NOTES ON LINEAR ALGEBRA OVER INTEGRAL DOMAINS. Contents. 1. Introduction 1 2. Rank and basis 1 3. The set of linear maps 4. 1.

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1 NOTES ON LINEAR ALGEBRA OVER INTEGRAL DOMAINS Contents 1. Introduction 1 2. Rank and basis 1 3. The set of linear maps 4 1. Introduction These notes establish some basic results about linear algebra over integral domains that are used in the classification of finitely generated modules over a principal ideal domain in Samuel s book on algebraic number theory. The assertions are very easy. They were prepared for Basic Algebra, 60210, discussion on modules in December Rank and basis Let R be a ring and let M be a R-module with submodules M 1 and M 2. Consider the R-module homomorphism, f : M 1 M 2 M, given by f(x,y) = x+y, for x M 1, y M 2. We set M 1 + M 2 = f(m 1 M 2 ) = {x 1 + x 2 : x 1 M 1,x 2 M 2 }. Since the image of R-module homomorphism is a submodule, it follows that M 1 +M 2 is a submodule of M. By Theorem in Ash, f : M 1 M 2 M is a R-module isomorphism if and only if: (1) M = M 1 +M 2, and (2) M 1 M 2 = 0. When f is an isomorphism, we say M = M 1 M 2, or M = M 1 +M 2 is direct. Recall that a R-module M is called free if M has a basis. The next result asserts that if M is a direct sum of two free modules, then M is free. Lemma 2.1. Let M = M 1 +M 2 be direct. Suppose S 1 = {v 1,...,v t } is a basis of M 1 and S 2 = {w 1,...,w u } is a basis of M 2. Then T := S 1 S 2 is a basis of M. In particular, if M 1 and M 2 are free, then M 1 + M 2 is free, and the number of elements in a basis of M 1 +M 2 is t+u. 1

2 2 Proof : We first show that T is linearly independent, so suppose that t i=1 a iv i + u j=1 b jw j = 0, a i,b j R. Set v = t i=1 a iv i and set w = u j=1 b jw j = 0. Note that v M 1 and w M 2. Then by assumption, v +w = 0, so v = w M 1 M 2 = 0. Since v = 0, t i=1 a iv i = 0, so a i = 0,i = 1,...,t, since S 1 is a basis, and hence linearly independent. Since w = 0, u j=1 b jw j = 0, so b j = 0,j = 1,...,u, since S 2 is linearly independent. Hence T is linearly independent. We now show that T generates M. Since M = M 1 +M 2, if m M, then m = v +w, with v M 1, and w M 2. Since S 1 generates M 1, v = t i=1 a iv i for some a i R. Since S 2 generates M 2, w = u j=1 b jw j, for some b j R. Then m = v +w = t i=1 a iv i + u j=1 b jw j, so T generates M. The last assertion follows from the definition of a free R-module. Definition 2.2. We say a R-module M is finitely generated if there exists a finite generating set S of M, or equivalently, if there is a finite subset S M such that if v M, then v = v i S a iv i, with a i R. By convention, the 0-submodule {0} is generated by the empty set, and an empty sum 0 i=1 r iv i = 0.. Let R be an integral domain and let F = Frac(R) be its fraction field. We regard R F via the injective map a a. Then it follows that 1 R n = {(a 1,...,a n ) : a i R} F n = {(α 1,...,α n ) : α i F}. Let V R n be a R-submodule. Let FV := {αv : α F,v V} F n. Lemma 2.3. FV is a F-subspace of F n. Proof : It suffices to show that if u 1,u 2 FV and λ F, then u 1 + u 2 FV and λu 1 FV. We set u 1 = α 1 v 1 and u 2 = α 2 v 2 FV, with v 1,v 2 V, and α 1 = a 1 b 1,α 2 = a 2 b 2 F, so a 1,a 2,b 1,b 2 R, and b 1,b 2 are both nonzero. Then α 1 v 1 +α 2 v 2 = a 1 v 1 + a 2 v 2 = 1 (a 1 b 2 v 1 +b 1 a 2 v 2 ) FV, since a 1 b 2 v 1 +b 1 a 2 v 2 V by definition of submodule. If λ F, then λ α 1 v 1 = (λ α 1 ) v 1 FV. Thus, FV is a F-subspace. Definition 2.4. Let V R n be a R-submodule. Then rk R (V) = dim F (FV).

3 Proposition 2.5. Let R be an integral domain, and let V R n be a R-submodule. (1) 0 rk R (V) n. (2) rk R (R n ) = n. (3) Let M 1,M 2 be R-submodules of R n and suppose M = M 1 + M 2 is direct. Then FM 1 FM 2 = 0, FM 1 +FM 2 = F(M 1 +M 2 ), and rk R (M) = rk R (M 1 )+rk R (M 2 ). (4) Let v 1,...,v r be a basis of a submodule M of R n. Then v 1,...,v r is a basis of FM, so rk R (M) = r. (5) Let v M be nonzero. Then {v} is a basis of R v, so rk R (R v) = 1. 3 Proof : (1) and (2): Since V R n, it follows from definitions that FV FR n = F n. Hence, by linear algebra over fields, 0 rk R (V) = dim F (FV) dim F (F n ) = n. For (3), we first show that FM 1 FM 2 = 0. For this, let α 1 v 1 FM 1, and α 2 v 2 FM 2, with (*) α 1 = a 1 b 1, α 2 = a 2 b 2, for v i M 1, v 2 M 2, a 1,a 2 R, and b 1,b 2 nonzero elements of R. If α 1 v 1 = α 2 v 2 FM 1 FM 2, then a 1 b 1 v 1 = a 2 b 2 v 2, so a 1b 2 v 1 = a 2b 1 v 2. Since 0, then a 1 b 2 v 1 = a 2 b 1 v 2 M 1 M 2 = 0. Since b 2 0, a 1 v 1 = b 2 1 a 1 b 2 v 2 = 0, so α 1 v 1 = 0. It follows that FM 1 FM 2 = 0. We next show that FM 1 + FM 2 = F(M 1 + M 2 ). Indeed, let v 1 M 1 and v 2 M 2. Then if α F, then α(v 1 +v 2 ) = αv 1 +αv 2 FM 1 +FM 2, so F(M 1 +M 2 ) FM 1 +FM 2. Conversely, let α 1 v 1 FM 1 and α 2 v 2 FM 2 with α 1,α 2 as in (*) above. Then α 1 v 1 +α 2 v 2 = 1 1 (a 1 b 2 v 1 +a 2 b 1 v 2 ) F(M 1 +M 2 ), since F and a 1 b 2 v 1 +a 2 b 1 v 2 M 1 +M 2. Given these observations, rk R (M) = rk R (M 1 +M 2 ) = dim F (F(M 1 +M 2 )) = dim F (FM 1 +FM 2 ). But from linear algebra over fields, if V 1 and V 2 are F-subspaces of F n, then dim F (V 1 +V 2 ) = dim F (V 1 )+dim F (V 2 ) dim F (V 1 V 2 ). Applying this with V 1 = FM 1 and V 2 = FM 2 gives assertion (3), since FM 1 FM 2 = 0. We now prove assertion (5). It is clear from the definition of R v that {v} generates R v. Let v R n is nonzero, then v = r 1 e r n e n for some r 1,...,r n R (here e 1,...,e n are standard basis vectors of R n.) Since v 0, some r i 0. If a v = 0 for a R, then ar 1 e 1 + +ar n e n = 0, so since {e 1,...,e n } is a basis of R n, ar i = 0. Since r i 0, it follows that a = 0. Hence, the set {v} is linearly independent. It follows easily that {v} is a basis of F v = FR v, so dim F (F v) = 1, and this completes the proof of (5).

4 4 We prove assertion (4) by induction on r, and note that the case r = 0 is trivial and the case r = 1 was proved as part of assertion (5). Let M 1 = Rv 1 + +Rv r 1. Then it follows easily that S 1 = {v 1,...,v r 1 } is a basis of M 1. By induction, S 1 is a basis of FM 1 and rk R (M 1 ) = r 1. Let M 2 = R v r. By (5), {v r } is a basis of M 2, and rk R (M 2 ) = 1. By Lemma 2.1, it follows that v 1,...,v r is a basis of M = M 1 +M 2. Note that M 1 M 2 = 0 since {v 1,...,v r } is a basis. Hence, FM 1 FM 2 = 0 by assertion (3), and FM = F(M 1 +M 2 ) = FM 1 +FM 2 = F v i by induction. This proves (4). Definition 2.6. Let M be a free finitely generated R-module with R-module isomorphism φ : M R n. Let M 1 M be a R-submodule. Then rk R (M 1 ) = rk R (φ(m 1 )). Proposition 2.7. Let M be a free finitely generated R-module with R-module isomorphism φ : M R n. Then (1) Let M 1,M 2 M be submodules. If M 1 M 2 = 0, then rk R (M 1 ) + rk R (M 2 ) = rk R (M 1 +M 2 ). (2) Let M 1 M be a free submodule with basis v 1,...,v r. Then rk R (M 1 ) = r. (3) If M 1 M is a submodule, then 0 rk R (M 1 ) n. (4) If v M is nonzero, then rk R (R v) = 1. Proof : For(1), notethatφ(m 1 ) φ(m 2 ) = φ(m 1 M 2 ) = 0. Thus, by(3)ofproposition 2.5, rk R (φ(m 1 +M 2 )) = rk R (φ(m 1 )+φ(m 2 )) = rk R (φ(m 1 ))+rk R (φ(m 2 )), and this implies (1). For (2), it follows easily from definitions that φ(v 1 ),...,φ(v r ) is a basis of φ(m 1 ), and now (2) follows from (4) of Proposition 2.5. Assertion (3) is clear by (1) of Proposition 2.5, and Assertion (4) is clear by (5) of Proposition 2.5. Note that a R-submodule M 1 of a free module M may not be generated by one element, but may still have rk R (M 1 ) = 1. For example, let R = F[x,y], the polynomial ring in two variables, and let M = R, which is free with basis {1}. Let M 1 = (x,y). Then M 1 R v for any v M 1 since (x,y) is not a principal ideal. But FM 1 = Fx+Fy = F, so rk R (M 1 ) = The set of linear maps We discuss the R-linear maps from a R-module M to N. As before, R is a commutative ring. Definition 3.1. Let R be a ring and let M,N be R-modules. Then Hom R (M,N) = {φ : M N : φ is a R module homomorphism}.

5 Remark 3.2. Let f Hom R (M,N). Let M 1 M be a R-submodule, and N 1 N be a R-submodule. (i) f(m 1 ) := {f(x) : x M 1 } is a R-submodule of N. (2) f 1 (N 1 ) := {x M : f(x) N 1 } is a R-submodule of M. (3) Let P M be a R-submodule. Then P M 1 is a R-submodule of M. All these assertions are routine to prove, and we leave them to the reader, aside from noting that some work can be saved by citing in Ash. We can deduce (3) from (2) using the inclusion map i : P M, and realizing that P M 1 = i 1 (M 1 ). Forf 1,f 2 Hom R (M,N), letf 1 +f 2 : M N bedefinedby(f 1 +f 2 )(x) = f 1 (x)+f 2 (x) for x M. For r R, let r f 1 : M N be defined by (r f 1 )(x) = r (f 1 (x)). Note that f 1 +f 2,r f 1 Hom R (M,N). Indeed, for the second assertion, let a R and x M, and compute (r f 1 )(a x) = r (f 1 (a x)) = f 1 (ra x) = f 1 (ar x) = ar (f 1 (x)) = a (r f 1 )(x). This requires that R is commutative, but f 1 + f 2 Hom R (M,N) even when R is noncommutative, and we leave the easy verification to the reader. Lemma 3.3. Hom R (M,N) is a R-module. We leave these assertions to the reader. They are quite easy. We consider the special case when N = R, viewed as a R-module using multiplication in R. We let M have basis x 1,...,x n, and recall that if m M, then we can write m = r 1 x 1 + +r n x n. For i = 1,...,n, we define p i : M R by the formula p i (r 1 x 1 + +r n x n ) = r i. Remark 3.4. For i = 1,...,n, then p i Hom R (M,R). Indeed, if m = n i=1 r ix i and n sum n i=1s i x i, then m+n = n i=1 (r i +s i )x i, so p i (m+n) = r i +s i = p i (m)+p i (n). We leave to the reader the verification that if s M, then p i (s m) = s p i (m). Remark 3.5. In the above, let M be the free module R n, with standard basis e 1,...,e n. Then p i (r 1,...,r n ) = r i is projection on the rth factor. Exercise 3.6. If R is a commutative ring and M is a free R-module with basis x 1,...,x n, then p 1,...,p n is a basis for the R-module Hom R (M,R) has basis p 1,...,p n as an R- module. It follows that Hom R (M,R) = R n as a R-module. We establish one further result concerning principal ideal domains. Lemma 3.7. Let R be a principal ideal domain, and let S = {I j } j J be a collection of ideals of R. Then S has a maximal element N; i.e., N S is an ideal, and if I j S, and N I j, then N = I j. Proof : We suppose that there does not exist a maximal element N. Then for every K S, there exists L S such that K L and K L. Pick an ideal I 1 S. By 5

6 6 assumption, there exists an ideal I 2 S such that I 1 I 2, but I 1 I 2. Similarly, for each j 1, there exists an ideal I j+1 S such that I j I j+1, but I j I j+1. Thus, I 1 I 2 I n... is an infinite ascending chain of proper inclusions of ideals. But R is a PID, so it satisfies the ascending chain condition on ideals by the PID theorem proved in class (or Ash, Theorem and Theorem 2.6.8).