Writing the previous post clarified my thinking and it allowed me to understand Mittag-Leffler modules better. Namely, condition (*) implies that a countably generated Mittag-Leffler module over an Artinian local ring R is a direct sum of finite R-modules. Hence an indecomposable, countably generated, not finitely generated R-module is not Mittag-Leffler.

An explicit example of this phenomenon is the following. Say R = k[a, b] where k is a field and a, b are elements with a^2 = ab = b^2 = 0 in R. Let M be the R-module generated by elements e_0, e_1, e_2, … subject to the relations b e_i = a e_{i + 1} for i ≥ 0. Then M is indecomposable as an R-module (nice exercise), hence not Mittag-Leffler. Now consider the R-algebra S = R[t]/(at – b). Then S ≅ M as R-modules via the map which sends e_i to t^i. Hence S is not Mittag-Leffler as an R-module.

Let’s return to the question I posed at the end of the previous post. Let R be a ring and S an R-algebra of finite presentation. In the Raynaud-Gruson paper they show that, *if S is also flat over R*, then the condition that S be Mittag-Leffler as an R-module is roughly a condition on the topology of the map Spec(S) —> Spec(R), namely of being “pure” which I will discuss in a future post. The simple example above shows that we cannot expect a similar result in the non-flat case. Thus, whereas I had at first thought that the Mittag-Leffler condition on S as an R-module would be a “mild” condition, now I think it is a very strong condition, and almost never satisfied in practice unless S is flat over R.