Lemma 15.100.1. Let $I$ be an ideal of a Noetherian ring $A$. Let $ K \xrightarrow {\alpha } L \xrightarrow {\beta } M $ be a complex of finite $A$-modules. Set $H = \mathop{\mathrm{Ker}}(\beta )/\mathop{\mathrm{Im}}(\alpha )$. For $n \geq 0$ let

\[ K/I^ nK \xrightarrow {\alpha _ n} L/I^ nL \xrightarrow {\beta _ n} M/I^ nM \]

be the induced complex. Set $H_ n = \mathop{\mathrm{Ker}}(\beta _ n)/\mathop{\mathrm{Im}}(\alpha _ n)$. Then there are canonical $A$-module maps giving a commutative diagram

\[ \xymatrix{ & & & H \ar[lld] \ar[ld] \ar[d] \\ \ldots \ar[r] & H_3 \ar[r] & H_2 \ar[r] & H_1 } \]

Moreover, there exists a $c > 0$ and canonical $A$-module maps $H_ n \to H/I^{n - c}H$ for $n \geq c$ such that the compositions

\[ H/I^ n H \to H_ n \to H/I^{n - c}H \quad \text{and}\quad H_ n \to H/I^{n - c}H \to H_{n - c} \]

are the canonical ones. Moreover, we have

$(H_ n)$ and $(H/I^ nH)$ are isomorphic as pro-objects of $\text{Mod}_ A$,

$\mathop{\mathrm{lim}}\nolimits H_ n = \mathop{\mathrm{lim}}\nolimits H/I^ n H$,

the inverse system $(H_ n)$ is Mittag-Leffler,

the image of $H_{n + c} \to H_ n$ is equal to the image of $H \to H_ n$,

the composition $I^ cH_ n \to H_ n \to H/I^{n - c}H \to H_ n/I^{n - c}H_ n$ is the inclusion $I^ cH_ n \to H_ n$ followed by the quotient map $H_ n \to H_ n/I^{n - c}H_ n$, and

the kernel and cokernel of $H/I^ nH \to H_ n$ is annihilated by $I^ c$.

**Proof.**
Observe that $H_ n = \beta ^{-1}(I^ nM)/\mathop{\mathrm{Im}}(\alpha ) + I^ nL$. For $n \geq 2$ we have $\beta ^{-1}(I^ nM) \subset \beta ^{-1}(I^{n - 1}M)$ and $\mathop{\mathrm{Im}}(\alpha ) + I^ nL \subset \mathop{\mathrm{Im}}(\alpha ) + I^{n - 1}L$. Thus we obtain our canonical map $H_ n \to H_{n - 1}$. Similarly, we have $\mathop{\mathrm{Ker}}(\beta ) \subset \beta ^{-1}(I^ nM)$ and $\mathop{\mathrm{Im}}(\alpha ) \subset \mathop{\mathrm{Im}}(\alpha ) + I^ nL$ which produces the canonical map $H \to H_ n$. We omit the verification that the diagram commutes.

By Artin-Rees we may choose $c_1, c_2 \geq 0$ such that $\beta ^{-1}(I^ nM) \subset \mathop{\mathrm{Ker}}(\beta ) + I^{n - c_1}L$ for $n \geq c_1$ and $\mathop{\mathrm{Ker}}(\beta ) \cap I^ nL \subset I^{n - c_2}\mathop{\mathrm{Ker}}(\beta )$ for $n \geq c_2$, see Algebra, Lemmas 10.51.3 and 10.51.2. Set $c = c_1 + c_2$.

Let $n \geq c$. We define $\psi _ n : H_ n \to H/I^{n - c}H$ as follows. Say $x \in H_ n$. Choose $y \in \beta ^{-1}(I^ nM)$ representing $x$. Write $y = z + w$ with $z \in \mathop{\mathrm{Ker}}(\beta )$ and $w \in I^{n - c_1}L$ (this is possible by our choice of $c_1$). We set $\psi _ n(x)$ equal to the class of $z$ in $H/I^{n - c}H$. To see this is well defined, suppose we have a second set of choices $y', z', w'$ as above for $x$ with obvious notation. Then $y' - y \in \mathop{\mathrm{Im}}(\alpha ) + I^ nL$, say $y' - y = \alpha (v) + u$ with $v \in K$ and $u \in I^ nL$. Thus

\[ y' = z' + w' = \alpha (v) + u + z + w \Rightarrow z' = z + \alpha (v) + u + w - w' \]

Since $\beta (z' - z - \alpha (v)) = 0$ we find that $u + w - w' \in \mathop{\mathrm{Ker}}(\beta ) \cap I^{n - c_1}L$ which is contained in $I^{n - c_1 - c_2}\mathop{\mathrm{Ker}}(\beta ) = I^{n - c}\mathop{\mathrm{Ker}}(\beta )$ by our choice of $c_2$. Thus $z'$ and $z$ have the same image in $H/I^{n - c}H$ as desired.

The composition $H/I^ n H \to H_ n \to H/I^{n - c}H$ is the canonical map because if $z \in \mathop{\mathrm{Ker}}(\beta )$ represents an element $x$ in $H/I^ nH = \mathop{\mathrm{Ker}}(\beta )/\mathop{\mathrm{Im}}(\alpha ) + I^ n\mathop{\mathrm{Ker}}(\beta )$ then it is clear from the above that $x$ maps to the class of $z$ in $H/I^{n - c}H$ under the maps constructed above.

Let us consider the composition $H_ n \to H/I^{n - c}H \to H_{n - c}$. Given $x, y, z, w$ as in the construction of $\psi _ n$ above, we see that $x$ is mapped to the cass of $z$ in $H_{n - c}$. On the other hand, the canonical map $H_ n \to H_{n - c}$ from the first paragraph of the proof sends $x$ to the class of $y$. Thus we have to show that $y - z \in \mathop{\mathrm{Im}}(\alpha ) + I^{n - c}L$ which is the case because $y - z = w \in I^{n - c_1}L \subset I^{n - c}L$.

Statements (1) – (4) are formal consequences of what we just proved. Namely, (1) follows from the existence of the maps and the definition of morphisms of pro-objects in Categories, Remark 4.22.5. Part (2) holds because isomorphic pro-objects have isomorphic limits. Part (3) is immediate from part (4). Part (4) follows from the factorization $H_{n + c} \to H/I^ nH \to H_ n$ of the canonical map $H_{n + c} \to H_ n$.

Proof of part (5). Let $x \in I^ cH_ n$. Write $x = \sum f_ i x_ i$ with $x_ i \in H_ n$ and $f_ i \in I^ c$. Choose $y_ i, z_ i, w_ i$ as in the construction of $\psi _ n$ for $x_ i$. Then for the computation of $\psi _ n$ of $x$ we may choose $y = \sum f_ iy_ i$, $z = \sum f_ i z_ i$ and $w = \sum f_ i w_ i$ and we see that $\psi _ n(x)$ is given by the class of $z$. The image of this in $H_ n/I^{n - c}H_ n$ is equal to the class of $y$ as $w = \sum f_ i w_ i$ is in $I^ nL$. This proves (5).

Proof of part (6). Let $y \in \mathop{\mathrm{Ker}}(\beta )$ whose class is $x$ in $H$. If $x$ maps to zero in $H_ n$, then $y \in I^ nL + \mathop{\mathrm{Im}}(\alpha )$. Hence $y - \alpha (v) \in \mathop{\mathrm{Ker}}(\beta ) \cap I^ nL$ for some $v \in K$. Then $y - \alpha (v) \in I^{n - c_2}\mathop{\mathrm{Ker}}(\beta )$ and hence the class of $y$ in $H/I^ nH$ is annihilated by $I^{c_2}$. Finally, let $x \in H_ n$ be the class of $y \in \beta ^{-1}(I^ nM)$. Then we write $y = z + w$ with $z \in \mathop{\mathrm{Ker}}(\beta )$ and $w \in I^{n - c_1}L$ as above. Clearly, if $f \in I^{c_1}$ then $fx$ is the class of $fy + fw \equiv fy$ modulo $\mathop{\mathrm{Im}}(\alpha ) + I^ nL$ and hence $fx$ is the image of the class of $fy$ in $H$ as desired.
$\square$

## Comments (0)