Course notes. for 600 algebra. Jon Brundan Department of Mathematics University of Oregon Eugene

Transcription

1 Course notes for 600 algebra Jon Brundan Department of Mathematics University of Oregon Eugene

2

3 Contents 0 Basic algebraic categories Categories Examples of categories Products and coproducts Functors and natural transformations *Additive categories* Groups Isomorphism theorems Cyclic and dihedral groups The symmetric group Free groups Group actions The Sylow theorems Semidirect products Solvable and nilpotent groups Jordan-Hölder theorem Rings Homomorphisms and ideals Maximal and prime ideals Euclidean domains and PIDs Unique factorization domains Localization of rings *Factorization in polynomial rings* Introduction to modules Modules, submodules and homomorphisms Direct products and direct sums Free modules Elementary divisors Structure theorems for modules over PIDs General theory of modules Short exact sequences Projectives and injectives Semisimple modules Tensor products Adjoint functors Morita equivalence

4 4 CONTENTS 5 Structure of rings Algebras Chain conditions Wedderburn structure theorems The Jacobson radical Topics in linear algebra Change of basis Jordan normal form Rational normal form *Simultaneous diagonalization* Tensor and symmetric algebras Determinants and the exterior algebra Fields Field extensions Transcendental extensions Splitting fields Normal extensions Separable extensions The fundamental theorem of Galois theory Radical extensions Finite fields Commutative algebra Localization revisited Completion The prime spectrum of a ring Ring extensions Hilbert s basis theorem Nullstellensatz

5 0 CONTENTS

6 Chapter 0 Basic algebraic categories 0.1 Categories There are some logical problems with the foundations of category theory. So we re going to need some basic notions of set theory, about which I want to be as vague as possible. In set theory, there are two basic notions: the notion of a class and the notion of is an element of,. It s best not to try to define these too carefully or you start running into paradoxes! A set is then a small class: formally, a set is defined to be a class that is an element of another class. Then there are some axioms giving you existence of various sets, such as the empty set, the power set of a set, unions, intersections, and so on. These axioms guarantee that there are at least enough sets to do anything reasonable with. Most of these axioms do not apply to classes though, preventing paradoxes like the set of all sets (its a class!). Now, a category C consists of a class ob(c ) of objects and for each pair A, B of objects, a set Hom C (A, B) of morphisms from A to B. (In case the class ob(c ) is actually itself a set, the category is called small.) We write simply f : A B to indicate that f is a morphism from A to B. So you should think of a morphism as simply an arrow from A to B indeed I ll often call a morphism an arrow to emphasize that part of the data of a morphism is the name of its source object and its destination object... Moreover, there is a given map Hom C (B, C) Hom C (A, B) Hom C (A, C) for each triple of objects A, B, C. We write the image of a pair (f, g) of morphisms under this map by f g and call it the composition of f and g. (Note wherever possible I write maps on the left.) In addition, the following axioms are imposed: (C1) for every object A, there exists a distinguished arrow id A : A A such that id A f = f and g id A = g for all other arrows f : B A and g : A B. (C2) given arrows f : A B, g : B C, h : C D, we have that h (g f) = (h g) f. For obvious reasons, (C1) is called the identity axiom and (C2) is called the associative axiom. Here s an immediate consequence of the axioms: if i : A A is an arrow such that i f = f and g i = g for every f : B A and g : A B, then i = id A. In other words, the identity morphism id A is unique. Proof: i = id A i = id A. In algebra, we will study various sorts of algebraic structure. The notion of category is the umbrella to keep the various structures under! A few more definitions. We say B is a subcategory of C if ob(b) ob(c ) and for every pair of objects A, B in B, Hom B (A, B) Hom C (A, B). A subcategory B of C is called a full subcategory if in fact Hom B (A, B) = Hom C (A, B) for every pair of objects A, B in B. Note to specify a full subcategory of C it is sufficient just to specify a subset of the objects: then one has no choice but to take all morphisms that make sense to obtain the morphisms in the subcategory. 1

7 2 CHAPTER 0. BASIC ALGEBRAIC CATEGORIES A morphism f : A B in a category C is called an isomorphism if there exists another morphism g : B A such that f g = id B and g f = id A. 0.2 Examples of categories So I now want to define the various categories that we re going to be working with The category sets of sets. The objects are all sets. The morphisms are just the functions between the sets. Recall a function f : A B between two sets means a subset f of A B = {(a, b) a A, b B} such that for every a A there exists a unique b B with (a, b) f (of course we always write f(a) = b instead of (a, b) f!) For example, the empty set is a set, hence an object in the category of sets. Given any other set A, Hom sets (, A) consists of exactly one morphism namely the function which in the set notation for functions is just the empty set! The category mon of monoids. A monoid is a set S with an associative operation S S S (i.e. writing the operation just as a product, a(bc) = (ab)c for all a, b, c S) such that there exists a (necessarily unique) element 1 S satisfying 1a = a1 = a for all a S. This element 1 is called the identity element. Note the associativity means that we can unambiguously write a 1 a 2... a n for the product of n elements of S: any way of interpreting this by putting brackets in gives the same outcome by associativity (Warning: this is harder to prove formally than you might think!). Given a monoid S, I can associate to S a category consisting of just one object and with the set of morphisms being precisely the set S underlying the monoid. The rule for composition of morphisms is just the multiplication in the monoid S. So: a monoid is exactly a category with one element. Turning this around, you can think of a category as a generalization of a monoid! Note the argument given in the previous section to prove that the identity morphism id is unique proves at the same time that the identity element 1 of a monoid S is uniquely determined. I haven t yet defined the category of monoids: I haven t told you the set of morphisms Hom mon (S, T ) between two monoids S and T. By definition, a morphism f : S T is a function such that f(ab) = f(a)f(b) for all a, b S and such that f(1 S ) = 1 T. Note the second axiom of a morphism here is not a consequence of the first!! Now you can check that with this definition of morphism, we do indeed obtain a category The category groups of groups. A group is a monoid G such that in addition, for every element g G there is a two-sided inverse g 1 G, i.e. an element such that g 1 g = 1 = gg 1. Note the inverse g 1 of g is uniquely determined: given another element g with the same property g g = 1 = gg, we have that g 1 = 1g 1 = (g g)g 1 = g (gg 1 ) = g 1 = g. A morphism of groups is the same as a morphism of monoids, and this gives us the category of groups. In fact, the category of groups is a full subcategory of the category of monoids (because morphisms mean the same thing and groups are just special monoids). You should notice that to test if a function f : G H is a morphism of groups, henceforth called a homomorphism, it suffices to check the multiplicative property that f(g 1 g 2 ) = f(g 1 )f(g 2 ). The other condition that f(1 G ) = 1 H is automatic. To see this, first observe that if gh = h for any g, h G then g = 1. Indeed, gh = h implies g = ghh 1 = hh 1 = 1. Hence, since f(1 G )f(1 G ) = f(1 G 1 G ) = f(1 G ), we deduce that f(1 G ) = 1 H. Now you can also show that for a group homomorphism f : G H, you always have that f(g 1 ) = f(g) The category ab of Abelian groups. This is the full subcategory of the category of groups consisting of all groups G such that gh = hg for all g, h G. Actually, if a group G is Abelian, we will generally write the operation additively, i.e. writing g + h instead of gh to help remind us that g + h = h + g. Also, we ll write the identity element of an Abelian group G not as 1 = 1 G but as 0 = 0 G.

8 0.3. PRODUCTS AND COPRODUCTS The category rings of rings. A ring is an Abelian group (written in the additive notation) together with an additional multiplication operation written in multiplicative notation satisfying the following axioms: (R1) a(b + c) = ab + ac, (b + c)a = ba + ca for all a, b, c R; (R2) (ab)c = a(bc); (R3) there exists a distinguished element 1 = 1 R R such that 1a = a = a1 for all a R. Actually, strictly speaking because I ve included (R3), we re always talking about unital rings. In Hungerford, and in other courses especially analysis, you will meet more general, not necessarily unital rings where axiom (R3) is dropped. Note: the identity element 1 R is uniquely determined (exercise). If R is a ring, we ll always write R = {a R a 0}, R = {a R there exists an element a 1 R such that aa 1 = 1 = a 1 a}. Elements of R are called units. I haven t defined the category of rings yet, because I haven t defined what a morphism of rings means. This is a morphism f : R S of Abelian groups such that f(ab) = f(a)f(b) for all a, b R, and moreover f(1 R ) = 1 S. (Hungerford calls this a unital homomorphism because of the requirement that 1 goes to 1: this is not in general implied by the multiplicative property). I should of course mention the ring Z of integers under usual multiplication and addition. Note for any ring R, there is a unique ring homomorphism Z R. Indeed, if f : Z R is a ring homomorphism, then we have no choice but to set f(0) = 0 R and f(1) = 1 R. Hence since f is a homomorphism, f(n) = 1 R R (n times) for n 1, and f( n) = f(n). We ll always write n.1 R for the element f(n). The other basic example of a ring is the zero ring, namely the ring with just one element, namely 0 = 1. In all other rings, we have that The category fields of fields. A field is a commutative ring, i.e. a ring F in which ab = ba for all a, b F, such that in addition, 1 0 and every non-zero element is a unit. The category of fields is a full subcategory of the category of rings. For example, the ring Z n of integers modulo n is a field if and only if n is prime The category vec(f ) of finite dimensional vector spaces over a field F. The objects are the finite dimensional vector spaces over F, the morphisms are the F -linear maps between vector spaces. There are of course many other examples of categories, we ll introduce the ones we need when we need them. You ll also see examples in other subjects, for example the category of topological spaces, morphisms being continuous maps. 0.3 Products and coproducts The language of category theory is ideally suited for defining general notions which arise again and again when studying examples. In this section, I want to introduce the notions of products and coproducts (a.k.a. direct sums) in a general categorical setting. We ll see them again when studying the category of groups, the category of rings, of modules, etc... so its useful to know there s some motivating theory justifying the terminology in the examples. You should by the way never get carried away by category theory: algebra is really all about examples category theory just provides us with a language to understand the examples in a unified way. Let me mention first the notions of initial and terminal objects in a category, since these are easy but illustrate the idea of categorical definitions quite well.

9 4 CHAPTER 0. BASIC ALGEBRAIC CATEGORIES An object I in a category C is called an initial object if for every object A in C, there s a unique morphism from I to A. An object T in a category C is called an terminal object if for every object A in C, there s a unique morphism from A to I. An object Z in a category C is called a zero object if its both an initial and a terminal object. Its an exercise to show that if an initial (resp. terminal, resp. zero) object exists (it may not!) then it is unique up to a unique (or canonical) isomorphism in C. For example, in the category of groups, the trivial group {1} is a zero object. In the category of rings, Z is an initial object and the zero ring {0} is a terminal object hence, the category of rings has no zero object because these are different. In the category of sets, the empty set is an initial object, and any one element set is a terminal object. Now we discuss the notion of a product in an arbitrary category. So let C be a category. Suppose that A i (i I) is a family of objects in C. A product of the A i is an object P C together with morphisms π i : P A i for each i I, such that (P) given any object Q and morphisms φ i : Q A i for each i I, there exists a unique morphism φ : Q P such that π i φ = φ i for each i I. You should convince yourself that a product of an empty set of objects means exactly the same as a terminal object in the category C. The definition of product is an example of a definition by a universal property. It is always the case that if something is defined by a universal property, then (if it exists) it is unique up to a canonical isomorphism. So we generally (abusing notation) say the product of the objects A i and denote it i I A i. That is providing of course such an object exists: whether it does depends on the particular category you re working in. Note I often refer to products as direct products out of habit... Now for some examples of products in specific categories: Suppose C is the category of sets. Given sets A i (i I), their product is the set i I A i (Cartesian product). The maps π i : i I A i A i in the definition of product are just the projections. You can check this really is the product in the category of sets. Note each π i is in fact surjective (this statement is equivalent to the Axiom of Choice in set theory!!!) Suppose C is the category of groups. Given groups G i (i I), their product is the Cartesian product i I G i of the underlying sets, made into a group by defining multiplication coordinatewise. The projections π i : i I G i G i are group homomorphisms, and this data satisfies the universal property to be the product in the category of groups. Exactly the same construction works in the category of Abelian groups, or in the category of rings (where one needs to define both multiplication and addition coordinatewise) Finally, suppose C is the category of fields. Consider the fields Z 2 and Z 3 of integers modulo 2 and 3 respectively. What could their product Z 2 Z 3 be? Well, it would have to be a field F which had field homomorphisms to the both of the fields Z 2 and Z 3. Hence, since field homomorphisms are always injective, we would have to have that both = 0 and that = 0 in F, which implies that 1 = 0 which is not allowed in a field. Hence, products do not exist in the category of fields. We turn to discussing coproducts. Let C be any category and A i (i I) be a family of objects in C. A coproduct of the A i is an object C C together with morphisms ι i : A i C for each i I, such that (C) given any object D and morphisms φ i : A i D for each i I, there exists a unique morphism φ : C D such that φ ι i = φ i for each i I. Observe the definition of coproduct is the definition of product but with all arrows reversed. Moreover, a coproduct of an empty set of objects means exactly the same as an initial object in the category C.

10 0.4. FUNCTORS AND NATURAL TRANSFORMATIONS 5 If a coproduct of the A i exists it is unique up to a canonical isomorphism. So we call it the coproduct and denote it by i I A i. Examples: In the category of sets, the coproduct is just the disjoint union of the sets, the maps ι i are the inclusions. (This is also the coproduct in the category of topological spaces where you define the open sets in the disjoint union to be disjoint unions of opens sets) Coproducts in the category of groups exist but are rather nasty things called free products and we will not discuss them here. But in the category of Abelian groups, coproducts are easy to understand and very important. Let {G i i I} be Abelian groups. Then, their coproduct i I A i is defined to be the set of elements (a i ) i I of the Cartesian product i I A i such that a i = 0 for all but finitely many i I. We usually denote the element (a i ) i I of i I A i as i I a i instead (morally the sum makes sense because we re adding up elements all but finitely many of which are zero!). Given two elements i I a i and i I b i of i I A i, their sum (remember the operation in an Abelian group is written additively!) is simply i I (a i + b i ) all but finitely many terms are again zero so this makes sense. Now we define the maps ι i : A i i I A i needed in the definition of coproduct. One has simply that ι i (a) = j I a j where a j = 0 for j i and a i = a. Then you need to check that the object we have defined together with these morphisms really is a coproduct in the category of Abelian groups. We will often write i I instead of i I A i and call it the direct sum of the Abelian groups A i instead of the coproduct. Notice that if the underlying set I is finite, then the object i I A i is exactly the same as the object i I A i: that is finite products and coproducts coincide for Abelian groups. In particular, if I = {1, 2}, we get that A 1 A 2 (the notation for product of just two objects) equals A 1 A 2 (the notation for sum of just two objects). You need to be flexible about which you write!! Coproducts do not exist (in general) in the category of rings. 0.4 Functors and natural transformations So categories encompass different sorts of algebraic structures. We should also allow maps between categories if we wish to be able to compare different algebraic structures. Let A, B be categories. A functor F : A B means: (F1) a rule assigning to each object A A an object F A B; (F2) a rule assigning to each arrow f : A B in A an arrow F f : F A F B in B. These rules should satisfy the following axioms: (F3) F (f g) = F f F g for all arrows f, g in A whose composition makes sense; (F4) F id A = id F A for each object A in A. I ll just give two examples of functors right now. We ll meet many more as we go along. First, let C be any category. Then, there s a functor Id C : C C defined to be the identity on objects and morphisms. It s called the identity functor. Second, define a functor F : groups sets by setting F G = G for a group G (i.e. F G is the same underlying set as G but we ve just forgotten that its a set with additional group structure). On a morphism f : G H, let F f be the same function, viewed just as a map of sets. A functor like this one F is called a forgetful functor: we re just calling the group a set and forgetting the extra group structure! Strictly speaking, people call a functor as I ve defined it above a covariant functor. There s also another notion of contravariant functor. Its almost the same thing, but given a morphism A i

11 6 CHAPTER 0. BASIC ALGEBRAIC CATEGORIES f : A B in A, a contravariant functor sends it to a morphism F f : F B F A in B (before it was F f : F A F B). The composition axiom (F3) becomes F (f g) = F g F f for all morphisms f, g whose composition makes sense. Informally, contravariant functors reverse the directions of arrows. I prefer usually to use the word functor for covariant functor. To eliminate the need for defining everything twice, once for covariant once for contravariant functors, I ll use a trick: the opposite of a category. So let C be a category. Define C op, the opposite category, as follows. The objects are the same. Moreover, given objects A, B, the set of morphisms Hom C op(a, B) is defined to be the set Hom C (B, A). Given arrows f : A B and g : B C in C op (i.e. arrows f : B A and g : C B in C ) the composite g f in the category C op is defined to be the same as the composite f g in the category C. So the category C op is really the same as the category C but with the direction of all arrows reversed. With this trick in mind, we then have that a contravariant functor F : A B is the same as an ordinary (covariant) functor F : A op B (or for that matter F : A B op ). For an example of a contravariant functor, consider the category vec(f ) of finite dimensional vector spaces over a field F. Given such a vector space V, we have the dual space V = Hom F (V, F ) of all linear functionals on V. If e 1,..., e n is a basis for V, then we obtain a basis for V by considering the dual basis f 1,..., f n. Here, f i is the linear map from V to F determined uniquely by the formula { 1 if i = j, f i (e j ) = δ i,j = 0 otherwise. So V is another vector space of the same dimension n as V itself. Now given a morphism θ : V W, i.e. a linear transformation between finite dimensional vector spaces, we can define the dual map θ : W V as follows. Take f W. So f is a linear map from W to F. Then, θ f is the element of V, i.e. the linear map from V to F, defined by (θ f)(v) = f(θ(v)) for all v V. Now we can define a contravariant functor D : vec(f ) vec(f ) by setting D(V ) = V for an object V and D(f) = f for a morphism f. Of course, you need to check things like D(f g) = D(g) D(f), i.e. (f g) = g f, to verify that this really is a contravariant functor. Alternatively, you can think of D as an ordinary (covariant) functor from vec(f ) op vec(f ) or vec(f ) vec(f ) op. Composing the functor D with itself, we obtain a functor D D : vec(f ) vec(f ). So (D D)(V ) = (V ) and (D D)(f) = (f ). Now, for each finite dimensional vector space V, there is a linear map (1) i V : V (V ) defined so that for each v V, i V (v) is the functional on V such that i V (v)(f) = f(v) for each f V. You should check that i V (e 1 ),..., i V (e n ) is simply the basis for V that s dual to the dual basis f 1,..., f n for V. In other words, i V : V (V ) is an isomorphism since it maps a basis to a basis. This example brings us to the next notion in category theory. Let A and B be two categories. They are called isomorphic if there exist (covariant) functors F : A B and G : B A such that F G = Id B and G F = Id A. The duality functor D : vec(f ) op vec(f ) defined above really ought to be an isomorphism of categories. The inverse functor vec(f ) vec(f ) op really ought to be just D itself again. But it is not!!!! Indeed, D D(V ) = (V ) which, although isomorphic to V is not equal to V. This is maybe an indication that the notion of isomorphism of

12 0.5. *ADDITIVE CATEGORIES* 7 categories is not the useful one. Instead we need something slightly weaker called equivalence of categories. Two categories A and B are said to be equivalent if there exist functors F : A B and G : B A such that F G = Id B and G F = Id A. (Almost the same as the definition of isomorphic categories in the previous paragraph but the functors are only = Id not = Id!) We do not understand this definition yet, because I have not defined what it means for functors to be isomorphic rather than equal. For this, we need the notion of natural transformation of functors. So finally, let F, G : A B be two functors. A natural transformation η : F G means the following data: for each object A A, a morphism η A : F (A) G(A) in the category B. These maps η A need to satisfy the following axiom: for each arrow f : A B in A, we have that η B F f = Gf η A. A natural transformation is called a natural isomorphism between the functors F and G if in addition each morphism η A is actually an isomorphism in the category B. If F and G are naturally isomorphic functors, we write F = G. Now let s go back to our example, the functor D : vec(f ) op vec(f ). I claim that D is an equivalence of categories (so that vec(f ) is equivalent to its opposite category). To prove this, consider the functor D : vec(f ) vec(f ) op. I claim that D D = Id (either way round). To prove this, we need a natural isomorphism between these two functors. But we wrote down a natural isomorphism i : Id D D of functors earlier (1) (you should check this of course). We will see more of these notions of natural transformations later on (chapter 5). It is good enough at this stage to think of natural transformations rather informally: it simply means that your definition doesn t depend on any specific information about the object you are considering. 0.5 *Additive categories* I want to include here one last categorical notion. This section is starred, which means we did not cover it in class and you shouldn t waste time on it when preparing for exams!! A category C is called an additive category if (A1) for all objects A, B C, Hom C (A, B) has the additional structure of an Abelian group (the operation written f + g : A B for arrows f, g : A B); (A2) addition distributes over composition, i.e. f (g+g ) = f g+f g and (f +f ) g = f g+f g for arrows f, f : B C and g, g : A B; (A3) each finite family of objects in C has both a product and a coproduct. Note axiom (A3) in particular implies that the empty family consisting of no objects in C has both a product and a coproduct. This means that C has both a terminal and an initial object. Actually, it would be enough in axiom (A3) just to require that C has an initial and a terminal object and that every pair of objects in C possesses a product and a coproduct. For finite families of objects in a category C satisfying axioms (A1) and (A2), there is a third notion of product called a biproduct which connects the two notions of product and coproduct discussed so far. Let X 1,..., X n be objects in C. A biproduct of X 1,..., X n, to be an object X together with maps p i : X X i, q i : X i X such that p j q i = δ i,j id Xi, n q i p i = id X. i=1 Here, δ i,j id Xi denotes the identity map X i X i in case i = j and the zero map X i X j (the zero in the Abelian group Hom C (X i, X j )) in case i j.

13 8 CHAPTER 0. BASIC ALGEBRAIC CATEGORIES Theorem. In a category C satisfying axioms (A1) and (A2) above, let X 1,..., X n, X be objects and p i : X X i, q i : X i X be maps such that p j q i = δ i,j id Xi. Then, the following are equivalent: (1) X together with the maps p i is a product of the X i ; (2) X together with the maps q i is a coproduct of the X i ; (3) X together with the maps p i and q i is a biproduct of the X i. Proof. We just need to prove (1) and (3) are equivalent. The equivalence of (2) and (3) is then immediate considering the opposite category to C instead. (1) (3). Suppose that X together with the p i is a product. Set φ = n i=1 q i p i. Then, p i φ = n p i q j p j = j=1 n δ j,i p j = p i. It follows from the uniqueness in the universal property of products that φ = id X, so X is a biproduct. (3) (1). Suppose we have an object Y and morphisms f i : Y X i for each i. To show that X is a product, we need to prove there exists a unique f : Y X such that p i f = f i for each i. Well, f = n i=1 q i p i f so if p i f = f i for each i, then f = n i=1 q i f i. In other words, there is no choice but to define n f = q i f i. Then, as required. p j f = i=1 j=1 n p j q i f i = f j i=1 If C is a category satisfying axioms (A1) and (A2), suppose that X together with maps p i : X X i is a product of objects X 1,..., X n. Fix j and defined f i : X j X i for each i. By the universal property of products, there exists a unique q j : X j X such that p i q j = δ i,j id Xi. Therefore by the theorem, X together with the maps q i : X i X is a coproduct of the X i. Making the same argument in the opposite category too, we obtain: Corollary. Let C be a category satisfying (A1) and (A2). Then any finite family of objects of C has a product if and only if it has a coproduct, in which case the two are isomorphic. In particular, this implies that in an additive category, initial objects and terminal objects are isomorphic, hence there is a zero object. Moreover, finite products and finite coproducts are isomorphic. The basic example of an additive category is the category ab of Abelian groups. We have already constructed finite products and coproducts (which happened to be isomorphic). The addition of two morphisms f, g : A 1 A 2 is defined by (f + g)(a) = f(a) + g(a), giving the Abelian group structure on the Hom sets. On the other hand, the category rings is not additive, even though it satisfies axioms (A1) and (A2): for, the initial object is Z and the terminal object is (0) so there is no zero object. Now let B and C be two additive categories. A functor F : B C is called an additive functor if F (f + g) = F f + F g for all arrows f, g : X Y in B. In other words, the set map Hom B (X, Y ) Hom B (F X, F Y ) induced by the functor F is actually a morphism of Abelian groups.

14 0.5. *ADDITIVE CATEGORIES* 9 Additive functors commute with finite products. Let F : A B be an additive functor between additive categories. If X together with maps p i : X X i is a product of finitely many objects X i in A, then F X together with the maps F p i : F X F X i is a product of the objects F X i in B. Proof. By the theorem (1) (3), there exist maps q i : X i X such that X together with these maps is a biproduct of the X i. The result of applying an additive functor to a biproduct is clearly a biproduct. Hence by the theorem (3) (1) it is a product. Restating this result for the opposite categories (which are also additive), you obtain the dual statement: Additive functors commute with finite coproducts. Let F : A B be an additive functor between additive categories. If X together with maps q i : X i X is a coproduct of finitely many objects X i in A, then F X together with the maps F q i : F X i F X is a coproduct of the objects F X i in B. These two general nonsense results are very useful. Actually, in the cases we need to apply them (basically tensor and hom functors) it is easy to prove the conclusion directly. But it gives you the flavour of the sort of general facts you can prove by developing category theory. The point really of introducing an additive category is that notions of image and kernel of a morphism can be defined in an additive category in an abstract way (though they may not exist in general). Then you go on to define something called an Abelian category, namely, an additive category with some extra axioms devised exactly so that there is a meaningful notion of exact sequence in such a category. For example, ab turns out to be an example of an Abelian category (and hence the name!). Fortunately for us, we will be able to avoid such abstraction since we will only consider one specific Abelian category in what follows, namely, the module category of a ring. In this particular category, it is easier to understand the notion of exact sequence without category theory.

15 10 CHAPTER 0. BASIC ALGEBRAIC CATEGORIES

16 Chapter 1 Groups 1.1 Isomorphism theorems Throughout the chapter, we ll be studying the category of groups. Let G, H be groups. Recall that a homomorphism f : G H means a function such that f(g 1 g 2 ) = f(g 1 )f(g 2 ) for all g 1, g 2 G. A homomorphism is called a monomorphism if it is injective, i.e. there exists a set function f : H G such that f f = id G. It s an epimorphism if its surjective, i.e. there exists a set function f : H G such that f f = id H. It s an isomorphism if it s bijective, i.e. both injective and surjective. Note this definition of isomorphism of groups is equivalent to the definition of isomorphism given earlier for an arbitrary category: luckily for groups a morphism is an isomorphism if and only if it is a bijection. We also talk about endomorphisms (homomorphisms from G to itself) and automorphisms (isomorphisms from G to itself). Given a homomorphism f : G H, we set im f = f(g) = {f(g) g G}, ker f = f 1 (1 H ) = {g G f(g) = 1 H }, its image and kernel. A subgroup of a group G is a non-empty subset K of G such that k 1 k 1 2 K for every k 1, k 2 K. This condition implies that K is itself a group in its own right, the operation being the restriction of the operation on G to K. We usually write K G to indicate that K is a subgroup of G. For example, im f H (and f is an epimorphism if and only if im f = H); ker f G (and f is a monomorphism if and only if ker f = {1 G }). Actually, ker f is more than just a subgroup of G: it is a normal subgroup. By definition, a subgroup N G is normal if any of the following equivalent conditions hold: (N1) gng 1 = {gng 1 n N} = N for all g G; (N2) gng 1 = {gng 1 n N} N for all g G; (N3) gng 1 N for all g G, n N; (N4) Ng = gn for all g G. We usually write N G to indicate that N is a normal subgroup of G. The sets Ng = {ng n N} and gn = {gn n N} appearing in (N4) are the right cosets and left cosets of N in G. Let s make this notion more precise. Let K be any subgroup of G. Define an equivalence relation on G called left congruence modulo K by g 1 g 2 if g1 1 g 2 K. The equivalence classes are called the left cosets of K in G: each equivalence class has the form gk for some (any) representative g G of the equivalence class. The set of all left cosets of K in G is denoted G/K. One also defines right cosets to be the equivalence classes of a relation called right congruence modulo K, namely, g 1 g 2 if g 1 g2 1 K. Right cosets look like Kg for g G, and the set of all 11

17 12 CHAPTER 1. GROUPS right cosets is denoted K\G. Note the map g g 1 induces a bijection between G/K and K\G, so it usually doesn t matter whether you choose work with left or right cosets. In case G is a finite group and K G, we let [G : K] = G/K denote the number of left cosets of K in G, the index of K in G. Now, all the left cosets have the same size, namely, K. Since equivalence classes partition G into disjoint subsets, we deduce: Lagrange s theorem. G = K [G : K]. Lagrange s theorem shows in particular that the order of any subgroup of a finite group G divides the order of the group. Now return to the case that N is a normal subgroup of G. Then, (N4) says that G/N = N\G, i.e. left cosets are the same as right cosets so we can just call them simply cosets. Define a multiplication on the set of cosets G/N by (g 1 N)(g 2 N) = (g 1 g 2 )N. The fact that this is well-defined depends on N being a normal subgroup. This multiplication gives G/N the structure of a group in its own right, called the quotient group of G by the normal subgroup N. Note that there is a canonical homomorphism π : G G/N, g gn, which is an epimorphism of G onto G/N with kernel exactly N. The group G/N together with the map π : G G/N has the following universal property: Universal property of quotients. Let N G and π : G G/N be the canonical epimorphism. Given any homomorphism f : G H with N ker f, there exists a unique homomorphism f : G/N H such that f = f π. Exercise. Let G be a group and H, K be subgroups. We can consider the subset HK = {hk h H, k K} of G. (1) HK is a subgroup of G if and only if HK = KH as subsets of G. (2) If either of H or K is a normal subgroup of G, then HK = KH, hence is a subgroup of G. G. (3) If in fact both H and K are normal subgroups of G, then HK is a normal subgroup of G. Now let me state the isomorphism theorems for groups. These are all really consequences of the universal property of quotients: you should be able to prove them for yourselves. First isomorphism theorem. Let f : G H and N = ker f. Then, N G and f factors through the quotient G/N to induce an isomorphism f : G/N im f. Second isomorphism theorem. Let K G, N G. Then, N KN, K N K and K/K N = KN/N. Third isomorphism theorem. Let K N G with K G, N G. Then, N/K G/K and G/N = (G/K)/(N/K). There is one other important result traditionally included with the isomorphism theorems: the lattice isomorphism theorem. First, recall that a relation on a set X is called a partial ordering (and X is called a partially ordered set) if for all x, y, z X (R1) x x; (R2) x y and y x implies y = x; (R3) x y and y z implies x z.

18 1.2. CYCLIC AND DIHEDRAL GROUPS 13 If in addition we have that for all x, y X one of x < y, x = y, x > y holds, then < is called a total order or a linear order (and X is called a totally ordered set). Now let X be a partially ordered set and A X. An element a X is called the join or least upper bound of A if a is the unique minimal element of {x X a x a A}. The join A may or may not exist: for instance there could be many such minimal elements or no minimal element at all. Similarly, an element a X is called the meet or greatest lower bound of A if a is the unique maximal element of {x X a x a A}. A partially ordered set is called a lattice if every pair of elements of X has both a join and a meet in X. A partially ordered set is called a complete lattice if every non-empty subset of X has both a join and a meet in X. Let G be any group and X be the set of all subgroups of G, partially ordered by inclusion. Then, X is a complete lattice. The meet of a set of subgroups of G is simply their intersection, also a subgroup of G. The join of a set of subgroups of G is the subgroup generated by the subgroups, that is, the intersection of all subgroups of G that contain all the given subgroups. More generally, given any subset A of G, we write A for the subgroup of G generated by A, namely, the intersection of all subgroups of G that contain A. In the special case A = {a} for some a G, we write simply a for the subgroup of G generated by the element a. As a variation, let G be any group and X be the set of all normal subgroups of G, partially ordered by inclusion. One checks that the intersection of a family normal subgroups is normal, and that the group generated by a family of normal subgroups is normal. Hence, X in this case is again a complete lattice, with join and meet being as for subgroups. Lattice isomorphism theorem. Let f : G G be an epimorphism with kernel K. Then, the map H f(h) gives an isomorphism between the lattice of subgroups (resp. normal subgroups) of G containing K and the lattice of subgroups (resp. normal subgroups) of G. (The inverse map sends a subgroup H of G to its preimage f 1 (H ) in G.) 1.2 Cyclic and dihedral groups It s nearly time for some examples of groups. But first, let G be any group and g G an element. Recall g denotes the subgroup of G generated by the element g. So, g = {1 G, g ±1, g ±2, g ±3,... }, and is called the cyclic subgroup generated by g. It may or may not be a finite group; if it is finite, its order is denoted o(g), the order of the element g. Note Lagrange s theorem implies that in a finite group, the order of every element divides the order of the group. A group G is called cyclic if G = g for some element g G. We then say that g is a generator of the cyclic group G. For example, the group (Z, +) of integers under addition is an infinite cyclic group generated by the integer 1 (the identity element of Z is the integer 0 of course we use the additive notation in this Abelian group!). Actually, any infinite cyclic group is isomorphic to the group (Z, +). We can classify all subgroups of the infinite cyclic group: Lemma. If H (Z, +), then H = n for some integer n 0. Proof. Take n > 0 minimal subject to the condition that n H (if no such n exists, then H = {0} and there is nothing to prove). Then, n H. Now take any m H, and write m = qn + r for integers q, r with 0 r < n. Then, r = m qn H, so by minimality of n we have that r = 0. Hence, m = qn, so that m n. This shows that H = n. Note that n m if and only if m n. It follows that the lattice of subgroups of Z is isomorphic to the opposite of the lattice of non-negative integers partially ordered by divides. The non-negative integer n corresponds under the isomorphism to the subgroup n. Given n N, we define the group (Z n, +), the group of integers modulo n under addition, to be the quotient group Z/ n. Thus, the elements of the group Z n are the cosets of n in Z, namely, {[0], [1],..., [n 1]} where [i] = {i + jn j Z}. We have that [i] + [j] = [i + j]. The group Z n is the cyclic group of order n: it is generated by the element [1] which has order n in Z n. Sometimes

19 14 CHAPTER 1. GROUPS we denote the cyclic group of order n instead by C n, to indicate that we re writing the operation multiplicatively instead. So C n = {1, x, x 2,..., x n 1 } where x is any element of C n of order n. By the lattice isomorphism theorem, the lattice of subgroups of Z n is isomorphic to the opposite of the lattice of divisors of n. The divisor d of n corresponds under the isomorphism to the cyclic subgroup of Z n generated by [d], which is a subgroup of order n/d. In other words: Lemma. The group Z n has a unique subgroup of order d for each divisor d of n, namely, the cyclic subgroup generated by [n/d]. Given groups G 1, G 2, their (external) direct product is the set G 1 G 2 (Cartesian product) with coordinatewise multiplication. Actually, G 1 G 2, with the obvious projections π i : G 1 G 2 G i is the product of G 1 and G 2 in the categorical sense. More generally given a family G i (i I) of groups, their product i I G i is simply their Cartesian product as sets with coordinatewise multiplication. In other words, the category of groups possesses arbitrary products (see (0.3.2)) Lemma. If n = st with (s, t) = 1, then Z n = Zs Z t. Proof. The element ([1] s, [1] t ) Z s Z t has order st as (s, t) = 1. Hence it generates a cyclic subgroup of Z s Z t of order st. Hence, it generates all of Z s Z t, which is therefore isomorphic to the cyclic group of order n. The Euler φ function is defined by φ(n) = #{x Z n o(x) = n} = #{1 k < n (k, n) = 1}. By the lemma, if n = st with (s, t) = 1, then φ(n) = φ(s)φ(t). It follows immediately that to compute φ(n) it suffices to know φ(p n ) for each prime power p n. In this special case, it is an exercise to show that ( φ(p n ) = p n 1 1 ). p Here s an important number theoretic fact about the Euler φ function: Lemma. For n N, n = d n φ(d). Proof. We have from Lagrange s theorem that n = Z n = d n (the number of elements of Z n of order d). Now if x Z n has order d, it generates a subgroup of order d, isomorphic to Z d. By Lemma 1.2.2, Z n has a unique such subgroup of order d. Hence, the number of elements of Z n of order d equals the number of elements of Z d of order d, namely φ(d). I end the section by introducing one other basic example of a finite group: the dihedral group D n of order 2n. This is probably easiest left as an exercise. So, start with the group O 2 of orthogonal linear transformations of the plane R 2. So O 2 consists of all rotations (det = +1) around the origin and all reflections (det = 1) in axes passing through the origin. Consider the elements f, g O 2 where g is reflection in the x-axis and f is counterclockwise rotation through the angle 2π/n. Obviously, f n = 1, g 2 = 1. Let D n be the subgroup of O 2 generated by f and g.

20 1.3. THE SYMMETRIC GROUP 15 (1) Show that the product f i g is reflection in the line at angle iπ/n to the x-axis. Hence, the transformations 1, f, f 2,..., f n 1, g, fg,..., f n 1 g are all distinct. (2) Now show that the elements {1, f, f 2,..., f n 1, g, fg,..., f n 1 g} form a subgroup of O 2. (3) Deduce that D n = {1, f, f 2,..., f n 1, g, fg,..., f n 1 g} is a group of order 2n. Picture: D n is the group of symmetries of a regular n-gon. It contains the cyclic group C n as a subgroup, namely, the subgroup {1, f, f 2,..., f n 1 } consisting of all rotations. (4) Find all normal subgroups of D n. (5) For any group G, its centre is defined as Determine Z(D n ). 1.3 The symmetric group Z(G) = {x G xy = yx for all y G}. We come now to one of the most important examples of a group. Let X be any set. Denote by A(X) the set of all bijective ( invertible ) functions from X to itself. Then, A(X) is a group with multiplication being composition of functions. The identity element of the group A(X) is the identity function id X : X X, x x. The group A(X) is called the symmetric group on X. In the special case that X is the set {1, 2,..., n}, we denote the group A(X) instead by S n and call it the symmetric group on n letters (even though they re numbers!). Obviously, S n = n! so S n is a finite group. Now let G be any group. Then, A(G) is also a group. Define λ : G A(G), g λ g where λ g A(G) is the function x gx (i.e. λ g is left multiplication by g ). Note λ is a group homomorphism. Moreover, it s injective, for if λ g = id G, then λ g (1 G ) = 1 G = g so ker λ = {1 G }. Hence, λ defines an isomorphism between G and im λ A(G). We ve proved: Cayley s theorem. Every group is isomorphic to a subgroup of the symmetric group A(X) for some set X. Every finite group, is isomorphic to a subgroup of S n for some n. By the way, you can also define ρ : G A(G), g ρ g, where ρ g is the function right multiplication by g. But ρ satisfies ρ(gh) = ρ(h)ρ(g), the wrong way round to be a homorphism. In fact, ρ is an antihomomorphism from G to A(G). Now let s discuss the finite group S n in more detail. So it s the group of all permutations of {1,..., n}. Let x 1,..., x a be distinct elements of {1,..., n}. Denote the permutation which maps x a x 1, x i x i+1 for each i = 1,..., a 1 and fixes all other points, by (x 1 x 2... x a ). This is called an a-cycle. A 2-cycle is called a transposition. Disjoint cycle notation. Every permutation f S n can be written as a product of disjoint cycles f = (x 1 x 2... x a )(y 1 y 2... y b ).... (z 1 z 2... z a ). Moreover, this representation of the permutation f is unique up to deleting 1-cycles and reordering the product (disjoint cycles commute!). (You should be quite familiar with using the disjoint cycle notation.) Now define a function sgn : S n {±1} as follows. Take g S n. Write g = c 1... c a as a product of disjoint cycles, where c i is an o i -cycle. Set a sgn(g) = ( 1) oi 1. So for example, sgn(1) = 1, sgn(t) = 1 for t a transposition,... i=1

21 16 CHAPTER 1. GROUPS Lemma. Every element g S n can be expressed as a product of transpositions. Proof. Using the disjoint cycle decomposition, it suffices to show that any a-cycle can be written as a product of transpositions. Then, for example, the cycle (12... n) equals (12)(23)(34)... (n 1 n) Theorem. sgn is a group homomorphism. Proof. We first prove: Claim. If s is a transposition and w S n is arbitrary, then sgn(sw) = sgn(w). To see this, say s = (a b) and w = c 1... c m as a product of disjoint cycles. If s and w are disjoint, the conclusion follows by definition. Else, without loss of generality, we may assume a appears in the cycle c 1. If b appears in c 1 too, then c 1 = (a x 1... x k b y 1... y l ) and (a b)c 1 = (b y 1... y l )(a x 1... x k ) which is a product of disjoint cycles. In then follows by definition of sgn that sgn((a b)w) = sgn(w). The other possibility is if b appears in another of the cycles, say c 2 (allowing 1-cycles). Now another similar explicit calculation with cycle notation expresses (a b)c 1 c 2 as a product of disjoint cycles, and the conclusion again follows using the definition of sgn. Now we can prove the theorem. We need to show that sgn(xw) = sgn(x) sgn(w) for any x, w S n. Write x = s 1... s m as a product of transpositions, applying Lemma 1.3.1, and proceed by induction on m, the case m = 1 being the claim. For m > 1, set y = s 1 x = s 2... s m. Then, by the claim, sgn(x) = sgn(y); by the induction hypothesis, sgn(yw) = sgn(y) sgn(w) = sgn(x) sgn(w). Hence, using the claim once more, sgn(xw) = sgn(s 1 (yw)) = sgn(yw) = sgn(x) sgn(w) Corollary. If w S n is written as a product of m transpositions, then sgn(w) = ( 1) m. We define the alternating group A n to be the kernel of the homomorphism sgn : S n {±1}. Providing n > 1, A n is a normal subgroup of S n of index 2. The elements of A n are called even permutations. A group G is called simple if it has no proper normal subgroups. For instance, the cyclic group C p where p is a prime is a simple group for Lagrange s theorem implies that C p has no proper subgroups at all. The goal in the remainder of the section is to prove that A n is simple for n 5. (On the other hand, A 4 is not simple, for it contains the Klein 4 group V 4 = {1, (12)(34), (13)(24), (14)(23)} as a normal subgroup of index 3.) As a first step to the goal, we need to understand the conjugacy classes of the group S n. In any group G, the conjugate of an element x G by g G is defined to be g x := gxg 1. The conjugacy class of x is the set G x := { g x g G}. The conjugacy classes of G partition it into disjoint subsets (because conjugacy classes are the equivalence classes of the equivalence relation is conjugate to ). Observe that a subgroup N G is a normal subgroup if and only if it is a union of conjugacy classes of G: so once we understand conjugacy classes we can quite easily test subgroups for normality Lemma. Take g, x S n and write as a product of disjoint cycles. Then, x = (a 1... a s )(b 1... b t )... gxg 1 = (ga 1... ga s )(gb 1... gb s )....