Lemma 110.9.1. 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 $R = k[t]$. 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 $n$th 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. 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.
$\square$

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## Comments (1)

Comment #8739 by Paolo Lammens on