# MODULES OVER A PID KEITH CONRAD

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1 MODULES OVER A PID KEITH CONRAD Every vector space over a field K that has a finite spanning set has a finite basis: it is isomorphic to K n for some n 0. When we replace the scalar field K with a commutative ring A, it is no longer true that every A-module with a finite generating set has a basis: not all modules have bases. But when A is a PID, we get something nearly as good as that: (1) Every submodule of A n has a basis of size at most n. (2) Every finitely generated torsion-free A-module M has a finite basis: M = A n for a unique n 0. (3) Every finitely generated A-module M is isomorphic to A d T, where d 0 and T is a finitely generated torsion module. We will prove this based on how a submodule of a finite free module over a PID sits inside the free module. Then we ll learn how to count with ideals in place of positive integers. 1. Preliminary results We start with three lemmas that have nothing to do with PIDs. Lemma 1.1. If A is a nonzero commutative ring and A m = A n as A-modules then m = n. Proof. The simplest proof uses a maximal ideal m in A. Setting M = A m and N = A n, if M = N as A-modules then it restricts to an isomorphism mm = mn and we get an induced isomorphism M/mM = N/mN. This says (A/m) m = (A/m) n as A-modules, hence also as A/m-vector spaces, so m = n from the well-definedness of dimension for vector spaces. Lemma 1.1 says all bases in a finite free module over any nonzero commutative ring have the same size, and we call the size of that basis the rank of the free module. Therefore the term rank means dimension when the ring is a field. 1 The proof of Lemma 1.1 remains valid for free modules with a basis of infinite cardinality, so the rank of a free module with an infinite basis is well-defined as a cardinal number. For our purposes, free modules of finite rank will be all we care about. Commutativity in Lemma 1.1 is important: there are noncommutative rings A such that A 2 = A as left A-modules. Lemma 1.2. Let A be a commutative ring and M be an A-module. Every surjective linear map onto a finite free module looks like projection out of a direct sum decomposition. That is, if f : M A n is A-linear and onto, then there is an A-module isomorphism h: M = A n ker f where h(m) = (f(m), ), making f the first component of h. Proof. Let A n = Ae 1 Ae n and pick m i M such that f(m i ) = e i. Let g : A n M by g(c 1 e c n e n ) = c 1 m c n m n 1 The word rank means something completely different in linear algebra the dimension of the image of a linear map. 1

2 2 KEITH CONRAD (i.e., g is A-linear and g(e i ) = m i ). So f(g(v)) = v for all v A n (just check at v = e i and use linearity). Define the function h: M A n ker f by h(m) = (f(m), m g(f(m))). Check that h is an isomorphism as Exercise 2. Any linear surjection f : M N of A-modules induces an isomorphism M/ ker f = N but there is usually not a direct sum decomposition M = N ker f. That is, we can t usually split off ker f as a direct summand of M. For example, the surjection f : Z Z/6Z doesn t come from an isomorphism Z = (Z/6Z) ker f. Lemma 1.2 says there is such a splitting when N = A n. This is a useful property of finite free modules. Lemma 1.3. Every finitely generated torsion-free module M over an integral domain A can be embedded in a finite free A-module. More precisely, if M 0 there is an embedding M A d for some d 1 such that the image of M intersects each standard coordinate axis of A d. Proof. Let K be the fraction field of A and x 1,..., x n be a generating set for M as an A-module. We will show n is an upper bound on the size of any A-linearly independent subset of M. Let f : A n M be the linear map where f(e i ) = x i for all i. (By e 1,..., e n we mean the standard basis of A n.) Let y 1,..., y k be linearly independent in M, so their A-span is isomorphic to A k. Write y j = n i=1 a ijx i with a ij A. We pull the y j s back to A n by setting v j = (a 1j,..., a nj ), so f(v j ) = y j. A linear dependence relation on the v j s is transformed by f into a linear dependence relation on the y j s, which is a trivial relation by their linear independence. Therefore v 1,..., v k is A-linearly independent in A n, hence K-linearly independent in K n. By linear algebra over fields, k n. From the bound k n, there is a linearly independent subset of M with maximal size, say t 1,..., t d. Then d j=1 At j = A d. We will find a scalar multiple of M inside of this. For any x M, the set {x, t 1,..., t d } is linearly dependent by maximality of d, so there is a nontrivial linear relation ax + d i=1 a it i = 0, necessarily with a 0. Thus ax d j=1 At j. Letting x run through the spanning set x 1,..., x n there is an a A {0} such that ax i d j=1 At j for all i, so am d j=1 At j. Multiplying by a is an isomorphism of M with am, so we have the sequence of A-linear maps M am where the last map is an isomorphism. d At j A d, j=1 2. Submodules of a finite free module Theorem 2.1. When A is a PID, any submodule of a free A-module of rank n is free of rank n. Proof. We may assume the free A-module is literally A n and will induct on n. The case where n = 1 is true since A is a PID: every A-submodule of A is an ideal, hence of the form Aa since all ideals in A are principal, and Aa = A as A-modules when a 0 since A is an integral domain. Say n 1 and the theorem is proved for A n. Let M A n+1 be a submodule. We want to show M is free of rank n + 1. View M A n+1 = A A n and let π : A A n A n be projection to the second component of this direct sum. Then N = π(m) A n is free of rank n by the induction hypothesis. Since π maps M onto N

3 and N is finite free, by Lemma 1.2 MODULES OVER A PID 3 M = N ker π M and ker π M = M (A 0). All submodules of A 0 = A are free of rank 1. Thus N ker π M is free of rank n + 1, so M is as well. Theorem 2.1 is false if A is not a PID, even for the A-module A itself. Nonexample 2.2. If A is not a PID because it has a nonprincipal ideal, that ideal is not a free A-module. If A is not a PID because it is not an integral domain, with xy = 0 where x and y are nonzero, then the principal ideal Ax is not a free A-module. Remark 2.3. Theorem 2.1 is true for non-finitely generated free modules too: any submodule of a free module over a PID is free. The proof allowing infinite bases uses Zorn s lemma. See [1, pp ]. Corollary 2.4. When A is a PID, every finitely generated torsion-free A-module is a finite free A-module. Proof. By Lemma 1.3, such a module embeds into a finite free A-module, so it is finite free too by Theorem 2.1. The term free in torsion-free module and free module means different things: a torsion-free module has no nonzero torsion elements (all elements have annihilator ideal (0) aside from the element 0), while a free module has a basis. So Corollary 2.4 is saying a finitely generated module over a PID that has no torsion elements admits a basis. Corollary 2.4 is false without the finite generatedness hypothesis. For example, Q is a torsion-free abelian group but it has no basis over Z: any (nonzero) free Z-module has proper Z-submodules (that is, proper subgroups) of finite index while Q does not. Corollary 2.5. Let A be a PID. If we have a tower of A-modules M M M with M = A n and M = A n then M = A n. Proof. Since M is free of rank n and M is a submodule, Theorem 2.1 tells us M = A m with m n. Using Theorem 2.1 on M as a submodule of M, M = A k with k m. By hypothesis M = A n, so k = n by Lemma 1.1. Thus m = n. Corollaries 2.4 and 2.5 are both generally false when A is not a PID. Nonexample 2.6. Let A = Z[ 5] and consider the tower of ideals 3Z[ 5] (3, 1 + 5) Z[ 5]. The bottom and top are principal ideals, so they are free A-modules of rank 1. The middle ideal (3, 1 + 5) is finitely generated and torsion-free, but is not principal and therefore is not a free A-module. (A nonzero ideal is a free module only when it is principal, since any two elements in an ideal are linearly related.) There is a convenient way of picturing any submodule of a finite free module over a PID: bases can be chosen for the module and submodule that are aligned nicely, as follows. Definition 2.7. If A is a PID, M is a finite free A-module, and M is a submodule of M, a basis {v 1,..., v n } of M and a basis {a 1 v 1,..., a m v m } of M with a i A {0} and m n is called a pair of aligned bases.

4 4 KEITH CONRAD Pictures will explain what alignment of bases means before we give the main theorem about them. Example 2.8. Let A = Z, M = Z[i] and take N = (1 + 2i)Z[i]. So M = Z + Zi and N = (1 + 2i)Z[i] = (1 + 2i)Z + (1 + 2i)Zi = Z(1 + 2i) + Z( 2 + i). The obvious Z-bases for M and N are {1, i} and {1 + 2i, 2 + i}. In Figure 1, we shade a box having each basis as a pair of edges and translate each box across the plane. The modules M and N are the intersection points of the networks of lines formed by the small and large boxes, respectively. The modules do not know about the lines, which only show us how a choice of basis gives a specific way to picture how the module is generated by the basis. Z[i] = Z(1 + 2i) + Zi, (1 + 2i) = Z(1 + 2i) + Z( 2 + i) Figure 1. Nonaligned bases for a modules and submodule To see a completely different picture of the same two modules, we use new bases: {1+2i, i} for M and {1 + 2i, 5i} for N. These are bases because of the relations ( ) ( ( ) ( ) ( ) ( ) 1 + 2i i i =, =, i 0 1) i 5i i where the two matrices are integral with determinant 1, so the vectors on both sides have the same Z-span. These new bases lead to Figure 2, where the parallelograms with each basis as a pair of edges is shaded and looks quite unlike the shaded boxes of Figure 1. Translating the parallelograms across the plane produces two new networks of lines (both sharing all the vertical lines) The intersection points are the same as before; make sure you

5 MODULES OVER A PID 5 can see the vertices of the large box from Figure 1 as intersection points of lines in Figure 2. In Figure 2 five of the parallelograms for M fill up of the parallelograms for N. These bases are aligned. Z[i] = Z(1 + 2i) + Zi, (1 + 2i) = Z(1 + 2i) + Z 5i Figure 2. Aligned bases for a module and submodule Theorem 2.9. Any finite free A-module M of rank n 1 and nonzero submodule M of rank m n admit a pair of aligned bases: there is a basis v 1,..., v n of M and nonzero a 1,..., a m A such that n m M = Av i and M = Aa j v j. i=1 Proof. A basis of M gives us coordinate functions for that basis, which are linear maps M A. Having a set of bases for M and M as in the theorem means there are a set of compatible coordinate systems on M and M. To motivate the main idea in the proof, first suppose the theorem is true and let s see what it tells us about linear maps ϕ: M A when they are restricted to M : ( m ) m ϕ(m ) = ϕ Aa j v j = Aa j ϕ(v j ) Aa Aa m. j=1 j=1 Write Aa Aa m = Ac, so ϕ(m ) Ac for every ϕ M. Moreover, writing c = x 1 a x m a m with x j A (the choice of x i s may not be unique, but fix such j=1

6 6 KEITH CONRAD a representation for c) and defining ψ : M A by ψ( n i=1 c iv i ) = m j=1 c jx j, we have ψ( m i=1 a iv i ) = c, so ψ(m ) = Ac. The set of all ideals ϕ(m ) as ϕ runs over linear maps M A has Ac as the unique maximal member in this set with respect to inclusion (this is not saying Ac is a maximal ideal!). Now we start over and define S to be the set of ideals ϕ(m ) where ϕ: M A is A-linear. This includes nonzero ideals; for example, let M have A-basis {e 1,..., e n }, so M = n i=1 Ae i. At least one coordinate function for this basis is not identically 0 on M, and the image of that coordinate function on M is a nonzero ideal in S. Any nonzero ideal in A is contained in only finitely many ideals since A is a PID, so S contains maximal members with respect to inclusion. Call one of these maximal members Aa 1, so a We know already that Aa 1 = ϕ 1 (M ) for some linear map ϕ 1 : M A. There s some v M such that a 1 = ϕ 1 (v ). Eventually we are going to show ϕ 1 takes the value 1 on M. Claim: For any linear map ϕ: M A, a 1 ϕ(v ). To show this, set ϕ(v ) = a ϕ A. Since A is a PID, Aa 1 + Aa ϕ = Ad for some d, so Aa 1 Ad. We have d = xa 1 + ya ϕ for some x, y A. Then so da (xϕ 1 + yϕ)(m ) S. Hence d = xϕ 1 (v ) + yϕ(v ) = (xϕ 1 + yϕ)(v ), ϕ 1 (M ) = Aa 1 Ad (xϕ 1 + yϕ)(m ). Since xϕ 1 + yϕ is a linear map M A it belongs to S, so maximality of ϕ 1 (M ) in S implies ϕ 1 (M ) = (xϕ 1 + yϕ)(m ) = Ad. Hence Aa 1 = Ad = Aa 1 + Aa ϕ, which implies a ϕ Aa 1, so a 1 a ϕ. With the claim proved, we are ready to build aligned bases in M and M. M = n i=1 Ae i for some basis {e 1,..., e n }, write v = c 1 e c n e n Letting for c i A. The ith coordinate function for this basis is a linear map M A taking the value c i at v, so c i is a multiple of a 1 by our claim. Writing c i = a 1 b i, n n v = c i e i = a 1 b i e i = a 1 (b 1 e b n e n ) = a 1 v 1, say. Then i=1 i=1 a 1 = ϕ 1 (v ) = ϕ 1 (a 1 v 1 ) = a 1 ϕ 1 (v 1 ), so ϕ 1 (v 1 ) = 1. We have found an element of M at which ϕ 1 takes the value 1. The module M can be written as Av 1 + ker ϕ 1 since for any v M v = ϕ 1 (v)v 1 + (v ϕ 1 (v)v 1 ). Also Av 1 ker ϕ 1 = {0} (check!). Thus M = Av 1 ker ϕ 1. Since M is free of rank n its submodule ker ϕ 1 is free and necessarily of rank n 1. 2 We anticipate Aa1 will be the unique maximal member of S by the argument at the start of the proof, but at the moment it is just some maximal member of S.

7 MODULES OVER A PID 7 How does M fit in this decomposition of M? 3 For any w M we have and the first term is So w ϕ 1 (w)v 1 M also. Therefore w = ϕ 1 (w)v 1 + (w ϕ 1 (w)v 1 ) ϕ 1 (w)v 1 ϕ(m )v 1 = (Aa 1 )v 1 = Aa 1 v 1 = Av M. M = (M Av 1 ) } {{ } =Aa 1 v 1 (M ker ϕ 1 ). So M = Av 1 ker ϕ 1 and M = Aa 1 v 1 (M ker ϕ 1 ). This last equation tells us M ker ϕ 1 is free of rank m 1 since M is free of rank m. If m = 1 then we re done. If m > 1, then we can describe how M ker ϕ 1 sits in ker ϕ 1 by induction on the rank: we have a basis v 2,..., v n of ker ϕ 1 and a 2,..., a m A {0} such that a 2 v 2,..., a m v m is a basis of M ker ϕ 1. Corollary Let A be a PID, M be finite free A-module, and M be a submodule of M. Then M and M have the same rank if and only if M/M is a torsion module. Proof. Let M have rank n. The module M is free of some rank m n. Using aligned bases for M and M, we can write n m M = Av i and M = Aa j v j i=1 with nonzero a j s. Then M/M = m j=1 A/(a j) n and only if m = n. j=1 i=m+1 3. Finitely generated modules A. This is a torsion module if As an application of aligned bases for a submodule of a finite free module, we describe the structure of every finitely generated module over a PID. Theorem 3.1. Let A be a PID. Every finitely generated A-module has the form F T where F is a finite free A-module and T is a finitely generated torsion A-module. Moreover, T = m j=1 A/(a j) for some m with nonzero a i. Proof. Let M be a finitely generated A-module, with generators x 1,..., x n. Define f : A n M by f(e i ) = x i. Then there is a surjective linear map A n M, so M is isomorphic to a quotient A n /N. As in the proof of Corollary 2.10, A n /N ( m = j=1 j)) A/(a A n m for some m n and nonzero a j s. The direct sum of the A/(a j ) s is a torsion module and A n m is a finite free A-module. Example 3.2. Taking A = Z, every finitely generated abelian group is isomorphic to Z n G where n 0 and G is a finite abelian group (a finitely generated torsion abelian group must be finite). Corollary 3.3. Every finitely generated torsion module over a PID A is a direct sum of cyclic torsion modules: it is isomorphic to A/(a 1 ) A/(a k ), where the a i s are nonzero. 3 Warning: if M = M1 M 2 and N M, usually N (N M 1) (N M 2). For example, consider M = Z 2 = Z(1, 0) Z(0, 1) with N = Z(1, 1).

8 8 KEITH CONRAD Some a i s might be units, making A/(a i ) = 0. Proof. Use the description of T in Theorem 3.1. Corollary 2.4 could be regarded as a consequence of Theorem 3.1: a finitely generated module is F T where F is finite free and T is torsion, and being torsion-free as well forces T = 0, so the module is free. When a finitely generated A-module is written as F T, where F is a finite free submodule and T is a torsion submodule, the choice of F is not unique but T is unique: T is the set of all elements in F T with nonzero annihilator ideal, which is a description that makes no reference to the direct sum decomposition. The best way to see F is not unique is by examples. Example 3.4. Using A = Z, Z Z/(2) has generating set {(1, 0), (0, 1)} and {(1, 1), (0, 1)}. Therefore it can be written as F 1 T 1, where F 1 = (1, 0) = Z and T 1 = 0 Z/(2), and also as F 2 T 2 where F 2 = (1, 1) = Z and T 2 = 0 Z/(2) = T 1. Example 3.5. It can be shown that every unit in Z[ 2] has the form ±(1 + 2) k some choice of sign ±1 and some integer k, so Z[ 2] is a finitely generated abelian group. The torsion subgroup of Z[ 2] is {±1}, while and (1 + 2) each generate different free subgroups that complement {±1}: Z[ 2] = {±1} (1 + 2) Z = {±1} ( (1 + 2)) Z This leads to two isomorphisms of Z[ 2] with Z Z/2Z, which identify different subgroups with Z (the powers of and the powers of (1 + 2)) but both identify the same subgroup {±1} with Z/(2). Example 3.6. It can be shown that Z[ 2, 3] = ±(1 + 2) Z (2 + 3) Z ( 2 + 3) Z, where the units 1 + 2, 2 + 3, and (with respective inverses 2 1, 2 + 3, and 3 2) have no multiplicative relations over Z: if (1 + 2) a (2 + 3) b ( 2 + 3) c = 1 for integer exponents a, b, and c, then a = b = c = 0. Therefore Z[ 2, 3] = Z 3 Z/2Z. Two examples of complements to ±1 in Z[ 2, 3] are 1 + 2, 2 + 3, and 1 + 2, (2 + 3), ( 2 + 3). While the free part of a direct sum decomposition is not unique, the rank of the free part is unique: writing M = F T with F finite free and T necessarily being the torsion submodule M tor of M, we have F = M/T = M/M tor so the rank of F is the rank of the finite free module M/M tor. The rank of a finitely generated module over a PID A is defined to be the rank of its free part, which is well-defined even though the free part itself is not. In down-to-earth terms, the rank is the largest number of linearly independent torsion-free elements (over Z, it is the largest number of independent elements of infinite order). Our focus here is describing the abstract structure of a finitely generated module over a PID, not proving a module over a PID arising somewhere in mathematics is finitely generated. The fact that certain abelian groups are finitely generated can be a major theorem (look up the Mordell Weil theorem) and a formula for the rank can be a major theorem or conjecture (see Dirichlet s unit theorem or the Birch and Swinnerton-Dyer conjecture). 4. Cardinality and Index over a PID A finite abelian group is the same thing as a finitely generated torsion module over Z, so finitely generated torsion modules over a PID are generalizations of finite abelian groups.

9 MODULES OVER A PID 9 Corollary 3.3 provides a method of defining a size for finitely generated torsion modules over a PID as an ideal that generalizes the size of a finite abelian group. Definition 4.1. Let T be a finitely generated torsion module over the PID A. Writing T = A/(a 1 ) A/(a m ), define the A-cardinality of T to be the ideal card A (T ) = (a 1 a 2 a m ). This is not the same thing as the annihilator ideal Ann A (T ) = {a A : at = 0}, just as in group theory the order of a finite abelian group is not the same thing in general as the least positive integer whose power kills everything in the group (the exponent of the group). A finitely generated torsion module can be written as a direct sum of cyclic modules in more than one way (including different numbers of cyclic components, e.g., Z/(4) Z/(5) = Z/(20)), so we really need to check A-cardinality is well-defined. But first we look at a few examples of A-cardinality. Example 4.2. If G is a finite abelian group and G = Z/(a 1 ) Z/(a m ), then card Z (G) = (a 1 a m )Z, whose positive generator is a 1 a m = G. Example 4.3. Let V = R 2, viewed as an R[X]-module with X acting on R 2 as the identity matrix. Then, as an R[X]-module, we have V is the direct sum of submodules Re 1 Re 2 = R[X]/(X 1) R[X]/(X 1), so cardr[x] (V ) = (X 1) 2 as an ideal in R[X]. Example 4.4. Let V = R 2, viewed as an R[X]-module with X acting on R 2 as ( ). Then X(e 1 ) = e 1 and X(e 2 ) = ( 1 1) = e1 + e 2, so {e 2, X(e 2 )} is a basis of R 2. Since X 2 (e 2 ) = 2e 1 + e 2 = 2(e 1 + e 2 ) e 2 = 2X(e 2 ) e 2, so (X 2 2X + 1)(e 2 ) = 0, as an R[X]-module V = R[X]/(X 2 2X +1) = R[X]/(X 1) 2. Therefore card R[X] (V ) = (X 1) 2, which is the same ideal as in the previous example. The difference between R[X]/(X 1) R[X]/(X 1) and R[X]/(X 1) 2 as R[X]-modules is similar to that between Z/(2) Z/(2) and Z/(4) as abelian groups, which are nonisomorphic but have the same size. Now we show card A (T ) is well-defined. You may want to skip the proof on a first reading. Theorem 4.5. If A/(a 1 ) A/(a m ) = A/(b 1 ) A/(b n ) as A-modules, where the a i s and b j s are nonzero, then (a 1 a 2 a m ) = (b 1 b 2 b n ) as ideals. Proof. If A is a field then the theorem is obvious since both ideals are (1), so we assume A is not a field: it has irreducible elements. For every irreducible π in A, we will show the highest power of π in a 1 a 2... a m and b 1 b 2 b n are equal. Therefore by unique factorization (every PID is a UFD) a 1 a 2... a m and b 1 b 2 b n are equal up to multiplication by a unit, so (a 1 a 2... a m ) = (b 1 b 2 b n ). Set T = A/(a 1 ) A/(a m ). For each irreducible π in A we look at the descending chain of modules T πt π 2 T π i T The quotient of successive modules π i 1 T/π i T is an A-module on which multiplication by π is 0, so this is an A/(π)-vector space. Since T is finitely generated, so is π i 1 T (multiply the generators of T by π i 1 ) and thus so is its quotient π i 1 T/π i T, so π i 1 T/π i T is a finitedimensional A/(π)-vector space. The dimensions dim A/(π) (π i 1 T/π i T ) will be the key. We will show the highest power of π in a 1 a 2... a m is i 1 dim A/(π)(π i 1 T/π i T ).

10 10 KEITH CONRAD Step 1: T = A/(a) is a cyclic torsion module. For i 1 we will show dim A/(π) (π i 1 T/π i T ) = { 1, if π i a, 0, otherwise. Since T is generated as an A-module by 1, π i 1 T/π i T is spanned as an A/(π)-vector space by π i 1, so π i 1 T/π i T has dimension 1 over A/(π). It is 0-dimensional if and only if π i 1 T = π i T, which is equivalent to π i 1 π i A + (a) = (π i, a). If π i a then (π i, a) (π i ), and obviously π i 1 (π i ), so π i 1 T/π i T is 1-dimensional over A/(π). If π i does not divide a then the greatest common divisor of π i and a is π j for some j i 1, and therefore π i 1 (π j ) = (π i, a). Thus dim A/(π) (π i 1 T/π i T ) = #{i 1 : π i a}, i 1 which is the highest power of π dividing a. Step 2: T is any finitely generated torsion module. Writing T = A/(a 1 ) A/(a m ), we have an A/(π)-vector space isomorphism π i 1 T/π i T m = π i 1 (A/(a k ))/π i (A/(a k )), so dim A/(π) (π i 1 T/π i T ) = k=1 m dim A/(π) (π i 1 (A/(a k ))/π i (A/(a k ))) k=1 so summing over all i 1 gives us on the right the sum of the highest powers of π in all a k s, which is the highest power of π in a 1 a 2 a m. Step 3: Comparing isomorphic cyclic decompositions. The number dim A/(π) (π i 1 T/π i T ), which does not depend on a cyclic decomposition, is unchanged if T is replaced by an isomorphic A-module because any A-module isomorphism T T induces an A/(π)-vector space isomorphism π i 1 T/π i T π i 1 T /π i T in a natural way. Therefore a 1 a 2... a m and b 1 b 2 b n are divisible by the same highest power of π, for all π. Example 4.6. We have card A (T ) = (1) if and only if each a i in a cyclic decomposition is a unit, which means T = 0. Example 4.7. For nonzero a A, card A (A/(a)) = (a). (This is trivial only after we know that A-cardinality is well-defined.) For nonzero a and b, card A (A/(ab)) = (ab) = (a)(b) = card A (A/(a)) card A (A/(b)). Corollary 4.8. If T is a finitely generated torsion module over A then card A (T ) Ann A (T ), Proof. It is obvious from a cyclic decomposition T = A/(a 1 ) A/(a m ) that multiplication by a 1 a 2 a m kills T, so a 1 a m Ann A (T ), and therefore card A (T ) Ann A (T ). Theorem 4.9. For two finitely generated torsion A-modules T 1 and T 2, card A (T 1 T 2 ) = card A (T 1 ) card A (T 2 ). Proof. Combine cyclic decompositions of T 1 and T 2 to get one for T 1 T 2.

11 MODULES OVER A PID 11 Some properties of card A (T ) that generalize results about orders of finite abelian groups (e.g., Cauchy s theorem) are in the exercises. The notion of index also generalizes from abelian groups to finitely generated modules over a PID, as an ideal. Definition If M is a finitely generated module over the PID A with submodule M such that M/M is a torsion module, we set the A-index of M in M to be the A-cardinality of their quotient: [M : M ] A = card A (M/M ). Example If M M then M = M if and only if [M : M ] A = (1), since M/M = 0 if and only if card A (M/M ) = (1). Example For nonzero a and b in A, [A 2 : aa ba] A = (ab) since A 2 /(aa ba) = A/(a) A/(b). Example Let A = Z[i], ( ) ( ) 1 0 M = Z[i] 2 = Z[i] + Z[i], 0 1 and ( ) ( ) ( ) ( ) M = Z[i] + Z[i] = Z[i]3 + Z[i](1 + 2i) i 0 1 Since M/M = Z[i]/(3) Z[i]/(1 + 2i), [M : M ] Z[i] = (3(1 + 2i)). Using different bases for these two modules, ( ) ( ) 3 1 2i M = Z[i] + Z[i] 1 + 2i 2 and ( ) ( ) 3 1 2i M = Z[i] + Z[i]3(1 + 2i), 1 + 2i 2 so M/M = Z[i]/(3(1 + 2i)). Thus again we compute [M : M ] Z[i] = (3(1 + 2i)). If M is a finite free A-module and M is a submodule, by Corollary 2.10 the A-index [M : M ] A is defined if and only if M and M have equal rank. This is also true in the general case (Exercise 7). When A = Z, [M : M ] Z is the subgroup of Z generated by the positive integer #(M/M ), which is the usual index [M : M ], so the A-index generalizes the index in group theory. Theorem Let M be a finitely generated A-module with submodules M M and assume M/M is a torsion module. Then Proof. In terms of A-cardinalities, this says [M : M ] A = [M : M ] A [M : M ] A. card A (M/M ) = card A (M/M ) card A (M /M ). The ideal card A (M/M ) is defined by hypothesis. Since M /M is a submodule of M/M and M/M is a quotient module of M/M, both are finitely generated torsion modules so card A (M/M ) and card A (M /M ) are both defined. Setting T = M/M and T = M /M, the identity we want to prove becomes This is Exercise 8. card A (T ) = card A (T ) card A (T/T ).

12 12 KEITH CONRAD Any finite-free Z-module M looks like Z n, and a submodule M also of rank n has finite index in M. We will prove a determinant formula for the index of M in M and then generalize to the case of an arbitrary PID in place of Z. Theorem Let M be a finite free Z-module with rank n and M be a submodule of M with rank n. Let x 1,..., x n be a basis of M and y 1,..., y n be a basis of M. Writing y j = n i=1 c ijx i with c ij Z, the index [M : M ] equals det(c ij ). Proof. Our proof will be an application of aligned bases in a finite free abelian group and finite-index subgroup: Theorem 2.9 with A = Z. The n-tuples x 1,..., x n and y 1,..., y n do not have the same Z-span (unless M = M), but morally they should have the same Q-span. To make this idea precise, we transfer the data of M and M into the vector space Q n. Let e 1,..., e n be the standard basis of Q n and set f j = n i=1 c ije i, using the same coefficients that express y 1,..., y n in terms of x 1,..., x n. Identify M with Z n by x i e i (extended by Z-linearity). This isomorphism identifies y i with f i, so M inside M is identified with the Z-span of f 1,..., f n inside Z n. Every Z-linear map Z n Z n is determined by where it sends e 1,..., e n. Let ϕ: Z n Z n be the Z-linear map determined by ϕ(e 1 ) = f 1,..., ϕ(e n ) = f n. Then ϕ(e j ) = n i=1 c ije i for j = 1,..., n, so (c ij ) is the matrix representation of ϕ with respect to the standard basis e 1,..., e n of Z n and ϕ(z n ) = Zϕ(e 1 ) Zϕ(e n ) = Zf 1 Zf n, so [M : M ] = [Z n : ϕ(z n )]. We will show [Z n : ϕ(z n )] = det(ϕ) = det(c ij ). The bases e 1,..., e n and f 1,..., f n of Z n are usually not aligned with each other. By Theorem 2.9, there is a set of aligned bases for Z n and its submodule ϕ(z n ): Z n = Zv 1 Zv n, ϕ(z n ) = Za 1 v 1 Za n v n, where the a i s are nonzero integers. Then which tells us Z n /ϕ(z n ) = n Z/a i Z, i=1 (4.1) [Z n : ϕ(z n )] = a 1 a 2 a n. We want to prove det(ϕ) = a 1 a 2 a n too. Any Z-linearly independent set of size n in Q n is a Q-basis of Q n, so all four sets {e i }, {ϕ(e i )}, {v i } and {a i v i } are Q-bases for Q n. For any two Q-bases of Q n there is a unique Q-linear map Q n Q n taking one basis to the other. The Q-linear map Q n Q n taking e i to ϕ(e i ) is the natural extension of ϕ: Z n Z n from a Z-linear map to a Q-linear map, so we will also call it ϕ (its matrix representation with respect to the standard basis of Q n is (c ij ), just like ϕ as a Z-linear map on Z n ). Consider the diagram of Q-linear maps Q n α Q n ϕ β Q n γ Q n where e i α v i ϕ β f i γ a i v i

13 This diagram commutes: ϕ = γ β α. Q n Q n, 4 (4.2) det(ϕ) = det(γ) det(β) det(α). MODULES OVER A PID 13 Taking determinants of these Q-linear maps Using the Q-basis {v i } of Q n, the matrix representation [β] is diagonal with a i s along its main diagonal, so det(β : Q n Q n ) = a 1 a 2 a n. What is det(α)? The map α identifies two Z-bases of the same Z-module Z n : α(c 1 e c n e n ) = c 1 v c n v n, c i Z. Therefore α is invertible as a Z-linear map of Z n to itself. Using {e i } as the basis in which a matrix for α is computed, the matrices of α: Q n Q n over Q and α: Z n Z n over Z are the same. Since an invertible Z-linear map has determinant ±1, A similar argument shows det(α: Q n Q n ) = det(α: Z n Z n ) = ±1. det(γ : Q n Q n ) = det(γ : ϕ(z n ) ϕ(z n )) = ±1 since γ identifies two Z-bases of ϕ(z n ). Feeding these determinant formulas into the right side of (4.2), (4.3) det(ϕ) = ±a 1 a 2 a n. Comparing (4.1) and (4.3), det(ϕ) = a 1 a 2 a n = [Z n : ϕ(z n )] = [M : M ]. Example In the ring Z[ 10] let a be the ideal (2+5 10, ). We will compute the index of a in Z[ 10] as abelian groups by finding Z-bases for Z[ 10] and a and then computing the determinant of the matrix expressing the second basis in terms of the first. The Z-bases do not have to be aligned to make the calculation. A Z-basis for Z[ 10] is {1, 10}. A Z-basis for a is { , }, but this requires verification because it is a stronger condition to generate a as a Z-module than to generate it as an ideal: the initial definition of a tells us that a = Z[ 10]( ) + Z[ 10]( ) = (Z + Z 10)( ) + (Z + Z 10)( ) = Z( ) + Z( ) + Z( ) + Z( ). For a to be spanned over Z by and , we need to write the other two Z-module generators in terms of them. After some linear algebra, we can do this: = 57( ) + 41( ), = 79( ) + 57( ). Therefore a = Z( ) + Z( ) and the numbers and are obviously Z-linearly independent, so they are a Z-basis of a. Under the isomorphism Z[ 10] Z 2 as abelian groups (Z-modules) by a + b 10 ( a b), the ideal a is identified with Z ( ( 2 5) + Z 4 ) 7 = ( )(Z2 ), so Theorem 4.15 tells us that the index of a in Z[ 10] is det( ) = 6 = 6. The absolute value is important: the index is not 6. 4 We are using Q-linear maps throughout because it is nonsense to talk about the determinant of a Z- linear map Z n ϕ(z n ) when ϕ(z n ) Z n : the different bases don t all have the same Z-span but they do all have the same Q-span.

14 14 KEITH CONRAD Theorem Let M be a finite free module over the PID A with rank n and M be a submodule with rank n. Let x 1,..., x n be a basis of M and y 1,..., y n be a basis of M. Writing y j = n i=1 c ijx i with c ij A, (det(c ij )) = [M : M ] A. In particular, det(c ij ) 0. Proof. Let K be the fraction field of A. If we run through the proof of Theorem 4.15 with A in place of Z, K in place of Q, and use aligned bases v 1,..., v n and a 1 v 1,..., a n v n for M and M, with all a i nonzero in A, so M/M = n i=1 A/(a i), then the proof of Theorem 4.15 shows det(c ij ) = det ϕ = ua 1... a n, where u = det α det γ is a unit in A. The K-linear operators α and γ on K n have unit determinant since they are also A-linear operators on A n and ϕ(a n ) sending a basis to a basis. By the definition of A-index, [M : M ] A = card A (M/M ) = (a 1 a 2 a n ). Corollary Let M be a finite free A-module with rank n and basis x 1,..., x n. For y 1,..., y n in M, write y j = n i=1 c ijx i with c ij A. Then y 1,..., y n is linearly independent if and only if det(c ij ) 0. Proof. If y 1,..., y n is a linearly independent set, then its A-span in M is a free A-submodule of rank n, so det(c ij ) 0 by Theorem Conversely, if det(c ij ) 0, then we want to show an A-linear relation n j=1 c jy j = 0 with c j A must have all c j equal to 0. Writing y j in terms of the x i s, ) n n n n 0 = c j y j = c ij x i = c ij c i x i, j=1 j=1 c j ( n i=1 so looking at the coefficients of x 1,..., x n tells us n j=1 c ijc i = 0 for all i. As a matrix equation this says c 11 c c n1 c 1 0 c 12 c c n2 c = 0.. c 1n c 2n... c nn 0 This is an equation in A n K n, where K is the fraction field of A. The matrix is the transpose of (c ij ). Since det((c ji ) ) = det(c ij ) 0 the vector vanishes, so all c i are 0. Corollary If M is a finite free module over the PID A with basis x 1,..., x n, a set of n elements y 1,..., y n in M is a basis of M if and only if the matrix (c ij ) expressing the y s in terms of the x s has unit determinant. Proof. If y 1,..., y n is a basis of M then (det(c ij )) = [M : M] A = (1) by Theorem 4.17, so det(c ij ) A. Conversely, if det(c ij ) A then y 1,..., y n is linearly independent by Corollary 4.18 and the A-index of Ay j in M is (det(c ij )) = (1), so Ay j = M. Example A set of n vectors v 1,..., v n in Z n is a basis of Z n if and only if the matrix (v 1 v 2 v n ) with the v s as the columns has determinant ±1, since this matrix expresses v 1,..., v n in terms of the standard basis of Z n. Exercises. 1. Let A be a commutative ring. If every submodule of every finite free A-module is a free A-module, show A is a PID. c n i=1 j=1

15 MODULES OVER A PID Let A be a commutative ring, M be an A-module, and f : M A n be an A- linear surjection. For the standard basis e 1,..., e n of A n, pick m i M such that f(m i ) = e i. Define the linear maps g : A n M by g(e i ) = m i for all i and h: M A n ker f by h(m) = (f(m), m g(f(m))) for all m M. Show h is an A-module isomorphism. (Hint on surjectivity: for (x, y) A n ker f let m = g(x) + y. Then h(m) = (x, y).) 3. Suppose A is a PID and π is irreducible in A. Inside A 2, set ( ) ( ) {( ) } 1 0 x M = A + A 0 π 2 = : y 0 mod π 2 y and ( ) ( ) π 1 N = A + A = 0 π {( ) } x : y 0 mod π, πx y mod π 2. y a) Find a basis {e 1, e 2 } of A 2 and a 1 and a 2 in A such that {a 1 e 1, a 2 e 2 } is a basis of N. (Such an aligned pair of bases obviously exists for A 2 and M.) b) Show there is no basis {e 1, e 2 } of A 2 and a 1, a 2, b 1, b 2 in A such that {a 1 e 1, a 2 e 2 } is a basis of M and {b 1 e 1, b 2 e 2 } is a basis of N. That is, the submodules M and N of A 2 do not admit bases simultaneously aligned with a single basis of A If M is a finitely generated module over a PID A and M is a submodule, is it always possible to align their decompositions into free parts and torsion parts: can we write M = F T and M = F T such that F and F are free, T and T are torsion, and F F and T T? If A is not a field, show the answer is no by picking an irreducible π in A and using M = A A/(π) and M = A(π, 1). (Hint: first show M is free.) 5. Let A be a PID. a) Prove an analogue of Cauchy s theorem from group theory: for a finitely generated torsion A-module T and irreducible π in A such that π divides card A (T ), meaning π divides a generator of the ideal card A (T ), show there is some t T with order π: the annihilator ideal Ann A (t) = {a A : at = 0} is πa. b) Let T and T be finitely generated torsion A-modules such that card A (T ) = card A (T ). If f : T T is an A-linear map, show f is one-to-one if and only if it is onto. (When A = Z this is the familiar statement that a homomorphism between finite abelian groups of equal size is one-to-one if and only if it is onto.) c) Show a finitely generated A-module M is a torsion module if and only if there is some a 0 in A such that am = 0. (This is false without a hypothesis of finite generatedness even for A = Z, since infinite torsion abelian groups exist.) d) For a finitely generated A-module M and submodule N, show M/N is a torsion module if and only if there is some a 0 in A such that am N. 6. Let V be a finite-dimensional vector space over a field K and f : V V be an K-linear operator. Viewing V as an K[X]-module where X acts on V as the map f, show card K[X] (V ) = (χ f (X)), where χ f is the characteristic polynomial of f. 7. For a pair of finitely generated A-modules M M, show M/M is a torsion module if and only if M and M have the same rank (that means the free parts of M and M have equal rank). This generalizes Corollary 2.10 in the case of free modules.

16 16 KEITH CONRAD 8. Let T be a finitely generated torsion module over the PID A and T be a submodule. Show card A (T ) = card A (T ) card A (T/T ). 9. Prove Corollaries 4.18 and 4.19 when A is an arbitrary integral domain, not necessarily a PID. References [1] J. Rotman, Advanced Modern Algebra, Prentice-Hall, Upper Saddle River, NJ, 2002.

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