Abstract Nonsense

Crushing one theorem at a time

Tensor Products Naturally Commute with Direct Limits

Point of Post: In this post we give a proof that, roughly, \varinjlim (M_\alpha\otimes_R N)\cong (\varinjlim M_\alpha)\otimes_R N, and moreover we show that this isomorphism is natural.

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In this post we begin the long succession of instances where the adjointness of Hom and tensor in conjunction with Yoneda’s lemma will be useful. In particular, we will show that tensor and direct limits naturally commute, a fact that shall be supremely useful in calculations involving direct limits (most prominently, coproducts).

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Tensor Commutes with Direct Limits

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Suppose that we have a directed system of right (A,R)-bimodules \left(\{M_\alpha\},\{f_{\alpha,\beta}\}\right) over some directed set (\mathcal{A},\leqslant) where, considering our recent discussion of directed systems as functors, we can consider this as a functor F:\mathcal{A}\to(A,R)\text{-}\mathbf{Mod}. Suppose we then had some (R,B)-bimodule N. Then, post-composing F with the tensor functor \bullet\otimes_R N gives us a functor \mathcal{A}\to (A,B)\text{-}\mathbf{Mod} which, upon inspection, looks like \left(\{M_\alpha\otimes_R N\},\{f_{\alpha,\beta}\otimes 1_N\}\right). What we’d now like to prove is that, as (A,B)-bimodules, \left(\varinjlim M_\alpha\right)\otimes_R N is isomorphic to \varinjlim (M_\alpha\otimes_R N) (where we are taking the direct limit over two different directed systems). The idea will go as follows. We’ll basically show using the adjointness of Hom and tensor and previous results on the commuting of Hom and direct limits that the covariant Hom functors associated to \left(\varinjlim M_\alpha\right)\otimes_R N and \varinjlim\left(M_\alpha\otimes_R N\right) are naturally isomorphic, and then appeal to Yoneda’s lemma to conclude that they are naturally isomorphic.

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Theorem: Let (\{M_\alpha\},\{f_{\alpha,\beta}\}) be a directed system of (A,R)-bimodules and N an (R,B)-bimodule. Then, if (\{M_\alpha\otimes_R N\},\{f_{\alpha,\beta}\otimes 1_N\}), denoted by \mathscr{D},  is the above discussed directed system of (A,B)-bimodules, then 

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\varinjlim (M_\alpha\otimes_R N)\cong\left(\varinjlim M_\alpha\right)\otimes_R N

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Proof: We note that for any (A,B) L we have the following set of natural isomorphism

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\begin{aligned}\text{Hom}_{(A,B)}\left(\left(\varinjlim M_\alpha\right)\otimes_R N,L\right) &\cong\text{Hom}_{(R,B)}\left(N,\text{Hom}_A(\varinjlim M_\alpha,L)\right)\\ &\cong\text{Hom}_{(R,B)}\left(N,\varprojlim\text{Hom}_A(M_\alpha,L)\right)\\ &\cong\varprojlim\text{Hom}_{(R,B)}\left(N,\text{Hom}_A(M_\alpha,L)\right)\\ &\cong \varprojlim\text{Hom}_{(A,B)}\left(M_\alpha\otimes_R N,L\right)\\ &=\text{Hom}_{(A,B)}\left(\varinjlim\left(M_\alpha\otimes_R N\right),L\right)\end{aligned}

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and so by Yoneda’s lemma we may conclude that \left(\varinjlim M_\alpha\right)\otimes_R N\cong\varinjlim\left(M_\alpha\otimes_R N\right) as desired. \blacksquare

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Using the exact same method and the analogous definitions one can prove that

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Theorem: Let \left\{N_\alpha\right\}_{\alpha\in\mathcal{A}} be a directed system of left (R,B)-bimodules and M a right (A,R)-bimodule. Then, if one turns \{M\otimes_R N_\alpha\} into a direct system in the obvious way then one has that M\otimes_R\left(\varinjlim M_\alpha\right)\cong\varinjlim\left(M\otimes_R N_\alpha\right).

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Ok, so the above is cool, for sure, but we always need to keep a little grounded. So, what does the above actually tell us for some particular direct limits? In particular, the most important direct limit, coproducts. To figure out what the above says about coproducts let’s simplify things and suppose we are dealing with some commutative unital ring R. Then, if we consider a set \left\{M_\alpha\right\}_{\alpha\in\mathcal{A}} of R-modules (thought of as (R,R)-bimodules in the obvious way) and an R-module N (same thing) then the above tells us that as R-modules we have that

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\displaystyle \left(\bigoplus_{\alpha\in\mathcal{A}}M_\alpha\right)\otimes_R N\cong\bigoplus_{\alpha\in\mathcal{A}}\left(M_\alpha\otimes_R N\right)

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but what does this isomorphism look like? In particular what do the simple tensors on the left hand side map to? It’s a little hard to see this, given all the abstract nonsense, but one can prove that under the (unique!) isomorphism as described above one has that

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(m_\alpha)\otimes n\mapsto (m_\alpha\otimes n)

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which is exactly what we’d expect.

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Note that the above completely tells us how tensoring modules with free modules works. Indeed, if we consider R^{\oplus\lambda} to be an (R,R)-bimodule in the obvious way (i.e. that r(r_\alpha)s=(rr_\alpha s)) then we have the following theorem:

Theorem: Let M be an (A,R)-bimodule, then M\otimes_R R^{\oplus\lambda}\cong M^{\oplus\lambda} as (A,R)-bimodules in such a way that m\otimes(r_j)\mapsto (mr_j).

Proof: We know that

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M\otimes_R R^{\oplus\lambda}\cong\left(M\otimes_R R\right)^{\oplus\lambda}\cong M^{\oplus\lambda}

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where, following how simple tensors move, we have that

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m\otimes(r_j)\mapsto (m\otimes r_j)\mapsto (mr_j)


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Obviously the same thing holds for on the right.

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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.

[4] Lang, Serge. Algebra. Reading, MA: Addison-Wesley Pub., 1965. Print.

[5] Grillet, Pierre A. Abstract Algebra. New York: Springer, 2007. Print.


January 19, 2012 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , ,


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