# Homogeneous ideals are contained in homogeneous prime ideals

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???

• 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

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.
Let $T$ be the set of all proper homogeneous ideals which contain $I$. Then, by Zorn's lemma, there exists a maximal element of $T$, say $P$. We 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 strictly contains $P$. It follows that $\langle a \rangle + P=S$, so $1=ax+p$ with $p\in P$. We thus get $b=abx+bp\in P$. Now use Showing that a homogenous ideal is prime. to conclude that $P$ is prime.