We know that $A$ can be given a total order structure compatible with its group structure (Algebra, Chapter II, § 11, no. 4, Lemma 2). Let $p$ be a prime ideal associated with $M$, the annihilator of an element $x \in M$, and let $(x_i)_{i \in \Delta}$ be the family of homogeneous components of $x$; let $i(1) < i(2) < \cdots < i(r)$ be the values of $i$ for which $x_i \neq 0$. Consider an element $a \in p$ and let $(a_i)_{i \in \Delta}$ be the family of its homogeneous components; we shall prove that $a, \in p$ for all $i \in A$, which will show that $p$ is a graded ideal.

We argue by induction on the number of indices $i$ such that $a, \neq 0$. Our assertion is obvious if this number is 0; if not, let $m$ be the greatest of the indices $i$ for which $a, \neq 0$; if we prove that $a, \in p$, the induction hypothesis applied to $a - a,$ will give the conclusion. Now, $ax = 0$; for all $j \in A$, using the fact that the homogeneous component of degree $m + j$ of $ax$ is 0, we obtain

we conclude that $a_m x_j$ is a linear combination of the $x_i$ of indices $i > j$. In particular therefore $a_m x_{i(r)} = 0$, whence, by descending induction on $n < r$, $a_m^{r-n+1} x_{i(n)} = 0$. Then $a_m^r x = 0$, whence $a_m^r \in p$ and, as $p$ is prime, $a, \in p$.

We now show that $p$ is the annihilator of a homogeneous element of $M$. Let us write $b_n = \operatorname{Ann}(x_{i(n)})$ for $1 \leq n \leq r$. For every homogeneous element $b$ of $p$ and all $n$ the homogeneous component of $bx$ of degree $i(n) + \deg(b)$ is $bx_{i(n)}$, hence $bx_{i(n)} = 0$ and therefore $b \in 6$; as $p$ is generated by its homogeneous elements, $p \subset 6$. On the other hand, clearly $\bigcap_{n=1}^r 6_n \subset \operatorname{Ann}(x) = p$; as $p$ is prime, there exists an $n$ such that $b_n \subset p$ (Chapter 11, § 1, no. 1, Proposition 1), whence $b_n = P = \operatorname{Ann}(x_{i(n)})$, which completes the proof.

With the notation of the proof of Proposition 1, consider the homomorphism obtained, by taking quotients, from the homomorphism $a \mapsto ax_{i(n)}$ of $A$ to $M$; the latter is a graded homomorphism of degree $i(n)$ and hence it gives on taking quotients a graded bijective homomorphism of degree $i(n)$ of $A/p$ onto a graded submodule of $M$.

It is sufficient to retrace the argument of § 1, no. 4, Theorem 1 taking on this occasion $\mathscr{G}$ to be the set of graded submodules of $M$ with a composition series with the properties of the statement; we conclude using the Corollary to Proposition 1.

It suffices to show that $\operatorname{Ass}(M) = \{p\}$ (§ 2, no. 1, Proposition 1). Let $q$ be a prime ideal associated with $M$; it is a graded ideal and it is the annihilator of a homogeneous element $x \neq 0$ of $M$ (no. 1, Proposition 1). For every homogeneous element $a$ of $q$, $ax = 0$ and hence the homothety of ratio $a$ on $M$ is not injective, whence $a \in p$. Conversely, let $b$ be a homogeneous element of $p$; there exists an integer $n > 0$ such that $b^n x = 0$, whence $b^n \in \operatorname{Ann}(x) = q$ and, as $q$ is prime, $b \in q$. As $p$ and $q$ are generated by their respective homogeneous element, $p = q$, which proves that $\operatorname{Ass}(M) \subset \{p\}$. As $M \neq \{0\}, \operatorname{Ass}(M) \# \varnothing$ (§ 1, no. 1, Corollary 1 to Proposition 2), whence $\operatorname{Ass}(M) = \{p\}$.

We know (loc. cit.) that the homogeneous elements of $p'$ (resp. $N'$) are just the homogeneous elements of $p$ (resp. $N$). Let $a$ be a homogeneous element of $p$; if $x$ is a homogeneous element of $M$, there exists an integer $n > 0$ such that $a^n x \in N$; as $a^n x$ is homogeneous, $a^n x \in N'$; as every $y \in M$ is the direct sum of a finite number of homogeneous elements, we conclude that there exists an integer $q > 0$ such that $a^q y \in N'$, so that the homothety with ratio $a$ in $M/N'$ is almost nilpotent.

Consider now a homogeneous element $b$ of $A - p'$; then $b \notin p$ since $b$ is homogeneous. Let $x$ be an element of $M$ such that $bx \in N'$ and let $(x_i)_{i \in \Delta}$ be the family of homogeneous components of $x$. As $N'$ is graded, $bx_i \in N'$ for all $i$, hence $bx_i \in N$ and, as $b \notin p$, we conclude that $x_i \in N$; as $x_i$ is homogeneous, $x_i \in N'$, whence $x \in N'$ and the homothety with ratio $b$ on $M/N'$ is injective. Proposition 4 then follows from Proposition 3 applied to $p'$ and $M/N'$.

(i) Let $Q'_i$ be the largest graded submodule of $M$ contained in $Q_i$. Then the $Q'_i$ are primary and $N = \bigcap_{i \in I} Q'_i$.

(ii) If the primary decomposition $N = \bigcap_{i \in I} Q_i$ is reduced, so is the primary decomposition $N = \bigcap_{i \in I} Q'_i$ and for all $i \in I$ the prime ideals corresponding to $Q_i$ and $Q'_i$ are equal.

(iii) If $Q_i$ corresponds to a prime ideal $p_i$ which is a minimal element of $\operatorname{Ass}(M/N)$, $Q_i$ is a graded submodule of $M$.

We have seen (no. 2, Proposition 4) that the $Q'_i$ are primary with respect to $M$ and $N \subset Q'' \subset Q_i$, which proves (i). Proposition 4 of no. 2 also shows that the prime ideal $p'_i$ corresponding to $Q'_i$ is the largest graded ideal contained in the prime ideal $p_i$ corresponding to $Q_i$. If the decomposition $N = \bigcap_{i \in I} Q_i$ is reduced, $p_i \in \operatorname{Ass}(M/N)$ for all $i$ (§ 2, no. 3, Proposition 4), hence $p_i$ is a graded ideal (no. 1, Proposition 1) and therefore $p'_i = p_i$; then $\operatorname{Ass}(M/N) = \bigcup_{i \in I} \{p'_i\}$ (§ 2, no. 3, Proposition 4), which proves that the decomposition $N = \bigcap_{i \in I} Q'_i$ is reduced (§ 2, no. 3, Proposition 4). Finally, if $p_i$ is a minimal element of $\operatorname{Ass}(M/N)$, then $p'_i = p_i$ since $p_i$ is graded (no. 1, Proposition 1), whence $Q'_i = Q_i$ by virtue of § 2, no. 3, Proposition 5.