MMATH18-201: Module Theory Lecture Notes: Modules Over PID

Home , Cyclic module, Free module, Torsion (algebra)

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 = {0} and M itself is the only submodule and we are done.

Assume now that M is a free module with basis {e1, e2, . . . , en}, 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 = {0}, then nothing to prove. So let N 6= {0}, then every element x of N can be uniquely written as

n X x = aiei, ai ∈ 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 . {0}, 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) {v2, v3, . . . , vm}, 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 ∈ R such that Im φ = hai.

If a = 0, then N = ker φ and we are done. Otherwise there exist some v1 ∈ 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 = {0}.

Clearly hv1i + ker φ ⊆ N, so let x be an arbitrary element of N, then Im φ = hai implies that there exist some r ∈ R such that (x)φ = ra. Now

(x − rv1)φ = (x)φ − (rv1)φ = ra − ra = 0, implies that x − rv1 ∈ ker φ so x ∈ hv1i + ker φ, i.e., N ⊆ hv1i + ker φ. If y ∈ hv1i ∩ ker φ, then y = rv1, for some r ∈ 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 = {0}. Hence the claim follows. As {v2, v3, . . . , vm} 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 {v1, v2, . . . , vm}, in particular

0 has a unique representation which implies that the set {v1, v2, . . . , vm} 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 {e1, e2, . . . , en}, 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 {e1, e2, . . . , en} 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 ψ 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

ψ 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

ψ 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) {x1, x2, . . . , xk}, where k ≤ n. Then verify that the set {(x1)φ, (x2)φ, . . . , (xk)φ} 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 {e1, e2, . . . , en} be the standard basis of R and {x1, x2, . . . , xn} 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 = {0}. Hence φ is an isomorphism and maps the linearly independent set {e1, e2, . . . , en} onto a linearly independent set. Therefore the spanning set

S = {x1, x2, . . . , xn} 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 ∈ 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 = {0} 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 ∈ R, so 0 ∈ tor M. Now for any x, y ∈ tor M there exist some non-zero elements r1, r2 ∈ 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 ∈ R

r1r2(rx + y) = (r1r2)rx + (r1r2)y = (r2r)r1x + r1(r2y) = (r2r)0 + r10 = 0,

4 i.e., rx + y ∈ 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 ∈ tor (M/ tor M) . Then there exist some non-zero element s ∈ R such that sx = 0, which implies that sx = 0, i.e., sx ∈ tor M. Consequently there exist some non-zero r ∈ R such that r(sx) = 0 =⇒ (rs)x = 0. Since R is an integral domain so rs 6= 0 and hence the above equation implies that x ∈ tor M, i.e., x = 0.

Deﬁnition. Let M be an R-module and x ∈ M. Then the annihilator of x, denoted by Ann x, is the set of all those elements of the ring R which kills x, i.e.,

Ann x = {r ∈ R | rx = 0}.

Deﬁnition. Let N be a submodule of an R-module M. Then the annihilator of N is the set Ann N = {r ∈ R | rx = 0 ∀ x ∈ N}.

Exercise 2.2. Show that Ann x and Ann N, as deﬁned above, are ideals of the ring R.

Exercise 2.3. Let M be an R-module. Show that an element x of M is torsion if and only if Ann x 6= {0}.

Exercise 2.4. Show that a cyclic R-module hxi is free if and only if x is a non-torsion element.

Theorem 2.5. Every free module over an integral domain is torsion free.

Proof. Let M be a free R-module, where R is an integral domain, and let X be a basis of M. To prove that M is torsion free, we need to show that tor M = {0}. Let y ∈ tor M, then there exist some non-zero r ∈ R such that ry = 0. Since

X is a basis of M, so there exist ﬁnitely many elements (say) x1, x2, . . . , xn ∈ X and r1, r2, . . . , rn ∈ R such that

y = r1x1 + r2x2 + ··· + rnxn.

Now ry = 0 implies that

(rr1)x1 + (rr2)x2 + ··· + (rrn)xn = 0,

5 which, in view of the linear independence of {x1, x2, . . . , xn} over R, implies that rr1 = rr2 = ··· = rrn = 0. Since R is an integral domain and r 6= 0, therefore from rri = 0, (1 ≤ i ≤ n), we have that each ri = 0, and hence y = 0.

Remark 2.6. The converse of the above result is not true in general. For example the Z-module Q, is torsion free but not a free Z-module.

Theorem 2.7. A ﬁnitely generated torsion free module over a PID is free.

Proof. Let M be a ﬁnitely generated torsion free module over a PID R. If M = {0}, then M is free with empty basis and we are done. So assume that

M 6= {0} and let S = {x1, x2, . . . , xn} be a generating set for M. Then n ≥ 1 and each xi 6= 0, (1 ≤ i ≤ n), which in view of the fact that M is torsion free implies that each singleton subset of S is linearly independent over R. Let X be the maximal linearly independent subset of S. On re-indexing (if necessary), we may assume that X = {x1, x2, . . . , xm} for some m ≤ n. Now if m = n, then S = X forms a basis of M are we are done. So assume that m < n, and let F be the submodule of M generated by X. Then F is free with basis X. By maximality of X, we have that for each i > m, the set X ∪ {xi} must be linearly dependent over R, i.e., there exist some ai(6= 0) ∈ R and ai1 , ai2 , . . . , aim not all zero in R such that m X aixi + aij xj = 0. (5) j=1

As ai 6= 0 for each i > m and R is an integral domain, so b = am+1am+2 ··· an 6= 0, which in view (5) and the fact that F is an R-module generated by x1, x2, . . . , xm, implies that

bxi ∈ F for each i ∈ {1, 2, . . . , n}. (6)

Since {x1, x2, . . . , xn} spans M, therefore from (6), we have that by ∈ M for every y ∈ M and consequently gives an R-module homomorphism f : M −→ F deﬁned by (y)f = by. Since M is a torsion free module, so for any y ∈ M, by = 0 implies that y = 0 (∵ b 6= 0), i.e., f is injective. Thus M is isomorphic to a submodule (Im f) of F , but F is a free module over a PID, so by Theorem 1.1, we have that Im f and hence M is a free R-module.

Remark 2.8. From the above two results we can now deduce that “Over a PID a ﬁnitely generated module is free if and only if it is torsion free”.

6 Corollary 2.9. Let M be a ﬁnitely generated module over a PID R. Then M is isomorphic to a direct sum of a torsion module and a free module.

Proof. Since R, being a PID, is an integral domain, so by Theorem 2.1, the set tor M is a submodule of M and the quotient M/ tor M is torsion free. Therefore we have a short exact sequence

0 −→ tor M −→α M −→π M/ tor M −→ 0, where α is the inclusion map and π the canonical projection. As quotient of a finitely generated module is finitely generated, so the torsion free module M/ tor M is finitely generated and therefore by Theorem 2.7 must be free. Now in view of the fact that every short exact sequence with thrid term free splits, we have that M ∼= tor M ⊕ M/ tor M and the result follows.

Deﬁnition. Let M be a ﬁnitely generated module over a PID. Then by rank of M we mean the rank of the free module M/ tor M, where tor M is the torsion submodule of M. The quotient M/ tor M is usually called the torsion free part of M, so the rank of M is same as the rank of the torsion free part of M.

Remark 2.10. If M is a free module over a PID R, then by Theorem 2.5, M is torsion free, i.e., tor M = {0} which implies that M/ tor M = M. Therefore for a ﬁnitely generated free module over a PID both the deﬁnition of ranks coincide.

Proposition 2.11. Let M be a module of a PID R, with presentation

f 0 −→ Rm −→ Rn −→ M −→ 0. (7)

Then the integer n − m equals the rank of M and is independent of the choice of the presentation.

Proof. Since a quotient of a finitely generated module is finitely generated, therefore by First Isomorphic Theorem, we have that M ∼= Rn/ ker f, is finitely generated. As R is a PID so tor M is submodule of M and M/ tor M is torsion free. Now if π is the canonical projection then we have a composition

f g = fπ : Rn −→ M −→π M/ tor M, and hence a short exact sequence

g 0 −→ ker g −→α Rn −→ M/ tor M −→ 0,

7 where α is the inclusion map. As M/ tor M is free (see Theorem 2.5), therefore the above sequence splits, i.e.,

Rn ∼= ker g ⊕ M/ tor M, which implies that

n = rank Rn = rank ker g + rank M/ tor M. (8)

By deﬁnition the rank of M is the rank of the free module M/ tor M, which is unique (∵ R is a commutative ring), so to prove the result, in view of (8), it is enough to prove that rank of the free submodule ker g of Rn is m. As g = fπ, so ker f ⊆ ker g which implies that

m = rank ker f ≤ rank ker g. (9)

Now for any x ∈ ker g

0 = (x)g = (x)fπ = ((x)f) π =⇒ (x)f ∈ tor M, so there exist some non-zero a ∈ R such that a ((x)f) = 0, which by linearity of f, implies that (ax)f = 0, i.e., ax ∈ ker f. So in particular if {x1, x2, . . . , xk} is a basis of ker g, then there exist non-zero elements (say) a1, a2, . . . , ak ∈ R such that aixi ∈ ker f for each i ∈ {1, 2, . . . , k}. Then the mapping φ : ker g −→ ker f given by k ! k X X (y)φ = rixi φ = ai(rixi) i=1 i=1 is an R-module homomorphism (verify). Moreover φ is injective, because if m X y = rixi ∈ ker φ, then (y)φ = 0 implies that i=1

k k X X ai(rixi) = (airi)xi = 0 i=1 i=1 which in view of the linear independence of {x1, x2, . . . , xk}, implies that

a1r1 = a2r2 = ··· = akxk = 0, now as R is an integral domain and ai 6= 0 for each 1 ≤ i ≤ k, the above equations implies that

r1 = r2 = ··· = rk = 0, i.e., y = 0.

8 Therefore, we have that ker g is isomorphic to a submodule of ker f and hence

rank ker g ≤ rank ker f = m. (10)

Thus from (9) and (10), we have that rank ker g = m and the result follows.

3 Primary Modules

We ﬁrst recall some deﬁnitions for commutative rings with unity.

Deﬁnition. Let R be a commutative ring and a, b ∈ R with a 6= 0. (1). a is said to divide b, denoted a | b, if there exist some c ∈ R such that b = ax. (2). The greatest common divisor (gcd) of a and b is a non-zero element d of R such that (i). d | a and d | b and (ii). if d0 | a and d0 | b, then d0 | d.

Deﬁnition. A non-zero element u of a commutative ring R is said to be a unit if there exist some v ∈ R such that uv = 1 or equivalently u is a unit if and only if u | 1. Two non-zero elements a, b of R are said to be associates if a = ub for some unit u in R or equivalently a | b and b | a.

Deﬁnition. A non-zero, non-unit element a of an integral domain R is called irreducible if whenever a = rs, for some r, s ∈ R, then either r or s is a unit, otherwise a is called reducible.

Deﬁnition. A non-zero non-unit element p of an integral domain R is said to be prime if whenever p | rs, for some r, s ∈ R, then either p | r or p | s

Exercise 3.1. Show that in an integral domain every prime element is irreducible. Does the converse holds? Justify.

Exercise 3.2. Show that in an PID a non-zero element is prime if and only if it is irreducible.

9 Deﬁnition. An integral domain R is called a unique factorization domain (UFD)

(i). if every non-zero, non-unit element a of R can be written as a product of

ﬁnite number of irreducibles, i.e., there exist irreducibles p1, p2, . . . , pn (not

necessarily distinct) in R such that a = p1p2 ··· pn and

(ii). if a = q1q2 ··· qm is another factorization of a into irreducibles, then m = n

and there exist a permutation σ of the set {1, 2, . . . , n} such pi and qσ(i) are associates.

Exercise 3.3. Show that every PID is a UFD.

Deﬁnition. Let M be a module over a PID R and p a prime element of R. The set n Mp = {x ∈ M | p x = 0, for some n > 0} , of all those elements of M which are annihilated by some positive power of p, forms a submodule of M called the p-primary submodule or p-primary component of M. If M = Mp for some prime element p of R, then we say that M is p-primary module or just a primary module.

Remark 3.4. A p-primary module is not q-primary for q 6= p, primes.

Theorem 3.5. Let R be a PID and M a torsion module over R. Then M can be expressed as a direct sum of p-primary modules for diﬀerent primes p of R in a unique way

M = ⊕Mp. (11) If M is ﬁnitely generated, then only ﬁnitely many terms occur in the direct sum.

Proof. For any prime p of R, the p-primary component

n Mp = {x ∈ M | p x = 0, for some n > 0}

P is the largest p-primary submodule of M, so Mp ⊆ M. For the other inclusion let x ∈ M, then as M is a torsion module so there exist some non-zero element a ∈ R such that ax = 0. Since R is PID, so by Exercise 3.3 there exist distinct irreducibles p1, p2, . . . , pk ∈ R and positive integers n1, n2, . . . , nk such that

n1 n2 nk a = p1 p2 ··· pk . (12)

10 Let

ni n1 ni−1 ni+1 nk qi = a/pi = p1 ··· pi−1 pi+1 ··· pk , (13) and hqii the ideal generated by qi for 1 ≤ i ≤ k. Then as R is a PID, therefore the sum of the ideals generated by qi must be principal (say)

hq1i + hq2i + ··· + hqki = hui, for some u ∈ R. (14)

Claim. The element u is a unit in R. Proof of claim. Suppose if possible that u is not a unit. Then as R is a UFD (see Exercise 3.3), so there exists some irreducible element (say) p in R such that p | u. From (14), we have that hqii ⊆ hui, i.e., qi ∈ hui, which implies that u | qi, for 1 ≤ i ≤ k. Now p | u together with u | qi, implies that p | qi, i.e.,

n1 ni−1 ni+1 nk p | p1 ··· pi−1 pi+1 ··· pk for each i ∈ {1, 2, . . . , k}. (15)

As irreducibles and primes are same in a PID (see Exercise 3.2), therefore from

(15) and the fact that the distinct irreducibles p1, p2, . . . , pk are pairwise coprime, n1 n2 nk n1 the prime p must divide exactly one of p1 , p2 , . . . , pk (say) p | p1 , which implies that p | p1. So p, p1 being irreducibles must be associates, which implies that there exist some unit v ∈ R such that

−1 p1 = vp =⇒ v p1 = p, i.e., p1 | p.

n2 n3 nk Therefore for i = 1, (15) now implies that p1 | p2 p2 ··· pk , i.e., p1 | pi for some i ∈ {2, 3, . . . , k}, which contradicts the fact that p1, p2, . . . , pk are pairwise coprime. Hence the claim follows.

From claim it follows that hui = R, which implies that 1 ∈ hq1i + hq2i + ··· + hqki, so there exist a1, a2, . . . , ak ∈ R such that

a1q1 + a2q2 + ··· + akqk = 1. (16)

Now let xi = aiqix, 1 ≤ i ≤ k, then using (16), we get that

x = x1 + x2 + ··· + xk and for each i ∈ {1, 2, . . . , k},(13) together with (12) implies that

ni ni ni pi xi = pi (aiqix) = (aipi qi) x = (aia)x = ai(ax) = 0,

11 ni i.e., pi annihilates xi, so xi ∈ Mpi and therefore

x = x1 + x2 + ··· + xk ∈ Mp1 + Mp2 + ··· + Mpk . P As x ∈ M was arbitrary, we have that M ⊆ Mp. Hence X M = Mp. T X It only remains to show that the above sum is direct. Let x ∈ Mp0 Mp.

p6=p0

Now x ∈ Mp0 implies that there exist some positive integer m such that

m p0 x = 0. (17) X For x ∈ Mp, there exist ﬁnitely many distinct primes (say) p1, p2, . . . , pn

p6=p0 diﬀerent from p0 such that x ∈ Mp1 + Mp2 + ··· + Mpn , i.e.,

x = x1 + x2 + ··· + xn, (18)

for some xi ∈ Mpi , and

mi pi xi = 0, mi > 0 for 1 ≤ i ≤ n. (19)

Since the primes p1, p2, . . . , pn are all distinct, so from (18) and (19) it follows

m1 m2 mn that a = p1 p2 ··· pn annihilates x, which in view of (17) implies that x is m0 m0 annihilated by the gcd of a and p0 . But gcd(a, p0 ) = 1, therefore we have that

1 · x = 0, which implies that x = 0. Hence the M = ⊕Mp.

Finally, if M is ﬁnitely generated (say) by x1, x2, . . . , xn, then as M is a torsion module there exist non-zero elements a1, a2, . . . , an ∈ R such that

a1x1 = a2x2 = ··· = anxn = 0.

Then a = a1a2 ··· an annihilates every generator xi of M and hence annihilates every element of M. Now M = ⊕Mp implies that a annihilate every p-primary submodule Mp occurring in the direct sum. But a can annihilate a p-primary module if and only if p | a and as R being a PID is a UFD, so a can have only ﬁnitely many irreducible and hence prime factors. Therefore only ﬁnitely many terms can occur in the direct sum.

m1 m2 mn Corollary 3.6. Let R be a PID and a(6= 0) ∈ R. If a = p1 p2 ··· pn is the factorization of a in terms of distinct prime elements of R and m1, m2, . . . , mn ∼ m1 m2 mn are positive integers, then R/hai = R/hp1 i ⊕ R/hp2 i ⊕ · · · ⊕ R/hpn i.

12 Proof. An arbitrary element of the R-module R/hai is of the form r + hai which on using scalar multiplication can be written as r (1 + hai), so if we denote by x = 1 + hai, then R/hai = hxi, i.e., R/hai is a cyclic module generated by x. Now as a ∈ hai, so ax = a(1 + hai) = a + hai = 0 + hai = 0 which implies that Annhxi = hai. Therefore hxi is a torsion module, and hence by Theorem 3.5 is a direct sum of ﬁnitely many p-primary submodules (say)

hxi = Mq1 + Mq2 + ··· + Mqk , for some prime q1, q2 . . . , qk ∈ R.

Now as hxi is cyclic, so by Corollary 1.4 each of the p-primary submodules Mqi must also be cyclic (say) generated by xi, therefore

hxi = hx1i + hx2i + ··· + hxki. (20)

As hxii is a qi-primary submodule, so there exists some positive integer ti > 0 ti such that qi xi = 0, 1 ≤ i ≤ k. Now if ei is the least positive integer such that ei qi xi = 0, then it follows that

ei Annhxii = hqi i, 1 ≤ i ≤ k.

From (20), we have that xi ∈ hxi, i.e., xi = rix, for some ri ∈ R, therefore

axi = a(rix) = (ari)x = (ria)x = ri(ax) = 0,

ei which implies that qi | a for each i ∈ {1, 2, . . . , k}. Since q1, q2, . . . , qk are distinct e1 e2 ek e1 e2 ek primes, so (q1 q2 ··· qk ) | a, which implies that a = (q1 q2 ··· qk ) c for some e1 e2 ek c ∈ R. If we denote q1 q2 ··· qk by b, then

a = bc and bxi = 0, 1 ≤ i ≤ k.

Consequently from (20), we have that bx = 0, i.e., b ∈ hai, so there exist some d ∈ R such that b = ad. Now b = ad together with a = bc implies that cd = 1, i.e., c is a unit or a and b are associates which in view of the factorization

m1 m2 mn e1 e2 ek a = p1 p2 ··· pn and b = q1 q2 ··· qk and the fact that every PID is a UFD, implies that k = n and after a suitable reordering (if necessary) qi and pi are associates. Hence we have that

ei mi Annhxii = hqi i = hpi i.

13 The mapping from R −→ hxii given by r 7−→ rxi is a surjective R-homomorphism with kernel Annhxii. Therefore by First Isomorphism Theorem, we have that

∼ mi hxii = R/hxii = R/hpi i, 1 ≤ i ≤ n.

Thus (20), now implies that

∼ m1 m2 mn R/hai = R/hp1 i ⊕ R/hp2 i ⊕ · · · ⊕ R/hpn i which proves the result.

Remark 3.7. For an alternate proof of the above corollary, let φ : R −→

m1 m2 mn R/hp1 i ⊕ R/hp2 i ⊕ · · · ⊕ R/hpn i be a mapping deﬁned by

m1 m2 mn (r)φ = (r + hp1 i, r + hp2 i, . . . , r + hpn i) , r ∈ R.

Then φ is an R-homomorphism, which is surjective by Chinese Remainder The-

mi mi orem and r ∈ ker φ ⇐⇒ r ∈ hpi i ∀ i ⇐⇒ pi | r ∀ i, which in view of 0 the fact that all pis are distinct primes, implies that a | r, i.e., ker φ = hai. The result now follows from the First Isomorphism Theorem.

4 Fundamental Theorem for ﬁnitely generated modules over PID

In this section, we ﬁrst state the analogue of the fundamental theorem for abelian groups for ﬁnitely generated modules over PID (without proof) and then discuss its various applications.

Theorem 4.1. Let M be a ﬁnitely generated module over a PID R. Then

∼ m M = R ⊕ R/ha1i ⊕ R/ha2i ⊕ · · · ⊕ R/hani, (21) where m is a non-negative integer and a1, a2, . . . , an are non-zero, non-units elements in R such that

ai | ai+1, for i = 1, 2, . . . , n − 1. (22)

Moreover the decomposition (21), subject to (22) is unique up to isomorphism in ∼ k the sense that if M = R ⊕ R/hb1i ⊕ R/hb2i ⊕ · · · ⊕ R/hbti, with bi | bi+1, 1 ≤ i ≤ t − 1, then k = m, t = n and after a suitable re-indexing (if necessary) ai and bi are associates.

14 Remark 4.2. The above result basically says that every ﬁnitely generated module over a PID is a direct sum of cyclic modules. The integer m is called the rank of M and a1, a2, . . . , an are called the invariant factors of M. Also from m decomposition (21), we have that tor M = R ⊕ R/ha1i ⊕ R/ha2i ⊕ · · · ⊕ R/hani and M/ tor M ∼= Rm.

Corollary 4.3. Let R be a PID and M a ﬁnitely generated R-module with decomposition (21). Then (i). M is torsion free if and only if n = 0, i.e., if and only if M is free.

(ii). M is a torsion module if and only m = 0 and then Ann M = hani.

Proof. Exercise.

Corollary 4.4. Let M be a ﬁnitely generated torsion module over a PIDR. ei,j Then M is a direct sum of primary modules R/hpi i, where pi are distinct primes of R and 0 ≤ ei,1 ≤ ei,2 ≤ · · · ≤ ei,n.

Proof. Since M is a torsion module therefore from Corollary 4.3, we have that

∼ M = R/ha1i ⊕ R/ha2i ⊕ · · · ⊕ R/hani, where a1, a2, . . . , an are non-zero, non-units in R such that

a1 | a2, . . . , ai | ai+1, . . . , an−1 | an. (23)

Now in view of (23), we see that ai | an for each 1 ≤ i ≤ n − 1, therefore if

e1,n e2,n ek,n an = p1 p2 ··· pk is the factorization of an in terms of distinct primes p1, p2, . . . , pk with positive integers e1,n, e2,n, . . . , ekn, then p1, p2, . . . , pk are the only primes which can occur i,j in the factorization of each ai. Hence there exist non-negative integers e , 1 ≤ i ≤ k, 1 ≤ j ≤ n such that

e1,n−1 e2,n−1 ek,n−1 an−1 = p1 p2 ··· pk . . . . e1,i e2,i ek,i ai = p1 p2 ··· pk . . . . e1,1 e2,1 ek,1 a1 = p1 p2 ··· pk

15 where 0 ≤ e1,1 ≤ · · · ≤ e1,i ≤ · · · ≤ e1,n−1 ≤ e1,n 0 ≤ e2,1 ≤ · · · ≤ e2,i ≤ · · · ≤ e2,n−1 ≤ e2,n ...... 0 ≤ ek,1 ≤ · · · ≤ ek,i ≤ · · · ≤ ek,n−1 ≤ ek,n

e1,i e2,i ek,i Since each R/haii is a torsion module, and ai = p1 p2 ··· pk , therefore Corol- lary 3.6 now implies that

∼ e1,i e2,i ek,i R/haii = R/hp1 i ⊕ R/hp2 i ⊕ · · · ⊕ R/hpk i proving the result.

From Theorem 3.5 and Theorem 4.1, we see that every finitely generated module over a PID is a direct sum of cyclic module. Next we investigate, when a sum of cyclic module is cyclic. For motivation recall that for abelian groups a direct sum of cyclic groups is cyclic if and only if their orders are finite and pairwise coprime. To prove an analogue of the above result for modules over PID, we first need to define the notion of order for modules Definition. Let M be a module over a PID R. An element x ∈ M is torsion iff the (principal) ideal Ann x of R is non-zero. If Ann x = hci, for some non-zero element c of R, then c is called the order of the element x. The order of x is unique upto a unit factor, for if d ∈ R is another order of x, then Ann x = hdi = hci implies that c and d are associates. Definition. A module M over a PID R is said to be bounded if there exists a non-zero element of R, which annihilates whole of M, and any such element is then called the annihilator of M. The set of all such annihilators forms an ideal (say) hai of R, and a is then called the bound of M. Clearly the bound of a module is unique upto a unit factor. Proposition 4.5. Let M be a bounded module over a PID R with bound a. Then a is the lcm of the orders of all elements of M. Proof. Since M is bounded with bound a 6= 0, so M is a torsion module. For any x ∈ M, let cx denote the order of x, i.e., Ann x = hcxi. As the bound annihilates whole of M, so ax = 0, which implies that a ∈ hcxi, i.e., cx | a. Now if d ∈ R is such that cx | d for all x ∈ M, then we have that dx = 0, ∀ x ∈ M, which implies that d ∈ Ann M = hai, i.e., a | d. Hence we have that a is the lcm of all the orders cx.

16 Lemma 4.6. Let M be a cyclic module over a PID R. The bound of M, if exists, is the order of its generator.

Proof. Suppose that the cyclic module M = hxi is bounded with bound a 6= 0, i.e., Ann M = hai. Now if order of x is c, i.e., Ann x = hci, then ax = 0 implies that a ∈ hci. As M = hxi, so for any y ∈ M, there is some r ∈ R such that y = rx and then

cy = c(rx) = (cr)x = (rc)x = r(cx) = 0, which implies that c annihilates every element of M, i.e., c ∈ Ann M = hai. Thus we have that Ann M = hai = hci = Ann x and the lemma follows.

Exercise 4.7. Let R be a PID and consider the R-module M = R/hci, for some non-zero element c of R. Show that M is bounded with bound c.

Theorem 4.8. Let M be a ﬁnitely generated module over a PID R. Then M is bounded if and only if M is a torsion module and there is then an element in M whose order is the bound of M.

Proof. Suppose ﬁrst that M is bounded with bound a. Then a annihilates whole of M, i.e., ax = 0 for every x ∈ R, which implies that M is a torsion module. Conversely assume that M is a torsion module. Since M is ﬁnitely generated, so by part (ii) of Corollary 4.3, we have that

∼ M = R/ha1i ⊕ R/ha2i ⊕ · · · ⊕ R/hani, and Ann M = hani. Hence M is bounded with bound an. Now the R-module

R/hani is cyclic generated by xn = 1 + hani and clearly (Why?)

Ann R/hani = Annhxni = hani, which implies that order of xn is an. Therefore there is an element xn in M whose order is the bound of M.

Remark 4.9. The above result fails to hold if M is not ﬁnitely generated, e.g.,

M = ⊕Zp, where the sum runs over all prime numbers p, is a torsion Z-module which is not bounded.

17 Theorem 4.10. Let M be a module over a PID R and assume that

M = M1 ⊕ M2 ⊕ · · · ⊕ Mn. (24)

Then M is bounded if and only if each Mi is bounded and the bound of M is the lcm of the bounds of Mi.

Proof. Suppose ﬁrst that M is bounded with bound a. Then a annihilates every element of M, in particular a annihilates every element of Mi, i.e., ax = 0 for every x ∈ Mi, 1 ≤ i ≤ n, which implies that each Mi is torsion and hence bounded in view of Theorem 4.8. Let bi denote the bound of Mi, i.e.,

Ann Mi = hbii, 1 ≤ i ≤ n.

Again as a annihilates every element of M and hence of Mi, we have that a ∈ 0 hbii, which implies that bi | a for each i = 1, 2, . . . , n. If l ∈ R be such that 0 0 0 bi | l , then l ∈ hbii = Ann Mi, i.e., l annihilates every element of Mi for each i = 1, 2, . . . , n. Now if x is an arbitrary element of M, then from (24), there exist some xi ∈ Mi, 1 ≤ i ≤ n such that x = x1 + x2 + ··· + xn and

0 0 0 0 l x = l x1 + l x2 + ··· + l xn = 0 + 0 + ··· + 0 = 0, which implies that l0 ∈ Ann M = hai, i.e., a | l0. Thus we have that a is the lcm of b1, b2, . . . , bn the bounds of the summands M1,M2,...,Mn.

Conversely suppose that each Mi is bounded with bound bi, i.e., Ann Mi = hbii, 1 ≤ i ≤ n. Then each bi is non-zero and since R is an integral domain, we have that c = b1b2 ··· bn 6= 0. Let x be an arbitrary element of M, then in view of

(24), there exist some xi ∈ Mi for 1 ≤ i ≤ n such that x = xi +x2 +···+xn. Now c = b1b2 ··· bn implies that c ∈ hbii = Ann Mi therefore cxi = 0 for 1 ≤ i ≤ n, and hence

cx = cx1 + cx2 + ··· + cxn = 0. As x in M was arbitrary the above equation implies that c annihilates whole of M, i.e., M is a bounded module.

Theorem 4.11. A direct sum of ﬁnitely many cyclic modules over a PID R is cyclic if and only if each of the modules are bounded with pairwise coprime bounds.

18 Proof. Suppose ﬁrst that the direct sum

M = M1 ⊕ M2 ⊕ · · · ⊕ Mn of cyclic modules Mi is cyclic, we need to show that each Mi is bounded and their bounds are pairwise coprime. By induction it is enough to prove that result for n = 2. Since M1,M2 are cyclic R-module, so

M1 = hx1i and M2 = hx2i.

Regarding R as a module over itself, the mappings φi : R −→ Mi, i = 1, 2 deﬁned by (r)φi = rxi, r ∈ R, are surjective R-homomorphisms, with ker φi = haii (∵ R is a PID). Now by First Isomorphism Theorem, we have that

∼ R/haii = Mi = hxii, i = 1, 2.

∼ Similarly for the cyclic R-module M = M1 ⊕ M2, we have that M = R/hai for some a ∈ R. Therefore

∼ R/hai = R/ha1i ⊕ R/ha2i. (25)

We now claim that a, a1 and a2 all are non-zero. Assume to the contrary that a = 0, then the left hand side of (25), is free with rank 1. Now if one of a1 and a2 is zero, then the right hand side of (25), is not torsion free and if both of a1 and a2 are zero, then the right hand side of (25) becomes a free module of rank 2. So in either case we arrive at a contradiction, therefore a cannot be zero, which implies that M is a bounded module and hence in view of Theorem 4.10, both M1 and M2 are bounded, i.e., annihilators of M1,M2 must be non-trivial.

But Ann Mi = haii, i = 1, 2, therefore a1 and a2 both are non-zero and the claim follows. In fact we can assume without loss of generality that a, a1, a2 are non- units. For if a is unit, then hai = R, which implies that the module R/hai is trivial and so both a1 and a2 must also be units, and we have nothing to prove.

Therefore a is a non-unit and if one of a1 or a2 is a unit, then either M = M1 or

M = M2 and again there is nothing to prove.

Let d = gcd(a1, a2). Then d | ai, implies that ha1i ⊆ hdi, i = 1, 2 and we have canonical surjective R-module homomorphisms

R/haii −→ R/hdi, i = 1, 2,

19 which gives rise to a surjective R-homomorphism ∼ f R/hai = R/ha1i ⊕ R/ha2i −→ R/hdi ⊕ R/hdi. Since quotient of a cyclic module over a PID is cyclic (see Corollary 1.4), we have that (R/hai) / ker f and hence, by First Isomorphism Theorem, R/hdi ⊕ R/hdi is a cyclic R-module. Therefore R/hdi ⊕ R/hdi = R/hci, for some c ∈ R. As d | d, so by uniqueness part of Theorem 4.1, the above equation is possible if R/hdi is the trivial module, i.e., R = hdi which implies that d must be a unit.

Hence the gcd of a1, a2 is a unit, i.e., a1, a2 are coprime. Conversely assume that

M = M1 ⊕ M2 ⊕ · · · ⊕ Mn, where each module Mi is cyclic (say) generated by xi (1 ≤ i ≤ n) and bounded with pairwise coprime bounds. We need to show that M is a cyclic module. To prove this we will show that M = hxi, where x = x1 + x2 + ··· + xn, which will follow once we show that xi ∈ hxi for each i ∈ {1, 2, . . . , n}. Let ai be the bound of Mi, then Ann Mi = Ann xi = haii and ai’s are pairwise coprime for i = 1, 2, . . . , n. Let bi = a1 ··· ai−1ai+1 ··· an, 1 ≤ i ≤ n, then as R is a PID, the ideal hb1i + hb2i + ··· + hbni is principal (say) generated by u. Arguing similarly as in the proof of Theorem 3.5 (see Claim), it can be shown that u is a unit in R. Therefore we have that

hb1i + hb2i + ··· + hbni = R.

As 1 ∈ R, so there exist some c1, c2, . . . , cn ∈ R such that

c1b1 + c2b2 + ··· + cnbn = 1.

By deﬁnition of bj’s, it follows that ai | bj for each i 6= j, which in view of the fact that ai is a bound of Mi, implies that (cjbj)xi = cj(bjxi) = 0, and hence for each i = 1, 2, . . . , n, we have that n ! X xi = 1 · xi = cjbj xi = (c1b1) xi + ··· + (cibi) xi + ··· + (cnbn) xi = (cibi) xi. j=1 Now on using the above equation, we get that

(cibi) x = (cibi)(x1 + ··· + xi + ··· + xn) = (cibi) xi = xi, which implies that xi ∈ hxi for each 1 ≤ i ≤ n and the result follows.

20 5 Exercises

1. Let R be an integral domain and let A and B be R-modules of rank m and n, respectively. Prove that the rank of A ⊕ B is m + n.

2. For any short exact sequence

0 −→ M 0 −→ M −→ M 00 −→ 0

of ﬁnitely generated modules over a principal ideal domain show that rank M = rank M 0 + rank M 00.

3. Every free module over an arbitrary integral domain is torsion free. Does the converse holds?

4. Let M be a module over integral domain R. Suppose x is a non-zero torsion element in M. Show that x, 0 are linearly dependent.

5. If M is ﬁnitely generated module over a PID R, describe the structure of M/torM.

6. Let R = Z[x] and let M = h2, xi be the ideal of R considered as a submodule of R-module R. Show that {2, x} is not a basis of M. Also show that the rank of M is 1 but M is not free.

7. Show that if R is an integral domain and M is any non-principal ideal of R, then M is torsion free of rank 1 but is not free R-module.

8. Let R be an integral domain, M be an R-module and N be a submodule of M. Suppose that M has rank n, N has rank r and the quotient M/N has rank s. Prove that n = r + s.

References

[1] P. M. Cohn, Classic Algebra, John Wiley & Sons Ltd., 2000.

[2] D. S. Dummit and R. M. Foote, Abstract Algebra (Third Edition), Wiley India Pvt. Ltd., 2011.