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Let $I$ be a homogeneous ideal of a graded ring $S$. 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???

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You have to assume $I \neq S$; likewise in the definition of $T$ only proper ideals are allowed. And by the way, you may assume $I=0$ in the first place (look at $S/I$). In your proof, you haven't shown that $\langle a \rangle + P$ is proper, and you haven't used $ab \in P$ yet. So the proof is not complete. –  Martin Brandenburg May 8 '13 at 7:59
thank you Matrin. –  Sang Cheol Lee May 8 '13 at 15:15
do you want another proof ? –  11156 Mar 30 at 8:09

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