## 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, $\Theta = \Psi$.