Adic completion is exact on finite modules over a Noetherian ring

For each $n\ge 0$, the given short exact sequence induces an exact sequence $0\to M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{n}M)\to M\mathrm{/}{I}^{n}M\to M^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}{I}^n{M}^{\mathrm{\prime }\mathrm{\prime }}\to 0.$ The first map is injective because its kernel is $(M^{\mathrm{\prime }}\cap I^{n}M)\mathrm{/}(M^{\mathrm{\prime }}\cap I^{n}M)$, and the second map is surjective because every class in $M^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}{I}^n{M}^{\mathrm{\prime }\mathrm{\prime }}$ lifts to a class in $M\mathrm{/}{I}^{n}M$.

The left inverse system $M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{n}M)$ has surjective transition maps, hence is Mittag-Leffler. Indeed, if $n\ge m$ and $x\in M^{\mathrm{\prime }}$, then the class of $x$ in $M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{m}M)$ is the image of the class of the same $x$ in $M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{n}M)\mathrm{.}$ Therefore every transition map $M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{n}M)\to M^{\mathrm{\prime }}\mathrm{/}(M^{\mathrm{\prime }}\cap I^{m}M)$ is surjective.

Taking inverse limits in the sequences of step 1.1 and using [L1] yields an exact sequence

By [L2], the middle and right inverse limits are $\widehat{M}$ and $\widehat{M^{\mathrm{\prime }\mathrm{\prime }}}$. For the left inverse limit, the induced filtration $M^{\mathrm{\prime }}\cap I^{n}M$ on $M^{\mathrm{\prime }}$ is equivalent to the intrinsic $I$-adic filtration $I^n{M}^{\mathrm{\prime }}$ by The filtration induced on a submodule is equivalent to its intrinsic ideal-adic filtration, so its completion is $\widehat{M^{\mathrm{\prime }}}$.

Substituting these identifications into step 3.1 gives $0\to \widehat{M^{\mathrm{\prime }}}\to \widehat{M}\to \widehat{M^{\mathrm{\prime }\mathrm{\prime }}}\to 0,$ which is the claimed exactness.