Finitely Generated Modules Over Principal Ideal Domains

Finitely generated modules over principal ideal domains

Matheus Frederico Stapenhorst June 6, 2018

1 Introduction

In this text, we establish the structure of finitely generated modules over principal ideal domains. Initially, we recall some properties of exact sequences. Next, we establish some results about free modules over principal ideal domains. In the last section we prove, step by step, the following theorem: Let A be a finitely generated module over a principal ideal domain R. 1. A is the direct sum of a free submodule F of finite rank and a finite number of cyclic torsion modules. The cyclic torsion summands (if any) are of orders r1, . . . , rt, where r1, . . . , rt are (not necessarily distinct) nonzero nonunit elements of R such that r1|r2| ... |rt. The rank of F and the list of ideals (r1),..., (rt) are uniquely determined by A. 2. A is the direct sum of a free submodule E of finite rank and a finite number of cyclic torsion s1 sk modules. The cyclic torsion summands (if any) are of orders p1 , . . . , pk , where p1, . . . , pk are (not necessarily distinct) prime elements in R and s1, s2 . . . , sk are (not necessarily distinct) s1 sk positive integers. The rank of E and the list of ideals (p1 ),..., (pk ) are uniquely determined by A.

This text is based on the book of Hungerford, pages 218-225 (see references). Throughout the text, R is a commutative ring with unity. Usually, it will also be a principal ideal domain.

2 Preliminaries

First, let’s recall some definitions. Definition 1. A partially ordered set is a nonempty set A together with a relation R on A × A which is reflexive, transitive and antissymetric. If R is a partial ordering of A, then we write a ≤ b instead of (a, b) ∈ R. Elements a, b ∈ A are said to be comparable provided a ≤ b or b ≤ a. A partial ordering of a set A such that any two elements are comparable is called a linear ordering of A. Definition 2. Let B be a nonempty subset of a partially ordered set (A, ≤). An element c ∈ B is a least element of B provided c ≤ b for every b ∈ B. If every nonempty subset of A has a least element, then A is said to be well ordered. Note that every well ordered set is linearly ordered, since for any a, b ∈ A, the subset {a, b} ⊂ A must have a least element. Now we are able to state a theorem that is equivalent to the Axiom of Choice. Theorem 1. The well ordering principle: If A is a nonempty set then there exists a linear ordering ≤ in A such that (A, ≤) is well ordered. Now we are able to prove the so called principle of transfinite induction Theorem 2. If B is a subset of a well ordered set (A, ≤) such that for every a ∈ A,

{c ∈ A : c < a} ⊂ B ⇒ a ∈ B then B = A

1 Proof. If A − B ≠ ∅ then there is a least element a ∈ A − B. By the deﬁnitions of least element and A − B we have that {c ∈ A : c < a} ⊂ B By hypothesis, we have that a ∈ B, which is absurd. Therefore, A = B.

Proposition 1. Let (A, ≤) be a well ordered set. The immediate successor of a ∈ A, if it exists, is the least element in the set {x ∈ A : a < x} and is denoted by a + 1. Then at most one element of A has no immediate successor. Proof. Let m ∈ A be such that m has no immediate successor. Then, since A is well ordered, we have that {x ∈ A : m < x} = ∅ Suppose now that there exists n ∈ A such that m ≠ n. Then, since A is well ordered, and by the construction of m we must have that n ≤ m. This implies that n < m and therefore

{x ∈ A : n < x} ̸= ∅.

Hence, since A is well ordered, n must have an immediate successor. We also recall the following theorems Theorem 3. If R is a ring and f : A → B is an R- module homomorphism and C is a submodule of Kerf, then there exists a unique R-module homomorphism f : A/C → B such that f(a+C) = f(a) for all a ∈ A, Imf = Imf and Kerf = Kerf/C. f is an R-module isomorphism if and only if f is an R-module epimorphism and C = Kerf. In particular,

A/kerf ≃ Imf.

Theorem 4. Let B and C be submodules of a module A over a ring R. There is an R-module isomorphism B/B ∩ C ≃ (B + C)/C. Proof. Let ϕ : B → B + C be given by ϕ(b) = (b + 0) and let ψ : B + C → (B + C)/C be given by ψ(b + c) = (b + c) + C. Then ψ ◦ ϕ : B → (B + C)/C is such that

ψ ◦ ϕ(b) = b + C

Hence, Ker(ψ ◦ ϕ) = B ∩ C and Im(ψ ◦ ϕ) = (B + C)/C. Therefore,

B/B ∩ C ≃ (B + C)/C.

The next theorem is a consequence of Proposition 2.10 in Atiyah. Theorem 5. Let R be a ring and

f g 0 A B C 0

α β γ f ′ g′ 0 A′ B′ C′ 0 be a commutative diagram of R-modules and R-modules homomorphisms such that each row is a short exact sequence. Then, 1. α, γ monomorphisms → β is a monomorphism 2. α, γ epimorphisms → β is a epimorphism 3. αγ isomorphisms → β is a isomorphism

f g Theorem 6. Let R be a ring and 0 A1 B A2 0 a short exact sequence of R-module homomorphisms. Then the following conditions are equivalent.

2 → 1. There is an R-module homomorphism h : A2 B with gh = 1A2 . → 2. There is an R-module homomorphism k : B A1 with kf = 1A1 . 3. the given sequence is isomorphic(with identity maps on A and A ) to the direct sum short ex- ⊕ 1 2 ⊕ i1 π2 act sequence 0 A1 A1 A2 A2 0 . In particular, B ≃ A1 A2.

A short exact sequence that satisﬁes one of the above conditions is said to be split. ⇒ − → → Proof. 1. 3. : We have two R module⊕ homomorphisms f : A1 B and h : A2 B, which induce a R-module homomorphism φ : A1 A2 → B given by φ((a1, a2)) = f(a1) + h(a2). Then, we can check that the following diagram is commutative ⊕ i1 π2 0 A1 A1 A2 A2 0.

1A1 φ 1A2 f g 0 A1 B A2 0

But then, by Theorem 5, we have that φ is a isomorphism. ⇒ − → → 2. 3.: Now we have two R module homomorphisms⊕ g : B A2 and k : B A1, which induce a R-module homomorphism ψ : B → A1 A2 given by ψ(b) = (k(b), g(b)). Then, we can check that the following diagram is commutative:

f g 0 A1 B A2 0.

1A1 ψ 1A2 ⊕ i1 π2 0 A1 A1 A2 A2 0.

Hence, ψ is an isomorphism by Theorem 5, and we are done. ⊕ 3. ⇒ 1.,2.: Now suppose that condition 3. holds. Then, we have an isomorphism φ : A1 A2 → B such that the following diagram commutes. ⊕ i1 π2 0 A1 A1 A2 A2 0.

1A1 φ 1A2 f g 0 A1 B A2 0 ⊕ We deﬁne h : A2 → B by h = φ ◦ i2 where i2 is the immersion from A2 into A1 A2. Then,

gh(a2) = g(φ(i2(a2))) = g(φ(0 + a2)) = 1A2 (π2(0 + a2)) = 1A2 (a2) = a2 → and therefore condition 1. was proven. To prove⊕ that condition 2. holds, we deﬁne k : B A1 by −1 ◦ k = φ π1 where π1 is the projection of A1 A2 onto A1 and check that kf = 1A1 . → → ◦ Corollary 1. If⊕f : A B and g : B A are R-module homomorphisms such that g f = 1A, then B = Im f Ker g.

Proof. By hypothesis, g must be an epimorphism. Therefore, the following sequence is split exact:

g 0 Ker g i B A 0 ⊕ Hence, by the preceding theorem B ≃ A Ker g. But also by hypothesis⊕ f must be an monomorphism, and therefore A ≃ Im f. Then, we conclude that B ≃ Im f Ker g.

3 Free modules over PID: Basic Properties

Theorem 7. Let F be a free module over a principal ideal domain R and G a submodule of F. Then G is a free R-module and rank G ≤ rank F Proof. See Hungerford’s book.

3 Corollary 2. Let R be a principal ideal domain. If A is a ﬁnitely generated R-module generated by n elements, then every submodule of A may be generated by m elements with m ≤ n.

Proof. Suppose that A is generated by {x1, x2, . . . , xn} and let B be a submodule of A. Let F be a free R− module generated by n elements with basis {f1, . . . , fn}. Let φ : F → A be the unique R−module homomorphism such that φ(fi) = xi for each 1 ≤ i ≤ n. Let E = {v ∈ F : φ(v) ∈ B}. It is easily checked that E is a submodule of F . By construction, φ|E : E → B is surjective. Also, by Theorem 7, we have that E is ﬁnitely generated by say {y1, y2, . . . , ym}with m ≤ n. Hence, {φ(y1), φ(y2), . . . φ(ym)} generates B, and therefore the desired result is proven. The next result introduces us to some very important concepts.

Theorem 8. Let A be a module over an principal ideal domain R and for each a ∈ A let Oa = {r ∈ R : ra = 0}. Let p ∈ R be prime. Then,

1. Oa is an ideal of R.

2. At = {a ∈ A : Oa ≠ 0} is a submodule of A. 3. For each a ∈ A there is an isomorphism of modules

R/Oa ≃ Ra = {ra : r ∈ R}.

i j 4. If p a = 0 then Oa = (p ) with 0 ≤ j ≤ i.

i j 5. If Oa = (p ) then p a ≠ 0 for all j such that 0 ≤ j < i. Proof. 1 and 2 are easy exercises. To prove 3, simply deﬁne ϕ : R → Ra by ϕ(r) = ra. Then, ϕ is an R−module homomorphism, which is also surjective. Then Theorem 3 guarantees that Ra ≃ R/Kerϕ = R/Oa. Now let’s prove item 4. Since R is a principal ideal domain and Oa is an i ideal of R, we have that Oa = (r) for some r ∈ R. By hypothesis (p ) ⊂ (r) and therefore there exists s ∈ R such that pi = rs. Therefore, the unique factorization theorem implies that r = pju j j for some unit u ∈ R and for some j ≤ i. Hence, Oa = (r) = (p u) = (p ), and the result is proven. j j i Finally, let’s prove 5. By absurd. suppose that p a = 0 for some j < i. Then p ∈ Oa = (p ). But then there exists s ∈ R such that pj = pis, but this is absurd since it contradicts unique factorization theorem.

The ideal Oa in Theorem 8 is called the order ideal of a ∈ A. The submodule At from the second item of the theorem above is called the torsion submodule of A. A is said to be a torsion module if A = At and to be torsion free if At = 0. Lemma 1. If F is a free R−module then F is torsion-free.

Proof. Suppose that ra = 0 for some 0 ≠ r ∈ R. Since F is free we have that a = r1f1 + . . . rnfn where {f1, . . . , fn} is a linearly independent subset of F . Hence,

0 = ra = rr1f1 + . . . rrnfn

Hence rri = 0 for each 1 ≤ i ≤ n and since r ≠ 0 it follows that ri = 0 for each i. Therefore, a = 0 and the result is proven. In the case where R is a principal ideal domain and A is an R−module, we have that the order ideal of a ∈ A is principal, and therefore Oa = (r) for some r ∈ R. In this case, we say that a has order r. The element r is unique only up to multiplication by a unit. The cyclic submodule Ra generated by a is said to be cyclic of order r. The proof of item 3 from Theorem 8 shows that a ∈ A has order zero if and only if Ra ≃ R. Now we are able to prove a very important result Theorem 9. A ﬁnitely generated torsion-free module A over a principal ideal domain R is free. Proof. We may assume A ≠ 0. Let X be a ﬁnite set of nonzero generators of A. If x ∈ X, then rx = 0(r ∈ R) if and only if r = 0, since A is torsion-free. Consequently, there is a nonempty subset S = {x1, . . . , xk} of X which is maximal with respect to the following property:

r1x1 + ... + rkxk = 0(ri ∈ R) ⇒ ri = 0 ∀i.

4 Then, the submodule F generated by S is a free R-module with basis S. If y ∈ X − S, then by maximality there exist ry, r1, . . . , rk not all zero such that

ryy + r1x1 + ... + rkxk = 0 and therefore ryy = −(r1x1 + ... + rkxk) ∈ F. ̸ Note that ry = 0, since otherwise ri = 0 for every i. But also note∏ that y is arbitrary and that X is finite. Then we conclude that the element r defined by r = ry is a well defined nonzero y∈X−S element of R. Also, for each y1 ∈ X − S we have that   ∏   ∈ ry1 = ry ry1 y1 F

y∈X−S,y≠ y1

By construction we conclude that rX = {rx : x ∈ X} ⊂ F . Since X generates A, we must have that rA ⊂ F as well. Now, consider the map f : A → Agiven by f(a) = ra. Then Imf = rA and since A is torsion free, we have that Kerf = {0}. Therefore, A ≃ rA ⊂ F . Therefore, A is free by Theorem 7.

4 The structure of ﬁnitely generated modules over PID’s

Theorem⊕ 10. If A is a ﬁnitely generated module over a principal ideal domain R, then A = At F , where F is a free R-module of ﬁnite rank and F ≃ A/At.

Proof. The quotient module A/At is torsion-free, since for each r ≠ 0 we have that

r(a + At) = At ⇒ ra ∈ At ⇒ (r1r)a = 0 ⇒ a ∈ At(r1 ≠ 0)

Also, A/At if ﬁnitely generated, because A is. By Theorem 9, we have that A/At is free of ﬁnite rank. Consequently, the exact sequence

0 → At → A → A/At → 0 is exact. But since A/At is⊕ a free module, it must be projective. Therefore, this sequence must be split-exact, and A ≃ At A/At. This isomorphism sends At onto At and sends A/At to some free subset F ⊂ A of ﬁnite rank. To clarify, we have the inclusion i : At → A, the projection → − → ◦ π : A A/At and some R module homomorphism h : A/At A such that π ⊕h = id. In the proof of Theorem 6, 1 ⇒ 3 we showed that the R− module homomorphism φ : At (A/At) → A given by φ(a, b) = i(a) + h(b) = a + h(b) is actually an isomorphism. Therefore, it is clear that A = Im i + Im h = At + F . However, if x ∈ Im i ∩ Im h, then x = h(a + At), and therefore π(x) = ∈ a + At. However, x⊕ Im i = At. Hence, 0 = π(x) = a + At. Hence, x = h(a + At) = h(0) = 0. Therefore, A = At F . It only remains to show that F is indeed free. Since A/At is a ﬁnitely generated free- module, it has a basis {x1, . . . , xn}. Hence, {h(x1), . . . , h(xn)} is a generator of Im h = F . Suppose that there exist ri ∈ R such that

r1h(x1) + . . . rnh(xn) = 0.

Then h(r1x1 + . . . rnxn) = 0. Hence, φ(0, r1x1 + . . . rnxn) = 0. But φ is an isomorphism, and therefore

r1x1 + . . . rnxn = 0.

But A/At is free. Hence, ri = 0 for each i. Therefore, F is free, and the result is now proven.

5 The last theorem is the first step in the direction of classifying finitely generated modules over principal ideal domains. The next result will show that every torsion module can be written as a sum of `‘ p primary modules”. Theorem 11. Let A be a torsion submodule over a principal ideal domain R and for each prime p ∈ R let A(p) = {a ∈ A : a has order of a power of p}. 1. A(p) is a submodule of A for each prime p ∈ R. ∑ 2. A = A(p) where the sum is over all primes p ∈ R. If A is finitely generated, only finitely many of the A(p) are nonzero.

m n Proof. Let’s prove the ﬁrst item. If a, b ∈ A(p) then Oa = (p ) and Ob = (p ). Let k = k i max{m, n}. Then p (a + b) = 0. Hence, by Theorem 8, Oa+b = (p ) for some 0 ≤ i ≤ k. There- m i fore, a + b ∈ A(p). Similarly, if a ∈ A(p) and r ∈ R, then p (ra) = 0. Hence, Ora = (p ) for some 0 ≤ i ≤ m and therefore ra ∈ A(p). Hence, A(p) is a submodule. Now let’s prove item 2. Let 0 ≠ a ∈ A with Oa = (r). Then, since R is a unique factor- n1 n2 nk ization domain, r = p1 p2 . . . pk with pi distinct primes in R and ni > 0 for each i. Let n1 ni−1 ni+1 nk ri = p1 . . . pi−1 pi+1 . . . pk . Then r1, . . . , rk are relatively prime and therefore there exist s1, . . . , sk ∈ R such that s1r1 + ... + skrk = 1R. ni ∈ Consequently a = 1Ra = s1r1a + ... + skrka. But pi siria = sira = 0, whence siria A(pi). Hence, every element of A me be written as a ﬁnite sum of elements each of them belonging to A(p) for some prime p ∈ R. Now let p ∈ R be prime and let A1 be the submodule of A generated by all A(q) with q ≠ p. Suppose that a ∈ A(p) ∩ A1. We will be done if we prove that a = 0. Since a ∈ A(p), we have that m there exists m ≥ 0 such that p a = 0. But since a ∈ A1, we have that a = a1 + ... + at where ai ∈ A(qi) for some primes q1, . . . , qt, all distinct from p. Thus, there are integers mi such that mi m1 m2 mt m1 m2 mt m qi ai = 0. Hence, q1 q2 . . . qt a = 0. But d = q1 q2 . . . qt and p must be relatively prime. m Hence, there are elements r, s ∈ R such that 1R = rp + sd. Consequently,

m a = 1Ra = rp a + sda = sda = 0 ∑ Hence, A = A(p). The last statement of the theorem is a consequence of the fact that a direct sum of modules with infinitely many nonzero summands cannot be finitely generated. Now we will proceed by determining the structure of finitely generated modules in which every element has order a power of a prime p. To do that, we will need the following lemma. Lemma 2. Let A be a module over a principal ideal domain R such that pnA = 0 but pn−1A ≠ 0 for some prime p ∈ R and positive integer n. Let a be an element of A of order pn. 1. If A ≠ Ra then there exists a nonzero b ∈ A such that Ra ∩ Rb = {0}. 2. There is a submodule C of A such that A = Ra ⊕ C. Proof. Let’s prove the first item. If A ≠ Ra then there exists c ∈ A − Ra. Note that pnc = 0. Let j j−1 j j be the least positive integer such that p c ∈ Ra. Therefore, p c ̸∈ Ra and p c = r1a for some k r1 ∈ R. Since R is a unique factorization domain, we have that r1 = rp for some k ≥ 0 and r ∈ R such that p ̸ |r. Consequently,

0 = pnc = pn−j(pjc) = pn−j+kra

n−j+k n This implies that p r ∈ Oa = (p ). Hence, pn−j+kr = spn, for some s ∈ R.

Since p ̸ |r, we must have by unique factorization theorem, that n − j + k ≥ n. Then, it follows that k ≥ j ≥ 1. Then let b := pj−1c − rpk−1a. The preceding argument shows that b ∈ A is a well deﬁned element of A . First, we note that b ≠ 0 because by construction pj−1c ̸∈ Ra. Also,

j k j pb = p c − rp a = p c − r1a = 0.

6 Also, if Ra ∩ Rb ≠ 0, then there exists s ∈ R such that sb ∈ Ra and sb ≠ 0. Since sb ≠ 0 but n n pb = 0, p does not divide s. Therefore , s and p are relatively prime and sx + p y = 1R for some x, y ∈ R. Thus, since pnA = 0,

n b = 1Rb = sxb + p yb = x(sb) ∈ Ra.

Consequently, pj−1c = b + rpk−1a ∈ Ra. If j − 1 ≠ 0, this contradicts the minimality of j. If j − 1 = 0, then c ∈ Ra, which again is absurd. Therefore, Ra ∩ Rb = 0. Now, let’s prove item 2. If A = Ra, let C = 0. If A ≠ Ra, then let S be the set of all submodules B of A such that Ra ∩ B = 0. By item (1), S ≠ ∅. Partially order S by set-theoretic inclusion. { } Let’s prove that every∪ chain in S has maximal element. Let Ci i∈I be a chain in S with respect ∈ ∩ to inclusion. Let C = i∈I Ci. By construction, C is a submodule of A. Also, if x Ra C then x ∈ Ci for some i ∈ I. Since Ci ∩ Ra = 0, it follows that x = 0. Hence, Ra ∩ C = 0 and C ∈ S. It is also clear that C is a maximal element of the chain {Ci}i∈I . Hence, we can aplly the Zorn’s Lemma to conclude that there exists a submodule C of A that is maximal in S. Consider now the quotient module (A/C). Since pnA = 0, we have that pn(A/C) = 0, whence pn(a + C) = 0. Since Ra ∩ C = 0 and pn−1a ≠ 0, we have that pn−1a ̸∈ C, which means that pn−1(a + C) ≠ 0 + C. Hence, a + C has order pn in the R-module (A/C). Observe that

pn(A/C) = 0 and pn−1(A/C) ≠ 0

Which tell us that A/C satisfy the hypothesis of this lemma. By the ﬁrst item of this lemma, if (A/C) ≠ R(a+C) then there exists d+C ∈ A/C such that d+C ≠ C and R(a+C)∩R(d+C) = C. Let E = Rd + C be another submodule of A. It is clear that C ⊂ E. But also, since Ra ∩ C = 0, we have that Ra ∩ E = 0. Indeed, if x ∈ E ∩ Ra, then x = rd + c = sa. Then rd + C = sa + C. Hence, r(d + C) = s(a + C). By construction, rd ∈ C and sa ∈ C. But then sa = 0. Hence, x = sa = 0 and Ra ∩ E = 0. Since d ̸∈ C, E is in S and properly contains C, which contradicts the maximality of C. Therefore, A/C = R(a + C) . Hence, if z ∈ A, then z + C = ra + C for some r ∈ R. This means that z = ra + c for some c ∈ C. Hence, A = Ra + C. But by construction Ra ∩ C = 0, and therefore A = Ra ⊕ C, and the result is proven. Theorem 12. Let A be a ﬁnitely generated module over a principal ideal domain R such that every element of A has order a power of some prime p ∈ R. Then A is a direct sum of cyclic R-modules n1 n of orders p , . . . p k respectively, where n1 ≥ n2 ≥ ... ≥ nk ≥ 1. Proof. The proof is by induction on the number r of generators of A. If r = 1 then A = Ra for some a ∈ A. Therefore, A is a cyclic R-module of order= order of a. Suppose now that r > 1. n1 m2 mr Then A is generated by elements a1, . . . , ar whose orders are p1 , p2 . . . , pr respectively. We may assume that n1 = max{n1, m2, m3, . . . , mr}.

n1 n1 But then, we see that if x ∈ A, then x = s1a1 + ... + srar, and therefore p x = p s1a1 + n1 n1 n1−1 ... + p srar = 0. Hence, p A = 0, but also p A ≠ 0. By item 2. of Lemma 2 there is a submodule C of A such that A = Ra1 ⊕ C. Let π : A → C be the canonical epimorphism given by π(ra1, c) = c. Since A is generated by a1, . . . , ar, C must be generated by π(a1), . . . , π(ar). But π(a1) = 0. Hence, C is generated by fewer than r elements. Consequently, the induction hypothesis implies that C is a direct sum of cyclic R-modules of orders pn2 , . . . , pnk respectively n2 n1 with n2 ≥ n3 ≥ ... ≥ nk ≥ 1. Thus, C contains an element of order p . Since p A = 0 and C is n1 a subset of A, we have that p C = 0. Therefore, n1 ≥ n2 (argue by contradiction and use item 5 of 8). The theorem is now proven.

In order to prove uniqueness of the decomposition, we will need the following lemmas.

Lemma 3. Let A, B and Ai (i ∈ I) be modules over a principal ideal domain R. Let r ∈ R and p ∈ R be prime. 1. rA = {ra : a ∈ A} and A[r] = {a ∈ A : ra = 0} are submodules of A.

2. R/(p) is a ﬁeld and A[p] is a vector space over R/(p).

7 3. For each positive integer n there are R-module isomorphisms

(R/(pn))[p] ≃ R/(p) and pm(R/(pn)) ≃ R/(pn−m)(0 ≤ m < n). ∑ ∑ ∑ ≃ ≃ ≃ 4. If A i∈I Ai, then rA i∈I rAi and A[r] i∈I Ai[r].

5. If f : A → B is an R-module isomorphism, then f : At ≃ Bt and f : A(p) ≃ B(p). Proof. Item 1. is clear, and is left as an exercise. Let’s prove item 2. Since R is a principal ideal domain, (p) is a maximal ideal of R, hence R/(p) must be a field. Let’s define a scalar multiplication on A[p]. Define φ : R/(p) × A[p] → A[p] by φ(r + (p), a) = ra. Let’s see if φ is indeed well defined: If r1, r2 ∈ R are such that r1 = r2 + ps for some s ∈ R, then

r1a = r2a + s(pa) = r2a + s0 = r2a.

By the preceding item, if a ∈ A[p] then ra ∈ A[p] for every r ∈ R. Therefore, φ is valued on A[p]. Hence, there is nothing wrong with our deﬁnition. It is an exercise to check that φ satisﬁes the desired module properties (φ(r + s(p), a) = φ(r + (p), a) + φ(s + (p), a), etc) so that A[p] is a vector space over R/(p). Now let’s go to item 3. Observe:

(R/(pn))[p] = {x ∈ R/(pn): px = 0} = {r + (pn) ∈ R/(pn): p(r + (pn)) = 0} = {r + (pn) ∈ R/(pn): pr ∈ (pn)} = {r + (pn) ∈ R/(pn): pr = pns for some s ∈ R} = {r + (pn) ∈ R/(pn): r = pn−1s for some s ∈ R} = {pn−1s + (pn): s ∈ R} = {s(pn−1 + (pn)) : s ∈ R}.

Therefore, (R/(pn))[p] is generated as an R−module by the single nonzero element pn−1 + (pn). Now, we deﬁne ϕ :(R/(pn))[p] → R/(p) by ϕ(spn−1 + (pn)) = s + (p). Let’s show that ϕ is well deﬁned. Indeed, if s, t ∈ R are such that spn−1 + (pn) = tpn−1 + (pn) then

(t − s)pn−1 = zpn for some z ∈ R. Hence p must divide t − s, and therefore there exists z ∈ R such that

t − s = pz.

Therefore, t − s ∈ (p) and ϕ is well deﬁned. By construction, ϕ is a surjective R− module homomorphism. ϕ is also injective. Indeed, if ϕ(spn−1 + (pn)) = 0, then s ∈ (p) and therefore spn−1 ∈ (pn). Hence, spn−1 + (pn) = 0, as we wanted to prove. Now, let’s show that

pm(R/(pn)) ≃ R/(pn−m)(0 ≤ m < n)

Indeed, note that

pm(R/(pn)) = {pmr + (pn): r ∈ R} = R(pm + (pn)).

Therefore, pm(R/(pn)) is precisely the submodule of R/(pn) generated by the element pm + (pn). But the order of pm + (pn) is exactly pn−m. Hence, by Theorem 8 we have that

R/(pn−m) ≃ R(pm + (pn)) = pm(R/(pn)).

Item 4. is left as an exercise. Let’s prove item 5. Let f : A → B be an R-module isomorphism. If x ∈ A(p) then pnx = 0. Hence, f(pnx) = 0 and therefore pnf(x) = 0. Hence, f(x) ∈ B(p). Therefore, f : A(p) → B(p). Since f is an isomorphism, the same argument shows that f −1 : B(p) → A(p). Therefore, A(p) ≃ B(p) and the isomorphism is given by the restriction of f. Now, let’s take y ∈ At. Then there exists r ∈ R, r ≠ 0 such that ry = 0. Hence, rf(y) = 0 and therefore −1 y ∈ Bt. Since f is an isomorphism, the same argument shows that f : Bt → At. Therefore, At ≃ Bt and the isomorphism is given by the restriction of f.

8 We just need one more lemma before proving the main theorem.

∈ n1 nk Lemma 4. Let R be a principal ideal domain. If r R factors as r = p1 . . . pk with p1, . . . , pk distinct primes and each ni > 0, then there is an R-module isomorphism

≃ n1 ⊕ ⊕ nk R/(r) R/(p1 ) ... R/(pk ). Consequently, every cyclic R-module of order r is a direct sum of k cyclic R-modules of orders n1 nk p1 , . . . , pk respectively. Proof. We will prove that if s, t ∈ R are relatively prime then R/(st) ≃ R/(s) ⊕ R/(t). Then, the ﬁrst part of this lemma follow by induction on the number of distinct primes in the prime decomposition of r. The last statement of the lemma is a consequence of the third item of Theorem 8. Indeed, if A is a R−module and a ∈ A has order r then the cyclic R-module Ra is isomorphic n1 ⊕ ⊕ nk to R/(r), which is isomorphic to a direct sum of the form R/(p1 ) ... R/(pk ). But each of ni ni these terms are cyclic R-modules, each of which are generated by 1R + (pi ), which has order pi . Hence, the last statement is proven. Let θ1 : R → R be given by θ1(x) = tx. θ1 is an R-module homomorphism that takes the ideal (s) onto (st). Since R is a principal ideal domain, θ1 is also injective. Hence, θ1 induces an R-module homomorphism φ1 : R/(s) → R/(st) given by φ1(x + (s)) = tx + (st). Similarly, there is an R-module homomorphism φ2 : R/(t) → R/(st) given by φ2(y + (t)) = sy + (st). Now, deﬁne the map α : R/(s) ⊕ R/(t) → R/(st) by

α(x + (s), y + (t)) = φ1(x + (s)) + φ2(y + (t)) = tx + sy + (st).

α is a well deﬁned R-module homomorphism. Since by hypothesis (s, t) = 1R (notation for greatest common divisor), there are elements u, v ∈ R such that su+tv = 1R. Let’s show that α is surjective. Let c ∈ R be any element of R. Then,

c = c1R = c(su + tv) = csu + ctv

Hence, α(cv + (s), cu + (t)) = tcv + scu + (st) = c + (st). Now, let’s prove that α is a monomorphism. We must show that

α(x + (s), y + (t)) = 0 + (st) ⇒ x ∈ (s) and y ∈ (t).

But by construction α(x + (s), y + (t)) = 0 + (st) ⇒ tx + sy ∈ (st).

Hence, tx + sy = λst for some λ ∈ R. But remember that su + tv = 1R. Hence,

txu + suy = λstu ⇒ txu + (1R − tv)y = λstu.

Hence, y = t(λsu + vy − xu) ∈ (t) Similarly, txv + svy = λstv ⇒ (1R − su)x + svy = λstv. Hence, x = s(λtv + ux − vy) ∈ (s), and the result is proven. Finally, we can prove the main theorem, but the hard work is over.

Theorem 13. Let A be a ﬁnitely generated module over a principal ideal domain R.

1. A is the direct sum of a free submodule F of ﬁnite rank and a ﬁnite number of cyclic torsion modules. The cyclic torsion summands (if any) are of orders r1, . . . , rt, where r1, . . . , rt are (not necessarily distinct) nonzero nonunit elements of R such that r1|r2| ... |rt. The rank of F and the list of ideals (r1),..., (rt) are uniquely determined by A.

9 2. A is the direct sum of a free submodule E of ﬁnite rank and a ﬁnite number of cyclic torsion s1 sk modules. The cyclic torsion summands (if any) are of orders p1 , . . . , pk , where p1, . . . , pk are (not necessarily distinct) prime elements in R and s1, s2 . . . , sk are (not necessarily distinct) s1 sk positive integers. The rank of E and the list of ideals (p1 ),..., (pk ) are uniquely determined by A.

The elements r1, . . . , rt in Theorem 13 are called the invariant factors of the module A. Also, s1 sk p1 , . . . , pk are called the elementary divisors of A. Proof. Initially, we will prove that the free modules E and F given above have ﬁxed rank. More precisely, we will prove that there is a unique nonnegative integer s such that the number of the summands isomorphic to R in any decomposition of A as a direct sum of cyclic R-modules is precisely s. Indeed, note that any such decomposition yields an isomorphism A ≃ H ⊕ F , where H is a torsion module and F is free of rank s. If i : H → H ⊕ F is given by i(h) = (h, 0) then i(H) is the torsion subgroup of (H ⊕F ). By Lemma 3 we must have At ≃ (H ⊕F )t = i(H). Consequently, (A/At) ≃ (H ⊕ F )/i(H) ≃ F . Therefore, any decomposition of A leads to the conclusion that A/At is free of rank s. But the rank of A/At does not depend on the particular decomposition of A. Hence, s is uniquely determined. Now, we will prove the existence of the decompositions. Because of Theorems 10, 11 and 12, we have that ∑ ∑ A = At ⊕ F = A(p) ⊕ F = (⊕Rai) ⊕ F = ⊕Rai ⊕ F.

si where each of the elements ai have orders pi , where the primes pi are not necessarily distinct and the positive integers si are also not necessarily distinct. Therefore, the existence part of item 2. is already proven. Let’s now prove the existence of the decomposition described in the ﬁrst item. To do that, rearrange the elementary divisors of A in the following way:

n11 n12 n1r p1 , p2 , . . . pr n21 n22 n2r p1 , p2 , . . . pr ......

nt1 nt2 ntr p1 , p2 , . . . pr where p1, . . . , pr are distinct primes and for each j = 1, 2, . . . , r, 0 ≤ n1j ≤ n2j ≤ ... ≤ ntr with some nij ≠ 0 and ﬁnally n1j ≠ 0 for some j. By what we have shown, we have that ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ A = (Ra11 Ra21 ... Ram11) (Ra12 Ra22 ... Ram22) ... (Ra1r Ra2r ... Ramr r) F.

≃ ξj where aij has order a positive power of pj. By Theorem 8, item 3, we have that Raij R/(pj ), where 0 < ξj ∈ {n1j, . . . , ntj}. Hence,   ∑t ∑r ≃  nij  ⊕ A R/(pj ) F i=1 j=1

nij 0 where some summands are zero, namely those with pj = pj = 1. For each i = 1, 2, . . . , t let ni1 ni2 ··· nir | | | mi = p1 p2 pr . By construction, m1 m2 ... mt and Lemma 4 implies that   ∑t ∑r ∑t ≃  nij  ⊕ ≃ ⊕ A R/(pj ) F R/(mi) F i=1 j=1 i=1

Hence, ∑t A ≃ R/(mi) ⊕ F. i=1 Therefore, the existence is proven for both items 1. and 2. Now, let’s prove uniqueness of the decomposition given in item 2. First of all, prime factorization in R is unique up to multiplication by unit. This causes no diﬃculties in Z since the only units are 1 and primes are deﬁned to be positive. In an arbitrary principal ideal domain R, however, an element a ∈ A may have order p and order q with p and q distinct primes. However, since Oa = (p) = (q), there must be a unit

10 u ∈ R such that q = pu. Hence, the uniqueness statements in items 1. and 2. deal with ideals rather than elements. Note that a ≠ 0 implies that Oa ≠ R and that a cyclic module Ra is free if and only if Oa = (0). Thus, the elements ri in item 1. are nonzero nonunits. Suppose now that

∑r ∑d ′ A ≃ R/(ni) ⊕ F and A ≃ R/(kj) ⊕ F i=1 j=1 ′ where each ni, kj is the power of a prime (diﬀerent∑ primes may occur) and F,F are free∑R-modules r ⊕ r of ﬁnite rank. First of all, the torsion subgroup of i=1 R/(ni) F is isomorphic∑ to i=1 R/(ni). ∈ r ⊕ Indeed, we will show that if x∑= s1 + (n1) + s2∑+ (n2) + ... + sr + (nr) + f ( i=1 R/(ni) F )t ∈ r r then f = 0 and therefore x i=1 R/(ni) = ( i=1 R/(ni))t. Suppose that

t(s1 + (n1) + s2 + (n2) + ... + sr + (nr) + f) = 0 for some t ∈ R, t ≠ 0. Then

ts1 + (n1) + ts2 + (n2) + ... + tsr + (nr) + tf = 0, Since we are dealing with a direct sum, we have that tf = 0, and hence f = 0, because F is free, as we wanted to prove. Therefore,

∑r ∑d At ≃ R/(ni) and At ≃ R/(kj) i=1 j=1 whence ∑r ∑d R/(ni) ≃ R/(kj). i=1 j=1 ∑ r But for each prime p, we have that ( i=1 R/(ni)) (p) is a direct sum of those R/(ni) such that ni is a power of p. Indeed, if A and B are R-modules such that A = A(p) and B = B(q) where p, q are distinct primes, then (A ⊕ B)(p) = A. Indeed, if pi(a + b) = 0 then pia = 0 and pib = 0. But then, by Theorem 8 item v) we have that b has order pj for some j ≤ i. But then b ∈ A(p) and therefore b ∈ A(p)∑∩ A(q), which means that b = 0 (see item 2. from Theorem 11). A similar r argument shows that ( i=1 R/(ni)) (p) is a direct sum of those R/(ni) such that ni is a power of p(it is clear that the direct sum of p- primary modules is also p-primary). Note also that the last equation and Lemma 3, item v) imply that ( )   ∑r ∑d   R/(ni) (p) ≃ R/(kj) (p). i=1 j=1

The preceding information allow us to conclude that

∑rp ∑dp ai ≃ cj ≤ ≤ ≤ ≤ ≤ R/(p ) R/(p )(1 a1 ... arp ; 1 c1 . . . cdp ). i=1 i=1 ∑ ∑ r r ≃ p ai We will first show that rp = dp. Since A(p) ( i=1 R/(ni)) (p) = i=1 R/(p ), Lemma 3 itens 3. and 4. gives that ( ) ∑rp A(p)[p] ≃ R/(pai ) [p] ≃ R/(p) ⊕ R/(p) ⊕ ... ⊕ R/(p), i=1 where the direct sum above is repeated rp times. Again, Lemma 3 item 2. asserts that A(p)[p] is a vector space over the field R/(p). The above equation implies that rp = dimR/(p)A(p)[p]. Since the dimension of A(p)[p] over R/(p) is fixed, we must have dp = rp and therefore r = d. It remains to prove that ai = ci for all i ≤ rp. Let ν(1 ≤ ν ≤ rp) be the first integer such that ai = ci for all aν ai i < ν and aν ≠ cν . We may assume that aν < cν . Since p (R/(p )) = 0 for ai ≤ aν , the first decomposition and Lemma 3 itens 3. and 4. imply that

∑rp ∑rp ∑rp ∑rp − paν A(p) ≃ paν R/(pai ) ≃ paν R/(pai ) ≃ paν R/(pai ) ≃ R/(pai aν ), i=1 i=1 i=ν+1 i=ν+1

11 with aν+1 − aν ≤ aν+2 − aν ≤ ... ≤ ar − aν . Clearly, there are at most r − ν nonzero summands. Similarly, since ai = ci for i < ν, and aν < cν , the second decomposition implies that

∑rp ∑rp ∑rp ∑rp − paν A(p) ≃ paν R/(pcj ) ≃ paν R/(pcj ) ≃ paν R/(pcj ) ≃ R/(pci aν ), j=1 j=1 j=ν j=ν ≤ − ≤ − ≤ ≤ − − with 1 cν aν cν+1 aν ... crp aν . Hence, there are at least r ν + 1 nonzero summands. Therefore, we have two decompositions of the group paν A(p) as a direct sum of cyclic R-modules of prime power order and the number of summands in the ﬁrst decomposition is less than the number of summands in the second. This is absurd by what we recently proved (that r = d, i.e, the number of summands must be the same). Hence, we must have ai = ci for all i. Finally, let’s prove the uniqueness of decomposition given in item 1. Suppose that

A ≃ R/(r1) ⊕ R/(r2) ⊕ ... ⊕ R/(rt) ⊕ F and that ′ A ≃ R/(k1) ⊕ R/(k2) ⊕ ... ⊕ R/(kd) ⊕ F ′ where r1 is a nonzero nonunit, r1|r2| ... |rt, k1 is a nonzero nonunit, k1|k2| ... |kd and F,F are free 0 R-modules. Each ri, kj has a prime decomposition and by inserting factors of the form p we may assume that the same (distinct) primes p1, . . . , pr occur in all the factorizations, say

a11 a12 a1r ai1 ai2 air r1 = p1 p2 . . . pr , . . . , ri = p1 p2 . . . pr for each 1 ≤ i ≤ t and similarly

c11 c12 c1r cj1 cj2 cjr k1 = p1 p2 . . . pr , . . . , kj = p1 p2 . . . pr for each 1 ≤ j ≤ d. Since r1|r2| ... |rt we must have that 0 ≤ a1m ≤ a2m ≤ ... ≤ atm for each 1 ≤ m ≤ r. Similarly, 0 ≤ c1m ≤ c2m ≤ ... ≤ cdm for each 1 ≤ m ≤ d. By Lemmas 3 and 4, we have that ∑t ∑r ∑t ∑d ∑d ∑r aij ≃ ≃ ≃ ≃ cij R/(pj ) R/(ri) At R/(ki) R/(pj ), i=1 j=1 i=1 i=1 i=1 j=1 where some summands may be zero. Hence, for each j = 1, . . . , r we have that

∑t ∑d aij ≃ ≃ cij R/(pj ) A(pj) R/(pj ) i=1 i=1 ∈ { } ≤ ≤ ≤ But since r1is not a unit and r∑1 is nonzero, there must exist j 1, . . . , r such that 1 a1j a2j ... ≤ a . Choosing this j, t R/(paij ) has exactly t nonzero summands. By the uniqueness tj i=1 ∑j d cij result of decompositions of item 2., i=1 R/(pj ) must have exactly t summands. Therefore, t ≤ d. Similarly, d ≤ t and therefore t = d. Using the same argument to conclude uniqueness in item 2., we have aij = cij for all i, j. Therefore, rj = ukj, where u ∈ R is a unit. Therefore, the theorem is proven.

References

Hungerford, T. “Algebra”, Springer, 2000.