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 $t(n + m - 1) + 1$st 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. $\square$

Comment #945 by correction_bot on

1. In the first paragraph of the proof, it looks like the induced map $\widetilde{\phi}: \widetilde{I} \to \widetilde{M}$ should be from $\widetilde{I^n}$.

2. In the expressions and replace $x_i$ with $x_{i(E)}$.

Comment #4848 by awllower on

A minor typo: in the first line of the proof it should be $I^{n+1}\subset I^{n}$ instead of the other way around.

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