Showing that a homogenous ideal is prime.

Solution 1:

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.