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MODULES OVER a PID Every Vector Space Over a Field K That Has a Finite

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MODULES OVER a PID Every Vector Space Over a Field K That Has a Finite 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 Kn 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 An has a basis of size at most n. (2) Every finitely generated torsion-free A-module M has a finite basis: M =∼ An for a unique n ≥ 0. (3) Every finitely generated A-module M is isomorphic to Ad ⊕ 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 two lemmas that have nothing to do with PIDs. Lemma 1.1. If A is a nonzero commutative ring and Am =∼ An as A-modules then m = n. Proof. The simplest proof uses a maximal ideal m in A. Setting M = Am and N = An, 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 a 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 A2 =∼ A as left A-modules. Lemma 1.2. 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 6= 0 there is an embedding M,! Ad for some d ≥ 1 such that the image of M intersects each standard coordinate axis of Ad. Proof. Let K be the fraction field of A and x1; : : : ; xn be a generating set for M as an A-module. We will show n is an upper bound on the size of each A-linearly independent n subset of M. Let f : A ! M be the linear map where f(ei) = xi for all i. (By e1; : : : ; en 1The word \rank" means something completely different in linear algebra { the dimension of the image of a linear map. 1 2 KEITH CONRAD n we mean the standard basis of A .) Let y1; : : : ; yk be linearly independent in M, so their k Pn A-span is isomorphic to A . Write yj = i=1 aijxi with aij 2 A. We pull the yj's back to n A by setting vj = (a1j; : : : ; anj), so f(vj) = yj. A linear dependence relation on the vj's is transformed by f into a linear dependence relation on the yj's, which is a trivial relation n by their linear independence. Therefore v1; : : : ; vk is A-linearly independent in A , hence K-linearly independent in Kn. By linear algebra over fields, k ≤ n. From the bound k ≤ n, there is a linearly independent subset of M with maximal size, Pd ∼ d say t1; : : : ; td. Then j=1 Atj = A . We will find a scalar multiple of M inside of this. For each x 2 M, the set fx; t1; : : : ; tdg is linearly dependent by maximality of d, so there is a Pd Pd nontrivial linear relation ax + i=1 aiti = 0, necessarily with a 6= 0. Thus ax 2 j=1 Atj. Letting x run through the spanning set x1; : : : ; xn there is an a 2 A − f0g such that Pd Pd axi 2 j=1 Atj for all i, so aM ⊂ j=1 Atj. Multiplying by a is an isomorphism of M with aM, so we have the sequence of A-linear maps d X d M ! aM ,! Atj ! A ; j=1 where the last map is an isomorphism. 2. Submodules of a finite free module First we will show submodules of a finite free module over a PID are finitely generated, with a natural upper bound on the number of generators. Theorem 2.1. When A is a PID, each submodule of a free A-module of rank n is finitely generated with at most n generators. Proof. A free A-module of rank n is isomorphic to An, so we may assume the free A-module is literally An. We will argue by induction on n. The case where n = 0 is trivial and 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. n n+1 Suppose n > 1 and the theorem is proved for all submodules of A . Let M ⊂ A be a submodule. We want to show M has at most n + 1 generators. View M ⊂ An+1 = An ⊕ A n n and let π : A ⊕ A A be projection to the first component of this direct sum. Let's look at the image and kernel of πjM , the restriction of π to M. Its image is π(M), which is a submodule of An and therefore has at most n generators by the inductive hypothesis, Pk n so π(M) = i=1 Ayi for some y1; : : : ; yk 2 A where k ≤ n. We can write yi = π(xi) for Pk ∼ some xi 2 M, so π(M) = i=1 Aπ(xi). And ker(πjM ) = M \ (0 ⊕ A), with 0 ⊕ A = A as A-modules. Every A-submodule of A has a single generator since A is a PID, so ker(πjM ) = Ax0 for some x0 2 M. Pk We will show M = i=0 Axi, so M has at most k + 1 generators and k + 1 6 n + 1. The Pk containment i=0 Axi ⊂ M is clear. For the reverse containment, pick an arbitrary x 2 M and the previous paragraph tells us π(x) = a1π(x1) + ··· + akπ(xk) = π(a1x1 + ··· + akxk) Pk Pk for some a1; : : : ; ak in A. Therefore x − i=1 aixi 2 ker(πjM ), so x − i=1 aixi = a0x0 for Pk Pk some a0 2 A. Thus x = a0x0 + a1x1 + ··· + akxk 2 i=0 Axi, so M ⊂ i=0 Axi. Next we will refine the previous theorem by showing a submodule of a finite free A- module is free (has a basis). The proof will be quite similar to the one we just gave, but it will not logically depend on it. MODULES OVER A PID 3 Theorem 2.2. When A is a PID, each submodule of a free A-module of rank n is free of rank 6 n. Proof. As before, since a free A-module of rank n is isomorphic to An, we can assume the free A-module we use is An. We'll induct on n. The case n = 0 is trivial and the case n = 1 follows from all submodules of A being (0) or principal with a nonzero generator, and Aa =∼ A as A-modules when a is nonzero in A since A is an integral domain. n Suppose n > 1 and the theorem is proved for all submodules of A . For a submodule M of n+1 n+1 n n n A , to show M is free of rank at most n+1 write A = A ⊕A and let π : A ⊕A A be projection to the first component. Since π(M) is a submodule of An, π(M) is free of rank 6 n by the inductive hypothesis. Case 1: π(M) = 0. Here M ⊂ 0 ⊕ A, so M is free of rank at most 1 since A is a PID. Case 2: π(M) 6= 0. Here π(M) is a nonzero submodule of An, so π(M) is free of positive Ld rank d ≤ n. Write a basis of π(M) as π(e1); : : : ; π(ed) where ei 2 M: π(M) = i=1 Aπ(ei). The elements e1; : : : ; ed in M are linearly independent since their images under π are: if P P aiei = 0 in M then apply π to get aiπ(ei) = 0 in π(M), so all ai are 0. Pd Pd For m 2 M, π(m) = i=1 aiπ(ei) for unique a1; : : : ; ad 2 A. Then π(m− i=1 aiei) = 0, Pd so m − i=1 aiei 2 ker(πjM ). We get inverse maps ( Pd d m 7−! (a1; : : : ; ad; m − i=1 aiei) M ! A ⊕ ker(πjM ) by Pd i=1 aiei + k − (a1; : : : ; ad; k) [ ∼ d and you should check both directions are linear. Therefore M = A ⊕ ker(πjM ) as A- modules, so M is a free A-module of rank d or d+1 (depending on whether or not ker(πjM ) is f0g), and d + 1 ≤ n + 1.
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