Principal Ideal Domains

Transcription

1 Principal Ideal Domains Waffle Mathcamp 2009 Last week, Ari taught you about one kind of simple (in the nontechnical sense) ring, specifically semisimple rings. These have the property that every module splits as a direct sum of simple modules (in the technical sense). This week, we ll look at a rather different kind of ring, namely a principal ideal domain, or PID. These rings, like semisimple rings, have the property that every (finitely generated) module is a direct sum of simple modules, though here we use simple in the nontechnical sense of easy to understand. However, while this property of modules was almost the definition of semisimple rings, for PIDs it is much less obvious, and the bulk of our time will be devoted to proving this classification of modules. This classification is very powerful, and its applications include both a complete classification of finitely generated abelian groups and a classification of matrices up to conjugation over C (or any algebraically closed field). The main example of a PID we will focus on is the integers Z, for which modules are just abelian groups. However, another important example will be k[x], the ring of polynomials in one variable over a field. Indeed, while k[x] and Z may look like fairly different rings at first, they are in fact very similar just by both being PIDs. 1 The first, most obvious difference between what I will do and what Ari did is that PIDs are by definition commutative. Thus throughout these notes, all rings will be assumed to be commutative. If a, b, c... R are elements of a ring, we let (a, b, c,...) denote the ideal they generate. More generally, if a, b, c... M are elements of an R-module, we let (a, b, c,...) denote the submodule they generate. 1 Generalities on commutative rings We start with some generalities on commutative rings that will lead up to the notion of a PID. Definition 1.1. A ring R is an integral domain (or domain, for short) if 0 1 and whenever a, b R and ab = 0, either a = 0 or b = 0. A ring is a field if 1 There is even mathematicians who are trying to make sense of an imaginary field with one element, or F 1, such that Z would be F 1 [x] (or perhaps a similar ring formed from F 1 that is not quite the same as F 1 [x]; I m not an expert on this so I might be getting the details wrong). If they can make enough sense of this, they may be able to use it to prove the Riemann hypothesis! 1

2 0 1 and every nonzero element is a unit, i.e. has a multiplicative inverse. If ab = 0 and a, b 0, we say a and b are zero-divisors. Thus a ring is a domain iff it has no zero-divisors. A unit u is never a zero-divisor, since ub = 0 implies b = u 1 ub = u 1 0 = 0. Hence every field is a domain. Fields will usually be denoted by k instead of R. 2 Note that in a domain, multiplication allows cancellation: if ac = bc and c 0, then a = b. Indeed, we have (a b)c = 0, so since c 0, a b = 0. Example 1.2. The integers Z are a domain but not a field. Example 1.3. The rationals Q, the reals R, and the complex numbers C are all fields. Example 1.4. Let C(R) be the ring of continuous functions f : R R (with pointwise addition and multiplication). Then R is not a domain. For example, we can construct continuous functions f and g such that f(x) = 0 for x [0, 1] and g(x) = 0 for x [2, 3] but f(1/2) = g(5/2) = 1. Then fg = 0 but f, g 0. Example 1.5. For any prime p, Z/(p), the integers mod p, is a field. This is a standard fact from number theory, that you can divide (except by 0) modulo a prime. Example 1.6. If n is not prime, Z/(n) is not a domain. Indeed, write n = ab where a, b ±1. Then modulo n, a and b are nonzero but ab 0. Example 1.7. Let R be a ring. Then we can form the ring R[x] of polynomials (in one variable) with coefficients in R. A polynomial is a formal expression a n x n + a n 1 x n a 1 x + a 0 where a i R, and polynomials are added and multiplied according to the familiar rules from elementary algebra. If f(x) = ak x k, the largest n such that a n 0 is the degree deg(f) of f, and we call a n the leading coefficient of f. 3 If R is a domain, then deg(fg) = deg(f) + deg(g) since the leading coefficients of f and g multiply to give a nonzero leading coefficient of fg. It follows that if R is a domain, so is R[x]. Polynomials with leading coefficient 1 are called monic. Definition 1.8. An ideal I in a ring R is prime if R/I is a domain and maximal if R/I is a field. The name prime comes from the fact that, as we saw in Examples 1.5 and 1.6, the ideal (n) Z is prime iff n is a prime number. The name maximal come from the following result. Proposition 1.9. An ideal I R is maximal iff there is no strictly larger ideal J such that I J R. 2 Blame the Germans for this. 3 We formally write deg(0) =. 2

3 Proof. Note that ideals J R containing I are equivalent to ideals of R/I. Indeed, let p : R R/I and q : R R/J be the canonical maps. If I J, then q(i) = 0 so we can factor q uniquely as fp for some f : R/I R/J, which is surjective since q is. We then obtain an ideal J = ker(f) in R/I. Conversely, if we start with J R/I, we can define an ideal J I as the kernel of the composition R R/I (R/I)/J. It is easy to check that these are inverse to each other. Thus it suffices to show that k = R/I is a field iff its only (proper) ideal is 0. If k is a field and I k is an ideal and a I is nonzero, then 1 = a 1 a I, from which it follows that I = k, a contradiction. Conversely, if k is not a field and a k is nonzero and not a unit, then 1 (a) so (a) is a nonzero proper ideal. It follows by the Axiom of Choice that any ideal is contained in a maximal ideal (see Exercise 1.14). Intuitively, this makes sense, because we can keep making an ideal bigger and bigger, and if we can t make it any bigger we ve hit a maximal ideal. 4 As a general principle, domains and fields are much nicer than arbitrary rings. A central idea of commutative algebra and algebraic geometry is that rings can often be understood by studying the collection of their maximal or prime ideals. We will not explore this in this class, but I think Mark will in his Commutative Algebra class next week. 1.1 Exercises Exercise Show that an ideal I R is prime iff whenever ab I, either a I or b I. Definition Let R be a ring. Then a sequence of homomorphisms 0 K i M p N 0 is a short exact sequence if i is injective, p is surjective, and ker(p) = Im(i). Equivalently, up to isomorphism, K is a submodule of M and N = M/K, with i and p the obvious maps. Exercise Let 0 K i M p N 0 be a short exact sequence of modules. Prove that the following are equivalent: 1. M = K N, with i : K K N and p : K N N the canonical maps. 2. There is a homomorphism q : M K such that qi = There is a homomorphism j : N M such that pj = 1. 4 This argument, when put into the right framework, is actually rigorous! Ask me about it or take my set theory class to learn how. 3

4 If these conditions hold, we say that the short exact sequence splits. The next exercise uses the following result, which is equivalent to the Axiom of Choice. Lemma 1.13 (Zorn s Lemma). Let P be a partially ordered set, and suppose that whenever C P is totally ordered, there is x P such that x c for all c C (briefly, every chain in P has an upper bound ). Then there is an element z P which is maximal: there is no x P such that x > z. Exercise (a): Let I R be an ideal. Show that there is a maximal ideal containing I. (Hint: To get upper bounds, use that an ideal is proper iff it does not contain 1). (b): Show that an element a R is contained in a maximal ideal iff it is not a unit. Exercise (a): Show that the polynomial ring R[x] satisfies the following universal property: a homomorphism from R[x] to another ring S is equivalent to a homomorphism from R to S together with a chosen element in S (the image of x). 5 (b): Show that an R[x]-module is equivalent to an R-module M together with an element of End R (M) (corresponding to multiplication by x). (Hint: An S-module structure on an abelian group M is the same as homomorphism from S to the (noncommutative) ring End Z (M).) Exercise (a): Let R be a domain. Let k be the set of formal expressions a/b where a, b R and b 0, and we say a/b = c/d if ad = bc. Show that with the usual addition and multiplication of fractions, k is a field. Show that a a/1 is an injective ring-homomorphism from R into k. We call k the field of fractions of R. (b): Show that a ring is a domain iff it is a subring of a field. Exercise Show that if a domain is finite, it is a field. 5 By equivalent, we mean that there is a naturally defined bijection between the set of homomorphisms R[x] S and the set of pairs (f, s) where f : R S and s S. 4

5 2 PIDs and prime factorization We are now ready to define PIDs. Definition 2.1. An ideal I in a ring R is principal if it is equal to (a) for some a R. A principal ideal domain (PID) is a domain in which every ideal is principal. Note that every ideal in a PID is trivially finitely generated, so a PID is Noetherian. Recall that this implies that a submodule of any finitely generated module is finitely generated. The most basic example of a PID is Z. Proposition 2.2. The integers Z are PID. Proof. Let I Z be an ideal; we want to show I is principal. If I = 0, this is trivial. If I 0, let n be the smallest positive integer in I (if m I, then m I, so some positive integer is in I). We claim that I = (n). For any m I, we can divide m by n with remainder to write m = qn + r where the remainder r satisfies 0 r < n. But then r = m qn I, so minimality of n forces r to be 0. Hence m = qn (n). Since m I was arbitrary, I = (n). It is important to note that this proof used only the fact that we have division with remainder and the remainder is strictly smaller than what you were dividing by in a sufficiently nice sense. For polynomials over a field in a variable x, we can do long division in base x to show that for any f, g k[x], g = qf + r where r has strictly lower degree than f. It follows by a nearly identical proof that k[x] is also a PID if k is a field. Proposition 2.3. Let k be a field. Then the polynomial ring k[x] is a PID. Proof. Let I k[x] be an ideal; we want to show I is principal. If I = 0, this is trivial. If I 0, let f I be a nonzero polynomial of minimal degree. We claim that I = (f). For any g I, we can divide g by f with remainder to write g = qf + r where the remainder r has strictly smaller degree than f. But then r = g qf I, so minimality of f forces r to be 0. Hence g = qf (f). Since g I was arbitrary, I = (f). In Exercise 2.10, you will generalize this to an abstract ring that has a good notion of division with remainder, called a Euclidean domain. Note that most rings, even most domains, are not PIDs. For example, in the ring k[x, y] = k[x][y] of polynomials in two variables with coefficients in a field, the ideal (x, y) cannot be generated by a single element (see Exercise 2.7). Similarly, in the ring Z[x] of polynomials in one variable over Z, the ideal (2, x) is not principal. Prime numbers and the unique factorization of integers into primes is extremely useful in studying the integers and number theory. In fact, these are not special to the integers, but work just as well in any PID. First, we need some definitions. 5

6 Definition 2.4. Let R be a domain and a, b R. Then a divides b, or a b, if there exists c R such that b = ac. Equivalently, a b if (b) (a). We say p R is irreducible if p is not a unit and whenever p = ab, either p a or p b. We say p is prime if the ideal (p) is a prime ideal. Note that if a b and b a, then (a) = (b). In this case, we can write a = ub and b = va. It follows that a = ub = uva, so by cancellatin uv = 1. Thus if a b and b a, a and b differ by multiplying by a unit, and the converse clearly also holds. In this case, we say a and b are associate. Whenever we talk about divisibility properties, everything is always up to units ; we will generally not care to distinguish an element from any of its associates. It follows from this discussion that if p is irreducible and p = ab then one of a and b is an associate of p and the other is a unit. If R = Z, then n and m are associate iff n = ±m, since the only units are ±1. By our earlier discussion, we see that this definition of prime is equivalent to the usual definition for integers, and is also equivalent to irreducibility. In fact, the equivalence of primeness and irreducibility holds in any PID. Irreducible elements of k[x] are called irreducible polynomials. Because degrees add when you multiply polynomials and 1 has degree 0, any unit must have degree 0. Since k is a field, every degree 0 polynomial is a unit, so the units are exactly k = k {0} k[x]. It follows that any degree 1 polynomial is irreducible: if f = gh and deg(f) = 1, then deg(g) = 0 or deg(h) = 0, so one of them must be a unit. If k is algebraically closed, then every polynomial factors into linear polynomials, so these are the only irreducible polynomials. However, for example, if k = R, then x is also irreducible (see Exercise 2.8). Theorem 2.5. Let R be a PID and let p R be nonzero. Then TFAE: 1. p is irreducible. 2. p is prime 3. The ideal (p) is maximal Proof. (1 3): Suppose p is irreducible. Then whenever a p, b is either a unit or an associate of p. Since (p) (a) iff a p, we get that (p) (a) only holds for (a) = (p) or (a) = R. Since R is a PID, every ideal is (a) for some a, so this says exactly that (p) is maximal in the sense of Proposition 1.9. (3 2): Every maximal ideal is prime. (2 1): Let p be prime and suppose p = ab. Then modulo (p), ab 0. Since R/(p) is a domain, it follows that either a 0 or b 0, i.e. a (p) or a (p). This is equivalent to either p a or p b, as desired. Note that the proof of (2 1) (or Exercise 1.10) shows that that for p prime, if p bc then p b or p c. By induction, it follows that if p b 1 b 2 b n, then p b i for some i. From this, we can prove that any PID has unique factorization into primes. This is a generalization of the Fundamental Theorem of Arithmetic for the case of Z. 6

7 First, we note that changing an element by a unit does not change its divisibility properties. If we allow ourselves to change factors by units, factorization is not even unique in Z. For example 6 = 2 3 = ( 2)( 3), where 2, 3, 2, and 3 are all prime. For the integers, we remedy this by arbitrarily choosing a representative of each associate class, namely the positive one. For polynomials over a field, we can similarly choose to use only monic polynomials (polynomials with leading coefficient 1) in our factorizations. For a general PID, we will just arbitrarily choose representatives of each associate class of primes, and call these the chosen primes. Equivalently, we are picking a generator for every maximal ideal. Theorem 2.6 (Unique Factorization). Let R be a PID. Then every nonzero a R can be factorized as up d1 1 pd2 2 pdn n, where u is a unit, the p i are chosen primes, and d i > 0. This factorization is unique up to permuting the p i. Proof. First, we prove existence of the factorization. If a is a unit, this is trivial. Otherwise, let P be any maximal ideal containing (a). Then P = (p 1 ) for some chosen prime p 1, and p 1 a since (a) (p 1 ). Write a = p 1 a 1. Repeat the argument with a 1 in place of a to write a 1 = p 2 a 2 if a 1 is not a unit. Continue by induction. If we ever get a n = u to be a unit, we are done, since a = p 1 a 1 = p 1 p 2 a 2 = = p 1 p 2 p n u. If we never get a unit, we get an infinite ascending chain of ideals (a) (a 1 ) (a 2 ) (a 3 ).... Let I = (a n ); then I is an ideal. But then I = (b) since R is a PID, and b (a n ) for some n. But then I = (a n ) (a n+1 ), so (a n ) = (a n+1 ), a contradiction. Hence the process must eventually stop with a n a unit, and we get the desired factorization. Now we prove uniqueness by induction on d i. If d i = 0 (i.e., n = 0), then a = u is just a unit, and uniqueness is obvious. Now suppose d i > 0 and a = up d1 1 pd2 2 pdn n = vq e1 1 qe2 2 qem m are two different factorizations. Then p 1 divides the product on the right-hand side and is prime, so p 1 must divide one of the factors. Since no two chosen primes are associate, this implies that p 1 = q i for some i. Cancelling these common factors, we get b = up d1 1 1 p d2 2 pdn n = vq e1 1 qe2 2 qei 1 qm em. We have decreased d i by one, so by induction the i factorization of b is unique, so these two factorizations are the same up to permutation. It follows that the two original factorization of a were the same up to permutation. 2.1 Exercises Exercise 2.7. Let k be a field (though any ring will do for the first part). Show that the ideal (x, y) k[x, y] is not principal. Conclude that k[x, y] is a domain but not a PID. 6 Exercise In fact, it can be shown that k[x, y] has unique factorization, showing that a domain with unique factorization (or UFD) need not be a PID. 7

8 (a): Show that if a polynomial f(x) k[x] is irreducible and of degree greater than 1, then f has no roots. (Hint: If f(a) = 0, show by division that (x a) f.) (b): Show that the polynomial x is irreducible as an element of R[x]. (c): Show that the polynomial x 4 4 is reducible as an element of Q[x], but has no roots. The next exercise gives another class of examples of PID which are important in more advanced commutative algebra. Exercise 2.9. A local ring is a ring R with a single maximal ideal P. By Exercise 1.14(b), this means that every element of R P is a unit. A discrete valuation ring (DVR) is a Noetherian local domain such that the maximal ideal P is principal. Let R be a DVR and let p be a generator for P. (a): Show that every nonzero element of R is of the form up n for u a unit. (Hint: Imitate the proof of existence of prime factorizations in a PID and use Noetherianness.) (b): Show that R is a PID, and in fact that every nonzero ideal is (p n ) for some n. (c): Let p Z be a prime, and let Z (p) be the set of rational numbers whose denominators are relatively prime to p. Show that Z (p) is a DVR with maximal ideal (p). (d): Let k be a field, and let k[[x]] = { n=0 a nx n : a n k} be the ring of formal power series in one variable with coefficients in k. Show that k[[x]] is a DVR with maximal ideal (x). (Hint: Show that a n x n is a unit if a 0 0, and use this to prove that (b) above holds for R = k[[x]] and p = x.) The example of Exercise 2.9(c) is generalized in Exercise Exercise A Euclidean domain is a domain R together with a function d : R {0} N satisfying d(ab) max(d(a), d(b)) and such that if a, b R and b 0, then there exist q and r such that a = qb + r and r = 0 or d(r) < d(a). For example, if R = Z we can let d(n) = n and if R = k[x] we can let d(f) = deg(f). Show that any Euclidean domain is a PID. Exercise Let R be a PID and a, b R. Then a greatest common divisor (gcd) of a and b is c R such that c a, c b, and if d a and d b then d c. We write c = gcd(a, b). (a): Show that c is a gcd of a and b iff c generates the ideal (a, b). Conclude that any two elements have a gcd, well-defined up to units. (b): Show that if c = gcd(a, b), then there exist r, s R such that c = ra + sb. Conversely, show that if c = ra + sb and c a and c b, then c = gcd(a, b). 8

9 (c): Suppose R is a Euclidean domain. Then show that the following algorithm, known as the Euclidean algorithm, computes gcd(a, b) and allows one to explicitly write it in the form ra + sb: Assume WLOG that d(b) d(a). If a b, stop and say the gcd is a. Otherwise, write b = qa + r with d(r) < d(a). Now repeat the algorithm, but replace b with r. The algorithm will eventually terminate because d(a) + d(b) decreases with each step. (To get an idea of how the algorithm works, you may want to try it with R = Z and a = 18, b = 26.) 7 7 The Greeks invented the Euclidean algorithm to prove that Z is a PID (though they didn t call it that!). Oddly, I have seen multiple references in which the example used to demonstrate the Euclidean algorithm is 18 and 26. 9

10 3 Modules over a PID In this section, R will always denote a PID, and primes will be taken to mean chosen primes. We can now get to the real meat classifying finitely generated modules over a PID. Note that a Z-module is the same as an abelian group, so this will also give a complete classification of finitely generated abelian groups. This is perhaps more impressive when you consider that is it generally accepted to be hopeless to even classify finite nonabelian groups. What about the case R = k[x]? In this case, by Exercise 1.15, an R-module is a k-module V (i.e., vector space) together with an endomorphism A End k (V ) (i.e., a linear transformation on V ). If (V, A) and (W, B) are k[x]-modules, a homomorphism between them is a k-linear map T : V W that intertwines with the endomorphisms: T A = BT (this just is the property that T (xv) = xt (v), since A and B correspond to multiplication by x). It follows that (V, A) and (V, B) are isomorphic iff there is a k-linear automorphism T of V such that T A = BT, or B = T AT 1. Thus by classifying k[x]-modules, we will classify linear transformations on vector spaces up to conjugation by linear automorphisms. Or, in simpler language, we will classify matrices up to conjugation. Let s now look at what the classification is. Definition 3.1. An R-module M is cyclic if it is generated by a single element. This generalizes the notion of a cyclic group. Note that if M is generated by x, then f(a) = ax is a surjective homomorphism from R to M. Hence cyclic modules are exactly those of the form R/I for some ideal I. Here is the main theorem of this section. Theorem (Classification of Modules over a PID). Let M be a finitely generated R-module. Then M is isomorphic to a direct sum of cyclic modules R n R/(q i ), where each q i = p di i is a power of a prime. This decomposition is unique up to permuting the factors. The number n is sometimes called the Betti number of M and the q i are the elementary divisors or torsion coefficients. There are several steps to proving this theorem. Basically, we can prove it separately for the summands R n and for the summands R/(q) corresponding to each prime. 3.1 Torsion modules Definition 3.2. Let M be an R-module and x M. Then the annihilator of x is (0/x) = {a R : ax = 0}. Note that (0/x) is an ideal. If x 0 and (0/x) 0, we say x is a torsion element. If (0/x) = (a), we say a is the period of x. If every (nonzero) element of M is torsion, we say M is a torsion module. If no element of M is torsion, we say M is torsion-free. 10

11 Note that the period of an element is only defined up to units; this should not cause any confusion. If x has period a, then (x) M is isomorphic to R/(a). Example 3.3. Let R = Z. Then a module is a finitely generated torsion module iff it is a finite abelian group. Indeed, it is clear that any element of a finite abelian group must be torsion (otherwise its multiples would all be different and there would be infinitely many of them). Conversely, if M is generated by x 1,..., x n with periods a 1,..., a n, then any element of M can be written as b 1 x b n x n where each b i satisfies 0 b i < a i, so M is finite. Example 3.4. Let R = k[x]. If (V, A) is an k[x]-module and v V is torsion, then (v) = k[x]/(f(x)) for some nonzero polynomial f, which (by changing by a unit) we may assume to be monic. If f(x) = x n + a n 1 x n , then in k[x]/(f), x n = a n 1 x n 1... is k-linearly dependent on the lower powers of x. We similarly can see that all higher powers of x can be spanned just by B = {1,..., x n 1 }. In fact, B is a basis for k[x]/(f) as a k-vector space, since every nonzero element of (f) has degree at least n, so no linear combination of elements of B can vanish in k[x]/(f). Thus if v is torsion, (v) is finite-dimensional. In fact, more generally, a k[x]- modules is finitely generated torsion iff it is finite-dimensional as a vector space, which you will prove in Exercise In this section, we will prove the classification theorem for torsion modules, which is the general classification theorem in the case where the Betti number is 0. First, we show that we can treat each prime separately when analyzing a torsion module. Definition 3.5. Let p R be a prime and M be an R-module. Then an element x M is p-torsion if its period is a power of p. If every element of M is p-torsion, we say M is a p-module. The name p-module generalizes the traditional term p-group for a group whose order is a power of p. Lemma 3.6. Let M be a finitely generated torsion R-module. For each p, let M p = {x M : x is p-torsion}. Then M p is a submodule of M, and M splits as the direct sum M p over all primes p. Proof. First, we show M p is a submodule. Let x, y M p, with (0/x) = (p n ) and (0/y) = (p m ) and WLOG m n. Then p n (x + y) = 0, so p n (0/x + y). Since p is prime, the only ideals containing (p n ) are (p k ) for k n. Hence the period of x + y is a power of p, so x + y M p. A similar argument shows that M p is closed under scalar multiplication. Now we show that M = M p. This consists of two statements: M = M p and M p q p M(q) = 0 for all p. That is, the M p generate all of M and they are linearly disjoint. We prove the second statement first. 11

12 Suppose x M p q p M(q) is nonzero; write x = y q for y q M(q). Let r be the product of the periods of the y q. Then we have rx = ry q = 0, so r (0/x). But x M p, so (0/x) = (p n ) for some n. Since r, as a product of powers of q for q p, is relatively prime to p, this is impossible. Hence M p q p M(q) = 0. Now we show that M = M p, i.e. the M p together generate M. Let x M be nonzero and factor the period of x as r = p d1 1 pdn n (ignoring units for convenience). Let a i = r/(p ni i ) and y i = a i x; i.e. multiply x by all of its period except the p i part. Then p ni i y i = rx = 0, and as above this implies y i M(p i ). Now consider the ideal I = (a 1,..., a n ) R. Since R is a PID, I = (a) for some a. We then have a a i for all i. By unique factorization, this implies a = 1, so I is all of R. In particular, 1 I = (a 1,..., a n ), so we can write 1 = b i a i for some b i R. But now we have ( ) x = bi a i x = b i y i. Since y i M p, we conclude that x M p, as desired. You can think of this result as analogous to splitting a semisimple module into its parts corresponding to each isomorphism class of simple modules (i.e., its Wedderburn components). Remember that we want to prove that we can split a module as a direct sum of modules of the form M/(q) where q is a prime power. By Lemma 3.6, it suffices to show this for p-modules, in which case every q will be a power of p. Here s the idea behind the proof. Supposing M is a p-module and we already know that M = R/(p di ), let d 1 be the largest of the d i. Then if we take any element x M of period p di, M will split as (x) M/(x). Why should we expect this? Consider the example of R = Z, p = 2 and M = Z/(4) Z/(2), for concreteness sake. Then, for example, if we pick the element x = (2, 0), which has period 2, the short exact sequence 0 (x) M M/(x) 0 does not split, because there is an element y = (1, 0) satisfying 2y = x, which clearly does not hold in the direct sum (x) M/(x) = Z/(2) Z/(2) Z/(2). On the other hand, if we had chosen an element x of period 4, it would have split. Indeed, this is obvious if x = (±1, 0). If, say, x = (±1, 1), note that (x) = {0, (1, 1), (2, 0), ( 1, 1)} is still disjoint from the Z/(2) = {0, (0, 1)}. Thus it seems that an obstacle to splitting is being able to divide x by p, and it will turn out that this is the only obstacle. This can t happen if the period of x is p d1, since if py = x, y would have period p d1+1, contradicting maximality of d 1. Thus the idea is, take an element x of M of maximal period, split M = (x) M/(x), and repeat by induction on M/(x). To guarantee that this induction works, we need a lemma. Lemma 3.7. Let M be a finitely generated p-module, and let x M have maximal period (i.e. period p d, where d is maximal). Let N = M/(x) and 12

13 f : M N be the canonical map. Then as a R/(p)-vector space, N/pN has smaller dimension than M/pM. Proof. First, note that M/pM and N/pN are vector spaces over the field R/(p); this is straightforward to check. Also, they are finite-dimensional since M and N are finitely generated as R-modules, and the same generators can be used. Now note that by maximality of the period of x, x pm. Also, note that f induces a natural surjective map F : M/pM N/pN. Since F (x) = 0 but x is nonzero as an element of M/pM, ker(f ) 0. Since dim(m/pm) = dim(n/pn) + dim(ker(f )), we re done. Thus when we say by induction, we will mean by induction on dim(m/pm). We can now prove the classification theorem for p-modules. Theorem 3.8. Let M be a finitely generated p-module. Then M is isomorphic to a direct sum M/(p di ). The numbers d i are uniquely determined up to permutation. Proof. We prove existence of the splitting by (strong) induction on dim(m/pm). Let x M have maximal period, say period p d. Such an x exists since M is finitely generated: if M is generated by elements x i with periods p di, then it is easy to see that every element of M has period at most p d where d is the greatest of the d i. By Lemma 3.7 and the induction hypothesis, we can write N = M/(x) as a direct sum R/(p di ). If we can show M = N (x), we re done, since (x) = R/(p d ). We thus want to lift generators of N to generators of M to induce a splitting. For this, we use the following lemma. Lemma 3.9. Let M be a p-module, and let x M have maximal period (i.e. period p d, where d is maximal). Let N = M/(x) and f : M N be the canonical map, and let ȳ N. Then there is y M such that f(y) = f(ȳ) and y has the same period as ȳ. Proof. Let p e be the period of ȳ and let z M be any lift of ȳ. Then f(p e z) = p e y = 0, so p e z (x); write p e z = cp n x, where p does not divide c and n d. Note that since x has period p d, p e z has period d n, so z has period d + e n. By maximality of d, d + e n d, so e n. Now let y = z p n e cx. Then f(y) = ȳ, and p e y = p e z p n cx = 0, as desired. Now let ȳ i be generators of N corresponding to the direct summands R/(p di ). Let y i M be lifts of ȳ i having the same period, as in Lemma 3.9. We then obtain a map r : N M given by r(ȳ i ) = y i ; this is well-defined since N is just a direct sum and y i has the same period as ȳ i. Furthermore, for each ȳ i, f(r(ȳ i )) = ȳ i. Since the ȳ i generate N, this implies that fr = 1. That is, r induces a splitting of the short exact sequence 0 (x) M N 0. Hence M = N (x), and we re done. Now we show uniqueness of the d i. If M = R/(p di ), it is clear that the dimension of M/pM is just the total number of d i. Furthermore, pm = R/(p d i 1 ). Thus the dimension of pm/p 2 M is the number of d i that are 13

14 greater than 1. Similarly, by looking at the dimension of p k M/p k+1 M for all k, we can see how many d i there are that are greater than k, for all k. This shows that the d i are uniquely determined by M, as desired. Corollary 3.10 (Classification of finitely generated torsion modules). Let M be a finitely generated torsion R-module. Then M is isomorphic to a direct sum R/(qi ), where each q i = p di i is a power of a prime. The q i are unique up to permutation. Proof. Existence of this splitting is immediate from Theorem 3.8 and Theorem 3.6. For uniqueness, note that if M = R/(q i ), then M p = p q i R/(q i ). Thus the uniqueness of the representation of M p in Theorem 3.8 gives the uniqueness for all of M Exercises Exercise Show that a k[x]-modules is finitely generated torsion iff it is finite-dimensional as a k-vector space. Exercise Let f(x) = x n + a n 1 x n a 0. Write down the matrix for multiplication by x in k[x]/(f(x)) with respect to the basis {1, x,..., x n 1 }. By Corollary 3.10, every linear transformation on a finite-dimensional vector space is a direct sum of matrices of this form for some choice of basis. Exercise Let a R and factor a = p ni i. Show that R/(a) = R/(p ni i ). Exercise Let M be a finitely generated torsion R-module. Show that M is isomorphic to a unique direct sum of cyclic modules R/(a i ) such that a i+1 a i for all i. These a i are called the invariant factors of M, and are sometimes more useful than the elementary divisors. (Hint: For each prime p, let n p be the highest power of p appearing as an elementary divisor. Let a 1 = p pnp. Define the other a i in a similar way, and use Exercise 3.13.) Exercise Let R be a ring and P R be a prime ideal. Define the localization of R at P to be the set R P of fractions a/s for s P, where we identify a/s and b/t if u(at bs) = 0 for some u P. Informally, we are allowing division by elements not in P. (a): Show that R P is a well-defined ring with ordinary addition and multiplication of fractions, and that a a/1 is a homomorphism from R to R P. (b): Show that if R is a domain and P = 0, we just get the field of fractions of R. Show that if R = Z and P = (p), we get the ring Z (p) of Exercise 2.9(c). (c): Show that R P is a local ring with maximal ideal generated by the image of P R under the map R R P. 8 (Hint: It suffices to show that anything not in this ideal is a unit.) 8 It can be shown that a localization of a Noetherian ring is Noetherian, and that if R is a Noetherian domain and P is a minimal nonzero prime ideal, then R P is a DVR. 14

15 (d): If M is an R-module, we can define the localization M P similarly as fractions x/s with x M and s R P, with the same identification. Show that if R is a PID and M is torsion, the localization M (p) is naturally isomorphic to the module M p of Lemma 3.6. (Hint: Localization preserves direct sums. Use Lemma 3.6, and show (M p ) (p) = M p and (M q ) (p) = 0 if q p.) Localization is an extremely important technique in commutative algebra, though we don t have time to discuss it in more depth here. Unless it turns out that we do, in which case I might talk about it some more on Saturday. 15

16 3.2 Torsion-free modules Now we look at the other extreme case of the classification, torsion-free modules. We want to show that any torsion-free module M is free, i.e. isomorphic to a direct sum of copies of R. Generators of M inducing such a direct sum representation are called a basis for M. More concretely, {x i } is a basis for M iff they generate M and a i x i = 0 implies a i = 0 for all i. Like bases for a vector space, this implies that any x M can be written as a i x i for unique a i. The size of the basis {x i } is called the rank of M (like the dimension of a vector space. Note that for any prime p, {x i } will also give a basis for the R/(p)-vector space M/pM. This implies that the rank of M is well-defined. The first step to proving all torsion-free modules are free is the following. Lemma Let M be a finitely generated free R-module, and let N M be a submodule. Then N is free. Proof. Let {x 1,..., x n } be a basis for M, let M r = (x 1,..., x r ) and let N r = N M r. We show by induction that N r is free; since N n = N this will imply N is free. For r = 0, N 0 = 0, so this is trivial. Now suppose N r is free, and let I be the set of a R such that some element of N r+1 has a as its coefficient for x r+1. Equivalently, identifying (x r+1 ) with R, I is the ideal (M r + N r+1 )/M r M r+1 /M r = (x r+1 ). If I = 0, then clearly N r+1 = N r, so N r+1 is free. Otherwise, since R is a PID, we can write I = (a) for a nonzero. Let w N r+1 be such that w = b 1 x b r x r + ax r+1. If y N r+1, then the x r+1 coefficient of y is ca for some c R, so y cw N r. Thus N r+1 = N r +(w). Furthermore, N r (w) = 0 clearly, since any nonzero multiple of w has nonzero x r+1 coefficient. Thus N r+1 = N r (w); since N r and (w) = R are free, so is N r+1. Note that this proof also shows that the rank of N is at most the rank of M. We can now prove that torsion-free modules are free. Theorem Let M be an R-module. Then M is free, i.e. isomorphic to R n for some n, and this n is unique. Proof. Uniqueness of n is just the well-definedness of rank of free modules. Now let {v 1,..., v n } M be a maximal linearly independent set, i.e. a maximal set such that a i v i = 0 implies a i = 0. Let N = (v 1,..., v n ); by linear independence, N is free. For any y M, there is some nonzero a M such that ay N. Indeed, if no such a existed, then y would be linearly independent from {v 1,..., v n }, contradicting maximality. In particular, let {y 1,..., y m } generate M and a i y i N, and let b = a i. Then by i N for all i, so bm N. But now multiplication by b is a homomorphism from M to N, and is injective since M is torsion-free. Thus M is isomorphic to a submodule of N. By Lemma 3.16 and freeness of N, M is free. 16

17 3.3 Exercises Exercise Let k be a field and let M be the ideal (x, y) k[x, y]. Show that M is a finitely generated torsion-free k[x, y]-module, but M is not free. (Hint: Show that I = (x, y) and J = (x 1, y) are both maximal ideals and k[x, y]/i = k[x, y]/j = k as rings. Then show that M/IM and M/JM have different dimensions as k-vector spaces, and that this is impossible for a free module. 9 ) Exercise Find an example of a (non-finitely generated) torsion-free Z- module which is not free. Exercise Find an example of a (non-finitely generated) torsion-free k[x]- module which is not free. Exercise Find an example of a (non-finitely generated) torsion Z-module which is not a direct sum of cyclic modules. Exercise Let R be a PID. Show that a submodule of any (not necessarily finitely generated) free R-module is free. (Hint: Let M be free and N M. Use Zorn s Lemma on the poset of linearly independent subsets of N which generate free submodules F N such that N/F is torsion-free. Show that a maximal such basis must generate N itself by imitating the proof of Lemma To show that every chain has an upper bound, show that if (F α ) are a chain of nested submodules of N such that N/F α is torsion-free, then N/ F α is torsion-free.) 3.4 Putting it all together We can now combine Corollary 3.10 for torsion modules and Theorem 3.17 for torsion-free modules to prove the full classification theorem. Theorem 3.23 (Classification of Modules over a PID). Let M be a finitely generated R-module. Then M is isomorphic to a direct sum of cyclic modules R n R/(q i ), where each q i = p di i is a power of a prime. This decomposition is unique up to permuting the factors. Proof. Let T = {x M : x is torsion}, the torsion part of M. Then T is a submodule of M; the argument is the same as the argument in Lemma 3.6 that M p is a submodule for p a prime. Clearly T is torsion, and T is finitely generated by Noetherianness, so Corollary 3.10 says it splits uniquely as R/(q i ) for q i prime powers. Let N = M/T, and let f : M N be the canonical map. Then N is torsion-free. Indeed, if x N is nonzero and a x = 0, let x M be such that f(x) = x. Then ax T, so b(ax) = 0 for some nonzero b. But then (ba)x = 0 so x T, contradicting the assumption that x 0. 9 Here IM is the submodule of M generated by products ix for i I and x M. 17

18 Thus N is torsion-free and hence free by Theorem If the short exact sequence 0 T M N 0 were to split, we would get the desired direct sum representation fo M. But a splitting map r : N M is easily constructed: let { x i } be a basis for N, and let r( x i ) = x i for some x i such that f(x i ) = x i. This is well-defined since N is free. We clearly have fr = 1, so we get a splitting M = N T = R n R/(q i ). Finally, we show uniqueness of the representation of M. Given M = R n R/(qi ), it is clear that T = R/(q i ) and N = R n. The uniqueness thus follows from the uniqueness of the q i and n given in Corollary 3.10 and Theorem

19 4 Applications to linear algebra The case of R = Z, in which we have a classification of finitely generated abelian groups, is already very powerful and has applications basically wherever abelian groups come up. Somewhat less obvious but no less powerful are the applications when R = k[x]. Recall that a k[x]-module is a k-vector space V together with a linear map A : V V, and that V is finitely generated and torsion iff it is finite-dimensional. In this case, we can choose a basis for V and think of A as a matrix. If we identify the entire module with just the matrix, we have that two n n. matrices correspond to isomorphic modules iff they are conjugate If V = W 1 W 2 as a k[x]-module, then the matrix A of V is the direct sum of the matrices B i of the W i. The direct sum matrix is the block diagonal ( ) B1 0 matrix. 0 B 2 Let s look at the simplest case, which is when k is algebraically closed, i.e. every polynomial over k can be factored into linear polynomials. The Fundamental Theorem of Algebra says that the complex numbers C are algebraically closed, so you can pretend k = C if you re not comfortable with an arbitrary algebraically closed field. As we noted earlier, in this case the primes of k[x] are exactly the linear polynomials x a for a k (we can assume they are monic since multiplying by a unit does not change things). The classification theorem then says that every finitely generated torsion module is a direct sum of modules V = k[x]/((x a) n ). What do the matrices for these modules look like? Well, a basis for V is given by {1, x a, (x a) 2,..., (x a) n 1 }. In this basis, it is easy to check that the matrix looks like: a a a a a a A matrix of this form is called a Jordan block. Note in particular that every Jordan block is lower triangular. 10 A matrix is said to be in Jordan normal form if it is a direct sum of Jordan blocks. The classification theorem for finitely generated torsion k[x]-modules thus gives: Corollary 4.1. Let k be an algebraically closed field. Then every square matrix over k is conjugate to a Jordan normal form matrix. That is, every linear transformation of a finite-dimensional k-vector space is in Jordan normal form with respect to some basis. Two matrices are conjugate iff their Jordan normal forms are the same up to permuting the Jordan blocks. 10 By convention, the order of the basis is often reversed, so that Jordan blocks are upper triangular instead of lower triangular. 19

20 Corollary 4.2. Let k be an algebraically closed field. Then every linear transformation of a finite-dimensional k-vector space is lower triangular 11 with respect to some basis. Proof. Jordan normal forms are lower triangular. Jordan normal forms enormously simplify much of linear algebra. For example, the determinant of a triangular matrix is just the product of its diagonal entries. If (V, A) = k[x]/((x a i ) n i ), the polynomials (x a i ) ni are called the elementary divisors of the linear transformation A on V. The product (x a i ) ni is called the characteristic polynomial χ A (x). Note that if we were working over Z instead of k[x], the product of the elementary divisors of a finitely generated torsion module would be exactly its order as a group. Thus in some sense we could consider the characteristic polynomial to measure the size of (V, A). For example, it is easy to see that deg(χ A ) = dim V. Proposition 4.3. Let k be algebraically closed and A be a matrix over k. The characteristic polynomial is χ A (x) = det(xi A), where xi A is a matrix with entries in k[x]. Proof. The matrix xi A over the ring k[x] is the same as the matrix of A, except that its diagonal entries are x a i instead of a i. Its determinant is thus (x ai ) ni = χ A (x). Recall that an eigenvalue of a linear map A is a scalar λ k such that λi A is not invertible, i.e. there exists a nonzero v V such that Av = λv. Note that the eigenvalues of A are exactly the roots of the polynomial χ A (x): λi A is not invertible iff 0 = det(λi A) = χ A (λ). What are the roots of χ A? Since χ A = i (x i a i ) ni for (x i a i ) ni the elementary divisors of A, the roots are the a i. This is also easy to see quite concretely from the Jordan normal form: if A is in Jordan normal form and Av = λv, explicit computations with a basis fairly easily show that λ is one of the a i. The following theorem is very trivial using Jordan normal forms. Theorem 4.4 (Cayley-Hamilton over Algebraically Closed Fields). Let A be a square matrix over an algebraically closed field k. Then χ A (A) = 0. Proof. We may assume A is in Jordan normal form, and we can split it into Jordan blocks. But on a (x a i ) ni Jordan block, it is easy to explicitly compute that (A a i I) ni = 0. Alternatively, note that A represents multiplication by x on the module k[x]/((x a i ) ni ), so clearly (A a i I) ni represents multiplication by 0. In any case, we get χ A (A) = i (A a ii) ni = 0. All of this is well and good, but what if our field is not algebraically closed? Then irreducible polynomials are not always linear, so we do not have Jordan 11 Or, as in the previous footnote, upper triangular 20

21 normal forms. However, we can embed k in a larger algebraically closed field K, and then often we can answer questions over k by working over K. For doing this, the invariant factors of Exercise 3.14 are often useful. Proposition 4.5. Let A and B be matrices over a field k and let K k be a larger field. Then if A and B are conjugate over K, they are conjugate over k. Proof. For any degree n polynomial f over k, let A(f) be the matrix for the module k[x]/(f) with respect to the basis {1, x,..., x n 1 } (cf. Exercise 3.12). Note that A(f) is also the matrix for K[x]/(f) with respect to the same basis. Now Exercise 3.14 says that there are unique monic polynomials {f i } such that f i+1 f i and A is conjugate to the block diagonal matrix A(f 1 ) A(f 2 ) 0 D = A(f n ) It follows that A and B are conjugate iff they have the same invariant factors f i. Also, the invariant factors of A are the same whether we think of it as a matrix over k or over K. Indeed, A is conjugate to D over k and hence also over K, and by uniqueness of D, the invariant factors are the same over both fields. Now suppose A and B are conjugate over K. Then they have the same invariant factors over K. But these are the same as the invariant factors over k, so A and B are also conjugate over k. Proposition 4.5 is quite useful. For example, we could have two real matrices which we know are diagonalizable over C and have the same (complex) eigenvalues. This implies that they are conjugate over C, so we can conclude that they are conjugate over R. We can also talk about characteristic polynomials over non-algebraically closed fields. The elementary divisors (i.e., the torsion coefficients of (V, A) as a k[x]-module) are no longer necessarily powers of linear polynomials, but we can still define χ A as the product of A s elementary divisors. Note that it is easy to see that the product of the invariant factors is the same as the product of the elementary divisors (since the elementary divisors are just the irreducible-power factors of the invariant factors). Since the invariant factors are invariant under extending the field, so is the product of the elementary divisors, i.e. χ A. We thus obtain: Proposition 4.6. Let k be any field and A be a matrix over k. The characteristic polynomial is χ A (x) = det(xi A), where xi A is a matrix with entries in k[x]. Proof. Let K be an algebraically closed field containing k. By the discussion above, we can compute χ A over K instead of over k. However, det(xi A) clearly also doesn t depend on what field we re working over. Thus the result follows by Proposition

22 This could also be proven by explicitly writing down the matrices A(f) and computing that det(xi A(f)) = f(x). By a similar method, we can prove the Cayley-Hamilton Theorem over any field. Theorem 4.7 (Cayley-Hamilton). Let A be a square matrix over a field k. Then χ A (A) = 0. Proof. Let K be an algebraically closed field containing k. Then by the discussion above, χ A is the same over k and over K. By Theorem 4.4, we have χ A (A) = 0 over K, and hence also over k. Let s look at some more concrete applications of all this theory. Theorem 4.8. Diagonalizable matrices are dense in M n (C). That is, for any complex matrix A and any ɛ > 0, there is a diagonalizable matrix B such that every entry of B A is smaller than ɛ. Proof. Since conjugation by any invertible matrix is continuous, we may assume A is in Jordan normal form. By modifying A by less than ɛ and only on the diagonal, we can obtain a triangular matrix B all of whose diagonal entries are distinct; call them b i. But then χ B (x) = (x b i ), and it follows that B s elementary divisors are x b i so its Jordan normal form is diagonal. Theorem 4.8 is sometimes useful when trying to prove a property of a continuous function of matrices, since it then suffices to check that property for diagonalizable matrices. As another application, let s classify all 2 2 real matrices( up to conjugation. ) a 0 Over C, the only possible Jordan normal forms are D ab = and T 0 b a = ( ) a 1. If D 0 a ab is conjugate to a real matrix, its trace a + b and determinant ab must be real, and it easily follows that a and b must be either be conjugate complex numbers or both be real. On the other hand, for any z ( = x + iy ) C, x y there does indeed exist a real matrix conjugate to D z z, namely. y x If T a is conjugate to a real matrix, similar considerations show that a R. Thus we have the following result: Theorem( 4.9. ) Every ( real ) 2 2 ( matrix) is conjugate to exactly one matrix of x 0 x y x 1 the form,, or (for x, y R). 0 y y x 0 x 4.1 Exercises Exercise Show that the matrices A =

23 and B = are conjugate over Q. (Hint: Compute their characteristic polynomials to show they have the same Jordan normal form over C (or to just show they have the same elementary divisors over Q)). ( ) ( ) Exercise Show that the matrices and are conjugate over Q but not over Z. Thus Proposition 4.5 does not hold for arbitrary rings. Exercise Show that a matrix (over an arbitrary field) is diagonalizable iff its elementary divisors all have degree 1. Exercise 4.13 (Cayley-Hamilton over an arbitrary ring). Let A be a square matrix over a ring R and define χ A (x) = det(xi A). (a): Assume R is a domain. Show that χ A (A) = 0. (Hint: Embed R in its field of fractions and use Theorem 4.7. (b): Show that χ A (A) = 0 even if R is not a domain. (Hint: Show that it suffices to prove this when R = Z[x ij ] and the entries of A are the variables x ij.) 23