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Proposition. The category $\Cmod$ has coproducts.
Proof. Let $(A\sal,M\sal)_{\al\in\Lambda}$ be a nonempty family in $\Cmod$. Let $A'$ be the direct limit of finite tensor products of $(A\sal)\sal$, i.e., $A'=\dlim\ten_{\al\in I}A\sal$, where $I\sst\Lambda$ is a finite subset. Given $\al\in\Lambda$, let $A'\sal:=\dlim\ten_{\be\in I\sal}A\sbe$, where $I\sal\sst\Lambda\bb\set{\al}$ is a finite subset. We claim that $A'\simeq A\sal\ten A'\sal\fall\al\in\Lambda$. To see this, first note that for every $\al\in\Lambda$, there is an obvious ring \hmm\ $\tau'\sal:A'\sal\dra A'$.
This induces a ring \hmm
$$\phi:A\sal\ten A'\sal\dra A',\qqq a\sal\ten x\mt\tau_{\set{\al}}(a\sal)\tau'\sal(x),$$
where $(\tau_I)_{I\sst\Lambda}$ is the direct limit for $A'$.
On the other hand, let $(\sigma_{I\sal})_{I\sal}$ be the direct limit for $A'\sal$.
Given a finite subset $I\sst\Lambda$, define a ring \hmm
$$\psi_I:A\sal\ten A'\sal\la \ten_{\al\in I}A\sal$$
as follows: if $\al\not\in I$, then $\psi_I(x):=1_{A\sal}\ten\sigma_I(x)$. If $\al\in I$, then
$$\psi_I(\ten_{\be\in I}a\sbe)=a\sal\ten\sigma_{I'\sal}(\ten_{\be\in I'\sal}a\sbe),$$
where $I'\sal:=I\bb\set{\al}$. It is easy to see that $\psi_I$ is compatible with inclusions of the finite subsets $I$. Hence, there is an induced ring \hmm
$$\hpsi:A\sal\ten A'\sal\dla A'.$$
It is easy to see that $\phi$ and $\hpsi$ are inverses of each other.
Now let $M'\sal:=M\sal\ten A'\sal$. It is then an $A'$-module, and we will show that $(A',\opl\sal M'\sal)$ is the coproduct of $(A\sal,M\sal)\sal$.
There is an abelian group \hmm
$$j\sal:M\sal\ra M'\sal,\qqq m\sal\mt m\sal\ten1'\sal.$$
Let $(i\sal)\sal$ be the coproduct for $\opl\sal M'\sal$. It is easy to see that
$$(\tau_{\set{\al}},i\sal\cc j\sal):(A\sal,M\sal)\dra(A',\opl\sal M'\sal)$$
is a morphism in $\Cmod$.
Let $(\phi\sal,\psi\sal)\sal$ be a family of morphisms in $\Cmod$, where
$$(\phi\sal,\psi\sal):(A\sal,M\sal)\ra(B,N).$$
Then there are induced ring \hmms\ $\hphi:A'\dra B$, $\hphi\sal:A'\sal\dra B$, and a unique abelian group \hmm
$$\hpsi\sal:M'\sal\dra N$$
\st $m\sal\ten x\mt\hphi\sal(x)\cdot\psi\sal(m\sal)$. Note that $\hpsi\sal\cc j\sal=\psi\sal$.
It is easy to see that
$$(\hphi,\hpsi\sal):(A',M'\sal)\ra(B,N)$$
is a morphism of modules, so there is an induced morphism of modules
$$(\hphi,\hpsi):(A',\opl\sal M'\sal)\dra(B,N).$$
It is easy to see that $\hpsi\sal$ is the unique abelian group \hmm\ \st for every $\al\in\Lambda$, $\hpsi\sal\cc j\sal=\psi\sal$ and $(\hphi,\hpsi\sal)$ is a morphism of modules. Hence, the uniqueness of $(\hphi,\hpsi)$ follows.