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Brandeis MATH 101A - MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 35

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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:(x : tM) := {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


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