Showing that a homogenous ideal is prime.

Question

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?

Answer 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.

Answer 2

$\def\p{\mathfrak{p}} \def\Deg{\operatorname{Deg}} \def\Z{\mathbb{Z}} $I found a direct proof for a $\mathbb{Z}$-graded ring. Following this, since my proof is direct, whereas Ted's one is by contrapositive, I will post it here in case it's useful for anyone, as this post seems to have had a lot of visits.

Let $A$ be a $\mathbb{Z}$-graded ring. For $f\in A$, decompose $f=\sum_{i\in\Z} f_i$ into its homogeneous components and define the set $$ \Deg f=\{i\in\Z\mid f_i\neq 0\} $$ of non-vanishing degrees of $f$. Define the length of $f$ as $$ \ell(f)=\max(\Deg f)-\min(\Deg f). $$

Let $\p\subset A$ be an ideal such that for all homogeneous $f,g\in A$ with $fg\in\p$, we have $f\in\p$ or $g\in\p$. We show that $\p$ must be prime. Suppose then $f,g\in A$ are arbitrary elements such that $fg\in\p$. We show $f\in\p$ or $g\in\p$ by induction on $\ell=\ell(f)+\ell(g)$. Case $\ell=0$ means that $f,g$ must be homogeneous, so the assertion follows from the assumption. Let now $\ell>0$ and suppose the result true for all values strictly less than $\ell$. Denote \begin{align*} m_1=\min(\Deg f),\qquad\quad&\ell_1=\ell(f),\\ m_2=\min(\Deg g),\qquad\quad&\ell_2=\ell(g), \end{align*} and write \begin{align*} f=\sum_{i=m_1}^{m_1+\ell_1}f_i,\qquad\quad g=\sum_{i=m_2}^{m_2+\ell_2}g_i. \end{align*} Then $$ I\ni fg=\sum_{i=m_1+m_2}^{m_1+m_2+\ell_1+\ell_2}\sum_{j+k=i}f_jg_k. $$ Since $I$ is homogeneous, the $m_1+m_2$ homogeneous component of $fg$ falls inside $I$, i.e., $f_{m_1}g_{m_2}\in I$. By the assumption and without loss of generality, we have $f_{m_1}\in I$. If we define $f'=f-f_{m_1}$, then $\ell(f')+\ell(g)<\ell(f)+\ell(g)=\ell$ and $$ f'g=fg-f_{m_1}g\in I. $$ Hence, by the induction hypothesis, $f'\in I$ or $g\in I$. If it were the former, this would imply $f=f'+f_{m_1}\in I$.