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

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30 MATH 101A: ALGEBRA I PART B: RINGS AND MODULES8. Finite generation and ACCI continued the discussion of finitely generated modules by talkingabout Noetherian modules and rings. One example lead to a discus-sion of the “restriction of scalars” functor and the basis theorem forfinite dimensional vector spaces over a field. This is a precursor to thefundamental theorem for f.g. modules over a PID and applications tomatrices.8.1. Noetherian rings and modules.8.1.1. definitions.Definition 8.1. If M is an R-module, an An ascending chain of sub-modules of M is an increasing sequence of submodules:N1⊆ N2⊆ N3⊆ · · · ⊆ MWe say that M satisfies the ascending chain condition (ACC) for sub-modules if every such sequence stops, i.e., if there is a k so that Nn= Nkfor all n ≥ k.Definition 8.2. A module M is Noetherian if it satisfies the ACC forsubmodules. A commutative ring R is called Noetherian if it satisfiesthe ACC for ideals. This is equivalent to saying that R is a Noetherianwhen considered as an R-module. (A noncommutative ring is calledleft-Noetherian if it satisfies the ACC for left ideals.)8.1.2. examples. The first example I gave was Z.Theorem 8.3. Z is Noetherian (as a ring and as a module over Z).I proved this twice. The first time, I used the properties of theintegers. The second time I used only the fact that Z is a PID.Proof. To show that Z is Noetherian we took an ascending chain:N1⊆ N2⊆ N3⊆ · · · ⊆ ZEach submodule Niis an ideal (or Z) generated by one element ni≥ 0and the condition Ni⊆ Ni+1is equivalent to saying that ni+1dividesni. In particular, ni+1≤ ni. So,n1≥ n2≥ n3≥ · · · .Since these numbers are bounded below (by 0), the sequence stops(becomes constant) at some point. Thus nk= nk+1= · · · . This isequivalent to saying thatNk= Nk+1= · · · .So, the ACC holds and Z is Noetherian. !MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 31Second proof. In the second proof, I took the union of the Niand calledit N∞N∞=∞!i=1Ni.This is a submodule of Z since it is closed under addition and scalarmultiplication: If a ∈ Niand b ∈ Njwhere i ≤ j then a + b ∈ Nj. Thesubmodule N∞is generated by one element n∞which is contained insome Nk. But then:(n∞) ⊆ Nk⊆ Nk+1≤ Nk+2≤ · · · ⊆ N∞= (n∞).So,Nk= Nk+1= Nk+2= · · ·as before. !The second proof works for any PID. This is Corollary 8.6 below.Some other easy examples of Noetherian rings and modules are thefollowing.(1) Any finite ring or module is Noetherian.(2) Any quotient of a Noetherian ring or module is Noetherian.(3) A finite product of Noetherian rings is Noetherian. (Proof:Every ideal in R×S has the form I × J where I is an ideal in Rand J is an ideal in S. In any ascending chain (In× Jn), bothsequences Inand Jnhave to stop.)8.1.3. theorem. Here is the general theorem about Noetherian modules.Theorem 8.4. An R-module is Noetherian M if and only if everysubmodule is finitely generated.Corollary 8.5. A commutative ring is Noetherian if and only if everyideal is finitely generated.Corollary 8.6. Every PID is Noetherian.Proof. Suppose first that M has a submodule N which is not finitelygenerated. Then, any finite subset of N will generate a proper sub-module.Let x1∈ N. Then Rx1! N. So, there is an x2∈ N, x2/∈ Rx1.Then Rx1+ Rx2! N. So, there is an x3∈ N\Rx1+ Rx2. Continuingin this way, we get a strictly increasing sequence of submodulesRx1! Rx1+ Rx2! Rx1+ Rx2+ Rx3! · · ·So, the ACC fails and M is not Noetherian.32 MATH 101A: ALGEBRA I PART B: RINGS AND MODULESConversely, suppose that every submodule of M is finitely generated.Then we have to show that any ascending chain:N1⊆ N2⊆ N3⊆ · · · ⊆ Mstops. To show this take the unionN∞=!NiThis is a submodule of M and is therefore finitely generated. Letx1, · · · , xnbe the generators. Then each xiis contained in some Nj. Ifk is the maximum of the indices j then all the xiwill be contained inNk. But then Nkmust equal N∞. So,Nk= Nk+1= Nk+2= · · ·and the ACC holds. So, M is Noetherian. !8.2. r estriction of scalars. I explained one example of a Noetherianmodule which used restriction of scalars and the basis theorem for f.g.vector spaces. I explained these tools after the example. So, I will dothe same thing here.Example 8.7. Let R be the polynomial ring R = F [T ] over a field F .Then a module M can be constructed as follows. Suppose that A is ann×n matrix with coefficients in the field F . Then M = Fnis a moduleif we define the action of a polynomial p(T ) ="riTi∈ R = F [T ] byp(T )x =#riAixthen M is a Noetherian R-module. To see this take any ascendingchain of submodules:N1⊆ N2⊆ N3⊆ · · · ⊆ MSince R contains the ground field F , every R-module is also a vectorspace over F . Thus every Niis a vector space over F . So, it has adimension di. Since M∼=Fn, its dimension is n. So,d1≤ d2≤ d3≤ · · · ≤ nThis implies that the sequence stops, i.e., there is a k ≥ 1 so thatdk= dk+1= · · ·But, every proper subspace of a vector space has smaller dimension, soNk= Nk+1= · · ·I.e., the ACC holds and M is Noetherian.MATH 101A: ALGEBRA I PART B: RINGS AND MODULES 33The key point is the realization that an R-module is also a vectorspace over F . This is called “restriction of scalars.” The definition isas follows.Definition 8.8. Suppose that either(1) S is a subring of R or(2) φ : S → R is a ring homomorphism. (We can let φ : S → R bethe inclusion map to make (1) a special case of (2).)Then, in either case, we get a restriction of scalars functorφ∗: R- Mod → S- Modgiven on objects by φ∗(M) = M considered as an S-module in thefollowing way:(1) When S ⊆ R we just restrict the action of R. So, for any a ∈ Sand x ∈ M we let ax = ax. This makes sense since a ∈ R.(2) In the general case we let sx = φ(s)x. In other words, theaction of S on M is given by the compositionSφ−→ Rα−→ EndZ(M)where α is the action of R on M.Just as φ∗(M) = M with a different action, φ∗(f : M → N) is alsothe same mapping f : M → N considered as a homomorphism ofS-modules.The other fact that we used in the last example was our knowledgeof linear algebra over any field. I decided to go over some of that.8.3. finite dimensional vector spaces. Recall that a module overa field F is the same as a vector space. So I called it V . Suppose Vis finitely generated with generators v1, v2, · · · , vn. In linear algebrawe say that these vectors span V and


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