MODULES OVER A PRINCIPAL IDEAL DOMAIN NIKLAS GASSNER 1. Elementary Notions To begin with the topic, we determine some basic notation and elementary definitions: Notation 1.1. For the following chapters, a ring is always commutative with unity. If R is a ring, A ⊂ R a subset of R, we denote (A) the ideal generated by A. Further, if M is an R-module, B ⊂ M a set, we denote by hBi the submodule generated by B. Definition 1.1. Let R be a ring, I ⊂ R an ideal. I is a principal ideal :() 9 a 2 R with I = (a). Remark 1.1. This means that I is principal () I = fra j r 2 Rg. Definition 1.2. A ring R is a principal ring :() every ideal I ⊂ R is principal. Definition 1.3. A ring R is a principal ideal domain (PID) :() R is a principal ring and an integral domain. As the title suggests, our main interest will be the study of PID's. We quickly recall the following result from Algebra I: Theorem 1.1. Let R be a principal ideal domain, r 2 R. Then • r is irreducible () r is a prime () (r) is a prime ideal () (r) is a maximal ideal. • R is a unique factorization domain. • R is a Noetherian ring. Example 1.1. (Z; +; ·) is a principal ideal domain Proof. With the help of the extended Euclidean algorithm, it can be shown that for any (ai)i2I , where ai 2 Z for all i 2 I, ((ai)i2I ) is generated by gcd((ai)i2I ). This example motivates the following definitions: Definition 1.4. Let R be a principal ideal domain, A ⊂ R a subset, c 2 R. Then gcd(A) ∼ c :() (A) = (c). Remark 1.2. Since generators in principal ideals are unique up to units, this is well- defined. 1 Seminar: Advanced Topics in Linear Algebra Niklas Gassner FS18 [email protected] Definition 1.5. Let R be a ring, M an R-module, N ⊂ M a submodule, T ⊂ M a subset of M. Then the quotient of N by T is defined as N :R T := fx 2 R j xt 2 N 8t 2 T g The special case 0 :R N := fx 2 R j xn = 0 8n 2 Ng and is called the annihilator of N, denoted ann(N). Proposition 1.1. Let R be a ring, M an R-module, N ⊂ M an R-submodule and P ⊂ M a subset. Then N :R P is an ideal of R. Remark 1.3. This especially implies that ann(M) is an ideal of R. 2. First Results Having determined some basic notions, we can begin the study of modules over principal ideal domains. Definition 2.1. Let R be a PID, M an R-module, N ⊂ M an R-submodule, r 2 R, m 2 M. • r is an order of N :() ann(N) = (r). • r is an order of m :() r is an order of hmi. Notation 2.1. If N is an R-submodule of M, we denote its order o(N). Proposition 2.1. Let R be a PID, M an R-module. If M = A ⊕ B, then o(M) = lcm(o(A); o(B)). Proof. Let o(M) = δ, o(A) = α, o(B) = β, lcm(α; β) = γ. Then δA = f0g, δB = f0g. Thus, α j δ, β j δ, particularily γ jδ. Further, γ 2 ann(A), γ 2 ann(B) =) γ 2 ann(A ⊕ B). It follows that δ j γ, so there exists u 2 R∗ with δ = uγ. 3. Cyclic Modules Our first step of studying modules over PID is to study the simplest type of modules. Definition 3.1. Let R be a ring, M an R-module, N ⊂ M an R-submodule. Then N is a cyclic submodule :() 9n 2 N such that N = hni. Remark 3.1. This implies that N is cyclic if and only if it is of the form f rn j r 2 Rg Theorem 3.1. Let R be a PID, hvi a cyclic R-module with annihilator (α) , N ⊂ hvi a submodule, β 2 R. Then 2 Seminar: Advanced Topics in Linear Algebra Niklas Gassner FS18 [email protected] (1) The function τ : R ! hvi r 7! rv is an R-epimorphism with kernel (α). Further, the map τ : R=(α) ! hvi r + (α) 7! rv is a well-defined R-module isomorphism. (2) N is a cyclic submodule. α (3) o(βv) = gcd(α,β) . Further, hβvi = hvi () gcd(o(v); β) ∼ 1 () o(βv) = o(v). Proof. (1) Remark 3.1 immediately implies that τ is surjective. Further, rv = 0 () r 2 ann(v), so ker(τ) = (α). Now, consider τ and let r, r be such that r + (α) = r + (α). Then r − r 2 (α) = ann(v). Thus τ(r) = rv = (r + (r − r))v = rv + (r − r)v = rv = τ(r), so τ is well-defined. Surjectivity follows from τ being surjective. Lastly, rv = 0 () r 2 (α) () r + (α) = 0 + (α), so injectivity follows. (2) Let I = fr 2 Rjrv 2 Ng. It can easily be seen that I is an ideal. Since R is a PID, there exists s 2 R such that I = (s). It follows that N = Iv = Rsv = < sv >. α (3) For r 2 R we have r(βv) = 0 () (rβ)v = 0 () α j rβ () gcd(α,β) j r. It follows α that r 2 ann(βv) () r 2 ( gcd(α,β) ). The second statement follows immediately. 4. Decomposition of Cyclic Modules The following result classifies the composition and decomposition of cyclic modules: Theorem 4.1. Let R be a PID, M an R-module, u1; :::; un 2 M. Then (1) If u1; :::; un have relatively prime orders, then o(u1 + ::: + un) = o(u1) · ::: · o(un) and hu1i ⊕ ::: ⊕ huni = hu1 + ::: + uni: (2) If o(v) = α1 · ::: · αn where the αi's are pairwise relatively prime, then there exist u1; :::; un 2 M with o(ui) = αi 8i such that v = u1 + ::: + un and thus hvi = hu1 + ::: + uni = hu1i ⊕ ::: ⊕ huni: 3 Seminar: Advanced Topics in Linear Algebra Niklas Gassner FS18 [email protected] 5. Free Modules over a Principal Ideal Domain Submodules of free modules are not always free. For example Z=4Z is free as Z=4Z- module, but 2Z=4Z is a submodule that is not free. However, over PID's this cannot happen. Theorem 5.1. Let R be a PID, M a free module over R, S ⊂ M an R-submodule of M. Then S is free with rank(S) ≤ rank(M). Theorem 5.2. Let R be a PID, M a free R-module of rank(M) = n < 1, S = fs1; :::; sng ⊂ M a spanning set for M. Then S is a basis for M. Proof. Let B = fb1; :::; bng be an R-basis of M and define τ : M ! M bi 7! si M is free, so M =∼ ker(τ) ⊕ im(τ) = ker(τ) ⊕ M. Further, ker(τ) is an R-submodule of τ, so by Theorem 5.1, ker(τ) is free of rank(ker(τ)) ≤ n. So rank(ker(τ)) + rank(M) = rank(M), which implies rank(ker(τ)) = 0, so ker(τ) = f0g. In Linear Algebra, we learn that for any K-vectorspace V and K-subspace S, a basis of S can be extended to a basis of V . There is a similar result for modules over PID's. Theorem 5.3. Let R be a PID, M a free R-module of rank n < 1, N ⊂ M a submodule of rank(N) = k ≤ n. Then there exist r1; :::; rk 2 Rnf0g and a basis B containing b1; :::; bk 2 M such that S = fr1b1; :::; rkbkg is an R-basis for N. 6. Torsion-free and Free Modules Definition 6.1. Let R be a ring, M an R-module. Then the torsion module of M is defined as Mtor := fm 2 M j 9r 2 R : rm = 0g: Remark 6.1. The torsion module of M is always a submodule of M. Theorem 6.1. Let R be a PID, M a finitely generated R-module. Then M is free () M is torsion-free. We can now classify the cyclic decomposition of modules over a PID. Theorem 6.2. Let R be a PID, M a finitely generated R-module. Then M = Mfree ⊕ Mtor where Mfree is a free module, unique upto isomorphism, and Mtor is unique. Remark 6.2. Since M is finitely generated, its summands Mtor and Mfree are as well. 4 Seminar: Advanced Topics in Linear Algebra Niklas Gassner FS18 [email protected] Proof. We will outline the proof: Mtor is a submodule of M, and M=Mtor is torsion-free as R-module. Since M is finitely generated, M=Mtor is finitely generated aswell. Thus, by Theorem 6.1, M=Mtor is free. We conclude M = Mtor ⊕ M=Mtor where M=Mtor is free. Definition 6.2. Let R be a PID, p 2 R be a prime. A p-primary module is an R-module n M with o(M) = p for some n 2 N. e1 en Theorem 6.3. Let R be a PID, M a torsion R-module with order µ = p1 · ::: · pn where the pi's are pairwise distinct nonassociate primes in R. Then (1) We have the following primary decomposition: M = Mp1 ⊕ ::: ⊕ Mpn µ ei where Mpi = ei M = fv 2 M j pi v = 0g is a pi-primary submodule of M 8i 2 pi f1; :::; ng.