# Showing that a homogenous ideal is prime.

I'm trying to read a proof of the following proposition:

Let $S$ be a graded ring, $T \subseteq S$ a multiplicatively closed set. Then a homogenous ideal maximal among the homogenous ideals not meeting $T$ is prime.

In this proof, it says

"it suffices to show that if $a,b \in S$ are homogenous and $ab \in \mathfrak{p}$, then either $a \in \mathfrak{p}$ or $b \in \mathfrak{p}$"

where $\mathfrak{p}$ is our maximal homogenous ideal. I don't know how to prove this is indeed sufficient. If I try writing general $a$ and $b$ in terms of "coordinates":

$a=a_0+...a_n$ where $a_0 \in S_0$, ... etc

then I can see it working for small $n$, but it seems to get so complicated I wouldn't know how to write down a proof. Is there a better way to attack this problem?

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Please see page 186 of people.fas.harvard.edu/~amathew/CRing.pdf – user38268 Jul 20 '12 at 7:27

We wish to prove: If $S$ is a $\mathbb{Z}$-graded ring and $\mathfrak{p}$ is a homogeneous ideal of $S$ satisfying $ab \in \mathfrak{p}$ implies $a$ or $b$ in $\mathfrak{p}$ for homogeneous $a$ and $b$, then $\mathfrak{p}$ is prime.
So take 2 general elements $a,b \in \mathfrak{p}$ and assume $ab \in \mathfrak{p}$ but neither $a$ nor $b$ is in $\mathfrak{p}$. Let $a = \sum a_d$ and $b = \sum b_d$ be their homogeneous decompositions. Since $a \not \in \mathfrak{p}$, then some $a_d \not \in \mathfrak{p}$, and since all but finitely many $a_d$ are 0, there exists a largest integer $d$ such that $a_d \not \in \mathfrak{p}$. Similarly, there exists a largest integer $e$ such that $b_e \not \in \mathfrak{p}$.
Since $ab \in \mathfrak{p}$ and $\mathfrak{p}$ is a homogeneous ideal, then all the components of $ab$ are in $\mathfrak{p}$. The $d+e$ component of $ab$ is $\sum a_i b_j$ where we sum over all pairs $(i,j)$ with $i+j = d+e$. But each such pair $(i,j)$, other than $(d,e)$, must have either $i>d$ or $j>e$, and hence (by the maximality of $d$ and $e$) we have $a_i b_j \in \mathfrak{p}$. Thus $a_d b_e \in \mathfrak{p}$ also, yet neither $a_d$ nor $b_e$ is in $\mathfrak{p}$, which contradicts the original assumption about $\mathfrak{p}$ for the homogeneous elements $a_d$ and $b_e$.
In short: If $a,b$ is a general counterexample for the primality of $\mathfrak{p}$, then $a_d, b_e$ is a homogeneous counterexample.