Lemma 110.9.1.slogan Completion is not an exact functor in general; it is not even right exact in general. This holds even when I is finitely generated on the category of finitely presented modules.
110.9 Completion is not exact
A quick example is the following. Suppose that R = k[t]. Denote M^\wedge the completion of an R-module with respect to I = (t). Let P = K = \bigoplus _{n \in \mathbf{N}} R and M = \bigoplus _{n \in \mathbf{N}} R/(t^ n). Then there is a short exact sequence 0 \to K \to P \to M \to 0 where the first map is given by multiplication by t^ n on the nth summand. We claim that 0 \to K^\wedge \to P^\wedge \to M^\wedge \to 0 is not exact in the middle. Namely, \xi = (t^2, t^3, t^4, \ldots ) \in P^\wedge maps to zero in M^\wedge but is not in the image of K^\wedge \to P^\wedge , because it would be the image of (t, t, t, \ldots ) which is not an element of K^\wedge .
A “smaller” example is the following. In the situation of Lemma 110.8.1 the short exact sequence 0 \to J \to R \to R/J \to 0 does not remain exact after completion with respect to I. Namely, if f \in J is a generator, then f : R \to J is surjective, hence R \to J^\wedge is surjective, hence the image of J^\wedge \to R is (f) = J but the fact that R/J is noncomplete means that the kernel of the surjection R \to (R/J)^\wedge is strictly bigger than J, see Algebra, Lemmas 10.96.1 and 10.96.10. By the same token the sequence R \to R \to R/(f) \to 0 does not remain exact on completion.
Proof. See discussion above. \square
Comments (2)
Comment #8739 by Paolo Lammens on
Comment #9338 by Stacks Project on