MMATH18-201: Module Theory Lecture Notes: Modules over PID Contents 1 Free modules over PID1 2 Torsion Modules4 3 Primary Modules9 4 Fundamental Theorem for finitely generated modules over PID 14 5 Exercises 21 References 21 In these notes we extend some of the results on abelian groups to modules over a principal ideal domain (PID). The results are from Section 10.6, Chapter 10 of [1]. 1 Free modules over PID An integral domain R is called a PID if every ideal of R is principal, i.e., generated by a single element. Since every integral domain is a commutative ring so if M is a finitely generated free R-module, then we know that any two basis of M has the same number of elements, therefore the number of basis element is independent of the choice of basis, we call this the rank of the free module M. Theorem 1.1. Let R be a PID and M a free R-module of rank n. Then every submodule of M is free with a basis having atmost n elements. Proof. We prove the result by induction on n. If n = 0, then empty set is the basis of M, so M = f0g and M itself is the only submodule and we are done. Assume now that M is a free module with basis fe1; e2; : : : ; eng, n > 0, and the result holds for all modules with basis having less than n elements. Let N be any submodule of M. If N = f0g, then nothing to prove. So let N 6= f0g, then every element x of N can be uniquely written as n X x = aiei; ai 2 R: (1) i=1 Keeping in mind the above expression of an arbitrary element x of N, we define a map φ : N −! R by (x)φ = a1. Regarding R as a module over itself, verify that φ is an R-module homomorphism. The kernel of φ consists of all those elements of N, whose expression of the form (1) does not contain the basis element e1, i.e., ker φ is a submodule of the free R-module ( he ; e ; : : : ; e i; if n > 1 2 3 n : f0g; if n = 1 Since F is free with basis having n−1 elements, so by induction hypothesis every submodule of F is free with basis having atmost n − 1 elements. In particular ker φ is a free R-module with basis (say) fv2; v3; : : : ; vmg, where m ≤ n: As φ is homomorphism, therefore Im φ is a submodule of the R-module R and hence an 1 ideal of R. Since R is a PID, so there exist some a 2 R such that Im φ = hai: If a = 0, then N = ker φ and we are done. Otherwise there exist some v1 2 N such that (v1)φ = a: We now claim that N = hv1i ⊕ ker φ. (2) To prove (2), we need to show that N = hv1i + ker φ and hv1i \ ker φ = f0g. Clearly hv1i + ker φ ⊆ N, so let x be an arbitrary element of N, then Im φ = hai implies that there exist some r 2 R such that (x)φ = ra. Now (x − rv1)φ = (x)φ − (rv1)φ = ra − ra = 0; implies that x − rv1 2 ker φ so x 2 hv1i + ker φ, i.e., N ⊆ hv1i + ker φ. If y 2 hv1i \ ker φ, then y = rv1; for some r 2 R and (y)φ = 0: Now (y)φ = 0 =) (rv1)φ = 0 =) ra = 0; since R is an integral domain and a 6= 0, so ra = 0 implies that r = 0 i.e., y = 0 and therefore hv1i \ ker φ = f0g: Hence the claim follows. As fv2; v3; : : : ; vmg is a basis of ker φ, so from (2) it follows immediately that every element of N can be uniquely written as an R-linear combination of fv1; v2; : : : ; vmg, in particular 0 has a unique representation which implies that the set fv1; v2; : : : ; vmg is R- linearly independent and hence forms a basis. Thus N has a basis having m ≤ n elements. Corollary 1.2. If R is a PID, then every submodule of the R-module Rn is free of rank atmost n. n Proof. The R-module R is free with basis fe1; e2; : : : ; eng, where ei is the n- tuple (0;:::; 1 :::; 0); having 1 at the i-th co-ordinate and zero elsewhere for 1 ≤ i ≤ n (called standard basis). The result now follows immediately from the theorem. The following result shows that every finitely generated module over a PID gives rise to a short exact sequence in a natural way. Proposition 1.3. Let M be a finitely generated module over a PID R. Then there is a representation (exact sequence) 0 −! Rm −! Rn −! M −! 0 for some positive integers m ≤ n: 2 Proof. Suppose M is generated as an R-module by the elements x1; x2; : : : ; xn n and let fe1; e2; : : : ; eng be the standard basis of the free R-module R . Then the mapping ei 7−! xi for each 1 ≤ i ≤ n extends to a surjective R-module homomorphism (say) φ between Rn and M (Why?). Now as ker φ is a submodule of the free R-module Rn, therefore by Corollary 1.2, must be a free module with basis having atmost n elements hence ker φ ∼= Rm for some m ≤ n: If is the composition of the isomorphism from Rm into ker φ and the inclusion map from ker φ into Rn, then we have that the following φ 0 −! Rm −! Rn −! M −! 0 is the required short exact sequence. Corollary 1.4. Let M be a finitely generated module over a PID R. If M has a generating set with n elements, then every submodule of M is also finitely generated by atmost n elements. Proof. Since M is a finitely generated over a PID, so by the Proposition 1.3, there is an exact sequence φ 0 −! Rm −! Rn −! M −! 0; where n is the number of elements in some generating set for M and m ≤ n: Now if N is any submodule of M, then (N)φ−1 is a submodule of Rn. Since R is a PID and Rn is a free R-module therefore by Corollary 1.2, (N)φ−1 is a free R-module with basis (say) fx1; x2; : : : ; xkg, where k ≤ n. Then verify that the set f(x1)φ, (x2)φ, : : : ; (xk)φg generates the submodule N of M and hence N is finitely generated. The above result may not hold in general as is evident from the following example. Example. Let R = K[X1;X2;:::] be the polynomial ring in infinitely many indeterminates over a field K. Then R regarded as a module over itself is finitely generated but the submodule hX1;X2;:::i is not finitely generated. Theorem 1.5. Let R be a PID and M a free R-module of rank n. Then every generating set of M consisting of n elements forms a basis of M. 3 n Proof. Let fe1; e2; : : : ; eng be the standard basis of R and fx1; x2; : : : ; xng a generating set of M. Then by Proposition 1.3 there is a short exact sequence φ 0 −! K = ker φ −!α Rn −! M −! 0; where α is the inclusion map and φ the surjective R-homomorphism mapping ei 7−! xi. Since every short exact sequence with third term free splits, we have that Rn ∼= M ⊕ K: (3) As R is a PID and Rn a free R-module of rank n, so by Corollary 1.2 the submodule K is free of rank atmost n. Now from (3), we have that rank Rn = rank M + rank K; which together with the fact that M is free of rank n, implies that rank of K is zero, i.e., K = f0g: Hence φ is an isomorphism and maps the linearly independent set fe1; e2; : : : ; eng onto a linearly independent set. Therefore the spanning set S = fx1; x2; : : : ; xng forms a basis of M. 2 Torsion Modules Definition. Let M be an R-module. An element x of M is called torsion if there is some non-zero r 2 R, such that rx = 0. If tor M denotes the set of all torsion elements of M, then M is said to be torsion free if tor M = f0g and is called a torsion module if tor M = M: Theorem 2.1. Let R be a non-trivial integral domain and M an R-module. Then tor M is a submodule of M and the quotient M= tor M is torsion free. Proof. Since r · 0 = 0 for every r 2 R, so 0 2 tor M. Now for any x; y 2 tor M there exist some non-zero elements r1; r2 2 R such that r1x = 0 = r2y: (4) As R is an integral domain, so r1r2 6= 0: Therefore, keeping in mind that R is a commutative ring and (4), we have that for any r 2 R r1r2(rx + y) = (r1r2)rx + (r1r2)y = (r2r)r1x + r1(r2y) = (r2r)0 + r10 = 0; 4 i.e., rx + y 2 tor M so tor M is a submodule of M. Now to prove that the quotient R-module M= tor M is torsion free, we need to show that 0 is the only torsion element of M= tor M. Let x 2 tor (M= tor M) : Then there exist some non-zero element s 2 R such that sx = 0, which implies that sx = 0, i.e., sx 2 tor M.