Proposition 9

Let A be a local ring, m its maximal ideal, M, N two A-modules and u : N \to M an A-homomorphism. M and N are given the m-adic filtrations and suppose that: (1) M satisfies property (GR); (2) \operatorname{gr}0(u) : \operatorname{gr}_0(N) \to \operatorname{gr}_0(M) is injective. Then \operatorname{gr}(u) : \operatorname{gr}(N) \to \operatorname{gr}(M) is injective, N and P = \operatorname{Coker}(u) satisfy property (GR) and $m^n N = \bar{u}^{-1}(m^n M)$ for every integer $n > 0$.*

It is immediately verified that the diagram

is commutative. As $\mathrm{gr}_0(A) = A/m$ is a field, the hypothesis implies that $1 \otimes \mathrm{gr}_0(u)$ is injective; as by hypothesis $\gamma_M$ is injective, so is $\gamma_M \circ (1 \otimes \mathrm{gr}_0(u))$. This implies first that $\gamma_N$ is injective and hence bijective and therefore that $\mathrm{gr}(u)$ is injective. The formula $\tilde{u}^1(m^nM) = m^nN$ is then a consequence of Theorem 1 (i).

Also, let us write $N' = u(N)$ and let $j : N' \to M$ be the canonical injection. If $p : M \to P = M/N'$ is the canonical homomorphism, then in the commutative diagram

the lower row is exact (no. 4, Proposition 2) and so is the upper row by virtue of Proposition 2 of no. 4 and the fact that $\mathrm{gr}_0(A)$ is a field. Moreover, $\mathrm{gr}(j)$ is injective (no. 4, Proposition 2) and hence $\mathrm{gr}_0(j)$ is injective, The first part of the argument applied to $j$ shows that $\gamma_{N'}$ is bijective; as $\gamma_M$ is also bijective by hypothesis, we conclude that $\gamma_P$ is bijective (Chapter I, § 1, no. 4, Corollary 2 to Proposition 2).