The Stacks project

Proof. Assume $M/IM \to N/IN$ is surjective. Then the map $M/I^ nM \to N/I^ nN$ is surjective for each $n \geq 1$ by Nakayama's lemma. More precisely, apply Lemma 10.20.1 part (11) to the map $M/I^ nM \to N/I^ nN$ over the ring $R/I^ n$ and the nilpotent ideal $I/I^ n$ to see this. Set $K_ n = \{ x \in M \mid \varphi (x) \in I^ nN\} $. Thus we get short exact sequences

We claim that the canonical map $K_{n + 1}/I^{n + 1}M \to K_ n/I^ nM$ is surjective. Namely, if $x \in K_ n$ write $\varphi (x) = \sum z_ j n_ j$ with $z_ j \in I^ n$, $n_ j \in N$. By assumption we can write $n_ j = \varphi (m_ j) + \sum z_{jk}n_{jk}$ with $m_ j \in M$, $z_{jk} \in I$ and $n_{jk} \in N$. Hence

This means that $x' = x - \sum z_ j m_ j \in K_{n + 1}$ maps to $x \bmod I^ nM$ which proves the claim. Now we may apply Lemma 10.87.1 to the inverse system of short exact sequences above to see (1). Part (2) is a special case of (1). If the assumptions of (3) hold, then for each $n$ the sequence

is short exact by Lemma 10.39.12. Hence we can directly apply Lemma 10.87.1 to conclude (3) is true. To see (4) choose generators $x_ i \in M$, $i = 1, \ldots , n$. Then the map $R^{\oplus n} \to M$, $(a_1, \ldots , a_ n) \mapsto \sum a_ ix_ i$ is surjective. Hence by (2) we see $(R^\wedge )^{\oplus n} \to M^\wedge $, $(a_1, \ldots , a_ n) \mapsto \sum a_ ix_ i$ is surjective. Assertion (4) follows from this. $\square$

Proof. Since $I$ is finitely generated, $I^ n$ is finitely generated, say by $f_1, \ldots , f_ r$. Applying Lemma 10.96.1 part (2) to the surjection $(f_1, \ldots , f_ r) : M^{\oplus r} \to I^ n M$ yields a surjection

On the other hand, the image of $(f_1, \ldots , f_ r) : (M^\wedge )^{\oplus r} \to M^\wedge $ is $I^ n M^\wedge $. Thus $M^\wedge / I^ n M^\wedge \simeq M/I^ n M$. Taking inverse limits yields $(M^\wedge )^\wedge \simeq M^\wedge $; that is, $M^\wedge $ is $I$-adically complete. $\square$

Proof. Say $I^ c Q = 0$. Then $Q/I^ nQ = Q$ for $n \geq c$. On the other hand, it is clear that $I^ nM \subset M \cap I^ nN \subset I^{n - c}M$ for $n \geq c$. Thus $M^\wedge = \mathop{\mathrm{lim}}\nolimits M/(M \cap I^ n N)$. Apply Lemma 10.87.1 to the system of exact sequences

for $n \geq c$ to conclude. $\square$

Proof. The module $I^ n M^\wedge $ is contained in $K_ n$. Thus for each $n \geq 1$ there is a canonical exact sequence

As $I^ nM^\wedge $ maps onto $I^ nM/I^{n + 1}M$ we see that $K_{n + 1} + I^ n M^\wedge = K_ n$. Thus the inverse system $\{ K_ n/I^ n M^\wedge \} _{n \geq 1}$ has surjective transition maps. By Lemma 10.87.1 we see that there is a short exact sequence

Hence $M^\wedge $ is complete if and only if $K_ n/I^ n M^\wedge = 0$ for all $n \geq 1$. $\square$

Proof. Let $x \in R^\wedge $ map to a unit $x_1$ in $R/I$. Then $x$ maps to a unit $x_ n$ in $R/I^ n$ for every $n$ by Lemma 10.32.4. Hence $y = (x_ n^{-1}) \in \mathop{\mathrm{lim}}\nolimits R/I^ n = R^\wedge $ is an inverse to $x$. Parts (2) and (3) follow immediately from (1). Part (4) follows from (1) and Lemma 10.19.1. $\square$

Proof. Note that $\mathop{\mathrm{lim}}\nolimits M/I^ nM = \mathop{\mathrm{lim}}\nolimits M/(f_1^ n, \ldots , f_ r^ n)M$ as $I^ n \supset (f_1^ n, \ldots , f_ r^ n) \supset I^{rn}$. An element $\xi $ of $\mathop{\mathrm{lim}}\nolimits M/(f_1^ n, \ldots , f_ r^ n)M$ can be symbolically written as

with $x_{n, i} \in M$. If $M \to \mathop{\mathrm{lim}}\nolimits M/f_ i^ nM$ is surjective, then there is an $x_ i \in M$ mapping to $\sum x_{n, i} f_ i^ n$ in $\mathop{\mathrm{lim}}\nolimits M/f_ i^ nM$. Then $x = \sum x_ i$ maps to $\xi $ in $\mathop{\mathrm{lim}}\nolimits M/I^ nM$. $\square$

Proof. Assume $M$ is $J$-adically complete and $I$ is finitely generated. We have $\bigcap I^ nM = 0$ because $\bigcap J^ nM = 0$. By Lemma 10.96.7 it suffices to prove the surjectivity of $M \to \mathop{\mathrm{lim}}\nolimits M/I^ nM$ in case $I$ is generated by a single element. Say $I = (f)$. Let $x_ n \in M$ with $x_{n + 1} - x_ n \in f^ nM$. We have to show there exists an $x \in M$ such that $x_ n - x \in f^ nM$ for all $n$. As $x_{n + 1} - x_ n \in J^ nM$ and as $M$ is $J$-adically complete, there exists an element $x \in M$ such that $x_ n - x \in J^ nM$. Replacing $x_ n$ by $x_ n - x$ we may assume that $x_ n \in J^ nM$. To finish the proof we will show that this implies $x_ n \in I^ nM$. Namely, write $x_ n - x_{n + 1} = f^ nz_ n$. Then

The sum $z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots $ converges in $M$ as $f^ c \in J^ c$. The sum $f^ n(z_ n + fz_{n + 1} + f^2z_{n + 2} + \ldots )$ converges in $M$ to $x_ n$ because the partial sums equal $x_ n - x_{n + c}$ and $x_{n + c} \in J^{n + c}M$. $\square$

Proof. Consider the system of maps $M/I^ nM \to M/J^{\lfloor n/d \rfloor }M$ and the system of maps $M/J^ mM \to M/I^{\lfloor m/c \rfloor }M$ to get mutually inverse maps between the completions. $\square$

Proof. Set $N = M/K$. By Lemma 10.96.1 the map $M = M^\wedge \to N^\wedge $ is surjective. Hence $N \to N^\wedge $ is surjective. It is easy to see that the kernel of $N \to N^\wedge $ is the module $\bigcap (K + I^ nM) / K$. $\square$

Proof. By Lemma 10.96.1 the map $M = M \otimes _ R R = M \otimes _ R R^\wedge \to M^\wedge $ is surjective. The kernel of this map is $\bigcap I^ nM$ hence zero by assumption. Hence $M \cong M^\wedge $ and $M$ is complete. $\square$

Proof. Let $x_1, \ldots , x_ n \in M$ be elements whose images in $M/IM$ generate $M/IM$ as a $R/I$-module. Denote $M' \subset M$ the $R$-submodule generated by $x_1, \ldots , x_ n$. By Lemma 10.96.1 the map $(M')^\wedge \to M^\wedge $ is surjective. Since $\bigcap I^ nM = 0$ we see in particular that $\bigcap I^ nM' = (0)$. Hence by Lemma 10.96.11 we see that $M'$ is complete, and we conclude that $M' \to M^\wedge $ is surjective. Finally, the kernel of $M \to M^\wedge $ is zero since it is equal to $\bigcap I^ nM = (0)$. Hence we conclude that $M \cong M' \cong M^\wedge $ is finitely generated. $\square$