Lemma 10.57.7. Let $S$ be a graded ring. Let $\mathfrak p \subset S$ be a prime. Let $\mathfrak q$ be the homogeneous ideal of $S$ generated by the homogeneous elements of $\mathfrak p$. Then $\mathfrak q$ is a prime ideal of $S$.

Proof. Suppose $f, g \in S$ are such that $fg \in \mathfrak q$. Let $f_ d$ (resp. $g_ e$) be the homogeneous part of $f$ (resp. $g$) of degree $d$ (resp. $e$). Assume $d, e$ are maxima such that $f_ d \not= 0$ and $g_ e \not= 0$. By assumption we can write $fg = \sum a_ i f_ i$ with $f_ i \in \mathfrak p$ homogeneous. Say $\deg (f_ i) = d_ i$. Then $f_ d g_ e = \sum a_ i' f_ i$ with $a_ i'$ to homogeneous par of degree $d + e - d_ i$ of $a_ i$ (or $0$ if $d + e -d_ i < 0$). Hence $f_ d \in \mathfrak p$ or $g_ e \in \mathfrak p$. Hence $f_ d \in \mathfrak q$ or $g_ e \in \mathfrak q$. In the first case replace $f$ by $f - f_ d$, in the second case replace $g$ by $g - g_ e$. Then still $fg \in \mathfrak q$ but the discrete invariant $d + e$ has been decreased. Thus we may continue in this fashion until either $f$ or $g$ is zero. This clearly shows that $fg \in \mathfrak q$ implies either $f \in \mathfrak q$ or $g \in \mathfrak q$ as desired. $\square$

There are also:

• 6 comment(s) on Section 10.57: Proj of a graded ring

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).