Proposition 6

Let d be an integer $\geq 1$.

(i) For every essential graded ideal p of A, $p \cap A^{(d)}$ is an essential graded prime ideal of $A^{(d)}$.

(ii) Conversely, for every essential graded prime ideal $p'$ of $A^{(d)}$, there exists a unique (necessarily essential) graded prime ideal $p$ of A such that $p \cap A^{(d)} = p'$.

(i) If $a \in A_k$ does not belong to $p_k$, $a^{kd}$ does not belong to $p_{kd}$ and hence $p \cap A^{(d)}$ is essential.

(ii) For all $n \geq 0$, the set $p \cap A$, must be equal to the set $a_n$ of $x \in A$, such that $x^d \in p'$. Let us show that $a = \bigoplus_{n \geq 0} a_n$ is a graded prime ideal; as $a_n = p'_n$ when $n$ is a multiple of $d$, since $p'$ is prime, this will prove the uniqueness of $p$. Now, if $x \in a_n$, $y \in a_m$, $(x - y)^{2d}$ is the sum of terms each of which is a product of $x^d$ or $y^d$ by a homogeneous element of degree $nd$ and hence $(x - y)^{2d} \in p'$ and, since $p'$ is prime, $(x - y)^d \in p'$ and therefore $a_n$ is a subgroup of $A$. As $p'$ is an ideal of $A^{(d)}$, $a$ is a graded ideal of $A$; finally, the relation $(xy)^d \in p'$ implies $x^d \in p'$ or $y^d \in p'$, which completes the proof by virtue of Proposition 4.

Let $A$ be a graded commutative ring with positive degrees and $p$ an essential graded prime ideal of $A$. The set $S$ of homogeneous elements of $A$ not belonging to $p$ is multiplicative and the ring of fractions $S^{-1}A$ is therefore graded canonically (Chapter II, § 2, no. 9) (note that there will in general be homogeneous elements $\neq 0$ of negative degree in this graduation). We shall denote by $A_{(p)}$ the subring of $S^{-1}A$ consisting of the homogeneous elements of degree 0, in other words the set of fractions $x/s$, where $x$ and $s$ are homogeneous of the same degree in $A$ and $s \notin p$. Similarly, for every graded $A$-module $M$, $S^{-1}M$ is graded canonically (loc. cit.) and we shall denote by $M_{(p)}$ the subgroup of homogeneous elements of degree 0, which is obviously an $A_{(p)}$-module.