Let $I$ be a homogeneous ideal of a graded ring $S$, $I\ne S$. I want to show that there exists a homogeneous prime ideal which contains $I$.
I proved the following:
Let $T$ be the set of all homogeneous ideals which contain $I$.
Then, by Zorn's lemma, there exists a maximal element of $T$, say $P$.
I will claim that $P$ is prime. Suppose that for homogeneous elements $a,b \in S$, $ab \in P$ but, $a\notin P$. Then $\langle a \rangle + P$ is a homogeneous ideal which contains $I$.
It contradicts by maximality of $P$ so, $a\in P$.
Is it right???
Best Answer
A different proof: $I$ is contained in a prime (maximal) ideal of $S$, say $P$. Since $I$ is homogeneous, $I\subseteq P^*$, where $P^*$ is the ideal generated by the homogeneous elements of $P$. Note that $P^*$ is also a prime ideal (see Bruns and Herzog, Lemma 1.5.6(a)) and homogeneous.