Karoubi envelope / idempotent completion of $R-Mod$ I have a question about motivation for building a Karoubi envelope or idempotent completion of a category $C$. A problem in a non-complete category is that it probably contains idempotent elements (that is $e: X\to X$ with e=e^2$) which do not split.
I understand the construction but I still haven't any intuition how to think about idempotent completion. If we think about commutative algebra and consider for a ring $R$ the category $C:=R-\mathrm{Mod}$ how can I think about the idempotent completion of $C$? Is $C$ complete? I think that in this case an idempotent $e: X \to X$ can only split if $X$ is projective. On the other hand not for all idempotents $e: X \to X$ is $X$ projective. Thus $R-\mathrm{Mod}$ isn't complete? What is its completion?
I also read somewhere that if we start with the category $C:=R-\mathrm{ModFree}$ of free $R$-modules, its completion adjuncts all projective $R$-modules. Why is it neccessary? Isn't $R-\mathrm{ModFree}$ already complete? I'm still confused with the role of projective modules in this context.
 A: Suppose that $M$ is a left $R$-module for some ring $R$. Let $e:M\rightarrow M$ be an $R$-endomorphism such that $e^2 = e$. Pick $m\in M$ and write
$$m = (e(m) - m) + e(m)$$
Note that $e(m) - m \in \mathrm{Ker}(e)$ and $e(m)\in \mathrm{Im}(e)$. Moreover, if for some $m\in M$ we have $m\in \mathrm{Im}(e)\cap \mathrm{Ker}(e)$, then
$$m = e(m) = 0$$
Thus $M = \mathrm{Im}(e)\oplus \mathrm{Ker}(e)$ and hence $e$ is a split idempotent. This means that $\mathrm{Mod}(R)$ is complete.
This also answers your second question. If $F$ is a free module and $e$ is an idempotent $R$-endomorphism of $F$, then $F=\mathrm{Im}(e)\oplus \mathrm{Ker}(e)$. Hence $\mathrm{Im}(e)$ is a projective module. Now
$$(F,e)\mapsto \mathrm{Im}(e)$$
induces an equivalence of completion of free $R$-modules with the category of projective $R$-modules.
A: *

*$R-\mathrm{Mod}$ is itself idempotent complete, so its own idempotent completion: Given a module homomorphism $e:M \rightarrow M$ that is idempotent, take $N:=\mathrm{Im}\,e$, $i:N \rightarrow M$ the inclusion and let $\tilde{e}: M \rightarrow N$ denote the morphism $e$ with changed codomain. Then $\tilde{e}i=\mathrm{id}_N$, which translates to the fact that that for $x \in M$ (so that $e(x)\in N$), $e(e(x))=e(x)$. Secondly, $i\tilde{e}=e$, which is obvious (this is how $\tilde{e}$ was defined, more or less).

(Aside: Note that in the module case, you can recognize idempotents as direct summands of modules. In the case above, $\mathrm{Im}\,e$ is complemented by $\mathrm{Ker}\,e$. Conversely, given a summand $N \leq_{\oplus} M,$ the projection of $M$ onto $N$ followed by inclusion to $M$ will give you an idempotent.)


*The remark in parentheses above should also tell you why you need to add projectives to the category of free modules to get something idempotent complete: Given a projective module $P \leq_{\oplus}F$ where $F$ is a free module, the projection onto $P$ followed by inclusion of $P$ into $F$ will give you an idempotent $e: F \rightarrow F$. To have this idempotent split, you need to have the projective $P$ and its inclusion to $F$ in your category.

