Derived complete modules

Example due to Bhargav Bhatt

Here's one attempted example showing that filtered colimits are not
 left exact. It's too simple, so I'm afraid I missed something!

Consider R = Z with I = (p). One has inclusions f_n:Z/p -----> Z/p^n
of I-complete R-modules given by multiplication by p^n. There is a
natural map from f_n to f_{n+1} given by the identity on the first
factor, and 'p' on the second. Now take the filtered colimit of these
inclusions in I-complete R-modules to get Z/p -----> 0 (as the derived
completion of Q_p/Z_p, as a complex, is Z_p[1]), which is not an
inclusion in this category.

Best wishes,
Bhargav