Abstract Nonsense

Crushing one theorem at a time

Quotient Modules


Point of Post: In this post we construct quotient modules and discuss some of the important theorems.

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Motivation

As always, when discussing a new algebraic object we always wish to discuss the quotient objects of the objects. So, of course, in this post we define the notion of quotient modules. Not only should the basic idea of what and how the quotient module should look be clear (since, in a universal algebra sort of sense, all quotient objects are the same) but this is really just a generalization of  the idea of a quotient space for vector spaces. Namely, we shall just take the quotient group and multiply scalars representative-wise. As in all algebraic theories we shall occupy a central role in our study, they are convenient tools which encompass the set of possibilities of homomorphic images of R-modules (i.e. the first isomorphism theorem).

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Quotient Module

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Let R be a ring and M a left R-module. For N\leqslant M consider the quotient group M/N (written additively as usual). We claim that M/N is a left R-module with the obvious multiplication r(m+N)=rm+N. Indeed, we prove this:

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Theorem: Let R be a ring and M a left R-module with N\leqslant M. Then, the quotient group M/N is a left R-module with R-multiplication defined by r(m+N)=rm+N. If M is unital then so is M/N. Moreover, the natural projection \pi:M\to M/N:m\mapsto m+N is an epimorphism.

Proof: It suffices to show that (r+s)(m+N)=r(m+N)+s(m+N), r((m+N)+(m'+N))=r(m+N)+r(m'+N), (rs)(m+N)=r(s(m+N)), and 1_R(m+N)=m+N (in the case that M is unital). These are pretty easy to show:

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\begin{aligned}(r+s)(m+N) &=(r+s)m+N\\ &=(rm+sm)+N\\ &=rm+N+sm+N\\ &=r(m+N)+s(m+N)\end{aligned}

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\begin{aligned}r((m+N)+(m'+N)) &=r((m+m')+N)\\ &=r(m+m')+N\\ &=(rm+rm')+N\\ &=rm+N+rm'+N\\ &=r(m+N)+r(m'+N)\end{aligned}

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(rs)(m+N)=(rs)m+N=r(sm)+N=r(sm+N)=r(s(m+N))

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and if M is unital then

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1_R(m+N)=1_Rm+N=m+N

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The fact that \pi is an epimorphism is obvious from definition. \blacksquare

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Remark: Note that we skirted over one small fact in the above proof, that the definition of R-multiplication is well-defined. But, this is clear for if m+N=m'+N then m-m'\in N and so by assumption that N\leqslant M this allows us to conclude that rm-rm'=r(m-m')\in N and so rm+N=rm'+N.

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As an initial application we can finally finish our proof about the equivalence of epimorphisms and surjective morphisms:

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Theorem: Let M,N be left R-modules. Then, a R-morphism f:M\to N is an epimorphism if and only if its surjective.

Proof: Suppose first that f is an epimorphism and let L=\text{im }f. Since L\leqslant N we have that we have the quotient module N/L. Consider then that we have that natural maps \pi,0:N\to N/L. Note though that by definition \pi\circ f=0\circ f=0 and since f is an epimorphism we may conclude that \pi=0. But, this implies that N/L=0 and so N=L. Surjectivity follows.

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The converse follows since any surjection allows right cancellation of any map, and so R-morphisms. \blacksquare

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To keep things in perspective we note that if we think back to the two most basic examples of modules, that quotients there are what we expect them to be. Firstly, we recall that \mathbb{Z}-modules are nothing more than abelian groups and submodules nothing more than subgroups, and it’s not hard to see that quotient \mathbb{Z}-modules are nothing more than quotient groups. Secondly, if we recall that we can canonically view every ring R as being a left R-module over itself with the submodules being precisely the ideals, in which case it’s not hard to see that the quotient modules are nothing more than the quotient rings. Some less familiar examples may come from the obvious fact that R\times\{0\} is a submodule of R^2 with quotient isomorphic to R. Or, given a commutative unital ring R we can look at the polynomial ring R[x] as a left R-module and note that (p(x)) is a submodule of R[x] from where we can form the quotient R[x]/(p(x)) which shall prove to have some interesting applications. In particular, from what we already know about vector spaces it’s not hard to see (we shall prove this in more detail at a later date) that if R is a field then R[x]/(p(x))\cong R^{\deg p(x)}.

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Anyways, knowing what we do about rings it’s not shocking that there is a universal characterization of quotient rings (i.e. a set of mapping properties of quotient rings which characterizes them amongst the set of all left R-modules. Indeed:

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Theorem:  Let R be a ring and M a left R-module with a submodule N\leqslant M. Then, for any R-morphism f:M\to L (where L is some R-morphism) with \ker f\subseteq N then there exists a unique morphism j:M/N\to L making the following diagram commute

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\displaystyle \begin{matrix}M & \overset{f}{\longrightarrow} & L\\ ^{\pi}\big\downarrow & \nearrow _{j} & \\ M/N & & \end{matrix}

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Conversely, if P is any other left R-module with some epimorphism e:M\twoheadrightarrow P satisfying this property (with e in place of \pi) then P\cong M/N (up to unique isomorphism).

Proof: This first part follows immediately from our discussion of triangle completions. Conversely, taking f=e:M\to P allows us to conclude from M/N‘s mapping properties that there is a mapping j:M/N\to P satisfying j\circ \pi=e. Similarly, taking f=\pi:P\to M/N allows us to conclude by P‘s mapping properties that there exists some map j':P\to M/N satisfying j'\circ e=\pi. From these two equations we know that e=j\circ\pi=j\circ j'\circ e. And, since e is an epimorphism we know that e is right cancellable and so j\circ j'=\text{id}_P. Similarly, we know that \pi=j'\circ e=j'\circ j\circ\pi and since \pi is an epi this tells us that j'\circ j=\text{id}_{M/N}. Thus, j:M/N\xrightarrow{\approx}P and so the conclusion follows. \blacksquare

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.

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November 7, 2011 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , , , , ,

3 Comments »

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