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AProject Example 1: Topics in Galois

Theory

Preamble

For this project, the student explored an area of mathematics that was outsidethe course, and was difficult. The information was available and well embeddedin the literature: however the concepts were advanced. In style therefore, thisproject most closely resembles the Hypergeometric Functions project outlined inSection 3.7.

Abstract

Galois Theory was originally formulated to determine whether the roots of arational polynomial could be expressed in radicals. The theory is based on theassociation of a group to each polynomial and the analysis of the group todetermine solubility in terms of radicals. We first study Galois Theory withthe approach of the author Evariste Galois, then in its modern form based onabstract algebra, particularly field theory, which allows a more general interpre-tation. We then outline the theory of soluble groups and give some examples.Finally, we consider the application of methods in Galois Theory to the threeclassical construction problems and the construction of n-gons.

151

152 Managing Mathematical Projects - With Success!

Introduction

Evariste Galois (1811–1832) was a French mathematician who was born nearParis. Although he died in tragic circumstances at 20 years old he had alreadyproduced work that would make him famous. Unfortunately during his lifetimeGalois suffered many setbacks and was considered to be a troublemaker by thegovernment. He also found himself unaccepted by the mathematical establish-ment. It was not until 1846 that Joseph Liouville published some of Galois’work in his Journal de Mathematiques [ED, p1]. Since then the importance ofGalois’ results has been realised and his work has provided the impetus formany developments in algebra.

In his Memoir on the Conditions for Solvability of Equations by RadicalsGalois was solving a problem that interested many mathematicians includingLagrange. The solution to the general quadratic equation has been known sinceBabylonian times. In the 16th century formulae were discovered for the cubicand quartic. The solution for the cubic equation x3 + px = q found by NiccoloFontana is

x =3

√q

2+

√p3

27+

q2

4+

3

√q

2−

√p3

27+

q2

4.

This is an expression built up from the coefficients of the equation using theusual arithmetical operators with the nth roots, or radicals [ST, p xiv]. If theroots of an equation can be expressed in this way it is said to be “soluble byradicals”. Mathematicians hoped that they would be able to find formulas forgeneral equations of degree n ≥ 5. Lagrange worked for many years on thesolution of general equations, finding improved solutions for both the cubicand quartic [ED, pp18–22]. He was unable to find a solution for the quinticand expressed doubts that a solution existed. His work was almost certainly asource of inspiration for Galois. In 1824 Abel proved that the general quinticwas not soluble by radicals so that a general formula did not exist but wasunable to find a method for identifying which specific are soluble. This questionwas answered by Galois comprehensively in a piece of brilliant and originalmathematics known as Galois Theory. This is the subject of this project andwe now give an outline of each section.

In Section A.1 we closely follow the English translation of Galois’ memoirby Harold M. Edwards in his book Galois Theory [ED]. The style, both of thetranslation and accompanying text, is traditional and often difficult to followso much of the material in this section is a modified interpretation. The booktakes a concrete rather than abstract approach to Galois Theory, and Edwardshas attempted to present the theory with as little modern algebra as possible,for the most part drawing only on the techniques available to Galois. Some

A. Project Example 1: Topics in Galois Theory 153

basic field theory is used in the text, particularly in the proofs, but we delaythe introduction of fields until Section A.2.

In Section A.2 we study Galois Theory in its modern form which makes useof the developments in abstract algebra over the last 150 years. Field theoryhas allowed mathematicians to interpret Galois Theory in a formal, abstractsense, suited to the more rigorous standards of definition and proof introducedat the turn of the century. The main source for this chapter is Ian Stewart’sGalois Theory [ST] which has a more accessible style than Edwards’ book. Inseparating the two approaches to Galois Theory we are able to appreciate thesimplicity of Galois’ approach which the more structured methods of abstractalgebra do not share although they allow Galois Theory to be applied to moregeneral areas than rational polynomials.

Groups play an essential part in Galois Theory and Galois’ work in thisarea provided a foundation for the study of group theory. This is expandedon in Section A.3, where we give an outline of the relevant theory and someexamples. Sources for this chapter include Stewart’s book and Allenby’s Rings,Fields and Groups [AL] which was useful for a more detailed approach andworked examples.

Finally, Section A.4 explores the applications of field theory to classicalgeometry, using the methods developed in our study of Galois Theory. Herewe combine elements from Stewart, Allenby and also John R. Durbin’s ModernAlgebra: An Introduction.

This project is intended to be an introduction to Galois Theory although,due to the depth of the subject and restrictions upon the project, it has notbeen possible to give more than an outline of many of the proofs. It is hopedhowever that it will provide the reader with the impetus for a more detailedstudy of the subject.

A.1 Galois’ Approach

A.1.1 Preparation

We begin this section by reviewing some facts from basic algebra.

A.1.1.1 Polynomials. We consider polynomials with rational coefficients un-less specifically declared otherwise. A polynomial of degree n is irreducible in Q

if it cannot be expressed as a product of two or more polynomials whose degreesare less than n but greater than 0. The polynomial of degree 0 is the constant


polynomial. The factorisation of a polynomial into irreducible polynomials isunique up to constant factors and the order in which the factors are written.If f(x) = anxn + . . . + a1x + a0 and an = 1 then f is a monic polynomial.

Since we are interested in the roots of a polynomial, when two polynomialsdiffer only by a constant factor we consider them to be essentially the same.This is demonstrated by the example

f(x) =13x2 − 1

3x − 2 =

13(x + 2)(x − 3)

g(x) = x2 − x − 6 = (x + 2)(x − 3).

So f = 13g but f and g have the same roots.

A.1.1.2 Some Methods of Checking Irreducibility. If f is a polynomial thatis irreducible in Z then f is also irreducible in Q. For example

let g(x) =13x2 +

13x + 6

and f(x) = x2 + x + 18 = 3g(x).

Now f is irreducible in Z so f is irreducible in Q.Eisensteins’s Irreducibility Criterion [ST, p20]Let

f(x) == anxn + . . . + a1x + a0, ai ∈ Z, i = 0, 1, . . . , n.

Suppose there is a prime q such that an, q|ai for i = 0, 1, 2, . . . , n − 1, andq2 a0 then f is irreducible in Q.

For example, if f(x) = x5 + 4x4 + 6x3 + 2x2 + 8x + 2 and we take q = 2then f is irreducible in Q.

Unfortunately not every polynomial can be treated with this method. How-ever, we can sometimes rewrite a polynomial so that the method will be suit-able. For example, f(x) = g(x)h(x) if and only if f(x + 1) = g(x + 1)h(x + 1).Hence f(x) is irreducible if and only if f(x + 1) is irreducible [ST,p21].

An example of this is f(x) = x4+x3+x2+x+1 where Eisenstein’s criteriondoes not apply. But f(x + 1) = x4 + 5x3 + 10x2 + 10x + 5 and is irreducible byEisenstein’s criterion with q = 5. Hence f(x) is irreducible in Q.

A.1.1.3 Roots of Polynomials. Each root of a polynomial is either simple ormultiple (repeated). We will want to work with polynomials that have onlysimple roots, but it is quite straightforward to detect simple roots.

If f(x) has roots α1, α2, . . . , αn which are roots of multiplicity m1,m2, . . . , mn

respectively then

f(x) = (x − α1)m1(x − α2)m2 . . . (x − αn)mn .


If we differentiate formally then

Df = (x − α1)m1−1(x − α2)m2−1 . . . (x − αn)mn−1∑

j

mj

∏i =j

(x − αi)

and it is clear that if, for any i, mi > 1 then f and Df have a common factor.Thus we can sayLemma 1 [ST,p85]

A polynomial has a multiple root if and only if f and Df have a commonfactor of degree ≥ 1. If f has no multiple roots then f and Df will be coprime,i.e. the highest common factor will be a polynomial of degree 0.

Any irreducible polynomial with rational coefficients must have only simpleroots since if it had a multiple root then f and Df would have a commonfactor of degree≥ 1 which can only be a multiple of f . But f is irreducible andDf has degree less than n so we must have Df = 0 which implies that f is aconstant.

A.1.1.4 Symmetric Polynomials. Symmetric polynomials play an essentialrole in Galois Theory as they allow us to express rational functions of the rootsof a given equation in terms of the coefficients of the equation. Consider ageneral quadratic equation f(x) = x2 + bx + c. If the roots of f are α and β

then

f(x) = (x − α)(x − β)

= x2 − (α + β)x + αβ.

We can immediately see that α + β = −b, αβ = c. The functions α + β andαβ are called symmetric polynomials since their values do not vary if α and β

are interchanged.A symmetric polynomial is any function in n indeterminates that is invari-

ant under any permutations of the indeterminates. For example, if the roots ofa polynomial are α1, α2, α3 then the following functions are symmetric polyno-mials in the roots:

α21 + α2

2 + α23, α1 + α2 + α3, α1α2 + α1α3 + α2α3, α1α2α3.

The last three examples are special cases of symmetric polynomials called ele-mentary symmetric polynomials. It is these functions that are directly relatedto the coefficients of the polynomial of which the indeterminates are the roots.


Let f(x) = xn +b1xn−1 +b2x

n−2 + . . .+bn = (x−r1)(x−r2)(x−r3) . . . (x−rn). The elementary polynomials are

σ1 = r1 + r2 + r3 + . . . + rn = −b1

σ2 = r1r2 + r1r3 + . . . + rn−1rn = b2

...

σn = r1r2r3 . . . rn = (−1)nbn

where σj = 0 for j > n. The general formula is σk = (−1)kbk.

Theorem 2 (Fundamental Theorem on Symmetric Polynomials) [ED,p9]Every symmetric polynomial in r1, r2, . . . , rn can be expressed as a polynomialin the elementary symmetric polynomials σ1, σ2, . . . , σn. Moreover, a symmet-ric polynomial with integer coefficients can be expressed as a polynomial inσ1, σ2, . . . , σn with integer coefficients. (Proof [ED, pp9–12].)

Let sk = rk1 + rk

2 + rk3 + . . .+ rk

n for k = 1, 2, 3, . . .. Then we have the generalformula [ED, p12]

sk − sk−1σ1 + sk−2σ2 − . . . + (−1)k−1s1σk−1 + (−1)kkσk = 0k = 1, 2, 3, . . .

We can also express the elementary symmetric polynomials in n − 1 indeter-minates in terms of elementary symmetric polynomials in n indeterminates.Let the indeterminates be r1, r2, r3, . . . , rn and let the elementary symmetricpolynomials in the n indeterminates be σk. Let the elementary polynomials inn − 1 indeterminates r1, r2, r3, . . . , rn−1 be τk. Then [ED, pp10–11]

r1 = σ1 − rn

r2 = σ2 − rnσ1 + r2n

r3 = σ3 − rnσ2 + r2nσ1 − r3

n

...

rn = σn−1 − rnσn−2 + . . . + (−1)n−1rn−1n

Finally, we note Lemma 1 from Galois’ Memoir [ED, p51]:

Lemma 3[ED, pp51–52]:An irreducible equation cannot have a root in common with a rational

equation without dividing it.Proof : Let g and h be polynomials and g be irreducible and let g and h have

a common root. Then using the Euclidean algorithm we can write d = Ag+Bh

where d is the greatest common divisor of g and h, and A and B are rational


polynomials. If we substitute into this equation the common root, r, then wehave d(r) = 0. Therefore d has degree > 0. Since d divides g and g is irreducible,d must be a non-zero multiple of g. Since d divides h it follows that g dividesh.

A.1.2 The Galois Resolvent

A.1.2.1 Forming a Galois Resolvent. With the preliminary information de-tailed in Section A.1.1 we first study Galois Theory through Galois’ originalapproach. The remainder of this chapter is drawn mostly from Edwards trans-lation of Galois’ Memoir [ED, Appendix 1, pp101–113] and accompanying text.As a starting point we have the “given” equation. Let this be

f(x) = xn + an−1xn−1 + an−2x

n−2 + . . . + a1x1 + a0.

This equation should be irreducible and have n distinct roots α1, α2, . . . , αn.In addition, the coefficients of f are assumed to be rational.

Our aim is to show whether or not the roots of f can be expressed in terms ofradicals. The only information available is the coefficients of the given equation.However, symmetric polynomials allow us to extract information about theroots from these coefficients.

Following earlier work by Lagrange ([AL, pp181–186] and [ED, pp18–22,pp32–35]) Galois constructed a function of the roots of the given equation.This function had particular properties and is now known as a Galois resol-vent. The conditions governing the function are stipulated by Galois’ LemmaII [ED, p102]:

Lemma 4: Given any equation with distinct roots a, b, c . . . , one can alwaysform a function V of the roots such that no two of the values one obtains bypermuting the roots in this function are equal. Further, the coefficients of thisfunction can be chosen as integers.

Proof :Let our Galois resolvent be

V0 = x1α1 + x2α2 + . . . + xnαn, xi ∈ Z.

We define a permutation of the roots by φ(αi) = αφ(i). Any permutation φ ∈ Sn

of the roots will give a new function of the roots which we will call a conjugate


of the resolvent, say V1. We shall denote the numerical value of any of theresolvent Vi by ti. So if φ = (123 . . . n) then

φ(V0) = x1α1 + x2α2 + . . . + xnαn = V1 V0 = V1

Note that we insist the value of each conjugate of the resolvent is different, notjust the formal representation. We wish to show that the n! conjugates havedistinct values so let

D =∏S,T

[x1 (S(α1) − T (α1)) + x2 (S(α2) − T (α2)) + . . .]2

where the product is over the n!(n! − 1)/2 pairs of unordered permutations S

and T in which S = T . If we regard the xi as variables then D is a polynomialin the xi with coefficients that are polynomials in the αi. D is symmetric inthe αi so the coefficients are symmetric in the αi. This means that D is apolynomial with known (rational) coefficients. Since D is a product of non-zeropolynomials it is non-zero. We can therefore choose the xi as integers so thatD = 0.

A.1.2.2 Using a Galois Resolvent. Once we have a suitably chosen Ga-lois resolvent we can use it in a series of calculations that yield informationabout the given equation f . We form the “equation for V ” by constructinga polynomial with the resolvent and its conjugates as its roots. Let these beV0 = t0, V1 = t1, . . . , Vn!−1 = tn!−1. Then the polynomial will be

F (X) = (X − t0)(X − t1) . . . (X − xn!−1).

Note that F is invariant under any permutation of the roots of f since thiswould only interchange the ti. So the coefficients of F must be symmetricpolynomials in the roots of f and are therefore known quantities. We cancalculate these quantities as shown in the following example. This example willbe used throughout sections A.1 and A.2 to demonstrate various aspects ofGalois Theory.

Let u(x) = x3 − 2. This is an irreducible polynomial with simple roots. Letthe roots of u be a, b, c. We form the Galois resolvent, choosing coefficients thatare integers. To make the calculations easier for this example, the coefficientshave been chosen with the prior knowledge that they yield different values ofthe resolvent for each conjugation. Normally, such a calculation would need tobe carried out without explicit values for the coefficient. Let the resolvent andits conjugates be

V0 = a + 2b + 3c, V1 = c + 2a + 3b, V2 = b + 2c + 3a

V3 = b + 2a + 3c, V4 = c + 2b + 3a, V5 = a + 2c + 3b.


Then the polynomial with the values of the resolvent and its conjugates asroots is

U(X) = (X − t0)(X − t1)(X − t2)(X − t3)(X − t4)(X − t5)

= X6 − (t0 + t1 + t2 + t3 + t4 + t5)X5

+ [(t0 + t1)(t2 + t3 + t4 + t5) + (t2 + t3)(t4 + t5)]X4

+ [t0t1 + t2t3 + t4t5]X4

− [t0t1(t2 + t3 + t4 + t5) + t2t3(t4 + t5) + t4t5(t2 + t3)]X3

+ [(t0 + t1)(t2t3 + t4t5 + (t2 + t3)(t4 + t5))] X3

+ [t2t3t4t5 + (t0 + t1)(t2t3(t4 + t5))]X2

+ [(t4t5(t2 + t3)) + t0t1((t2 + t3)(t4 + t5) + t2t3 + t4t5)]X2

− [(t0 + t1)t2t3t4t5 + t0t1(t2t3(t4 + t5) + t4t5(t2 + t3))] X

+ t0t1t2t3t4t5.

If we expand the coefficients of U in terms of the roots a, b and c we find thatthey are symmetric polynomials in the roots. For example the terms for X5

and X4 are12(a + b + c)X5[

58(a2 + b2 + c2) + 122(ab + ac + bc)]X4.

Using the symmetric polynomials and elementary symmetric polynomials asdefined in Section A.1 we have the following information:

σ1 = σ2 = 0, σ3 = 2

s1 = s2 = s4 = s5 = 0, s3 = 6, s6 = 12.

So our polynomial becomes

U(X) = X6 + t0t1t2t3t4t5

= X6 + 108

which is irreducible in Q.In general the polynomial F will not always be irreducible but will decom-

pose into irreducible factors so that F (X) = G1(X)G2(X)G3(X) . . . Gs(X).Note that since the roots of F are the conjugates of the resolvent then eachof the factors must have some of the conjugates as its roots, and each factorhas roots distinct from the roots of each other factor. We choose G1 to be thefactor that has V0 = t0 as a root.


A.1.2.3 Expressing roots in terms of the Galois Resolvent. We now move onto Galois Lemma III [ED, p103]

Lemma 5: When the function V is chosen as indicated above, it will havethe property that all the roots of the given equation can be expressed as ratio-nal functions of V .

Proof : First, Galois considered the conjugates of the resolvent that held one,arbitrarily chosen, root fixed and permuted all the others. There are (n−1)! ofthese conjugates and we use them to construct a polynomial with them in thesame manner we used for F . V is always one of these conjugates that holds aroot fixed. Let the others be V1, V2, . . . then

Fa(X) = (X − t0)(X − ti)(X − ti+1) . . . (X − ti+s).

Fa is symmetric in the roots that have been permuted. It can be shown that anyone root of a given equation can be expressed in terms of elementary symmetricpolynomials in all the other roots [ED, pp10–11]. So we can eliminate from Fa

all but the fixed root, and we can say

Fa(X) = Fa(X, α1) where α1 is the fixed root.

Replace α1 with the indeterminate Y and substitute t0 for X so that we haveFa(t0, Y ) and f(Y ) as polynomials in Y with coefficients that are rationalfunctions in t0. We find the greatest common divisor d(Y ) of Fa(t0, Y ) andf(Y ). Now (Y −α1) divides Fa(t0, Y ) and f(Y ) so that d(Y ) = A(Y )Fa(t0, Y )+B(Y )f(Y ). Setting Y = 0 gives

γ(−α1) = A(0)Fa(t0, 0) + B(0)f(0)

α1 = h1(t0)

where h1 is a polynomial in t0 with rational coefficients. We can perform asimilar process for each of the roots in turn. If we take the conjugates of theresolvent that hold another of the roots fixed we form another equation, sayα2 = h2(t0). This is most clearly demonstrated by returning to the exampleu(x) = x3 − 2 with roots a, b, c. We see that V0 and V5 leave a fixed:

U0 = (X − t0)(X − t5)

= X2 − (t0 + t5)X + t0t5.

Now t0 + t5 = 2a + 5(b + c) and t0t5 = a2 + 6(a2 + b2) + 5a(b + c) + 13bc. If τi

are the elementary symmetric polynomials in b and c then we have

τ1 = b + c = −a, τ2 = bc = a2, b2 + c2 = τ21 − 2τ2 = −a2.


So Ua(X, a) = X2 + 3aX + 3a2. The greatest common divisor of Ua and u is d

whered(Y ) =

13(2t20Y + t30 − 6).

Setting d(Y ) = γ(Y −α1) and Y = 0 we find γ =2t203

and a = h1(t0) =6 − t302t20

and similarly we can show b = h2(t0) = − 6t20

, c = h3(t0) =6 + t302t20

.

Since the conjugates of the resolvent are obtained by a permutation of theroots of the given equation f we should expect that the roots can be expressedin terms of all of these conjugates. We have Galois’ Lemma IV [ED, p104]:

Lemma 6:Suppose one has formed the equation for V , and that one has taken one

of its irreducible factors, so that V is the root of an irreducible equation. LetV, V ′, V ′′, . . . be the roots of this irreducible equation. If a = f(V ) is one of theroots of the given equation, f(V ′) will also be a root of the given equation.

Proof :In the notation we are using we have

f(X) = G1(X)G2(X)G3(X) . . . Gs(X)

where t0 is a root of G1 which has deg. G1 roots taken from the ti. So ifwe have α1 = h1(t0) then perform the permutation θ = (123 . . . n) we haveα2 = h1(θ(t0)).

Returning to our example, we have a = h1(t0) =6 − t302t20

. The permutation (132)

takes t0 to t2 and a to c so c = h1(t2) =6 − t322t22

.

A.1.3 The Galois Group

A.1.3.1 Finding the Galois Group. We come now to a fundamental part ofGalois Theory. Any given equation has a particular group associated with itand in this section we show how Galois found this group. Firstly we quoteGalois’ Proposition 1 [ED, p104]:


Theorem 7:Let an equation be given whose m roots are a, b, c, . . . . There will always be

a group of permutations of the letters a, b, c, . . . which will have the followingproperty:

1. That each function invariant under the substitutions of this group will beknown rationally,

2. Conversely, that every function of the roots which can be determined ra-tionally will be invariant under these substitutions.

We have F (X) = G1(X)G2(X)G3(X) . . . Gs(X). Let deg G1 be m so theroots of G1 are m of the ti including t0. The given equation f has n roots thatcan all be represented as functions of the ti. We can construct a table as below:

h1(t0), h2(t0), h3(t0), . . . hn(t0)h1(ti), h2(ti), h3(ti), . . . hn(ti)

......

......

...h1(tj), h2(tj), h3(tj), . . . hn(tj).

This table has m rows representing the m conjugates of the resolvent and n

columns representing the roots of the given equation f. This table gives us theGalois group since each row represents one arrangement of the roots. We candemonstrate this by using our example u(x) = x3 − 2 once again. The table offunctions of the ti is

h1(t0), h2(t0), h3(t0)h1(t1), h2(t1), h3(t1)h1(t2), h2(t2), h3(t2)h1(t3), h2(t3), h3(t3)h1(t4), h2(t4), h3(t4)h1(t5), h2(t5), h3(t5)

which givesa b c

c a b

b c a

b a c

c b a

a c b.

These permutations are those contained in the whole of the symmetric groupS3. So the Galois group is for u in S3.

Proof of Theorem 7:This breaks down into several parts. We need to show:


i) the entries in the table are all roots of f ,

ii) no root appears in a row more than once;

iii) points 1 and 2 regarding invariant and rational functions;

iv) closure of the group represented in the table.

i) f(hi(X)) is a polynomial in X with rational coefficients. Now h1(t0) = α1 sof(h1(t0)) = 0. This shows that t0 is a root of f(h1(X)). But t0 is also a root ofG1, an irreducible equation. By Lemma 3 we must have G1 divides f(h1). Soevery root of G1 is also a root of f(h1). Similarly, every root of G1 is a root off(h2), f(h3), etc. So all the entries in the table are roots of f [ED, p52].

ii) Let αm and αn be two roots of f so that hm(t0) = αm and hn(t0) = αn.Suppose

hm(ti) = hn(ti)

hm(ti) − hn(ti) = 0

where ti is a root of G1. Then ti is a root of hm(X)−hn(X). So by Lemma 3 G1

divides hm(X)−hn(X) and t0 is a root also. But then hm(t0) = hn(t0), αm =αn and the roots of f are distinct. So we cannot have hm(ti) = hn(ti) and theentries of each row of the table are distinct.

iii) Let P (α1, . . . , αn) be a rational polynomial in the n roots of f , suchthat P is invariant under any permutation of the roots. Since each root can beexpressed as a function of the Galois resolvent t0 we have

P (α1, . . . , αn) = P (h1(t0), . . . , hn(t0)).

But since P is invariant under the permutations of the roots we must have

P (h1(t0), . . . , hn(t0)) = P (h1(ti), . . . , hn(ti)) for all ti roots of G1.

Let (h1(ti), . . . , hn(ti)) = h(ti). So we can say

P (h(t0)) =1m

[P (h(t0)) + P (h(ti)) + . . .] where m = degG1

But this is a symmetric polynomial in the roots of G1 and can therefore beexpressed in terms of the rational coefficients of G1. So

P (h(ti)) ∈ Q ⇒ P (α1, . . . , αn) ∈ Q.

Conversely, assume that P (α1, . . . , αn) ∈ Q so that P (h(ti)) ∈ Q. NowP (h(X)) is a polynomial with rational coefficients so P (h(X)) − P (h(t0))is also a rational polynomial which has a root t0. Therefore by Lemma 3P (h(X)) − P (h(t0)) is divisible by G1 and has the roots of G1 in common.


But then P (h(ti)) = P (h(t0)) for any ti a root of G1. So P is invariant underany permutation of the roots of f .

iv) To prove closure we show that no matter which row arrangement webegin with we obtain the same group of permutations. Let s and t be twodifferent Galois resolvents for f , let φ be a permutation belonging to the Galoisgroup of f and let α1, α2 be the roots of f such that α2 = φ(α1).

Suppose α1 = hs(s) and α1 = ht(t) then α2 = φ(hs(s)) = hs(φ(s)). Wewant to show that φ(ht(t)) = α2.

Now

hs(s) = ht(t)

hs(s) − ht(t) = 0 0 ∈ Q

φ(hs(s) − ht(t)) = 0

hs(φ(s)) = ht(φ(t))

α2 = ht(φ(t)).

Since the elements of the group are independent of the choice of Galoisresolvent, they must also be independent of the choice of which conjugate ofthe resolvent we begin with.

A.1.3.2 Galois’ Examples of Groups. Galois highlights two examples ofgroups. The first is the general polynomial of degree n where the coefficientsare not explicitly known but are rational. In this case the only functions thatcan be determined rationally are the symmetric polynomials in the roots. Ga-lois states, without proof, that the Galois group of a general polynomial ofdegree n is the symmetric group Sn containing n! permutations of the roots.We delay discussion of this point until Section A.2, accepting for the momentthe statement without proof.

The second example given is the cyclotomic equation of degree n

xn − 1x − 1

= xn−1 + xn−2 + . . . + x + 1 where n is prime.

Galois shows that the group for this polynomial consists of n permutationsonly. We will use this group as an example in Section A.3 which concentrateson soluble groups.


A.1.4 Soluble Equations and Soluble Groups

A.1.4.1 Adjoining Quantities to an Equation. At this point it is worth re-viewing our progress so far. For our given equation f we can find a Galoisresolvent, say V0, and form an equation F of degree n! such that

F (X) =n!−1∏i=0

(X − ti)

= G1(X)G2(X)G3(X) . . . Gs(X)

where the Gi are irreducible polynomials with rational coefficients and rootsti. We choose G1 to be the polynomial with one of its roots t0. By expressingeach root of f as a function of t0 we can construct a table which leads us tothe Galois group of f . Importantly, the number of rows in the table, which isdeg G1, is the order of the Galois group.

We now move on to the analysis of the Galois group. Galois was interestedin how the group changed as certain quantities were adjoined to the givenequation f . If we wish to express the roots of f in radicals we will adjoin therequired radicals, which will all be of smaller degree than n. In fact, we needonly adjoin radicals of prime order or, if we want the jth root where j is nota prime, then we decompose j into the product of primes. Say j = j1j2j3 . . .

then we can construct a j1th root of a j2th root of a j3th root, etc. So we areonly interested in adjoining pth roots where p is a prime.

We are interested in adjoining sufficient quantities to the given equation f

to allow us to decompose it into linear factors. If this is possible then the equa-tion is obviously soluble by radicals. We have Galois’ Proposition II [ED, p106]:

Theorem 8:If one adjoins to a given root r of an auxiliary irreducible equation,

1. Either the group of the equation will not be changed, or it will be parti-tioned into j groups where j divides p, each belonging to the given equationrespectively when one adjoins each of the roots of the auxiliary equation;if p is prime then the partition is into 1 or p groups.

2. These groups will have the remarkable property that one will pass fromone to the other in applying the same substitution of letters to all thepermutations of the first.

The auxiliary irreducible equation that Galois had in mind is of the form xp = k,the solution of which gives us primitive pth roots of k. Let these roots ber, αr, α2r, . . . , αp−1r where α is a pth root of unity. Gauss’ work had shownthat pth roots of unity are expressible in radicals for any prime p (although


Edwards disagrees [ED, p27]). So Galois always assumed that pth roots of unityhad already been adjoined to the given equation.

Proof :Galois does not give full proof, but Edwards’ detailed proof [ED, pp59-

61] is outlined here. If G1 remains irreducible then the Galois group of thegiven equation f remains unchanged. Otherwise, G1 reduces to the product ofirreducible factors G1(X) = H1(X)H2(X) . . . Hv(X). The new Galois group isof order deg H1 and consists of the arrangements of the roots of f expressed asfunctions of those conjugates of V0 that are roots of H1. Again we must chooseH1 to be the factor of G1 that has V0 = t0 as a root. Since we have adjoined tof the root r of xp = k we can represent H1 as H1(X, r). Edwards shows thatby substituting αir for i = 1, . . . , p − 1 into H1 we can form the product

h(x) = H1(X, r)H1(X, αr) . . . H1(X, αp−1r).

Since G1 and H1 have deg H1 roots in common and coefficients that are rationalfunctions of r then by Lemma 3 G1(X) = H1(X, αir)Q(X, r) where Q is apolynomial with coefficients that are also rational functions of r. We can alsoshow that

G1(X) = H1(X, αir)Q(X, αir) for i = 1, . . . , p − 1.

So we can say G1(X)p = h(X)q(X) and by Lemma 3

1 =[

h(X)G1(X)j

] [q(X)

G1(X)p−j

].

So h(X) =constant×G1(X)j for some integer j. This gives us deg h = pdegH1 = jdeg G1. The index of the new Galois group in the old Galois groupis deg G1/deg H1 and this, by the above equation, is p/j. If p is prime theneither j = 1 or p so the index is 1 or p. Obviously, the value of 1 relates to G1

remaining irreducible. To show the second part of Theorem 8 we first note thatwhen j is 1 we have

constant × G1(X) = H1(X, r)H1(X, αr) . . . H1(X, αp−1r).

The roots of H1(X, r) are a subset of the roots of G(X), as are the roots ofH1(X, αr). Let the two roots of H1(X, r) be t0 and t1 and a root of H1(X, αr)be t2. A row of the group table for H1 is

α1 = h1(t0), α2 = h2(t0), . . . , αn = hn(t0).

If S is a permutation belonging to the new Galois group and S(t0) = t2 then

S(α1) = h1(t2), S(α2) = h2(t2), . . . , S(αn) = hn(t2)


is a row of the table for H1(x, αr). Edwards shows that if we apply S to the rowof the table which contains functions of r1 then we obtainh1(t3), h2(t3), . . . , hn(t3) where t3 is also a root of H1(X, αr). So by applyingthe same permutation we move from the new Galois group to another subsetof the old Galois group, relating to another factor of G1. A subgroup withthis property is known as a normal subgroup. A normal subgroup partitions agroup into conjugacy classes. In the above context these conjugacy classes arethe subsets of the old Galois group relating to each irreducible factor H1 of G1.As a reminder, a subgroup B of a group A is normal if and only if for a ∈ A

and b ∈ B we have a−1ba ∈ B.The process of adjoining quantities is continued until the Galois group of

f has been reduced to the identity permutation only. At this stage we haveadjoined all the necessary radicals to f and F has reduced to linear factors.Then the equation of f is soluble for its roots.

This shows that if f is soluble by radicals then its Galois group will reduce by aseries of subgroups, each a normal subgroup of prime index in its predecessor.Let the Galois group of f be B0. Ignoring any groups that are identical to theirpredecessors, we have the series

B0B1B2. . .Bv = e where |Bi−1|/|Bi| = p with p prime and i = 1, . . . , v.

We call a group soluble if it has such a series of subgroups.Consider our example u(x) = x3 − 2. If we adjoin a primitive cube of unity,

y, then decompose into two factors:

U(X) = (X3 − 2(75y2 + 69y + 72))(X3 − 2(69y2 + 75y + 72))

= V1(X)V2(X).

The irreducible factor V1 has Galois group A3 which is the alternating groupof degree 3. A3 is a normal subgroup of S3 of index 2. If we now adjoin z, areal cube root of 2, then U can be expressed in linear factors since the roots ofu are a = z, b = yz, and c = y2z. This gives

U(X) =(X − (z + 2yz + 3y2z))(X − (y2z + 2z + 3yz))(X − (yz + 2y2z + 3z))

×(X − (yz + 2z + 3y2z))(X − (y2z + 2yz + 3z))(X − (z + 2y2z + 3yz)).

So the series of subgroups for the Galois group of u is S3 A3 e where e is anormal subgroup of A3 with index 3.

Galois’ Propositions III and IV [ED, p107] can be dealt with fairly briefly.Proposition III states:


Theorem 9:If one adjoins to an equation all the roots of an auxiliary equation, the

groups in Theorem II [Theorem 8 here] will have the further property thateach group contains the same substitutions.

Our choice of r, as a solution of the auxiliary equation, is completely ar-bitrary so we must necessarily obtain the same partitioning of the group foreach root r, r′, rn, . . . , rp−1 of the auxiliary equation. If we adjoin all these rootsthen rather than splitting the Galois group of f into conjugacy classes, eachrelating to one irreducible factor of G1, we have the same normal subgroup ofthe Galois group applying to each factor.

Theorem 10:If one adjoins to an equation the numerical value of a certain function of

its roots, the group of the equation will be reduced in such a way as to containno permutations other than those which leave this function fixed.

A.1.4.2 Finding Soluble Groups and Soluble Equations. Proposition V [ED,p108] asks the question which is the motivation behind Galois Theory–“Inwhich case is an equation solvable by simple radicals?”

We know that the group of the equation must have a series of normalsubgroups of index p which relate to the adjunction of pth roots and the seriesmust end with the identity permutation (a fuller discussion of soluble groupsfollows in Section A.3). Galois gives a rather compact proof of this [ED, p109].Edwards expands this proof considerably covering some points missed by Galois[ED, pp61–64].

Galois gives the example of the general quartic equation [ED, p109]. Thegroup has 24 elements and the adjunction of a square root decomposes thegroup into two subsets of order 12. The new Galois group is A4 which is anormal subgroup of S4 of index 2. Then a cube root is adjoined and A4 splitsinto 3 subsets. The new Galois group is e, (12)(34), (13)(24), (14)(23) whichis a normal subgroup of A4 of index 3. We then adjoin another square root togive a Galois group e, (12)(34) which is a normal subgroup of the previousgroup, of index 2. Finally we adjoin another square root to reduce the Galoisgroup to the identity permutation. Therefore the general quartic is soluble byradicals.

More interesting are the general equations of degree ≥ 5. Consider thegeneral quintic equation. The Galois group is S5 so we need to find whether S5

is soluble. Here we quote a theorem which Edwards has derived from Galois’memoir [ED, p91]


Theorem 11:If an irreducible equation has prime degree p and is solvable by radicals,

then the roots of the equation can be ordered a1, a2, . . . , ap in such a way thatthe permutations s of the Galois group are all of the form s(ai) = ari+s whereai is defined for all integers i by setting ai = aj for i ≡ j(modp), where r and s

are integers, and where r = 0(modp). The proof can be found in Edwards [ED,pp 91–93].

If the general quintic equation is soluble by radicals then, by the abovetheorem the symmetric group S5 should contain only permutations of the forms(ai) = ari+s. If we are working modulo 5 then there are 5 values of s butonly 4 values of r that we can use. This gives 20 possible permutations butS5 has 120 elements. Therefore the general quintic equation is not soluble byradicals [ED, p91]. This does not mean that every quintic is insoluble, but itis not possible to express the roots in terms of a formula in the coefficients ofthe equation using arithmetical operators and radicals.

Since S5 is a subgroup of every symmetric group Sn for n ≥ 5 no symmetricgroup is soluble for n ≥ 5 [ED, p90]. This means that there is no formula forthe solution of general equations of degree greater than the general quartic.

A.2 The Modern Approach

A.2.1 Field Extension

A.2.1.1 Constructing Field Extensions. For this section we shall assume thatthe reader understands the basics of ring, field and group theory althoughsome theorems will be quoted (without proof) to highlight their importance. Auseful basic text is Allenby’s Rings, Fields and Groups [AL] This text covers allthe preparatory work for modern Galois Theory but stops short of the theoryitself. We will also need to refer to irreducibility, multiple roots and symmetricpolynomials, all of which were introduced in Section A.1.1.

In Section A.1 we considered polynomials with rational coefficients, i.e.polynomials over the field of rational numbers Q. Modern algebra provides aframework for Galois theory in which it can be applied to a wider subject areathan rational polynomials although most of our examples in this section willstill be rational polynomials.

Through the rest of this chapter we fix a general field K (not necessarilyQ). However we need to know whether or not K is “like” Q in one importantsense. Recall


Theorem 1 [ST, pp 3–4]

1. The prime subfield of a field K is the intersection of all subfields of K.

2. Every prime subfield is isomorphic either to the field Q of rationals or thefield Zp of integers modulo a prime number p.

3. If the prime subfield of K is isomorphic to Q we say K has characteristic 0.If the prime subfield of K is isomorphic to Zp we say K has characteristicp.

4. If K is a subfield of L then K and L have the same characteristic.

Most of the examples we will see in this section have characteristic 0 and mostof the theory is independent of the characteristic. However, there are sometheorems that are sensitive to the characteristic and these will be highlightedwhere they occur.

Galois used the concept of adjoining roots to a given equation and thistranslates easily into the theory of field extensions. If f(x) = (x2 + 1)(x2 − 5)then f is a polynomial over Q since its coefficients are in Q. We say the factorsof f are “irreducible over Q” if there are no elements in Q that would allow usto reduce f any further. If we “extend” Q to include i =

√−1 we denote thefield generated (as a subfield of C say) as Q(i) which reads “Q adjoined withi”, then

f(x) = (x + i)(x − i)(x2 − 5)

as a polynomial over Q(i). Similarly

f(x) = (x + i)(x − i)(x −√

5)(x +√

5)

as a polynomial over Q(i,√

5). We denote this as a field extension by Q(i,√

5) :Q if we want to make clear what the “ground” field is, being the one we startedwith. We can classify with

Definition 2 [ST, pp33-34]

1. A simple extension is an extension L : K having the property that L =K(α) for some α ∈ L.

2. Let K(α) : K be a simple extension. If there exists a non-zero polyno-mial p over K such that p(α) = 0 then α is an algebraic element overK and the extension is called a simple algebraic extension. Otherwise α

is transcendental over K and K(α) : K is called a simple transcendentalextension.


For example, the polynomial f(x) = anxn + . . . + a1x + a0 is an element of asimple transcendental extension of Q, that is, the field of all rational expressionsin x. If there is a polynomial p over Q such that p(x) = 0 then p = 0.

Field extensions are an important part of Galois Theory since the way webuild a field extension, to allow us to solve an equation for its roots, is inti-mately related to the structure of the Galois group of the equation.

Theorem 3[ST, p35,45,47]:

1. Let L : K be a field extension and suppose that α ∈ L is algebraic over K.Then the minimum polynomial of α over K is the unique monic polynomialm over K of smallest degree such that m(α) = 0.

2. The degree of an extension L : K is denoted [L : K]. The degree of a fieldextension L : K is the dimension of L considered as a vector space over K.

3. If K, L, M are fields and K ⊆ L ⊆ M then [M : K] = [M : L][L : K].

4. Let K(α) : K be a simple extension. If it is transcendental then [K(α) :K] = degm where m is the minimum polynomial of α over K.

Examples

1. Consider C : R. We can construct C by adjoining 1, i to R as everyelement of C is of the form x+ iy where x, y ∈ R. Then 1, i forms a basisfor C over R so [C : R] = 2.

2. The extension K(x) : K where x is transcendental over K has infinitedegree since the elements 1, x, x2, . . . are linearly independent and forma basis for K(x) over K.

Finally in this section we have the following:Lemma 4 [ST, pp47–48]

1. A finite extension is one whose degree is finite.

2. An extension L : K is algebraic if every element of L is algebraic over K.

3. L : K is a finite extension if and only if L is algebraic over K and thereexist finitely many elements α1, . . . , αs ∈ L such that L = K(α1, . . . , αs).

A.2.1.2 Splitting Fields. Although formal field theory did not exist in Galois’time he would have had an intuitive notion of fields. When he wrote “. . . a pthroot of unity, α, is included among the quantities that have already been ad-joined to the equation.”[ED, p108], this would be in modern terminology Q(α).The concept of a splitting field is quite simple. We construct field extensionsover our ground field until we are able to express the given equation in linear


factors and, therefore, find the roots of the equation. The splitting field willcontain all the roots of f .

For example, in Section A.1 we found that the roots of u(x) = x3 − 2 werez, yz, y2z where y is a primitive cube root of unity and z the real cube rootof 2. We first construct Q(y) : Q but find that u does not completely reduce.We then construct Q(y, z) : Q(y) so that u(x) = (x− z)(x− yz)(x− y2z). Oursplitting field for u is Q(y, z).

Theorem 5 [ST,p78–79]: The field Σ is a splitting field for the polynomialf over the field K if K ⊆ Σ and

1. f splits into linear factors over Σ.

2. Σ = K(α1, . . . , αn) where α1, . . . , αn are the roots of f in Σ.

In the example u(x) = x3 − 2, if we construct the field Q(z) first then one rootof u lies in Q(z). But Q(z) : (Q) is not a normal field extension as the otherroots of u are not in Q(z) and u does not split Q(z).

Definition 7 [ST, p83]:An irreducible polynomial f over a field K is separable over K if it has no

multiple roots in a splitting field. This means that in any splitting field f takesthe form

f(x) = k(x − α1) . . . (x − αn)

where the αi are all different. An irreducible polynomial over a field K isinseparable over K if it is not separable over K. For example, let K0 = Zp

where p is prime and let K = K0(u) where u is transcendental over K0. If

f(x) = xp − u ∈ K[x]

then let L be a splitting root for f over K. Let τ be a root of f in L so thatτp = u. We have

(x − τ)p = xp +(

p

1

)xp−1(−τ) + . . . + (−τ)p.

But p divides(

p

1

)for 0 < r < p and any multiple of p in K is zero. So

(x − τ)p = xp − τp = xp − u = f(x).If σp − u = 0 then (σ − u)p = u so σ = τ and all the roots of f in L are

equal [ST, p85].In Section A.1 we showed that any irreducible polynomial with rational co-

efficients had only simple roots. This result holds for any irreducible polynomialover a field K where K has characteristic 0. However the result is different forfields of characteristic p > 0.


Proposition 8 [ST, p86]:If K is a field of characteristic 0 then every irreducible polynomial f over

K is inseparable if and only if

f(x) = krxrp + . . . + xp + k0 k0, . . . , kr ∈ K.

If f is defined as above then

Df = rpkrxrp−1 + (r − 1)pkr−1x

(r−1)p−1 + . . . + pk1xp−1.

Since every multiple of p is zero in a field of characteristic p > 0 the resultDf = 0 does not imply that f is a constant. However, if f has terms that arenot raised to a power that is a multiple of p then Df will contain terms whichare identically zero only when their coefficient is zero.

The last result in this section is

Lemma 9 [ST, p86]Let L : K be a separable algebraic extension and let M be an immediate

field. Then M : K and L : M are separable.

A.2.2 The Galois Group

A.2.2.1 Automorphisms. If we are able to construct a finite separable normalextension for a given equation we can relate the Galois group of the equation tothe field extension through the operation of mappings on the field extension. Wecan define a mapping from one field to another that satisfies certain conditions.If θ : K → L is such a map from the field K to the field L and x, y ∈ K then

θ(x + y) = θ(x) + θ(y)

θ(xy) = θ(x)θ(y).

We shall be concerned with automorphisms which are maps of this type. We areparticularly interested in automorphisms which fix our ground field K whichare defined as follows

Theorem 10 [ST, p72]

1. If K and L are two fields with K ⊆ L and θ : L → L is an automorphismsuch that θ(k) = k for all k ∈ K then θ is a K-automorphism of L.

2. The set of all K-automorphisms of L forms a group under composition ofmaps.


3. The Galois group Γ (L : K) of the extension L : K is the group of allK-automorphisms of L under composition of maps.

Thus we have a connection between a field extension and its Galois group.

Proof(of(2))If α and β are K-automorphisms of L then αβ is also an automorphism.

For k ∈ K

αβ(k) = α(k) ⇒ αβ is a K-automorphism.

If e is the identity then e is a K-automorphism since e(k) = e

α−1 is an automorphism of L and k = α−1α(k) = α−1(k) ⇒ α−1

is an automorphism of L.

Let us return again to the example u(x) = x3 − 2. Suppose the field extensionis Q(y, z) : Q then we have the following possible Q-automorphisms of Q(y, z).

θ : θ(y) = y θ(z) = yz, θ3 = e

φ : φ(y) = y2 φ(z) = z, φ2 = e

We can construct the following table which gives all possible Q-automorphismsof Q(y, z)

(Q)-automorphism Effect on y Effect on z

c y z

θ y yz

θ2 y y2z

φ y2 z

φθ2 = θφ y2 yz

φθ = θ2φ y2 y2z

So Γ (Q(x, y) : Q) = e, θ, θ2, φ, θφ, θ2φ. This group is isomorphic to S3 whichis the Galois group of u that we found in Section A.1.

A.2.2.2 Subfields and Subgroups. The elements of the Galois group relate tothe splitting field by defining subfields. Each subfield has its own automorphismin the larger field and these also form a group. The automorphisms of a splittingfield are generally the permutations of roots such as we considered in SectionA.1.2. Subfields may be considered to be intermediate stages in the adjunctionof roots although the process is somewhat different in this context.


Definition 11 [ST, pp73–75]If L : K is a field extension let M be an “intermediate” field such that

K ⊆ M ⊆ L. To each intermediate field M we associate a group M∗ whichcontains all the M -automorphisms of the field L. Γ (L : M) is a subset of theGalois group Γ (L : K). Now K∗ is the group of all K-automorphisms of L

and is therefore the whole Galois group, whilst L∗ contains only the identityautomorphism. If M ⊆ N then M∗ ⊇ N∗ since any automorphism fixing theelements of N will fix the elements of M .

Working from the other direction, given a subgroup of Γ (L : K) we canassociate with it a set of fixed elements of the field L. If H is a subgroup ofΓ (L : K) then H denotes the set of all elements in L fixed by the automorphismsin H.

The relationship between the subgroups and subfields is given by

Lemma 12 [ST, p74]If H is a subgroup of Γ (L : K) then H is a subfield of L containing K. We

call H the fixed field of H. If H ⊆ G then H⊇G since any elements fixed by G

will be fixed by H. If M is an intermediate field and H is a subgroup of theGalois group then M ⊆ M∗ and H ⊆ H∗.

The example u(x) = x3 − 2 has the following fixed fields and related sub-groups:

Subgroup of Γ (Q(y, z) : Q) Fixed fielde Q(y, z)

e, θ2φ Q(yz)e, θφ Q(yz(y + z))e, φ Q(z)

e, θ, θ2 Q(y)e, θ, θ2, φ, θφ, θ2φ Q

A more complicated example is the field extension Q(√

2,√

3,√

5) : Q. Let thisbe L : Q. Let

r1 =√

2, r2 = −√

2, r3 =√

3, r4 = −√

3, r5 =√

5, r6 = −√

5

then the Q-automorphisms can be represented as permutations on the ri. If θ

is a permutation then θ(ri) = rθ(i). Our Galois group is then

Γ (L : Q) = e, (12), (34), (56), (12)(56), (34)(56), (12)(34)(56).Some proper subgroups of Γ (L : Q) are

B = e, (12)(34), (12)(56), (12)(56), C = e, (12), (34), (12)(34),D = e, (12), (56), (12)(56), E = e, (34), (56), (34)(56),F = e, (12)(34), G = e, (12)(56), H = e, (34)(56),R = e, (12), S = e, (34), T = e, (56), U = e, (12)(34)(56).


The fixed fields associated with these subgroups are

B = Q, C = Q(√

5), D = Q(√

3), E = Q(√

2),

F = Q(√

5,√

6), G = Q(√

3,√

10), H = Q(√

2,√

15),

R = Q(√

3,√

5), S = Q(√

2,√

5), T = Q(√

2,√

3),

U = Q(√

6,√

10,√

15).

We also have e = K and Γ (K : Q)ˆ = Q.In summary we have

Definition 13 [ST, p74]If F denotes the set of intermediate fields, and G the set of subgroups of

the Galois group then we have two maps

∗ : F → Gˆ : G → F

which reverses inclusions and satisfies Lemma 12.

A.2.2.3 Group Order and Degree of Field Extension. There is also a rela-tionship between the degree of a field extension and the order of the group ofautomorphisms associated with the extension.

Theorem 14 [ST, p91]

1. Let G be a finite subgroup of the group of automorphisms of a field K, andlet K0 be the fixed field of G. Then [K : K0] = |G|.

2. If G is the Galois group of the finite extension L : K and H is a finitesubgroup of G then [H:K] = [L : K]/|H|.

Consider the example of Q(√

2,√

3,√

5) : Q = L : Q detailed above.

[L : Q] = [L : Q(√

2,√

3)][Q(√

2,√

3) : Q(√

2)][Q(√

2) : Q]

= 2.2.2 = 8.

Let Z = γ(L : Q) then |Z| = 8. We have Z ⊃ S = e, (34) and |S| = 2. Nowthe fixed field of S is S = Q(

√2,√

5) and [S:Q]=[Q(√

2,√

5) : Q(√

2)][Q(√

2) :Q] = 2.2 = 4. So 4 = [S : Q] = [L : Q]/|S| = 8/2 = 4.

Another example is supplied by the field extension for u(x) = x3 − 2.Let G = Γ (Q(y, z) : Q) and H = e, θ, θ2 so |H| = 3.Also [Q(y, z) : Q] = [Q(y, z) : Q(y)][Q(y, z) : Q] = 3.2 = 6.

The fixed field of H is H = Q(y), [ H : Q] = 2. So

2 = [ H : Q] = [Q(y, z) : Q]/|H| = 6/3 = 2.


In the above two examples the degree of the field extension that gives thesplitting field is equal to the order of the Galois group of the extension. Thisis formally stated with

Theorem 15 [ST, p100]If L : K is a finite separable normal extension of degree n then there are

precisely n distinct K-automorphism of L, so that |Γ (L : K)| = n.Suppose that a separable extension L : K is not normal. It can be shown

[ED, pp98–102] that a ‘normal closure’ that enlarges the extension L : K canbe constructed, and that Theorem 15 will still hold for L : K. A normal closureis simply an extension N of L such that N : K is normal, and N is the smallestsuch extension of L which is normal over K.

A.2.3 Applying Galois Theory

A.2.3.1 The Galois Correspondence. We can now give a formal statement ofthe relationship between a field extension and its Galois group. This is knownas the Galois correspondence and is the fundamental theorem of Galois theoryin its modern algebra setting.

Theorem 16 [ST pp104–105]If L : K is a finite separable normal field extension of degree n, with Galois

group G and if F ,G, ∗, ˆ are defined as in Section A.2.2 then

1. The Galois group has order n.

2. The maps ∗ andˆare mutual inverses and set up an order reversing 1 − 1correspondence between F and G.

3. If M is an intermediate field then [L : M ] = |M∗| and [M : K] = |G|/|M∗|.4. An intermediate field M is a normal extension of K if and only if M∗ is a

normal subgroup of G.

5. If an intermediate field M is a normal extension of K then the Galois groupof M : K is isomorphic to the quotient group G/M∗.

We can show that the Galois correspondence of Q(√

2,√

3,√

5) : Q = L : Qas follows (refer to Section A.2.2 for the details):

1. [L : Q] = 8 = |Z|.2. L = e, Q(

√2,√

3) = T , Q(√

2) = E, Q = Z, e = L∗, T =Q(

√2,√

3)∗, E = Q(√

2), Z = Q∗.


3. [L : Q(√

2,√

3)] = 2 = |T |, [Q(√

2,√

3) : Q] = 4 = |Z|/|T |, [L :Q(

√2)] = 4 = |E|, [Q(

√2) : Q] = 2 = |Z|/|E|.

4. Z is abelian so every subgroup of Z is normal.Therefore Q(

√2) : Q, Q(

√2,√

3) : Q(√

2) and L : Q(√

2,√

3) are allnormal extensions.

5. Let V = (12), (12)(34), (12)(56), (12)(34)(56). Then Z/E = E, V which is isomorphic to Γ (Q(

√2) : Q) = R = e, (12). Let

W = (12), (12)(56), X = (34), (34)(56), Y = (12)(34), (12)(34)(56)then Z/T = T, W, X, Y is isomorphic to Γ (Q(

√2,√

3) : Q) = C =e, (12), (34), (12)(34).

The field extension for u(x) = x3 − 2 also satisfies the Galois correspondence,with Q(y) being an intermediate field. Q(y)∗ = e, θ, θ2 ∼= A3 and A3 is anormal subgroup of S3.

A.2.3.2 Radical Extensions. Although field extensions are the ‘adjunction ofquantities’ that Galois proposed, the modern expression of Galois Theory re-lies on starting with a splitting field and analysing the behaviour of the Galoisgroup as elements of the splitting field are fixed. In practice, as we are unawareof the elements in the splitting field, we rely entirely on studying the nature ofthis field and its associated Galois group. That is, ensuring that our splittingfield satisfies the conditions in Theorem 16 and analysing the solubility of thegroup. However, we have not yet imposed any conditions on the em type ofelements by which we extend the ground field. If we wish to show that a poly-nomial can be solved, then we must have a “radical extension”.

Definition 17 [ST, p128]An extension L : K is radical if L = K(α1, . . . , αn) where for each i =

1, . . . , m there exists an integer n(i) such that

αn(i)i ∈ K(α1, . . . , αi−1) i ≥ 2.

The elements αi are said to form a radical sequence for L : K.For example if L : Q(

√2,√

3) then (√

3)2 = 3 ∈ Q(√

2).So a radical extension is nothing more than a series of adjunctions of pth

roots as detailed by Galois. It is at this point we concentrate on fields of charac-teristic 0 since the proofs of some of the following items contain this assumption.


Theorem 18[ST, pp129–130]

1. If L : K is a radical extension and M is a normal closure of L : K thenM : K is radical.

2. If K has characteristic 0 and L : K is normal and radical then Γ (L : K) issoluble.

3. If K is a field of characteristic 0 and K ⊆ L ⊆ M where L : K is a solublegroup.

Consider Q(√

2,√

3,√

5) : Q. This is, by definition, a radical extension. There-fore we should expect its Galois group, Z, to be soluble. This is confirmed inSection A.1.4

Q(√

2,√

3) : Q is less obvious. We know that y and z are expressible inradicals so the splitting field must be a radical extension. In general, however,we may not know whether an extension is radical. In such a case we can turnto a theorem which is almost the converse of Theorem 18.

Theorem 19[ST, p146]Let K be a field of characteristic 0 and let L : K be a finite normal extension

with soluble Galois group G. Then there exists an extension R of L such thatR : K is radical.

So, since we have already shown that S3 is soluble, we can state the exten-sion Q(y, z) : Q is radical and u is soluble by radicals.

For fields of characteristic p > 0 we must impose an additional condition toensure that a group implies an equation soluble by radicals. In addition to theadjoining of radical elements as defined in Definition 17 we must also adjoinelements α such that αp−α is in the given field where p is the characteristic. Theadjunction of such elements extends the proof of Theorem 19 to allow inclusionof fields of characteristic p > 0. Otherwise, we may have some polynomials withsoluble Galois group that are not soluble by radicals.

A.2.3.3 The General Polynomial of degree n. We have so far considered onlyfinite algebraic extensions. To tackle the general polynomial equation

g(x) = xn − σ1xn−1 + . . . + (−1)nσn

(where the coefficients are written in terms of the elementary symmetry poly-nomials σk) we will need to use a transcendental extension.

Definition 20[ST, p139]

1. An extension L : K is finitely generated if L = K(α1, . . . , αn) where n isfinite. This is true whether the αi are algebraic or transcendental over K.


2. If t1, . . . , tn are transcendental elements over a field K, all lying insidesome extension L of K then they are independent if there is no non-trivialpolynomial p over K (in n indeterminates) such that p(t1, . . . , tn) = 0 inL.

3. Let t1, . . . , tn be independent transcendental elements over K. The sym-metric group Sn can be made to act as a group of K-automorphisms ofK(t1, . . . , tn) by defining θ(ti) = tθ(i) for all θ ∈ Sn. Distinct elements giverise to distinct K-automorphisms.

As we would expect, the fixed field of Sn consists of the quotients of the sym-metry polynomials in the ti. In particular it contains the elementary symmetricpolynomials σk, so if F is the fixed field of Sn we have F = K(σ1, . . . , σn).

Lemma 21[ST, p143]With the above notation, σ1, . . . , σn are independent transcendental ele-

ments over K.Now, if we consider the ti to be the n roots of the general polynomial of

degree n over K(σ1, . . . , σn) where

g(x) = xn − σ1xn−1 + . . . + (−1)nσn.

Thus we have

Theorem 22[ST, p143]For any field K let g be the general polynomial of degree n over K and let

Σ be a splitting field for g over K(σ1, . . . , σn). Then the roots t1, . . . , tn of g

in Σ are independent transcendental elements over K, and the Galois group ofΣ : K(σ1, . . . , σn) is the full symmetric group Sn.

Corollary 23The general polynomial of degree n has Galois group Sn which is not soluble

for n ≥ 5.

A.3 Soluble Groups

A.3.1 Normal Subgroup Series

In Section A.1 we showed that a soluble group has a series of normal subgroupsof index p (for p prime) and this series terminates with the identity element.


A more general definition is available which does not confine solubility to theaction of adjoining pth roots.

Definition 1 [ST, pp 114–115]A group B is soluble if it has a finite series of subgroups

B = B0 ⊇ B1 ⊇ . . . ⊇ Bn = e

called a composition series, such that

1. Bi+1 Bi for i = 0, . . . , n − 1;

2. Bi/Bi+1 is abelian for i = 0, . . . , n − 1.

Subgroup normality is not transitive so Bi Bi−1 Bi−2 = Bi Bi−2.Note that if we compare this to Galois’ definition of a soluble group then

|Bi/Bi+1| = p for a prime number p. Therefore Bi/Bi+1 is abelian since allgroups of prime order are cyclic and hence abelian.

We know that from the examples in Section A.1.2 that the decompositionof a group into subgroups is not necessarily unique. If we have more than oneway of composing a series of normal subgroups, we need to be sure that eachdifferent series satisfies the definition of a soluble group. The Jordan–HolderTheorem gives

Theorem 2[AL, p263]Let B be a finite group with composition series

B = B0 ⊇ B1 ⊇ . . . ⊇ Bn = e

and B = C0 ⊇ C1 ⊇ . . . ⊇ Cm = e.

Then m = n and the r quotient groups Bi/Bi+1 can be put in 1–1 correspon-dence with the r quotient groups Ci/Ci+1 in such a manner that the correspond-ing groups are isomorphic. We will not necessarily have Bi/Bi+1

∼= Ci/Ci+1

but all the quotient groups can be paired off. Provided we can construct oneseries that satisfies Definition 1 then every series will also satisfy the definition.

A.3.2 Normal Subgroups

As we have seen in Section A.3.1 the solubility of a group depends on thenormality of the subgroups. Abelian groups are useful since every subgroupof an abelian group is normal, and the quotient of an abelian group is alwaysnormal. This means that every abelian group is soluble and therefore everycyclic group is soluble since they are all abelian.


In Section A.1 we mentioned that one of Galois’ examples of a solubleequation was the cyclotomic equation

xp − 1x − 1

= xp−1 + . . . + x + 1

where p is prime. It is clear that one of the roots of xp − 1 is 1 so we areinterested only in the roots of xp−1 + . . .+x+1. This is an irreducible equationwith simple roots. Note that if a is one of these roots then all the roots can beexpressed as ai for i = 1, . . . , p − 1. So the splitting field for this equation isQ(a) and the Q-automorphisms are

φ1 : a → a

φ2 : a → a2

φ3 : a → a3

......

φp−1 : a → ap−1.

The Galois group is abelian since for any two Q-automorphisms we have

φi(a)φj(a) = aiaj = ai+j = ajai = φj(a)φi(a)

so the Galois group is soluble, Q(a) is a radical extension and all pth roots ofunity are expressible in radicals when p is prime. Since we have already shownthat any nth root, where n is not prime, can be constructed by using pth rootswhen p is a prime number, this shows that the cyclotomic equation xn − 1 issoluble by radicals for any n. The following theorem is useful when we havepartial knowledge of the structure of a group.

Theorem 3 [ST, p116]Let G be a group, H a subgroup of G, and N a normal subgroup of G.

1. If G is soluble then H is soluble;

2. If G is soluble then G/N is soluble;

3. If N and G/N is soluble then G is soluble.

Every subgroup H of a group G, where H has index 2 in G is a normal subgroup.A particular family of such subgroups is the alternating group An which hasindex 2 in the symmetric group Sn. So An is always a normal subgroup Sn.However, for n ≥ 5 this information, as we might expect, is not much help.This leads us on to our next section.


A.3.3 Simple Groups

Theorem 4 [ST, pp117–118]

1. A group G is simple if its only normal subgroups are e and G.

2. A soluble group is simple if and only if it is cyclic of prime order.

3. If n ≥ 5 then the alternating group An of degree n is simple.

The importance of this last result is highlighted by

Corollary 5 [ST, p119]The symmetric group Sn of degree n is not soluble if n ≥ 5.

ProofIf Sn were soluble then An would be soluble. But if n ≥ 5 An is simple

and of order n!/2, which is not prime for n ≥ 5, and therefore not soluble. Thisshows that for n ≥ 5 Sn is not soluble.

A.3.4 p-Groups

These are another example of soluble groups. We have

Lemma 6[ST, p121]:

1. Let p be a prime. A finite group B is a p-group if its order is a power of p.

2. If B is a finite p-group of order pn then B has a series of normal subgroups

B = B0 ⊇ B1 ⊇ . . . ⊇ Bn = e

such that |Bi| = pn−i for all i = 0, . . . , n.

3. Every finite p-group is soluble.

Proof(of (3))Every quotient group has order

|Bi|/|Bi+1| = pn−i/pn−(i+1) = p

which is prime. Therefore every quotient group is cyclic and abelian. As everysubgroup is normal, this implies B is soluble.


The theory of p-groups was extended by Sylow who gave the following theorem

Theorem 7 [ST, p123]Let B be a finite group of order par where p is prime and does not divide

r. Then

1. B possess at least one subgroup of order pa;

2. all such subgroups are conjugate in B;

3. any p-subgroup of B is contained in one of order pa;

4. the number of subgroups of B of order pa leaves remainder 1 on divisionby p.

If the Sylow p-subgroup P of B is a normal subgroup of B, and B/P is abelian,then G is soluble, since P is soluble.

S5 is a group of order 120 = 23.3.5. So S5 should have Sylow p-subgroupsfor p = 2, p = 3 and p = 5. Let P2 be a Sylow 2-subgroup of S5 then |P2| = 8.Using similar notation we also have |P3| = 3 and |P5| = 5.

Allenby gives a more detailed version of part 4 in Theorem 7:

4 The number of conjugates of a Sylow p-subgroup of B divides |B|/pa andis congruent to 1(modp)

Consider S3 which has order 6 = 2.3. S3 has one Sylow 3-subgroup which isA3.

S3/3 = 2 and 1|2 1 = 1(mod3).

S3 has a Sylow 2-subgroup which is X = e, (12) where X ∼= e, (13) ∼=e, (23)

S3/2 = 3 and 3|3 3 = 1(mod2).

Both A3 and X are normal in S3.The example from Section A.2 of Z = Γ (Q(

√2,√

3,√

5) : Q) is an exampleof a p-group since |Z| = 8 = 23. So, as we expected, Z is a soluble group.

A.4 Geometrical Constructions

In this section we study an interesting application of field theory to classicalgeometry. Since ancient Greek times mathematicians have been interested inthe construction of geometrical figures using only a straight edge and compass.With this restriction we are allowed two operations:


1. drawing a line through two given points;

2. drawing a circle with centre at one point and passing through a secondgiven point.

Many figures have proved to be constructible under these restrictions but notall. Three problems were of particular interest to the ancient Greeks and subse-quent generations of mathematicians over the following centuries. These prob-lems are [DU, p264]:

1. The Duplication of the Cube–given a cube construct another cube withtwice the volume of the given cube.

2. The Trisection of an Arbitrary Angle–some angles can be trisected but amethod is required for the trisection of any angle.

3. The Quadrature of the Circle–given a circle construct the square that hasthe same area as the given circle.

Not until the 19th century was it proved that these three problems have nosolution. The “impossibility proofs” depend on algebraic techniques rather thatgeometry. We shall use field theory, particularly that of field extensions, to provethe impossibility of these three constructions.

A.4.1 Constructible Points

We need to find a way of expressing geometrical construction in algebraic terms.We can assume that we are given at least two points. Let the set of givenpoints be P0 ⊂ R2 and let p1, p2 ∈ P0. Then we can use our two operations toconstruct:

1. a line through p1 and p2;

2. a circle with centre p1 and radius p1p2;

3. a circle with centre p2 and radius p1p2.

We add points to P0 by adjoining the intersection points of such lines andcircles. Let r1 be one of these intersection points. Then r1 is constructible inone step from P0. We define a point r in R2 to be constructible from P0

if there is a finite sequence r1, r2, . . . , rn = r of points in R2 such that foreach i = 1, 2, . . . , n the point ri is constructible in one step from the set P0 ∪r1, . . . , ri−1[ST, p52].

Let K0 be the subfield of R generated by the x and y coordinates of thepoints in P0. Let ri be an intersection point with coordinates (xi, yi). Then


Ki is the subfield of R generated by adjoining xi and yi with Ki−1 i.e. Ki =Ki−1(xi, yi). So

K0 ⊆ K1 ⊆ . . . ⊆ Kn ⊆ R.

With this notation we have

Lemma 1 [ST, p53]xi and yi are roots in Ki of quadratic polynomials over Ki−1.

Outline proofThe point ri is an intersection point of either two lines, two circles, or a line

and a circle all constructed from points of Pi−1. If ri is the intersection pointof

1. Two lines then xi, yi ∈ Ki−1 which implies xi, yi ∈ Ki. Also xi and yi arecertainly roots of quadratics polynomials over Ki−1;

2. Two circles then we obtain quadratic expressions for x and y with coeffi-cients in Ki−1;

3. A circle and a line then again we have quadratic expressions for x and y

with the same result as in (2) above.

The following theorem is the method by which we will prove the impossibilityof the three constructions mentioned at the start of the section.

Theorem 2 [ST, pp54–55]If r = (x, y) is constructible from a subset P0 of R2 and if K0 is the sub-

field of R generated by the coordinates of the points of P0, then the degrees[K0(x) : K0] and [K0(y) : K0] are powers of 2.

ProofIf ri is the intersection of two lines then xi, yi ∈ Ki−1 and [Ki−1(xi) :

Ki−1] = [Ki−1(yi) : Ki−1] = 1If ri is the intersection of two circles or of a circle and a line then xi and yi

are the roots of quadratic polynomials over Ki−1. If the quadratic polynomialsare reducible then the result is the same as for the intersection of two lines. Ifthe quadratic polynomials for xi and yi are irreducible then

[Ki−1(xi) : Ki−1] = [Ki−1(yi) : Ki−1] = 2.

So we can write

[Ki−1(xi, yi) : Ki−1] = [Ki−1(yi) : Ki−1(xi)][Ki−1(xi) : Ki−1] = 1, 2 and 4


If [Ki : Ki−1 is a power of 2 then by induction [Kn : K0] is a power of 2.But

[Kn : K0(x)] = [K0(x) : K0] = [Kn : K0]

so [K0(x) : K0] is also a power of 2. Similarly, [K0(y) : K0] is a power of 2 [ST,p55].

A.4.2 Impossibility Proofs

We can assume that our given points P0 include a pair of perpendicular axesand the unit circle so that (0, 0), (1, 0) ∈ P0. The following three proofs showthat it is not possible to construct these figures using only straight edge andcompass.

1. Duplication of the Cube Without loss of generality, we can assume thatthe given cube is the unit cube and that one side is the segment of the x-axis(0, 0) to (1, 0). Hence we wish to construct the cube with volume 2 units andwith length of side α such that α3 = 2. If we could construct the point (α, 0)then [Q(α) : Q] would be a power of 2. But α has a minimum polynomial x3−2over Q which is irreducible over Q. So [Q(α) : Q] = 3 and therefore (α, 0) isnot constructible [ST, pp55–56].

2. Trisection of an Arbitrary Angle We will show that the angle π/3 can-not be trisected. We wish to construct the angle π/9 which is equivalent toconstructing the point (α, 0) where α = cos(π/9). Recall that for any angleA, cos A = 4 cos3(A/3) − 3 cos(A/3), so if A is π/3 this gives

4 cos3(π/9) − 3 cos(π/9) = 1/2

8α3 − 6α − 1 = 0.

Now 8x3 − 6x − 1 is irreducible over Q therefore Q(α) : Q = 3. This contra-diction shows that (α, 0) cannot be constructed ([DU, p265] and [ST, p56]).

3. Quadrature of the Circle Without loss of generality we can assume thatthe given circle is the unit circle. So we wish to construct the square with areaπ. This is equivalent to constructing the point (

√π, 0). If we can construct

(√

π, 0) then we can certainly construct (π, 0) (constructing the square of anumber is demonstrated in Section A.1.3) so that [Q(π) : Q = 2m for somem ≥ 0. But π is not algebraic over Q (by Lindermann’s Theorem) and therefore[Q(π) : Q = 2m which shows that we cannot construct (π, 0) and therefore byimplication (

√π, 0) [ST, p57].


A.4.3 Performing Algebraic Operations by Construction

We can now give demonstrations of constructions where we are carrying outan algebraic operation using the method of geometrical construction.

1 Constructing the square of a number Given the point a we construct a2 bydrawing the line from (0, a) to (1, 0) then constructing the line parallel to thisline and passing through (a, 0). By similar triangles the point (0, b) at whichthe line intersects the vertical axis is a2 since

a

1=

b

a.

2 Constructing the square root of a number: Given the point b we construct√b by drawing the circle which has diameter (−1, 0) to (b, 0). This circle will

intersect the vertical axis at (0, a). By the intersecting chords theorem a2 = 1×b

so that a =√

b.

A.4.4 Regular n-gons

In 1796, at the age of 18, Gauss found a construction by straight edge andcompass for the regular 17-gon. This was the first new such polygon for 2,000years. Prior to this it was known how to construct regular n-gons for only thefollowing values of n [AL, p174]:

2r, 2r.3, 2r.5, 2r.15.

In 1801 Gauss gave the following theorem:

Theorem [ST, p169]The regular n-gon is constructible by straight edge and compass if and only

if n = 2rp1 . . . ps where r and s are integers greater than 0 and p1 . . . ps are oddprimes of the form pi = 22n

+1 for positive integers ri (proof: [ST, pp169–170]).The pi are known as Fermat numbers since they were discovered by Fermat

in the 17th century. If Fn = 22n

+ 1 then F0 = 3, F1 = 5, F2 = 17, F3 =257, F4 = 65, 537. These are all primes and Fermat conjectured that Fn isprime for all n. This was disproved in 1732 by Euler when he showed that F5

is not a prime. This leads us to

Proposition [ST, p170]The only primes p < 1040,000 for which the regular p-gon is constructible

are 2, 3, 5, 17, 257, 65537.


Figure A.1 Constructing the square of a number

If we wish to construct a p-gon where p is prime we can make use of thecyclotomic equation

xp − 1x − 1

= xp−1 + . . . + x + 1.

The roots of the right hand side of the equation are ak where ak = cos(2kπ/p)+i sin(2kπ/p).

As an example we will find a construction for the regular pentagon. If weassume that the pentagon has centre (0, 0) and its first vertex is at (1, 0) thenwe wish to construct the angle θ = 2π/5 which is equivalent to constructing


Figure A.2 Constructing the square root of a number

the point (α, 0) where α = cos θ.The primitive roots of x5 − 1 are a1, a2, a3, a4 where a1 +a2 +a3 + a4 = −1

(an elementary symmetric polynomial). We also need

a5−k + ak = cos(5 − k)θ + cos kθ + i sin(5 − k)θ + i sin kθ

= 2 cos(2π) cos[(5 − 2k)π

5

]+ 0

= 2 cos(2kπ/5)

= 2 cos kθ.

So a1 + a4 = 2 cos θ and a2 + a3 = 2 cos 2θ.


Putting these values into the elementary symmetric polynomial we have

2 cos θ +2 cos 2θ = −1

⇒ 4 cos2 θ + 2 cos θ − 1 = 0

⇒ α2 + 2α − 1 = 0.

Taking the positive root of this quadratic we have

α =√

5 − 14

.

In the demonstrations in Section A.1.2 we showed how to construct the squareroot of a number so, with the construction of rational numbers and one squareroot, we can construct the angle θ = 2π/5 to give the second vertex of thepentagon.

Gauss gave the following algebraic expression for constructing the 17-gon[ED, p32]:

cos(2π/17) = − 116

+116

√17 +

116

√34 − 2

√17

+18

√17 + 3

√17 −

√34 − 2

√17 − 2

√34 + 2

√17.

This can be derived from the cyclotomic equation of degree 17 by forming equa-tions of the roots similar to the one we used for the pentagon above althoughmore complicated [ST, pp171–173]. Research into constructions of n-gons hascontinued and methods have been found for the 257-gon. There has been someresearch into the 65,537-gon, with Richelot publishing a paper as early as 1832,although Stewart is not clear on whether this resulted in a satisfactory construc-tion [ST, p170]. It is difficult to imagine the complexity of solving trigonomet-ric equations in the 65,536 primitive roots of cyclotomic equation, particularlywithout the advanced computer packages now available to us.


Bibliography

[AL ] Allenby, R. B. J. T. Rings, Fields and Groups: An Introduction to Ab-stract Algebra Edward Arnold, London 1983

[DU ] Durbin, John R. Modern Algebra: An Introduction (3rd Edition). JohnWiley & Sons, New York, 1992

[ED ] Edwards, Harold M. Galois Theory Springer-Verlag, New York, 1984.

[GA ] Gaal, Lisl Classical Galois Theory with Examples. Markham Publishing,Chicago, 1971

[ST ] Stewart, Ian Galois Theory (2nd edition) Chapman and Hall, London,1989.

[WA ] van der Waerden, B. L. Algebra: Volume 1. Translated by Fred Blum andJohn R Schulenberger. Frederick Ungar Publishing, New York, 1970.


Epilogue

The assessors for this project were unanimous in awarding this student a first.The overall standard of the mathematics was very high, and the student obvi-ously understood the sometimes quite difficult material. This was obvious fromboth the oral presentation as well as at the viva-voce. The style of the presen-tation is mature and the student’s judgement on what to include and what toleave as a reference is nigh perfect. The layout is good (although this is lessobvious here as it has had to change to fit the LATEX style of this text) andleaves the reader no doubt that the student had read and understood the booksreferred to.

The only criticisms were the odd technical lapse and typographical error(many have been left in–maybe you can spot them).

The student took the responsibility to read and understand about GaloisTheory. This is a commitment and involves being willing to devote a greatdeal of time to reading, re-reading and understanding books that can be quiteadvanced. It also meant doing many examples. Therefore, much time needs tobe put aside which is consistent with the amount the project counts (one sixth ofthe final year in this instance). The self confidence of the student is also obviousfrom the writing, and this indicates a high level of mathematical ability.

BProject Example 2: Algebraic Curves

Preamble

This projects takes further a topic with which mathematics students are famil-iar, namely graphs of functions. As in Appendix A, the subject is well embeddedin the literature, but here the technicalities are less. There are places where thestudent has to branch out into the unknown (for the student), but much of theproject is putting what is reasonably familiar into more rigorous mathematicalterms.

Abstract

This project is an investigation into the theory of algebraic curves. We startby considering the basic definition and characteristics of an algebraic curve.We consider how an algebraic curve lies in a projective space and this leadson to studying the intersection of two curves. The properties of some simplecurves and how they intersect is considered in more detail including the groupstructure of elliptic curves. We also investigate some results from complex anal-ysis and subsequently use them to define Riemann surfaces and explain theirconnection with algebraic curves. This leads to more advanced theories fromcomplex function theory such as integration on a curve. The topology of alge-braic curves as subsets of C2 is also considered.

195


Introduction

Algebraic curves have a history of over 2000 years and it is inevitable that theGreeks made the first developments and discoveries. The Greeks had a limitedalgebraic knowledge but good geometric methods and to Euclid a circle was notdefined by an equation but by the locus of all points having an equal distancefrom a fixed point. Greek mathematics was virtually ignored in Western Europeuntil it was rediscovered in the Renaissance period. The Arabs were responsiblefor reintroducing Greek mathematics into Western society and also broughtwith them a sophisticated algebraic knowledge and a good algebraic notation.

In the late seventeenth century a more algebraic approach was being appliedto algebraic curves and a large number of mathematicians published papers onthe subject. These included Newton who made a study of cubic curves andidentified and classified 72 different cases. By the nineteenth century it wassoon apparent that considering curves as subsets of C2 was easier than lookingat curves as subsets of R2 as a curve in the Euclidean plane can degenerate somuch that it no longer resembles a curve.

In the nineteenth century mathematicians such as Riemann were interestedin more abstract geometrical spaces and this led to his theory of Riemann sur-faces and their connection with algebraic curves. Research into algebraic curvescontinues up to the present day particularly in the area of particle physics. Herethe ‘lifetime’ of a particle is represented by a ‘string’ in four dimensional space-time. This can consequently be represented by a Riemann surface.

Algebraic curves is one of the most varied and beautiful subjects in mathe-matics. This project does not have a specific aim and the discussion is relativelyinformal, but it does give an insight into some of the more interesting and ap-pealing results that have arisen. As one can imagine, for a subject that hasexisted for such a long time, the theory of algebraic curves is vast and involvesa lot of interplay from different areas of mathematics such as algebra, topol-ogy, complex analysis and geometry. It would therefore be foolish to try andincorporate every idea in the subject: consequently a lot of very elegant andcompelling theories have been missed out; these include the Riemann–Rochtheorem and an in-depth investigation into singularities.

In Section B.1 we show that an algebraic curve is defined by a polynomialand look at why we consider curves in C2. We also define singularities andconsider the multiplicity of such points. In Section B.2 we investigate how twocurves intersect by considering one of the most famous theorems in the studyof algebraic curves: Bezout’s theorem. Section B.3 applies results from previoussections to some simple curves namely conics and cubics. We shall see that acubic curve gives rise to some amazing properties such as the additive groupstructure. Section B.4 is a resume of results from complex analysis that we need

B. Project Example 2: Algebraic Curves 197

to be familiar with for the final chapters.In Section B.5 we investigate the topology of curves in C2 which is a space

that has four real dimensions. We find that these objects are topological surfacesand we consequently introduce the idea of an abstract Riemann surface torepresent an algebraic curve. Section B.6 is a less rigid investigation into moreadvanced ideas involving algebraic curves and complex analysis which henceleads on to topics from complex function theory.

Throughout the course of the project three books were heavily relied uponfor the theorems and subsequent proofs. These were Frances Kirwan ComplexAlgebraic Curves, Bak and Newman Complex Analysis and R. J. Walker, Al-gebraic Curves and particularly Bak and Newman in Section B.4 and Kirwanin Sections B.5 and B.6.

B.1 Basic Definitions and Properties

In this section we will look at some basic properties of algebraic curves. We alsoinvestigate the concept of a projective space and why it is useful to considercurves in the projective plane as well as curves in C2.

B.1.1 Complex Algebraic Curves and Real AlgebraicCurves

A complex algebraic curve C in C2 is defined by

C = (x, y) ∈ C2 : P (x, y) = 0where P (x, y) is a polynomial in two variables with complex coefficients. Allcomplex algebraic curves are subsets of R2 namely

C = (x, y) ∈ R2 : P (x, y) = 0where P (x, y) has real coefficients.

For every real algebraic curve there is an associated complex algebraic curvedefined by the same polynomial. Investigating curves in the complex space is farsimpler than only considering curves in the Euclidean space R2. If we work withcomplex numbers every polynomial factorises completely in C (not the case inR). This result from complex analysis, known as the fundamental theorem ofalgebra (see B.4.31), allows us to completely factorise any polynomial, so for


simplicity in the remainder of the project we will only usually consider curvesas subsets of C2.

The degree of a curve C is given by the degree of the polynomial that definesC. Hence the curve defined by P (x, y) = x(x2 + 1) has degree 3.

The polynomial P (x1, . . . , xn) = 0 is homogeneous of degree d if and only if

P (tx1, . . . , txn) = td(x1, . . . , xn)

for all t, x1, . . . , xn and some d, i.e. P is homogeneous if every one of its termsare of the same degree d.

A polynomial is irreducible if it has no factors apart from constants andscalar multiples of itself. A curve defined by P (x, y) is irreducible if P (x, y) isirreducible.

So far we have only considered curves in C2, but a curve in C2 is nevercompact as it always has branches going to infinity. We can compactify a curvein C2 by adding in the “points at infinity” and getting a projective space. Itcan be beneficial to consider curves in a projective space, for example considertwo parallel lines which lie in C2. They will of course never intersect but ifwe add a point at infinity to each line they meet at a distinct point (albeit at“infinity” for this example). Consequently any two lines in a projective spacewill meet at a distinct point. For this reason it is useful to consider curves in aprojective space studying their algebraic properties.

B.1.2 Projective Spaces

B.1.2.1 Complex Projective Plane. Consider the set of all possible complexlines which pass through the origin of C3. It can be shown that the set ofall such lines form a complex surface (4 real dimensions). The surface that isformed is known as the complex projective plane and denoted by CP2. Thisidea can be generalised to give a projective space for any complex dimensionn.

The complex projective space CPn of dimension n is the set of complexone dimensional subspaces of the vector space Cn+1. When n = 2 we get thecomplex projective plane CP2 as above.

To prove results algebraically we need a notation to specify points in CP2.Clearly a line through the origin of C3 is uniquely determined if we choose anypoint (save the origin) which lies on that line. So a point p in CP2 consists ofthe unique line in C3 which passes through (0, 0, 0) and (a, b, c) where a, b andc are not all zero. The notation for such a point p is given by [a, b, c].

We can define the points in CP2 by

CP2 = [a, b, c] : (a, b, c) ∈ C3 − 0


and also [a, b, c] = [d, e, f ] if and only if a = λd, b = λe and c = λf for someλ ∈ C−0. So far we have used the theory of three dimensional complex spaceto develop the idea of a projective space. However we want to use this space tostudy objects which have two complex dimensions, specifically algebraic curves.So we must find a way of associating figures in CP2 with figures in C2.

Suppose we have a projective figure which lies in CP2. We can place aplane in C3 making sure it does not pass through the origin. The correspond-ing figure in C2 are the points of the figure that pierces the plane. This is okayprovided that each of the projective figures pierces the plane, but any pointof CP2 which consists of a line through the origin parallel to the plane thatdoes not pierce the plane. Such a point is called an ideal point for that plane.In general a projective figure can be represented by the points which pierce aplane in C3 together with some or all the ideal points for that plane. Using thisidea we could have defined the complex projective plane as simply the usualcomplex plane with “ideal points at infinity”, one ideal point for each complexline through the origin. In the following example we will show how points inC2 can be identified with points in CP2.

Example 1.22 Consider the subset S of points in CP2 which pierce the planez = 1:

S = [x, y, z] : z = 0.Each point (x, y) ∈ C2 can be identified with the point (x, y, 1) ∈ S. Conversely,[x, y, z] = [x/z, y/z, 1] ∈ S can be identified with (x/z, y/z) ∈ C2. This gives a1–1 correspondence between C2 and S. So C2 can be regarded as a subset ofCP2.

B.1.2.2 Projective Transformations. A geometry consists of a space togetherwith a group of transformations that act on that space: projective geometry isno different. Since the points of CP2 consist of lines through the origin of C3 aprojective transformation maps straight lines on to straight lines. Consequentlytransformations in CP2 must be linear transformations of C3. If we define themap Φ:C3−0 → CP2 by Φ(a, b, c) = [a, b, c] then we can define a hyperplanein CP2 as the image of V −0 under Φ where V is a two-dimensional complexsubspace of C3.

Proposition 1.24 Given four distinct points p0, p1, p2 and q of CP2 no threeof which lie on a hyperplane, there is a projective transformation which takesP0 to [1, 0, 0], p1 to [0, 1, 0], p2 to [0, 0, 1] and q to [1, 1, 1].

Proof Let u0, u1, u2 and v be elements of C − 0 whose image under Φ are


p0, p1, p2 and q. So u0, u1, u2 form a basis of C3. There is therefore a uniquelinear transformation α of C3 taking u0, u1, u2 to the standard basis(1, 0, 0), (0, 1, 0), (0, 0, 1). Now α(v) = (λ0, λ1, λ2) where λ0, λ1, λ2 are non-zero complex numbers.

Hence the composition of α with a 3 × 3 matrix⎡⎣ 1/λ0 0 00 1/λ1 00 0 1/λ2

⎤⎦defines a projective transformation taking p0 to [1/λ1, 0, 0] = [1, 0, 0], p1 to[0, 1/λ2, 0] = [0, 1, 0], p2 to [0, 0, 1/λ3] = [0, 0, 1] and q to [1, 1, 1]. So when weare investigating a certain point on a curve we can assume that under a suitableprojective transformation that the point can be a point of reference, namely[1, 0, 0], [0, 1, 0], or [0, 0, 1].

B.1.3 Affine and Projective Curves

Now we can define a projective curve in CP2 by a homogeneous polynomialP (x, y, z) in x, y and z. Thus a projective curve C is defined by

C = [x, y, z] ∈ CP2 : P (x, y, z) = 0.Algebraic curves in C2 as defined in Section B.1.1 will be called affine curvesto distinguish them from projective curves. Affine and projective curves areclosely related and it was shown that points in C2 can be identified with pointsin CP2. This idea can be extended for affine and projective curves.

Let P (x, y, z) = 0 be a curve in CP2 which does not have z = 0 as a com-ponent. The curve has an associated non-homogeneous polynomial P (x, y, 1) ofthe same degree. Consequently an affine curve P (x, y) in C2 can be identifiedwith zdP (x/z, y/z) in CP2 where d is the degree of the polynomial.

B.1.4 Singular Points

A singular point is basically a point where the curve does not look “smooth”.This idea can be best illustrated by looking at the sketches of algebraic curvesin R2 given at the end of this section.

Mathematically, for an affine curve C a singular point or singularity is where

∂P

∂x(a, b) = 0 =

∂P

∂y(a, b)


where P (x, y) = 0 defines C and (a, b) ∈ C.Similarly for a projective curve, we have

∂P

∂x(a, b, c) =

∂P

∂y(a, b, c) =

∂P

∂z(a, b, c) = 0, where (a, b, c) ∈ C.

If a curve does not have any singularities it is termed non-singular i.e. x2 +y2 − 1 = 0. The multiplicity of an affine curve defined by P (x, y) at a point(a, b) is the smallest possible integer m such that

∂mP

∂xi∂yj(a, b) = 0 for some i ≤ 0, j ≤ 0 where i + j = m.

Similarly for a projective curve the multiplicity of a point (a, b, c) is the smallestpossible integer m such that

∂mP

∂xi∂yj∂zk(a, b, c) = 0 where i + j + k = m.

The point is non-singular if its multiplicity is one. If this is the case it has aunique tangent line at (a, b). A point is a double point, triple points, etc. Asingular point is ordinary if the first non-vanishing terms expanded about thispoint have no repeated factors, it thus has m distinct tangent lines at (a, b).We will finish this section by looking at some types of singular points in R2.Example 1.41

P (x, y) = x3 − x2 + y2 see Figure B.1.

The origin is a double point with two distinct tangents. This type of doublepoint is called a node.

Example 1.42P (x, y) = x3 − y2 see Figure B.2.

This double point is not ordinary as it has only one tangent line that runs alongthe x-axis. This type of singularity is known as a cusp. Example 1.43

P (x, y) = (x4 + y4)2 − x2y2 see Figure B.3.

This is not an ordinary point of multiplicity four as it has two tangents at rightangles running along the x and y axis.

Example 1.44

P (x, y) = (x4 + y4 − x2 − y2)2 − 9x2y2 see Figure B.4.

Here we have an ordinary point of multiplicity four with four distinct tangentswhich pass through the origin.


Figure B.1 The curve x3 − x2 + y2 = 0

B.2 Intersection of Two Curves and Points ofInflection

This section investigates whether two polynomials in x, y and z have any com-mon solutions, i.e. how do two projective curves intersect? We shall also lookat points of inflection on a curve.

B.2.1 Bezout’s Theorem

The number of points of intersection between two projective curves defined bythe polynomials P (x, y, z) and Q(x, y, z) is given by Bezout’s Theorem. If P is


Figure B.2 The curve x3 − y2 = 0

of degree n and Q is of degree m then P and Q have precisely nm commonsolutions. To prove this we first need the following algebraic result.

B.2.1.1 Resultant of Two Polynomials. We can rearrange P and Q so that

P (x, y, z) = a0(y, z) + a1(y, z)x + a2(y, z)x2 + . . . + an(y, z)xn

and

Q(x, y, z) = b0(y, z) + b1(y, z)x + b2(y, z)x2 + . . . + bm(y, z)xm

where ai, bi ∈ C.The resultant R(y, z) of P and Q is given by the determinant of


Figure B.3 The curve (x4 + y4)2 − x2y2 = 0

⎡⎢⎢⎢⎢⎢⎢⎣

a0(y,z) a1(y,z) ... an(y,z) 0 0 ... ... 00 a0(y,z) a1(y,z) ... an(y,z) 0 ... ... 0

......

......

......

......

...0 0 ... 0 a0(y,z) a1(y,z) ... ... an(y,z)

b0(y,z) b1(y,z) ... ... ... bm(y,z) 0 ... 0

......

......

......

......

...0 ... 0 b0(y,z) b1(y,z) ... ... ... bm(y,z)

⎤⎥⎥⎥⎥⎥⎥⎦R(y, z) is thus a polynomial in y and z of degree nm.

If P (x, y, z) and Q(x, y, z) are non-constant homogeneous polynomials wecan arrange it so that

P (1, 0, 0) = 0 = Q(1, 0, 0)

which ensures that they have the same degree when regarded as polynomialsin x with coefficients in y and z. It follows that P (x, y, z) and Q(x, y, z) have


Figure B.4 The curve (x4 + y4 − x2 − y2)2 − 9x2y2 = 0

a non-constant common factor if and only if R(y, z) = 0 and it always can bearranged so that this condition is true. Using this result we can now prove theweak form of Bezout’s theorem

Theorem 2.12

If two projective curves of degrees n and m have more than nm common points,then they have a common component.

Proof [Walker 50, 3.1 p59]

Let C and D be two projective curves of degrees n and m defined by the


polynomials P (x, y, z) and Q(x, y, z) that have more than nm common points.We can select any set of nm + 1 of these points and join each selected pair bya line. As there is only a finite set of such lines then there is a point p not onany of the lines or on C or D. Under a suitable projective transformations wecan choose this point to be p = [1, 0, 0]. Consequently P and Q can be put inthe form

P (x, y, z) = a0(y, z) + a1(y, z)x + a2(y, z)x2 + . . . + an(y, z)xn

and

Q(x, y, z) = b0(y, z) + b1(y, z)x + b2(y, z)x2 + . . . + bm(y, z)xm

where ai, bj ∈ C. The resultant R(y, z) of P and Q is either zero or a homo-geneous polynomial of degree nm in y and z. We deduce that R(b, c) = 0 ifand only if there is an a such that P (a, b, c) = 0 = Q(a, b, c), that is to saythe coordinates b, c common to P and Q satisfy R(y, z) = 0. However each ofthe nm + 1 points has a different value for the ratio b : c since no pair of themare collinear with [1, 0, 0]. Hence R(y, z) = 0 and so C and D have a commonfactor, i.e. two curves intersect at nm points at most.

In section B.1 we defined the multiplicity of a curve. We use a similar ideato define the intersection multiplicity of two curves and use it to show that thenumber of intersections counted properly is precisely nm, the strong case ofBezout’s theorem.

Theorem 2.13

If C and D are two projective curves of degrees n and m in CP2 which have nocommon component then they have precisely nm points of intersection countedproperly; i.e. ∑

p∈C∩D

Ip(C,D) = nm.

The intersection multiplicity can be defined by using the resultant of two poly-nomials as defined in Section B.2.1. It follows that from our definition of theresultant that the intersection multiplicity Ip(C,D) for two projective curvesC and D at a point p has the following properties:

* Ip(C, D) = Ip(D, C).

* Ip(C, D) = ∞ if the point p lies on a common component of C and D

otherwise Ip(C,D) is a non-negative integer.

* Ip(C, D) if and only if the point p does not lie on C ∩ D.

* Two distinct lines meet with intersection multiplicity one.


* If C1 and C2 are defined by the homogeneous polynomials P (x, y, z)and Q(x, y, z) and C is defined by R(x, y, z) = P (x, y, z)Q(x, y, z) thenIp(C, D) = Ip(C1, D) + Ip(C2, D).

* If C and D are defined by P (x, y, z) and Q(x, y, z) of degrees n and m andE is defined by PQ + R where R(x, y, z) is homogeneous of degree m − n

then Ip(C,D) = Ip(C, E).

From this we can define the intersection multiplicity Ip(C, D) for some p =C ∩D as the largest k where (bz− cy)k divides the resultant R(y, z) and it canbe shown that k is uniquely determined. Using this definition we can prove thestrong case of Bezout’s theorem.

Proof of Theorem 2.13 [Kirwan 92, 3.1 p62] Under a projective transfor-mation we can choose a point p = [1, 0, 0] such that p /∈ C ∪ D. p does notlie on a line containing two distinct points of C ∩ D and p does not lie on atangent line to C or D at any point of C ∩ D. If C and D are defined by thehomogeneous polynomials P (x, y, z) and Q(x, y, z) then the resultant R(y, z)is a homogeneous polynomial of degree nm in y and z.

The resultant R(y, z) can be expressed as a product of linear factors

R(y, z) =k∏

i=1

(ciz − biy)ei where ci, ei ∈ C

and ei = nm = e1 + e2 + . . . + ek.There exists a unique complex number ai such that C∩D = pi : i ≤ i ≤ k

where pi = [ai, bi, ci] which leads to Ipi(C, D) = ei and hence∑p∈C∩D

Ip(C, D) = nm

proving the result.

B.2.2 Points of Inflection on a Curve

A point p on a curve C in CP2 is a point of inflection of C when the tangentto C at p is not a component of C and meets C at p with a multiplicity ofat least three. We want a way to compute such points: for this we use theHessian curve. If C is defined by the homogeneous polynomial P (x, y, z) then


the Hessian H(x, y, z) of C is given by the determinant of the 3 × 3 matrix⎡⎣ Pxx Pxy Pxz

Pyx Pyy Pyz

Pzx Pzy Pzz

⎤⎦ .

When H vanishes at a point of C then that point is an inflection point of C,for example if [a, b, c] lies on C and H(a, b, c) = 0 then [a, b, c] is a point ofinflection (or flex) of C. The second derivatives of P (x, y, z) are of degree d−2.It follows that H must be a polynomial of degree 3(d−2) in x, y and z. We caneasily deduce from this and from Bezout’s theorem that a non-singular curveof degree greater than or equal to three has at least one point of inflection, i.e.when d = 2, H is a polynomial of degree zero which no a, b, c could satisfy.When H = 0 or a constant then the curve that leads to this is defined as havingno Hessian.

B.3 Conics and Cubics

To illustrate some of the results that were investigated in Section B.2 it wouldbe useful to apply them to some simple curves. It can be easily seen that curvesof order one, namely lines intersect at only one point and that every point onthe line is a point of inflection, that is to say its second derivative vanishes.

B.3.1 Conics

A conic is a curve of degree 2 in CP2. Any non-singular projective conic C inCP2 is equivalent under a projective transformation to the conic

x2 = yz.

B.3.2 Cubics

Cubics are curves of degree 3 in CP2. Cubics are the first class of curve wheresome more interesting properties occur. Firstly we will show that a non-singularcubic can be written in the standard form y2 = g(x).


Theorem 3.21

By a proper choice of coordinates any non-singular cubic C can be put inthe form

y2 = g(x) (1)

where g(x) is a cubic polynomial with distinct roots.

Proof [Walker 50, p72]

To prove this we need the result from Section B.2 that every non-singularcurve of degree leq3 has at least one point of inflection. Under a projectivetransformation we can choose C to have a point of inflection at [0, 0, 1], whichreduces (1) to

x3 + h(x, y) = 0 (2)

where h(x, y) is of degree 2. Also h(x, y) must include a term ay2 (a not zero);if not [0, 0, 1] would be a singularity. Solving (2) for y we get

y = αx + β +√

g(x).

If we make the linear transformation y′ = y − αx − β and x′ = x we obtain

y′ + αx′ + β = αx′ + β +√

g(x).

Squaring and dropping the dashes we get the required form y2 = g(x).

If we had a singular cubic then g(x) would not have distinct roots. If this werethe case under a suitable projective transformation g(x) = x2(x+1) for a nodalcubic or g(x) = x3 for a cuspidal cubic. We can now investigate one of the mostfamous properties of cubics.

B.3.2.1 Inflections on a Cubic. We have the following theorem,

Theorem 3.23

A non-singular cubic C has 9 distinct points of inflection with the propertyevery line joining two of them contains a third.


Proof [Walker 50, 6.6, p 72]

Under the linear transformation x′ = ax + b and y′ = y, y2 = g(x) can bewritten

F = y2 − x3 − ax2 − bx = 0

with b(a2 − 4b) = 0 since x3 + ax2 + bx = 0 has distinct roots. We first look forpoints in the finite plane. The Hessian of F is given by

H = (y2 + bx)(3x + a) − (ax + b)2.

Eliminating y between F and H gives

p(x) = 3x4 + 4ax3 + 6bx2 − b2 = 0.

So p(x) has 4 distinct roots and p′(x) = 12(x3 + ax2 + bx) and the resultant ofp and p′ is

b4(a2 − 4b)2 = 0.

So for each x value there are two y values. None of these values of x satisfyx3 +ax2 + bx = 0. So this makes eight points of inflection plus a flex at infinity(i.e. at [0, 0, 1] from the assumption made in Theorem 3.21) which makes a totalof nine. If we choose two points of inflection in the finite plane to be [1, a, b] and[1, a,−b] say, then there must be a line that passes through these two points.If then we take the flex at infinity namely [0, 0, 1] then a line which containsthe first two points must also contain this point. We can arrange the ninepoints of inflection into an array such that three points in any row, column ordiagonal lies on a line in CP2. We now look at the result that connects conics,cubics, and Bezout’s Theorem. Let C and D be projective cubics defined by thepolynomials P (x, y, z) and Q(x, y, z) and assume C and D meet at exactly ninepoints p1, . . . , p9. Now let E be another projective cubic defined by R(x, y, z)where E contains the points p1, . . . , p8. If C,D and E are linearly independentthen there is a curve C(λ, µ, ν) defined by

λP (x, y, z) + µQ(x, y, z) + νR(x, y, z) = 0 for some λ, µ, ν

which passes through the points p1, . . . , p8 and any two arbitrary points q andr say. We will show that this in turn leads to a contradiction and therefore ifE contains p1, . . . , p8 then E contains p9. By Bezout’s Theorem, no line cancontain four of these points as such a line would be a common component of C

and D. Similarly no cubic can contain seven of these points. Now if p6, p7 andp8 lie on a line L then p1 . . . , p5 lie on a unique conic Q. This conic is uniquebecause if two conics lie on five common points they must have a commoncomponent (Bezout’s Theorem), the remaining components can only have oneintersection not on this line, and hence the other four points must be collinear.


Figure B.5 The array in CP2

Now let q ∈ L and r /∈ L ∪ Q. C(λ, µ, ν) contains p1, . . . , p8 and λ, µ, ν arechosen so that it contains q and r. So C(λ, µ, ν) must have L as a component,the other component is therefore Q. Hence C(λ, µ, ν) = L ∪ Q, but this isimpossible since r /∈ L∪Q. Now if p1, . . . , p6 lie on a conic Q then p7, p8 lie ona line L. If we choose C(λ, µ, ν) to vanish at the points q ∈ Q and r /∈ L ∪ Q

then this again leads to a contradiction since C(λ, µ, ν) = L ∪ Q. Finally ifno three of p1, . . . , p8 lie on a line and no six on a conic, we let L be the linewhere p1, p2 lie and Q be the unique conic on which p3, . . . , p7 lie. If we takeq ∈ L and r ∈ L then this again leads to a contradiction since p8 /∈ Q and


C(λ, µ, ν) = L∪Q as before. This exhausts all possibilities so we can concludethat the curve C(λ, µ, ν) does not pass through p1, . . . , p8, q and r for any choiceof q and r. So E is linearly dependent on C and D, and hence E must containp9 as well. We can use this result in the next section.

B.3.3 Additive Group Structure on a Cubic

Theorem 3.31

Given any non-singular projective cubic C in CP2 and a point of inflectionp0 on C there is a unique additive group structure on C such that p0 is thezero element and three points of C add up to zero if and only if the intersectionof C with some line in CP2 (allowing for some multiplicities).

Proof [Kirwan 92 3.38, p 77]

Firstly we have to show that additions and inverse operations are uniquelydetermined. Now −p0 = p0 since p0 is a point of inflection and if p = p0 then−p is the third point of intersection with some line in CP2 through p and p0.So additive inverses are uniquely determined. Now let p and q be any points ofC then p + q = −r where r is the third point of intersection of C with a linein CP2 through p and q (if p = q) or the tangent to C at p (if p = q). So theaddition operation is uniquely determined.

Now we have to show that there is an additive group structure on C withp0 as the zero element. Commutativity, namely p + q = q + p, follows from thedefinition of the addition operation. Similarly p + (−p) = p0 follows from thedefinition of additive inverses. We have to deduce that p0 is the zero element.

For any p ∈ C such that p = p0 we have p + p0 = −r where r is the thirdpoint of intersection of C with the line in CP2 through p and p0. This pointr is not p0, so −r is the third point of intersection of C with the line in CP2

through r and p0, which is the point p. Thus, p+p0 = p if p = p0 and p+p0 = p0

since p0 is a point of inflection, so p0 is the zero element.To prove associativity we let p, q, r ∈ C and L1 is the line in CP2 that

meets C at p, q,−(p+q) and similarly we let the following be lines in CP2 thatmeet C at

L2 : p0, p + q,−(p + q), L3 : r, p + q,−((p + q) + r), M1 : q, r,−(q + r),

M2 : p0, q + r,−(q + r) and M3 : p, q + r,−(p + (q + r)).


Figure B.6

Now let D and E be the reducible curves D = L1 ∪ M2 ∪ L3 and E =M1 ∪ L2 ∪ M3. C meets L1 at [p, q,−(p + q)] and therefore D meets C at

p0, p, q, r, p + q, q + r,−(p + q),−(q + r),−((p + q) + r)

and by a similar deduction we find that E meets C at

p0, p, q, r, p + q, q + r,−(p + q),−(q + r),−(p + (q + r)).

We have to show that (p+q)+r = p+(q+r). We can see that C and D meet atnine points and that E contains eight of them. From the argument previouslyin the section, if two projective curves C and D intersect at nine points andif a projective curve E contains eight of them, then it also must contain theninth. So −(p+(q +r)) and −((p+q)+r) must describe the same point, hence(p+ q)+ r = p+(q + r) proving associativity. Note that we used p0 as the zeroelement.


In fact we could have used any point of C as long as we constructed ourgroup around that point using the proviso that any three points of intersectionwith C add up to zero if they are three points of intersection of C with CP2.

B.4 Complex Analysis

For the final sections we need some standard results from complex analysis.We consider the definition of holomorphic functions and their properties andCauchy’s closed curve theorem. Although some proofs are given not every resultis proved rigorously and the reader may refer to any introductory book oncomplex analysis for the details, e.g. [Bak and Newman 82].

We will use the notation z = x+iy to represent a complex number and f(z)denotes a complex function. D(z0 : r) is the open disc of radius r > 0 centredat z0 (i.e. D is an open subset of C, or alternatively all neighbourhoods of z0

contain such a disc.)

B.4.1 Holomorphic Functions and Entire Functions

A function f is only differentiable at z if∂f

∂xand

∂f

∂yare continuous at z and

it satisfies the Cauchy–Riemann equations:

∂f

∂x= i

∂f

∂y.

Consequently it can be shown thatdf

dz=

∂f

∂x. If D is an open subset of C

and a complex number f is differentiable at every z ∈ D then the functionis holomorphic in D. If the first derivatives exist then it can be shown thatall subsequent derivatives exist. The term analytic is often used to describe aholomorphic function.

Example 4.11

The function f(x + iy) = x2 − y2 + 2ixy is holomorphic, but the functionf(x + iy) = x2 − y2 − 2ixy is not holomorphic.

Functions such as polynomials that are everywhere differentiable are calledentire functions. An entire function is therefore infinitely differentiable.


B.4.2 Closed Curve Theorem and Line Integrals

B.4.2.1 Line Integrals. If we let f(t) = u(t) + iv(t) be any continuouscomplex-valued function of real variable t, then we can write∫ b

a

f(t)dt =∫ b

a

u(t)dt + i

∫ b

a

v(t)dt, where a ≤ t ≤ b.

Now we let z(t) = x(t) + iy(t) where the curve determined by z(t) is piecewisedifferentiable and we set z′(t) = x′(t)+ iy′(t). If x and y are continuous on [a, b]and are continuously differentiable on some partition of [a, b], we can define aline integral by ∫

C

f(z)dz =∫ b

a

f(z(t))z′(t)dt

where C is a piecewise smooth curve parametrised by z(t).If two curves C1 and C2 are smoothly equivalent then it turns out that∫

C1

f =∫

C2

f.

If we define −C by z(b + a − t) (i.e. C traversed in the opposite direction) wefind that ∫

−C

f =∫

C

f.

If f is the complex derivative of a holomorphic function F say, where F isholomorphic on C then∫

C

f(z)dz = F (z(b)) − F (z(a)).

If f is entire, then f is everywhere the derivative of a holomorphic function,namely there exists an entire F such that F ′(z) = f(z) for all z (proof of this isnot given). We will use this and previous results to prove the first incarnationof the closed curve theorem.

B.4.2.2 Closed Curve Theorem. A curve is closed if its initial and end pointscoincide. If a curve C is given by z(t) and a ≤ t ≤ b then if z(a) = z(b) C isclosed. We will now show that the integral along a closed curve is zero.


Theorem 4.23

If f is entire and C is a smooth closed curve then∫C

f(z)dz = 0.

Proof [Bak and Newman 82, 4.16, p 51]

Since f is entire there exists an f(z) = F ′(z) where F is an entire function.Hence ∫

C

f(z)dz =∫

C

F ′(z)dz = F (z(b)) − F (z(a)).

Since C is closed, z(a) = z(b) and F (z(b)) = F (z(a)) thus∫C

f(z)dz = 0.

This can be related to Stokes’ theorem by writing∫

C

f(z)dz as∫C

(udx − vdy) =∮ (

−dv

dx− du

dy

)= 0

by the Cauchy–Riemann equations.

B.4.2.3 Cauchy Integral Formula. If f is entire and a is some given complexnumber and C is given by

C : Reiθ, 0 ≤ θ ≤ 2π R > |a|then

f(a) =1

2πi

∫C

f(z)z − a

dz.

The proof of this theorem is not given here (see [Bak and Newman 82, 5.5, p55]).

B.4.3 Liouville’s Theorem and Fundamental Theorem ofAlgebra

B.4.3.1 Liouville’s Theorem. An application of the Cauchy integral formulais Liouville’s theorem. This says that if f is holomorphic and bounded in C,then f is a constant. To prove this we simply calculate f(a) − f(b) using theformula where a and b are complex numbers and C is a circle centred at zerowith R > (|a||b|). We note that we can take R as large as we want so f(a)−f(b)approaches zero as R tends to infinity, i.e. f(a) − f(b) which implies that f isa constant.


B.4.3.2 Fundamental Theorem of Algebra. Every non-constant polynomialwith complex coefficients has a zero in C.

Proof [Bak and Newman 82, 5.12, p 59]

Let P (z) be a polynomial. If P (z) = 0 for all z then we can write f(z) = 1/P (z)which is an entire function. If P (z) is not constant, then P → ∞ as z → ∞ sof is bounded. If f is bounded, then f is constant by Liouville’s theorem. HenceP must be constant which contradicts the original assumption.

B.4.4 Properties of Holomorphic Functions

If f is holomorphic in an open disc D(α : r) and α ∈ D(α : r) it can be shownthat there exists functions F and G which are holomorphic in D such that

F ′(z) = f(z) and G′(z) =f(z) − f(a)

z − a.

If C is a closed curve in D then∫C

f(z)dz =∫

C

f(z) − f(a)z − a

dz

and ∫C

f(z) − f(a)z − a

dz =∫

C

G′(z)dz = G(z(b)) − G(z(a)).

Using this and the result that∫

C

dz

z − a= 2πi (proof [Bak and Newman 82,

5.4, p 55]) gives us Cauchy’s integral formula for functions holomorphic in thedisc D

f(a) =1

2πi

∫CP

f(z)z − a

dz, 0 < p < r

where CP : α + peiθ, 0 < θ < 2π.

Cauchy’s theorem leads to an integral representation of derivatives of f ,namely

f (n)(a) =n!2πi

∫CP

f(z)(z − a)n+1

dz.

This in turn leads to a power series representation for holomorphic functionsin D(α : r)

f(z) =∞∑

n=0

Cn(z − a)n.


We can choose α ∈ D(α : r) and set p > 0 such that |a − α| < p < r. If|z − α| < |a − α| then

f(z) =1

2πi

∫CP

f(w)w − z

dw,

and using the result that

1w − α

+z − α

(w − α)2+

(z − α)2

(w − α)3+ . . .

converges to 1/(w − z) throughout CP we get the following convergent seriesexpansion for f :

f(z) =1

2πi

∫CP

f(w)[

1w − α

+z − α

(w − α)2+

(z − α)2

(w − α)3+ . . .

]dw,

= C0(p) + C1(p)(z − α) + C2(z − α)2 + . . .

whereCn(p) =

12πi

∫CP

f(z)(z − α)n+1

dz.

As f is infinitely differentiable at α and Cn(p) =f (n)(α)

n!for each p, 0 < p < r

then

Cn =f (n)(α)

n!=

12πi

∫CP

f(z)(z − α)n+1

dz

for all z ∈ D(α : r).

B.4.4.1 Mean Value Property. A consequence of Cauchy’s theorem is themean value property of holomorphic functions. This is where f(α) is equal tothe mean value of f taken around the boundary of a disc centred at α, that isto say

f(α) =12π

∫ 2π

0

f(α + reiθ)dθ.

To prove this we simply reformulate Cauchy’s integral formula and set a = α.

B.4.4.2 Morera’s Theorem. The converse of Cauchy’s theorem is Morera’stheorem which states that if f is continuous in some open subset of C, D sayand if ∫

f = 0

then f is holomorphic in D. The proof of this theorem is simply to observethat we can define an indefinite integral F of f and then note that f is thederivative of the holomorphic function f .


B.4.5 General Cauchy Closed Curve Theorem

A function f can be holomorphic on a closed curve C and yet∫

C

f = 0, for

example ∫|z|=1

1zdz = 2πi.

For the general case of Cauchy’s closed curve theorem we want to find wherethe theorem is valid.

B.4.5.1 Simply Connected Domains. The function f(z) = 1/z is the punc-tured plane, i.e. there is a ‘hole’ at z = 0. We want to define a region with no“holes”. For this we use the definition of a simply connected domain. A domainis simply connected if every closed curve in the region can be contracted to apoint of C. Clearly a rectangle is simply connected since every closed curve inthe rectangle can be contracted to a point by means of affine transformations.Simply connected domains are best illustrated by simple examples (see below).

Example 4.52

The annulus A = z : 1 < |z| < 3 is not simply connected (see FigureB.7)

Example 4.53

The infinite strip S = z : −1 < Im(z) < 1 is simply connected (see Fig-ure B.8)

We can now define the general case of Cauchy’s theorem.

B.4.5.2 General Closed Curve Theorem. If f is holomorphic and lies in asimply connected domain D and C is a continuous smooth curve in D then∫

C

f = 0.

B.4.6 Isolated Singularities and Removable Singularities

A complex function f has an isolated singularity at z0 if f is holomorphic ina deleted neighbourhood of z0 but is not necessarily holomorphic at z0, i.e.f(z) = 1/(z − 4) has a singularity at z0 = 4.


Figure B.7

Now suppose that f has a singularity at z0. If there exists a holomorphicfunction g holomorphic at z0 too so that f(z) = g(z) for all z in some deletedneighbourhood of z0 then we can say that f has a removable singularity at z0,for example

f(z) =

sin z z = 20 z = 2

has a removable singularity at z0 = 2.

If f can be written in the formA(z)B(z)

where A and B are holomorphic and

A(z0) = 0 and B(z0) = 0 then we say that f has a pole at z0. If f does nothave a pole or removable singularity at z0 then it is said to have an essentialsingularity at z0.


Figure B.8

B.4.7 Laurent Expansions

In section B.4.4 we saw that functions holomorphic in a disc could be repre-sented by a power series. A Laurent expansion is a power series which involvesnegative powers of z also and which converges to a holomorphic function in anannulus R1 < |z − z0| < R2, that is

f(z) =∞∑−∞

ak(z − z0)k.

Conversely, if f is holomorphic in the annulus then it has a convergent powerseries as above with the coefficients ak given by

ak =1

2πi

∫C

f(z)(z − z0)k+1

dz

where C = C(z0 : R) and R is chosen so that R1 < |z| < R < R2.

B.4.8 Residue Theorem

If γ is a circle surrounding z0 and f(z) =∞∑−∞

Ck(z − z0)k in a deleted

neighbourhood of z0 that contains γ then∫

γ

f = 2πiC−1. Thus the coeffi-

cient C−1 has a special importance and it is called the residue of f at z0.If f has a pole of order one at z0 and can be written A(z0)/B(z0) then


C−1 =Res(f : z0) = limz→z0

(z − z0)f(z) =A(z0)B′(z0)

.

Example 4.81

1(z4 − 1)

has a pole at z0 = i and hence Res(1

(z4 − 1); i) =

14i3

=i

4.

We can now generalise the Cauchy closed curve theorem further to includefunctions with singularities. To do this we must first consider the windingnumber of a closed curve. The winding number is given by

n(γ, a) =1

2πi

∫γ

dz

z − a

where γ is a closed curve and a does not lie on γ. It can be shown that forany closed curve the winding number is an integer. (See the examples in FigureB.9). If we fix γ and vary a then the winding number is always a continuous

Figure B.9

function. If γ is a semi-circle traversed anti-clockwise then

n(γ, a) =

1 a inside semi-circle0 a outside semi-circle.

B.4.8.1 Cauchy’s Residue Theorem. If f is a holomorphic function in a sim-ply connected domain except at isolated singularities z1, z2, . . . , zm and if γ isa closed curve not intersecting any singularities then∫

γ

f = 2πi

m∑k=1

n(γ, zk)Res(f, zk).


For proof see [Bak and Newman 82, 10.5, p 110]. We define f as meromorphic ina domain if f is holomorphic there except at isolated poles. A useful applicationof the residue theorem is the evaluation of definite integrals using contourintegral techniques. Suppose we have an integral of the form∫ ∞

−∞

P (x)Q(x)

dx

where P and Q are polynomials. This will make sense if Q(x) = 0 and ifdegQ − degP ≥ 2, we then take∫ ∞

−∞

P (x)Q(x)

dx = limR→∞

∫ R

−R

P (x)Q(x)

dx

and we want to estimate this for large R. Let CR be the closed curve from −R

to R of the real line. By the residue theorem we get

Figure B.10

∫CR

P (z)Q(z)

dz = 2πi∑

k

Res(P

Q; zk)

where zk are poles of Q contained in CR. It can be shown that when R → ∞we get ∫ ∞

−∞

P (x)Q(x)

dx = 2πi∑

k

Res(P

Q; zk).


Example 4.83 ∫ ∞

−∞

dx

x4 + 1= 2πi

∑k

Res(1

z4 + 1; zk)

and we have poles at z1 = eπi/4 and z2 = e3πi/4. Hence

Res(

1z4 + 1

; eπi/4

)=

14z3

1

= −18(√

2 + i√

2)

and

Res(

1z4 + 1

; e3πi/4

)=

14z3

2

= −18(√

2 − i√

2)

thus ∫ ∞

−∞

dx

x4 + 1=

π√

22

.

B.4.9 Conformal Mapping

If f is holomorphic in some domain D which contains the point z0 and iff ′(z0) = 0 then f is conformal at z0. This can be interpreted as the anglebetween any two curves crossing at z0 being preserved under f . For examplef(z) = ez has a non-zero derivative at all points and is hence conformal. Underf the lines x =constant and y =constant must remain orthogonal. Under f thelines x =constant map to circles centred at the origin and y =constant maps torays away from the origin (see Figure B.11). If f is a 1–1 holomorphic function

Figure B.11

in some domain D then f−1 exists in f(D); also f and f−1 are conformal in


D and f(D). A 1–1 holomorphic mapping is called a conformal mapping. Thisidea of a holomorphic map under some complex function will be used in thenext section.

B.5 Topology and Riemann Surfaces

In the first chapter we looked at sketches of algebraic curves in R2. Howeverwe cannot sketch curves in C2 which has four real dimensions, but curves inC2 do have a topology and we can get accurate topological pictures.

B.5.1 Topology of Complex Algebraic Curves

Consider the standard form for the non-singular cubic which we investigated inB.3.2.1, namely y2 = g(x) where g(x) is a cubic polynomial. Under a suitableprojective transformation g(x) can be written g(x) = x(x − 1)(x − λ). If weadd a point at infinity then a non-singular cubic C is given by

C = y2 = x(x − 1)(x − λ) ∪ ∞where C is a subset of CP2 and λ ∈ C − 0. We want to investigate thetopology of this object so consider the mapping

π : C → CP2 defined by π[x, y, z] → [x, z] and ∞ → [1, 0].

This type of mapping is called a branched covering (i.e a double covering ofC). The points which coincide in C are called branch points. In affine coor-dinates this is defined by (x, y) → x which is a 2–1 map corresponding toy = ±√

x(x − 1)(x − λ). Under stereographic projection CP1 is homeomor-phic to the Riemann sphere. If we consider y as a function of x on CP1 thenthis has two values outside [0, 1, λ,∞]. Now we can cut CP1 along the two pathsfrom 0 to 1 and from λ to ∞. This causes the double cover of C to fall apartinto two pieces with y being single-valued on each sheet. We now open up thecuts. So a non-singular cubic in CP2 is topologically a torus. More generally itcan be shown that any non-singular algebraic curve in CP2 is homeomorphicto a sphere with g handles where g is the genus of the curve and related to thedegree d of the curve by the degree-genus formula

g =12(d − 1)(d − 2).

What is the topology of singular projective curves? We will look at what


Figure B.12

Figure B.13

happens at the end of this section. Thus a non-singular projective curve istopologically a surface in CP2. However this type of surface has more structureand we can do complex analysis on it. This type of surface is called a Riemannsurface.


Figure B.14

B.5.2 Riemann Surfaces

To define a Riemann surface we first need to show what we mean by a topologi-cal surface. A topological surface is a space S which is locally homeomorphic toC or R2. A homeomorphism φ : U → V between an open subset U of a surfaceS and an open subset of V of C is called a chart (see Figure B.15). A collection

Figure B.15

of all charts on S is called an atlas Φ where Φ = φα : Uα → Vα : α ∈ A andhence for all α ∈ A S is the union of all Uα.


If we have the intersection of two open subsets in S then

φα : Uα → Vα and φβ : Uβ → Vβ

are two charts and

φα(Uα ∩ Uβ) → Vα and φβ(Uα ∩ Uβ) → Vβ .

We can now construct the homeomorphism

φαβ = φαoφ−1β : Vβ → Vα

which maps between two open subsets of C. This is known as the transitionfunction (see Figure B.16). A surface is a Riemann surface if the transition

Figure B.16

functions are holomorphic (see section B.4.1) whenever they are defined. Theatlas is then called a holomorphic atlas.

Example 5.21

The most obvious example of a Riemann surface is the Riemann sphereCP1 = C ∪ ∞. We let U = CP1 − ∞ and V = CP1 − 0 and wedefine

φ : U → C by φ[x, y] = x/y and ψ[x, y] = y/x

(see Figure B.17). So φ and ψ form a holomorphic atlas on CP1 and hence

φoψ−1 = ψoφ−1 : C − 0 → C − 0


Figure B.17

are the transition functions defined by

z → 1z

where z ∈ C are identified with [z, 1] ∈ CP1.We can now return to algebraic curves and show that a projective curve

C − Sing(C) defined by P (a, b, c) in CP2 forms a holomorphic atlas and ishence a Riemann surface. Suppose that [a, b, c] ∈ C and also

P (a, b, c) = 0 and∂P

∂y(a, b, c) = 0.

By Euler’s relation

a∂P

∂x(a, b, c) + b

∂P

∂y(a, b, c) + c

∂P

∂z(a, b, c) = 0

which implies that a = c = 0 and hence that a = b = c = 0 which is impossibleby the definition of C − Sing(C) as [a, b, c] would be a singularity. So eithera = 0 or c = 0. Assume that c is not equal to zero, then as P is homogeneouswe can write

∂P

∂y(a/c, b/c, 1) = c−(d−1) ∂P

∂y(a, b, c) = 0

where d is the degree of the P (x, y, z). The implicit function theorem appliedto P (x, y, 1) reveals that there are open neighbourhoods V and W of a/c andb/c in C and a holomorphic function g : V → W such that if x ∈ V and y ∈ W

thenP (x, y, 1) = 0 ⇔ y = g(x).


If V and W are small then

U = [x, y, z] ∈ C : z = 0, x/z ∈ V, y/z ∈ W = [x, y, 1] ∈ Cx ∈ V, y ∈ Wis an open neighbourhood of [a, b, c] in C−Sing(C). The map φU → V definedby φ[x, y, z] = x/z has inverse w ∝ [w, g(w), 1]. Similarly if [a, b, c] ∈ C and wemake the suppositions

∂P

∂y(a, b, c) = 0 = a or

∂P

∂x(a, b, c) = 0 or

∂P

∂z(a, b, c) = 0

then by similar methods we can find a homeomorphism φU → V from an opensubset V of C where φ[x, y, z] has one of these forms

z/x, y/z, z/y, x/y, y/x.

The inverses have the following forms:

w → [1, g(w), w], [g(w), w, 1], [g(w), 1, w], [w, 1, g(w)] or [1, w, g(w)].

Now consider the homeomorphism φα[x, y, z] = x/z which has inverse [w, g(w), 1]and φβ [x, y, z] = y/x which has inverse [1, w, g(w)]. Hence,

φα : Uα → Vα and φβ : Uβ → Vβ

where Uα and Uβ are open neighbourhoods of [a, b, c] intersecting in C −Sing(C) and Vα and Vβ are open subsets of V . We can construct transitionfunctions

φαβ = φαoφ−1β : Vα → Vβ and φβα = φβoφ−1

α : Vβ → Vα.

See Figure B.18, hence φαβ = 1/g(w) and φβα = g(w)/w.If we define g : V → C to be holomorphic then the transition functions are

holomorphic. Similarly if we adopt this method for our other forms of φ thenwe get transition functions to be one of the following:

w → w, 1/w, g(w), 1/g(w), w/g(w), g(w)/w

where g is holomorphic and does not vanish where the transition function isdefined. Consequently we get a holomorphic atlas on C − Sing(C) and henceC − Sing(C) is a Riemann surface.


Figure B.18

Figure B.19

B.5.3 Degeneration of a Cubic

In this section we want to investigate what happens as the non-singular cubicdegenerates into an irreducible cubic with a node (see Figure B.19). Considera curve C described by C = y2 = x3 + x2 − c where c is real and small.As c → 0, C degenerates into a cubic with an ordinary double point. Toinvestigate what happens we again consider the mapping π : C → CP1

namely a branch covering. In affine coordinates this corresponds to the mapy = ±√

x2(x + 1) − c which has three branch points, a, b, c say, plus a pointat infinity. We know that CP1 = C ∪ ∞ so we represent the branch pointson C by Figure B.20. At c = 0 under π we get the map corresponding toy = ±√

(x + 1) which has two branch points at −1 and ∞. So as c tends to


Figure B.20

zero, four branch points tend towards two branch points, see Figure B.21. Itfollows that when c = 0 we get the Riemann sphere identified with two points.From Section B.5.1, we know that a non-singular cubic is topologically a torusand we can represent the branch points a, b, c and ∞ on the edge of a torus.Topologically a singular cubic with a node can be represented by contractingthe meridian a, b to a point. This is topologically equivalent to the Riemannsphere identified with two points, Figure B.22. As c → 0 we get a vanishingcycle for a cubic, i.e. a circle around the meridian of a torus contracting to apoint also represented in the next figure (Figure B.23) by the real cone. Thesepictures although topologically correct do not represent how the curve lies inC2. In the case of the cubic with a node in R2 the singular point is two linespassing through the origin. However the complex algebraic curve at the sin-gular point consists of two planes intersecting transversely in C2 which is notpossible to draw adequately.

We also notice that the genus of the curve changes from one to zero. Ingeneral the genus g of a curve of degree d with ordinary singular points ofmultiplicity mi is

g =12(d − 2)(d − 1) −

∑ 12mi(mi − 1).


Figure B.21

Figure B.22

B.5.4 Singularities and Riemann Surfaces

We have investigated the topology of singular curves but what happens at thesingularities? It can be shown for any singular projective curve C in CP2 wecan construct a surjective map π : R → C where R is a compact Riemannsurface. The singular points of C are the image of a finite set of points in R


Figure B.23

under π, and on the complement of this set π defines a holomorphic bijection

π : R − π−1(Sing(C)) → C − Sing(C).

We call this a resolution of singularities.

B.6 Further Topics

In this final section we briefly look at some other ideas which lead on directlyfrom what we have learnt in complex analysis. The discussion is informal andproofs are hence omitted.


B.6.1 The Weierstrass Function

We let L = nw1 + mw2 : n,m ∈ Z where L is a lattice in C which inherits anatural group structure of C, see Figure B.24. There is a meromorphic function

Figure B.24

in C defined by

℘(z) = z−2 +∑

w∈L−0[(z − w)−2 + w−2]

with derivative℘′(z) =

∑w∈L

−2(z − w)−3

where ℘(z) is called the Weierstrass function associated with the lattice L andw = nw1 + mw2, i.e. we can choose any n and m to obtain w.

It can be shown that ℘ satisfies the identity

℘′(z) = 4℘(z)2 − g2℘(z) − g3

where

g2 = g2(L) = 60∑

w∈L−0w−4, and g3(L) = 140

∑w∈L−0

w−6.

Using this identity we can define a non-singular projective curve CL by

y2z − 4x3 + g2xz2 + g3z3 = 0

with g2 and g3 as before.


We know that the lattice L has an additive group structure and is a sub-group of C so we can form a quotient group

C/L = L + α : L ∈ Z.We can investigate the topology of C/L by the map π : C → C/L. The effectof this map is to “glue” appropriate etches of the parallelogram together. HenceC/L is topologically a torus.

It can also be shown that the homeomorphism φ : C/L → CL defined by

φ(L + z) =

[℘(z), ℘′(z), 1] z /∈ L

[0, 1, 0] z ∈ L

is holomorphic. So we can associate a lattice L in C to a non-singular cubiccurve CL in CP2 and conversely it can be shown that every non-singular cubiccurve comes from some CL.

If we are given CL how do we get the lattice L? To do this we need tointegrate along a smooth path in a Riemann surface.

B.6.2 Differential Forms on a Riemann Surface

To define integration on a curve we need to show what we mean by a smoothpath. A piecewise smooth path in a Riemann surface S is a continuous map γ

from a closed interval [a, b] in R to S such that if φ : U → V is a holomorphicchart on an open subset U of S and [c, d] ⊆ γ−1(U) then φoγ : [c, d] → V isa piecewise smooth path in the open subset V of C. We have a closed path ifγ(a) = γ(b).

In order to integrate in S we need to define the concept of meromorphicand holomorphic differentials. A meromorphic function on a Riemann surfaceS is a function f : S → CP1 which is holomorphic and is not identically ∞ onany connected component of S.

Let f and g be meromorphic functions on an open subset of a Riemannsurface S. We define the symbol fdg as a meromorphic differential on S. If f

and g are meromorphic functions on S then we say that fdg = fdg if and onlyif for every holomorphic chart φ : U → V on an open subset U of S we get

(foφ−1)(goφ−1)′ = (foφ−1)(goφ−1)′.

We say that the meromorphic differential fdg has a pole at φ(p). If fdg has nopoles then it is defined as a holomorphic differential.


The integral of a holomorphic differential on S along a piecewise smoothpath γ : [a, b] → S is given by∫

γ

fdg =∫ b

a

foγ(t)(goγ)′dt

where t is a parameter independent of the choices of fdg.If the Riemann surface is C then∫

γ

fdg =∫

γ

f(z)g′(z)dz

from the normal form in complex analysis.

B.6.2.1 Abelian Integrals. An Abelian integral is an integral of the form∫γ

fdg

where f and g are rational functions on C − Sing(C) and γ is a piecewisesmooth path in C−Sing(C) not passing through the poles of the meromorphicdifferential fdg. We assume that C is not the line at infinity defined by z = 0and we take the rational function g to be [x, y, z] ∝ x/z and we hence workin affine coordinates [x, y, 1] and write dx for dg. Consequently, f becomes arational function R(x, y) and the integral is written∫

γ

fdg =∫

γ

R(x, y)dx.

A special case of an Abelian integral is given by the integral over the curve CL

as defined in Section B.6.1 where CL is associated with the lattice L.We define γ1 and γ2 to be closed smooth paths in C/L, see Figure B.25.

Any closed path on C/L can be obtained by using combinations of γ1 and γ2

and we can traverse these paths as many times as we want. It can be shownthat there is a meromorphic differential η such that

L = n

∫γ1

η + m

∫γ2

η = nw1 + mw2 where n,m ∈ Z.

In Section B.6.1 we defined the homeomorphism φ : C/L → CL and if we workin affine coordinates η is a rational function and it can therefore be shown that

L =∫

γ

y−1dx

where γ is a piecewise smooth curve in CL. Hence given any non-singularprojective cubic we get a lattice L in C. As C has a natural additive groupstructure it makes sense that the cubic has a group structure as discussed inSection B.3.3.


Figure B.25

B.6.3 Abel’s Theorem

In Section B.6.1 we saw that a complex torus C/L is biholomorphic to thecubic CL in CP2. Using this result and our investigation into differential formsleads to Abel’s theorem for cubics which gives an addition formula for ellipticintegrals.

Theorem 6.41

If t, v, w ∈ C then t + v + w ∈ L if and only if there is a line in CP2 whoseintersection with CL consists of points φ(L + t), φ(L + v) and φ(L + w) whereφ is the homeomorphism defined in Section B.6.1.

Equivalently, if p, q, r lie in CL then

L +∫ p

[0,1,0]

y−1dx +∫ q

[0,1,0]

y−1dx +∫ r

[0,1,0]

y−1dx = L + 0.

Proof [Kirwan 92, 6.23, p 155].


Bibliography

1 [Bak and Newman 82] Bak J. and Newman D.J., Complex Analysis, Un-dergraduate Texts in Mathematics Springer–Verlag, Berlin, 1982.

2 [Beardon 84] Beardon A.F., A Primer on Riemann Surfaces Lond. MathSoc Lecture Notes 78, Cambridge, 1984

3 [Brieskorn and Knorrer 86] Brieskorn E. and Knorrer H., Plane AlgebraicCurves, Birkhauser-Verlag, Basel 1986.

4 [Kirwan 92] Kirwan F., Complex Algebraic Curves, Lond. Math Soc. Stu-dent Texts 23, Cambridge, 1992.

5 [Reid 88] Reid M., Undergraduate Algebraic Geometry, Lond. Math Soc.Student Texts 12, Cambridge, 1988.

6 [Siegal 71] Siegal C.L., Topics in Complex Function Theory, Vols I and II,Wiley Classics Library, New York, 1971.

7 [Walker 50] Walker R.J., Algebraic Curves Princeton University Press,Princeton, 1950.


Epilogue

This project report reads quite well, if not as well as that in Appendix A. Thematerial is more accessible, indeed the chapters on complex variable could formpart of the second year curriculum. In this instance this particular student couldnot take any modules on complex analysis other than elementary first year ma-terial on complex numbers. Some of the later ideas involving topology, Riemannsurfaces and the Weierstrass function are quite advanced. The assessors for thisproject were considering the possibility of a good second class degree or perhapsgoing even higher. Then came the oral examination where two assessors askedthe student questions on the written project report.

This was conducted in a friendly supportive way, but when it came to sayinganything sensible, at the oral examination the student was completely tonguetied. There was no evidence that the student understood what had been written;even the more elementary questions were left unanswered. There are severalreasons why this could happen. The nasty reason is that someone else reallyhas written the project! If this can be proved, or the student confesses, a zeromark has to be awarded. For this project, this was not the case. Alternativelythe student could be very nervous and, in the stressful situation of an oral,completely dry up. The student could be swept along during the year by anover enthusiastic supervisor, so much so that even the student thinks (s)heunderstands everything until the crunch comes and (s)he finds all questionsat the oral baffling. A third reason could be that the student could be too laidback, and there is usually sufficient time between submitting a project reportand the oral to forget everything, or at least put it beyond recall amongst allthe other stuff needed for finals that are looming large. Whatever the reason,the student did him/herself no justice at all at the oral. The compromise markwas given in the mid 50s, corresponding to a lower second class degree. Theexternal examiner was probably surprised the mark was this low until he readthe reports from the assessors. It is very disappointing for both student andsupervisor (first assessor) when this happens.

CProject Example 3: Water Waves on a

Sloping Beach

Preamble

In contrast to the first two appendices, this project example is very much ap-plied mathematics. The mathematics is much more elementary, but there isthe possibility of relating it to the real world. This project thus resembles theboomerang project outlined in Section 3.5 or the hurricane project of Section3.6. Although the equations governing water waves are not in the course, theirsolution as presented here could be easily understood by second or even firstyear undergraduates. In order to get high marks in such a project therefore,it is necessary to either do some advanced mathematics, e.g. stochastic waves,non-linear waves, or to relate the simple mathematics to a real world problemand to understand the limitations of the model, how it might be extended, pos-sible numerical work, etc. It will be seen that unfortunately this project doesnot really do this.

C.1 Abstract

In this project I have studied the basic Surface Water Wave equations to obtaina linear model for the Free Surface and Velocity Potential. From then on right

241


running wave equations were used and from these the wave velocity and motionof the particle are calculated which could be used to illustrate a breaking wavein shallow water. After this wave rays were studied to be able to look at thedirection of the waves on a sloping beach and show refraction of the wave crestsas they approach the shore line.

C.2 Introduction

Waves on the ocean are generated by wind, which you can see as when there arehigh winds the sea is rough and on calm days the sea can be almost flat. Winddislodges the surface of the water and gravity endeavours to put it back toits original state, nature tries to remain in equilibrium. These waves generatedcontinue travelling on the open ocean, varying in size depending on conditions,they are dampened if opposing the direction of the wind or can grow andtravel faster when going with the wind. They will obviously eventually meetsuch obstacles as land!

Something very interesting happens as the waves approach land, i.e. a beach,they refract and the wave crests turn parallel to the shoreline. You will alsonotice that the waves break on the beach.

I have read various books on this subject and found that Ocean EngineeringWave Mechanics by M.E. McCormick and An Introduction to Water Waves byG.D. Crapper to be the most helpful in my understanding of the subject.

To begin with we will calculate a wave equation for Linear Surface Waves,the free surface is modelled by η at some position x and time t, and z = 0 isassumed to be the still water level as illustrated in Figure C.1.

C.3 Surface Waves

The Free Surface is defined by

z = η(x, t). (C.1)

The flow is assumed to be irrotational

∇× V = 0. (C.2)

Therefore the fluid potential V can be represented by a potential function φ

V = ∇φ. (C.3)

C. Project Example 3: Water Waves on a Sloping Beach 243

Figure C.1 Travelling surface wave. (Adapted from Ocean EngineeringWave Mechanics by M.E. McCormick, John Wiley and Sons Inc. Reprintedwith permission.)

Continuity states that∇.V = 0. (C.4)

From Equation (C.3), (C.4) becomes

∇2φ. (C.5)

That is Laplaces Equation.

C.3.1 The Current

Suppose we impose a uni-directional current

u = Ui. (C.6)

This current produces additional velocity potential φ of the form

φ1 = Ux + φ (C.7)

∴ ∂φ1

∂x= U. (C.8)

However here it will be assumed that there is no current.


C.3.2 The Boundary Conditions

Before we can solve Laplaces Equation there are some boundary conditions weneed to impose.

Suppose that S is a surface in a fluid, moving with the fluid, and fluidalready in S remains there.

Expressing S as the equation

S(x, z, t) = 0. (C.9)

Also particles in S remain there with varying x, z and t which gives us

DS

Dt= 0 (C.10)

where D denotes differentiation following the fluid.Now with our free surface (C.1), S can be written as

S = η(x, t) − z = 0. (C.11)

The Chain Rule states that

Ds

Dt=

∂s

∂t+

∂s

∂x

dx

dt+

∂s

∂z

dz

dt(C.12)

dx

dtand

dz

dtare the components of the velocity V (u, , v, w) (v = 0).

Therefore Equations (C.11) and (C.12) gives

Ds

Dt=

∂s

∂t+ V .∇s = 0 (C.13)

⇒ ∂η

∂t+ u

∂η

∂x− w = 0 (C.14)

withu =

∂φ

∂xw =

∂φ

∂z. (C.15)

With no current and the second set of Equations (C.15), (C.14) gives

∂η

∂t=

∂φ

∂zat z = η. (C.16)

This is called the Kinematics Surface ConditionOn the sea bed

z = −h(x). (C.17)

Therefore Equation (C.13) gives

u∂h

∂x+ w = 0 at z = −h (C.18)


and once again assuming no current, on the sea bed, (C.18) gives

∂φ

∂z= 0 at z = −h. (C.19)

This is the sea bed boundary condition.Euler’s Equation of motion for inviscid flow is

∂V

∂t+ (V .∇)V =

−1ρ

∇p. (C.20)

From continuity, (C.4), the second term on the L.H.S of (C.20) is zero. Hydro-static balance is

p = ρg(η − z) (C.21)

and recall that∇p = (

∂p

∂x,∂p

∂y,∂p

∂z). (C.22)

Therefore hydrostatic balance implies

−1ρ

∂p

∂x= −g

∂η

∂x(C.23)

and Euler’s Equation of motion becomes

∂V

∂t= −g

∂η

∂x. (C.24)

Representing V as in (C.3), (C.24) can be written as

∂

∂x(∂φ

∂t) = −g

∂η

∂x(C.25)

⇒∫

∂

∂x

∂φ

∂tdx =

∫−g

∂η

∂x(C.26)

∂φ

∂t= −gη + const. (C.27)

Taking the constant of integration to be zero, we obtain

η = −1g

∂φ

∂tat z = η. (C.28)

This is called the Dynamic Surface Condition.To avoid difficulties when solving Laplaces Equation, with the imposed

boundary conditions, at z = η, it is assumed that the mean surface level isat z = 0.


C.3.3 A Separable Solution of Laplace’s Equation

It is assumed that φ has a separable solution of the form

φ = X(x)Z(z). (C.29)

Applying Laplaces Equation to this gives us

1X

d2X

dx2=

−1Z

d2Z

dz2= −k2. (C.30)

We can now solve for Zd2Z

dz2= −k2 (C.31)

of which a general solution can be found to be

Z = C sinh(kz) + D cosh(kz). (C.32)

Using the boundary condition on the sea bed, (C.19), we have

dZ

dz= 0 at z = −h (C.33)

where C and D are constants. Using condition (C.33) we can find that

C = Dsinh(kh)cosh(kh)

. (C.34)

Therefore we now have Z(z) as

Z(z) = D cosh(kz + kh) (C.35)

Similarly for Xd2X

dx2+ k2X = 0. (C.36)

We get the general solution

X(x) = A sin(kx) + B cos(kx) (C.37)

where A and B are constants.Now using the two surface conditions (C.16) and (C.28) we can obtain

∂φ

∂z+

1g

∂2φ

∂t2= 0 at z = 0. (C.38)

From this we can find an expression for T (t)

TDk sinh(kh) +1g

d2T

dt2D cosh(kh) = 0 (C.39)


Making

ω =√

gk tanh(kh) (C.40)d2T

dt2+ ω2T = 0 (C.41)

with the general solution being

T (t) == E sin(ωt) + F cos(ωt). (C.42)

E and F are constants Letting B and F equal zero and putting together ourX, (C.35), Z, (C.31), and T , (C.42) gives us the velocity potential

φ = A sin(ωt) sin(kx) cosh(kz + kh). (C.43)

The constants A,D and E are written as the one new constant A.Now that we have a solution for φ we can use the Dynamic Surface condition,

(C.28), to obtain an equation for the free Surface, η.

η = −Aµ cosh(kh) sin(kx) cos(ωt)g

. (C.44)

Letting

a = −Aω cosh(kh)g

. (C.45)

Using this in Equation (C.44) we have

η = a sin(kx) cos(ωt). (C.46)

Other solutions of this but with different combinations of cosine and sine areobtained when we allow, say, B and E to equal zero, (or A and E, or B andF )

η = a sin(kx) sin(ωt) (C.47)

η = a cos(kx) cos(ωt) (C.48)

η = a cos(kx) sin(ωt). (C.49)

Because the solutions we have obtained are for linear waves we can use theproperty of Superposition to add together two of our µ’s to make left and rightrunning waves.

Left running wave η = a sin(kx + ωt) (C.50)

Right running wave η = a cos(kx − ωt). (C.51)

We then have corresponding Velocity Potentials

φ = A cosh(kx + kh) cos(kx + ωt) (C.52)

φ = A cosh(kz + kh) sin(kx − ωt). (C.53)

We shall only be looking at right running waves.


C.4 [No Title]

C.4.1 The Velocity of the Waves

If the coordinate system is allowed to move with the argument of the cosine inEquation (C.51), then

kx − ωt = const. (C.54)

So we can see that the differentiation of this is equal to zero which leads us tothe wave velocity, c

c =dx

dt=

ω

k(C.55)

and using our value of ω from (C.40)

c =

√g tanh(kh)

k. (C.56)

C.4.2 The Group Velocity of the Waves

This is the velocity of a group of waves with the same amplitude, a, but slightlyvarying wave number, k, and frequency. Take two waves with waves numbers k

and δk, and frequencies with ω and δω. Then their combined velocities wouldbe

cg =δω

δk(C.57)

and as δk and δω tend to zero,

cg =dω

dk. (C.58)

Then differentiating ω from (C.40) we have

cg =gkhsech2(kh)

2ω. (C.59)

C.4.3 The Motion of the Particles

Remembering that the velocity components of V are

u =∂φ

∂xv = 0 w =

∂φ

∂z. (C.60)


Now that we have our velocity potential we obtain solutions to these

u =agk cosh(kz + kh) cos(kx − ωt)

ω cosh(kh)(C.61)

w =agk sinh(kz + kh) sin(kx − ωt)

ω cosh(kh). (C.62)

The particles can take on new coordinates, (α, β), with the origin at z = 0 andsome point x = x0. So now u and w vary with time and we have

u =dα

dtw =

dβ

dt. (C.63)

We can integrate u and w with respect to t to find expressions α and β

α =agk sin(kx0 − ωt)

−ω2(C.64)

β =agk tanh(kh) cos(kx0 − ωt)

−ω2. (C.65)

Equations (C.64) and (C.65) can be rearranged, squared and added togetherto give

α2(kag

ω2

)2 +β2(

kag tanh(kh)ω2

)2 = 1. (C.66)

Thus from this we can see that the motion of the particles is an Ellipse whichis affected in shape by the depth of the water, h In very deep water we have alarge value for h, i.e. as h ⇒ ∞ tanh(kh) ⇒ 1 So we can see from Figure C.2that the particles travel in a circle in deep water However in shallow water withdecreasing h, i.e. as h ⇒ 0 tanh(kh) ⇒ 0 The particles travel in an Ellipsewhich will become flatter, as shown in figure C.2 until they are eventuallytravelling in a flat line, as you see with waves on a beach.

C.4.4 Breaking Waves in Shallow Water

The velocity of a wave in shallow water can be approximated, for small h

tanh(kh) ≈ kh. (C.67)

Thenω = k

√gh. (C.68)

Thus the velocity of the wave c, is approximated using equations (C.56) and(C.67) to

c =√

gh. (C.69)


Figure C.2 Motion of the particles

Now the horizontal velocity component of V, u in shallow water is approximatedalso, once again for small h

cosh kh ≈ 1 (C.70)

and so using (C.68), (C.70) our u from (C.59) becomes

u =ag cos(kx − ωt)√

gh. (C.71)

We can see that the velocity of the waves, in (C.69) decreases as the depthdecreases but the velocity of the particle in (C.71) increases with decreasingdepth. Therefore there is a point at which the velocity of the particles willbegin to be faster than that of the wave.

When the velocities are equal (C.69 = C.71)

a cos(kx − ωt) = h. (C.72)

If this occurs at the top of a wave, the crest, i.e. cos(kx − ωt) = 1. Then thewave will break at, from (C.72), h = a As u becomes greater then c the wavewill spill over. The linear solution of η that we have obtained gives the crest


Figure C.3 The motion of a particle in shallow water. (Adapted fromOcean Engineering Wave Mechanics by M.E. McCormick, John Wiley andSons Inc. Reprinted with permission.)

a sinusoidal shape, however as we can see from real waves on the beach it isevident that the crest takes on a slightly more pointed tip, as shown in FigureC.4 A non-linear solution for η would model this much better.

Figure C.4 The crest of a wave. (Adapted from Ocean Engineering WaveMechanics by M.E. McCormick, John Wiley and Sons Inc. Reprinted withpermission.)


C.5 [No Title]

C.5.1 Plane Waves

A plane wave is a 2 dimensional wave not now propagating along the x axis butin any direction, as shown in Figure C.5 We now have the coordinate system

Figure C.5

(x1, x2, z) and the wave number k is represented as a vector k, where

k = (k cos θ, k sin θ) = (k1, k2). (C.73)

From this we need to find a new expression for kx in our Free Surface.From Figure C.5 we can see that

x1 = x cos θ (C.74)

x2 = x sin θ (C.75)

which gives us x in terms of x1 and x2

x = x1 cos θ + x2 sin θ. (C.76)

Therefore we havekx = x1k1 + x2k2 = k.x. (C.77)

The Free Surface of a plane wave is now expressed as

η = a cos(k.x − ωt) (C.78)

and Velocity Potential

φ = A cosh(kz + kh) sin(k.x − ωt) (C.79)


with corresponding vector wave velocity

c =ω

k2k. (C.80)

C.5.2 Wave Rays

The equations obtained so far have been for an infinite train of uniform waves,but in reality this is not always the case. We are now going to allow small andslow changes in, wave amplitude, a, frequency,

ω

2π, wave number, k, and depth,

h. The mean surface level previously assumed to be zero is also now a variable,say, b. So our depth is now

d = b + h. (C.81)

So our Free Surface changes again to become

η = b + cos Φ (C.82)

where the phase functionΦ = k.x − ωt (C.83)

and

φ = A cosh(kz + kd) sin(k.x − ωt) (C.84)

ω =√

gk tanh(kd). (C.85)

We can find expressions for k1, k2 and ω from the phase function (C.83)

k1 =∂Φ

∂x1k2 =

∂Φ

∂x2ω =

∂Φ

∂t. (C.86)

We can show that Φ holds for the slowly varying changes in the wave traintalked about on the previous page. Using the Taylor Expansion near a fixedpoint (x10, x20, t0)

Φ(x1, x2, t) = Φ(x10, x20, t0)+(x1−x10)∂Φ

∂x1

∣∣∣∣0

+(x2−x20)∂Φ

∂x2

∣∣∣∣0

+(t− t0)∂Φ

∂t

∣∣∣∣0

(C.87)(neglecting higher derivative terms). Using Equations (C.84) and (C.85) it canbe shown that

Φ = k10x1 + k20x2 − ω0t. (C.88)

Thus Φ hold locally.


It is assumed that the second derivatives of Φ are continuous, so Equations(C.84) give us

∂k1

∂x2=

∂k2

∂x1(C.89)

∂k1

∂t+

∂ω

∂x1= 0 (C.90)

∂k2

∂t+

∂ω

∂x2= 0. (C.91)

The depth is now a slowly varying function of x1, x2 and t. So ω can be repre-sented as a function of (k1, k2, x1, x2, t)

ω(k, d) = Ω(k1, k2, x1, x2, t). (C.92)

From the group velocity cg =dω

dkthe vector group velocity could be assumed

to be

cg =(

∂Ω

∂k1,∂Ω

∂k2

)= (cg1, cg2). (C.93)

Using Equations (C.90) and (C.92)

∂ω

∂x1=

∂Ω

∂x1+

∂Ω

∂k1

∂k1

∂x1+

∂Ω

∂k2

∂k2

∂x1= −∂k1

∂t. (C.94)

Substituting Equation (C.89) and the vector group velocity (C.93), (C.94) canbe written as

− ∂Ω

∂x1=

∂k1

∂t+ cg1

∂k1

∂x1+ cg2

∂k1

∂x2. (C.95)

Now the total derivative

Dk1

Dt=

∂k1

∂t+ (V.∇)k1 (C.96)

Dk1

Dt=

∂k1

∂t+ u1

∂k1

∂x1+ u2

∂k1

∂x2. (C.97)

You can see from comparing (C.97) and (C.95) that the L.H.S. of (C.95) is alsoa total derivative but moving with the vector group velocity, cg and not theparticle velocity. We have

dx1

dt=

∂Ω

∂k1(C.98)

dx2

dt=

∂Ω

∂k2(C.99)


These define lines everywhere that are parallel to the vector group velocity andare called Rays

And on these rays from Equation (C.95)

dk1

dt=

∂k1

∂t+

dx1

dt

∂k1

∂x1+

dx2

dt

∂k1

∂x2= − ∂Ω

∂x1. (C.100)

And from deducing∂ω

∂x2and

∂ω

∂tusing the same method as in (C.94) we have

on the rays that

dk1

dt= − ∂Ω

∂x1(C.101)

dk2

dt= − ∂Ω

∂x2(C.102)

dω

dt=

∂Ω

∂t. (C.103)

From (C.92) the rays can also be written as

dx1

dt=

∂ω

∂k

k1

k(C.104)

dx2

dt=

∂ω

∂k

k2

k(C.105)

which is the form we will use in the next section.These rays represent the direction of travel of the waves and the wave crests

are perpendicular to this.

C.5.3 The Waves Approaching a Beach

Using the Ray equations established previously we can predict what will happento the waves as the depth of water decreases, i.e. at a sloping beach. The figurebelow shows the sea in the x1 less than zero region and the shore is the x2

axis. To begin with we will look at the waves in the deep water, at x1 = −∞,before they are affected by the sloping beach. We shall also assume again thatthe variation in the mean surface level, b, is negligible and is zero. We shallassume there is no variation on depth, so Ω (C.92) is independent of x1, x2,and t which from (C.102) gives ω as a constant on the rays and so a constanteverywhere. And in deep water using (C.65) we have

ω2 = gk. (C.106)

The previous assumption also tells us that from (C.100) and (C.101), k1 and k2

are constant on the rays in deep water. Dividing the ray equations in (C.103)


Figure C.6 Wave rays in deep water

we havedx2

dx1=

k2

k1. (C.107)

And as k1 and k2 are constants we can integrate this∫dx2 =

∫k2

k1dx1 (C.108)

x2 =k2

k1x1 + x3 (C.109)

where x3 is the constant of integration. Equation (C.107) shows us that thewaves in deep water travel in straight lines The wave crests are perpendicularto the rays so for the crests (C.105) can be written as

dx2

dx1= −k1

k2(C.110)

which is integrated to give

x2 = x4 − k1

k2x1 (C.111)

where x4 is the constant of integration.These equations (C.107) and (C.109) tells us that in deep water the waves

are not parallel to the shore (as the wave number is never zero), as shown inFigure C.7. But when standing on a beach the waves more often than not runup parallel to the beach, as in Figure C.9, so now we look at the wave rayswhen in shallow water.


Figure C.7 Direction of waves in deep water

C.5.4 Wave rays in shallow water

We have established that ω is constant on these rays, so

ω2 = gk tanh(kd) = const. (C.112)

Thus as we approach the beach and the depth decreases, k must increase tokeep ω2 a constant, so if k increases the wave velocity

c =ω

kwill decrease, i.e. the waves will slow down

and the wave length

λ =2π

kwill also decrease, shortening the waves.

We can see from this that as the wave nears the shore it will slow down andturn parallel to the shore, refraction of the waves.

We can write the depth of the water when it is shallow as a fraction of x1

where the slope has a gradient of γ

d = −γx1. (C.113)

From the shore, i.e. along the x2 axis, the waves look the same, thus

∂

∂x2= 0. (C.114)


So from (C.101)

dk2

dt= 0 (C.115)

⇒ k2 = const = m. (C.116)

Now thatk2 = m ⇒ k1 =

√k2 − m2 (C.117)

and Equation (C.107) is now

dx2

dx1=

m√k2 − m2

(C.118)

For small d

tanh(kd) ∼ kd. (C.119)

Using (C.119) and (C.113), (C.112) can be written as

ω2 = gk2(−γx1). (C.120)

So k can be written as a function of x1, and letting x1 = −|x1|

k2 =ω2

gγ|x1| . (C.121)

Thus the differential equation (C.118) is

dx2

dx1=

m√ω

gγ|x1| − m2. (C.122)

And if |x1| is small enough

− dx2

d|x1| = m

√gγ|x1|

ω2. (C.123)

We can integrate (C.123) to give

x2 = x5 −2m

√gγ|x1| 323ω

(C.124)

(where x5 is a constant of integration)Once again we have for the crests

dx2

d|x1| =1m

√ω2

gγ|x1| . (C.125)

Also integrated gives

x2 =2ω|x1| 12m√

gγ− x6 (C.126)


Figure C.8

(where x6 is the constant of integration)We now have from (C.124) and (C.126) the direction of travel of the waves

and shape of the crests as ω, m, γ, g, x5 and x6 are all constants. So we can drawthe basic shape of these curves. Figure C.8 shows the basic form the wave raysand crests take approximated by (C.124) and (C.126) with varying constants ofintegration. This figure would obviously vary depending on the other constants.Comparing Figure C.8 to Figure C.9, real waves on a beach you can see thatthe solutions obtained are quite a good approximation to the actual waves.

C.6 [No Title]

C.6.1 Conclusion and Discussion

The solutions obtained, although only fairly simple approximations, I feel aregood approximations to the direction of travel of waves on a beach, as can beseen by comparing Figures C.8 and C.9, even with the assumptions made.

I feel that the next thing to go on to here would be to look at a non-linear


Figure C.9 Wave refraction at West Hampton Beach, Long Island. (Re-produced from Shore Protection Manual, Volume 1, Third edition, 1977, USGovernment Printing Office.)

Wave Theory, with which we could model the breaking waves on the beach toshow the peaks the crests form. Stokes Theory for non-linear waves suggestsletting the wave equations be represented as a series of small perturbation, thehigher the order the more accurate an approximation is obtained.

There are many aspects of water waves that can be studied further and inmore detail and I have only really touched the surface! Such things as the Vshape pattern of waves generated by a moving object on the surface, whetherthe object is a streamlined speed boat or a lump of wood, or Kelvin’s Theoryof the wave pattern created by a moving ship. Another interesting wave is thesolitary wave, a single ‘bump’ that travels uniformly.


References

1. Crapper, G.D. Introduction to Water Waves Ellis Horwood, Chichester,1984.

2. McCormick, M.E. Ocean Engineering Wave Mechanics John Wiley andSons Inc., New York, 1973.

3. LeBlond, P.H. and Mysak, L.A. Waves in the Ocean Elsevier, Amsterdam,1978.

4. James, P.W. MATH305 Course Notes

5. Stoker, J.J.Water Waves Interscience, New York, 1957.


Epilogue

This student was one of the many that came to a lecturer and wanted to do“something about sailing”! The fundamental problem with this project is itslack of scope. Initially, the student and lecturer decided on trying to model themovement of floating bodies, hoping to treat the effects of wind on a sail ina rudimentary way. However, it soon became clear that this was not going tohappen, and pragmatism restricted the project student to the study of waterwaves. Even given this, the project still failed to come up to expectations. Onlyelementary linear water wave theory was covered; there were no generalisationseven though there was some relevant course material on hand. Something suchas a description of the Gerstner wave, perturbation theory as used by Civil En-gineers, even perhaps Cnoidal waves would have at least raised the level a littleabove the mundane, but no. This project ended up both limited in scope andshort on relating water wave theory to lecture course material on inviscid flowor waves.

The style of the project itself was also rather immature. There were mistakesin the English (most of which remain), slips in the equations (most of whichremain) and errors in the numbering of equations (all of which have been elim-inated through the use of LATEX). The layout of the project here is better thanthe original, again due to the use of LATEX.

At the oral examination, things did not improve. The student was very quietand gave a lack-luster performance. This seemed not to be due to nerves, moreto a lack of preparation. It was almost as if the student realised that the projectwas irretrievably poor and had written it off. The most elementary questionswere left unanswered.

There was enough here for a pass. The mathematics though elementarywas (mostly) correct. The content though meagre was just acceptable. This stu-dent was unanimously but with some regret awarded a third class mark forthis project. For the record, the student went on to gain an upper second onthe strength of examination performance. Perhaps there really was a deliberatechannelling of effort towards exams in this case.

Index

acceleration, Coriolis, 108advice, 36aerodynamic, 46aerofoil, 47, 51Airy’s differential equation, 130algebraic curves, 195algorithm, 82, 88angular momentum, 47appeals, 19applied mathematics, 2, 45, 62argument, 73assessment, 2, 11, 106astronomy, 94axioms, 37

background, non-historical, 60Bessel’s equation, 63biography, 22, 31, 37bipartite graph, 88Black–Scholes equation, 96Boolean algebra, 101boomerang, 44, 45boomerangs, 46, 51branch points, 64

car density, 126car velocity, 125case studies, 10, 11, 24, 105case study, 128Cauchy integral formula, 133characteristic curve, 127characteristics, 123characteristics, method of, 122cheating, 1circulation, 51

co-operation, 88communication, 73complex potential, 46complex variable theory, 46, 128computer algebra packages, 33computer failure, 91computer programs, 28computing, 28, 65, 88concepts, 129confluent hypergeometric equation, 63conics, 94constructive feedback, 32content, 66content marks, 18continuity equation, 124continuous assessment, 1contour integral solutions, 128convergence, 143convolution, 58copying, 22, 35, 40Coriolis acceleration, 48, 56, 106Coriolis parameter, 56, 109coursework, 5, 105credits, 9, 21, 75cross-ratio, 40

Davidon–Fletcher–Powell, 136, 139debate, 73Desargues’ Theorem, 38diffusion, 78, 80diffusion equation, 79Dirac-δ function, 90discussion, 73dividing into groups, 76Duffing’s equation, 116

263


dynamic meteorology, 53dynamical systems, 116

earth, 107eddy viscosity, 112, 113Ekman equations, 113, 114, 147Ekman flow, 114Ekman layer, 55, 57elliptic integral, 63Euclid, 37Euler, 141, 142Euler constant, 145Euler’s equations, 48, 49Eulerian view, 110examinations, 2Excel, 136explanation, 84explanation, not enough, 58extension, 132extensions, 115external examiner, 23

feedback, 81Fields Medal, viifinal report, 21, 81, 94final year tutor, 23financial mathematics, 96fluid dynamics, 52fluid mechanics, 46, 110formal lectures, 10formative, 12FORTRAN, 28, 99Fourier series, 89, 116Fourier transform, 89, 90Fourier transforms, 89friction, 56, 111frisbee, 45frisbees, 51Frobenius method, 59, 62

Galois Theory, 151general manager, 90geometric thinking, 43geometry, 35, 38geophysical fluid dynamics, 52geostrophic balance, 55, 113geostrophic flow, 57geostrophic wind, 57gradient, 138grammar checkers, 33graphs, 82group project, 7, 11, 24, 74, 89, 95group project presentation, 18group velocity, 248

group work, 74groups, 78guidelines, 21

handout, 7, 74, 78, 89, 92, 93handshaking lemma, 83Hessian matrix, 138Hill’s equation, 120historical research, 78history of mathematics, 102, 141homework, 11homogeneous coordinates, 38Hungarian algorithm, 84hurricane, 52–54, 56hypergeometric equation, 59hypergeometric function, 59, 60, 62

independent assessor, 18individual, 88individual project, 3, 11, 32industrial problem, 10infinite series, 143instant insanity, 84integral operations, 129interests, 31interim report, 12, 13, 31, 91inviscid flow, 50iterative scheme, 138

job interviews, 35Joukowski transform, 46, 50

Lagrangian mechanics, 92Lagrangian view, 110Laguerre’s equation, 64Laplace transform, 58Laplace transforms, 89, 128Laplace’s linear equation, 128Latex, 33layout glitches, 95Legendre’s equation, 132, 133levels of difficulty, 22lift, 51line at infinity, 38linear waves, 120

management, 10manual, 88MAPLE, 33, 99, 134, 136mathematical ability, 9, 75mathematical accuracy, 22mathematical biography, 22mathematical content, 16mathematical modelling, 65

Index 265

mathematics education, 29Mathieu’s equation, 120, 148mechanics, 44, 92Microsoft Word, 33mini-projects, 5, 82model, 54moderated, 24moderating, 23module leader, 9, 75, 80, 82, 91modules, 105moment of inertia, 47, 49

Navier Stokes’ equation, 111networks, 83Newton–Raphson method, 135, 136, 139Newtonian viscous fluid, 112Nobel Prize, viinon-linear first order differential

equations, 116non-linear optimisation, 141non-linear oscillations, 116normal modes, 97numerical methods, 79numerical solutions, 99numerical techniques, 94

objectives, 36ocean surface dynamics, 106open book examination, 24operational research, 135opinion, 73optimisation, 134, 137oral examination, 40oral presentation, 23orbits, 93overlap, 85

Pappus’ theorem, 41, 42parametric excitation, 120partial differential equations, 55, 77, 134partition, 77peer assessment, 74, 81, 91performance, 92perturbation method, 118perturbation solutions, 118precession, 50preference, 77preliminary work, 5presentation, 66, 80, 84, 88, 105presentation marks, 18presentation skills, 16presentations, 91presented material, 146presenting, 74

pressure gradients, 112Principle of Duality, 43problem classes, 105production problems, 95Project Example, 151, 195, 241project examples, 27, 66project failure, 28project list, 70project report, 21project supervisor, 18, 32project team, 79project titles, 35projectiles, 45projective geometry, 44projectivity, 42projects, 34proof, 40–42, 142proofs, 35pure mathematics, 34, 36, 67

quality assessment, 76quality assurance, 80quality audit, 76

rate of change following the fluid, 111real problem, 58references, 33, 36report writing, 31research paper, 10Reynolds stress, 111Riemann P function, 59rigid body, 46rotating fluids, 55rotating frame, 109

sandwich year, 3Schlafli integrals, 133, 134scholarly, 65scope, 35, 45, 53, 60, 90second assessor, 91second marker, 23, 76second order differential equation, 59selecting a project, 29shock, 127shock wave, 122skills, 65special functions, 59spell checkers, 33stability analysis, 119staff interests, 30steepest descent, 138strategies, 89stress in a fluid, 111student assignment, 139


student handbook, 31subharmonic oscillations, 118submission deadline, 32supervisor, 12

technical level, 16thermodynamics, 52, 55, 57third assessor, 19, 23tops and gyroscopes, 100torque, 48traditional lecture course, 105traffic flow, 120traffic jam, 124traffic wave, 125turbulence, 111

validation, 69velocity potential, 247verbal presentation, 13, 18, 94vocational subjects, 3vortex shedding, 51

water waves, 98, 249, 259wave rays, 256Whittaker’s confluent hypergeometric

function, 60word processing, 33writing skills, 28writing style, 84writing timetable, 32

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