Proposition 5

Let $A$ be a torsion-free commutative group, $A$ a graded Noetherian ring of type $A$, $M$ a graded $A$-module of type $M$, $N$ a graded submodule of $M$ and $N = \bigcap_{i \in I} Q_i$ a primary decomposition of $N$ in $M$.

(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.