The inverse limit of modules is a module

Let R be a ring. Let I be a nonempty partially ordered set, and let \{M_i\}_I be a family of left R-modules. In a previous exercise we constructed the inverse limit \varprojlim M_i of the abelian groups M_i. Verify that \varprojlim M_i is naturally a left R-module, that the projections \mu_i : \varprojlim M_i \rightarrow M_i are module homomorphisms, and that \varprojlim M_i satisfies an appropriate universal property.


Let I be a partially ordered set and let \{M_i\}_I be a family of left R-modules. Suppose we have, for all i \leq j, an R-module homomorphism \mu_{j,i} : M_j \rightarrow M_i such that (1) \mu_{j,i} \circ \mu_{k,j} = \mu_{k,i} whenever i \leq j \leq k and (2) \mu_{i,i} = 1 (the identity map) for all i \in I.

The inverse limit of the M_i (with respect to the \mu_{i,j}) is the subset \varprojlim M_i = \{ (m_i) \in \prod_I M_i \ |\ \mu_{i,j}(a_j) = a_i\ \mathrm{whenever}\ i \leq j \}. We showed previously that \varprojlim M_i is an abelian group. In fact, we claim that \varprojlim M_i is a submodule of \prod_I M_i. To that end, let r \in R and (a_i) \in \varprojlim M_i; since \mu_{j,i}(r \cdot a_j) = r \cdot \mu_{j,i}(a_j) = r \cdot a_i whenever i \leq j, r \cdot (a_i) = (r \cdot a_i) \in \varprojlim M_i. Moreover, since \mu_{j,i}(0) = 0 for all i \leq j, 0 \in \varprojlim M_i, so that \varprojlim M_i is nonempty. Thus by the submodule criterion \varprojlim M_i is a left R-module as we would expect.

Moreover, for all i \in I, we have \mu_i(r \cdot (a_k)) = \mu_i((r \cdot a_k)) = r \cdot a_i = r \cdot \mu_i((a_k)). Thus the projections \mu_i are R-module homomorphisms. If the M_i are all unital, then \varprojlim M_i is unital since 1 \cdot (a_i) = (1 \cdot a_i) = (a_i). Certainly we have \mu_{j,i} \circ \mu_j = \mu_i for all i \leq j.

We claim that \varprojlim M_i is universal with respect to this property in the following sense. If N is a left R-module and \psi_i : N \rightarrow M_i a family of R-module homomorphisms such that, whenever i \leq j, we have \mu_{j,i} \circ \psi_j = \psi_i, then there exists a unique R-module homomorphism \Psi : N \rightarrow \varprojlim M_i such that \mu_i \circ \Psi = \psi_i for all i \in I. That is, there is a unique homomorphism \Psi such that the following diagram commutes for all i \leq j.

Universal property of inverse limits of modules

By the universal property of inverse limits of abelian groups, there exists a group homomorphism \Psi : N \rightarrow \varprojlim M_i such that \mu_i \circ \Psi = \psi_i for all i \in I. Specifically, \Psi(n) = (\psi_i(n)). It remains to be seen that \Psi is a module homomorphism. To that end, let r \in R and n \in N. Then \Psi(r \cdot n) = (\psi_i(r \cdot n)) = (r \cdot \psi_i(n)) = r \cdot (\psi_i(n)) = r \cdot \Psi(n). So \Psi is in fact a module homomorphism.

For uniqueness, suppose \Theta is another R-module homomorphism such that \mu_i \circ \Theta = \psi_i for all i. Then in particular \Theta is a group homomorphism, and by the universal property of inverse limits of abelian groups, \Theta = \Psi.

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