NUMBER THEORY VIA ALGEBRA AND GEOMETRY

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1 NUMBER THEORY VIA ALGEBRA AND GEOMETRY DANIEL LARSSON CONTENTS 1. Introduction 2 2. Rings Definition and examles Ideals and subrings 6 4. Integral domains Homomorhisms, quotient rings and the first isomorhism theorem UFD s, PID s and Euclidean domains The Gaussian integers Polynomials Fields Definition and examles Fields of fractions Field extensions Field extensions Algebraic extensions, transcendental extensions Simle and finitely generated extensions Algebraic closure Finite fields The main theorem The Frobenius morhism Algebraic number fields Algebraic numbers Norms, traces and conjugates Algebraic integers and rings of integers Integral bases Comuting rings of integers Examles Quadratic number fields Ring of integers of quadratic number fields The Ramanujan Nagell theorem Dirichlet s unit theorem Roots of unity Units in number fields Dedekind domains A few imortant remarks The main theorem on Dedekind domains 50 1

2 2 D. LARSSON 13. Extensions, decomosition and ramification Ramification and decomosition Consequences for quadratic number fields Consequences for some non-quadratic number fields Cyclotomic number fields Cyclotomic fields Galois theory of number fields Gauss sums and Quadratic Recirocity Cubic recirocity Arithmetic and Geometry Affine n-sace Projective n-sace Algebraic curves Cubic and ellitic curves The grou structure of an ellitic curve The grou law Points of finite order The Nagell Lutz theorem Mordell s Theorem and Conjecture Gauss Class Number Problem and the Riemann hyotheses: A Historical Survey Gauss class number roblem Quadratic fields and forms What does this have to do with the class number roblem? Zeta functions and the Riemann hyothesis Back to quadratic fields Aendix A: Linear algebra Vector saces and bases Mas Dual saces Oerations on mas Linear equations and inverses Modules Vandermonde determinants Aendix B: Chinese remainder theorem INTRODUCTION These set of notes are the Lecture Notes accomanying my class in Number theory at Usala University, Sring terms 2008 and The notes begin with some very basic ring and field theory in order to set the stage, and continues to more advanced toics successively and robably in a rather stee uwards slant from an extremely soft and cosy start. I allow myself a few digressions in the text that are not art of the syllabus but that I feel are in a sense art of the required know-of for asiring mathematicians. In this case I am mainly refering to the sections concerning the Gauss class number roblem and the Ramanujan Nagell theorem. These are then not formally art of the course and will not

3 3 be discussed in an exam. But I strongly encourage readers to at least read through these arts to get an idea of the beauty that lies within. Also, some arts are more abstract and technical, mainly the section on rings of integers. The roofs here are rather difficult but I felt that if I didn t include them in the course I would be cheating (which I don t like). Therefore, I don t require the students to learn this material, but rather to have this as a fall-back solution if the later results feel a bit hollow and imroerly motivated. In the same sirit, I include Aendices with notions from linear algebra over rings for easy reference. As a twist to this course I added a section on ellitic curves, a toic that, without a doubt, will be art of every course on number theory that ever will be given anywhere on the lanet, or elsewhere (this is a foretelling on my art). A home-assignment will be given where the students are to learn the basics of ellitic curves over finite fields, so that they immediately can understand the basic ideas and quickly learn the techniques of ellitic curve crytograhy and (large) integer factorization using ellitic curves. This will surely be a worthwhile effort for every mathematically inclined student. 2. RINGS I will assume that everyone knows what an abelian grou is Definition and examles. We begin with the following definition. Definition 2.1. Let R be a binary set with two closed oerations, + (addition) och (multilication). Then R = (R, +, ) is a ring if addition and multilication is comatible according to the following axioms: Rng1: The oeration + makes R into an abelian grou, that is, a + b = b + a for all a,b R. Rng2: There is an element 1 such that 1 a = a 1 = a for all a R, or in other words, a multilicative unit, often simly called a one. Rng3: The multilication is associative: Rng4: The multilication is distributive: a (b c) = (a b) c, a,b,c R. a (b + c) = a b + a c and (b + c) a = b a + c a, a,b,c R. Remark 2.1. Strictly seaking, what we have defined here should be called an associative ring with unity (or associative, unital ring) The most general definition includes only Rng1 och Rng4. There are many examles of structures satisfying only these two axioms. However, for us it is enough to state the definition in the above, more restrictive, way. Note! From now on we write multilication in the usual fashion a b or simly ab Examles. Examle 2.1. The easiest and most obvious (and arguably the most imortant) examles are of course the following - Z, the ring of integers; - Q, the ring of rational numbers; - R, the ring of real numbers; - C, the ring of comlex numbers.

4 4 D. LARSSON Convince yourselves that these are indeed rings! In fact, they are even commutative. Examle 2.2. Another, extremely imortant examle of a ring is Z/ n, the ring of integers modulo n. Recall that this is the set of congruence classes of elements modulo division by n, and can be reresented by {0,1,2,...,n 1}, the remainders after division with n. Informally, one writes Z/ n := {0,1,2,...,n 1}, imlicit being that addition and multilication are allowed. Examle 2.3. Another examle is Z[ n] := {a + b n a,b Z} for n a square-free integer. We will see a lot of examles like this in the future, since number-theoretic alications to rings often involve examles like this one. So far, every examle has been commutative. Let s round off this examle-listing with a non-commutative number system. Examle 2.4 (Quaternions). Recall that the comlex numbers C was formed by, to R, add an imaginary number i := 1. As you no doubt remember C := {a + bi a,b R}. The ring C has one further roerty that we haven t discussed yet (but will, in the next section): every non-zero comlex number has an inverse 1 A question that was asked during around if one could add yet another really imaginary element (not equal to 1), j, so that one gets a new ring with the roerty that every non-zero element has an inverse. This can be shown to be imossible! However, 1843, W.R. Hamilton realized that if one adds two more elements to C, the element j and another k, then every non-zero element has an inverse! But, alas, one loses one imortant thing: commutativity. The definition is as follows. Definition 2.2. The set of all numbers H := {z = a + bi + cj + dk a,b,c,d R, i 2 = j 2 = 1 1, ij = ji = k} is a non-commutative ring, where all elements 0 has an inverse. An element of H is called a quaternion and H is called the ring of quaternions. I strongly suggest that you do the following exercise. Exercise 2.1. Try out a few comutations and check that H is indeed a non-commutative ring. (You don t have to check that it is a ring. This is obvious! Why?) To define an inverse recall how inverses are comuted in C and emulate that. In order to do this you will have to define a suitable notion of conjugate quaternion. Notice that R C H and that we double the dimension in each ste: C is twodimensional as a vector sace over R and H, is four-dimensional. The natural follow-u question is: is it ossible to add more imaginary elements and get larger number systems? The answer is yes, and this was given only a few months after Hamilton s discovery by J.T Graves and later by A. Cayley. But adding just one element is not sufficient (just as it was not sufficient to add just one to C to get H): one has to add 1 One thing, however that is lost when assing from R to C (that is by adding i) is order: it is meaningless to ask which one of two comlex numbers is the biggest. The only thing one can, meaningfully say is which of them has the greatest distance from the origin.

5 5 four new ones! The result is the eight-dimensional octonions, denoted by O. The roblem is now that another roerty is lost: namely, associativity! (Recall the remark after the definition of a ring.) After the octonions comes the sixteen-dimensional sedonions, S. This time there are non-zero elements a and b such that ab = 0. Such elements are called zero-divisors (we will see more on this soon). One can continue this indefinitely with doubling dimensions in each ste. The objects in this nested sequence of rings are called Cayley Dickson algebras (CDA s) R C H O S } {{ }. CDA For those finding this fascinating I recommend the Wikiedia-article on the Cayley Dickson construction. 3. We begin with the following rather lengthy definition. Definition 3.1. Let R be a ring (commutative with unity). - An element 0 a R is a zero-divisor if there is a 0 b R such that ab = 0. - An inverse to 0 a R is an element a 1 such that aa 1 = a 1 a = 1. An element is called invertible if it has an inverse. - The characteristic, char(r), of R is the least number n N such that na = a + a + a = 0 }{{} for all a R. n times If no such least n exists, we ut char(r) = 0. Examle 3.1. In Z/ 6 = {0,1,2,3,4,5}, 2 is a zero-divisor since 2 3 = 6 = 0. The characteristic of Z/ 6 is six. The invertible elements are 1 and 5 which are their own inverses (check!). On the other hand, Z has no zero-divisors and characteristic zero. The only invertible elements in Z are ±1 and they are also their own inverses. Theorem 3.1. Let R be a ring and a,b,c R := R \ {0}. Then, (i) 0a = a0 = 0, (ii) the unity of R is unique, (iii) ( a)b = a( b) = (ab), and (iv) if a multilicative inverse to a exists, it is unique. Proof. These are simle: (i) 0a = (0 + 0)a = 0a + 0a = 2 0a 0a = 0 and similarly we have a0 = 0. (ii) The roof of this is exactly as for grous: 1 = 1 1 = 1. (iii) What one needs to observe is that to show (iii) all you need to show is that the terms involved are all inverses ab. Then the result follows since additive inverses are unique (the additive structure is a grou structure, remember). Indeed, ab + ( a)b = (a + ( a))b = (a a)b = 0b = [from (i)] = 0. The others are exactly the same and are left to you. (iv) Suose there are b,c R such that ab = ac = 1. Then ab ac = 0 a(b c) = 0 so multily with one of the inverses to a from the left: ba(b c) = 0 1(b c) = 0 b = c.

6 6 D. LARSSON Direct roducts. There are several ways of constructing new rings out of old ones. Here is one useful examle. Let {R i }, i I, where I is some index set, be a collection of rings. The direct roduct of {R i }, written i I R i is the set of all sequences R i := {(r i ) r i R i }. i I When I is finite (as all our examles will be), one usually write this as R 1 R 2 R n. In this case the ring structure is given by comonent-wise addition and multilication, (r 1,...,r n ) + (r 1,...,r n) := (r 1 + r 1,...r n + r n), (r 1,...,r n )(r 1,...,r n) := (r 1 r 1,...r n r n), and the zero element and unity are resectively, (0,0,...,0), (1,1,...,1). It is easy to rove (do this!) that R 1 R n is a ring when all the R i s are rings. On the other hand, it is a little more subtle to show the ring axioms for infinite index sets. The statement is nonetheless true for general index sets Ideals and subrings. From now on we will only deal with commutative rings. This makes life a lot easier since non-commutative rings are often very strange creatures and one has to be very tongue-in-cheek when dealing with them so as not to fall into the tra of thinking commutatively. But for basic number theory the commutative theory suffices. Definition 3.2. Let R be a (commutative) ring. An ideal in R is a subgrou i R such that ri i for all r R. This means that for all r R and i i, ri i. An ideal is called roer if i R and trivial if i = 0. A subring is a subgrou S of R such that ab S for a,b S and such that 1 S. Notice the difference. Lemma 3.2. If 1 i for some ideal i then i = R. Proof. For every r R, r = r1 i so R i. The other inclusion is clear from definition of course. Examle 3.2. In Z, the ideals are on the form nz := {nz z Z}, for some n Z. How do we rove this? First of all, any subgrou S of a cyclic grou (like Z) is cyclic so S = nz as grous 2. Clearly nz is stable under multilication of Z in the sense that for a nz, xa nz, for x Z. Hence nz is an ideal. What are the subrings? A subring has to be a subgrou so they must then also be on the form nz for some n Z. But a subring has to include 1 and so n = 1. This means that there is only one subring, namely, Z itself. 2 If this doesn t ring a bell, here is a direct argument. Let G be a grou generated by a G in the sense that for g G, g = a k for some k Z and let S be a roer subgrou. Every element s S can be written as s = a l for some l Z. Let d be the smallest integer dividing all the l s as s ranges through S. Then S is generated by a d. Notice that if d = 1 then S = G, so by the assumtion of roerness we have, d > 1.

7 7 Examle 3.3. The only ideal of Q is Q itself, so there are no roer ideals. Indeed, let i Q be an ideal. Then for q i, we have q 1 q = 1 i and so by the lemma above, i = Q. This is a general henomena for rings where all non-zero elements have inverses. As an exercise, find a subring (there are infinitely many)! You will see examles of this later. Definition 3.3. Let R be a ring. - An ideal given as f := {r f r R} is called a rincial ideal. - A roer ideal R is called a rime ideal if ab imlies that a or b. - A roer ideal m R is called a maximal ideal if, for a an ideal such that m a R then a = m. Examle 3.4. In Z all ideals are rincial n = nz. An ideal in Z is rime if and only if n is a rime; an ideal in Z is maximal if and only if it is rime Ideal generation. The easiest and by far the most common way of constructing ideals is by generation: Definition 3.4. The ideal i is generated by { f i f i i,i I}, where I is some index set, if a i = a = r i f i, for r i R. J I J finite We write this as i = { f i f i i,i I}. Check that this is an ideal! The ideal is called finitely generated if I <, that is, if the number of f i s are finite. Notice that if I = 1 then we get a rincial ideal Oerations on ideals. Suose that i and j are two ideals of R. We make the following definitions: i + j := {i + j i i, j j} ir := { ir i i,r R, finite sum} ij := { i j i i, j j, finite sum}. Theorem 3.3. The sets i + j, ir and ij are all ideals of R. Proof. Exercise! 4. INTEGRAL DOMAINS For number theorists the following definition is of utmost imortance. Definition 4.1. A ring with no zero-divisors is called an integral domain or simly a domain. Examle 4.1. The ring Z is an integral domain. In fact, Z is the reason for integral in the name. It was the first and most natural ring to study. On the other hand, Z/ 6 is not an integral domain since 2 3 = 0. Theorem 4.1. A ring R is an integral domain if and only if 0 is a rime ideal. Proof. Exercise!

8 8 D. LARSSON Theorem 4.2. A ring R is an integral domain if and only if it satisfies the cancellation roerties: if a 0 then ab = ac = b = c, and ba = ca = b = c. Proof. Suose first that R is a domain. That ab = ac is equivalent to 0 = ab ac = a(b c) and since R does not have any zero-divisors and a 0, we get b c = 0. In the same manner one shows the right-handed version. Conversely, if ab = ac b = c, suose that xy = 0 with x 0. This is equivalent to xy = x0 and hence y = 0. Theorem 4.3. If every non-zero element in R has an inverse then R is an integral domain. Proof. Let ab = 0 with a 0. Multily with a 1 from the left and the result follows. Theorem 4.4. Every non-zero element in a finite domain is invertible. Proof. Let R := {0,1,a 2,,a n }. Pick an element 0 b R. We want to fin a c R such that bc = cb = 1. Multily all elements in R with b from the left. Then we get: {0,b,ba 2,ba 3,ba n }. Every element in R aears in this set exactly once. This is because, ba i = ba k a i = a k since R is an integral domain. This means that 1 has to aear somewhere in this set. Hence ba j = 1 for some (secific) a j. The same argument alies to the right-handed case. Theorem 4.5. The zero-divisors in Z/ n are exactly the elements m 0 that are not relatively rime to n, that is, gcd(m,n) = d > 1. Proof. First if gcd(m, n) = d > 1, m 0, then m(n/d) = (m/d)n = 0 in Z/ n since the right-hand-side is a multile of n. This means that m is a zero-divisor in Z/ n since neither m nor n/d is zero. On the other hand, if gcd(m,n) = 1, m 0, and mk = 0 = tn. Therefore, n has to divide mk. But since m and n are relatively rime, we have n k, which is equivalent to k = 0 in Z/ n. Corollary 4.6. The ring Z/, where is a rime, has no zero-divisors. Corollary 4.7. The ring Z/ n is a domain and every non-zero element is invertible if and only if n is a rime. Proof. If n is not a rime there are zero-divisors 3. On the other hand, if n =, it follows from Theorem 4.4 that every element is invertible, and from the revious corollary that Z/ has no zero-divisors. This shows, incidentally, that Z/ is a field. More on this later Homomorhisms, quotient rings and the first isomorhism theorem. A very imortant rincile in the hilosohy of modern mathematics is that mathematical objects are to a very large extent governed by their mas to other objects of the same category (e.g., grous or rings). Hence we need to define what is to be meant by a ma between rings. Definition 4.2. A set-theoretical ma of rings φ : R S is a ring homomorhism or ring morhism if it is a homomorhism of the underlying abelian grous, i.e., φ(a + b) = φ(a) + φ(b), and if φ resects multilication in the sense that in addition to φ(1 R ) = 1 S. φ(ab) = φ(a)φ(b), 3 Also, we note that zero-divisors have no inverses: ab = 0 = b = 0 and zero is not invertible.

9 9 Definition 4.3. Let R φ S be a ring morhism and for s S denote φ 1 (s) := {r R φ(r) = s}. This is called the fiber over s or the inverse image of s. Put ker(φ) := φ 1 (0) = {r R φ(r) = 0}, the kernel to φ, and im(φ) := {s S r R, φ(r) = s}, the image to φ. Notice that since φ is a grou homomorhism we have φ(0) = 0. Theorem 4.8. For R φ S a ring morhism, we have (i) ker(φ) is an ideal in R. (ii) im(φ) is a subring in S but not necessarily an ideal. Proof. The roof is rather easy: (i) Let b ker(φ) R, a S. Take r R. We want to show that rb ker(φ) to show that ker(φ) is an ideal. But this is obvious since φ(rb) = φ(r)φ(b) = φ(r) 0 = 0, and so rb ker(φ). (ii) That im(φ) is a subring follows immediately from the definition. Think through this! Definition 4.4. We can make the following definitions in analogy with the corresonding notions from grou theory. - A ring morhism φ : R S is an injection (or one-to-one ) if kerφ = {0}; this is written φ : R S; - φ is a surjection (or onto ) if imφ = S, written φ : R S; and an isomorhism ( one-to-one, onto ) if it is both injective and surjective. We say that R and S are isomorhic if there is an isomorhism R φ S and it is customary to write this as R S or R S. Examle 4.2. Here are some easy examles. - Reduction mod n is a ring morhism Z Z/ n. - The inclusion Z Q is a ring morhism. - Comlex conjugation is a ring morhism C C. This is in fact an isomorhism. Isomorhisms between a ring and itself are called automorhisms. The set of all ring morhisms between rings R and S is denoted by Hom(R,S) (to remind you of homomorhism ). If S = R then we ut End(R) := Hom(R, R). Ring morhisms from R to itself is called endomorhisms. Examle 4.3. The ma n : Z Z, n z nz, is a grou homomorhism, but not a ring homomorhism since it is not multilicative: n (ab) nanb. So there are a lot fewer ring morhisms than grou morhisms.

10 10 D. LARSSON Quotient rings. Let i be an ideal of R. Introduce a relation on R as follows: a i b a b i, for all a,b R. Theorem 4.9. The relation i is an equivalence relation on R. Proof. We need to show three things: - Reflexivity: a i a for all a R. This is clear since 0 i. - Symmetry: a i b b i a. This follows since a b i (b a) i and since i is a subgrou a i a i. - Transitivity: a i b and b i c imlies a i c. This follows by a b i and b c i which, by adding (once again using that i is a subgrou), gives a b + b c = a b i. Definition 4.5. The coset (i.e., an equivalence class) of a R under this equivalence relation is denoted a + i or sometimes [a] or ā. The set of cosets is denoted Theorem The oerations R/i := {a + i a R}. (a + i) + (b + i) := (a + b) + i (a + i)(b + i) := ab + i defines a ring structure on R/i, with zero element i and unity 1 + i. There is also a canonical surjective ring morhism π : R R/i with ker π = i, defined by π(r) = r + i. Definition 4.6. The ring defined by this theorem is called the quotient (or factor) ring modulo i. What this essentially means is that we set all elements in i to zero, i.e., we are killing the kernel kerπ = i since, in R/i the zero element is 0 + i = i. Proof. We have to check that the definitions are well-defined, i.e., that choosing different reresentatives from each coset gives the same result. Let us rove this in the case of multilication, the additive counter-art being comletely analogous and therefore left to the reader. So, suose we are given x + i = a + i and y + i = b + i. This means that x = a + i x and y = b + i y, for i x,i y i. Hence, (x + i)(y + i) = xy + i = (a + i x )(b + i y ) + i = ab + ai y + i x b + i x i y + i = = ab + i = (a + i)(b + i), where we have used that ai y,i x b,i x i y i. Notice that this uses in a essential way that i is an ideal. The rest of the ring axioms follow since the ring oerations are induced from R. For the last statement, π(r + s) = (r + s) + i = (r + i) + (s + i) = π(r) + π(s), π(rs) = rs + i = (r + i)(s + i) = π(r)π(s), and clearly given r + i R/i, π(r) = r + i so π is certainly surjective. Examle 4.4. The canonical examle is the following. Every subset of Z of the form nz = {nz z Z} is an ideal of Z. It is a fact (that can be deduced from the next subsection) that Z/ n Z/nZ.

11 11 We will return to this examle after the following theorem, subsequent discussion and roof The first isomorhism theorem. The following theorem is arguably the most useful theorem in all ring theory. Theorem 4.11 (First isomorhism theorem). Let φ : R S be a ring homomorhism. Then the following diagram commutes: φ R S π φ R/kerφ, that is, φ = φ π, and induces an isomorhism R/kerφ imφ. Remark 4.1. First of all, what does it mean for a diagram to commute? Well, intuitively it means that what ath you take from a articular oint to any other (following the directions of the arrows) is not imortant. In the above examle it means that going from R to R/i and then to S is the same as going from R to S directly. Another way of saying this is that φ factorizes through R/ ker φ as φ = φ π. Proof. The roof of this theorem is surrisingly simle. First of all we define π by π(r) := r + kerφ. Then φ can be defined as φ(r + kerφ) := φ(r). Notice that all the mas are now ring morhisms. Clearly, φ = φ π, but we have to check that φ is well defined. So suose that r + kerφ = s + kerφ r s = f kerφ r = s + f. Hence, φ(r + kerφ) = φ(r) = φ(s + f ) = φ(s) + φ( f ) = φ(s) = φ(s + kerφ). To see that we have the isomorhism R/kerφ imφ, note that φ is a surjection onto its image (by definition!) and so since im φ = imφ the same goes for φ. Also, ker φ = {0}, so ker φ is injective, and therefore an isomorhism. This theorem gives a more recise meaning to killing the kernel! Examle 4.5. Let s continue the examle given right before this subsection. We have the ring morhism reduction mod n : Red n : Z Z/ n given by Z r r, where r is the reduction (remainder) modulo n. The kernel of this ma is all the multiles of n: ker(red n ) = nz. Therefore, we get the following commutative diagram: Red n Z Z/ n π Red n Z/nZ. Clearly, Red n : Z Z/ n is a surjection so im(red n ) = Z/ n and the isomorhism from the theorem becomes, Z/nZ im(red n ) = Z/ n. Examle 4.6. The rojections R S r R r S R,S

12 12 D. LARSSON are ring homomorhisms (check this!). Consider the morhism r R : R S R. The kernel of this is clearly S. Hence the theorem tells us, (R S) / S R, and similarly, by using r S, (R S) / R S. So in this sense, / is really like a division, if is viewed as a multilication. This is robably the origin of the name quotient ring and the notation / Ideals and ring morhisms. Now we come to another, very imortant result: how do ideals behave under ring morhisms. We make the following definition. Definition 4.7. Let φ : R S be a morhism of rings and i an ideal of R and j an ideal of S. Then, i := φ(i)s = { i φ s i φ φ(i),s S} finite is the extended ideal of i in S and j := φ 1 (j) = {r R φ(r) j} the contracted ideal of j in R. Sometimes these are denoted is and j R, resectively. I will robably use the first of these alternate notations but hardly the second. Theorem With notation as above, i and j are ideals of S and R, resectively. Proof. We rove the second, leaving the first to you as an exercise (do it!). Suose, j j and r R. Then φ(r j) = φ(r)φ( j) j since φ(r) S, φ( j) j (by definition) and j is an ideal. In the same manner, j + j j, when j, j j, φ( j + j ) = φ( j) + φ( j ) j since j is an ideal. Theorem Let φ : R S be a ring morhism between two rings with ideals i R and j S. Then (4.1) (4.2) (j ) = φ(φ 1 (j)) = imφ j (i ) = φ 1 (φ(i)) = kerφ + i. Proof. We divide the roof as in the statement of the theorem. (4.1) We need to show two inclusions: φ(φ 1 (j)) imφ j and φ(φ 1 (j)) imφ j. Let s start with the first. Take a φ(φ 1 (j)). Obviously a imφ. Since φ 1 (j) is the set of elements maing into j, we obviously also have a j. Hence a imφ j. For the other inclusion, assume a imφ j. Then, φ 1 (a) is a set of elements maing to a imφ j under φ, so a φ(φ 1 (j)). Hence the (4.1) is roved. (4.2) As above, we need to rove two inclusions: φ 1 (φ(i)) kerφ + i and φ 1 (φ(i)) kerφ + i. Take a kerφ + i. Then φ(a) φ(i), and so a φ 1 (φ(s)) φ 1 (φ(i)) since this is the set of elements maing to φ(i) under φ. To show the other inclusion, take a φ 1 (φ(i)). Then φ(a) φ(i) so φ(a) = φ(b) for some b i. This is equivalent to φ(a b) = 0 and so a b kerφ meaning that a = b + k, with k kerφ. But b i so the inclusion is roved, thereby comleting the whole roof.

13 13 This theorem has the following remarkable corollaries. Corollary If φ : R S is a surjection, then (j ) = φ(φ 1 (j)) = j. On the other hand if φ : R S is an injection, then (i ) = φ 1 (φ(i)) = i. Proof. Obvious. Corollary Let i be an ideal of R. Then there is a bijective corresondence between ideals of R/i and ideals j of R, such that i j. The corresondence is given by i j j/i := π(j) = { j + i j j}. Proof. First of all it is clear that every ideal in R/i must be on the form B/i = {b +i s B} for some subset B of R. However, if B/i is an ideal then (r +i)(b+i) should be in B/i for all r + i R/i. We have, (r + i)(b + i) = rb + i, and so rb must be in B, i.e., B must be an ideal. Furthermore, it is clear that B i. Hence every ideal in R/i is on the form j/i for some ideal j i. Conversely, given an ideal j R we get an ideal j/i R/i by rojection. Suose k 1 := j 1 /i = j 1 /i =: k 2. Then (k 1 ) = k 2 by the revious corollary. But this means that ((k 1 ) ) = k 2 and so j 1 = j 2 and the corresondence is also one-toone UFD s, PID s and Euclidean domains. In this subsection all rings are assumed to be integral domains even if it is not exlicitly stated Prime elements and irreducible elements. Definition 4.8. Let D be an integral domain and a,b D. - We say that b divides a, written (as usual) a b, if there is a c D such that a = bc; a is then called a divisor of b. - A divisor of 1 is called a unit 4. The set of units in a ring is a multilicative grou (check!) and denoted U(D). - Two elements a and b such that a = bu au 1 = b, for u U(D), are called associates. Association is an equivalence relation (check!). - An element a D is irreducible if any factorization a = bc imlies that either b or c is a unit. Otherwise, a is called reducible. - An element D is called a rime (element) if ab imlies a or b. Remark 4.2. Be sure to searate the similarly named, but distinctly different, notions unit and unity. This is notorious source of confusion and frustration. The following simle remark is imortant enough to earn its way as a full-fledged theorem. Theorem Let D be an integral domain. Then (i) a b b a ; (ii) a b and b a a = ub for a unit u U(D) a = b. Proof. Exercise! Theorem Any rime element is irreducible (but in general, not the other way around) and D is rime if and only if is a rime ideal. 4 Notice that this is equivalent to an invertible element. To be honest, I don t know why there are two names for the same concet. My lingering feeling is that the designation unit is a bit more number-theoretical. But I could be wrong.

14 14 D. LARSSON Proof. Let be a rime and assume that it is reducible = ab where neither a nor b is a unit. Then clearly ab but a and b (since if, for instance, a then a = k for some k, and so = kb; since D is a domain, we can cancel to obtain bk = 1, but b was not a unit so we have a contradiction) so we have a contradiction to being rime. Suose now that is a rime and let ab. This is equivalent to ab = c for some c D and this in turn equivalent to ab and since is rime a or b. Now use Theorem 4.16 (i). Conversely, suose that is a rime ideal and ab. This imlies that a or b. In either case a = c or b = c and we are done. This theorem is the reason why rime ideals are called rime ideals Unique factorization domains. Definition 4.9. A domain D is called a unique factorization domain, abbreviated UFD, if UFD1. Every element can be written as a roduct of irreducibles, and UFD2. this factorization is unique u to multilication by a unit. Let me comment on the last condition. Let n 1 1 n r r = q m 1 1 qm s s be two factorizations of a into irreducibles. Then the condition says that r = s and i = u i j q j for some i and j and u i j U(D). Theorem 4.18 (The fundamental theorem of Arithmetic). The ring of integers Z is a unique factorization domain. In general it is rather difficult to rove that a given integral domain admits unique factorization. Look u the roof of the above theorem in your elementary algebra books. Theorem If UFD1 in Definition 4.9 holds, then UFD2 is equivalent to the statement every irreducible element is rime. Proof. Assume first that factorization is unique u to a unit. So let be an irreducible element and suose ab. We need to show that a or b. We begin by assuming that a and b are also irreducible and a. We have that c = ab and,a,b are irreducible so c is not a unit, since otherwise = (c 1 a)b, a factorization into irreducibles. This imlies that c = ak for some k D and so ak = ab a(k b) = 0, imlying that b since D is a domain. Let us return to the general case. Let a = α 1 α n and b = β 1 β m. Since the factorization is unique (u to units) the roduct of a and b must be α 1 α n β 1 β m. So if ab then must divide a roduct of irreducibles and so must divide (at least) one of those. Hence divides a or b. Assume now that every irreducible element is rime and let n 1 1 n r r = q m 1 1 qm s s be two factorizations of a into irreducibles. This tells us that q 1 (for instance) divides both sides. Since every irreducible element is rime we see that all the i s and q j s are rime and q 1 n 1 1 n r r imlies that q 1 i for some i and so q 1 = s i, but since q 1 is irreducible s U(D). Continuing this way inductively the result follows.

15 Princial ideal domains. Definition A domain D is called a rincial ideal domain, abbreviated PID, if all ideals are rincial, i.e., generated by one element. Theorem Let D be a PID. Then, is irreducible is rime. In addition, in a PID, every rime ideal is maximal. Proof. The imlication has already been demonstrated (as a general fact). For the reverse imlication, let be irreducible and let ab with b /. We then have that ab = c. Since is irreducible and b / we have a = c 1, i.e., a. For the last statement, observe that if is rime and q, for q / U(D), then = cq and since is irreducible, c U(D) and so = q. Theorem Every PID is a UFD. Proof. Let D be a PID and S D be the set of elements which cannot be written as a roduct of irreducibles. Assume that S is non-emty and take a S. Then a = a 1 b 1, a 1,b 1 / U(D). This means that a a 1 and a b 1. If both a 1,b 1 / S then this would imly that a is a roduct of irreducibles so assume that a 1 S. The same argument alies with a relaced by a 1 to get another a 2 such that a a 1 a 2. Continuing this rocess we get an infinite sequence of strict ideal inclusions: a a 1 a 2 a n. The infinite union i=1 a i is an ideal (check this!), and since D is a PID there is an element a D such that a = i=1 a i. Obviously, a i=1 a i and so there is a 1 j < such that a a j. But this means that a i = a a j a j+1 a = a i, i=1 i=1 which is a contradiction and hence S must be emty. To show that the factorization is unique we use Theorem So let be an irreducible element and ab and a. Since is maximal by Theorem 4.20,,a = D and so 1,a. This means that there are α,β D such that aα + β = 1. Multily this equation with b, to get abα + bβ = b. Since divides the left hand side, it divides b and by Theorem 4.19, the factorization is unique. The roof of this theorem shows, incidentally, that PID s belong to a large and imortant class of rings which we now define. Definition A ring R (not necessarily a domain 5 ) is called Noetherian 6 if it satisfies the ascending chain condition on ideals: for any ascending chain of ideals there is an N such that for all k N, Theorem Any PID is Noetherian. a 1 a 2 a n, a k = a k+1 = a k+2 =. 5 Or, commutative for that matter, although, in the non-commutative case one has to be a little more recise. 6 In honor of Emmy Noether, She was one of the big ioneers of abstract algebra. For more info, see her biograhy on htt://www-grous.dcs.st-and.ac.uk/.

16 16 D. LARSSON Integral domains that doesn t have the unique factorization roerty. Definition Let D be an integral domain. A (weak) norm on D is a function such that (i) Nrm(ab) = Nrm(a)Nrm(b), and (ii) Nrm(a) = 1 a U(D). The definition imlies: Nrm : D N 0 := N {0}, Lemma If Nrm is a norm on D, then Nrm(0) = 0. Proof. Let a D be arbitrary. Then Nrm(0) = Nrm(a 0) = Nrm(a)Nrm(0), which is equivalent to Nrm(0)(Nrm(a) 1) = 0. Since a was arbitrary, and N, as a subset of Z does not have any zero-divisors, we conclude that Nrm(0) = 0. This norm-function is convenient tool to show that certain elements are irreducible. Examle 4.7. Let D = Z[ 5] = {a + b 5 a,b Z}. Let us show that 2,3 and 1 ± 5 are irreducible. First of all we need to define a norm on Z[ 5]. We do this as Nrm(z) = z z = (a + b 5)(a b 5) = a 2 + 5b 2, for z = a + b 5. Check for yourselves that this is indeed a norm. If 2 is not irreducible we have 2 = ab, where a,b are not units. This means that Nrm(a)Nrm(b) = Nrm(2) = 4. Since a,b are not units, Nrm(a) = Nrm(b) = 2. Assume a = α + β 5. Then 2 = Nrm(a) = α 2 + 5β 2, which is clearly imossible for integers α and β. Hence 2 is irreducible. A similar argument shows that 3 is irreducible. Suose now that 1 ± 5 is reducible, i.e., 1 ± 5 = uw, u,w / U(Z[ 5]). Then Nrm(u)Nrm(w) = Nrm(1 ± 5) = = 6. Since 6 = 2 3 and u and w are not units either u is 2 or 3 and w the other ossibility. But in that case u or w must be units because, as we showed above, it is imossible to find non-units with norm 2 or 3. Hence 1 ± 5 is irreducible. If 2 and 3 were associates then 2 = 3u, with u U(Z[ 5]). This would imly that Nrm(2) = Nrm(3)Nrm(u) = Nrm(3) 1, and this is clearly false (we also see that associates have the same norm). The same argument alies to the other cases excet for and 1 5. What are the units of Z[ 5]? An element u D is a unit if and only if Nrm(u) = 1, which in this case is equivalent to, a 2 + 5b 2 = 1, for u = a + b 5. This is clearly only ossible if b = 0 and so a 2 = 1, showing that u = ±1. Hence U(Z[ 5]) = { 1,1}. So, 1 5 = ±1 (1 + 5) is clearly not an otion and hence, 2,3,1 ± 5 are irreducible, non-associates. Now, we see that 6 = 2 3 = (1 5)(1 + 5), and so we see that 6 has two different (i.e., they don t differ by a unit) factorizations into irreducibles! This means that Z[ 5] is not a UFD, and hence not a PID either. Ok, you might say, what on earth is so interesting about (strange) rings such as Z[ 5]? If you also say that it is uninteresting to find integer solutions to equations of the tye y 2 + a = x 3, a Z (the above case being a = 5), then, fine, I agree, in that case studying

17 17 rings such as Z[ 5] could be seen as a bit artificial. But, on the other hand, then your interest in number theory may be, at best, suerficial. The above examle indicates, for instance, that the equation y = x 3 might not have any, non-trivial, integer solutions 7. Also, as we will very soon see, these kind of rings are very imortant in number theory. Clearly, unique factorization in number theory would be a something extremely desirable. But as this examle shows this is not always (or most often even) ossible to achieve. However, and here is where things become beautiful, even though unique factorization of elements is not always ossible, unique factorization of ideals into rime ideals is! Unfortunately, I don t have time to go into this in detail, but we will come back to it briefly. Theorem If Nrm(a) =, where is a rime, then a is irreducible. Proof. Suose a is reducible a = bc, b, c / U(D). Then, = Nrm(a) = Nrm(bc) = Nrm(b)Nrm(c). Since is a rime, Nrm(b) (or Nrm(c)) must be. But this imlies that Nrm(c) = 1 and so c is a unit, contrary to assumtion. Hence, a is irreducible Euclidean domains. Now we will introduce yet another tye of norm, or rather valuation. Definition Let E be a domain. Then a function v : E N 0 is called an Euclidean valuation if the following are satisfied. Eu1. There is a division algorithm with resect to v. That is, for each air a,b E, b 0, there are q,r E such that a = bq + r, with v(r) < v(b), or, r = 0. Eu2. v(a) v(ab) for all E a,b 0. A domain E with a Euclidean valuation is called a Euclidean domain. Examle 4.8. The following are examles of Euclidean domains. - Z, with v(a) := a ; - Z[ ±2] = {a + b 2 a,b Z}, with v(a + b ±2) = a 2 ± 2b 2 ; - The Gaussian integers, Z[ 1] := {a + b 1 a,b Z}, with v(a + b 1) = a 2 + b 2 ; - Z[ 14] = {a+b 14 a,b Z}. This examle is highly non-trivial, roved It is not true that any Z[ n] is Euclidean, even if n is rime (see below). So what is so secial about Euclidean domains? Well, we have: Theorem Euclidean domain = PID = UFD. Proof. Only the first imlication needs a roof since we have already roved the second one. For the first imlication let i be an ideal of an Euclidean domain. Then let s be a smallest element, with resect to v, of i. Such an element exists since imv N 0. Then i = s (check!). Hence we have unique factorization in Euclidean domains, something desirable in number theory. 7 Fix x and factor the left-hand-side to so what I mean.

18 18 D. LARSSON Examle 4.9. At the moment of writing, the following facts are known (according to Wikiedia, which I kind of trust on this issue) concerning Z[ d]: it is Euclidean for d = 1, 2, 3, 7, 11, and hence a PID; it is not Euclidean for d = 19, 43, 67, 163, but is in fact a PID for these d s. Quite an amazing result! Examle Notice that the above theorem shows that Z[ 5] is neither a Euclidean domain, nor a PID, since it is not a UFD. It is natural to wonder (somewhat stunned, I hoe) why are all these nearby cases so different!?. Frankly, I don t think anyone knows. But after all, this is number theory, the science of simle questions and hard (at best) answers The Gaussian integers. In order to fully motivate the abstract theory that I have covered so far, and that will be taken to even greater aarently nonsensical heights in a while, I thought it rudent to insert an examle to show how all this actually has something to do with number theory, the de facto toic of this course. Theorem Any odd rime number (i.e., 0) is the sum of two squares if and only if 1 (mod4), that is, = a 2 + b 2 1 (mod4). Proof. A very convenient way to rove this is by using the Gaussian integers Z[ 1]. Recall that Z[ 1] = {a + b 1 a,b Z}. We have already observed that Z[ 1] is Euclidean (see Examle 4.8), and so, consequently, both a PID and UFD. Moreover, there is a norm Nrm : Z[ 1] Z defined by Nrm(z) := z z = a 2 + b 2, where z = a + b 1 and z = a b 1. So = a 2 + b 2, when considered in Z[ 1], factorizes as = (a + b 1)(a b 1). Since we will be dealing both with rimes in Z and rimes in Z[ 1], we follow tradition and refer to rimes in Z as rational rimes. Suose a rational rime considered as an element of Z[ 1], factors as = xy, where x,y / U(Z[ 1]), that is, x and y are not units (see Remark below). Then Nrm() = Nrm(x)Nrm(y), and Nrm() = 2 = Nrm(x)Nrm(y) = 2. An element x in a normed ring is a unit if and only if Nrm(x) = 1 and since we assumed that x and y are not units we must have that Nrm(x) = Nrm(y) =. This means in articular that, for x = a + b 1, Nrm(x) = a 2 + b 2 =. Now, assume that 1 (mod4) = 4n+1, for some n Z. By quadratic recirocity the congruence z 2 1 (mod ) have a solution for = 4n + 1 (check this!). This means that there is a t Z such that (t 2 + 1). But in Z[ 1], t = (t + 1)(t 1) so (t + 1) or (t 1) if it were to be a rime. However, none of these are elements of Z[ 1]: for instance, t/ + 1/ / Z[ 1]. Hence cannot be a rime in Z[ 1], and so by the receeding aragrah = a 2 + b 2.

19 19 The other imlication, namely, = a 2 +b 2 1 (mod4) follows by elementary reasoning. Indeed, not both a and b can be even because then would be even, contradicting the assumtion. Similarly, if a and b are both odd then would once again be even. So we must have that one is odd and one is even, and this imlies that 1 (mod4) (check it!). Remark 4.3. A very imortant remark here is that even though Sec(Z) is irreducible, it can become reducible in an extension of Z. This is one of the basic realization of the number theorists of the 19 th century. In fact, this can always be given as general good advice: if something can t be solved, can it be solved in some larger context? Lemma The units of Z[ 1] are {1, 1, 1, 1}. Proof. Use the characterization: z U(R) Nrm(z) = 1. Theorem The rimes, modulo multilication by units, of Z[ 1] are exactly the elements on the forms: (i) = 1 + 1; (ii) = a + b 1 with a 2 + b 2 = and 1 (mod4), where is a rational rime 2; (iii) = a rational rime such that 3 (mod4). Recall that, since Z[ 1] is a PID, the irreducible elements are exactly the rime elements. Proof. First we rove that the elements listed are indeed rime. If factored as xy, we would have = Nrm() = Nrm(x)Nrm(y) (recall that the norm of an irreducible element is a rational rime). This immediately imlies that Nrm(x) (for instance) has norm 1 and hence is a unit. Therefore, the elements in (i) and (ii) are indeed rimes. For the third case, note that a factorization, = xy would imly 2 = Nrm(x)Nrm(y) and so = Nrm(x) = Nrm(y) = a 2 + b 2. But this is equivalent to 1 (mod4) and contradicts the assumtion of (iii). We still need to rove that any Z[ 1] is associated to one (and only one) of the forms (i) (iii). To do this, assume that Nrm() factors as Nrm() = = 1 n, with the i s rime in Z[ 1]. From this we see that 1 n and so i for some 1 i n. This means that Nrm() Nrm( i ) Nrm() 2 i, so either Nrm() = i or Nrm() = 2 i. In the first case we get that = a + b 1 with a 2 + b 2 = i so is on the form (ii), or if i = 2, is associated with 1 + 1, i.e., (i). In the second case, since i i = a we get that 2 i = Nrm( i ) = Nrm()Nrm(a) and so Nrm(a) = 1 a U(Z[ 1]). This would then mean that i 3 (mod4) since otherwise i = 2 or i 1 (mod4) which would mean that i = (a + b 1)(a b 1) and this is imossible when is a rime, so we must have that i 3 (mod4).

20 20 D. LARSSON FIGURE 1. The geometry of Z[ 1] Corollary Every rational rime Z decomoses in Z[ 1] into rimes as = (a + b 1)(a b 1) if 1 (mod4), or remains rime in Z[ 1] if 3 (mod4). If = 2 then we see that 2 = (1+ 1) 2 1 is associated to the square of (1 + 1). so See Figure 1 for the geometric interretation of this result Polynomials. We could define rings of olynomials on any level of abstraction. It can be well argued that the more abstract the more general, but for the urose of number theory this is a bit shooting over the target. For this reason, we will take a more hands-on aroach, sacrificing absolute rigor and utmost generality in favor of real mathematics. So we make the following definition. Definition Let A be a ring. Then a olynomial over A is a formal sum a 0 + a 1 z + a 2 z a n z n, with a i A, 0 n <. The set of all olynomials in an indeterminate z is denoted by A[z]. We want to make this into a ring. This we do with the following definitions. Let (z), q(z) A[z]. Then we define, (z) + q(z) = ( z + + n z n ) + (q 0 + q 1 z + + q m z m ) = = ( 0 + q 0 ) + ( 1 + q 1 )z +, and (z)q(z) = ( z + + n z n )(q 0 + q 1 z + + q m z m ) = ( = i q j )z k n+m( k = k j q j )z k. i+ j=k k=0 j=0 n+m k=0 Notice that, if we want A[z] to be a (commutative) ring, distributivity forces us to have the above multilication. We also make the convention that z 0 = 1 A[z] = 1 A. In this way A becomes a subring of A[z]. Theorem The above definitions endow A[z] with the structure of a commutative ring with unity, and A can in a natural way be considered a subring of A[z]. Proof. Exercise!

21 21 It would be a serious negligence of duty if I didn t mention the ring of formal ower series over A, denoted A[[z]], at this oint (although we will robably don t need it). This is defined to be the set of elements of the form i=0 a i z i = a 0 + a 1 z + a 2 z a n z n +, a i A, with the same addition and multilication as above. However, we need to observe that the sum giving the coefficient of z i is always a finite sum, so the definition makes sense. Also, we disregard all questions of convergence. This is indeed a formal construction Iterated olynomial rings. Since A was only suosed to be a ring and A[z] was a ring we can consider the iterate A[z][w]. This is obviously then the ring of olynomials z 0 + z 1 w + z 2 w z n w n, with z i A[z], 0 n <. Convince yourselves that we have the equality A[z][w] = A[z,w], that is, any (w) A[z][w] can be written as i j z i w j, with i j A, 0 i, j <. i, j In this way we can continue to iteratively construct olynomial rings A[z] = A[z 1,z 2,...,z m ], in several indeterminates, z := {z 1,z 2,...,z m } (think through this!). From now on we will however focus our attention on the one-determinate case The degree-function on A[z]. There is a degree-function deg : A[z] N 0 defined by deg (z) = deg( z + + n z n ) := n, and deg(0) =. Observe that some author s define deg(0) =. The difference is a matter of taste more than anything. We note one immediate consequence of deg: Theorem If D is a domain then deg(q) = deg + degq and the ring D[z] is also a domain. Proof. The first statement follows from the definition of a roduct of two olynomials, observing that the highest coefficient is not zero, since deg and q degq are non-zero (and not zero-divisors). Assume 0,q D[z] with q = 0. Then = deg(0) = deg(q) = deg + degq. For this last sum to be either deg or degq (or both) must be and the only olynomial satisfying this is the zero olynomial. Hence or q (or both) is zero. A olynomial f is called irreducible if f = gh imlies that either g or h is a unit. If it is not irreducible, it is said to be reducible. We have the following useful criterion: The Eisenstein criterion. Theorem Let f (z) Z[z], f (z) = f 0 + f 1 z + + f n z n. Suose there is a rime Z such that f n, a i for 0 i n 1, and 2 f0 2. Then f (z) is irreducible over Q.

22 22 D. LARSSON D[z] is Euclidean. The following theorem is well-known to you, at least when formulated differently. Theorem The ring F[z], for F a field (i.e., a domain where all non-zero elements are invertible), is a Euclidean domain with deg as Euclidean valuation. Proof. We know from high school that there is a division algorithm on F[z] (at least when F is the field of real numbers R). That deg(q) deg() is obvious for,q 0. This result holds for more general coefficient rings (other than fields) but then the degree function is not the right choice for an Euclidean valuation. As a result we have the first statement of the following theorem. Theorem Let D[z] be a olynomial ring over a domain D. Then, (i) if D is a field D[z] is a PID and hence a UFD; (ii) if D is a UFD, D[z] is a UFD. Proof. The roof of the first statement follows from Theorems 4.33 and For the roof of the second I refer you to more secialized abstract algebra literature D[z,w] is not Euclidean. The reason is simly because it is not a PID. There are ideals in D[z,w] which cannot be generated by one element. One examle: i := z,w Reduction modulo ideals in A. We have seen how to define quotient rings A/i modulo an ideal i. Given a olynomial A[z] there is a ring homomorhism A[z] (A/i)[z], reduction mod i, where we reduce the coefficients modulo i. Examle Let 3+5z 6z 2 +z 3 Z[z]. Reducing this olynomial mod 5 Z yields the olynomial 3 + 4z 2 + z 3 Z/ 5 [z]. This is a useful technique in number theory. In fact we have the following useful result: Theorem Let f (z) Z[z]. If the reduction mod n, f (z) Z/ n [z], of f is irreducible, where deg( f ) = deg( f ), then f is irreducible over Z. Proof. Suose f is reducible: f = gh. Then the ring morhism Z Z/ n yields a ring morhism Z[z] Z/ n [z] maing f f. Therefore, f = gh = ḡ h, and so, since deg( f ) = deg( f ), deg(ḡ) = deg(g) and deg( h) = deg(h) and so f is also reducible. 5. FIELDS Now we begin to aroach the heart of the subject, namely algebraic number fields. But first some more generalities Definition and examles. Definition 5.1. A field F is an integral domain where every non-zero element is invertible. We will often denote fields by blackboard letters, F,E,J,L, etc. The one excetion, and the examle most imortant to us, is algebraic field extensions. Traditionally, these are written in an ordinary roman (italic) font. Examle 5.1. The following are examles of fields: - Q, R and C; - F := Z/, where is rime;