Introduction To Algebraic Number Theory Part III

Introduction to Algebraic Number TheoryPart IIIA. S. MosunovUniversity of WaterlooMath CirclesNovember 21st, 2018

RECALLIIIWe learned how to generalize rational integers and today wewill look at the generalization of rational numbers.h i We looked at quadratic rings, like Z[ 2] or Z 1 2 3 , and today we will look at more general rings, like Z[ 3 2].Also, we learned that in certain algebraic rings the uniquefactorization can fail. For example, in Z[ 5]: 2 · 3 (1 5)(1 5).Today we will see how the unique factorization can be fixedwith the theory of ideals.

GENERALIZING RATIONAL INTEGERS

Algebraic Numbers and Their Minimal PolynomialsIA number α is called algebraic if there exists a non-zeropolynomial f (x) with rational coefficients such that f (α) 0.Otherwise it is called transcendental.IFor each algebraic number α there exists the unique minimalpolynomialf (x) cd x d cd 1 x d 1 . . . c1 x c0 .This polynomial satisfies the following five properties:1.2.3.4.5.If (α) 0;c0 , c1 , . . . , cd Z;cd 0;gcd(c0 , c1 , . . . , cd ) 1;f (x) has the smallest degree d among all polynomialssatisfying the conditions 1), 2), 3) and 4).We say that an algebraic number α has degree d, denoteddeg α, if its minimal polynomial has degree d.

Algebraic Numbers and Their Minimal PolynomialsI Example. Considerthenumber2. This number is 2algebraic, since 2 is a root of the polynomial f (x) x 2.In fact, f (x) is the minimal polynomial of 2. Note that it isalso a root off1 (x) 0,f2 (x) 21 x 2 1,f3 (x) x 2 2,f4 (x) x 3 3x 2 2x 6,f5 (x) 6x 2 12.However, none of these polynomials satisfy the definition of aminimal polynomial.

EXERCISES

ExercisesInp o2 3 find aExercise 1. For each α 0, 1/2, i,non-zero polynomial such that f (α) 0 and then determinean upper bound on deg α.IExercise 2. Prove that every rational number has degree 1.IExercise 3. Prove that every quadratic irrational has degree2. In other words, show that every number α of the forma b d, where a, b, d Q and d 6 r 2 for any r Q, satisfiessome non-zero polynomial f (α) 0 of degree 2 and does notsatisfy any polynomial of degree 1.

Number FieldsILet α be an algebraic number of degree d. The setQ(α) {ad 1 α d 1 . . . a1 α a0 : ad 1 , . . . , a1 , a0 Q}is called a number field generated by α.IExample. Gaussian rationals:Q(i) {a bi : a, b Q},where i is a root of x 2 1 0.IExample. Here is the first example of a cubic field: 333Q( 2) {a b 2 c 4 : a, b, c Q}.IEvery field is also a ring: you can add, subtract and multiplythere. However, the division by a non-zero element is nowallowed as well.

Number FieldsIExample. In order to divide two Gaussian rationals, we usethe trick called multiplication by a conjugate. For example,(8 i)(1 i)9 7i9 78 i i.1 i(1 i)(1 i) N(1 i) 2 2In particular, we see that this number is not a Gaussianinteger, so 1 i does not divide 8 i.IExercise 4. Consider the ring of Eisenstein rationals Q(ω),where ω 2 ω 1 0. The number a b bω is called aconjugate of a bω. Note thatN(a bω) a2 ab b 2 (a bω)(a b bω).4 5ωUse multiplication by a conjugate to compute 1 2ωandDetermine whether 1 2ω 4 5ω or 1 2ω 1 4ω.1 4ω1 2ω .

Rings of IntegersIIIAn algebraic number α is an algebraic integer if the leadingcoefficient of its minimal polynomial is equal to 1. Example. The numbers 2, 1 2 3 are algebraic integersbecause their minimal polynomials are x 2 2 and x 2 x 1,respectively. Example. The number cos 2πis not an algebraic integer7because its minimal polynomial is 8x 3 4x 2 4x 1.IFact: The set of all algebraic numbers forms a field, denotedby Q. The set of all algebraic integers forms a ring.ILet α be an algebraic integer. Then the set of all algebraicintegers of Q(α) is called the ring of integers of Q(α). It isdenoted by O.IThe ring O inside a number field Q(α) is a naturalgeneralization of the ring Z inside the field Q.

Rings of IntegersIIII Exercise 5. Show that Q( 5) Q( 5 k) for any integer k. Exercise 6. Show that Z[ 2] is the ring of integers of Q( 2)by proving that every a b 2, where either a or b is not aninteger, necessarily has a minimal polynomial whose leadingcoefficient is greater than 1. Exercise7. Show that Z[ 5] is not the ring of integers of Q( 5) by finding a b 5 Q( 5), where either a or b isnot an integer, whose minimal polynomial has leadingcoefficient equal to 1.Conclusion. The ring of integers O always containsZ[α] {ad 1 α d 1 . . . a1 α a0 : ad 1 , . . . , a1 , a0 Z}but need not be equal to it. Determining the ring of integersof a given number field can be quite difficult.

The Norm MapIEvery number field Q(α) admits a multiplicative norm.IExample. For n a positiverational number, consider z a b n Q( n). Then z a b ni is a complexnumber and its conjugate is z a b n.We define N(z) z 2 zz. Then the multiplicativity of Nfollows from the properties of an absolute value.IExample. Consider the ring of Gaussian integers Z[i]. Thenthe conjugate of a bi is a bi, and soN(a bi) a bi 2 (a bi)(a bi) a2 b 2 . IExercise 8. Let µ 1 2 7 . Determine the conjugate ofa bµ in Z[µ]. Write down the norm map on Z [µ].

General Fields and NormsIMore generally, if α is an algebraic number andf (x) cd x d . . . c1 x c0its minimal polynomial, then the number c0 /cd is precisely thenorm of α.IExample. If a,polynomial of a b are 2integers, the2 minimal2 ). Therefore the normnumbera b 2ax (a 2b2isx on Z[ 2] is N(a b 2) a2 2b 2 .IExample. The norm on 333Z[ 2] {a b 2 c 4 : a, b, c, Z}is 33N(a b 2 c 4) a3 6abc 2b 3 4c 3 .

DETOUR

Detour: Abel-Rufini TheoremIISo far, we have been working with algebraic numbers like0, 23 , i, 1 2 3 , etc. These numbers can be expressed inradicals, i.e. they can be written in terms of addition,subtraction, multiplication, division and root extraction.Degree 2. The solutions to ax 2 bx c 0 are b b 2 4ac b b 2 4acand.2a2aIDegree 3. Cardano’s formula (1545): one of the roots ofx 3 px q isssrr2333qqpqq2 p3x .24272427IDegree 4. There is an analogous formula for degree 4, seethe Wikipedia article on “Quartic function”.Question. Can all algebraic numbers be expressed in radicals?I

Detour: Abel-Rufini TheoremIAnswer: No. This is asserted by the Abel-Ruffini Theorem.In 1799 Paolo Ruffini made an incomplete proof and in 1824Niels Henrik Abel provided a complete proof.IExample. The roots of x 5 x 1, such asα 1.1673039782614 . . . ,are not expressible in radicals.IIt also follows from the Abel-Ruffini Theorem that for everyrational number r the numbers sin(r π) and cos(r π) areexpressible in radicals.IExample.s1 cos482 π 2 rq 2 2 3.

Detour: Abel-Rufini TheoremFigure: Paolo Ruffini (left) and Niels Henrik Abel (right)

FIXING UNIQUE FACTORIZATION

IdealsII Recall how the unique factorization fails in Z[ 5]. We willexplain how to fix it by introducing ideals.Let Q(α) be a number field and let O be its ring of integers.A subset I of O is called an ideal if1. 0 I ;2. If α, β I then α β I ;3. If α I and β O then αβ I .IIIThe most important property is 3: an ideal I absorbsmultiplication by the elements of O.If there exists α O such that I {αβ : β O} then I iscalled a principal ideal and it is denoted by I (α). Thenumber α is called the generator of I .Example. Consider an ideal (2) in Z. We have(2) {2n : n Z}, so the ideal (2) consists of all evennumbers. Further, for any 2k (2) and any n Z we have2kn even, so 2kn (2). Therefore (2) absorbs multiplicationby the elements of Z.

IdealsIILet I and J be ideals of O. We say that I divides J, denotedI J, if I J.An ideal I is called prime if1. I 6 O;2. For any α, β O such that αβ I either α I or β I .IExercise 9. Prove that (0) and O are ideals of O.IExercise 10. Show that (3), (5) and (6) are ideals in Z.Prove that (3) (6) and (5) - (6). Prove that (3) and (5) areprime ideals and (6) is not a prime ideal.IA ring O where every ideal is principal is called the PrincipalIdeal Domain (PID).IFor rings of integers of number fields, the UniqueFactorization Domain and the Principal Ideal Domain isthe same thing.

Ideal ArithmeticIEvery ideal has generators and there are finitely many ofthem. For α1 , . . . , αn O, we use the notation(α1 , . . . , αn ) {a1 α1 . . . an αn : a1 , . . . , an O}IIIto denote the ideal generated by α1 , . . . , αn .Example. Note that in Z we have (4, 6) (2). Example. In Z[ 5] there is an ideal (2, 1 5), which isnot a principal ideal.Addition. If I , J are ideals in O then we can compute theirsum, which is also an ideal:I J {α β : α I , β J}.IMultiplication. If I (α1 , . . . , αm ), J (β1 , . . . , βn ) are idealsin O then we can compute their product, which is an ideal:IJ (α1 β1 , α1 β2 , . . . , αm βn 1 , αm βn ).

Unique Factorization of IdealsI(Special case of) Dedekind’s Theorem. Every ideal I of Ocan be written uniquely (up to reordering) as the product ofprime ideals.IExample. In Z we have (6) (2)(3). Example. Though in Z[ 5] we have 6 2 · 3 (1 5)(1 5),Iso unique factorization fails, the unique factorization of idealsholds: (6) (2, 1 5)2 (3, 1 5)(3, 1 5)(2) (2, 1 5)2 (3) (3, 1 5)(3, 1 5)(1 5) (2, 1 5)(3, 1 5)(1 5) (2, 1 5)(3, 1 5).

Open Problems in Algebraic Number TheoryIThere are many big open problems in algebraic numbertheory, but we will present only two of them.IAn integer d is squarefree if it is not divisible by a perfectsquare 1. For example, 6 is squarefree but 12 is not because4 12.IGauss’s class number problem. There are infinitely manysquarefree integers d 0 such that the ring of integers of areal quadratic field Q( d) is a UFD.IThe Cohen-Lenstra Heuristics. In 1993–84, Cohen andLenstra gave a heuristic argument that “approximately”75.446% of real quadratic fields are UFD’s. There is a lot ofcomputational evidence that their conjecture is true, but whyit is true is still unknown.

DETOUR

Detour: Kummer’s Progress on Fermat’s Last TheoremIFermat’s Last Theorem. For every n 3 the equationx n y n z n has no solutions in positive integers x, y , z.IThis “theorem” was stated without proof by Fermat in 1670and proved by Andrew Wiles and Richard Taylor in 1995.IIt is sufficient to prove the theorem for n 4 (done byFermat) and for every n that is an odd prime.IIf p is an odd prime, then there exists an algebraic integer ζpof degree p 1 whose minimal polynomial isx p 1 x p 2 . . . x 1.The roots of this polynomial are ζp , ζp2 , . . . , ζpp 1 .IThis number is called the primitive p-th roof of unity, as itsatisfies ζpp 1.

Detour: Kummer’s Progress on Fermat’s Last TheoremINote that for every prime p we can writep 1z p x p y p (x y ) (x ζpi y ).i 1Therefore we factored x p y p over Z[ζp ].IThis reminds us of Euler’s idea for solving y 2 x 3 2! IfZ[ζp ] is a UFD then each numberx y , x ζp y , . . . , x ζpp 1 yis a perfect p-th power.IIn 1847 Gabriel Lamé outlined the proof of Fermat’s LastTheorem based on this method. Liouville pointed out that hispremise that Z[ζp ] is a UFD is false.IUsing this method, in 1850 Ernst Kummer proved that FLT istrue for all regular primes.

Detour: Kummer’s Progress on Fermat’s Last TheoremITo understand the statement of Kummer’s Theorem we needto introduce just two more definitions.ITwo ideals I and J of O are equivalent, written I J, ifthere are α, β O such that (α)I (β )J.IIdeals that are equivalent to each other form an equivalenceclass. The number of equivalence classes of O is always finiteand it is called the class number, denoted by h(O).IThe ring of integers O is a UFD if and only if h(O) 1.IExample. In Z we have (2) (3) because (3)(2) (2)(3).The class number of Z is 1. Example. In Z[ 5] we have (2, 1 5) (3, 1 5).The class number of Z[ 5] is 2, so it is not a UFD.I