Abstract Nonsense

Crushing one theorem at a time

The Splitting Lemma (For Modules)

Point of Post: In this post we discuss the splitting lemma for modules, explain its significance, and some of its consequences.

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In our last post that the existence of a short exact sequences 0\to M\to N\to L\to 0 indicates that N is “put together” in some way from M,L. We noticed that, theoretically N, could be just M\oplus L which represents the simplest way N could be “put together” from M,L. It makes sense then that we would like to know when this is actually true. Namely, under what conditions in addition to the existence of a short exact sequence 0\to M\to N\to L\to 0 do we need to be able to conclude that N\cong M\oplus L? This shall be the subject of our discussion here.

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Splitting Lemma

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Given a short exact sequence 0\to M\xrightarrow{f} N\xrightarrow{g} L\to 0 we call N an extension of L by M. We say that the sequence splits if there exists a chain isomorphism c from this short exact sequence to the natural short exact sequence 0\to M\xrightarrow{\iota} M\oplus L\xrightarrow{\pi_L} L\to 0 (where the direct sum of modules is defined in the obvious way, it will be discussed more later) such that the maps M\to M and L\to L are the identity. This says that not only are the two chains isomorphic, but they are as naturally isomorphic as you’d like, in the sense that the translate to go from M\to L (in the sense discussed in the post) is just identity maps!  So, we shall now discuss what conditions need to be placed on an exact sequence so that it splits:

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Theorem (The Splitting Lemma): Let M,N,L be left R-modules for some ring R and C:0\to M\xrightarrow{f}N\xrightarrow{g}L\to0 a short exact sequence. Then, the following are equivalent

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\displaystyle \begin{aligned}&\mathbf{(1)}\quad C\text{ splits}\\&\mathbf{(2)}\quad \text{There exists }f^{\leftarrow}\in\text{Hom}_R(N,M)\text{ such that }f^{\leftarrow}\circ f=\text{id}_M\\ &\mathbf{(3)}\quad\text{There exists }g^{\leftarrow}\in\text{Hom}_R(L,N)\text{ such that }g\circ g^{\leftarrow}=\text{id}_L\end{aligned}

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Proof: We proceed in the following progression \mathbf{(1)}\Leftrightarrow \mathbf{(2)} and \mathbf{(2)}\Leftrightarrow\mathbf{(3)} but we arrange it in the way that is the most natural.

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We start with \mathbf{(1)}\implies\mathbf{(2)}. Suppose that our chains are isomorphic, and let \phi:N\xrightarrow{\approx} M\oplus L be the guaranteed isomorphism.Define a map f^{\leftarrow} by f^{\leftarrow}=\pi_M\circ\phi. To see what’s going on better we have the following diagram:

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\displaystyle \begin{matrix}0 & \to & M & \xrightarrow{f} & N & \xrightarrow{g} & L & \to & 0\\ & & {^{1_M}}\big\updownarrow & \nwarrow{^{\pi_M}} & {^{\phi}}\big\downarrow & & ^{1_L}\big\updownarrow & & \\ 0 & \to & M & \xrightarrow{\iota_M} & M\oplus L & \xrightarrow{\pi_L} & L & \to & 0\end{matrix}

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where we make no claims (as of right now) that everything is commutative with \pi_M (the natural inclusion). That said, f^{\leftarrow} is a morphism, being the composition of two morphisms, and f^{\leftarrow}\circ f=\pi_M\circ\phi\circ f=\pi_m\circ\iota_M\circ 1_M=1_M. The conclusion follows.

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We now take a go at \mathbf{(2)}\Leftrightarrow\mathbf{(3)}. Suppose we are given this function f^{\leftarrow}:N\to M. Define g^{\leftarrow}:N\to M as follows. Since g is an epimorphism there exists, for each \ell\in L some n\in N such that g(n)=\ell, define g^{\leftarrow}(\ell)=n-f(f^{\leftarrow}(n)). To see that this is well-defined, in the sense that if g(n')=g(n) then n-f(f^{\leftarrow}(n))=n'-f(f^{\leftarrow}(n')) we note that if g(n)=g(n') then n-n'\in\ker g and so by exactness n-n'\in\text{im }f and since f^{\leftarrow}\circ f=1_M this tells us that f\circ f^{\leftarrow}\circ f=f and so evidently f(f^{\leftarrow}(n-n'))=n-n' and so

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This is a morphism since if \ell=g(n) and k=g(n') then r\ell=g(rn) and k=g(n') and so r\ell+k=g(rn+n') and so

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\begin{aligned}g^{\leftarrow}(r\ell+k) &=rn+n'-f(f^{\leftarrow}(rn-n'))\\ &=r(n-f(f^{\leftarrow}(n)))+(n'-f(f^{\leftarrow}(n'))\\ &=rg^{\leftarrow}(\ell)+g^{\leftarrow}(k)\end{aligned}

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since g\circ g^{\leftarrow} is clearly 1_L one can see that \mathbf{(2)}\implies\mathbf{(3)} follows. Conversely, suppose we are given a g^{\leftarrow}:L\to N with g\circ g^{\leftarrow}=1_L. Let n\in N be arbitrary, note that

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so that n-g^{\leftarrow}(g(n))\in\ker g, but since C is exact this implies that n-g^{\leftarrow}(g(n))\in\text{im }f, and since f is a monomorphism this implies that there exists a unique m\in M with f(m)=n-g^{\leftarrow}(g(n)), so define f^{\leftarrow}(n)=m. To see that this is a morphism we note that f(m)=n-g^{\leftarrow}(g(n)) and f(m')=n'-g^{\leftarrow}(g(n')) then rf(m)+f(m')=rn-n'-g^{\leftarrow}(g(rn+n')) (where we made use of the fact that g,g^{\leftarrow} are morphisms) and so rf(m)+f(m')=f(rm+m'). Lastly, we need to prove that f^{\leftarrow}\circ f=1_M, or equivalently f(m)=f(m)-g^{\leftarrow}(g(f(m)) but since \text{im }f=\ker g this is clear. The conclusion follows.

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Finally, we prove \mathbf{(2)}\implies\mathbf{(1)}. To do this we define \phi:N\to M\oplus L by n\mapsto (f^{\leftarrow}(n),g(n)). This is a morphism since each coordinate map is a morphism. To see that \phi is a monomorphism we note that \phi(n)=\phi(n') implies f^{\leftarrow}(n)=f^{\leftarrow}(n') and g(n)=g(n'). From the second of these equalities we gather that n-n'\in\ker g and since C is exact this implies that n-n'\in\ker f and so as previously noted f(f^{\leftarrow}(n-n'))=n-n', but f(f^{\leftarrow}(n-n'))=f(f^{\leftarrow}(n)-f^{\leftarrow}(n'))=f(0)=0 and so n=n'. To see that \phi is surjective we let (m,\ell)\in M\oplus L be arbitrary. Since g is surjective we know that there exists n\in N such that g(n)=\ell and since f^{\leftarrow} is surjective (since it has a right inverse) we may find n'\in N with f^{\leftarrow}(n')=m. So, set x=n+f(f^{\leftarrow}(n'-n)). Note then that

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since C is exact. Thus, \phi(x)=(f^{\leftarrow}(x),g(x))=(m,n). Since (m,n) was arbitrary the surjectivity follows. It only remains to show that the diagram associated with this association forms a chain isomorphism (i.e. to show that the relevant diagram commutes). This amounts to showing that \phi\circ f=\iota_M\circ 1_M and 1_L\circ g=\pi_L\circ \phi. But, this is clear since, for example \phi(f(m))=(f^{\leftarrow}(f(m)),g(f(m))=(m,0)=\iota(m) and \pi_L(\phi(n))=\pi_L(f^{\leftarrow}(n),g(n))=g(n). Combining all this gives the desired conclusion. \blacksquare

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I feel like I’m obligated to comment on the motivation for the above proof. I had seen a proof the splitting lemma a while ago, and had completely forgotten it. So, I decided (as I do with most important stuff) to try to prove this on my own. I realized that if one is thinking about how to go about this proof the “wrong way” one is doomed for a headache. In particular, a lot of the above created functions (i.e. the creation of f^{\leftarrow}) seem awfully contrived. In fact, the opposite is true. Intuitively being chain isomorphic to the direct sum chain allows us to really think of f as \iota_M and g as \pi_L. So, I first found how to express something in the bottom chain and then transferred it up top, using the analogous maps (or transfer maps if necessary). For example, noting that (0,\ell)=(m,\ell)-\iota_M(\pi_M(m,\ell))  for any (m,\ell) is what allowed me to figure out that g^{\leftarrow} (which should be equal to \iota_L) should be n-f(f^{\leftarrow}(n)). I hope this makes this seeming magic-of-a-theorem less mysterious.

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[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.


November 4, 2011 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , , ,


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