The Stacks project

Proof. We have $M \to M/I^ nM \to M_ n$. Taking the limit we get $M \to M^\wedge \to M$. Hence $M$ is a direct summand of $M^\wedge $. Since $M^\wedge $ is $I$-adically complete by Lemma 10.96.3, so is $M$. $\square$

Proof. By Lemma 10.98.1 we see that $M$ is $I$-adically complete. Since the transition maps are surjective, the maps $M \to M_ n$ are surjective. Consider the inverse system of short exact sequences

defining $N_ n$. Since $M_ n = M_{n + 1}/I^ nM_{n + 1}$ the map $N_{n + 1} + I^ nM \to N_ n$ is surjective. Hence $N_{n + 1}/(N_{n + 1} \cap I^{n + 1}M) \to N_ n/(N_ n \cap I^ nM)$ is surjective. Taking the inverse limit of the short exact sequences

we obtain an exact sequence

Since $M$ is $I$-adically complete we conclude that $\mathop{\mathrm{lim}}\nolimits N_ n/(N_ n \cap I^ nM) = 0$ and hence by the surjectivity of the transition maps we get $N_ n/(N_ n \cap I^ nM) = 0$ for all $n$. Thus $M_ n = M/I^ nM$ as desired. $\square$

Proof. Pick $r$ and homogeneous elements $x_{1, 1}, \ldots , x_{1, r} \in N_1$ of degrees $d_1, \ldots , d_ r$ generating $N_1$. Since the transition maps are surjective, we can pick a compatible system of homogeneous elements $x_{n, i} \in N_ n$ lifting $x_{1, i}$. By the graded Nakayama lemma (Lemma 10.56.1) we see that $N_ n$ is generated by the elements $x_{n, 1}, \ldots , x_{n, r}$ sitting in degrees $d_1, \ldots , d_ r$. Thus for $m \leq n$ we see that $N_ n \to N_ n/I^ m N_ n$ is an isomorphism in degrees $< \min (d_ i) + m$ (as $I^ mN_ n$ is zero in those degrees). Thus the inverse system of degree $d$ parts

stabilizes as indicated. Let $N$ be the graded $A$-module whose $d$th graded part is this stabilization. In particular, we have the elements $x_ i = \mathop{\mathrm{lim}}\nolimits x_{n, i}$ in $N$. We claim the $x_ i$ generate $N$: any $x \in N_ d$ is a linear combination of $x_1, \ldots , x_ r$ because we can check this in $N_{d - \min (d_ i), d}$ where it holds as $x_{d - \min (d_ i), i}$ generate $N_{d - \min (d_ i)}$. Finally, the reader checks that the surjective map $N/I^ nN \to N_ n$ is an isomorphism by checking to see what happens in each degree as before. Details omitted. $\square$

Proof. By convention graded rings are in degrees $\geq 0$ and graded modules may have nonzero parts of any degree, see Section 10.56. The map $\varphi $ exists because $\mathop{\mathrm{lim}}\nolimits G_ n$ is a module over $A'$ as $G_ n$ is annihilated by $I^ n$. Another useful thing to keep in mind is that we have

where a subscript ${\ }_ d$ indicates the $d$th graded part.

Injective. Let $x \in M_ d$. If $x \mapsto 0$ in $\mathop{\mathrm{lim}}\nolimits G_{n, d}$ then $x \otimes 1 = 0$ in $M \otimes _ A A'$. Then we can find a finitely generated submodule $M' \subset M$ with $x \in M'$ such that $x \otimes 1$ is zero in $M' \otimes _ A A'$. Say $M'$ is generated by homogeneous elements sitting in degrees $d_1, \ldots , d_ r$. Let $n = d - \min (d_ i) + 1$. Since $A'$ has a map to $A/I^ n$ and since $A \to A/I^ n$ is an isomorphism in degrees $\leq n - 1$ we see that $M' \to M' \otimes _ A A'$ is injective in degrees $\leq n - 1$. Thus $x = 0$ as desired.

Surjective. Let $y \in \mathop{\mathrm{lim}}\nolimits G_{n, d}$. Choose a finite sum $\sum x_ i \otimes f'_ i$ in $M \otimes _ A A'$ mapping to $y$. We may assume $x_ i$ is homogeneous, say of degree $d_ i$. Observe that although $A'$ is not a graded ring, it is a limit of the graded rings $A/I^ nA$ and moreover, in any given degree the transition maps eventually become isomorphisms (see above). This gives

Thus we can write

with $f_{i, j} \in A_ j$, $f_ i \in A_{d - d_ i}$, and $g'_ i \in A'$ mapping to zero in $\prod _{j \leq d - d_ i} A_ j$. Now if we compute $\varphi _ n(\sum _{i, j} f_{i, j}x_ i) \in G_ n$, then we get a sum of homogeneous elements of degree $< d$. Hence $\varphi (\sum x_ i \otimes f_{i, j})$ maps to zero in $\mathop{\mathrm{lim}}\nolimits G_{n, d}$. Similarly, a computation shows the element $\varphi (\sum x_ i \otimes g'_ i)$ maps to zero in $\prod _{d' \leq d} \mathop{\mathrm{lim}}\nolimits G_{n, d'}$. Since we know that $\varphi (\sum x_ i \otimes f'_ i)$ is $y$, we conclude that $\sum f_ ix_ i \in M_ d$ maps to $y$ as desired. $\square$