Karoubi envelope / idempotent completion of $R-Mod$

Question

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.

Answer 1

1. $R-\mathrm{Mod}$ 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.)

2. 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.

Answer 2

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.