Proposition 1

Let $A$ be a torsion-free commutative group, $A$ a graded ring of type $A$ and $M$ a graded $A$-module of type $A$. Every prime ideal associated with $M$ is graded and is the annihilator of a homogeneous element of $M$.

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.