Proposition 1

Let $A$ be a commutative ring, 3 an ideal of $A$ and $M$ an $A$*-module*. The following conditions are equivalent: (a) For all $n \geq 1$, $\operatorname{Tor}_1^A(A/\mathfrak{g}^n, M) = 0$. (b) For all $n \geq 1$, the canonical homomorphism

is bijective.

Moreover these conditions imply : (c) The canonical homomorphism $\gamma_M : \operatorname{gr}(A) \otimes_{\operatorname{gr}_0(A)} \operatorname{gr}_r(M) \to \operatorname{gr}(M)$ is bijective. Conversely, if $\mathfrak{g}$ is nilpotent, (c) implies (a) and (b).

The equivalence of (a) and (b) follows from the exact sequence

Consider next the diagram

where we note that $(\mathfrak{g}^n/\mathfrak{g}^{n+1}) \otimes_A (M/\mathfrak{g}M)$ is canonically identified with $(\mathfrak{g}^n/\mathfrak{g}^{n+1}) \otimes_{A/\mathfrak{g}} (M/\mathfrak{g}M)$. This diagram is commutative by definition of $\gamma_n$ and its rows are exact. If (b) holds, $0$, and $\theta_{n+1}$ are bijective and so therefore is $\gamma_n$ by definition of cokernel, hence (b) implies (c). Conversely, assuming that $\mathfrak{g}$ is nilpotent, let us show that (c) implies (b); we shall argue by descending induction on $n$, since $\mathfrak{g}^n \otimes_A M = \mathfrak{g}^n M = 0$ for $n$ sufficiently large. Suppose then that in diagram (1), $\gamma_n$ and $\theta_{n+1}$ are bijective; then so is $\theta_n$ by virtue of Chapter I, § 1, no. 4, Corollary 1 to Proposition 2.

(C) The implication (ii) $\Rightarrow$ (iv)

If (ii) holds, so does (ii') by Remark 1; Proposition 1 then shows that $\gamma_M$ is an isomorphism. On the other hand, we already know that (ii) implies (iii) and hence $M/\mathfrak{g}M$ is a flat $(A/\mathfrak{g})$-module, which completes the proof that (ii) implies (iv).

Remark (2). Proposition 1 shows that, if $\mathfrak{g}$ is nilpotent, (iv) implies (iii); taking account of Remark 1, we have therefore proved in this case that (i), (ii), (iii) and (iv) are equivalent.

(D) The equivalence (iv) $\Leftrightarrow$ (v)

For all $n \geq 1$, $M/\mathfrak{g}^n M$ has a canonical $(A/\mathfrak{g}^n)$-module structure. If it is filtered by the $(\mathfrak{g}/\mathfrak{g}^n)$-adic filtration, it is immediate that $\mathrm{gr}_m(M/\mathfrak{g}^n M) = \mathrm{gr}_m(M)$ if $m < n$ and $\mathrm{gr}_m(M/\mathfrak{g}^n M) = 0$ if $m \geq n$. For all $k \geq 1$, let $A_k = A/\mathfrak{g}^k$, $\mathfrak{g}_k = \mathfrak{g}/\mathfrak{g}^k$, $M_k = M/\mathfrak{g}^k M$; let (iv), (resp. $(v)_k$) denote the assertion derived from (iv) (resp. (v)) by replacing $A, \mathfrak{g}, M$ by $A_k, \mathfrak{g}_k, M_k$. It follows from what has just been said that (iv) is equivalent to "for all $k \geq 1$, $(iv)_k$" and obviously (v) is equivalent to "for all $k \geq 1$, $(v)_k$". Then it will suffice to establish the equivalence $(iv)_k \Leftrightarrow (v)_k$ for all $k$ or also to show that (iv) $\Leftrightarrow$ (v) when $3$ is nilpotent. Now (Remark 2) we have seen that in that case (iv) is equivalent to (i). As $M/\mathfrak{g}^n M$ is isomorphic to $M \otimes_A (A/\mathfrak{g}^n)$, (i) implies (v) (Chapter I, § 2, no. 7, Corollary 2 to Proposition 8); moreover clearly (v) then implies (i). We have therefore shown the equivalence (iv) $\Leftrightarrow$ (v) in all cases and also that of all the properties of the theorem in the case where $3$ is nilpotent.

(E) The implication (v) $\Rightarrow$ (i) when $A$ is Noetherian and $M$ ideally Hausdorff

It is sufficient to prove that for every ideal $a$ of $A$ the canonical mapping $j : a \otimes_A M \to M$ is injective (Chapter I, § 2, no. 3, Proposition 1). Let $x \in \mathrm{Ker}\, j$; $a \otimes_A M$ is Hausdorff with the $\mathfrak{g}$-adic topology, it suffices to verify that, for every integer $n > 0$, $x \in \mathfrak{g}^n(a \otimes_A M)$. Let $f : \mathfrak{g}^n a \to a$ be the canonical injection; it suffices to show that $x \in \mathrm{Im}(f \otimes 1_M)$; for if $b \in \mathfrak{g}^n$, $a \in a$ and $m \in M$, the image under $f \otimes 1$, of the element $(ba) \otimes m$ of $(\mathfrak{g}^n a) \otimes_A M$ is the element $(ba) \otimes m = b(a \otimes m)$ of $a \otimes_A M$ and hence $\mathrm{Im}(f \otimes 1_M) \subset \mathfrak{g}^n(a \otimes_A M)$. By virtue of Krull's Theorem (§ 3, no. 2, Theorem 2), there exists an integer $k$ such that $a_j = a \cap \mathfrak{g}^k \subset \mathfrak{g}^n a$; if $i : a_j \to a$ is the canonical injection, it will then be sufficient to show that $x \in \operatorname{Im}(i \otimes 1_M)$. Now, denoting by $p : a \to a/a_k$ and $h : a/a_k \to A/\mathfrak{g}^k$ the canonical mappings, there is a commutative diagram

in which the first row is exact. It suffices to prove that $x \in \operatorname{Ker}(p \otimes 1_M)$ and, as $x \in \operatorname{Ker} j$ by hypothesis, it will suffice to verify that the mapping $h \otimes 1_1$, is injective. Now, it may also be written (Algebra, Chapter II, § 3, no. 6, Corollary 3 to Proposition 6)

and, as $h$ is injective and, by (v), $M/\mathfrak{g}^k M$ is a flat $(A/\mathfrak{g}^k)$-module, this completes the proof.