Lemma 10.98.1. Let I \subset A be a finitely generated ideal of a ring. Let (M_ n) be an inverse system of A-modules with I^ n M_ n = 0. Then M = \mathop{\mathrm{lim}}\nolimits M_ n is I-adically complete.
10.98 Taking limits of modules
In this section we discuss what happens when we take a limit of modules.
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
Lemma 10.98.2. Let I \subset A be a finitely generated ideal of a ring. Let (M_ n) be an inverse system of A-modules with M_ n = M_{n + 1}/I^ nM_{n + 1}. Set M = \mathop{\mathrm{lim}}\nolimits M_ n. Then M/I^ nM = M_ n and M is I-adically complete.
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
0 \to N_ n \to M \to M_ n \to 0
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
0 \to N_ n/(N_ n \cap I^ nM) \to M/I^ nM \to M_ n \to 0
we obtain an exact sequence
0 \to \mathop{\mathrm{lim}}\nolimits N_ n/(N_ n \cap I^ nM) \to M^\wedge \to M
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
Lemma 10.98.3. Let A be a Noetherian graded ring. Let I \subset A_+ be a homogeneous ideal. Let (N_ n) be an inverse system of finite graded A-modules with N_ n = N_{n + 1}/I^ n N_{n + 1}. Then there is a finite graded A-module N such that N_ n = N/I^ nN as graded modules for all n.
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
\ldots = N_{2 + d - \min (d_ i), d} = N_{1 + d - \min (d_ i), d} = N_{d - \min (d_ i), d} \to N_{-1 + d - \min (d_ i), d} \to \ldots
stabilizes as indicated. Let N be the graded A-module whose dth 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
Lemma 10.98.4. Let A be a graded ring. Let I \subset A_+ be a homogeneous ideal. Denote A' = \mathop{\mathrm{lim}}\nolimits A/I^ n. Let (G_ n) be an inverse system of graded A-modules with G_ n annihilated by I^ n. Let M be a graded A-module and let \varphi _ n : M \to G_ n be a compatible system of graded A-module maps. If the induced map
\varphi : M \otimes _ A A' \longrightarrow \mathop{\mathrm{lim}}\nolimits G_ n
is an isomorphism, then M_ d \to \mathop{\mathrm{lim}}\nolimits G_{n, d} is an isomorphism for all d \in \mathbf{Z}.
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
\bigoplus \nolimits _{d \in \mathbf{Z}} \mathop{\mathrm{lim}}\nolimits G_{n, d} \subset \mathop{\mathrm{lim}}\nolimits G_ n \subset \prod \nolimits _{d \in \mathbf{Z}} \mathop{\mathrm{lim}}\nolimits G_{n, d}
where a subscript {\ }_ d indicates the dth 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
A = \bigoplus \nolimits _{d \geq 0} A_ d \subset A' \subset \prod \nolimits _{d \geq 0} A_ d
Thus we can write
f'_ i = \sum \nolimits _{j = 0, \ldots , d - d_ i - 1} f_{i, j} + f_ i + g'_ i
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
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