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.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: