**Proof.**
Setting $M = N/K$ we find that part (1) follows from part (2). Let $0 \to K \to N \to M \to 0$ be as in (2). For each $n$ we get the short exact sequence

\[ 0 \to K/(I^ nN \cap K) \to N/I^ nN \to M/I^ nM \to 0. \]

By Lemma 10.87.1 we obtain the exact sequence

\[ 0 \to \mathop{\mathrm{lim}}\nolimits K/(I^ nN \cap K) \to N^\wedge \to M^\wedge \to 0. \]

By the Artin-Rees Lemma 10.51.2 we may choose $c$ such that $I^ nK \subset I^ n N \cap K \subset I^{n-c} K$ for $n \geq c$. Hence $K^\wedge = \mathop{\mathrm{lim}}\nolimits K/I^ nK = \mathop{\mathrm{lim}}\nolimits K/(I^ nN \cap K)$ and we conclude that (2) is true.

Let $M$ be as in (3) and let $0 \to K \to R^{\oplus t} \to M \to 0$ be a presentation of $M$. We get a commutative diagram

\[ \xymatrix{ & K \otimes _ R R^\wedge \ar[r] \ar[d] & R^{\oplus t} \otimes _ R R^\wedge \ar[r] \ar[d] & M \otimes _ R R^\wedge \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K^\wedge \ar[r] & (R^{\oplus t})^\wedge \ar[r] & M^\wedge \ar[r] & 0 } \]

The top row is exact, see Section 10.39. The bottom row is exact by part (2). By Lemma 10.96.1 the vertical arrows are surjective. The middle vertical arrow is an isomorphism. We conclude (3) holds by the Snake Lemma 10.4.1.
$\square$

## Comments (4)

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