Lemma 28.25.1. Let A be a ring. Let I \subset A be a finitely generated ideal. Let M be an A-module. Then there is a canonical map
\mathop{\mathrm{colim}}\nolimits _ n \mathop{\mathrm{Hom}}\nolimits _ A(I^ n, M) \longrightarrow \Gamma (\mathop{\mathrm{Spec}}(A) \setminus V(I), \widetilde{M}).
This map is always injective. If for all x \in M we have Ix = 0 \Rightarrow x = 0 then this map is an isomorphism. In general, set M_ n = \{ x \in M \mid I^ nx = 0\} , then there is an isomorphism
\mathop{\mathrm{colim}}\nolimits _ n \mathop{\mathrm{Hom}}\nolimits _ A(I^ n, M/M_ n) \longrightarrow \Gamma (\mathop{\mathrm{Spec}}(A) \setminus V(I), \widetilde{M}).
Proof.
Since I^{n + 1} \subset I^ n and M_ n \subset M_{n + 1} we can use composition via these maps to get canonical maps of A-modules
\mathop{\mathrm{Hom}}\nolimits _ A(I^ n, M) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ A(I^{n + 1}, M)
and
\mathop{\mathrm{Hom}}\nolimits _ A(I^ n, M/M_ n) \longrightarrow \mathop{\mathrm{Hom}}\nolimits _ A(I^{n + 1}, M/M_{n + 1})
which we will use as the transition maps in the systems. Given an A-module map \varphi : I^ n \to M, then we get a map of sheaves \widetilde{\varphi } : \widetilde{I^ n} \to \widetilde{M} which we can restrict to the open \mathop{\mathrm{Spec}}(A) \setminus V(I). Since \widetilde{I^ n} restricted to this open gives the structure sheaf we get an element of \Gamma (\mathop{\mathrm{Spec}}(A) \setminus V(I), \widetilde{M}). We omit the verification that this is compatible with the transition maps in the system \mathop{\mathrm{Hom}}\nolimits _ A(I^ n, M). This gives the first arrow. To get the second arrow we note that \widetilde{M} and \widetilde{M/M_ n} agree over the open \mathop{\mathrm{Spec}}(A) \setminus V(I) since the sheaf \widetilde{M_ n} is clearly supported on V(I). Hence we can use the same mechanism as before.
Next, we work out how to define this arrow in terms of algebra. Say I = (f_1, \ldots , f_ t). Then \mathop{\mathrm{Spec}}(A) \setminus V(I) = \bigcup _{i = 1, \ldots , t} D(f_ i). Hence
0 \to \Gamma (\mathop{\mathrm{Spec}}(A) \setminus V(I), \widetilde{M}) \to \bigoplus \nolimits _ i M_{f_ i} \to \bigoplus \nolimits _{i, j} M_{f_ if_ j}
is exact. Suppose that \varphi : I^ n \to M is an A-module map. Consider the vector of elements \varphi (f_ i^ n)/f_ i^ n \in M_{f_ i}. It is easy to see that this vector maps to zero in the second direct sum of the exact sequence above. Whence an element of \Gamma (\mathop{\mathrm{Spec}}(A) \setminus V(I), \widetilde{M}). We omit the verification that this description agrees with the one given above.
Let us show that the first arrow is injective using this description. Namely, if \varphi maps to zero, then for each i the element \varphi (f_ i^ n)/f_ i^ n is zero in M_{f_ i}. In other words we see that for each i we have f_ i^ m\varphi (f_ i^ n) = 0 for some m \geq 0. We may choose a single m which works for all i. Then we see that \varphi (f_ i^{n + m}) = 0 for all i. It is easy to see that this means that \varphi |_{I^{t(n + m - 1) + 1}} = 0 in other words that \varphi maps to zero in the t(n + m - 1) + 1st term of the colimit. Hence injectivity follows.
Note that each M_ n = 0 in case we have Ix = 0 \Rightarrow x = 0 for x \in M. Thus to finish the proof of the lemma it suffices to show that the second arrow is an isomorphism.
Let us attempt to construct an inverse of the second map of the lemma. Let s \in \Gamma (\mathop{\mathrm{Spec}}(A) \setminus V(I), \widetilde{M}). This corresponds to a vector x_ i/f_ i^ n with x_ i \in M of the first direct sum of the exact sequence above. Hence for each i, j there exists m \geq 0 such that f_ i^ m f_ j^ m (f_ j^ n x_ i - f_ i^ n x_ j) = 0 in M. We may choose a single m which works for all pairs i, j. After replacing x_ i by f_ i^ mx_ i and n by n + m we see that we get f_ j^ nx_ i = f_ i^ nx_ j in M for all i, j. Let us introduce
K_ n = \{ x \in M \mid f_1^ nx = \ldots = f_ t^ nx = 0\}
We claim there is an A-module map
\varphi : I^{t(n - 1) + 1} \longrightarrow M/K_ n
which maps the monomial f_1^{e_1} \ldots f_ t^{e_ t} with \sum e_ i = t(n - 1) + 1 to the class modulo K_ n of the expression f_1^{e_1} \ldots f_ i^{e_ i - n} \ldots f_ t^{e_ t}x_ i where i is chosen such that e_ i \geq n (note that there is at least one such i). To see that this is indeed the case suppose that
\sum \nolimits _{E = (e_1, \ldots , e_ t), |E| = t(n - 1) + 1} a_ E f_1^{e_1} \ldots f_ t^{e_ t} = 0
is a relation between the monomials with coefficients a_ E in A. Then we would map this to
z = \sum \nolimits _{E = (e_1, \ldots , e_ t), |E| = t(n - 1) + 1} a_ E f_1^{e_1} \ldots f_{i(E)}^{e_{i(E)} - n} \ldots f_ t^{e_ t}x_{i(E)}
where for each multiindex E we have chosen a particular i(E) such that e_{i(E)} \geq n. Note that if we multiply this by f_ j^ n for any j, then we get zero, since by the relations f_ j^ nx_ i = f_ i^ nx_ j above we get
\begin{align*} f_ j^ nz & = \sum \nolimits _{E = (e_1, \ldots , e_ t), |E| = t(n - 1) + 1} a_ E f_1^{e_1} \ldots f_ j^{e_ j + n} \ldots f_{i(E)}^{e_{i(E)} - n} \ldots f_ t^{e_ t}x_{i(E)} \\ & = \sum \nolimits _{E = (e_1, \ldots , e_ t), |E| = t(n - 1) + 1} a_ E f_1^{e_1} \ldots f_ t^{e_ t}x_ j = 0. \end{align*}
Hence z \in K_ n and we see that every relation gets mapped to zero in M/K_ n. This proves the claim.
Note that K_ n \subset M_{t(n - 1) + 1}. Hence the map \varphi in particular gives rise to an A-module map I^{t(n - 1) + 1} \to M/M_{t(n - 1) + 1}. This proves the second arrow of the lemma is surjective. We omit the proof of injectivity.
\square
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