Modules Over a Principal Ideal Domain

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 :⇐⇒ ∃ a ∈ R with I = (a). Remark 1.1. This means that I is principal ⇐⇒ I = {ra | r ∈ R}. 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 ∈ 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)i∈I , where ai ∈ Z for all i ∈ I, ((ai)i∈I ) is generated by gcd((ai)i∈I ). This example motivates the following deﬁnitions: Definition 1.4. Let R be a principal ideal domain, A ⊂ R a subset, c ∈ 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 := {x ∈ R | xt ∈ N ∀t ∈ T } The special case

0 :R N := {x ∈ R | xn = 0 ∀n ∈ N} 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 ∈ R, m ∈ 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 = {0}, δB = {0}. Thus, α | δ, β | δ, particularily γ |δ. Further, γ ∈ ann(A), γ ∈ ann(B) =⇒ γ ∈ ann(A ⊕ B). It follows that δ | γ, so there exists u ∈ 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 :⇐⇒ ∃n ∈ N such that N = hni. Remark 3.1. This implies that N is cyclic if and only if it is of the form { rn | r ∈ R} Theorem 3.1. Let R be a PID, hvi a cyclic R-module with annihilator (α) , N ⊂ hvi a submodule, β ∈ 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 ∈ ann(v), so ker(τ) = (α). Now, consider τ and let r, r be such that r + (α) = r + (α). Then r − r ∈ (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 ∈ (α) ⇐⇒ r + (α) = 0 + (α), so injectivity follows. (2) Let I = {r ∈ R|rv ∈ N}. It can easily be seen that I is an ideal. Since R is a PID, there exists s ∈ R such that I = (s). It follows that N = Iv = Rsv = < sv >. α (3) For r ∈ R we have r(βv) = 0 ⇐⇒ (rβ)v = 0 ⇐⇒ α | rβ ⇐⇒ gcd(α,β) | r. It follows α that r ∈ ann(βv) ⇐⇒ r ∈ ( 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 ∈ 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 ∈ M with o(ui) = αi ∀i 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 < ∞, S = {s1, ..., sn} ⊂ M a spanning set for M. Then S is a basis for M.

Proof. Let B = {b1, ..., bn} 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) = {0}. 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 < ∞, N ⊂ M a submodule of rank(N) = k ≤ n. Then there exist r1, ..., rk ∈ R\{0} and a basis B containing b1, ..., bk ∈ M such that S = {r1b1, ..., rkbk} 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 := {m ∈ M | ∃r ∈ R : rm = 0}. 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 ∈ R be a prime. A p-primary module is an R-module n M with o(M) = p for some n ∈ 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 = {v ∈ M | pi v = 0} is a pi-primary submodule of M ∀i ∈ pi {1, ..., n}. (2) The primary decomposition is unique up to the order of summands, that is if

M = M q1 ⊕ ... ⊕ M qm (1) is another primary decomposition of M, then n = m and ∀i ∈ {1, ..., n} ∃k ∈

{1, ..., m} such that Mpi = M qk . (3) If N is another R-module with primary decomposition

N = Nq1 ⊕ ... ⊕ Nqm (2) ∼ ∼ then M = N ⇐⇒ n = m and ∀i ∈ {1, ..., n} ∃k ∈ {1, ..., m} such that Mpi = Nqk .

7. Decomposition of Primary Modules As a consequence of the previous result, we start to study primary modules. Lemma 7.1. Let R be a PID, M an R-module, p ∈ R be a prime, S ⊂ M an R-submodule of M. Then (1) pM = {0} =⇒ M is a vectorspace over the field R/(p) with scalar-multiplication R/(p) × M → M (r + (p), v) 7→ rv and this scalar-multiplication is well-defined. (2) S(p) = {v ∈ S | pv = 0} is an R-submodule of M and if M = S ⊕ T , then M (p) = S(p) ⊕ T (p). With this, we can characterize the cyclic decomposition of primary modules. Theorem 7.1. Let R be a PID, M a primary finitely generated torsion R-module with order pe. Then (1) there exist v1, ..., vn ∈ M, e1, ..., en ∈ N such that M = hv1i ⊕ ... ⊕ hvni with e ann(hvii) = (p i ) ∀i and ann(hv1i) ⊂ ... ⊂ ann(hvni) (that is en ≤ ... ≤ e1 = e).

5 Seminar: Advanced Topics in Linear Algebra Niklas Gassner FS18 [email protected]

f (2) if M = hu1i ⊕ ... ⊕ humi is another cyclic decomposition with ann(huii) = (q i ) for some prime q ∈ R and ann(hu1i) ⊂ ... ⊂ ann(humi), then n = m, p ∼ q, ann(huii) = ann(hvii) and ei = fi ∀i ∈ {1, ..., n}. (3) if N is a ﬁnitely generated torsion R-module with cyclic decomposition

N = hu1i ⊕ ... ⊕ huni

such that ann(hu1i) ⊂ ... ⊂ ann(humi), then N is isomorphic to M ⇐⇒ m = n and ∀i ∈ {1, ..., n} ann(huii) = ann(hvii). We found a very simple example for such a decomposition: Example 7.1. Let M = Z/4Z × Z/2Z × Z/3Z as Z-module in a natural way. Then ord(M) = 12, which can be decomposed as 22 · 3. We start oﬀ with the primary decomposition: Let M2 := {(z1, z2, 0) | z1, z2 ∈ Z}, M3 := {(0, 0, r) | r ∈ Z}. It can easily be seen that M2, M3 are primary Z-submodules of M with annihilators 4Z and 3Z. Further, M = M2 ⊕ M3. Now, for the cyclic decomposition of the primary modules, note that M2 = h(1, 0, 0)i ⊕ h(0, 1, 0)i, where ann(h(1, 0, 0)i) = 4Z ⊂ 2Z = ann(h(0, 1, 0)i), and M3 = h(0, 0, 1)i. From Theorem 6.3, we know we can decompose an R-torsion module into p-primary modules, which, by Theorem 7.1, can again be decomposed into cyclic modules. Definition 7.1. Let R be a PID, M a ﬁnitely generated torsion R-module with order e1 en µ = p1 · ... · pn and primary decomposition

M = Mp1 ⊕ ... ⊕ Mpn .

Let, for all i ∈ {1, ..., n}, Mpi have the cyclic decomposition

Mpi = hvi,1i ⊕ ... ⊕ hvi,ki i. ei,j with annihilators ann(hvi,ji) = (pi ). ei,j Then the elements pi are called the elementary divisors of M. 8. Summary For a torsion R-module M over a PID R we have: • M can be decomposed into a direct sum of primary submodules, and this decomposition is unique up to the order of the summands. • Each primary submodule can be decomposed into a direct sum of cyclic submodules, such that the annihilators the cyclic submodules form an ascending chain. This decomposition and the chain of annihilators is unique. • Two R-modules M and N are isomorphic if and only if they have the same annihilator chains.

References [1] Roman, S., Advanced linear algebra, 3rd edition, Springer 2008. [2] Markus P. Brodmann, Kommutative Algebra, ausgearbeitet von Roberto Boldini und Fred Rohrer, accessible at: http://www.math.uzh.ch/index.php?file&key1=36386