Lemma 104.9.1. Let $R$ be a ring and let $I \subset R$ be a finitely generated ideal. The category of $I$-adically complete $R$-modules has kernels and cokernels but is not abelian in general.

## 104.9 The category of complete modules is not abelian

Let $R$ be a ring and let $I \subset R$ be a finitely generated ideal. Consider the category $\mathcal{A}$ of $I$-adically complete $R$-modules, see Algebra, Definition 10.95.2. Let $\varphi : M \to N$ be a morphism of $\mathcal{A}$. The cokernel of $\varphi $ in $\mathcal{A}$ is the completion $(\mathop{\mathrm{Coker}}(\varphi ))^\wedge $ of the usual cokernel (as $I$ is finitely generated this completion is complete, see Algebra, Lemma 10.95.3). Let $K = \mathop{\mathrm{Ker}}(\varphi )$. We claim that $K$ is complete and hence is the kernel of $\varphi $ in $\mathcal{A}$. Namely, let $K^\wedge $ be the completion. As $M$ is complete we obtain a factorization

Since $\varphi $ is continuous for the $I$-adic topology, $K \to K^\wedge $ has dense image, and $K = \mathop{\mathrm{Ker}}(\varphi )$ we conclude that $K^\wedge $ maps into $K$. Thus $K^\wedge = K \oplus C$ and $K$ is a direct summand of a complete module, hence complete.

We will give an example that shows that $\mathop{\mathrm{Im}}\not= \mathop{\mathrm{Coim}}$ in general. We take $R = \mathbf{Z}_ p = \mathop{\mathrm{lim}}\nolimits _ n \mathbf{Z}/p^ n\mathbf{Z}$ to be the ring of $p$-adic integers and we take $I = (p)$. Consider the map

where the left hand side is the $p$-adic completion of the direct sum. Hence an element of the left hand side is a vector $(x_1, x_2, x_3, \ldots )$ with $x_ i \in \mathbf{Z}_ p$ with $p$-adic valuation $v_ p(x_ i) \to \infty $ as $i \to \infty $. This maps to $(x_1, px_2, p^2x_3, \ldots )$. Hence we see that $(1, p, p^2, \ldots )$ is in the closure of the image but not in the image. By our description of kernels and cokernels above it is clear that $\mathop{\mathrm{Im}}\not= \mathop{\mathrm{Coim}}$ for this map.

**Proof.**
See above.
$\square$

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