TITLE: Showing that a homogenous ideal is prime.
QUESTION [17 upvotes]: 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 homogeneous ideal maximal among the homogeneous ideals not meeting $T$ is prime.
In this proof, it says

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

where $\mathfrak{p}$ is our maximal homogeneous ideal. I don't know how to prove this is indeed sufficient. If I try writing general $a$ and $b$ in terms of "coordinates":
$$a=a_0+\cdots+a_n$$ where $a_0 \in S_0,$ 
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?

REPLY [25 votes]: 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.