Direct Limit of Both Rings and Their Modules

commutative-algebralimits-colimitsmodules

Suppose we have a directed set $\langle I,\leq\rangle$, with a direct system $\langle A_i,f_{ij}\rangle$ of rings and a direct system $\langle M_i,g_{ij}\rangle$ of abelian groups, such that each $M_i$ is an $A_i$ module via $h_i:A_i\times M_i\to M_i$. Then suppose that these actions are compatible with the direct systems, so $$g_{ij}(h_i(a,m))=h_j(f_{ij}(a),g_{ij}(m))$$
Then do we have an action $$\mathop{\lim_{\longrightarrow}}A_i\times\mathop{\lim_{\longrightarrow}}M_i\to\mathop{\lim_{\longrightarrow}}M_i$$ which is determined by these systems?

Here is what I have tried so far:

Let $$\begin{align*}A&=\mathop{\lim_{\longrightarrow}}A_i\\
M&=\mathop{\lim_{\longrightarrow}}M_i\end{align*}$$

Then does the following define an action $h:A\times M\to M$?


Any $a\in A$, $m\in M$ must have some representative $a_i\in A_i$, $m_j\in M_j$. If $i\leq j$, set $a_j=f_{ij}(a_i)$, if $j\leq i$ then set $m_i=g_{ji}(m_j)$. In either case, we may now assume we have $a_k\in A_k$ and $m_k\in M_k$. Then define $h(a,m)$ to be $h_k(a_k,m_k)$ modulo the equivalence relation on the direct limit.


If so, is there a "nice" way of showing this? I'm struggling at the moment fiddling with the definitions directly, I was hoping for something possibly using the universal properties of colimits, but since we are working in two different categories I can't immediately see how to do this.

Any help would be much appreciated.

Best Answer

You can work in a monoidal category if you add some assumptions on how $\otimes$ behaves with respect to direct limits. In the specific situation of rings and modules, you want to work in the monoidal category $\mathbf{Ab}$ with tensor product the usual tensor product $\otimes_{\mathbb Z}$.

Indeed in this situation a ring is just a monoid in this category, and modules over this ring are just modules over this monoid.

So your question becomes : let $(C,\otimes)$ be a monoidal category, $(A_i)$ a directed system of monoids, $(M_i)$ a directed system such that $M_i$ is an $A_i$-module in a compatible way, that is, the following diagrams commute whenever $i\leq j$ : $$\require{AMScd}\begin{CD}A_i\otimes M_i @>>> M_i\\ @VVV @VVV \\ A_j\otimes M_j @>>> M_j\end{CD}$$

Assume that both direct limits (which are actually colimits) exist, let's call them $A,M$. Under what conditions is there an $A$-module structure on $M$ that is compatible with our direct system ?

First of all, one has to see when $A$ carries a monoid structure. That is, we want a map $A\otimes A\to A$. It's not clear how to do that, except if $\otimes$ commutes with direct colimits. In this case, what we want amounts to a map $\varinjlim_{(i,j)\in I\times I} A_i\otimes A_j \to A$, and then note that $\Delta = \{(i,i)\in I^2\}$ is cofinal in $I^2$ so we just need a map $\varinjlim_i A_i\otimes A_i\to A$ and this is easy to find, and it's easy to check that we get a monoid in the end.

So with this extra assumption, $A$ has a monoid structure. When can we then get an $A$-module structure on $M$ ? Well our assumption is enough : we want a map $A\otimes M\to M$, which again by the extra assumption boils down to a map $\varinjlim_i A_i\otimes M_i \to M$, which is again easy to find, and to check that this does provide a module structure on $M$; which is quite obviously compatible with the module structures on $M_i$.

Now how does this relate to your example ? Well as I said, a ring is nothing but a monoid in $(\mathbf{Ab},\otimes_\mathbb Z)$, and over such a ring a module is well just a module.

Moreover, since everything is nice and algebraic, one can check that the various colimits over a directed system computed in the various categories all agree (that is, if you compute $\varinjlim_i A_i$ in $\mathbf{Ring,Ab, Set}$, you get the same result up to a forgetful functor; and if you compute $\varinjlim_i M_i$ in $\mathbf{Ab,Set}$, the same thing happens), therefore there is no need to worry about that.

Finally, $\otimes_\mathbb Z$ in $\mathbf{Ab}$ does preserve directed colimits, in fact it preserves all colimits, as it is a left adjoint, so the extra assumption is satisfied.

You could probably also do this directly in $\mathbf{Set}$, because $\times$ does commute with directed colimits for the same reasons, but it would be longer and more tedious because a ring is not "just a monoid in $(\mathbf{Set},\times)$", you have more things to write down.

This is the (a) categorical approach, but of course in such a situation it can also be nice to have a more down-to-earth description - and here it's particularly easy to find :

if you have $x\in A, m\in M$, then $x$ is in the image of $A_i$ for some $i$, $m$ in the image of $M_j$ for some $j$; take $k\geq i,j$ then $x$ is in the image of $A_k$, $m$ in the image of $M_k$, you can then compute $xm$ in $M_k$ and take its image in $M$. You easily check that this doesn't depend on any of the choices made because of the compatibility condition (this is exactly what we're doing when defining the module structure abstractly via the colimit - the thing I haven't checked is that everything is compatible, but that's easy)

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