MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 359. Modules over a PIDThis week we are proving the fundamental theorem for finitely gener-ated modules over a PID, namely that they are all direct sums of cyclicmodules. The proof will be in stages. On the first day I decomposeda module into a torsion and torsion-free part. The presentation was alittle disorganized so that the steps do not follow one after the otherbut rather the other way around, i.e., to prove (1) we need to prove (2)and to prove (2) we need to prove (3), etc. I call this the “motivationalorder.” At the end we will go over the lemmas and put them in correctlogical order.9.1. torsion and torsion-free. Suppose that R is a PID and M isa finitely generated R-module. The main example I talked about wasR = Z in which case M = G is a f.g. abelian group.Definition 9.1. M is torsion-free if ann(x) = 0 for all x != 0 in M.Definition 9.2. M is torsion if ann(x) != 0 for all x != 0 in M .For example, R itself is torsion-free and R/(a) is torsion. In the caseR = Z, Znand Q are torsion-free additive groups. However, Q is notfinitely generated. The finitely generated torsion abelian groups areexactly the finite abelian groups.The first decomposition theorem is the following.Theorem 9.3. Every f.g. module over a PID is a direct summand ofa torsion module and a torsion-free module:M∼=tM ⊕ fMwhere tM is torsion and fM is torsion free.The second theorem tells us what the torsion-free part looks like:Theorem 9.4. A f.g. module over a PID is torsion-free if and only ifit is free:fM∼=Rn.9.1.1. torsion submodule. I used two lemmas to show that the secondtheorem implies the first theorem. During the class we decided thatthese two lemmas hold over any domain. First I need a definition.Definition 9.5. Suppose that M is a module over a domain R. Thenthe torsion submodule of M is defined to be the set of all elements ofM with nonzero annihilator ideal:tM := {x ∈ M | ann(x) != 0}36 MATH 101A: ALGEBRA I PART B: RINGS AND MODULESLemma 9.6. tM is a submodule of M provided that R is a domain.Proof. We need to show that tM contains 0 and is closed under additionand scalar multiplication.(1) 0 ∈ tM since ann(0) = R.(2) If x, y ∈ tM then there are nonzero elements a, b ∈ R so thatax = by = 0. Then ab(x + y) = 0. So, x + y ∈ tM.(3) If x ∈ tM and r ∈ R then ann(rx) ⊇ ann(x) is nonzero.In the second step we need to know that ab != 0. !Lemma 9.7. The quotient M/tM is torsion-free provided that R is adomain.Proof. Suppose not. Then there is a nonzero element x + tM in M/tMand a != 0 in R so that ax + tM is zero, i.e., ax ∈ tM. But this meansthere is a nonzero element b ∈ R so that bax = 0. But ba != 0. So,x ∈ tM which is a contradiction. !It was in this proof that I mentioned the fraction notation:(tM : x) := {a ∈ R | ax ∈ tM}.Next, I showed that the second theorem (Theorem 9.4) implies thefirst (Theorem 9.3).Proof of Theorem 9.4. Let fM = M/tM. Since this is free, there is asection s : fM → M of the projection map M → M/tM. Thenj ⊕ s : tM ⊕ fM → Mis an isomorphism where j : tM → M is the inclusion map. !Next, we need to prove that f.g. torsion-free modules are free andthat f.g. torsion modules are direct sums of cyclic modules. I intendto use “purity” to do both.9.1.2. pure submodules.Definition 9.8. We say that a submodule N ⊆ M is pure if wheneverx ∈ M and a ∈ R with ax ∈ N there exists z ∈ N so that az = ax. Inother words: “If an element of N is divisible by a ∈ R in M then it isdivisible by a in N.”The point is that pure submodules are direct summands in f.g. mod-ules over PID’s. However, using this fact is a little tricky as we sawthe next day.MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 37On the second day I pointed out that if P is a pure submodule of af.g. module M over a PID, then P is a direct summand. However, wecannot use this fact to prove the main theorem because it uses the maintheorem, namely that f.g. modules are direct sums of cyclic modules.However, we use the main theorem on a module with a fewer numberof generators than M. So, this can be used in an inductive proof of themain theorem. The theorem is:Theorem 9.9. Suppose that P is a pure submodule of a f.g. moduleM over a PID R and M/P is a direct sum of cyclic modules. Then Pis a direct summand of M .Proof. Suppose that M/P is a direct summand of cyclic modules. Theneach summand is generated by some element xi+ P with annihilator(ai). This means that (xi: P ) = (ai). In other words, aixi∈ P . SinceP is pure, there is an element zi∈ P so that aizi= aixi. But thenxi+ P = (xi− zi) + Pand ai(xi−zi) = 0. So si(xi+P ) = xi−zigives a lifting si: R/(ai) → Mof the direct summand R/(ai) of M/P to M and, together, these givean isomorphism!"si#⊕ j :"R/(ai) ⊕ P≈−→ Mwhere j : P → M is the inclusion map. !To find the pure submodule, I need to take a maximal cyclic sub-module of M. This exists because M is Noetherian. So that is next.9.2. submodules of free modules.Theorem 9.10. Suppose that M = Rnis a free R-module with ngenerators where R is a PID. Then any submodule of M is free with nor fewer generators.Proof. This is by induction on n. If n = 1 then M = R and thesubmodules are either R or an ideal Rx. But R is a domain. So, eitherx = 0 or ann(x) = 0. So, Rx is free with 0 or 1 generator.Now suppose that n ≥ 2 and N is a submodule of Rn. Then wewant to show that N∼=Rmwhere m ≤ n. Let e1, · · · , enbe the freegenerators of M = Rn. Let Rn−1denote the free submodule of Mgenerated by e1, · · · , en−1. Then, by induction on n, we have:N ∩ Rn−1∼=Rm−138 MATH 101A: ALGEBRA I PART B: RINGS AND MODULESwhere m ≤ n. If N = N ∩ Rn−1we are done. Otherwise, let J be theset of all encoordinates of all elements of N. I.e.,J =$a ∈ R | (∃x ∈ N) x = aen+n−1%i=1aiei&We get at least one nonzero element in J since N is not containedin Rn−1. Then it is easy to see that J is an ideal (or all of R) sinceit is closed under addition and scalar multiplication and is nonempty.Therefore, J = (b) is generated by one element b != 0. (So, every a ∈ Jhas the form a = rb where r ∈ R is unique.)By definition, there is an element x0∈ …