Completion commutes with finite direct sums I have been reading a couple of different places, that completion of module (or the $I$-adic completion to be more exact) commutes with finite direct sums. I was told, that it follows from the fact, that completion is an additive functor, but as I haven't worked with functors before, that doesn't make much sense to me, and I was wondering if it could be proven in another way (maybe from using the definition of completion only). I've been trying to work it around my self, but doesn't seem to get anywhere.
 A: Let $R$ be a ring and $I$ an ideal of $R$. Let $M$ and $N$ be two $R$-modules.
The question is to show that there is an isomorphism $$\underbrace{\left(\varprojlim M / (I^n M)\right)\oplus\left(\varprojlim N / (I^n N)\right)}_{\text{Call this $A$}}\simeq \varprojlim (M \oplus N)/ (I^n (M \oplus N))$$
One pedestrian way of showing this is by the universal property of inverse limits.
First step: we need to define projections $\pi_n:A\to (M \oplus N)/ (I^n (M \oplus N))$.
An element of $A$ is a collection of elements $x_n\in M/(I^nM)$ compatible$^{*}$ with the inverse system, together with a likewise compatible collection of elements $y_n\in N/(I^nN)$ ($^{*}$the $x_n$ are projections of each other by the relevant maps). For such an $x\oplus y\in A$ define $$\pi_n(x\oplus y)={x_n\oplus y_n}\in (M \oplus N)/ (I^n (M \oplus N)).$$
This makes sense since $$\left(M / (I^n M)\right)\oplus\left( N / (I^n N)\right)\simeq (M \oplus N)/ (I^n (M \oplus N)).$$
It is straightforward to check that the $\pi_n$ are compatible with the inverse system $(M \oplus N)/ (I^n (M \oplus N))$ (all relevant diagrams commute).
Second step: for every $R$-module $P$ and every set of maps $f_n:P\to (M \oplus N)/ (I^n (M \oplus N))$ compatible with the inverse system, there is a unique map $F: P\to A$ such that all diagrams made of $F$, $f_n$, $\pi_n$ commute.
For an element $p\in P$, consider the collection of all first coordinates of all $f_n(p)$, together with the collection of all second coordinates of the $f_n(p)$: these collections together form an element of $A$, that we call $F(p)$. It is again straightforward to check that the relevant diagrams commute.
Finally, to see that $F$ is unique assume that there is another map $F'$ satisfying the same conditions. Then for all $n$, $\pi_n \circ F=\pi_n \circ F'$. This implies that $F=F'$ because if $x\oplus y\in A$ and $x'\oplus y'\in A$ are such that for all $n$, $x_n\oplus y_n=x'_n\oplus y'_n$, then $x=x'$ and $y=y'$.
