## The direct limit of modules is a module

Let $R$ be a ring. In a previous exercise, we constructed the direct limit $\varinjlim_I M_i$ of abelian groups indexed by a directed, partially ordered set $I$. Verify that if the $M_i$ are left $R$-modules, then $\varinjlim_I M_i$ is naturally a left $R$-module, that the natural injections $M_i \rightarrow \varinjlim_I M_i$ are $R$-module homomorphisms, and that $\varinjlim_I M_i$ satisfies an appropriate universal property.

Let $I$ be a nonempty, directed, partially ordered set and let $\{M_i\}_I$ be a family of left $R$-modules indexed by $I$. Suppose that for all $i,j \in I$ with $i \leq j$, there exists an $R$-module homomorpism $\rho_{i,j} : M_i \rightarrow M_j$ such that (1) $\rho_{j,k} \circ \rho_{i,j} = \rho_{i,k}$ whenever $i \leq j \leq k$ and (2) $\rho_{i,i} = 1$ (the identity mapping) for all $i \in I$.

Let $X = \bigcup_I M_i \times \{i\}$, and define a relation $\sigma$ on $X$ by $(a,i) \sigma (b,j)$ if and only if there exists $k \geq i,j$ such that $\rho_{i,k}(a) = \rho_{j,k}(b)$.

Previously, we showed that $\sigma$ is an equivalence relation and that via the operator $[(a,i)]_\sigma + [(b,j)]_\sigma = [(\rho_{i,k}(a) + \rho_{j,k}(b),k)]$ (where $k$ is any element such that $k \geq i,j$) the set $X/\sigma$ is an abelian group which we denote $\varinjlim_I M_i$. Moreover, for each $i \in I$, the mapping $\rho_i : M_i \rightarrow \varinjlim_I M_i$ given by $\rho_i(a) = [(a,i)]_\sigma$ is a group homomorphism.

Define an action of $R$ on $\varinjlim M_i$ by $r \cdot [(a,i)]_\sigma = [(r \cdot a, i)]_\sigma$. We claim that this action is well defined and makes $\varinjlim M_i$ into a left $R$-module.

1. (Well-defined) Suppose $[(a,i)]_\sigma = [(b,j)]_\sigma$. Then there exists $k \geq i,j$ such that $\rho_{i,k}(a) = \rho_{j,k}(b)$. Now $r \cdot \rho_{i,k}(a) = r \cdot \rho_{j,k}(b)$, and since the $\rho_{i,j}$ are module homomorphisms, $\rho_{i,k}(r \cdot a) = \rho_{j,k}(r \cdot b)$. Thus $[(r \cdot a,i)]_\sigma = [(r \cdot b,j)]_\sigma$, and we have $r \cdot [(a,i)]_\sigma = r \cdot [(b,j)]_\sigma$. Hence this action is well-defined.
2. Now let $[(a,i)]_\sigma, [(b,j)]_\sigma \in \varinjlim M_i$ and let $r \in R$. Let $k \geq i,j$. Then we have $r \cdot \left( [(a,i)]_\sigma + [(b,j)]_\sigma \right) = r \cdot [(\rho_{i,k}(a) + \rho_{j,k}(b), k)]_\sigma$ $= [(r \cdot (\rho_{i,k}(a) + \rho_{j,k}(b)), k)]_\sigma$ $= [(r \cdot \rho_{i,k}(a) + r \cdot \rho_{j,k}(b), k)]_\sigma$ $= [(\rho_{i,k}(r \cdot a) + \rho_{j,k}(r \cdot b), k)]_\sigma$ $= [(r \cdot a, i)]_\sigma + [(r \cdot b, j)]_\sigma$ $= r \cdot [(a,i)]_\sigma + r \cdot [(b,j)]_\sigma$.
3. Let $[(a,i)]_\sigma \in \varinjlim M_i$ and $r,s \in R$. Then $(rs) \cdot [(a,i)]_\sigma = [(rs \cdot a, i)]_\sigma$ $= [(r \cdot (s \cdot a), i)]_\sigma$ $= r \cdot [(s \cdot a, i)]_\sigma$ $= r \cdot (s \cdot [(a,i)]_\sigma)$.
4. Let $[(a,i)]_\sigma \in \varinjlim M_i$ and let $r,s \in R$. Note that $i \geq i$. Then $(r+s) \cdot [(a,i)]_\sigma = [((r+s) \cdot a, i)]_\sigma$ $= [(r \cdot a + s \cdot a, i)]_\sigma$ $= [(r \cdot \rho_{i,i}(a) + s \cdot \rho_{i,i}(b), i)]_\sigma$ $= [(r \cdot a, i)]_\sigma + [(s \cdot a, i)]_\sigma$ $= r \cdot [(a,i)]_\sigma + s \cdot [(a,i)]_\sigma$.

So $\varinjlim M_i$ is a left $R$-module. Further, suppose that $R$ has a 1 and the $M_i$ are all unital. Then $1 \cdot [(a,i)]_\sigma = [(1 \cdot a, i)]_\sigma = [(a,i)]_\sigma$, so that $\varinjlim M_i$ is unital. Finally, note also that for all $i \in I$, $r \in R$, and $m \in M_i$, we have $r \cdot \rho_i(m) = r \cdot [(m,i)]_\sigma$ $= [(r \cdot m,i)]_\sigma$ $= \rho_i(r \cdot m)$. Thus the injections $\rho_i$ are all $R$-module homomorphisms. Certainly we have $\rho_j \circ \rho_{i,j} = \rho_i$ for all $i,j \in I$ with $i \leq j$.

We claim that $\varinjlim M_i$, together with the $\rho_i$, is universal with respect to this property in the following sense: If $N$ is a left $R$-module, and for each $i \in I$, $\psi_i : M_i \rightarrow N$ a module homomorphism, such that $\psi_j \circ \rho_{i,j} = \psi_i$ whenever $i \leq j$, then there exists a unique $R$-module homomorphism $\Psi : \varinjlim M_i \rightarrow N$ such that $\psi_i = \Psi \circ \rho_i$ for all $i$. That is, there exists a unique $R$-module homomorphism $\Psi$ such that the following diagram commutes for all $i \leq j$.

Universal property of the direct limit of modules

We will first prove existence and then uniqueness.

A group homomorphism $\Psi : \varinjlim_I M_i \rightarrow N$ given by $\Psi([(a,i)]_\sigma) = \psi_i(a)$ exists such that $\Psi \circ \rho_i = \psi_i$ for all $i \in I$ by the universal property of direct limits of abelian groups. It remains to be seen that $\Psi$ is a module homomorphism- that is, that it preserves scalars. To that end, let $r \in R$ and let $[(a,i)]_\sigma \in \varinjlim M_i$. Then $r \cdot \Psi([(a,i)]_\sigma) = r \cdot \psi_i(a) = \psi_i(r \cdot a)$ $= \Psi([(r \cdot a,i)]_\sigma) = \Psi( r \cdot [(a,i)]_\sigma)$; so $\Psi$ is indeed an $R$-module homomorphism.

For uniqueness, suppose that $\Theta : \varinjlim M_i \rightarrow N$ is another $R$-module homomorphism such that $\Theta \circ \rho_i = \psi_i$ for all $i$. In particular, $\Theta$ is a group homomorphism. By the universal property of direct limits of abelian groups, $\Theta = \Psi$, so that $\Psi$ is unique as desired.