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.

We shall see more generally that for every finitely generated $A$-module $N$, $N \otimes_A M$ is Hausdorff with the $3$-adic topology. For $N_{(B)} = N \otimes_A B$ is a finitely generated $B$-module and the $B$-module $N \otimes_A M$ is canonically identified with $N_{,,} \otimes_B M$ by virtue of the associativity of the tensor product. Let $\mathfrak{L}$ be the Jacobson radical of $B$; as $\mathfrak{g} B$ is contained in $\mathfrak{L}$, the $\mathfrak{g}$-adic topology on $N \otimes_A M$ is therefore identified with a finer topology than the $\mathfrak{L}$-adic topology on $N_{(B)} \otimes_B M$; but this latter topology is Hausdorff since $N_{,,} \otimes_B M$ is a finitely generated $B$-module (no. 1, Example 1), whence the conclusion.

Condition (a) implies (b) by Chapter I, §2, no. 7, Corollaries 2 and 3 to Proposition 8 and the fact that $M/\mathfrak{g} M$ is isomorphic to $M \otimes_B (B/\mathfrak{g} B)$. Suppose condition (b) holds; to show that $M$ is a flat $B$-module, we shall apply Theorem 1 of no. 2 with $A$ replaced by $B$ and $\mathfrak{g}$ by $\mathfrak{g} B$. It will therefore be sufficient to show that the canonical mapping $f : \mathfrak{J}B \otimes_B M \to \mathfrak{J}M$ is injective. Let $f_1$ be the canonical mapping $\mathfrak{J} \otimes_A B \to \mathfrak{J}B$ and $f_2$ the canonical isomorphism $\mathfrak{J} \otimes_A M \to (\mathfrak{J} \otimes_A B) \otimes_B M ; f \circ (f_1 \circ 1_M) \circ f_2$ is the canonical mapping $f' : \mathfrak{J} \otimes_A M \to \mathfrak{J}M$, as is easily verified. Now $f'$ is an isomorphism since $M$ is a flat $A$-module, whilst $f_1$ is an isomorphism because $B$ is flat over $A ; f$ is then an isomorphism.

Let $p : A/\mathfrak{J} \to B/\mathfrak{J}B$ be the canonical homomorphism; the $(A/\mathfrak{J})$-module structure on $M/\mathfrak{J}M$ derived by means of $p$ from its $(B/\mathfrak{J}B)$-module structure is isomorphic to that on $M \otimes_A (A/\mathfrak{J})$. Then it follows that, if $M$ is a flat $A$-module, $M/\mathfrak{J}M$ is a flat $(A/\mathfrak{J})$-module and hence also a flat $(B/\mathfrak{J}B)$-module if $p$ is an isomorphism; we have thus proved that (c) $\Rightarrow$ (b) in that case.

We know in fact that $\hat{A}$ is a Noetherian ring ($§ 3$, no. 4, Proposition 8) and a flat $A$-module ($§ 3$, no. 4, Theorem 3), that $\mathfrak{J} \hat{A} = \hat{\mathfrak{J}}$ ($§ 2$, no. 12, Proposition 16) and that the canonical homomorphism $A/\mathfrak{J} \to \hat{A}/\hat{\mathfrak{J}}$ is bijective ($§ 2$, no. 12, Proposition 15); Proposition 3 can therefore be applied.

As $B$ with the $2$-adic topology is a Zariski ring, $\hat{B}$ is a faithfully flat $B$-module ($§ 3$, no. 5, Proposition 9) and $M$ is canonically isomorphic to $M \otimes_B \hat{B}$ ($§ 3$, no. 4, Theorem 3); it is immediately verified that this canonical isomorphism is an isomorphism of the $A$-module structure on $\hat{M}$ onto the $A$-module structure on $M \otimes_B \hat{B}$ derived from that on $M$. Applying Proposition 4 of Chapter I, $§ 3$, no. 2 with $R$ replaced by $B$, $S$ by $A$, $E$ by $\hat{B}$, $F$ by $M$, we see that for $M$ to be a flat $A$-module, it is necessary and sufficient that $\hat{M}$ be a flat $A$-module. Moreover, $\hat{M}$ is a finitely generated $\hat{B}$-module and $\mathfrak{J} \hat{B}$ is contained in $\mathfrak{Q} = \mathfrak{Q} \hat{B}$ and hence in the Jacobson radical of $\hat{B}$ ($§ 3$, no. 4, Proposition 8); therefore $\hat{M}$ is an ideally Hausdorff $A$-module with respect to $\mathfrak{J} \hat{A}$ (Proposition 2). Conditions (b) and (c) are therefore equivalent by the Corollary to Proposition 3.