# 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 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?

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.

• Interestingly, this result turns out to be false for gradings over arbitrary monoids. A counterexample for $\mathbb{Z}/2$-gradings can be found here. Dec 13, 2022 at 15:19

$$\def\p{\mathfrak{p}} \def\Deg{\operatorname{Deg}} \def\Z{\mathbb{Z}}$$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 $$f=f'+f_{m_1}\in I$$.