DIHEDRAL QUINTIC FIELDS WITH A POWER BASISbyMelisa Jean LavalleeB.Sc., The University of British Columbia, 2006A THESIS SUBMITTED IN PARTIAL FULFILLMENT OFTHE REQUIREMENTS FOR THE DEGREE OFMASTER OF SCIENCEinThe College of Graduate Studies(Interdisciplinary Studies)THE UNIVERSITY OF BRITISH COLUMBIA(Okanagan)August 2008c Melisa Jean Lavallee, 2008ABSTRACTCryptography is de?ned to be the practice and studying of hiding informationand is used in applications present today; examples include the security of ATMcards and computer passwords ([34]). In order to transform information to make itunreadable, one needs a series of algorithms. Many of these algorithms are based onelliptic curves because they require fewer bits. To use such algorithms, one must ?ndthe rational points on an elliptic curve. The study of Algebraic Number Theory, andin particular, rare objects known as power bases, help determine what these rationalpoints are. With such broad applications, studying power bases is an interestingtopic with many research opportunities, one of which is given below.There are many similarities between Cyclic and Dihedral ?elds of prime degree;more speci?cally, the structure of their ?eld discriminants is comparable. Since theexistence of power bases (i.e. monogenicity) is based upon ?nding solutions to theindex form equation - an equation dependant on ?eld discriminants - does this implymonogenic properties of such ?elds are also analogous?For instance, in [14], Marie-Nicole Gras has shown there is only one monogeniccyclic ?eld of degree 5. Is there a similar result for dihedral ?elds of degree 5? Thepurpose of this thesis is to show that there exist in?nitely many monogenic dihedralquintic ?elds and hence, not just one or ?nitely many. We do so by using a well-known family of quintic polynomials with Galois group D5. Thus, the main theoremgiven in this thesis will con?rm that monogenic properties between cyclic and dihedralquintic ?elds are not always correlative.iiContentsABSTRACT................................................................... iiContents....................................................................... iiiList of Tables .................................................................. viList of Figures ................................................................. viiList of Symbols ................................................................ viiiAcknowledgements ............................................................. xDedication ..................................................................... xiCoauthorship Statement ....................................................... xiiIntroduction ................................................................... 1Chapter 1. Algebraic Preliminaries............................................ 31.1. Classic Theorems ...................................................... 31.2. Algebraic Numbers..................................................... 41.2.1. Minimal Polynomial............................................... 41.2.2. Conjugates of an Algebraic Number ............................... 81.3. Algebraic Number Fields............................................... 91.3.1. Representation of Algebraic Number Fields........................ 10Chapter 2. Introduction to Galois Theory..................................... 112.1. Galois Groups of Polynomials - Overview .............................. 122.2. Irreducible Polynomials ................................................ 152.3. Quadratic Polynomials................................................. 182.4. Cubic Polynomials ..................................................... 20iii2.4.1. Irreducible Cubic Polynomials..................................... 202.4.2. Reducible Cubic Polynomials...................................... 222.5. Quartic Polynomials ................................................... 232.5.1. Irreducible Quartic Polynomials ................................... 232.5.2. Reducible Quartic Polynomials .................................... 282.6. Quintic Polynomials ................................................... 302.6.1. Irreducible Quintic Polynomials ................................... 312.6.2. Reducible Quintic Polynomials .................................... 402.7. Summary .............................................................. 462.8. Finding the Galois Group of an Algebraic Number Field .............. 47Chapter 3. Integral and Power Bases.......................................... 503.1. Integral Bases.......................................................... 503.1.1. Minimal Integers .................................................. 533.1.2. Examples of Finding Integral Bases ............................... 563.2. Field Discriminant ..................................................... 613.3. Power Bases ........................................................... 653.3.1. Index Form ....................................................... 67Chapter 4. Cyclic and Dihedral Fields of Prime Degree ....................... 694.1. Field Discriminants .................................................... 694.2. Abelian and Cyclotomic Fields......................................... 74Chapter 5. Cyclic and Dihedral Quintic Fields with a Power Basis ............ 765.1. Some Important Lemmas .............................................. 765.2. Proof of Main Result................................................... 805.3. Further Results ........................................................ 82Chapter 6. Number Fields without a Power Basis ............................. 916.1. Ideals.................................................................. 916.2. Common Index Divisors................................................ 93Chapter 7. Conclusion and Future Work ...................................... 967.1. Future Research ....................................................... 96iv7.2. Conclusion............................................................. 98References ..................................................................... 99vList of Tables2.1 Group Table for the Cyclic Group of Order 2 ............................... 202.2 Group Table for the Cyclic Group of Order 3 ............................... 212.3 Group Table for the Symmetric Group of Order 6........................... 442.4 Summary of the Possible Galois Groups for Irreducible Polynomials......... 47viList of Figures2.1 Subgroups of S3 ............................................................ 45viiList of SymbolsZ Set of integers, 3Q Set of rationals, 3D(f) Discriminant of a polynomial f(x), 8N( ) Norm of , 8Tr( ) Trace of , 8[K : Q] Degree of K over Q, 10Aut(K) Collection of automorphisms of K, 11AutQ(K) Collection of automorphisms of K which ?x Q, 11Gal(K=Q) Galois group of K over Q, 11C Set of complex numbers, 12Gal(f(x)) Galois group of the splitting ?eld of f(x) 2 F[x] over F, 12Sn Symmetric group of order n, 13E L Sub?elds, 14jG : Hj Index of H in G, 45Q( = 1; 2;:::; n) Normal closure of K = Q( ), 47OK Ring of integers in the number ?eld K, 50d(K) Field discriminant of the ?eld K, 61ind( ) Index of , 63ind( ) = 1 Index form equation, 67viiii(K) Field index of K, 67m(K) Minimal Index of K, 67&n Primitive nth root of unity, 74Q(&n) Cyclotomic Field, 75Ra;b(x) Kondo family of quintics, 76ixAcknowledgementsGod has blessed me with an amazing network of family, friends and professors whohave been consistently supportive and encouraging throughout my post-secondaryexperience; my thanks go to these wonderful people.Thanks to my committee of brilliant professors: Dr. Blair Spearman, Dr. QiduanYang, Dr. Shawn Wang, Dr. Heinz Bauschke and Dr. Yong Gao. Your guidance andconstructive criticism proved invaluable and contributed greatly to the completion ofthis thesis.Furthermore, I wish to acknowledge and express my appreciation to Dr. BlairSpearman. Your ability to see the potential in others and the con?dence you instillin them has contributed immensely to the furtherance of my education and as a resultthis thesis.The research for this paper was supported by a scholarship from the NaturalSciences and Engineering Research Council of Canada.xDedicationI would like to dedicate this paper to my mother. Your strength and perseverancenever cease to amaze me.xiCoauthorship StatementThe main result of this thesis is from the paper [18]. All authors of [18] con-tributed equally to this paper. Using MAPLETM, I was responsible for ?ndingsolutions to the index form equation of our problem. I also contributed in editingthis paper.xiiIntroductionAlthough there are in?nitely many integral bases of the ring of integers of analgebraic number ?eld, seldom are these power bases. When a power basis is found, itscorresponding algebraic number ?eld is said to be monogenic. Even now, identifyingmonogenic ?elds is still an active area of research.Firstly, we know that all quadratic ?elds are monogenic. Also, much research hasbeen conducted on cubic and quartic ?elds. Dedekind gave an example of a non-monogenic cubic ?eld (see [7]). However, there are in?nitely many cyclic cubic ?eldsas well as non-cyclic cubic ?elds that are monogenic (see [10] and [28], respectively).In regards to quartic ?elds, we now know that there are in?nitely many pure quartic?elds ([12]), bicyclic quartic ?elds ([22]) and dihedral quartic ?elds ([15]) that aremonogenic. However, the cyclic quartic case is still open.On the other hand, higher degrees can make these problems quite challenging.Numerical methods are limited to lower degrees, usually at most 6. Thanks toMarie-Nicole Gras ([14]), we do know that there is only one monogenic cyclic quintic?eld. The essence of this thesis is to show by way of comparison that there are anin?nite number of monogenic dihedral quintic ?elds. However, a large portion isdedicated to establishing the foundations of the theory as well as important resultsneeded to prove this focal point.The contents of this thesis are divided into seven chapters. Chapter 1 serves as anintroduction to basic properties of Algebraic Number Theory. Chapter 2 gives somebasic theory of Galois Theory. It also gives methods on ?nding the Galois group ofa polynomial up to and including degree ?ve. Chapter 3 is devoted to the study ofintegral bases and in particular, power bases; it also discusses the Index Form andIndex Form Equation and related properties. Chapter 4 compares ?eld discriminantsof cyclic and dihedral ?elds of prime degree. In Chapter 5, the tools developed inearlier chapters are used to prove our main result (that there are an in?nite amount1of monogenic dihedral quintic ?elds). It also discusses related results to this proof.Chapter 6 is concerned with ideals and common index divisors. Speci?cally, itstudies properties of ?elds that do not contain a power basis. Lastly, Chapter 7 givesnumerous examples of future research in this particular area of study.2Chapter 1Algebraic Preliminaries1.1. Classic TheoremsWe begin this chapter with two classic theorems that will be quoted throughoutthis thesis.We begin with what is known as the Rational Root Theorem, Rational ZeroTheorem or Rational Root Test (pg. 214-215, [20]).Theorem 1.1. Consider the polynomial equationanxn + an 1xn 1 + ::: + a1x + a0 = 0with integer coe? cients. Let a0 and an be nonzero. Then each rational solutionx of this equation can be written in the form x = pq where p and q satisfy the twoproperties: p is an integer factor of the constant term a0, and q is an integer factor of the leading coe? cient an:We will be using this theorem to determine if a given polynomial has any rationalroots (and hence any linear factors).Now we state Gauss? Lemma (Proposition 5, pg. 303, [9]).Theorem 1.2. If a polynomial with integer coe? cients is irreducible over Z (theset of integers) then it is also irreducible over Q (the set of rationals).We will be looking at problems where we need to determine if a given polynomialis irreducible over Q. We can then use Gauss? Lemma to show a polynomial isirreducible over Q by showing it is irreducible over Z.31.2. Algebraic NumbersDe?nition 1.1. is an algebraic number if it is a root of a non-constant monicpolynomial f(x) with rational coe? cients (i.e. f(x) 2 Q[x]).De?nition 1.2. If the coe? cients in the above polynomial are integers (i.e. f(x) 2Z[x]), then is called an algebraic integer.The term "monic" means the coe? cient of the highest power of x is equal to 1.Example 1.1. Let f(x) = x2 12. It has roots x = 1p2. Since f is monic andhas coe? cients in Q, 1p2 are both algebraic numbers.Example 1.2. Let f(x) = x2 2. It has roots x = p2. Since f is monic andhas coe? cients in Z, p2 are both algebraic integers.1.2.1. Minimal PolynomialDe?nition 1.3. The minimal polynomial of an algebraic number (or integer) is the monic polynomial A(x) 2 Q[x] (resp. Z[x]) of least degree which has as aroot.We now introduce a simple property of the minimal polynomial.Lemma 1.1. If is an algebraic number then the minimal polynomial of dividesevery polynomial of which is a root.Proof. Let A(x) be the minimal polynomial of . Suppose p(x) 2 Q[x] such thatp( ) = 0. By the division algorithm, there exists polynomials q(x) and r(x) in Q[x]such thatp(x) = A(x) q(x) + r(x)where deg(r(x)) < deg(A(x)). Then0 = p( ) = A( ) q( ) + r( ) = 0 + r( ) )r( ) = 0:If r(x) 6= 0 then we can ?nd a suitable constant multiplier such that r(x) is monic.Then deg(r(x)) < deg(A(x)) would contradict the minimality of A(x) (as r( ) = 0).Thus, r(x) = 0 and so p(x) = A(x) q(x) )A(x) j p(x). 4We now give two examples on ?nding the minimal polynomial.Example 1.3. Let = p2 + p3. We want to ?nd the minimal polynomial of .Squaring we obtain 2 = 5 + 2p6so that 2 5 = 2p6:Squaring 2 5 yields 4 10 2 + 25 = 24:Thus, is a root of the monic quartic polynomialf(x) = x4 10x2 + 1 2 Z[x]:This shows that is an algebraic integer.We will show that this polynomial f(x) is A(x) - the minimal polynomial of .First we show that this polynomial is irreducible over Z.Assume by way of contradiction that f(x) = x4 10x2 + 1 is reducible over Z. Firstof all, if f(x) were to have a linear factor, then there is a rational root, say x1 = pqwhere p;q 2 Z, such that f(x1) = 0. Then by the Rational Root Theorem, p j 1 andq j 1. So our choices for x1 are 1. Since f( 1) = 8 6= 0, f(x) has no linearfactors in Q[x] and hence in Z[x]. Thus, f(x) factors into two irreducible quadraticfactors. So f(x) = x4 10x2 + 1 = (x2 + ax + b) (x2 + cx + d) where a;b;c;d 2 Z.Equating coe? cients yields the following equations:a + c = 0d + b + ac = 10ad + bc = 0bd = 1.From the ?rst equation we have c = a and so the second equation becomes d+b a2 = 10.From the last equation we have(1) b = 1 and d = 1(2) b = 1 and d = 15So, b+d = 2. Thus, from the revised second equation, we have a2 = 12 or 8, whichis a contradiction. Thus, f(x) is irreducible over Z and by Gauss?Lemma, f(x) isirreducible over Q.By Lemma 1.1, the minimal polynomial of must divide f(x) and hence must be afactor of f(x) over Z. This implies that f(x) must be the minimal polynomial of = p2 + p3 since f(x) does not factor over Z. Thus,A(x) = x4 10x2 + 1.Example 1.4. Let =p2 + p2. We want to ?nd the minimal polynomial of .Squaring we obtain 2 = 2 + p2so that 2 2 = p2:Squaring 2 2 yields 4 4 2 + 4 = 2:Thus, is a root of the monic quartic polynomialf(x) = x4 4x2 + 2 2 Z[x]:This shows that is an algebraic integer.By Gauss?Lemma, we know that if f(x) is irreducible over Z then it is irreducibleover Q. Thus, we will now show that f(x) is irreducible over Z. So we assume,by way of contradiction, that f(x) factors over Z. First of all, if f(x) were to havea linear factor, then there is a rational root, say x1 = pq where p;q 2 Z, such thatf(x1) = 0. Then by the Rational Root Theorem, p j 2 and q j 1. So our choices forx1 are 2. Since f( 2) = 2 6= 0, f(x) has no linear factors in Q[x] and hence inZ[x]. Since we assumed f(x) factors over Z, f(x) must factor as a product of twoquadratic polynomials in Z[x]:x4 4x2 + 2 = (x2 + ax + b) (x2 + cx + d);6where a;b;c;d 2 Z. Equating coe? cients, we geta + c = 0d + b + ac = 4ad + bc = 0bd = 2.From the ?rst equation we have c = a and so the second equation becomes d+b a2 = 4. From the last equation we have(1) b = 2 and d = 1(2) b = 2 and d = 1(3) b = 1 and d = 2(4) b = 1 and d = 2.So, b + d = 3. Thus, from the revised second equation, we have a2 = 7 or 1. Buta2 = 7 is impossible. If a2 = 1 then0 = ad + bc= ad ba= a(d b)= d b,which implies d = b, also impossible (see above possible values for b and d). Thus,f(x) is irreducible over Z and hence over Q (later on we shall see Eisenstein? Crite-rion which can also be used to determine irreducibility).By the Lemma, the minimal polynomial of must divide f(x) and hence must be afactor of f(x) over Z. This implies that f(x) must be the minimal polynomial of =p2 + p2 since f(x) does not factor over Z.Note that in Example 1.4, is also a root of x5 x4 4x3 + 4x2 + 2x 2 but weare looking for the polynomial of least degree.Now the work done in the above example can be simpli?ed immensely with theuse of computer software programs like MAPLETM. In MAPLETM, we need only7use the MinimalPolynomial() command to determine the minimal polynomial of analgebraic element.The de?nition of the minimal polynomial A(x) implies that A(x) must be irre-ducible. Otherwise, we would be able to factor A(x) and one of the factors wouldhave as a root and hence contradict the minimality of A(x).While on the topic of polynomials, we will introduce a de?nition which will beneeded later on in this paper.De?nition 1.4. The discriminant of a polynomial f(x) with roots 1 = ; 2;:::; nis de?ned byD(f) =Y1 i<j n( i j)2:That is, the discriminant of a polynomial is equal to the product of the squares of thedicurrency1erences of the polynomial? roots. As a note, we sometimes use the symbol D torefer to the discriminant of f.With the above de?nition, we are now able to de?ne the discriminant of an alge-braic number.De?nition 1.5. The discriminant of an algebraic number is equal to the dis-criminant of its minimal polynomial and is denoted as D( ).1.2.2. Conjugates of an Algebraic NumberDe?nition 1.6. Let be an algebraic number and let A(x) be its minimal poly-nomial. Then the conjugates of are all the roots of A(x), including itself.Notation 1.1. We write 1 = ; 2;:::; n to denote the n conjugates of wheren is equal to the degree of A(x). We say is an algebraic number of degree n.De?nition 1.7. If A(x) is the minimal polynomial of an algebraic number ofdegree n then the norm of , denoted as N( ), is the product of the conjugates of ; the trace of , denoted as Tr( ), is the sum of the conjugates of . That is,N( ) = 1 2 n and Tr( ) = 1 + 2 + ::: + n.8Example 1.5. The minimal polynomial of = p2 is x2 2. Thus, the conjugatesof are itself and p2. So we have N( ) = p2 p2 = 2 and Tr( ) =p2 + ( p2) = 0.1.3. Algebraic Number FieldsDe?nition 1.8. An algebraic number ?eld is a sub?eld of C of the form Q( 1; 2;:::; n)where 1; 2;:::; n are algebraic numbers.That is, an algebraic number ?eld is a ?eld K obtained from the ?eld of rationalnumbers Q by adjoining ?nitely many algebraic numbers 1; 2;:::; n. This ?eld isthe smallest ?eld that contains Q and the algebraic numbers 1; 2;:::; n.Example 1.6. Q(p2;p3) is an algebraic number ?eld as both p2 and p3 arealgebraic numbers.Now we will state a classic theorem known as the Primitive Element Theorem tosimplify the representation of ?elds like Q(p2;p3). The proof can be found on page595, [9].Theorem 1.3. If K = Q( 1; 2;:::; n) is an algebraic number ?eld then thereexists an algebraic number such that K = Q( ).This means that a composite ?eld can be expressed in the form Q( ), where isan algebraic number, and Q( ) is an algebraic number ?eld formed by constructingall possible numbers from and the set of rationals using the operators +; ; ; .Example 1.7. To demonstrate the Primitive Element Theorem, we will show thatQ(p2;p3) = Q( ) where is the algebraic number p2 + p3.By Example, 1.3, is an algebraic number with minimal polynomial f(x) = x4 10x2 + 1 2 Q[x].Now, Q( ) Q(p2;p3) as p2 + p3 2 Q(p2;p3) .Now, p2 = 92 + 12 3 2 Q( ) and p3 = 112 12 3 2 Q( ) so that Q(p2;p3) Q( ):Thus Q( ) = Q(p2;p3) and so the theorem holds.Remark 1.1. The expression K = Q( ) is not unique as Q( ) = Q( ) =Q(1 + ) = :::etc.91.3.1. Representation of Algebraic Number FieldsWe can also view K = Q( ) as an n-dimensional vector space over Q as seen in thefollowing result (Theorem 6.1.3, pg. 110, [2]).Theorem 1.4. Let K = Q( ) be an algebraic number ?eld, where is an algebraicnumber. Let A(x) be the minimal polynomial of with degree n. Then everyelement of K is uniquely expressible in the form a0 + a1 + ::: + an 1 n 1, wherea0;:::;an 1 2 Q. Thus, K = a0 + a1 + ::: + an 1 n 1 j a0;:::;an 1 2 Q .De?nition 1.9. K = Q( ) is an n-dimensional vector space over Q with basis 1; ;:::; n 1 and the degree of K over Q is n. We denote the degree of K overQ as n = [K : Q].Algebraic number ?elds are classi?ed according to their degree. So K is called aquadratic ?eld if n = 2, cubic if n = 3, quartic if n = 4, quintic if n = 5, etc.We can also term algebraic number ?elds as being pure or non-pure.De?nition 1.10. K = Q( ) is pure if is a root of the polynomial xn a wherea 2 Q and n is the degree of the minimal polynomial of . If K is not pure, then wesay K is non-pure.Algebraic number ?elds can also be classi?ed using Galois theory. We will studythis in detail in the following chapter.10Chapter 2Introduction to Galois TheoryWe begin this chapter with some basic de?nitions that will be used to establishthe foundation of Galois Theory.De?nition 2.1. An isomorphism of a ?eld K with itself is called an automor-phism of K. The collection of automorphisms of K is denoted Aut(K).Notice that every ?eld will have at least one automorphism - the identity map.De?nition 2.2. An automorphism 2 Aut(K) is said to ?x an element 2 Kif ( ) = . If F K then an automorphism is said to ?x F if it ?xes all theelements of F. That is, (a) = a, 8a 2 F.De?nition 2.3. Let K be an algebraic number ?eld. Let AutQ(K) be the collec-tion of automorphisms of K which ?x Q.Since Q K, any automorphism of K ?xes Q. This is because ?xes 1,the identity, and Q can be built up from 1 using +; ; and . Hence, Aut(K) =AutQ(K):If K is not a ?eld extension of Q then AutQ(K) is a subgroup of the group Aut(K).De?nition 2.4. An algebraic number ?eld K is separable if for each algebraicnumber 2 K, the minimal polynomial of over K has no repeated roots.De?nition 2.5. An algebraic number ?eld K is normal if for each algebraic num-ber 2 K, the minimal polynomial of over K factors entirely in K[x]. That is, itcompletely factors into linear factors over K.De?nition 2.6. An algebraic number ?eld is Galois if it is both separable andnormal. The group AutQ(K) is called the Galois group of K and is denoted asGal(K=Q).112.1. Galois Groups of Polynomials - OverviewDe?nition 2.7. Consider a polynomial f(x) 2 Q[x]. The splitting ?eld L ofthe polynomial f(x) over Q is the smallest sub?eld of C (set of complex numbers)containing Q and all the roots of f(x).De?nition 2.8. A polynomial f(x) over Q is called separable if it has no multipleroots (i.e. all its roots are distinct).We now state an important property of the splitting ?eld of a separable polynomial(Corollary 6, pg. 562, [9]).Corollary 2.1. If L is the splitting ?eld over Q of a separable polynomial f(x)then L is Galois.Because L is Galois, we are able to ?nd its Galois group. This leads us to thefollowing de?nition.De?nition 2.9. If f(x) is a separable polynomial over Q, then the Galois groupof f(x) over Q is the Galois group of the splitting ?eld of f(x) over Q and is denotedas Gal(f(x)).Now it is important to note that the above corollary implies the splitting ?eld ofany polynomial over Q is Galois. Why? Clearly, the splitting ?eld of f(x) is thesame as the splitting ?eld of the product of the irreducible factors of f(x) (i.e. thepolynomial obtained by removing multiple factors). Now this latter polynomial isseparable as it is the product of distinct irreducible polynomials (and so the factorsnever have zeros in common). Then using the above corollary on this polynomial,we get the result. From this, we can also conclude that every polynomial over Q hasa Galois group.Now we consider Galois theory from a permutational perspective. Given a poly-nomial, we wish to ?nd its Galois group. First we introduce a short de?nition.De?nition 2.10. An algebraic equation is one of the form P = 0 where P is apolynomial with coe? cients in Q.12The main idea of Galois theory is to consider those permutations of the rootsof our polynomial having the property that any algebraic equation satis?ed by theroots is still satis?ed after the roots have been permuted. These permutations forma group and is called the Galois group of the polynomial over Q. Thus, the elementsof Gal(f(x)) are permutations that act on the roots of f(x). So, Gal(f(x)) is asubgroup of Sn - the group of all permutations of the set f1;2;:::;ng. Now, we saysubgroup because we may not have all permutations of the roots of f(x) as elementsof Gal(f(x)). SojGal(f(x))j j jSnj:How does this correspond to the splitting ?eld of the polynomial f(x)? If f(x) isa separable polynomial over Q and if L denotes its splitting ?eld over Q, then weknow that Gal(f(x)) = Gal(L=Q). But Gal(L=Q) consists of automorphisms fromL to itself that ?x Q. Thus, any permutation of the roots which respects algebraicequations as described above gives rise to an automorphism of L over Q, and viceversa. That is, the automorphisms of L correspond to elements of Gal(f(x)):We will now do a small example to illustrate the above concept. This exampleparallels the "second example" in [33].Example 2.1. Consider the polynomial x4 5x2 + 6 = (x2 3)(x2 2).We wish to describe the Galois group of this polynomial over the ?eld of rationalnumbers. The polynomial has four roots:(1) A = p2(2) B = p2(3) C = p3(4) D = p3There are 24 possible ways to permute these four roots, but not all of these permu-tations are members of the Galois group. The members of the Galois group mustpreserve any algebraic equation with rational coe? cients involving A, B, C and D.One such equation is:A + B = 0:13Therefore the permutation (A;B;C;D) ! (A;C;B;D) is not permitted because ittransforms the valid equation A+B = 0 into the equation A+C = 0 ,p2+p3 = 0,which is invalid.Another equation that the roots satisfy is (A + B)2 + (C + D)2 = 0.This will exclude further permutations, such as(A;B;C;D) !(A;C;B;D):Continuing in this way, we ?nd that the only permutations satisfying both equationssimultaneously are:(1) (A;B;C;D) !(A;B;C;D) (trivial permutation)(2) (A;B;C;D) !(B;A;C;D)(3) (A;B;C;D) !(A;B;D;C)(4) (A;B;C;D) !(B;A;D;C).Thus, the Gal(f(x)) = V = Z2 Z2 (the Klein 4-group).Now showing the above four permutations satisfy every possible algebraic equa-tion satis?ed by A;B;C and D is less obvious and requires the theory of symmetricpolynomials. Also, as we increase the number of roots of our given polynomial, thismethod becomes less e? cient. We can see that calculating the Galois group of apolynomial is a nontrivial task. However, with technology, the task no longer seemsdaunting. For example, MAPLETM will calculate the Galois group of a polynomialup to degree 9. We use the galois() command to do so.We now introduce an important property needed in order to examine the Galoisgroup of polynomials, in particular, those of degree 3. For proof, see Theorem 14,pp. 523-524, [9].Proposition 2.1. The multiplicative property of ?elds states that if E L Fwhere E, L and F are ?elds and " " denotes "sub?eld", then [F : E] = [F : L] [L : E].Recall [F : E] denotes the degree of F over E.The next corollary is important when we examine the Galois group of polynomials.We refer the reader to pp. 51 of [3] for more details and the proof.14Corollary 2.2. Let K be a ?nite extension of the ?eld Q. If K is Galois thenjAutQ(K)j = [K : Q]:As a result, we know that AutQ(L) = Gal(L=Q) and since L is Galois (Corollary2.1),[L : Q] = jGal(L=Q)j .Lastly, we now present a proposition (Proposition 34, pp. 611, [9]) that will beencompassed in the theorems presented in these next sections.Proposition 2.2. The Galois group of f(x) 2 Q[x] is a subgroup of An if andonly if the discriminant D(f) is a square in Q.2.2. Irreducible PolynomialsDe?nition 2.11. A polynomial p(x) in Q[x] is called irreducible over Q if it isnon-constant and cannot be represented as the product of two or more non-constantpolynomials from Q[x].A useful test for irreducibility is Eisenstein? Criterion.Proposition 2.3. Suppose we have the following polynomial with integer coe? -cients:f(x) = anxn + an 1xn 1 + ::: + a1x + a0.Assume that there exists a prime number p such that p divides each ai for i 6= n p does not divide an p2 does not divide a0.Then f(x) is irreducible over Q.Proof. Suppose f(x) = anxn + an 1xn 1 + ::: + a1x + a0 2 Z[x] satis?es the aboveconditions for some prime p.By way of contradiction, assume f is reducible over Q: That is,f(x) = g(x) h(x)15where g(x) and h(x) are non-constant polynomials over Q.The reduction map mod p is given by' : Z[x] !Zp[x]and is a homomorphism. Thus,'(f) = '(g) '(h):But'(f) = '(anxn + an 1xn 1 + ::: + a1x + a0)= xnas p divides each ai for i 6= n and where an (mod p).So'(g) '(h) = xnand so'(g) = xkand'(h) = xn kfor some 0 < k < n, where (mod p) and since Zp[x] is a unique factorizationdomain.This implies p divides the constant term c1 of g(x) and the constant term c2 of h(x).That is, (c1 = pmc2 = plwhere m;l 2 Z. Since f(x) = g(x) h(x), the constant term a0 of f(x) isa0 = c1 c2= (pm)(pl)= p2(ml):Thus, p2 j a0 which is a contradiction. 16Example 2.2. Consider g(x) = 3x4 + 15x2 + 10.Let p = 5. Clearly, 5 j ai for i = 0 :::3. Also, 5 - 3 and 52 - 10:Thus, g(x) is irreducible over Q by Eisenstein? Criterion.We give another Theorem that may be helpful in trying to prove a polynomial isirreducible (see Proposition 12, pp. 309, [9]).Theorem 2.1. If f(x) is a nonconstant monic polynomial and f(x) factors overZ then f(x) factors over Zp for all primes p.Remark 2.1. The contrapositive of this theorem is very useful in showing a poly-nomial f(x) is irreducible. We need only ?nd one prime p for which f(x) does notfactor mod p to show f(x) is irreducible.Example 2.3. Let f(x) = x5 5x + 12: Thenf(x) x5 + 2x + 5 (mod 7)which does not factor over Z7. Hence, f(x) is irreducible.We now present one last theorem that may help in proving the irreducibility of apolynomial.Theorem 2.2. Let f(x) 2 Q[x]. f(x) is reducible if and only if f(x + a) isreducible for a 2 Z.Proof. For a 2 Z,f(x) is reducible , f(x) = g(x) h(x), f(x + a) = g(x + a) h(x + a), f(x + a) is reducible,where g(x) and h(x) are non-constant polynomials over Q. We illustrate this Theorem with the same polynomial used in Example 2.3.Example 2.4. Let f(x) = x5 5x + 12: As is, we cannot use Eisenstein?Criterion on this polynomial. However, if we make the translation x = y 2 we17obtainf(y 2) = y5 10y4 + 40y3 80y2 + 75y 10:Now we can use Eisenstein? Criterion with p = 5 to deduce f(y 2) is irreducible.Hence, by the contrapositive of Theorem 2.2, f(x) is also irreducible.The focus of the following sections is to present detailed methods for ?nding theGalois group of a polynomial of at most degree 5. For the most part, we are mainlyconcerned with ?nding the Galois group of an irreducible polynomial. Although westill present the cases of ?nding the Galois group of the reducible polynomial, we doso in less detail.2.3. Quadratic PolynomialsRecall that the Galois group of a polynomial f(x) 2 Q[x] is de?ned as the groupof automorphisms of the splitting ?eld of f that leave Q ?xed. To construct thissplitting ?eld, we merely adjoin to the elements of Q the roots of f(x). So considerthe general quadratic polynomial f(x) = x2 + bx + c 2 Q[x], the roots of which are b pb2 4c2 :It is easy to show that the discriminant of a quadratic polynomial of the formf(x) = x2 + bx + c is D(f) = D = b2 4c.To form the splitting ?eld L of f(x), we need to adjoin the two roots above toQ. However, both roots are of the form m + npD for some m;n 2 Q. Thus,Q[ b+pb2 4c2 ; b pb2 4c2 ] = Q[pD]. So we need only adjoin the number pD to theset of rationals Q to get the splitting ?eld L of f(x). That is, L = Q[pD]: Notethat Q[pD] =nm + npD j m;n 2 Qo(use the fact that pD is an algebraic numberand Theorem 1.4) and we see that both roots of f(x) are in this ?eld. So we nowconsider Gal(f(x)) = Gal(L=Q).Now there are two possibilities for Gal(L=Q).(1) If D is the square of a rational number, say D = r2 where r 2 Q, thenpD = r implies L = Q. This implies that f(x) is reducible over Q (as itsroots are in Q).18But there is only one group of automorphisms of Q that leave Q ?xed andsatisfy the above - it is the identity mapping. So Gal(f(x)) = Gal(Q=Q)which consists of a single element - namely the identity automorphism.(2) If D is not the square of a rational number, then the splitting ?eld L is morecomplicated and is of the form Q[pD] =nm + npD j m;n 2 Qo, for somenon-square D. Since D is not a square, this implies that f(x) is irreducibleover Q.Now Gal(L=Q) possesses one non-trivial automorphism given by (m + npD) = m npD where m;n 2 Q:That is, is the mapping that maps one root of f(x) to the other (and visaversa).To show that is an automorphism, we note that it is one-to-one and ontoand we have the relations ((m + npD) + (r + spD)) = ((m + r) + (n + s)pD)= (m + r) (n + s)pD= (m npD) + (r spD)= (m + npD) + (r + spD))and ((m + npD) (r + spD)) = ((mr + nsD) + (ms + nr)pD)= (mr + nsD) (ms + nr)pD= (m npD) (r spD)= (m + npD) (r + spD))where m;n;r;s 2 Q.Thus, is an automorphism that ?xes Q and hence in Gal(L=Q).We see that the composition of with itself yields the identity I. Thus,Gal(L=Q) consists of the two automorphisms I and and we have the fol-lowing group operation table (Table 2.1):19Table 2.1. Group Table for the Cyclic Group of Order 2 I I I IThus, Gal(f(x)) = Z2.In conclusion, there is only one possible Galois group for irreducible quadraticpolynomials: the group isomorphic to Z2 - also known as the cyclic group of order 2.2.4. Cubic PolynomialsNow we consider the general cubic polynomial f(x) = x3 + ax2 + bx + c 2 Q[x]with roots r1, r2 and r3. Then we haveD = (r1 r2)2(r1 r3)2(r2 r3)2;where D is the discriminant of f(x).Now thanks to computer software programs like MAPLETM, we have a formulafor D in terms of the coe? cients of f(x):D = a2b2 4b3 4a3c 27c2 + 18abc:It can be done by hand but the algebra can be challenging.2.4.1. Irreducible Cubic PolynomialsFirst we assume that f(x) is irreducible. Now in order to determine Gal(f(x)), weconsider two cases for D.(1) D = n2 for some n 2 Q. That is, D is equal to a square number.Then, any permutation on the roots of f(x) must also satisfy(r1 r2)(r1 r3)(r2 r3) = nand in particular,(r1 r2)(r1 r3)(r2 r3) = n:20The following permutations do not satisfy the above algebraic equation: (r1;r2;r3) !(r2;r1;r3) (r1;r2;r3) !(r1;r3;r2) (r1;r2;r3) !(r3;r2;r1)So we ?nd that the only permutations satisfying both of the equations (r1 r2)(r1 r3)(r2 r3) = n simultaneously are: (r1;r2;r3) !(r1;r2;r3) (trivial permutation) (r1;r2;r3) !(r2;r3;r1) (r1;r2;r3) !(r3;r1;r2)If the above permutations correspond to I, 1 and 2, respectively, then weobtain the following group operation table (Table 2.2):Table 2.2. Group Table for the Cyclic Group of Order 3 I 1 2I I 1 2 1 1 2 I 2 2 I 1Thus, Gal(f(x)) = A3 = Z3 (the cyclic group/alternating group of order 3).As before, using an argument based on symmetry of polynomials willshow the above three permutations satisfy every possible algebraic equationsatis?ed by r1, r2 and r3 and hence are indeed in Gal(f(x)).(2) D 6= n2 for any n 2 Q. That is, D is not equal to a square number.Now we consider the splitting ?eld of f(x) - the smallest ?eld containing Qand the roots of f(x). It is L = Q(r1;r2;r3).Since pD is the product of the dicurrency1erences of the roots of f(x), that is,pD = (r1 r2)(r1 r3)(r2 r3), thenpD 2 L = Q(r1;r2;r3):But pD = Q. Thus, Q(pD) =nm + npD j m;n 2 Qoand is a sub?eld ofL. So, [Q(pD) : Q] = 2 (since pD is not a square, the minimal polynomialof pD over Q is x2 D and then use Theorem 1.4).21Using the multiplicative property, we get[L : Q] = [L : Q(pD)] [Q(pD) : Q]= [L : Q(pD)] 2and so 2 j [L : Q].Since f(x) is irreducible, we can form the algebraic number ?eld K = Q(r1)which has the form fa + br1 + cr21 j a;b;c 2 Qg with f1;r1;r21g as a basis(see Theorem 1.4). So [K : Q] = 3. Since r1 2 L, we must have K L =Q(r1;r2;r3):Now using the multiplicative property again yields[L : Q] = [L : K] [K : Q]= [L : K] 3and so 3 j [L : Q].Since L can have at most degree 6 (using Corollary 2.2 and Gal(L=Q) S3),we must have [L : Q] = 6 and by Corollary 2.2 again, Gal(L=Q) = S3.Note that we could have used any of the roots of f(x) to form the ?eldK and still yield the same result. Also, it is very important that f(x) isirreducible else [K : Q] may not be 3.2.4.2. Reducible Cubic PolynomialsNow if f(x) = x3 + ax2 + bx + c is reducible over Q, then it splits either into threelinear factors or into a linear factor and an irreducible quadratic.(1) f(x) splits into three linear factors over Q.Then the three roots of f(x) are in Q and so adjoining the roots of f(x) toQ results in Q itself. This implies that Gal(f(x)) = Gal(Q=Q), the trivialGalois group.(2) f(x) splits into a linear factor and an irreducible quadratic over Q.Then adjoining the root of the linear factor to Q results in Q itself (as thatroot is in Q). So ?nding the splitting ?eld of f(x) is equivalent to ?ndingthe splitting ?eld of its quadratic factor. Since this factor is irreducible over22Q, its roots are not in Q and so the discriminant of this polynomial is not asquare. Thus, Gal(f(x)) = Z2 (see quadratic polynomial section).Example 2.5. Consider the irreducible polynomial f(x) = x3 3x + 1. Usingthe formula for D, its discriminant is equal to 81, a square.Thus, Gal(f(x)) = A3.Example 2.6. Consider the irreducible polynomial f(x) = x3 2. Using theformula for D, its discriminant is equal to 108, not a square.Thus, Gal(f(x)) = S3.Example 2.7. Consider the polynomial f(x) = x3 2x+1. It has roots 1;p52 12and p52 12 and factors into (x 1) (x2 + x 1) over Q.Thus, Gal(f(x)) = Z2.Check: If we take the discriminant of this quadratic polynomial, we obtainD(x2 + x 1) = ((p52 12) ( p52 12))2 = 5which is not a square.Remark 2.2. Note that for the last example, we could have also used the formulafor the discriminant of a quadratic polynomial to obtain D(x2 + x 1) = 5.In conclusion, there are only two possible Galois groups for irreducible cubic poly-nomials: the group isomorphic to Z3 (also known as the cyclic group C3 or theAlternating group A3 of order 3) and the group isomorphic to S3 (also known as thesymmetric group of order 6).2.5. Quartic PolynomialsNow we consider the general quartic polynomial f(x) = x4 + ax3 + bx2 + cx+ d 2Q[x] with roots r1, r2, r3 and r4. Also, let L be the splitting ?eld of f(x) over Q.2.5.1. Irreducible Quartic PolynomialsFirst we assume that f(x) is irreducible over Q. Recall that Gal(f(x)) is a subgroupof S4 - the symmetric group on 4 letters which has order 4! = 24.23De?nition 2.12. The cubic resolvent of f(x) is the polynomialr(x) = x3 bx2 + (ac 4d)x (a2d 4bd + c2)whose roots aret1 = r1r2 + r3r4t2 = r1r3 + r2r4t3 = r1r4 + r2r3.Denote the splitting ?eld of r(x) as E.Remark 2.3. The discriminant of f(x) and r(x) are the same. That is,D =Yi<j(ri rj)2 =Yi<j(ti tj)2or equivalently,D = 4b3A + b2B2 + 18dBA B3 27A2where A = a2d + c2 4bd and B = ac 4d.Theorem 2.3. Suppose f(x) is irreducible over Q, r(x) its resolvent with splitting?eld E, and D the discriminant of f(x). Then(1) Gal(L=Q) = S4 if and only if r(x) is irreducible over Q and D 6= n2 for anyn 2 Q. In this case, Gal(L=Q) is known as the symmetric group of order24.(2) Gal(L=Q) = A4 if and only if r(x) is irreducible over Q and D = n2 forsome n 2 Q. In this case, Gal(L=Q) is known as the alternating group oforder 12.(3) Gal(L=Q) = V = Z2 Z2 if and only if r(x) splits into linear factors overQ. In this case, Gal(L=Q) is known as the Klein 4-group of order 4.(4) Gal(L=Q) = C4 if and only if r(x) has exactly one root t in Q and g(x) =(x2 tx + d) (x2 + ax + (b t)) splits over E. In this case, Gal(L=Q) isknown as the cyclic group of order 4.(5) Gal(L=Q) = D4 if and only if r(x) has exactly one root t in Q and g(x) doesnot split over E. In this case, Gal(L=Q) is known as the dihedral group oforder 8.24Remark 2.4. For further details and derivations of the Galois groups of quarticpolynomials, we refer the reader to the paper written by Luise-Charlotte Kappe andBette Warren [16].Example 2.8. Let f1(x) = x4 4x3 + 4x2 + 6.In order to use the above theorem, we ?rst must determine if f1(x) is irreducibleover Q. Using Eisenstein? Criterion with p = 2, we conclude that indeed f1(x) isirreducible over Q.Now r(x) = x3 4x2 24x + 0 = x (x2 4x 24). Thus, r(x) is reducible over Qand this eliminates cases 1 and 2 of the theorem. Since x2 4x 24 is irreducibleover Q, r(x) does not split into three linear factors and hence eliminates case 3.Now the one and only rational root of r(x) is t = 0. So g(x) = (x2 tx + d) (x2 +ax + (b t)) = (x2 + 6) (x2 4x + 4) = (x2 + 6) (x 2)2.We now need to ?nd the splitting ?eld E of r(x). The roots of r(x) are 0 and 2 2p7.Thus, E = Q(0;2 2p7) = Q(p7). So E has the form E = a + bp7 j a;b 2 Q :Now clearly g(x) does not split completely over E (as two roots are p6i = Q(p7)).Thus, Gal(f1(x)) = D4 (case 5).Example 2.9. Let f2(x) = x4 + 4x3 + 6x2 + 4x + 2.Using p = 2, by Eisenstein? Criterion, f2(x) is irreducible over Q.Now r(x) = x3 6x2 + 8x + 0 = x (x2 6x + 8) = x (x 4) (x 2). Since r(x)splits into three linear factors (case 3), Gal(f2(x)) = V .Example 2.10. Let f3(x) = x4 + x3 + x2 + x + 1.Note that f3(x) is known as a cyclotomic polynomial of degree 4 and is irreducibleover Q.We have r(x) = x3 x2 3x + 2 = (x 2) (x2 + x 1). So r(x) has exactly oneroot in Q and it is t = 2. The other roots of r(x) are 1 p52 . Thus the splitting ?eldof r(x) is E = Q(p5).Now g(x) = (x2 2x+1) (x2 +x 1) = (x 1)2 (x2 +x 1). The roots of g(x) are1 and 1 p52 . Clearly, the roots of g(x) are in E = Q(p5) and so g(x) splits over E(case 4). In fact, r(x) and g(x) both have the same splitting ?eld E = Q(p5).Thus, Gal(f3(x)) = C4.25We now consider one more example that is more complicated than the above.Example 2.11. Let f4(x) = x4 + px + p for each prime p.Using p = p, Eisenstein? Criterion implies that f4(x) is irreducible over Q.Now r(x) = x3 4px p2. By Gauss?Lemma, if r(x) is irreducible over Z then itis irreducible over Q.If r(x) factors over Z then it factors as a product of either three linear factors or alinear factor and an irreducible quadratic factor. Either case, there is a linear factorpresent. Thus, r(x) is reducible over Z if and only if there is at least one integerroot.By the Rational Root Theorem, the possible roots are the divisors of p2. They are 1, p and p2. Nowr(1) = 1 4p p2r( 1) = 1 + 4p p2r(p) = p3 4p2 p2 = p2 (p 5)r( p) = p3 + 4p2 p2 = p2 (3 p)r(p2) = p6 4p3 p2 = p2(p4 4p 1)r( p2) = p6 + 4p3 p2 = p2( p4 + 4p 1).Thus, r(x) is reducible if and only if p = 3 or p = 5. So r(x) is irreducible if andonly if p 6= 3 and p 6= 5.Now we haveD(f4(x)) = 72p4 + 64p3 27p4 = p3 (99p 64) < 0.Thus, D(f4(x)) is not equal to a square in Q. So, Gal(f4(x)) = S4 if p 6= 3;5 (case1).If p = 3 then r(x) is reducible and r(x) = x3 12x 9 = (x+3) (x2 3x 3). Nowr(x) has exactly one rational root t = 3. So g(x) = (x2 + 3x + 3) (x2 + 3) whichhas roots p3i and 3 p3i2 . Thus g(x) has no real roots. Since the splitting ?eldof r(x) is E = Q( 3; 3 p212 ) = Q(p21), g(x) does not split over E (case 5). So,Gal(f4(x)) = D4 if p = 3.If p = 5 then r(x) is reducible and r(x) = x3 20x 25 = (x 5) (x2 +5x+5). Nowr(x) has exactly one rational root t = 5. So g(x) = (x2 5x + 5) (x2 5) which has26roots p5 and 5 p52 . Since the splitting ?eld of r(x) is E = Q(5; 5 p52 ) = Q(p5),the roots of g(x) are in E and so g(x) splits over E (case 4). So, Gal(f4(x)) = C4if p = 5.In conclusion we have the following:Gal(f4(x)) =8>><>>:D4 if p = 3C4 if p = 5S4 otherwise.Now Cases 4 and 5 in the above theorem can be di? cult to use. Determiningwhether or not a polynomial splits over a ?eld is not always an easy problem. Thus,we now introduce an alternate method in dicurrency1erentiating between Cases 4 and 5 forthe irreducible quartic.Theorem 2.4. Suppose f(x) is irreducible over Q with roots r1, r2, r3 and r4.Let K be the algebraic number ?eld formed by adjoining Q with any one of the rootsof f(x), say r1. That is, K = Q(r1). Assume Gal(f(x)) = D4;C4 or V . Then wecan ?nd integers a;b;c such that K = Q(pa + bpc) and if b 6= 0(1) Gal(f(x)) = V if and only if a2 b2c = k2 for some integer k.(2) Gal(f(x)) = C4 if and only if a2 b2c = ck2 for some integer k.(3) Gal(f(x)) = D4 if and only if a2 b2c 6= k2 or ck2 for any integer k.Remark 2.5. For further details and derivations regarding this alternate methodfor the above irreducible quartic cases, we refer the reader to the paper written byLuise-Charlotte Kappe and Bette Warren [16]. For more information regarding the?eld K = Q(pa + bpc) and the above example, we refer the reader to the paperwritten by James Huard, Blair Spearman and Kenneth Williams [15].Remark 2.6. An obvious question arises from this theorem: what happens ifb = 0?If Gal(f(x)) = D4;C4 or V , then we can ?nd integers a;b;c such that K = Q(pc;pa + bpc)(see [15]).This means if b 6= 0 then pa + bpc 2 K implies a + bpc 2 K. So, pc 2 K. Thus,K = Q(pc;pa + bpc) = Q(pa + bpc) if b 6= 0.27If b = 0 then K = Q(pc;pa + bpc) = Q(pc;pa).Now this case is examined in the reducible quartic case when the polynomial reducesinto two irreducible quadratic factors (see Theorem 2.5). We will see if b = 0 thenGal(f(x)) = Z2 if a c = square in QandGal(f(x)) = V if a c 6= square in Q.Now the obvious di? culty to this alternative method is trying to ?nd the integersa;b;c such that K = Q(pa + bpc). However, with the aid of MAPLETM, thismethod is quite appealing.Example 2.12. Let f(x) = x4 4x3 + 4x + 1.Using the solve() command in MAPLETM, we ?nd the roots of f(x):1+p22 +p10 + 4p22 ;1+p22 p10 + 4p22 ;1 p22 +p10 + 4p22 ;1 p22 p10 + 4p22 :We now form the algebraic number ?eld K = Q(1+p22 +p10+4p22 ) = Q(p10 + 4p2)and thus takea = 10b = 4c = 2:So a2 b2c = 68 6= k2 or 2k2 for any integer k. Thus, Gal(f(x)) = D4.2.5.2. Reducible Quartic PolynomialsNow lets take a look at the case when the general quartic polynomial f(x) = x4 +ax3 + bx2 + cx + d with roots r1, r2, r3 and r4 is reducible over Q.(1) If f(x) splits into a product of a linear factor and an irreducible cubic factor,then Gal(f(x)) will be equal to the Galois group of the irreducible cubicfactor which we determined how to ?nd in the previous section. There aretwo possibilities for Gal(f(x)) in this case: Z3 or S3.28(2) If f(x) splits into a product of four linear terms then Gal(f(x)) = Gal(Q=Q)and thus is the trivial Galois group.(3) We now consider the case when f(x) splits into a product of two irreduciblequadratic factors whose discriminants are D1 and D2.In order to fully understand this case, we need the following Corollary (Corol-lary 22, pp. 593, [9]).Corollary 2.3. Let K1 and K2 be Galois ?elds both containing a ?eld Fwith K1 \ K2 = F. ThenGal(K1K2=F) = Gal(K1=F) Gal(K2=F):We note that K1K2 stands for the composite ?eld of K1 and K2; it isthe smallest ?eld containing Q, K1 and K2. This leads us to the followingtheorem regarding this case:Theorem 2.5. Let f(x) be a quartic polynomial with rational coe? cientsand suppose that f(x) = f1(x) f2(x) over Q where f1(x) and f2(x) areirreducible quadratics. Let D1 and D2 denote the discriminants of f1(x) andf2(x), respectively. Then we haveGal(f(x)) = Z2 if D1D2 = a square in QandGal(f(x)) = Z2 Z2 = V if D1D2 6= a square in Q.Proof. Since both f1(x) and f2(x) are irreducible quadratics, their splitting?elds are Q(pD1) and Q(pD2), respectively. The splitting ?eld of f(x) isthen L = Q(pD1;pD2).If D1D2 = square in Q then D1D2 = n2 for some n 2 Q. So, pD1 D2 =pD1 pD2 = n , pD1 = npD2. Thus, Q(pD1) = Q(pD2) and soL = Q(pD1). So, [L : Q] = 2 which implies jGal(L=Q)j = 2. But the onlygroup of order 2 is Z2. Thus,Gal(f(x)) = Z2 if D1D2 = square in Q:29If D1D2 6= square in Q then Q(pD1) 6= Q(pD2). LetE = Q(pD1) \ Q(pD2):By de?nition of intersection, E Q(pD1) and E Q(pD2): However,Q(pD1) 6= Q(pD2) and neither is a proper subset of the other (as bothhave degree 2), so E < Q(pD1). The multiplicative property of ?eldsimplies[Q(pD1) : Q] = [Q(pD1) : E] [E : Q]or equivalently,2 = [Q(pD1) : E] [E : Q].Thus, [E : Q] = 1 or 2. But [E : Q] = 2 implies Q(pD1) = Q(pD2) whichis a contradiction. Thus, [E : Q] = 1 and so E = Q(pD1) \ Q(pD2) = Q.So by Corollary 2.3 (using F = Q), we getGal(Q(pD1;pD2)=Q) = Gal(Q(pD1)=Q) Gal(Q(pD2)=Q):But Gal(Q(pD1)=Q) = Z2 and Gal(Q(pD2)=Q) = Z2 (as f1(x) and f2(x)are irreducible quadratics). Thus,Gal(f(x)) = Z2 Z2 = V if D1D2 6= square in Q: In conclusion, there are only ?ve possible Galois groups for irreducible quarticpolynomials: the group isomorphic to S4, the group isomorphic to A4, the groupisomorphic to V , the group isomorphic to C4 and the group isomorphic to D4.2.6. Quintic PolynomialsNow we consider the general quintic polynomial f(x) = x5 + px3 + qx2 + rx + s 2Q[x] with roots r1, r2, r3, r4 and r5. Also, let L be the splitting ?eld of f(x) overQ. Note that we simpli?ed the general quintic polynomial g(x) = x5 + mx4 + nx3 +ux2 + vx + w 2 Q[x] to the polynomial f(x) given above by making the substitutiony = x + m5 ,x = y m5 in g(x). Now Gal(g(x)) = Gal(f(x)) as translations over Qdo not acurrency1ect the Galois group over Q.302.6.1. Irreducible Quintic PolynomialsFirst we assume that f(x) is irreducible over Q.De?nition 2.13. The sextic resolvent of f(x) is the polynomialf20(x) = x6 + 8rx5 + (2pq2 6p2r + 40r2 50qs)x4 ++( 2q4 + 21pq2r 40p2r2 + 160r3 15p2qs 400qrs + 125ps2)x3+(p2q4 6p3q2r 8q4r + 9p4r2 + 76pq2r2 136p2r3 + 400r4 50pq3s + 90p2qrs 1400qr2s + 625q2s2 + 500prs2)x2+( 2pq6 + 19p2q4r 51p3q2r2 + 3q4r2 + 32p4r3 + 76pq2r3 256p2r4+512r5 31p3q3s 58q5s + 117p4qrs + 105pq3rs + 260p2qr2s 2400qr3s 108p5s2 325p2q2s2 + 525p3rs2 + 2750q2rs2 500pr2s2 + 625pqs3 3125s4)x+(q8 13pq6r + p5q2r2 + 65p2q4r2 4p6r3 128p3q2r3 + 17q4r3+48p4r4 16pq2r4 192p2r5 + 256r6 4p5q3s 12p2q5s + 18p6qrs+12p3q3rs 124q5rs + 196p4qr2s + 590pq3r2s 160p2qr3s 1600qr4s 27p7s2 150p4q2s2 125pq4s2 99p5rs2 725p2q2rs2 + 1200p3r2s2+3250q2r2s2 2000pr3s2 1250pqrs3 + 3125p2s4 9375rs4).In the particular case when f(x) = x5 + ax + b, the resolvent simpli?es to:f20(x) = x6 + 8ax5 + 40a2x4 + 160a3x3 + 400a4x2+(512a5 3125b4)x + (256a6 9375ab4).Recall that Gal(f(x)) is a subgroup of S5 - the symmetric group on 5 letters whichhas order 5! = 120.31Using MAPLETM we can ?nd the polynomial discriminant of f(x) as a function ofits coe? cients. It isD = 560p2r2qs 72p4rsq 630prsq3 + 2250q2s2r 1600qsr3 + 16p4r3 128p2r4 ++256r5 + 3125s4 + 108p5s2 + 144pr3q2 900p3rs2 + 2000pr2s2 3750pqs3 ++825p2q2s2 + 16p3q3s 4p3q2r2 + 108q5s 27q4r2:We now break down the problem in ?nding Gal(f(x)) based on the existence of arational root of the resolvent f20(x).Theorem 2.6. Suppose f(x) = x5 + px3 + qx2 + rx + s 2 Q[x] is irreducibleover Q with resolvent f20(x) given by the above formula. Also, suppose f20(x) has norational roots. Let D be the discriminant of f(x). Then(1) Gal(f(x)) = S5 if and only if D 6= n2 for any n 2 Q. In this case, Gal(f(x))is known as the symmetric group of order 120.(2) Gal(f(x)) = A5 if and only if D = n2 for some n 2 Q. In this case,Gal(f(x)) is known as the alternating group of order 60.Example 2.13. Let f1(x) = x5 + 3x 3.Using p = 3, by Eisenstein? Criterion, f1(x) is irreducible over Q.Now since f1(x) is of the form x5 +ax+b, we can use the simpli?ed resolvent formula.So,f20(x) = x6 + 24x5 + 360x4 + 4320x3 + 32400x2 128709x 2091501.Now using MAPLETM and the factor() command, we deduce that f20(x) has no ra-tional roots. Equivalently, using the Rational Root Theorem, the possible rationalroots are the divisors of 2091501. However, one can easily show that none of thesedivisors are roots of f20(x).Thus, Gal(f1(x)) = S5 or A5. To determine which is the correct group, we mustnow calculate D - the discriminant of f1(x). It is D = 315333: This can be doneusing the formula given for the discriminant of a quintic polynomial or by using thediscrim() command in MAPLETM.32Now pD = p315333 = 9p3893 and hence, D is not a square in Q. Thus,Gal(f1(x)) = S5.Example 2.14. Let f2(x) = x5 + 20x + 16.In this example, we cannot use Eisenstein? Criterion as the only prime p dividing allcoe? cients of f2(x), other than the leading one, is p = 2. However, p2 cannot dividethe constant term which does not hold.If we make the translation x = y 1 we obtainf2(y 1) = y5 5y4 + 10y3 10y2 + 25y 5:Now using Eisenstein? Criterion with p = 5, we deduce f2(y 1) is irreducible. Thus,by the contrapositive of Theorem 2.2, f2(x) is also irreducible. Note that we can useMAPLETM and its irreduc() or factor() command. The irreduc() command returnstrue if the polynomial is irreducible over Q and false if the polynomial is reducibleover Q. We conclude that f2(x) is indeed irreducible.Now since f2(x) is of the form x5 +ax+b, we can use the simpli?ed resolvent formula.So,f20(x) = x6 + 160x5 + 16000x4 + 1280000x3 + 64000000x2+1433600000x + 4096000000.Now using MAPLETM and the factor() command, we deduce that f20(x) has no ratio-nal roots. Equivalently, using the Rational Root Theorem, the possible rational rootsare the divisors of 4096000000. However, one can show, with a little bit of work, thatnone of these divisors are roots of f20(x). Thus, Gal(f2(x)) = S5 or A5. To deter-mine which is the correct group, we must now calculate D - the discriminant of f2(x).It is D = 1024000000. This can be done using the formula given for the discriminantof a quintic polynomial or by using the discrim() command in MAPLETM.Now pD = p1024000000 = 32000 2 Q and hence D is a square. Thus, Gal(f2(x)) =A5.Now in order to use this next theorem, we need to be able to calculate d(K) -the ?eld discriminant. We will de?ne this value and examine this quantity in detail33in Chapter 3. For the examples proceeding this theorem, we will assume the givenvalue of the ?eld discriminant is true.Theorem 2.7. Suppose f(x) = x5 + px3 + qx2 + rx + s 2 Q[x] is irreducibleover Q with resolvent f20(x) given by the previous formula. Also, suppose f20(x) hasexactly one rational root. Let D be the discriminant of f(x) and let d(K) be the ?elddiscriminant of the ?eld Q(ri), where ri is any of the roots of f(x). Then(1) Gal(f(x)) = F20 if and only if D 6= n2 for any n 2 Q. In this case,Gal(f(x)) is known as the Frobenius or metacyclic group of order 20.(2) Gal(f(x)) = D5 if and only if D = n2 for some n 2 Q and d(K) is not equalto the fourth power of an odd integer. In this case, Gal(f(x)) is known asthe dihedral group of order 10.(3) Gal(f(x)) = C5 if and only if D = n2 for some n 2 Q and d(K) is equal tothe fourth power of an odd integer. In this case, Gal(f(x)) is known as thecyclic group of order 5.Remark 2.7. For further details and derivations regarding the irreducible quinticcase, we refer the reader to the paper written by D. S. Dummit [8].Remark 2.8. We will prove later that ?eld discriminants distinguish between theGalois groups D5 and C5.In fact, for cyclic quintic polynomials, its ?eld discriminant is always a fourth power,denoted as f4 for some f 2 Z. We call f4 the conductor of the ?eld.For dihedral quintic polynomials, its ?eld discriminant is of the form d2 f4 wheref 2 Z. In this case, d is the ?eld discriminant of the quadratic sub?eld of the normalclosure of K (normal closure will be de?ned later). So, the ?eld discriminant of adihedral polynomial is not a fourth power with the single exception of d2 = 24. Thisis the case when the quadratic sub?eld is Q(p 1).The topic of conductors and ?eld discriminants will be examined later in greater detail.Example 2.15. Let f3(x) = x5 + 15x + 12.34Using p = 3, by Eisenstein? Criterion, f3(x) is irreducible over Q. Now since f3(x)is of the form x5 + ax + b, we can use the simpli?ed resolvent formula. So,f20(x) = x6 + 120x5 + 9000x4 + 540000x3 + 20250000x2+324000000x.We can easily see that x = 0 is a rational root of f20(x). So we use Theorem 2.7 todeduce the Galois group of f3(x).We will ?nd the discriminant of f3(x) using the formula given in the beginning of thissection. We get pD = p259200000 = 7200p5.So D is not a square and thus Gal(f3(x)) = F20.Example 2.16. Let f4(x) = x5 5x + 12.In this example, we cannot use Eisenstein? Criterion as there is no prime p dividingall coe? cients (excluding the leading one) of f4(x). Using Example 2.3, we concludedthat f4(x) was indeed irreducible over Q. Now since f4(x) is of the form x5 +ax+b,we can use the simpli?ed resolvent formula. So,f20(x) = x6 40x5 + 1000x4 20000x3 + 250000x2 66400000x + 976000000.Now using MAPLETM and the factor() command, we deduce that f20(x) factors intof20(x) = (x 40) (x5 + 1000x3 + 20000x2 + 1050000x 24400000)and thus has one rational root, x = 40. Once again, we use Theorem 2.7 to deducethe Galois group of f4(x).We will ?nd the discriminant of f4(x) using the formula given in the beginning of thissection. We get pD = p64000000 = 8000.So D is a square and we now need to ?nd d(K) - the ?eld discriminant of the ?eldformed by adjoining Q with one of the roots of f4(x). We will show this in detail in35the following chapter. In this example,d(K) = 1000000 = (2)6 (5)6.Since d(K) is not the fourth power of an odd integer, Gal(f4(x)) = D5.Example 2.17. Let f5(x) = x5 110x3 55x2 + 2310x + 979.Using p = 11, by Eisenstein? Criterion, f5(x) is irreducible over Q. Now since f5(x)is not of the form x5 + ax + b, we cannot use the simpli?ed resolvent formula. So,we need to use the general formula to obtainf20(x) = x6 + 18480x5 + 47764750x4 580262760000x3 1796651418959375x2+2980357148316659375x 360260685644469671875.Now using MAPLETM and the factor() command, we deduce that f20(x) factors intof20(x) = (x + 9955) (x5 + 8525x4 37101625x3 210916083125x2+303018188550000x 36188918698590625)and thus has one rational root, x = 9955. Once again, we use Theorem 2.7 todeduce the Galois group of f5(x).We will ?nd the discriminant of f5(x) using the formula given in the beginning of thissection. We getpD = p1396274566650390625 = 1181640625.So D is a square and we now need to ?nd d(K) - the ?eld discriminant of the ?eldformed by adjoining Q with one of the roots of f5(x). We will show this in detail inthe following chapter. In this example,d(K) = 14641 = (11)4.Since d(K) is the fourth power of an odd integer, Gal(f5(x)) = C5.Distinguishing between Galois groups D5 and C5 requires ?nding the ?eld discrim-inant; this is no easy task. Thus, we provide the reader with an alternate method36to make this distinction. The basis of this method is formed from the followingCorollary (Corollary 41, pp. 641, [9]).Corollary 2.4. For any prime p not dividing the discriminant of f(x) 2 Z[x],each factorization of f(x) mod p exposes an element of the Galois group with thatcycle structure.We now present the method for distinguishing between the Galois groups D5 andC5.Theorem 2.8. Let f(x) 2 Z[x] be a monic irreducible quintic polynomial. Let pbe a prime and writef(x) (d1)n1 (d2)n2 (dr)nr (mod p)to denote the fact that f(x) factors modulo p into r distinct irreducible factors ofdegrees d1;:::;dr and multiplicities n1;:::;nr; respectively. The following procedurehas been given to determine Gal(f(x)):(1) If there exists a prime p1 such that f(x) (2)(3) (mod p1) then Gal(f(x)) =S5.(2) If there exists a prime p2 such that f(x) (1)(1)(3) (mod p2) and case 1does not hold then Gal(f(x)) = A5.(3) If there exists a prime p3 such that f(x) (1)(4) (mod p3) and cases 1 and2 do not hold then Gal(f(x)) = F20.(4) If there exists a prime p4 such that f(x) (1)(2)(2) (mod p4) and cases 1;2and 3 do not hold then Gal(f(x)) = D5.(5) If for every prime q either f(x) (1)(1)(1)(1)(1) (mod q) or f(x) 5(mod q) then Gal(f(x)) = C5.Remark 2.9. For further details and derivations regarding this alternate methodfor the irreducible quintic case, we refer the reader to the paper written by BlairSpearman and Kenneth Williams [27].Firstly, the reader should notice that the above method (as well as its precedingcorollary) is de?ned for polynomials over Z. However, given an irreducible quintic37polynomial over Q, we can multiply f(x) by scalars to produce a function f(x) that isover Z. Since both f(x) and f(x) have the same roots, they have the same splitting?eld and thus the same Galois group. So the theorem is applicable to our generalquintic polynomial with rational coe? cients.Also, we should mention that we do not consider p to be those primes foundin the prime factorization of the polynomial discriminant (as seen in the precedingcorollary). This is because such primes produce unpredictable factorizations.We will now illustrate the use of this theorem. We will be using the Factor()mod p command in MAPLETM to determine whether the function factors modulop. It should be noted that we are using this theorem to only distinguish betweenthe dihedral and cyclic Galois groups of an irreducible quintic polynomial. So westill must calculate the sextic resolvent and determine the existence of rational roots.However, now we do not need to include the calculation of the ?eld discriminant (anot so easy procedure).Example 2.18. Let f1(x) = x5 5x + 12.Recall from Example 2.16 that the sextic resolvent f20(x) had one rational root. Thisimplies from Theorem 2.7 that the Galois group of f1(x) is either F20;D5 or C5. Wealso found that the polynomial discriminant was a square thus restricting the Galoisgroup to either D5 or C5. So we can now use Theorem 2.8 to determine which oneit is.First we ?nd the prime factorization of the polynomial discriminant of f1(x). It isD = (2)12 (5)6.This implies that we should not choose p to be 2 or 5. Thus, lets try p = 3.Using the notation given in the theorem, factoring f1(x) mod 3 yieldsf1(x) = (1)(2)(2).Thus, from the theorem, we haveGal(f1(x)) = D5:38We now illustrate an example on which you could determine the Galois groupusing only Theorem 2.8.Example 2.19. Let f2(x) = x5 21x + 12.The prime factorization of its polynomial discriminant isD = (2)14 (3)4 (739):So we should avoid choosing p to be 2;3 and 739. So lets try p = 5: We getf2(x) = (5)which does not help us. So lets try p = 7 to getf2(x) = (1)(4):This tells us that we could have cases 1;2 or 3. Trying p = 11 yieldsf2(x) = (1)(2)(2):This implies we have cases 1;2;3 or 4 which yields no new information. Tryingp = 13 yieldsf2(x) = (2)(3):This implies we have case 1. ThusGal(f2(x)) = S5.Now Theorem 2.8 has its ?aws. It is not always practical to use this theorem todetermine the Galois group of an irreducible quintic polynomial. The theorem onlyguarantees the existence of a prime p. Theoretically, we could use the theorem to?nd the Galois group but we may be searching through a countless number of primesto ?nd it. Thus, using Theorems 2.6 and 2.7 may be more e? cient. However, withthis said, the theorem is best used to dicurrency1erentiate between the Galois groups D5 andC5, especially if the ?eld discriminant is not known.392.6.2. Reducible Quintic PolynomialsNow let? consider the case where the general quintic polynomial f(x) = x5 + px3 +qx2 + rx + s 2 Q[x] with roots r1, r2, r3, r4 and r5 is reducible over Q.(1) If f(x) splits into a product of ?ve linear terms then Gal(f(x)) = Gal(Q=Q)and thus is the trivial Galois group (as its roots are all in Q).(2) If f(x) splits into a product of two irreducible quadratic factors and onelinear factor, then Gal(f(x)) will be equal to the Galois group of the twoirreducible quadratic factors which we determined how to ?nd in the quarticsection. There are thus two possibilities for Gal(f(x)) in this reducible case:Z2 or V .(3) If f(x) splits into a product of three linear factors and one irreducible qua-dratic factor, then Gal(f(x)) will be equal to the Galois group of the irre-ducible quadratic factor which we determined how to ?nd in the quadraticsection. There is thus only one possibility for Gal(f(x)) in this case: Z2.(4) If f(x) splits into a product of an irreducible quartic factor and one linearfactor, then Gal(f(x)) will be equal to the Galois group of the irreduciblequartic factor which we determined how to ?nd in the quartic section. Thereare thus ?ve possibilities for Gal(f(x)) in this reducible case: S4, A4, V , C4or D4.(5) If f(x) splits into a product of two linear factors and one irreducible cubicfactor, then Gal(f(x)) will be equal to the Galois group of the irreduciblecubic factor which we determined how to ?nd in the cubic section. Thereare thus two possibilities for Gal(f(x)) in this case: Z3 or S3.(6) Now we consider the case when f(x) splits into a product of an irreduciblequadratic factor and an irreducible cubic factor.Theorem 2.9. Let f(x) be a quintic polynomial with rational coe? cientsand suppose that f(x) = f1(x) f2(x) over Q where f1(x) is an irreduciblequadratic and f2(x) is an irreducible cubic. Let D1 and D2 denote thediscriminants of f1(x) and f2(x), respectively. Then we haveGal(f(x)) = Z2 Z3 = Z6 if D2 = a square in Q,40Gal(f(x)) = S3 if D1 D2 = a square in Q,Gal(f(x)) = C2 S3 if both D2 and D1 D2 6= a square in Q.Proof. Firstly, since f1(x) is an irreducible quadratic, we know D1 is not asquare in Q. The splitting ?eld of f1(x) is L1 = Q(pD1) and Gal(f1(x)) =Z2:If D2 = a square in Q then D2 = n2 for some n 2 Q. Then from thecubic section we know jGal(f2(x))j = 3 (i.e. Gal(f2(x)) = Z3). So from themultiplicative property of ?elds, any sub?eld of the splitting ?eld of f2(x)must have degree dividing 3. Thus, this splitting ?eld, call it L2, has onlythe trivial sub?elds: itself and Q. So,L1 \ L2 = Q.So using Corollary 2.3, Gal(f(x)) = Gal(f1(x)) Gal(f2(x)) and soGal(f(x)) = Z2 Z3 if D2 = a square in Q.If D1 D2 = a square in Q then D1 D2 = n2 for some n 2 Q. Equivalently,D2 = n2D1 . But D1 6= a square in Q and so D2 6= a square in Q. Thenfrom the cubic section we know jGal(f2(x))j = 6 (i.e. Gal(f2(x)) = S3). Sofrom the multiplicative property of ?elds, any sub?eld of the splitting ?eld off2(x) must have degree dividing 6 which implies there is a quadratic sub?eld(i.e. a sub?eld of order 2).We will now use a result (from pp.345, [6]) that states the discriminant D2of f2(x) can be written in the unique form dk2, d;k 2 Z, where d is the ?elddiscriminant of the quadratic sub?eld of the splitting ?eld of f2(x): This41meansD1 D2 = n2) D1 dk2 = n2) D1 d = a square in Q)pD1 2 Q(pd)) Q(pD1) Q(pd) aspD1 generates Q(pD1)) Q(pD1) = Q(pd) as both ?elds have degree 2) Q(pD1) splitting field of f2(x)as Q(pd) is the quadratic sub?eld of the splitting ?eld of f2(x):So the splitting ?eld of f2(x) contains Q(pD1) - the splitting ?eld of f1(x).Therefore, the splitting ?eld of f(x) is the splitting ?eld of f2(x) (that is, theroots of f(x) are in the splitting ?eld of f2(x)). So the Galois group of f(x)is the Galois group of f2(x). Thus,Gal(f(x)) = S3 if D1 D2 = a square in Q:Assume D1 D2 6= a square in Q and D2 6= a square in Q: Then from thecubic section D2 6= a square in Q implies jGal(f2(x))j = 6 (i.e. Gal(f2(x)) =S3). Now let E = L1 \ L2 - the intersection of the splitting ?elds of f1(x)and f2(x); respectively. The only sub?elds of L1 = Q(pD1) are itself andQ. Thus, E = Q(pD1) or Q.However, using the same result from [6] as before, D1 D2 6= a square in Qimplies D1 dk2 6= n2 for any n 2 Q. That is, D1 d 6= a square in Q. So42we haveD1 d 6= a square in Q)pD1 pd = Q)pD1 = Q(pd)) Q(pD1) * Q(pd) aspD1 generates Q(pD1)) Q(pD1) 6= Q(pd)) Q(pD1) * splitting field of f2(x) (see remark at the end of proof).If Q(pD1) * L2 then E 6= Q(pD1). Thus, E = Q. So using Corollary 2.3,we getGal(f(x)) = Z2 S3 if both D2 and D1 D2 6= a square in Q: 43Remark 2.10. In the above proof, we have Gal(f2(x)) = S3.Now we want to show why the quadratic sub?eld of f2(x) is unique. Firstly,we wish to ?nd the subgroup diagram of the group S3. Recall that S3 is thepermutation group on the set f1;2;3g. We will ?rst list the permutations ofthis set and assign to each a Greek letter for a name. Let = 1 2 31 2 3!= 1 = 1 2 32 3 1!= (1 2 3) = 1 2 33 1 2!= (1 3 2) = 1 2 31 3 2!= (2 3) = 1 2 33 2 1!= (1 3) = 1 2 32 1 3!= (1 2):We will now show the multiplication table for S3 (Table 2.3).Table 2.3. Group Table for the Symmetric Group of Order 6 Note that = and = 2 (recall for permutations you read right toleft).Now we want to ?nd all the possible subgroups of S3. Since the orderof S3 is 6, any subgroup must have order dividing 6. Thus, the possible44orders for the subgroups of S3 are 1;2;3 and 6: The subgroup of order 1 isthe trivial subgroup f g. Remembering that each subgroup has to containthe identity permutation and must be closed under function composition,we get the following subgroups of S3:h i = f g = 1h i = f ; ; g = h ih i = f ; g , 2 = f ; 2gh i = f ; g ,h i = f ; gh i = f ; gh ; i = f ; ; ; ; ; g = S3:Using these subgroups, we can make a subgroup diagram for S3 (see Figure2.1). When using the Fundamental Theorem of Galois Theory, we invert thesubgroup diagram so it corresponds to the sub?eld diagram of a given ?eld.This is why Figure 2.1 is inverted.Figure 2.1. Subgroups of S3Next we introduce a simple de?nition.De?nition 2.14. Let G be a ?nite group with subgroup H. Then theindex of H in G is de?ned to bejG : Hj = order of Gorder of H = jGjjHj :45Since both G and H are ?nite groups, jGj (or jHj) is the number of ele-ments in G (resp. H). Given a ?eld K that is a ?nite extension of Q,the Fundamental Theorem of Galois Theory states that there is a one-to-onecorrespondence between the intermediate ?elds of K (i.e. ?elds E satisfyingQ E K) and subgroups of the Galois group G of K over Q. That is,8>>>>>>><>>>>>>>:sub?elds Eof Kcontaining FKjEjF9>>>>>>>=>>>>>>>; !8>>>>>>><>>>>>>>:subgroups Hof Galoisgroup G1jHjG9>>>>>>>=>>>>>>>;This theorem implies that the dimension of the intermediate ?eld E over Qwill equal the index of a subgroup of Gal(K=Q). Recall Gal(f2(x)) = S3.Thus, the quadratic sub?eld of the splitting ?eld of f2(x) corresponds to thesubgroup of S3 that has index 2 (and hence order 3). Since there is onlyone subgroup of S3 that has order 3 (see Figure 2.1), the quadratic sub?eld ofthe splitting ?eld of f2(x) is unique. Hence, Q(pd) is the unique quadraticsub?eld of the splitting ?eld of f2(x). Now Q(pD1) is a quadratic ?eld; sinceit is not equal to Q(pd), it is not a quadratic sub?eld of the splitting ?eld off2(x). This is the result we want.In conclusion, there are ?ve possible Galois groups of the irreducible quintic poly-nomial: the group isomorphic to C5, the group isomorphic to D5, the group isomorphicto F20, the group isomorphic to A5 and the group isomorphic to S5.The latter three groups prove challenging when determining whether the ?eld ismonogenic or not. Numerical methods used in the cubic and quartic cases that tookonly minutes to perform on the fastest PC? could take several hours when solvingone of these quintic cases. In this paper, we focus on the former cases - cyclic anddihedral quintic ?elds.2.7. SummaryThe following is a summary of the possible Galois groups for irreducible polyno-mials of the speci?ed degree (Table 2.4).46Table 2.4. Summary of the Possible Galois Groups for Irreducible PolynomialsQuadratic Cubic Quartic QuinticC2 S3 S4 S5C3 A4 A5V = Z2 Z2 F20C4 C5D4 D52.8. Finding the Galois Group of an Algebraic Number FieldGiven an algebraic number ?eld K, we can always ?nd an algebraic number such that K = Q( ) (recall the Primitive Element Theorem). Now given any alge-braic number , we can always ?nd its minimal polynomial A(x). Performing thesetwo calculations - ?nding and its minimal polynomial - is not always easy to do;nevertheless, it can be done.In the previous sections we looked at how to ?nd the Galois group of polynomialsof degree ?ve or less. Using this knowledge, we can now apply it to the problemof classifying an algebraic number ?eld according to the Galois group of its normalclosure (de?ned below).De?nition 2.15. The normal closure of K = Q( ) is Q( = 1; 2;:::; n) where = 1; 2;:::; n are the roots of the minimal polynomial of . That is, adjoin andits conjugates to Q to obtain the normal closure of K.Thus the normal closure of K is the smallest ?eld containing K and has theproperty that any element and its conjugates are also in this ?eld (the latter propertyis termed normal).Recall a ?eld is Galois if it is both separable and normal. We usually do not havethe latter property. Thus, if given an algebraic number ?eld that is not normal, wethen have to consider its normal closure. Because its normal closure is normal, wecan ?nd its Galois group. On the other hand, if an algebraic number ?eld is normal,then it is equal to its normal closure and you can ?nd the Galois group of the ?elddirectly.Now in order to classify ?elds according to their Galois group, we need to ?nd theGalois group of the normal closure of K over Q; in general, this is a hard problem.47We must adjoin and its conjugates to Q to obtain the normal closure of K. Nowthe roots of the minimal polynomial A(x) of are itself and its conjugates. Thus,the splitting ?eld of A(x) is Q( = 1; 2;:::; n). So Gal(A(x)) = Gal(Q( = 1; 2;:::; n)=Q). This means that the Galois group of the minimal polynomial of is the same as the Galois group of the normal closure of K (because the splitting?eld of the minimal polynomial of is equal to the normal closure of K). So ?ndingthe Galois group of the normal closure of an algebraic number ?eld K = Q( ) isequivalent to ?nding the Galois group of the minimal polynomial of :As a note, the Galois group of an algebraic number ?eld refers to the Galois groupof the ?eld? normal closure. The reader should note this when we refer to the Galoisgroup of an algebraic number ?eld as Gal(K=Q): The steps to ?nd the Galois groupof an algebraic number ?eld are outlined below:1: Find an algebraic number such that K = Q( ):2: Find the minimal polynomial A(x) of :3: Find the Galois group of A(x):4: Then Gal(Q( = 1; 2;:::; n)=Q) = Gal(A(x)):Like before, we will only be looking at algebraic number ?elds of degree ?ve orless and, in particular, of degree ?ve. We can now see why we focused on the caseswhere we were ?nding the Galois group of an irreducible polynomial f(x). If weare ?nding the Galois group of an algebraic number ?eld, then we are dealing withminimal polynomials which are always irreducible.Example 2.20. Find the Galois group of the algebraic number ?eld Q(p2;p3).That is, ?nd the Galois group of the normal closure of Q(p2;p3):We have seen this ?eld before and have shown Q(p2;p3) = Q(p2 + p3) (Example1.7). Thus, = p2 + p3:Now we must ?nd the minimal polynomial of . From Example 1.3, we found thatthe minimal polynomial of was x4 10x2 + 1. Thus,A(x) = x4 10x2 + 1.48Now we must ?nd the galois group of A(x). Since A(x) is a quartic, we examine itscubic resolvent. We getr(x) = x3 + 10x2 4x 40 = (x + 10) (x 2) (x + 2).Thus, r(x) is reducible over Q and moreover, splits into linear factors over Q. So byTheorem 2.3 (case 3), Gal(A(x)) = V , the Klein 4-group, and soGal(Q(p2;p3)=Q) = V:Example 2.21. Find the Galois group of the algebraic number ?eld Q(p2 + p2):In this example, our algebraic number ?eld is already of the form Q( ), where =p2 + p2: We must ?nd the minimal polynomial of . We did so in Example 1.4.The minimal polynomial of isA(x) = x4 4x2 + 2:Now we must ?nd the Galois group of A(x). Since A(x) is a quartic, we examine itscubic resolvent. We getr(x) = x3 + 4x2 8x 32 = (x + 4) (x2 8):Thus, r(x) is reducible over Q. This implies that Gal(A(x)) = C4;D4 or V . Becauseour algebraic number ?eld is of the form Q(pa + bpc), we will use the alternatemethod (Theorem 2.4) to ?nd the Galois group. We calculatea2 b2c = 22 122 = 2 12 = c k2 where k = 1 2 Z:Thus, Gal(A(x)) = C4 and soGal(Q(q2 + p2=Q) = C4:49Chapter 3Integral and Power Bases3.1. Integral BasesDe?nition 3.1. The set of all algebraic integers that lie in the algebraic number?eld K is denoted by OK. In fact, OK forms a ring and thus is called the ring ofintegers of the algebraic number ?eld K.Remark 3.1. In fact, OK is an integral domain.So the ring of integers of an algebraic number ?eld K consists of those numbersin K whose monic minimal polynomials have integer coe? cients.De?nition 3.2. An integral basis of an algebraic number ?eld K of degree n is aset f 1;:::; ng of n algebraic integers in K such that every element in OK can bewritten uniquely in the form c1 1 + ::: + cn n where ci 2 Z for i = 1;::;n.Note that a basis of OK is an integral basis of K. Thus, if we are able to ?nd anintegral basis of K then we know that OK consists of all integral linear combinationsof the elements in the integral basis of K.Now we can always make a slight adjustment in an integral basis of K so that 1 = 1 (Theorem 13, pp. 36-38, [19]). Another result worth mentioning is that everyalgebraic number ?eld has an integral basis (Theorem 2.10, pp. 55, [23]). However,these bases are hard to ?nd. We can explicitly compute such a basis but it is nowstandard for computer algebra systems such as MAPLETM to have built-in programsto do this.We ?rst will present a theorem that will give us a basis of the ring of integers for aquadratic ?eld. Afterwards, we discuss how to ?nd integral bases for ?elds of higherdimension.50Theorem 3.1. Let K be the quadratic ?eld Q(pm) such that m is a squarefreeinteger not equal to 1. If t 2 Z then OK is given byOK =(Z + Zpm = fa + bpm j a;b 2 Zg if m 6= 4t + 1Z + Z(1+pm2 ) = fa + b(1+pm2 ) j a;b 2 Zg if m = 4t + 1:Proof. We prove this theorem using a dual-inclusion argument. First we show part of the dual-inclusion argument.Assume m 6= 4t + 1 for any t 2 Z: Let = a + bpm 2 Z + Zpm where a;b 2 Z.Then = a bpm: So the minimal polynomial of isf(x) = (x (a + bpm)) (x (a bpm))= x2 2ax + (a2 mb2):Since a;b;m 2 Z, f(x) 2 Z[x]. Thus 2 OK. So Z + Zpm OK:Now assume m = 4t + 1 for some t 2 Z: Let = a + b(1+pm2 ) 2 Z + Z(1+pm2 ) wherea;b 2 Z: Then = a + b(1 pm2 ). So the minimal polynomial of isg(x) = (x (a + b(1 +pm2 )) (x (a + b(1 pm2 ))= x2 + ( 2a b)x + (a2 + ab b2t) as m = 4t + 1:Since a;b;t 2 Z, g(x) 2 Z[x]. Thus 2 OK: So Z + Z(1+pm2 ) OK. So we haveshown part of the dual-inclusion argument.Now we show the part of the dual-inclusion argument. Let 2 OK: We mustshow is in either Z+Zpm or Z+Z(1+pm2 ). Since OK K, 2 K = Q(pm). So = a + bpm where a;b 2 Q.If b = 0 then is an algebraic integer if and only if = a 2 Z and the result holds.Assume b 6= 0: The minimal polynomial of isf(x) = (x (a + bpm)) (x (a bpm))= x2 2ax + (a2 mb2)51where a;b 2 Q. Since 2 OK, this implies f(x) 2 Z[x]. So we conclude that(2a 2 Za2 mb2 2 Z:Thus, (2a)2 4(a2 mb2) 2 Z (as Z is closed under addition and multiplication).Simplifying yields(2b)2m 2 Z:Since m is squarefree (by assumption), 2b 2 Z. So we now have 2a 2 Z and 2b 2 Z.Thus = a + bpm ) = 12(c + dpm) where c;d 2 Z:Case #1: c and d are both evenThen = 12(2c1 + 2d1pm) = c1 + d1pm for some c1;d1 2 Z.Thus, 2 Z + Zpm and the result holds.Case #2: c is even and d is oddThen = 12(2c1 + (2d1 + 1)pm) = c1 + d1pm +pm2 for some c1;d1 2 Z.Now the minimal polynomial of d1pm is x2 d21m 2 Z[x]. So d1pm 2 OK. But and c1 are both in OK and since OK is closed under addition, it follows thatpm2 2 OK:This implies m4 2 OK. But m is squarefree and hence m4 = Z.But Q\ OK = Z. Since m4 2 Q and m4 2 OK this implies m4 2 Z; which contradictsthe above. Therefore, this case is not allowable.Case #3: c is odd and d is evenThis case is also not allowable using a similar argument to Case #2.Case #4: c and d are both oddThen = 12((2c1 +1)+(2d1 +1)pm) = c1 +d1pm+ 12 +pm2 for some c1;d1 2 Z. Nowwe have seen that d1pm 2 OK and we know and c1 are both in OK: By closure,12 +pm2 2 OK. So the minimal polynomial of12 +pm2 must be in Z[x] and soh(x) = (x (12 +pm2 )) (x (12 pm2 ))= x2 x + 1 m4 2 Z[x]:52So we require 1 m4 2 Z. That is,1 m4 = n for some n 2 Z,m = 4t + 1 for some t 2 Z:Now recall that = 12((2c1 + 1) + (2d1 + 1)pm). We wish to show 2 Z+ Z(1+pm2 )(as m = 4t + 1): So we set12((2c1 + 1) + (2d1 + 1)pm) = u + v(1 + pm2 )and show u and v are in Z: The above equation yields,(u = c1 d1 2 Zv = 2d1 + 1 2 Z:Thus, = (c1 d1) + (2d1 + 1)(1+pm2 ) 2 Z + Z(1+pm2 ) where m = 4t + 1 and so theresult holds and the proof is complete. 3.1.1. Minimal IntegersWe ?rst need to go over some theory regarding minimal integers before we give anexplicit method used to ?nd an integral basis of K. Further information regardingproofs of theorems and derivations can be found on pp. 160-170, [2].First, let K be an algebraic number ?eld of degree n and let 2 OK such thatK = Q( ). So by Theorem 1.4, every 2 OK can be expressed in the form = a0 + a1 + a2 2 + ::: + an 1 n 1where a0;a1;:::;an 1 2 Q and are uniquely determined by and . If k 2f0;1;2;:::;n 1g such thatak 6= 0 and ak+1 = ak+2 = ::: = an 1 = 0(which implies = a0 + a1 + a2 2 + ::: + ak k) then is called an integer of degreek in .A speci?c case is when a1 = a2 = ::: = an 1 = 0 (i.e. k = 0). This implies that = a0 and we say that is an integer of degree 0 in .53Now for k 2 f0;1;2;:::;n 1g, de?ne the set Sk bySk = ak 2 Q j a0 + a1 + ::: + ak k 2 OK for some a0;a1;:::;ak 1 2 Q .So we haveS0 = fa0 2 Q j a0 2 OKg = ZandZ Skfor k = 0;1;2;:::;n 1. Also, it can be shown that Sk has a least positive element,say a k. Since S0 = Z, we know a 0 = 1.De?nition 3.3. With the proceeding notation, any integer of K that is of theforma0 + a1 + a2 2 + ::: + ak 1 k 1 + a k k,where a0;a1;:::;ak 1 2 Q, is called a minimal integer of degree k in .Now the following theorem gives the form of the element a k.Theorem 3.2. For k 2 f0;1;2;:::;n 1ga k = 1dkfor some dk 2 N.Now there exists a property to the numbers dk and it is that each dk 1 divides itssuccessor dk (for k = 1;2;:::;n 1).Theorem 3.3. For k = 1;2;:::;n 1dk 1 j dk.We are now able to give the form of an integer of degree k in for k = 0;1;2;:::;n 1 and as a result, the form of a minimal integer of degree k in .Theorem 3.4. If is an integer of degree k in then there exists a0;a1;:::;ak 2Z such that = a0 + a1 + a2 2 + ::: + ak 1 k 1 + ak kdk .54In particular, if is a minimal integer of degree k in then there exists a0;a1;:::;ak 1 2Z such that = a0 + a1 + a2 2 + ::: + ak 1 k 1 + kdk .Remark 3.2. The reader should note that the ai? in the ?rst expression aregenerally not the same as the ones in the second expression.We have now reached the climax of this theory to obtain a method of ?nding anintegral basis for an algebraic number ?eld of degree n.Theorem 3.5. Let K be an algebraic number ?eld of degree n. Let 2 OK suchthat K = Q( ). For k = 0;1;2;:::;n 1 let k be a minimal integer in of degreek. Thenf 0; 1;:::; n 1gis an integral basis for K.Remark 3.3. So in order to ?nd an integral basis for an algebraic number ?eldof degree n, we need only ?nd a minimal integer of each degree up to n 1.We now come to one last theorem which will also be useful in ?nding these integralbases.Theorem 3.6. Let K be an algebraic number ?eld of degree n. Let 2 OK suchthat K = Q( ). For k = 0;1;2;:::;n 1 let k be a minimal integer in of degreek of the form k = ak0 + ak1 + ak2 2 + ::: + akk 1 k 1 + kdkwhere ak0;ak1;:::;akk 1 2 Z and 0 = d0 = 1. Thend0d1 dn 1 = ind( ),di j di+1andd2(n i)i j D( )for i = 0;1;:::;n 1.55Remark 3.4. The notation ind( ) means the index of . We shall see more ofthis in a later section. Currently, we are more interested in the second property.Also, recall that D( ) is the discriminant of the algebraic integer which is equal tothe polynomial discriminant of the minimal polynomial of .There may be many values of di that yield an algebraic integer. However, sincewe are ?nding a minimal integer k, we choose the biggest di. That is, the minimalinteger k is one whose denominator is largest yet still satis?es the above theorem.3.1.2. Examples of Finding Integral BasesAs stated before, we know that every algebraic number ?eld has an integral basis.However, explicitly ?nding these bases are di? cult; with the help of computer softwareprograms like MAPLETM, the problem becomes quite easy.Before we do an example of ?nding an integral basis, we need to be able todetermine whether a given element of K is in OK. Now this means, we needto ?nd the minimal polynomial of . If this polynomial has integer coe? cientsthen 2 OK. For the integral basis calculations, given an algebraic number ?eldK = Q( ) where 2 OK, we know for any 2 K that = f( ) for some polynomialf 2 Q[x]. We illustrate how to ?nd the minimal polynomial of via an example.We note that these calculations were performed on MAPLETM as the algebra canprove very challenging.Example 3.1. Let f(x) = x5 5x + 12 and let K = Q( ) where is a root off(x). That is, 2 K whose minimal polynomial is f(x). Given = 2, determinewhether or not is in OK:In order to do this, we must ?nd the minimal polynomial A(x) of . Now the degreeof A(x) must be less than or equal to the degree of K and in fact, must divide thedegree of K (see Remark 3.5): Thus, deg(A(x)) = 5. So letA(x) = x5 + ax4 + bx3 + cx2 + dx + e56where a;b;c;d;e 2 Q. We know that is a root of its minimal polynomial A(x) andsoA( ) = ( 2)5 + a( 2)4 + b( 2)3 + c( 2)2 + d( 2) + e = 0or equivalently,A( ) = 132 5 + a16 4 + b8 3 + c4 2 + d2 + e = 0:Thus, A( ) is a polynomial of which is a root. So letA(x) = 132x5 + a16x4 + b8x3 + c4x2 + d2x + e:So by Lemma 1.1, f(x) j A(x). Or, equivalently,A(x) 0 mod f(x):So we now compute the remainder of A(x) divided by f(x) and set its coe? cientsequal to zero. Using MAPLETM, we obtain the following:a = 0b = 0c = 0d = 516e = 38:Thus the minimal polynomial of isA(x) = x5 516x + 38 = Z[x]:Thus, = OK.Remark 3.5. Since 2 K we adjoin to Q to get the following sub?eld diagram:KjQ( )jQ57Using multiplicative property of ?elds, we get[K : Q] = [K : Q( )] [Q( ) : Q]:That is,[Q( ) : Q] j [K : Q] .But deg(A(x)) = [Q( ) : Q] and so the degree of A(x) divides the degree of K.We now illustrate via example how to ?nd an integral basis for an algebraic number?eld K by using Theorems 3.5 and 3.6 (this method is also provided in detail withproof on pp. 36-38, [19]). For e? ciency, the search for algebraic integers in thisexample was done using MAPLETM.Example 3.2. Let K = Q( ) where is a root of f(x) = x5 5x+12. We wantto ?nd an integral basis for K. Firstly, f(x) is irreducible. We can check this byusing the factor() command in MAPLETM. We also showed f(x) was irreducible inExample 2.3. So K is a quintic ?eld (i.e. n = 5). In fact, we have seen in Example2.16 that K is a quintic ?eld with the galois group of the normal closure of K equalto D5. The discriminant of f(x) isD = (2)12 (5)6.Thus,D( ) = (2)12 (5)6.By Theorems 3.5 and 3.6, an integral basis for K has the form 1; a10 + d1; a20 + a21 + 2d2 ;a30 + a31 + a32 2 + 3d3 ;a40 + a41 + a42 2 + a43 3 + 4d4 where the positive integers d1;d2;d3;d4 satisfyd1 j d2 j d3 j d4,(d1)8 j (2)12 (5)6,(d2)6 j (2)12 (5)6,(d3)4 j (2)12 (5)6,(d4)2 j (2)12 (5)6.58Firstly, we have d1 = 1 or 2. Trying d1 = 2 yields 1 = a10 + 2where a10 = 0 or 1 (see Remark 3.6). However, neither 2 nor +12 is an algebraicinteger of K = Q( ) (method to determine this is seen in last example). Since each k 2 OK, we must have d1 = 1. So a10 = 0 and 1 = 2 OK:Secondly, we have d2 = 1;2;4;5;10 or 20. We ?rst try the prime possibilities for d2.Trying d2 = 5 yields 2 = a20 + a21 + 25where 0 a20 4 and 0 a21 4. Like before, we loop through all the possiblecombinations of a20 and a21 to see if any yield an algebraic integer in K; we get none.Thus, we now know that d2 = 10 or 20 also yield no algebraic integers (see Remark3.7). Trying d2 = 2 yields no algebraic integers and so d2 = 4 yields no algebraicintegers. Thus, d2 = 1 (and so a20 = a21 = 0) which implies 2 = 2.Thirdly, d3 = 1;2;4;5;8;10;20 or 40. We ?rst try the prime possibilities for d3.Trying d3 = 5 yields 3 = a30 + a31 + a32 2 + 35where 0 a30 4, 0 a31 4 and 0 a32 4. From these possible choices, we?nd that there are no algebraic integers. Thus, we know that d3 = 10;20 or 40 alsoyield no algebraic integers (similar argument as Remark 3.7). So now we try d3 = 2which yields 3 = a30 + a31 + a32 2 + 32where 0 a30 1, 0 a31 1 and 0 a32 1. Looping through these possiblechoices, we ?nd an algebraic integer 3+ 2 . Since d3 = 4 and 8 yield no algebraicintegers, d3 = 2 and so 3 = 3 + 2 .59Now we consider d4. This is the hardest of the four cases as it yields the greatestnumber of possibilities to test. We have 27 non-trivial choices for d4. Since d3 = 2,by Theorem 3.3, d4 6= 5. Using a similar argument as Remark 3.7, we can excludethe possible choices of d4 that are divisible by 5. So we now try the other primechoice, that is, d4 = 2. We ?nd an algebraic integer 4+ 3+ 2+ 2 . However, d4 = 4also yields an algebraic integer 4+ 3+ 2+ 4 . Since none of the other larger choices ford4 yield an algebraic integer, it follows that d4 = 4 and 4 = 4 + 3 + 2 + 4 .So an integral basis for K is 1; ; 2; 3 + 2 ; 4 + 3 + 2 + 4 :Remark 3.6. When choosing to examine the possible choices for each di, weignore the trivial case di = 1 and examine the others ?rst, in particular, the primepossibilities of di.Also, we need only consider the values of ak0;ak1;:::;akk 1 mod dk. For example,when trying the case d2 = 5, we consider only values mod 5 for a20 and a21. SinceOK is a ring, we can subtract ocurrency1 integers or integer multiples of and still remain inOK. Choosing these appropriately has the ecurrency1ect of reducing the coe? cients modulothe denominator.Remark 3.7. We found that d2 = 5 yields no algebraic integers. That is, 2 = a20 + a21 + 25 =2 OK.By way of contradiction, assume that 2 = a20 + a21 + 210 2 OK.That is, assume d2 = 10: Since OK is closed under multiplication, 2 = 2 a20 + a21 + 210 =a20 + a21 + 25 2 OKwhich contradicts our earlier result. Thus, d2 6= 10. We can do a similar argumentfor d2 = 20.60Now this method yields itself nicely to computer software programs. Since thereis only a ?nite number of possibilities, MAPLETM can easily loop through the choicesfor ak0;ak1;:::;akk 1 to determine whether an algebraic integer is revealed. Smallerdegrees are easy to do by hand but require many calculations.Thus, from now on, we use MAPLETM to determine an integral basis of an al-gebraic number ?eld. We do so by using the integral_basis() command. Thiscommand computes an integral basis for an algebraic number ?eld K (and hence abasis of OK). If we de?ne K by Q( ) where 2 OK and has minimal polynomialA(x) then we pass in the polynomial A(x) in the integral_basis() command. Notethat in order to use this command, one must ?rst open the with(numtheory) package.3.2. Field DiscriminantWe ?rst describe the discriminant of a set of elements of an algebraic number ?eldK.De?nition 3.4. Let K be an algebraic number ?eld of degree n. If f 1;:::; ngis a subset of the ?eld K then we may de?ne a symmetric integral matrix M whose(i;j)-entry is de?ned to be Tr( i j). The discriminant of f 1;:::; ng is thende?ned as ( 1;:::; n) = det(M):The proof of the next proposition can be found on page 15, [5] (Proposition 2.3).Proposition 3.1. ( 1;:::; n) 6= 0 if and only if f 1;:::; ng is a basis of Kover Q.The proof of the next theorem can be found on page 127, [2] (Theorem 6.4.4).Theorem 3.7. Let K be an algebraic number ?eld of degree n.(1) If 1;:::; n 2 K then ( 1;:::; n) 2 Q.(2) If 1;:::; n 2 OK then ( 1;:::; n) 2 Z.De?nition 3.5. Let K be an algebraic number ?eld of degree n. If f 1;:::; ngis an integral basis of the ?eld K then the ?eld discriminant d(K) of K is de?ned asd(K) = ( 1;:::; n):61Remark 3.8. It is important to note that the value of the discriminant is inde-pendent of the integral basis that you choose (see Proposition 2.4, pg 16, [5]).Since the ?eld discriminant d(K) = ( 1;:::; n) where f 1;:::; ng is an inte-gral basis, it follows from Theorem 3.7, d(K) 2 Z.Now ?nding the ?eld discriminant is not an easy task. Using the above de?nition,it would require ?nding the trace of the product of two algebraic numbers a ?nitenumber of times. However, in order to do this, we ?rst need to ?nd an integralbasis of K which, as seen previously, is not very easy to do by hand. We ?rst willexamine the ?eld discriminant of a quadratic ?eld then give a method to ?nd the?eld discriminant of a ?eld K of higher dimension that does not depend on the tracefunction.Theorem 3.8. Let K be a quadratic ?eld Q(pm) such that m is a squarefreeinteger not equal to 1. Then the ?eld discriminant d(K) of K is given byd(K) =(4m if m 6= 4t + 1m if m = 4t + 1where t 2 Z.Proof. From Theorem 3.1, we haveOK =(Z + Zpm = fa + bpm j a;b 2 Zg if m 6= 4t + 1Z + Z(1+pm2 ) = fa + b(1+pm2 ) j a;b 2 Zg if m = 4t + 1:That is, an integral basis of OK is f1;pmg if m 6= 4t + 1 or f1; 1+pm2 g if m = 4t + 1:Using the de?nition of the ?eld discriminant, we wish to form the matrix M. Assumem 6= 4t + 1 and let 1 = 1 and 2 = pm. Then f 1; 2g is a basis of OK. So toform the matrix M we wish to ?nd Tr( i j). Since K has degree 2, we haveTr( 1 1) = Tr(1) = 1 + 1 = 2Tr( 1 2) = Tr(pm) = pm + ( pm) = 0Tr( 2 1) = Tr(pm) = pm + ( pm) = 0Tr( 2 2) = Tr(m) = m + m = 2m:62SoM ="2 00 2m#and so d(K) = det(M) = 2 2m 0 = 4m:Assume m = 4t + 1 and let 1 = 1 and 2 = 1+pm2 . Then f 1; 2g is a basis of OK.So to form the matrix M we wish to ?nd Tr( i j). So we haveTr( 1 1) = Tr(1) = 1 + 1 = 2Tr( 1 2) = Tr(1 +pm2 ) =1 + pm2 +1 pm2 = 1Tr( 2 1) = Tr(1 +pm2 ) =1 + pm2 +1 pm2 = 1Tr( 2 2) = Tr((m + 1) + 2pm4 ) =(m + 1) + 2pm4 +(m + 1) 2pm4 =m + 12 :SoM ="2 11 m+12#and so d(K) = det(M) = 2(m+12 ) 1 = m. This completes the proof. We now will look at the ?eld discriminant of ?elds of higher dimension. Firstwe recall two dicurrency1erent discriminants: the discriminant of a polynomial f(x) and thediscriminant of an algebraic number ?eld K. The latter was described in De?nition3.5. The former was equal to the product of the squares of the dicurrency1erences of theroots of f(x). Using these two discriminants, we introduce the following de?nition.De?nition 3.6. The index of an algebraic number is de?ned byind( ) =sD(f)d(K)where f is the minimal polynomial of and d(K) is the ?eld discriminant of thealgebraic number ?eld K formed by adjoining Q with .The reader may recognize the notation ind( ). We ?rst introduced this notion inTheorem 3.6. In fact, we will now make use of the ?rst property that was found inthat theorem. Recalld0d1 dn 1 = ind( )63where d0;d1; ;dn 1 are the denominators present in an integral basis for the alge-braic number ?eld K. Now we saw that ?nding an integral basis proved di? cult with-out the aid of a computer software program. However, with the use of MAPLETM,we can ?nd an integral basis using only the integral_basis() command.Also, ?nding a polynomial discriminant is quite straightforward. We have pro-vided the reader with a number of formulas to do so that vary according to thedegree of the polynomial. Thus, using De?nition 3.6, we can now calculate the ?elddiscriminant d(K) of the algebraic number ?eld K. We haved(K) = D(f)(ind( ))2 .Now we provide the reader with two examples illustrating the calculation of the?eld discriminants used in Examples 2.16 and 2.17, respectively.Example 3.3. Let f(x) = x5 5x + 12:If is a root of f(x) then let K = Q( ). We wish to ?nd the ?eld discriminant ofK. First we calculate the discriminant of f(x) to getD = (2)12(5)6.Now we need to ?nd an integral basis for K. We did so in Example 3.2 and found 1; ; 2; 3 + 2 ; 4 + 3 + 2 + 4 to be an integral basis for K. Thusind( ) = d0d1d2d3d4 = 1 1 1 2 4 = (2)3:So the ?eld discriminant for K isd(K) = (2)12(5)6((2)3)2 = (2)6(5)6.Example 3.4. Let f(x) = x5 110x3 55x2 + 2310x + 979.If is a root of f(x) then let K = Q( ). We wish to ?nd the ?eld discriminant ofK. The discriminant of f(x) isD = (5)20(11)4.64Using MAPLETM, we ?nd 1; + 45 ; 2 + 3 + 2125 ; 3 + 2 2 + 18 + 104125 ; 4 + 3 + 16 2 + 86 + 521625 to be an integral basis of K. Thusind( ) = d0d1d2d3d4 = 1 5 25 125 625 = (5)10:So the ?eld discriminant of the ?eld K isd(K) = (5)20(11)4((5)10)2 = (11)4.3.3. Power BasesAs stated before, an integral basis for an algebraic number ?eld K is a basis forOK. Moreover, there is a special case in which the basis is known as a power basis.These speci?c bases are very rare and also hard to ?nd.De?nition 3.7. Let K be an algebraic number ?eld (not necessarily equal toQ( )) of degree n. If there exists an element 2 OK such that 1; ; 2;:::; n 1 is an integral basis for K then K is said to be monogenic and the integral basis 1; ; 2;:::; n 1 is called a power basis for K.Now we can ?nd power bases based on the index of the element as seen in thefollowing theorem.Theorem 3.9. Let K be an algebraic number ?eld of degree n. Let 2 OK suchthat K = Q( ). Then 1; ; 2;:::; n 1 is an integral basis for K if and only ifind( ) = 1.Thus, if ind( ) = 1 then 1; ; 2;:::; n 1 is a power basis for K.We now introduce one more theorem that will help us in the search for powerbases.Theorem 3.10. Let K be an algebraic number ?eld of degree n. Let 2 OKsuch that K = Q( ). If D( ) is squarefree then 1; ; 2;:::; n 1 is a power basisfor K.65Remark 3.9. Theorem 3.10 is a consequence of Theorem 2.17, pp. 48, [30].Note that in Theorems 3.9 and 3.10, is a generator of K.Remark 3.10. The proofs for Theorems 3.9 and 3.10 are found on pp. 146-147,[2].We also remind the reader that a number is said to be squarefree if its prime decom-position contains no repeated factors. Thus, all primes are trivially squarefree.We now provide the reader with an example illustrating the use of Theorem 3.10.We shall use these theorems, in particular Theorem 3.9, in a later chapter.Example 3.5. Let f(x) = x3 + x + 1.We can use the Rational Root Theorem to show f(x) is irreducible over Q. Let Kbe the algebraic number ?eld Q( ) where is a root of f(x). Thus, K has degree 3.Since f(x) 2 Z[x], 2 OK.Now we must calculate D( ). We recall that the discriminant of an algebraic integeris equal to the polynomial discriminant of its minimal polynomial. Using MAPLETM,we getD( ) = D(f) = 31:Since D( ) is squarefree, by Theorem 3.10, 1; ; 2 is a power basis for K. Thus,K is monogenic.We must warn the reader when using Theorem 3.9. If ind( ) 6= 1 then thisdoes not mean K does not admit a power basis. It just means that using the speci?calgebraic integer , we do not have a power basis. Thus, in determining whether ornot K is monogenic, we want to examine a typical element in OK and examine itsindex (rather than a speci?c element). In order to use the theorem, we impose therestriction that the element in OK must be a generator of K. We will examine thismethod in detail in the following subsection.663.3.1. Index FormSuppose K is an algebraic number ?eld with integral basis f1; 2; 3;:::; ng . Let be a typical element in OK: Then we can write = X1 + X2 2 + X3 3 + ::: + Xn nfor some X1;X2;:::;Xn 2 Z. As a note, in order to use Theorem 3.9, we needK = Q( ). That is, must be a generator of K. Now in order to compute ind( );?rst we must ?nd the minimal polynomial A(x) of (see Example 3.1). Secondly, we?nd its polynomial discriminant D(A). Lastly, we ?nd the ?eld discriminant d(K)of K to getind( ) = sD(A)d(K) :The above equation will be a function of the n variables X1;X2;:::;Xn and is knownas the index form. Because power bases are equivalent if they dicurrency1er by an integer (pp.80, [23]), we can actually drop the X1 term in = X1 + X2 2 + X3 3 + ::: + Xn n.This then would leave us an index form that is a function of n 1 variables.In order to determine monogenicity of K, we want to solve the equation (over Z)ind( ) = sD(A)d(K) = 1for X2;:::;Xn. This equation is known as the index form equation. Solving thisequation is best suited for a computer software program. We illustrate, in detail,this method in a later chapter.We now introduce two more de?nitions and theorems to complete this chapter.De?nition 3.8. The index of K isi(K) = gcd find( ) j is a generator of Kg :We sometimes refer to i(K) as the ?eld index of K.De?nition 3.9. The minimal index of K ism(K) = min find( ) j is a generator of Kg :67Theorem 3.11. Let K be an algebraic number ?eld. Then m(K) = 1 if and onlyif K possesses a power basis.Theorem 3.12. Let K be an algebraic number ?eld such that K possesses a powerbasis. Then i(K) = 1.Remark 3.11. For proofs and further information regarding the above two theo-rems, see pp. 178-179, [2].The reader should note that the converse of the latter theorem is not true. How-ever, this theorem does have its advantages. The contrapositive states that ifi(K) 2 then K does not contain a power basis.68Chapter 4Cyclic and Dihedral Fields of Prime DegreeAs stated before, we will mainly concentrate on quintic ?elds, in particular, dihe-dral quintic ?elds. However, there are many similarities between cyclic and dihedralquintic ?elds. Thus, we ?rst examine cyclic and dihedral ?elds of prime degree andthen move on to the speci?c case of degree 5.4.1. Field DiscriminantsThe question was asked, "How many cyclic or dihedral ?elds have the same ?elddiscriminant?" This problem prompted researchers to examine the ?eld discriminantin detail, looking for any pattern or structure. In doing so, we have the followingresults ([21]).Theorem 4.1. Let p be an odd prime and K be a ?eld containing Q of degree psuch that the galois group of K is Cp (i.e. the cyclic group of order p). Then the?eld discriminant of K is of the formd(K) = fp 1where f 2 Z. We call f the conductor of K.Remark 4.1. In fact, if the galois group of the normal closure of K is Cp thenthe conductor f of K must be greater than 1 (see pp. 58-66, [23]).Now using this theorem, we see that for cyclic quintic ?elds (i.e. p = 5), its ?elddiscriminant is of the formd(K) = f4where f 2 Z and f > 1. Now we have the veri?cation of the ?rst part of Remark2.8.69Once again, there is a parallel between cyclic and dihedral ?elds of prime degree.So it comes as no surprise that there is a similar theorem for the ?eld discriminant ofa dihedral ?eld of prime degree.Theorem 4.2. Let p be an odd prime and K be a ?eld containing Q of degree psuch that the galois group of the normal closure of K is Dp (i.e. the dihedral groupof order 2p). Then the ?eld discriminant of K is of the formd(K) = fp 1 dp 12where f 2 Z and d is the ?eld discriminant of the quadratic sub?eld of the normalclosure of K. We call f the conductor of K.Remark 4.2. The quadratic sub?eld of K = Q( ) can be found using the methoddescribed on page 156 in [35]. Also, in [21] (pp. 835), the de?nition of the conductorf of a ?eld K with the galois group of the normal closure of K being Dp implies fmust be greater than zero.Now using this theorem, we see that for dihedral quintic ?elds (i.e. p = 5), its?eld discriminant is of the formd(K) = f4 d2where f 2 Z and f > 0. Now we have the veri?cation of the second part of Remark2.8.We now provide an example illustrating both theorems.Example 4.1. Let K be the ?eld formed by adjoining one of the roots of f(x) =x5 5x + 12 to Q.In Example 2.16, we foundGal(f(x)) = D5andd(K) = 1000000 = (2)6 (5)6:Since K is a quintic ?eld with the dihedral group of order 10 as the galois group of itsnormal closure, we know that d(K) = f4 d2: To determine the value of the conductor70f, we must ?rst calculate d. In order to ?nd d, we must work out the generator of thequadratic sub?eld of K. Recall the method can be found in [35]. Omitting details,we give the reader an overview on how to use this method via MAPLETM.First we must de?ne our polynomial f(x). Now substitute the value X + x for x inf(x). We getf(X + x) = (X + x)5 5X 5x + 12:Using the resultant() command, compute the resultant of the two polynomials f(x)and f(X + x) with respect to the indeterminate x. Factor this result to obtainX5(X10+10X8+125X6+500X4+2500X2+4000)(X10 10X8 75X6+1500X4 5500X2+16000):Based on this method, the quadratic sub?eld is of the form Q(p (constant term))where the constant term is the constant present in the factored resultant (note thatthere will always be a constant term present). We have two constants in our resultantand it does not matter which one you use as both yield the same result. That is,Quadratic sub?eld of K = Q(p 4000)= Q(p (2)5(5)3)= Q(20p 2 5)= Q(p 10):Since 10 = 4( 3) + 2 6= 4t + 1 for any t 2 Z, by Theorem 3.8, we have d = 40:Thus, f = 5 as d(K) = f4 d2:Example 4.2. Let K be the ?eld formed by adjoining one of the roots of f(x) =x5 110x3 55x2 + 2310x + 979 to Q.In Example 2.17, we foundGal(f(x)) = C5andd(K) = 14641 = (11)4.Thus, the conductor of K is 11.71Remark 4.3. In the dihedral case, more work is needed to determine the value off and d if there is more than one possible choice of f.Now consider the following scenario. Given a quintic ?eld K, we wish to ?nd thegalois group of its normal closure. Using Theorem 2.7 and assuming the discriminantof the irreducible polynomial de?ning K is a square, we deduce the group is eitherC5 or D5: We wish to know when f4 = d2 f4 where f is the conductor of thecyclic quintic ?eld and f is the conductor of the dihedral quintic ?eld. That is,when could we not distinguish between the cyclic and dihedral quintic cases using?eld discriminants? We answer this question in the next theorem.Theorem 4.3. Let K be a ?eld with the galois group of the normal closure of Keither C5 or D5: Then the ?eld discriminant of K will distinguish between the Galoisgroups C5 and D5:Proof. By way of contradiction, assume there exists f;f 2 Z such that f4 = d2 f4where d is the ?eld discriminant of the quadratic sub?eld of the normal closure of K.Thus,d2 = l4 where l 2 Z:Hence,d = n2 where n 2 Z:But by Theorem 3.8,d =(4m if m is not of the form 4t + 1m if m = 4t + 1where m is a squarefree integer. So d 6= m which implies d = 4m where m 6= 4t + 1:Thus,4m = n2, m = (n2)2, n = 2 as m is squarefree, m = 1 as m 6= 4t + 1:So d = 4.72On pp. 831, [21], the author states that for cyclic ?elds of degree p we havef = pe q1 qtwhere p is an odd prime, e = 0 or e = 2, t 0 and the qi are pairwise distinct rationalprimes satisfyingqi 1 (mod p)for i = 1:::t. We will now use this for p = 5 to getf = 5e q1 qtwhere e = 0 or e = 2, t 0 and the qi are pairwise distinct rational primes satisfyingqi 1 (mod 5):Since d = 4,f4 = d2 f4, f4 = ( 4)2 f4, 42 j f4, 4 j f2, 2 j f:However, 2 2 (mod 5) and hence a contradiction. In regards to the question posed in the beginning of this section, we know that thenumber of cyclic ?elds of prime degree p that share the same conductor f = pe q1 qt(and hence ?eld discriminant) is equal to(p 1)t+w 1where w = 12e. Thus, as the number of prime factors increase, the number of cyclic?elds of prime degree p that share the same conductor also increases.73Mayer also gives an expression for the number of dihedral ?elds of prime degreep that share the same conductor in [21], but it is more complicated than the aboveand beyond the scope of this paper.Now recall that there is only one monogenic cyclic quintic ?eld ([14]). As seen,there is a parallel between the ?eld discriminants of cyclic and dihedral ?elds ofprime degree. Since monogenicity is based on ?eld discriminants (i.e. the index formequation), this leads us to the following question.Question: If there is only one monogenic cyclic ?eld of degree 5, thencan we deduce that there is only one or ?nitely many monogenic dihedral?eld of degree 5?This question is the motivation for this thesis and will be answered in the nextchapter4.2. Abelian and Cyclotomic FieldsThe main focus of this section is to provide the reader with more insight intothe structure of the conductor of a cyclic quintic ?eld. We begin with a few basicde?nitions.De?nition 4.1. An algebraic number ?eld K containing Q is called abelian if Kis both normal and separable over Q and Gal(K=Q) is an abelian group.Recall that a group is abelian if its binary operation is commutative.Example 4.3. The only abelian Galois group for a quintic ?eld is the cyclic Galoisgroup Z5. Thus, any algebraic ?eld K that is both normal and separable containingQ with Gal(K=Q) = Z5 is abelian.We now discuss a few basic principles from complex analysis. First, let n 2 N.Then, over C, there are n distinct roots to the polynomial xn 1; these roots arecalled the nth roots of unity. The collection of these roots form a cyclic group undermultiplication over C (pp. 539, [9]). We call the generator of the group of all nthroots of unity a primitive nth root of unity. We denote such a generator as n.Geometrically, one can ?nd n = e2 in = cos(2 n ) + isin(2 n ):74De?nition 4.2. The ?eld Q( n) is called the cyclotomic ?eld of nth roots of unity.This now leads us to a well-known theorem used in the analysis of abelian ?eldsK containing Q. It is known as the Kronecker-Weber Theorem.Theorem 4.4. Let K be an abelian algebraic number ?eld containing Q. ThenK Q( n) for some n 2 N.It is clear that K is then contained in in?nitely many cyclotomic ?elds. Nowin terms of quintic ?elds, we know that the only abelian algebraic number ?elds Kcontaining Q must have Galois group isomorphic to Z5. That is, for quintic ?elds, wecan only apply the Kronecker-Weber Theorem to cyclic quintic ?elds. So assumingK containing Q is a cyclic quintic ?eld, the least such n such that K Q( n) iscalled the conductor f of the cyclic quintic ?eld.Now we can see why the conductors of cyclic and dihedral ?elds with prime degree(in particular, degree 5) have dicurrency1erent structure (see Remark 2.8). The conductoris based on the structure of the algebraic number ?eld. Cyclic ?elds are abelianwhereas dihedral ?elds are not. Such study of abelian ?elds and their structure isknown as class ?eld theory.75Chapter 5Cyclic and Dihedral Quintic Fields with a Power BasisWe have now reached the main focal point of this paper. It is in this chapter thatwe prove the main result of the abstract. We wish to show that there exist in?nitelymany dihedral quintic ?elds with a power basis. Now that we have discussed thetheory needed to prove this statement, we see that this is no trivial problem. Firstly,we have already seen that integral bases are hard to ?nd. Also, power bases arevery rare so to say that there are in?nitely many particular quintic ?elds that have apower basis seems unlikely. This is one reason why this problem is so interesting.5.1. Some Important LemmasWe now introduce an important parametric family of quintic polynomials whichwas studied by A. Brumer [4] and T. Kondo [17], independently.De?nition 5.1. The parametric family of quintics studied by T. Kondo is thefollowing:Ra;b(x) = x5 + (a 3)x4 + (b a + 3)x3 + (a2 a 1 2b)x2 + bx + awhere a;b 2 Z.Remark 5.1. For our purposes, we set a = 1; in the above expression, to getFb(x) = x5 2x4 + (b + 2)x3 (2b + 1)x2 + bx + 1where b 2 Z.We begin our study on the irreducibility of Fb(x):Lemma 5.1. Fb(x) is irreducible over Q for all b 2 Z.Proof. Firstly, if Fb(x) is irreducible over Zp, where p is a prime, then Fb(x) isirreducible over Z (recall contrapositive of Theorem 2.1). If we let p = 2 then, using76MAPLETM, we obtain the following:Fb(x) x5 + bx3 + x2 + bx + 1 mod 2.We can break down Fb(x) mod 2 into cases (b is even or odd) to getFb(x) =(x5 + x2 + 1 (mod 2) if b 0 mod 2x5 + x3 + x2 + x + 1 (mod 2) if b 1 mod 2.Finally, using MAPLETM and the Factor() mod 2 command, we see that both x5 +x2 + 1 and x5 + x3 + x2 + x + 1 are irreducible over Z2. Thus, Fb(x) is irreducibleover Z for all b 2 Z. So by Gauss?Lemma, Fb(x) is irreducible over Q. Now we look at the polynomial discriminant of Fb(x).Lemma 5.2. The polynomial discriminant of Fb(x) isD(Fb(x)) = (4b3 + 28b2 + 24b + 47)2:Proof. Using MAPLETM and the discrim() command, the result follows. Remark 5.2. We note that the cubic polynomial 4b3+28b2+24b+47 is irreducibleover Q.We now wish to determine the Galois group of Fb(x): We ?rst narrow down thechoices of Gal(Fb(x)) to two possibilities.Lemma 5.3. Gal(Fb(x)) = Z5 or D5.Proof. In order to use the theory of irreducible quintics, we ?rst must rewrite Fb(x)of the form x5 + px3 + qx2 + rx + s by making the substitution Fb(x + 25). We thenmust ?nd the Galois group of Fb(x + 25). Since translations do not acurrency1ect the Galoisgroup of a polynomial, the result will give us the Galois group of Fb(x).77We ?rst determine if the resolvent sextic for Fb(x+ 25) has any rational roots. Theresolvent sextic isf20(x) = 13814697265625(125x 579 + 230b)(648144531250bx3 + 4191965234375bx2+8455564403125bx + 30517578125x5 + 299873046875x3 + 2930874765625x2+12425749097500x 30637508969600b + 86669921875x4 40098357698500b2 85449218750bx4 + 413085937500b2x3 + 61035156250b3x3 29180050973750b3 15391108715625b4 + 446085937500b2x2 363164062500b3x2 151367187500b4x2 5671380700000b5 1077349609375b6 65917968750b7 + 3417867046875b2x+1665324531250b3x + 1521822656250b4x + 532031250000b5x+30517578125b6x 12808965183149).Thus, the resolvent sextic of Fb(x+ 25) has one rational root 579125 46b25 . So we now useTheorem 2.7 to deduce that Gal(Fb(x+25)) = F20;Z5 or D5. So, Gal(Fb(x)) = F20;Z5or D5. But by Lemma 5.2,D(Fb(x)) = (4b3 + 28b2 + 24b + 47)2 = square in Z:However, D(Fb(x)) = D(Fb(x+ 25)) and so Gal(Fb(x+ 25)) F20. Thus, Gal(Fb(x)) F20 and the result follows. Based on the polynomial discriminant of Fb(x), we can deduce the exact Galoisgroup of Fb(x).Lemma 5.4. If 4b3 + 28b2 + 24b + 47 is squarefree then Gal(Fb(x)) = D5.Proof. Assume 4b3 + 28b2 + 24b + 47 is squarefree.By way of contradiction, assume Gal(Fb(x)) D5. Then by Lemma 5.3, Gal(Fb(x)) =Z5: Let 2 C be a root of Fb(x) and set K = Q( ). Since Fb(x) is irreducible (Lemma5.1) and Gal(Fb(x)) = Z5, K is a quintic ?eld with the galois group of its normalclosure being Z5. Hence, the ?eld discriminant d(K) of K is of the form f4 for some78f > 1; f 2 Z (see Remark 4.1). So(4b3 + 28b2 + 24b + 47)2 = D(Fb(x)) (by Lemma 5.2)= g2 d(K) (by the index form and where g 2 Z)= g2 f4 (by above explanation).So we have4b3 + 28b2 + 24b + 47 = g f2:But by assumption, 4b3 + 28b2 + 24b + 47 is squarefree and so f = 1. This is acontradiction as f > 1. Therefore, Gal(Fb(x)) = D5: We now introduce another Lemma before we prove the abstract.Lemma 5.5. If 4b3 + 28b2 + 24b + 47 is squarefree then the quadratic sub?eld ofthe splitting ?eld of Fb(x) isQ(p 4b3 28b2 24b 47):Proof. Using Lemma 5.4, the proof is deduced from [29] (pp. 4748). We now state an important theorem, also known as Erd?s?Theorem ([11]).Theorem 5.1. Let f(x) 2 Z[x] a polynomial whose coe? cients have highest com-mon factor 1 (f(x) is thus called primitive). Let l 3 denote the degree of f(x).Also assume that the leading coe? cient in f(x) is positive and that f(x) has no fac-tors repeated (l 1) times. Lastly, assume the following condition holds: if l is apower of 2 then there exists some n such thatf(n) 6= 0 (mod 2l 1):Then there are in?nitely many positive integers n for which f(n) is (l 1)-th powerfree.In this paper, we are interested in the case when l = 3.Lemma 5.6. There are in?nitely many integers b such that 4b3 + 28b2 + 24b + 47is squarefree.79Proof. Let f(x) = 4x3 + 28x2 + 24x + 47 2 Z[x]. Then f(x) is primitive asgcd(4;28;24;47) = 1 (as 47 is prime) and all coe? cients are positive (and hencethe leading one is positive). Since f(x) is irreducible over Q, f(x) has no factorsrepeated 2 times.Let l = 3. Since l is not a power of 2 we do not have to ?nd an n such that f(n) 6= 0(mod 22). So by Erd?s?Theorem, there are in?nitely many positive integers n forwhich f(n) is squarefree and the result follows. We now have the tools to prove the central theorem of this paper.5.2. Proof of Main ResultTheorem 5.2. There are in?nitely many integers b such that the dihedral quintic?elds Q( ), where 5 2 4 + (b + 2) 3 (2b + 1) 2 + b + 1 = 0,are distinct and monogenic.This theorem then implies that there are in?nitely many dihedral quintic ?eldswith a power basis.Proof. Recall Fb(x) = x5 2x4 + (b + 2)x3 (2b + 1)x2 + bx + 1 and is irreducible(Lemma 5.1). By Lemma 5.2, D(Fb(x)) = (4b3 + 28b2 + 24b + 47)2. By Lemma5.6, there are in?nitely many integers b such that 4b3 + 28b2 + 24b + 47 is squarefree.Thus, by Lemma 5.4, Gal(Fb(x)) = D5 for in?nitely many integers b. Now for eachsuch b, let b be a root of Fb(x) and set Kb = Q( b): Then each Kb is a dihedralquintic ?eld (as Gal(Fb(x)) = D5 and Fb(x) is irreducible).Since each Kb is a dihedral quintic ?eld, the ?eld discriminant of Kb is of the formd(Kb) = f4b d2bwhere db is the ?eld discriminant of the quadratic sub?eld of the splitting ?eld ofFb(x) and fb (2 N) is the conductor of Kb. By Lemma 5.5, the quadratic sub?eld isQ(p 4b3 28b2 24b 47):80Now we will use Theorem 3.8 to determine the ?eld discriminant db of this quadraticsub?eld.Firstly, m = 4b3 28b2 24b 47 = 4( b3 7b2 6b 12) + 1: Since m issquarefree by assumption (and hence m is squarefree), we haved(Q(p 4b3 28b2 24b 47)) = 4b3 28b2 24b 47and sodb = 4b3 28b2 24b 47:Thus,d(Kb) = (4b3 + 28b2 + 24b + 47)2 f4b :Now by Lemma 5.2, D(Fb(x)) = (4b3 + 28b2 + 24b + 47)2 so thatd(Kb) = D(Fb(x)) f4b :Now recall the index form:ind( b) =sD(Fb(x))d(Kb) ,(ind( b))2 = D(Fb(x))d(Kb) :Since (ind( b))2 2 Z+, d(Kb) divides D(Fb(x)) over Z. This implies f4b = 1 and sofb = 1 (as fb 2 N). Thus,d(Kb) = (4b3 + 28b2 + 24b + 47)2 = D(Fb(x)):Consequently,ind( b) = 1:So by Theorem 3.9, f1; b; 2b; 3b; 4bg is a power basis for the ?eld Kb. Hence, each?eld Kb is monogenic. So we have shown that there are in?nitely many integers bsuch that the dihedral quintic ?elds Q( ), where 5 2 4 + (b + 2) 3 (2b + 1) 2 + b + 1 = 0,are monogenic. We now need to show that these ?elds are distinct.Recall that two ?elds are equal if they have the same ?eld discriminant. Sinced(Kb) = (4b3 + 28b2 + 24b + 47)2; we wish to know how many possible values of b81satisfy the given equation for ?xed k 2 Z:(4k3 + 28k2 + 24k + 47)2 = (4b3 + 28b2 + 24b + 47)2;or equivalently,4b3 + 28b2 + 24b + 47 = (4k3 + 28k2 + 24k + 47):Firstly, since k is ?xed and 4b3 + 28b2 + 24b + 47 is a degree 3 polynomial in terms ofb, there can be at most three solutions for the equation4b3 + 28b2 + 24b + 47 (4k3 + 28k2 + 24k + 47) = 0and three solutions for the equation4b3 + 28b2 + 24b + 47 + (4k3 + 28k2 + 24k + 47) = 0:Thus, there are at most six solutions for a given integer k. Recall that we have anin?nite sequence of the integers b such that 4b3 + 28b2 + 24b + 47 is squarefree. Sowe can choose an in?nite subsequence of this sequence such that the ?elds Kb havedicurrency1erent discriminants (as we can have at most six duplications). Thus, we will havean in?nite number of distinct ?elds Kb such that 4b3 + 28b2 + 24b + 47 is squarefree.To conclude, if 4b3 + 28b2 + 24b + 47 is squarefree then the dihedral quintic ?eld Kbhas the power basis f1; ; 2; 3; 4g where we have written for b. This concludesour proof. Remark 5.3. In comparison to the result found in [14], this theorem impliesthat there is not just one or ?nitely many monogenic dihedral ?elds of degree 5. Infact, there are in?nitely many.5.3. Further ResultsWe will now show that the ?eld Kb possess more than one power basis.Claim 5.1. The ?eld Kb possesses at least two more power bases.82Proof. We will show that Kb has the power bases f1;'i;'2i ;'3i ;'4i g for i = 1;2 where'1 = b (b + 1) 2 + 3 4and'2 = (2b + 1) (b + 2) 2 + 2 3 4:Using MAPLETM and the argument of Example 3.1, we ?nd the minimal polynomialsof '1 and '2 to bef1(x) = x5 + x4 + (b + 3)x3 + (b + 4)x2 + 3x + 1andf2(x) = x5 4bx4 + (6b2 2b 1)x3 + ( 4b3 + 6b2 + 4b + 2)x2+(b4 6b3 5b2 4b 2)x + (2b4 + 2b3 + 2b2 + 2b + 1),respectively. We can also use MAPLETM to show the discriminant of these polyno-mials is (4b3 + 28b2 + 24b + 47)2.Since f1(x) and f2(x) are in Z[x] (as b 2 Z), '1;'2 2 OK and soD('1) = D(f1(x)) = (4b3 + 28b2 + 24b + 47)2andD('2) = D(f2(x)) = (4b3 + 28b2 + 24b + 47)2:Thus D(f1(x)) = D(Fb(x)) and D(f2(x)) = D(Fb(x)) and hence are equal to d(Kb).We know that the minimal polynomials of '1 and '2 both have degree 5. Thus[Q('1) : Q] = 5 and [Q('2) : Q] = 5: Since Q('1) and Q('2) are both sub?elds ofQ( ) and [Q( ) : Q] = 5, we must have Q('1) = Q( ) = Q('2). Therefore, both '1and '2 are generators of Kb.Thus, because '1 and '2 are generators of Kb; we haveind('1) =sD(f1(x))d(Kb) = 183andind('2) =sD(f2(x))d(Kb) = 1:So by Theorem 3.9, f1;'i;'2i ;'3i ;'4i g for i = 1;2 are power bases of Kb. Remark 5.4. We found '1 and '2 by using a similar method seen in Claim 5.2(see below) with various values for b. By inspection, we were then able to build upthe formulas for '1 and '2:Claim 5.2. The ?eld Kb possesses an additional eight power bases if b = 0:Proof. Now consider the special case when b = 0.Let 0 be a root of the polynomialF0(x) = x5 2x4 + 2x3 x2 + 1:We know that f1; 0; 20; 30; 40g is a power basis of the ?eld K0 = Q( 0) (as 4b3 +28b2 +24b+47 = 47 is squarefree). In order to ?nd additional power bases, we mustset up and solve the index form equation. This is best suited for a computer softwareprogram so we just list the results.Firstly, we take a typical element in OK, say , and ?nd its minimal polynomial A(x).Since f1; 0; 20; 30; 40g is a power basis, we have = a + b 0 + c 20 + d 30 + e 40where a;b;c;d;e 2 Z. As stated before, we could drop the "a" term in : If wechoose not to, this term will disappear from the expression D(A(x)) yielding the sameresult.Using the argument of Example 3.1, we ?nd the minimal polynomial of . We then?nd and factor the discriminant of A(x), denoted as D(A(x)) using the discrim()command. We knowD(A(x)) = (d(K0)) (ind( ))2:From before,d(Kb) = (4b3 + 28b2 + 24b + 47)284and sod(K0) = (47)2:Thus,ind( ) = rD(A(x))472which is an equation in the four variables b;c;d and e (as D(A(x)) no longer containsthe variable a).Now we wish to ?nd solutions to the above index form equation by looping throughsmall values of b;c and d and seeing if any yield integral values for e. We chose toloop through the integers in the interval [ 10;10] and we got the following results:ind( ) = 1 )8>><>>:e = 2e = 0e = 2andind( ) = 1 )8>><>>:e = 1e = 0e = 1:Now using each value of e, we substitute e back into the corresponding index formequation (i.e. ind( ) = 1 or ind( ) = 1). We then loop through small values of band c looking for integral values of d. Once again, we chose the interval [ 10;10] toget:ind( ) = 1 )8>>>>>>><>>>>>>>:e = 2 )d = 5e = 0 )8>><>>:d = 1d = 0d = 1e = 2 )d = 585andind( ) = 1 =)8>>>>>>>>>>>>><>>>>>>>>>>>>>:e = 1 =)(d = 1d = 2e = 0 =)8>><>>:d = 1d = 0d = 1e = 1 =)(d = 1d = 2:Using the same procedure, we ?nd values for both b and c: We summarize theseresults as follows:ind( ) = 1 )8>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>:e = 2 )d = 5 )c = 7 )b = 6e = 0 )8>>>>>>>>>><>>>>>>>>>>:d = 1 )c = 1 )(b = 1b = 0d = 0 )(c = 1 )b = 1c = 1 )b = 1d = 1 )c = 1 )(b = 0b = 1e = 2 )d = 5 )c = 7 )b = 686andind( ) = 1 =)8>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>:e = 1 =)8>>>>><>>>>>:d = 1 )(c = 1 )b = 0c = 0 )b = 0d = 2 )c = 2 )(b = 1b = 2e = 0 =)8>>>>>>>>>><>>>>>>>>>>:d = 1 )(c = 0 )b = 1c = 2 )b = 1d = 0 )c = 0 )(b = 1b = 1d = 1 )(c = 2 )b = 1c = 0 )b = 1e = 1 =)8>>>>><>>>>>:d = 1 )(c = 0 )b = 0c = 1 )b = 0d = 2: )c = 2 )(b = 2b = 1:87Now we substitute each solution into (remembering that we can drop the "a"term). Disregarding those solutions that are the negation of another solution andthe solutions '1 and '2 we obtain the following values of (letting = 0).'1 = 3 4;'2 = 2 2 2 + 2 3 4;'3 = + 3;'4 = 2 2 + 3;'5 = 6 7 2 + 5 3 2 4;'6 = 2 3;'7 = 2 + 3;'8 = 2:Hence, we have the additional eight power bases f1;'i;'2i ;'3i ;'4i g for 1 i 8. Now the reader might wonder how we knew the Kondo family with a = 1 wouldyield monogenic ?elds. Why did we choose that speci?c a? Is there a way of knowinga parametric ?eld will yield a power basis without going through the long process ofsolving the index form equation?To answer these questions, we illustrate using our family - the Kondo family. The?rst step is to determine the Galois group. Kondo claimed his family of polynomi-als was dihedral under certain restrictions and we con?rmed this already. UsingMAPLETM, we now calculate and factor the polynomial discriminant of the Kondofamily. We obtaina2(4b3 b2 + 30b2a + 14ba 4a b2a2 + 34ba2 + 91a2 24a3b 40a3 4a4 + 4a5)2:Recall for a dihedral ?eld K the ?eld discriminant is of the formd(K) = f4 d2where d is the ?eld discriminant of the quadratic sub?eld of the normal closure of Kand f is called the conductor. Now in order for our ?eld to have a power basis, we88needa2(4b3 b2 +30b2a+14ba 4a b2a2 +34ba2 +91a2 24a3b 40a3 4a4 +4a5)2 = f4 d2(that is, the index of a generator of K is equal to 1): Now from the above equationwe see that we should let a = 1 in order to get rid of the a2 term out in front.Substituting a = 1 into this equation yields(4b3 + 28b2 + 24b + 47)2 = f4 d2:We now want f = 1 so that d = (4b3 + 28b2 + 24b + 47) (which is indeed trueby [29]). But d is the ?eld discriminant of a quadratic ?eld Q(pm) where m issquarefree and hence must be one of the two forms:d =(4m if m is not of the form 4t + 1m if m = 4t + 1:Now looking at d = (4b3+28b2+24b+47) we see that d must be odd (as it is the sumof an even number and an odd number) and hence cannot be of the form 4m. Thus,d is of the form m and so d must be squarefree. We thus require 4b3 +28b2 +24b+47to be squarefree. Thanks to Erd?s, we know that there are in?nitely many integers bsuch that 4b3 +28b2 +24b+47 is squarefree. Hence, we know we will have an in?nitenumber of monogenic ?elds.As a note, a similar theorem as Erd?s?holds true for quadratics. However, forquartics, a parallel theorem has yet to be proved (although most suspect it to betrue).We have discussed both cyclic and dihedral ?elds and the forms of the their ?elddiscriminants. What happens if we ?nd a parametric family of polynomials andthe Galois group is neither of these two? How do we then compare our polynomialdiscriminant to our ?eld discriminant if we do not know the structure of the ?elddiscriminant? These cases can be done but require more experimentation. Firstdetermine the Galois group and polynomial discriminant as before and try to sim-plify the discriminant as much as possible by eliminating one of the parameters (forinstance, if given the previous example, choosing a = 1 would help). Then using the89integral_basis() command in MAPLETM and trying values for the remaining parame-ter, see which ones yield a power basis. Hopefully, there will be in?nitely many valuesthat do. As before, this experimentation helps con?rm your suspicions that you dohave a parametric family yielding monogenic ?elds but is in no way a substitution ofa formal proof.90Chapter 6Number Fields without a Power BasisBecause power bases are so rare, there are plenty of algebraic number ?elds thatdo not have a power basis (i.e. not monogenic). Now we have seen that an algebraicnumber ?eld K is not monogenic if the index form of all generators of K cannot equal 1. But this can be a di? cult condition to use for ?elds of degree six or more as thenumber of variables increase as the degree of the ?eld increases. Thus, we wish toavoid solving this equation and rather ?nd certain properties of the ?eld that allowus to conclude the ?eld is not monogenic.6.1. IdealsWe begin with a few basic de?nitions and theorems.De?nition 6.1. Let K be an algebraic number ?eld and OK its ring of integers.An ideal I of OK is a subring of OK with the property8a 2 OK, aI I:An ideal I of OK is called a proper ideal of OK if f0g I OK.So we say an ideal I of OK is closed under external multiplication by elements ofOK:De?nition 6.2. A proper ideal I of OK is called a prime ideal ifa;b 2 OK and ab 2 I )a 2 I or b 2 I:In OK, every ideal I of OK is ?nitely generated. HenceI = h 1;:::; ki91for some 1;:::; k 2 OK. In fact, for ideals of OK; every ideal can be generated bytwo of its elements (this is known as the two-generator property). Ideals that can begenerated by only one element are known as principal ideals.Now we know that any integer can be written as a unique factorization into primes(Fundamental Theorem of Arithmetic). However, this property does not always holdwhen working over OK - the ring of integers. In truth, it is rare for this property tohold over OK for some algebraic number ?eld K. Here is an example.Example 6.1. Let K = Q(p5i): Clearly, 1 p5i 2 K.Then 1 p5i 2 OK as their minimal polynomial is x2 2x+6 2 Z[x]. We will showthat 1 + p5i is an irreducible over OK.By way of contradiction, assume 1 + p5i is reducible over OK. Since K is degree 2,N(1 + p5i) = (1 + p5i)(1 p5i) = 6.If 1+p5i is reducible then its nontrivial factors would have norms either 2 or 3. Wewill show there are no such elements of OK with such norms. By Theorem 3.1,OK = Z + Zp 5 = fa + bp 5 j a;b 2 Zg:Let = a + bp 5 be a typical element of OK (where a;b 2 Z). ThenN( ) = (a + bp 5)(a bp 5) = a2 + 5b2:If a2 + 5b2 = 2 then b = 0 and so a2 = 2, a contradiction as a 2 Z.If a2 + 5b2 = 3 then b = 0 and so a2 = 3, a contradiction as a 2 Z. Thus 1 + p5i isirreducible over OK. A similar argument can be used to show 1 p5i is irreducibleover OK: Now6 = 2 3 = (1 + p5i) (1 p5i):So there exists two factorizations of 6 2 OK into irreducibles. The above equationimplies 2 j (1+p5i) (1 p5i). We now show that 2 is not prime by a contradictionargument. By way of contradiction, assume 2 j (1 + p5i). Then taking the norm ofeach side yieldsN(2) j N(1 + p5i) ,4 j 6;92a contradiction. Similarly, 2 - (1 p5i). Thus, 2 is not prime. A similar argumentcan be made for 3. Thus, these factorizations are not into primes but irreducibles.When working with OK, we would like to have a unique factorization propertythat holds for any element in OK. This is the reason for working with ideals. Thisleads us to a very important theorem.Theorem 6.1. Let K be an algebraic number ?eld. Then every proper ideal ofOK can be expressed uniquely up to order as a product of prime ideals.Remark 6.1. The proof of this theorem is found on pp. 200-202, [2].Factoring ideals into prime ideals proves a di? cult problem. In [1], the authorsgive a method to ?nd the explicit decomposition of the principal ideal hpi in a cubic?eld, where p is a prime. Rather than go into the details of this method, we list twomajor consequences that the reader should be aware of.(1) The prime numbers p which divide the ?eld discriminant of K are preciselythose primes for which the prime ideal factorization of hpi contains at leastone prime ideal divisor to an exponent greater than 1.(2) Certain prime ideals give rise to common index divisors.The latter consequence is seen in the following section.6.2. Common Index DivisorsOne reason for an algebraic number ?eld K not to have a power basis is becausethe index of every element in OK is divisible by some ?xed prime p. Recall the indexform equationind( ) =sD(A)d(K) = 1where is a typical element in OK. If there exists a ?xed prime p j ind( ) for all 2 OK then ind( ) 6= 1 (as p - 1). This implies that K has no power bases.Such a prime p is called a common index divisor. The common index divisors of analgebraic number ?eld K are precisely the factors of the ?eld index i(K). It is a wellknown result that p < n where n is the degree of the algebraic number ?eld K (seepp. 186, [2]). Thus, to be a common index divisor of a cubic ?eld, p must be 2.93We will do an example illustrating a method to ?nd a common index divisor whenthere exists one.Example 6.2. Let f(x) = x3 x2 2x 8 and let be a root of f(x). LetK = Q( ).Using MAPLETM, we see that K is a cubic ?eld with Galois group S3. Next we ?ndan integral basis for K: Using MAPLETM, we get 1; ; 2 + 2 :Now we let be a typical element in OK: = a + b + c( 2 + 2 )where a;b;c 2 Z. We wish to ?nd the minimal polynomial A(x) of and ?nd itspolynomial discriminant (see Example 3.1). Its discriminant (in factored form) isD(A) = 503( 2c3 + 3bc2 + 5b2c + 2b3)2:Next we need to ?nd the ?eld discriminant of K using the element : Using a similarargument as in Example 3.3 we getind( ) = 2 and D(f) = 2012where f(x) is the minimal polynomial of . This impliesd(K) = D(f)ind( )2 = 201222 = 503.Thus,ind( ) =sD(A)d(K) = 2c3 + 3bc2 + 5b2c + 2b3 :But 2c3 + 3bc2 + 5b2c + 2b3 bc2 + b2c 0 mod 2:Thus,2 j ind( )and so 2 is a common index divisor. Thus, K does not possess a power basis.94Remark 6.2. Firstly, the reader must not get the algebraic integers and confused. is a speci?c element of OK such that K = Q( ) and represents atypical element of OK. Secondly, arriving to the conclusionbc2 + b2c 0 mod 2is not so obvious. We see that bc2 + b2c = bc(c + b). Now an argument by cases willsu? ce. If either b or c is even, then bc2 + b2c 0 mod 2: So we need only considerthe case when b and c are both odd. Then b + c is even and so bc(c + b) 0 mod 2and the result follows.From the above example, we can ?nd that 2c3 + 3bc2 + 5b2c + 2b3 = 2for (b;c) = (1;0) so thatm(K) = min 2c3 + 3bc2 + 5b2c + 2b3 : 2c3 + 3bc2 + 5b2c + 2b3 6= 0 and b;c 2 Z = 2andi(K) = gcd 2c3 + 3bc2 + 5b2c + 2b3 : 2c3 + 3bc2 + 5b2c + 2b3 6= 0 and b;c 2 Z = 2:As i(K) > 1, K does not possess a power basis. We also see that the prime factorof i(K) was indeed the common index divisor of K.95Chapter 7Conclusion and Future Work7.1. Future ResearchWe ?rst introduce the following two de?nitions.De?nition 7.1. A polynomial is called homogeneous if all its terms have the samedegree.Example 7.1. Let f(x) = x6 + 2x3y3 + 9x2y4. Then f(x) is a homogeneouspolynomial of degree 6, in 2 variables; the sum of the exponents in each term isalways 6.De?nition 7.2. Let F 2 Z[X;Y ] (i.e. F has two variables X and Y ): Also, letF be an irreducible homogeneous polynomial of degree n 3. Let m be a non-zerointeger. Then the following equation is known as a Thue equation:F(X;Y ) = m in X;Y 2 Z.Remark 7.1. We can assume the leading coe? cient of F is 1; otherwise, if theleading coe? cient of F is "a", we multiply the equation by an 1, replace either (aX;Y )or (X;aY ) (depending on which variable is the leading term) with (X;Y ) and get therequired condition.For example, consider the Thue equation 3x6 + 2x3y3 + 9x2y4 = 7 then multiply bothsides by 35 to get36x6 + 2 35x3y3 + 359x2y4 = 357:Now we replace (3x;y) with (X;Y ) to getX6 + 2 32X3Y 3 + 339X2Y 4 = 357;an equivalent Thue equation with leading coe? cient equal to 1.96In 1909, Axel Thue ?rst proved that this equation has only ?nitely many integersolutions ([31]). However, Thue? method did not enable one to ?nd the solutions.In 1989, Nikos Tzanakis and Benne de Weger gave a practical method for solving aThue equation by bounding the size of the solution ([32]). The main tool of theirmethod is the lattice reduction algorithm (LLL-algorithm). Other algorithms arealso available to help one solve a Thue equations (see [24] and [25]).In order to determine the monogenicity of a given ?eld, one must ?rst solve anindex form equation. Oftentimes, in cubic and quartic ?elds, these equations can bereduced to simpler types of Diophantine equations, in particular, Thue equations. Ifthe ?eld given is indeed monogenic, with some di? culty, we are then able to ?nd acomplete list of the power bases of this ?eld using those algorithms mentioned above.However, one must keep in mind, that although these algorithms exist to solve Thueequations, it does not mean they are "good", in terms of practicality and e? ciency(see [26]).For quintic ?elds and some ?elds of higher degree, methods exist for resolving theindex form equation into other Diophantine equations called unit equations ([13]).Depending on the Galois group, such methods can be very time consuming. However,they allow us to ?nd complete solutions to the index form equation.In [13], the author states that the number of power bases of a given ?eld is small.We suspect that the list of power bases given in Claim 5.2 is complete. For futureresearch, one might try to prove that this is indeed true. In order to do so, onewould have to use those resolution methods discussed above to reduce the index formequation into simpler types of Diophantine equations. Once reduced, solving theseequations, in principle, is possible.In this paper, we have shown an in?nite family of dihedral quintic ?elds aremonogenic. For future research, one might continue exhibiting in?nite families ofmonogenic ?elds of given Galois group and degree. As mentioned in the introduction,it is not known if there are in?nitely many monogenic cyclic quartic ?elds. Theresearch potential is great in this area of mathematics and the study of power basesis a long term project with considerable scope for sub-projects.977.2. ConclusionAfter building the necessary groundwork, we showed that there are in?nitely manydihedral quintic ?elds that are monogenic and thus, not just one or ?nitely many; thisresult was based on a parametric family of quintic polynomials studied by A. Brumerand T. Kondo, independently. Much work was needed to prove this result, however,experimentation with MAPLETM con?rmed our conjecture. We left the reader witha challenging problem to ?nd a complete list of power bases for the case b = 0.As an introduction, we have discussed various ?elds of given degree and Galoisgroup that also possess power bases; our result is a continuum of these studies. Thereis much research that can be done in this area of mathematics. Finding parametricfamilies that yield monogenic ?elds are hard to come by and constructing your ownfamily proves even more di? cult. However, there is an abundant number of ?eldsof higher degree and given Galois group unexplored and so the research potentialis great. For instance, our focus was on quintic ?elds of which there are only ?vepossible Galois groups. For octic ?elds (degree 8) there are ?fty possible Galoisgroups, much of which have not been analyzed; with the increase in technology, theability to study higher degrees becomes possible and seems less daunting. May thereader be encouraged to study power bases for themselves and delve into this greatsubject.98References[1] S. Alaca, B. K. Spearman and K. S. Williams. Explicit Decomposition of a Ratio-nal Prime in a Cubic Field. International J. of Math. and Math. Sciences (2006),Vol. 2006, Article ID 17641, pp. 1-11.[2] S. Alaca and K. S. Williams. Introductory Algebraic Number Theory. UnitedKingdom: Cambridge Univ. Press, 2004.[3] A. Baker. Notes for 4H Galois Theory 2003-2004. Home page. U. of Glasgow.19 June. 2007 <http://www.maths.gla.ac.uk/~ajb>.[4] A. Brumer, preprint.[5] R. Chapman. Algebraic Number Theory - Summaryof Notes. Home page. U. of Exeter. 10 Sept. 2007<http://www.secamlocal.ex.ac.uk/people/stacurrency1/rjchapman/rjc.html>.[6] H. Cohen. A Course in Computational Algebraic Number Theory. Germany:Springer?Verlag, 1993.[7] R. Dedekind. ?ber den Zusammenhang zwischen der Theorie der Ideale und derTheorie der h?heren Kongruenzen. Abh. Kgl. Ges. Wiss. G?ttingen (1878), Vol.23, pp. 1-23.[8] D. S. Dummit. Solving Solvable Quintics. Math. of Computation (Jul., 1991),Vol. 57, No. 195, pp. 387-401.[9] D. S. Dummit and R. M. Foote. Abstract Algebra. 3rd ed., John Wiley & SonsInc., 2004.[10] D. S. Dummit and H. Kisilevsky. Indices in Cyclic Cubic Fields. Number Theoryand Algebra, Academic Press (1977), pp. 29-42.[11] P. Erd?s. Arithmetic Properties of Polynomials. J. London Math. Soc. (1953)Vol. 28, pp. 416-425.99[12] T. Funakura. On Integral Bases of Pure Quartic Fields. Math. J. of OkayamaUniv. (1984), Vol. 26, pp. 27-41.[13] I. Ga?l. Diophantine Equations and Power Integral Bases. Boston: Birkh?user,2002.[14] M.-N. Gras. Non Monog?n?it? de L? nneau des Entiers des Extensions Cycliquesde Q de Degr? Premier l 5. J. Number Theory (1986), Vol. 23, pp. 347-353.[15] J. G. Huard, B. K. Spearman and K. S. Williams. Integral Bases for QuarticFields with Quadratic Sub?elds. J. Number Theory (1995), Vol. 51, pp. 87-102.[16] L.-C. Kappe and B. Warren. An Elementary Test for the Galois Group of aQuartic Polynomial. American Math. Monthly (Feb., 1989), Vol. 96, No. 2, pp.133-137.[17] T. Kondo. Some Examples of Unrami?ed Extensions over Quadratic Fields. Sci.Rep. Tokyo Woman? Christian Univ. (1997), No. 120-121, pp. 1399-1410[18] M. J. Lavallee, B. K. Spearman, K. S. Williams and Q. Yang. Dihedral Quintic?elds with a power basis. Math. J. of Okayama Univ. (2005), Vol. 47, pp. 75?79.[19] D. A. Marcus. Number Fields. 3rd ed., Springer-Verlag, 1977.[20] A. R. Marshall. Precalculus: Functions and Graphs. 1st ed., Addison-WesleyPub. Co., 1990.[21] D. C. Mayer. Multiplicities of Dihedral Discriminants. Math. of Computation(April, 1992), Vol. 58, No. 198, pp. 831-847.[22] T. Nakahara. 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The 2-Power Degree Sub?eldsof the Splitting Fields of Polynomials with Frobenius Galois Groups. Comm.Algebra (2003), Vol. 31, pp. 4745-4763.[30] I. Stewart and D. Tall. Algebraic Number Theory and Fermat? Last Theorem.3rd ed., Massachusetts: A K Peters, 2002.[31] A. Thue. ?ber Ann?herungswerte algebraischer Zahlen. J. Reine Angew Math(1909), 135, pp. 284-305.[32] N. Tzanakis and B. de Weger. On the Practical Solution of the Thue Equation.J. Number Theory (1989), Vol. 31, pp. 99-132.[33] ? ? ? -. Galois Theory. Wikipedia. 19 June. 2007<http://en.wikipedia.org/wiki/Galois_theory>.[34] ? ? ? -. Cryptography. Wikipedia. 29 August. 2008<http://en.wikipedia.org/wiki/Cryptography>.[35] C. J. Williamson. Odd Degree Polynomials with Dihedral Galois Groups. Diss. Uof California, Berkeley, 1989.101
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Dihedral quintic fields with a power basis Lavallee, Melisa Jean 2008
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Title | Dihedral quintic fields with a power basis |
Creator |
Lavallee, Melisa Jean |
Publisher | University of British Columbia |
Date | 2008 |
Date Issued | 2008-11-14T20:10:42Z |
Description | Cryptography is defined to be the practice and studying of hiding information and is used in applications present today; examples include the security of ATM cards and computer passwords ([34]). In order to transform information to make it unreadable, one needs a series of algorithms. Many of these algorithms are based on elliptic curves because they require fewer bits. To use such algorithms, one must find the rational points on an elliptic curve. The study of Algebraic Number Theory, and in particular, rare objects known as power bases, help determine what these rational points are. With such broad applications, studying power bases is an interesting topic with many research opportunities, one of which is given below. There are many similarities between Cyclic and Dihedral fields of prime degree; more specifically, the structure of their field discriminants is comparable. Since the existence of power bases (i.e. monogenicity) is based upon finding solutions to the index form equation - an equation dependant on field discriminants - does this imply monogenic properties of such fields are also analogous? For instance, in [14], Marie-Nicole Gras has shown there is only one monogenic cyclic field of degree 5. Is there a similar result for dihedral fields of degree 5? The purpose of this thesis is to show that there exist infinitely many monogenic dihedral quintic fields and hence, not just one or finitely many. We do so by using a well- known family of quintic polynomials with Galois group D₅. Thus, the main theorem given in this thesis will confirm that monogenic properties between cyclic and dihedral quintic fields are not always correlative. |
Extent | 530363 bytes |
Subject |
Dihedral quintic fields Power bases Algorithms Algebraic number theory |
Genre |
Thesis/Dissertation |
Type |
Text |
File Format | application/pdf |
Language | eng |
Collection |
Electronic Theses and Dissertations (ETDs) 2008+ |
Date Available | 2008-11-14 |
Provider | Vancouver : University of British Columbia Library |
DOI | 10.14288/1.0066793 |
Degree |
Master of Science - MSc |
Program |
Interdisciplinary Studies |
Affiliation |
Graduate Studies, College of (Okanagan) |
Degree Grantor | University of British Columbia |
Graduation Date | 2008-11 |
Campus |
UBCO |
Scholarly Level | Graduate |
URI | http://hdl.handle.net/2429/2788 |
Aggregated Source Repository | DSpace |
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