . .
What can go wrong with this? Sadly there are a couple of nasty things that can
happen.
(1) If you choose, or reach, a value ai that is a stationary point of the function
then you have an immediate problem. The whole process stops as you would
otherwise have a 0 in the denominator.
(2) In the last case you might just decide to pick a new value near that stationary
point. But that is dangerous. Sometimes the process can get stuck near a
stationary point and just cycle between two values.
Example Let g(x) = x32 + 1 and suppose a0 = 1. (Yes, I know this function doesn’t have a
zero but it gives you the idea. Better example coming up soon.)
a1 = 1 –
g(1)
g′(1)
= 1 –
13
– 1
23
= -1
a2 = -1 –
g(-1)
g′(-1)
= -1 –
13
+ 1
–
23
= 1 = a0
and the values cycle between 1 and -1.
Exercise The better example is the one contained in the Wikipedia article on this stuff. Let
h(x) = x3 – 2x + 2
First show that h(x) has a solution near -1.8, but that if we start with a0 = 0 then we
cannot find it.
This function mentioned on the page http://en.wikipedia.org/wiki/Newton’s method, accessed 20th December 2011.
60 Durand-Kerner Method
§10.3 Halley’s Method
You have met Halley before, he’s the same person that Halley’s comet is named after. This
method is very similar to Newton’s method and has the same good and bad points. The
advantage is that it is faster than Newton’s method (if it works.)
an+1 = an –
2f(an)f′(an)
2f′(an)2 – f(an)f′′(an)
Obviously this involves even more work that Newton’s method.
§10.4 Durand-Kerner Method
This method only works for polynomials, and dates from 1960.
Good points Finds all the zeros at once, can find complex zeros.
Bad points Requires complex arithmetic (even if all solutions are real), each step requires
a lot of work unless done on a computer, is touchy about multiple roots.
For simplicity, suppose we have a cubic. We can generalise to higher degree easily. Let
k(x) = x3 + 2×2 + 3x + 4
and let p0, q0 and r0 be approximate solutions to k(x) = 0. The following will be better
approximations to the solutions.

p1 = p0 –	k(p0) (p0 – q0)(p0 – r0) k(q0) (q0 – p0)(q0 – r0) k(r0) (r0 – p0)(r0 – q0)
q1 = q0 –
r1 = r0 –

Choosing the starting values is relatively easy. You just pick a complex number α that isn’t
real and isn’t a root of unity, then set
p0 = 1
q0 = α
r0 = α2
Repeatedly applying this algorithm will eventually produce all three solutions to the cubic.
Exercise Write down the formula that would work for a quartic equation.
Example To find the zeros of k(x) = x3 + 2×2 + 3x + 4 we take α = 1 + i and start with
p0 = 1, q0 = α = 1 + i and r0 = α2 = 2i
We now apply the algorithm until the values settle down. All of the calculations below were
Lecture 10 Abstract Algebra and Equations (2012) 61
done to 20 decimal places, but we only report 5 to save space.
i pi qi ri

0 1

1 5 – 2i

1 + i -6 – i

2i -1 + 3i

2 2.42623 – 1.49180i -3.65454 – 1.09276i -0.77169 + 2.58457i
3 1.03040 – 1.17765i -2.44528 – 0.87517i -0.58512 + 2.05282i
4 0.160443 – 1.11049i -1.75242 – 0.54423i -0.40803 + 1.65472i
5 -0.30567 – 1.41345i -1.47195 – 0.09769i -0.22238 + 1.51114i
6 -0.15555 – 1.56049i -1.67490 + 0.01396i -0.16955 + 1.54652i
7 -0.17443 – 1.54700i -1.65092 + 0.00016i -0.17465 + 1.54683i

8 -0.17469 – 1.54687i 9 -0.17469 – 1.54687i

-1.65063 -1.65063

-0.17469 + 1.54687i -0.17469 + 1.54687i

From this point on the values do not change and so these are the three solutions we were
seeking.
Example We will now find approximate zeros for the polynomial f(x) = x5 + x + 1. We
will use α = 1 + i again and so start with

p0 = 1,	q0 = α,	r0 = α2,	s0 = α3	and	t0 = α4
The steps of the algorithm will look like the following.
pn+1 = pn –	f(pn)

(pn – qn)(pn – rn)(pn – sn)(pn – tn)
The table below shows the iterations, again to save space we only display a few of the digits
that were used in the working.
i pi qi ri si ti
0 1 1 + i 2i -2 + 2i -4
1 1.07 + 0.01i 1.15 + 1.05i 1.19 + 1.80i 0.20 + 0.31i -3.61 – 3.16i
2 1.43 – 0.03i 2.04 + 0.34i -1.50 + 1.32i 0.17 + 0.19i -2.13 – 1.83i
3 2.06 – 0.37i 0.31 + 0.45i -1.15 + 1.13i 0.09 + 0.18i -1.31 – 1.38i
4 1.20 – 0.38i 0.75 + 0.17i -0.79 + 0.86i -0.30 + 0.46i -0.86 – 1.11i
5 0.66 – 0.78i 1.05 + 0.82i -0.49 + 0.70i -0.59 + 0.18i -0.64 – 0.91i
6 0.85 – 0.71i 0.88 + 072i -0.50 + 0.91i -0.73 – 0.09i -0.51 – 0.84i
7 0.88 – 0.76i 0.88 + 0.75i -0.50 + 0.87i -0.76 + 0.00i -0.50 – 0.87i
At this point the values are stable and so the approximate solutions are
-0.75488, 0.87744 ± 0.74486i and – 0.5 ± 0.86602i
If you remember your complex numbers well then you should recognise the last two. They
seem to be the complex cube roots of unity, namely -12 ± √23i, and so they are zeros of the
polynomial x2 + x + 1. If this is true then our quintic actually factorises
f(x) = x5 + x + 1 = (x2 + x + 1)(x3 – x2 + 1)
A fact we are unlikely to have noticed unless we found the solutions.
62 Durand-Kerner Method
Exercise Find approximate solutions to the polynomial we didn’t solve last time. That is
x4 + 2×3 + 3×2 + 5x – 4
Note As mentioned before, this method can have problems with multiple roots. The denominator has terms like pn – sn, if p and s are heading for the same value then we can get 0
in the denominator. Because these are approximations this doesn’t always lead to a problem,
so if it occurs try a different starting point, you might just avoid the problem second time
around. (But probably won’t, as we’ve seen convergence is quite fast.)
Computer Note If you are using Wolfram Alpha, or Mathematica or a similar program
then just you can just get the program to solve the polynomial for you. If you are writing your
own program you could use Durand-Kerner to solve polynomials for you. More importantly,
Microsoft Excel can be convinced to work with complex numbers so you can solve polynomials
just using Excel. Be warned that Excel doesn’t work with complex numbers easily so this is
not pleasant to do. For the record, from 2007 the functions like IMPRODUCT are always
available, check the help files to find the others you need. Before 2007, you needed to load a
toolpack to be able to use them.
§10.5 Why Does Durand-Kerner Work?
Suppose we continue looking at a cubic polynomial. Let P, Q and R be the zeros of our
polynomial and so
f(x) = (x – P)(x – Q)(x – R)

x – P =	f(x)
(x – Q)(x – R)
P = x –	f(x) (x – Q)(x – R)

So, if we knew Q and R, then we can find P immediately. Of course we could find P more
easily using the constant term (hope you remember how). If we only have approximations
for Q and R, say we know q and r are close to those solutions then we get an approximation
to P, call it p

p ≈ x –

f(x) (x – q)(x – r)

And we get the equivalent equations for q and r. What is important is that P, Q and R
are fixed points for these equations, if we put them in the right hand sides then we get
exactly those values back. The hope is that if we start close enough to the solutions then by
applying those formulae repeatedly then we get better approximations. By doing some truly
impressive calculus it can be shown that ‘close enough’ here means ‘almost anywhere’! That
is, as long as you take a complex number, α, that isn’t real and isn’t a root of unity and start
with 1, α, α2, . . . then you’ll get to the solutions, usually quite quickly.
Notice how this relies upon the fact that we started with a polynomial. If we don’t
have a polynomial then we don’t know how many zeros we are looking for, and even if we
do, we don’t get a nice relationship between them.

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