Abstract Nonsense

Crushing one theorem at a time

Internal Direct Sum of Modules


Point of Post: In this post we discuss the notion of the internal directs sum of modules, giving the usual characterizations, etc. We also give a standard application to short exact sequences.

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Motivation

We have seen that the category of modules has notions of coproducts, namely given a set \left\{M_\alpha\right\}_{\alpha\in\mathcal{A}} of modules, we can form the module \displaystyle \bigoplus_{\alpha\in\mathcal{A}}M_\alpha which is the submodule of the product \displaystyle \prod_{\alpha\in\mathcal{A}}M_\alpha of \mathcal{A}-tuples with finite support. That said, from experience dealing with vector spaces, we often see that coproducts sometimes come up more naturally as “internal direct sums”. Namely, a module M may be able to be written as the sum of submodules \{M_\alpha\} in a “unique way” (i.e. unique representation). In this case, M will be isomorphic to the coproduct over the set \{M_\alpha\} of modules, and so we don’t really get “new theory”. That said, often times we want to work with actual equalities: “M is the internal direct sum” opposed to “M is isomorphic to the coproduct”.

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Internal Direct Sum of Modules

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Let R be a ring and M a left R-module. Suppose that\{M_\alpha\}_{\alpha\in\mathcal{A}} is a family of submodules of M. We clearly have, for each \alpha\in\mathcal{A}, the inclusion maps \iota_\alpha:M_\alpha\to M. We see then by definition of coproduct that we are afforded a guaranteed map \displaystyle f:\bigoplus_{\alpha\in\mathcal{A}}M_\alpha\to M. We say that M is the internal direct sum of \{M_\alpha\}_{\alpha\in\mathcal{A}} if f is an isomorphism. Since, in an obvious sense, internal direct sums and coproducts are the same thing we shall use the same symbol \oplus to denote both. While it should be clear from context which we actually mean, the tipoff shall often come in the form of equality or isomorphism. For example, if one writes \displaystyle M=\bigoplus_{\alpha\in\mathcal{A}}M_\alpha then its clear that we mean that M is the internal direct sum of this family of modules (which are submodules).

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Of course, while the above is a nice, succinct way of putting internal direct sums, it is often not how one actually thinks about the concept. No, really one often thinks about a module as being an internal direct sum of a family if the following characterization holds:

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Theorem: Let M be a left R-module and \left\{M_\alpha\right\}_{\alpha\in\mathcal{A}} a family of submodules. Then, M is the internal direct sum of \{M_\alpha\} if and only if every element of M can be written uniquely as a sum \displaystyle \sum_{\alpha\in\mathcal{A}}m_\alpha with m_\alpha\in M_\alpha with all but finitely many m_\alpha=0.

Proof: The fact that \displaystyle \sum_{\alpha\in\mathcal{A}}M_\alpha=M is equivalent to the statement that the canonical map \displaystyle f:\bigoplus_{\alpha\in\mathcal{A}}M_\alpha\to M is surjective, and the uniqueness of this representation is equivalent to the injectivity of f. \blacksquare

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For the case when we are dealing with only two submodules, this takes the following particularly nice form:

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Theorem: Let R be a ring and M a left R-module. Then, if N,L\leqslant M then M=L\oplus N if and only if M=L+N and L\cap N is trivial.

Proof: Since we have a priori that M=L+N it suffices to show that uniqueness of representation is equivalent the triviality of N\cap L. Suppose first that we have uniqueness of representation. If x\in M\cap N, then we can write it as x=0+x where we are thinking about 0\in L and x\in N and x=x+0 where x\in L and 0\in N. From uniqueness we may conclude that x=0. Conversely, if L\cap N is trivial and if \ell+n=\ell'+n' then \ell-\ell'=n'-n and so clearly \ell-\ell',n'-n\in N\cap L and so \ell-\ell'=n'-n=0 and so uniqueness of representation follows. \blacksquare

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In fact, it’s not hard to see that this generalizes to the following:

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Theorem: Let M be a left R-module and \left\{M_\alpha\right\}_{\alpha\in\mathcal{A}} a family of submodules. Then, M is the internal direct sum of \{M_\alpha\}_{\alpha\in\mathcal{A}} if and only if \displaystyle \sum_{\alpha\in\mathcal{A}}M_\alpha=M and \displaystyle M_\alpha\cap\sum_{\beta\ne\alpha}M_\beta is trivial for each \alpha\in\mathcal{A}.

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Relation to Short Exact Sequences

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We can use this notion of internal direct sums to be able to write the central module of a split short exact sequence in a nice way, in terms of the image and kernel of certain maps. First we note the following:

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Theorem: Let R be a ring and M,N left R-modules. If f:M\to N and g:N\to M are R-maps such that f\circ g=\text{id}_N then M=\text{im }g\oplus\ker f.

Proof: By the above theorems it suffices to show that \text{im }g\cap\ker f is trivial and \text{im }g+\ker f=M. To see the first, suppose that x\in\text{im }g\cap\ker f. Since x\in\text{im }g we can find n\in N with x=g(n). But, note then that says that f(g(n))=f(x) but, since x\in\ker f we see that f(x)=0 and since f\circ g=\text{id}_N we see that f(g(n))=n and so n=0. Thus, x=g(n)=g(0)=0. Since x was arbitrary the triviality of \text{im }g\cap\ker f follows. To see that M=\text{im }g+\ker f we let x\in M be arbitrary. Note then that x=(x-g(f(x)))+g(f(x)). Clearly g(f(x))\in\text{im }g and since f(x-g(f(x))=f(x)-f(g(f(x))=f(x)-f(x)=0 we see that x-g(f(x))\in\ker f and so clearly x\in \text{im }g+\ker f. Since x was arbitrary the conclusion follows. \blacksquare

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Of course, applying this in conjunction with the splitting lemma gives us the following:

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Theorem: Let 0\to A\xrightarrow{h}B\xrightarrow{f}C\to0 be a split short exact sequence. If g:C\to B is the guaranteed backmap for f then B\cong \ker f\oplus\text{im }g.

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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 20, 2011 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , , ,

1 Comment »

  1. […] since is free, we know this sequence splits and so there exists some backmap and we know that and since is free we know that where if and if . We then let […]

    Pingback by Submodules of Free Modules Need Not be Free Unless Ring is a PID (pt. II) « Abstract Nonsense | November 21, 2011 | Reply


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