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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