Proposition 5

Let $a = \bigoplus_{i \geq 0} a_i$ be a graded ideal of $A$ and $n_0$ an integer $> 0$. For there to exist a graded prime ideal $p = \bigoplus_{i \geq 0} p_i$ such that $p_{n_0} = a,$ for $n \geq n_0,$ it is necessary and sufficient that, for pair of homogeneous elements $x, y$ of degrees $\geq n_0,$ the relation $xy \in a$ implies '"$x \in a$ or $y \in a$". If there exists $n \geq n_0$ such that $a_n \neq A_n,$ the prime ideal satisfying the above conditions is unique.

The condition of the statement is obviously necessary. If $a_n = A_n$ for all $n \geq n_0,$ clearly every prime ideal containing $m$ is a solution to the problem; there may therefore be several prime ideals which solve the problem; however, any two of these ideals are obviously equivalent. Suppose then that there exists a homogeneous element $a \in A_d$ (with $d \geq n_0$) not belonging to $a_d$. Let $p$ be the set of $x \in A$ such that $ax \in a$. Clearly $p$ is an ideal of $A$; as the homogeneous components of $ax$ are the products by $a$ of those of $x$ and $a$ is a graded ideal, $p$ is a graded ideal; moreover, $1 \notin p$ and hence $p \neq A$. To prove that $p$ is prime, it is sufficient to show that, if $x \in A, \ y \in A,$ satisfy $x \notin p$ and $y \notin p,$ then $xy \notin p$ (Proposition 4). Then $ax \notin a_{m+d}, ay \notin a_{n+d},$ whence by hypothesis

we conclude that $axy \notin a_{m+n+d}$ since $xy \notin p$. Finally, if $n \geq n_0$ and $x \in A_n,$ the conditions $x \in a,$ and $ax \in a_{n+d}$ are equivalent by hypothesis and hence $p \cap A_n = a_n$ which completes the proof of the existence of the graded prime ideal $p$ which solves the problem. If also $p'$ is another graded prime ideal of $A$ such that $p' \cap A_n = a_n$ for $n \geq n_0,$ then $a \notin p'$ and $ax \in p'$ for all $x \in p,$ whence $p \subset p'$ since $p'$ is prime. On the other hand, if $x$ is a homogeneous element of degree $n \geq 0$ of $p',$ $ax$ is homogeneous of degree $n + d \geq n_0$ and therefore belongs to $p' \cap A_{n+d} = a_{n+d}$, whence by definition $x \in p$, which shows that $p' \subset p$ and finally $p' = p$.