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. $\square$

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