## Tag `00JO`

Chapter 10: Commutative Algebra > Section 10.56: Proj of a graded ring

Lemma 10.56.2. Let $S$ be a $\mathbf{Z}$-graded ring containing a homogeneous invertible element of positive degree. Then the set $G \subset \mathop{\mathrm{Spec}}(S)$ of $\mathbf{Z}$-graded primes of $S$ (with induced topology) maps homeomorphically to $\mathop{\mathrm{Spec}}(S_0)$.

Proof.First we show that the map is a bijection by constructing an inverse. Let $f \in S_d$, $d > 0$ be invertible in $S$. If $\mathfrak p_0$ is a prime of $S_0$, then $\mathfrak p_0S$ is a $\mathbf{Z}$-graded ideal of $S$ such that $\mathfrak p_0S \cap S_0 = \mathfrak p_0$. And if $ab \in \mathfrak p_0S$ with $a$, $b$ homogeneous, then $a^db^d/f^{\deg(a) + \deg(b)} \in \mathfrak p_0$. Thus either $a^d/f^{\deg(a)} \in \mathfrak p_0$ or $b^d/f^{\deg(b)} \in \mathfrak p_0$, in other words either $a^d \in \mathfrak p_0S$ or $b^d \in \mathfrak p_0S$. It follows that $\sqrt{\mathfrak p_0S}$ is a $\mathbf{Z}$-graded prime ideal of $S$ whose intersection with $S_0$ is $\mathfrak p_0$.To show that the map is a homeomorphism we show that the image of $G \cap D(g)$ is open. If $g = \sum g_i$ with $g_i \in S_i$, then by the above $G \cap D(g)$ maps onto the set $\bigcup D(g_i^d/f^i)$ which is open. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 12785–12791 (see updates for more information).

```
\begin{lemma}
\label{lemma-Z-graded}
Let $S$ be a $\mathbf{Z}$-graded ring containing a homogeneous
invertible element of positive degree. Then the set
$G \subset \Spec(S)$ of $\mathbf{Z}$-graded primes of $S$
(with induced topology) maps homeomorphically to $\Spec(S_0)$.
\end{lemma}
\begin{proof}
First we show that the map is a bijection by constructing an inverse.
Let $f \in S_d$, $d > 0$ be invertible in $S$.
If $\mathfrak p_0$ is a prime of $S_0$, then $\mathfrak p_0S$
is a $\mathbf{Z}$-graded ideal of $S$ such that
$\mathfrak p_0S \cap S_0 = \mathfrak p_0$. And if $ab \in \mathfrak p_0S$
with $a$, $b$ homogeneous, then
$a^db^d/f^{\deg(a) + \deg(b)} \in \mathfrak p_0$.
Thus either $a^d/f^{\deg(a)} \in \mathfrak p_0$ or
$b^d/f^{\deg(b)} \in \mathfrak p_0$, in other words either
$a^d \in \mathfrak p_0S$ or $b^d \in \mathfrak p_0S$.
It follows that $\sqrt{\mathfrak p_0S}$ is a $\mathbf{Z}$-graded
prime ideal of $S$ whose intersection with $S_0$ is $\mathfrak p_0$.
\medskip\noindent
To show that the map is a homeomorphism we show that
the image of $G \cap D(g)$ is open. If $g = \sum g_i$
with $g_i \in S_i$, then by the above $G \cap D(g)$
maps onto the set $\bigcup D(g_i^d/f^i)$ which is open.
\end{proof}
```

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