Proposition 7

*Suppose that $\Delta = \mathbf{Z}$. With the hypotheses on $\mathbf{A}$ and $\mathbf{M}$ of Proposition 6, consider the graded $\mathbf{A}_0$-module $\mathbf{N} = \mathbf{M}/\mathfrak{J}_0 \mathbf{M}$ and suppose the following conditions hold: (i) each of the $N_\lambda$ considered as an $\mathbf{A}_0$-module admits a basis $(y_{i\lambda})_{i \in I_\lambda}$; (ii) the canonical homomorphism $\mathfrak{J}_0 \otimes_{\mathbf{A}} \mathbf{M} \to \mathbf{M}$ is injective.

Then $\mathbf{M}$ is a graded free $\mathbf{A}$-module (no. 2, Remark 3) and, to be precise, if $x_{i\lambda}$ is an element of $M_\lambda$ whose image in $N_\lambda$ is $y_{i\lambda}$, the family $(x_{i\lambda})_{(i, \lambda) \in I}$ (where $I$ is the set the sum of the $I_\lambda$) is a basis of $\mathbf{M}$.*

We know (no. 2, Remark 3) that there is a graded free $\mathbf{A}$-module $\mathbf{L}$ (of graduation $(L_\lambda)$) and a surjective homomorphism $p : \mathbf{L} \to \mathbf{M}$ of degree 0 such that $p(e_{i\lambda}) = x_{i\lambda}$ for all $(i, \lambda) \in I$ ($(e_{i\lambda})_{(i, \lambda) \in I}$ being a basis of $\mathbf{L}$ consisting of homogeneous elements $e_{i\lambda} \in L_\lambda$). It follows from the above Remark that $p$ is surjective. Consider the graded $\mathbf{A}$-module $\mathbf{R} = \mathrm{Ker}(p)$ and note that $R_\lambda = \{0\}$ for $\lambda < \lambda_0$ by definition; we need to prove that $\mathbf{R} = \{0\}$ and by Proposition 6 it will suffice to show that $\mathfrak{J}_0 R = R$. Consider the commutative diagram (§ 3, no. 6, Proposition 5)

where $j$ is the canonical injection, $a, b, c$ deriving from the canonical injection $\mathfrak{J}_0 \to A$ (§ 3, no. 4, Proposition 4); it must be shown that $a$ is surjective. Note that, as $L$ is free, $b$ is injective (§ 3, no. 7, Corollary 6 to Proposition 7) and $c$ is injective by hypothesis. Then let $t$ be an element of $R$ and $t$ its class in $R / \mathfrak{J}_0 R$; then there is an exact sequence (§ 3, no. 6, Proposition 5 and Corollary 2 to Proposition 6)

where $j$ and $\bar{p}$ derive from $j$ and $p$ when passing to the quotients and $\bar{p}$ is by hypothesis a bijection; then $j(\bar{t}) = 0$, in other words $j(t) \in \mathfrak{J}_0 L$. Then there is an element $z \in \mathfrak{J}_0 \otimes L$ such that $j(t) = b(z)$; as $p(b(z)) = 0$, $c((1 \otimes p)(z)) = 0$ and, as $c$ is injective, $(1 \otimes p)(z) = 0$. In other words, $z$ is the image of an element $t' \in \mathfrak{J}_0 \otimes R$ under $1 \otimes j$ and then $j(a(t')) = b(z) = j(t)$; as $j$ is injective, this implies $t = a(t')$.

We shall show later (Commutative Algebra, II, § 3, no. 2, Proposition 5) how this proposition can be extended to non-graded modules.