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 r1jr2j ::: jrt. 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) 2 R. Elements a; b 2 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 2 B is a least element of B provided c ≤ b for every b 2 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 2 A, the subset fa; bg ⊂ 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 2 A, fc 2 A : c < ag ⊂ B ) a 2 B then B = A 1 Proof. If A − B =6 ; then there is a least element a 2 A − B. By the definitions of least element and A − B we have that fc 2 A : c < ag ⊂ B By hypothesis, we have that a 2 B, which is absurd. Therefore, A = B. Proposition 1. Let (A; ≤) be a well ordered set. The immediate successor of a 2 A, if it exists, is the least element in the set fx 2 A : a < xg and is denoted by a + 1. Then at most one element of A has no immediate successor. Proof. Let m 2 A be such that m has no immediate successor. Then, since A is well ordered, we have that fx 2 A : m < xg = ; Suppose now that there exists n 2 A such that m =6 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 fx 2 A : n < xg 6= ;: 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 2 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 : B + C ! (B + C)=C be given by (b + c) = (b + c) + C. Then ◦ ϕ : B ! (B + C)=C is such that ◦ ϕ(b) = b + C Hence, Ker( ◦ ϕ) = B \ C and Im( ◦ 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 0 g0 0 A0 B0 C0 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- L 1 2 L 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 moduleL homomorphisms f : A1 B and h : A2 B, which induce a R-module homomorphism ' : A1 A2 ! B given by '((a1; a2)) = f(a1) + h(a2). Then, we can check that the following diagram is commutative L i1 π2 0 A1 A1 A2 A2 0: 1A1 ' 1A2 f g 0 A1 B A2 0 But then, by Theorem 5, we have that ' is a isomorphism. ) − ! ! 2. 3.: Now we have two R module homomorphismsL g : B A2 and k : B A1, which induce a R-module homomorphism : B ! A1 A2 given by (b) = (k(b); g(b)). Then, we can check that the following diagram is commutative: f g 0 A1 B A2 0: 1A1 1A2 L i1 π2 0 A1 A1 A2 A2 0: Hence, is an isomorphism by Theorem 5, and we are done. L 3. ) 1.,2.: Now suppose that condition 3. holds. Then, we have an isomorphism ' : A1 A2 ! B such that the following diagram commutes. L i1 π2 0 A1 A1 A2 A2 0: 1A1 ' 1A2 f g 0 A1 B A2 0 L We define h : A2 ! B by h = ' ◦ i2 where i2 is the immersion from A2 into A1 A2. Then, gh(a2) = g('(i2(a2))) = g('(0 + a2)) = 1A2 (π2(0 + a2)) = 1A2 (a2) = a2 ! and therefore condition 1. was proven. To proveL that condition 2. holds, we define k : B A1 by −1 ◦ k = ' π1 where π1 is the projection of A1 A2 onto A1 and check that kf = 1A1 . ! ! ◦ Corollary 1. IfLf : 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 L Hence, by the preceding theorem B ' A Ker g. But also by hypothesisL f must be an monomor- phism, 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 fx1; x2; : : : ; xng and let B be a submodule of A. Let F be a free R− module generated by n elements with basis ff1; : : : ; fng. Let ' : F ! A be the unique R−module homomorphism such that '(fi) = xi for each 1 ≤ i ≤ n. Let E = fv 2 F : '(v) 2 Bg. It is easily checked that E is a submodule of F . By construction, 'jE : E ! B is surjective. Also, by Theorem 7, we have that E is finitely generated by say fy1; y2; : : : ; ymgwith m ≤ n. Hence, f'(y1);'(y2);:::'(ym)g 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 2 A let Oa = fr 2 R : ra = 0g.