
Lemma 10.56.3 (Topology on Proj). Let $S = \oplus _{d \geq 0} S_ d$ be a graded ring.

1. The sets $D_{+}(f)$ are open in $\text{Proj}(S)$.

2. We have $D_{+}(ff') = D_{+}(f) \cap D_{+}(f')$.

3. Let $g = g_0 + \ldots + g_ m$ be an element of $S$ with $g_ i \in S_ i$. Then

$D(g) \cap \text{Proj}(S) = (D(g_0) \cap \text{Proj}(S)) \cup \bigcup \nolimits _{i \geq 1} D_{+}(g_ i).$
4. Let $g_0\in S_0$ be a homogeneous element of degree $0$. Then

$D(g_0) \cap \text{Proj}(S) = \bigcup \nolimits _{f \in S_ d, \ d\geq 1} D_{+}(g_0 f).$
5. The open sets $D_{+}(f)$ form a basis for the topology of $\text{Proj}(S)$.

6. Let $f \in S$ be homogeneous of positive degree. The ring $S_ f$ has a natural $\mathbf{Z}$-grading. The ring maps $S \to S_ f \leftarrow S_{(f)}$ induce homeomorphisms

$D_{+}(f) \leftarrow \{ \mathbf{Z}\text{-graded primes of }S_ f\} \to \mathop{\mathrm{Spec}}(S_{(f)}).$
7. There exists an $S$ such that $\text{Proj}(S)$ is not quasi-compact.

8. The sets $V_{+}(I)$ are closed.

9. Any closed subset $T \subset \text{Proj}(S)$ is of the form $V_{+}(I)$ for some homogeneous ideal $I \subset S$.

10. For any graded ideal $I \subset S$ we have $V_{+}(I) = \emptyset$ if and only if $S_{+} \subset \sqrt{I}$.

Proof. Since $D_{+}(f) = \text{Proj}(S) \cap D(f)$, these sets are open. Similarly the sets $V_{+}(I) = \text{Proj}(S) \cap V(E)$ are closed.

Suppose that $T \subset \text{Proj}(S)$ is closed. Then we can write $T = \text{Proj}(S) \cap V(J)$ for some ideal $J \subset S$. By definition of a homogeneous ideal if $g \in J$, $g = g_0 + \ldots + g_ m$ with $g_ d \in S_ d$ then $g_ d \in \mathfrak p$ for all $\mathfrak p \in T$. Thus, letting $I \subset S$ be the ideal generated by the homogeneous parts of the elements of $J$ we have $T = V_{+}(I)$.

The formula for $\text{Proj}(S) \cap D(g)$, with $g \in S$ is direct from the definitions. Consider the formula for $\text{Proj}(S) \cap D(g_0)$. The inclusion of the right hand side in the left hand side is obvious. For the other inclusion, suppose $g_0 \not\in \mathfrak p$ with $\mathfrak p \in \text{Proj}(S)$. If all $g_0f \in \mathfrak p$ for all homogeneous $f$ of positive degree, then we see that $S_{+} \subset \mathfrak p$ which is a contradiction. This gives the other inclusion.

The collection of opens $D(g) \cap \text{Proj}(S)$ forms a basis for the topology since the standard opens $D(g) \subset \mathop{\mathrm{Spec}}(S)$ form a basis for the topology on $\mathop{\mathrm{Spec}}(S)$. By the formulas above we can express $D(g) \cap \text{Proj}(S)$ as a union of opens $D_{+}(f)$. Hence the collection of opens $D_{+}(f)$ forms a basis for the topology also.

First we note that $D_{+}(f)$ may be identified with a subset (with induced topology) of $D(f) = \mathop{\mathrm{Spec}}(S_ f)$ via Lemma 10.16.6. Note that the ring $S_ f$ has a $\mathbf{Z}$-grading. The homogeneous elements are of the form $r/f^ n$ with $r \in S$ homogeneous and have degree $\deg (r/f^ n) = \deg (r) - n\deg (f)$. The subset $D_{+}(f)$ corresponds exactly to those prime ideals $\mathfrak p \subset S_ f$ which are $\mathbf{Z}$-graded ideals (i.e., generated by homogeneous elements). Hence we have to show that the set of $\mathbf{Z}$-graded prime ideals of $S_ f$ maps homeomorphically to $\mathop{\mathrm{Spec}}(S_{(f)})$. This follows from Lemma 10.56.2.

Let $S = \mathbf{Z}[X_1, X_2, X_3, \ldots ]$ with grading such that each $X_ i$ has degree $1$. Then it is easy to see that

$\text{Proj}(S) = \bigcup \nolimits _{i = 1}^\infty D_{+}(X_ i)$

does not have a finite refinement.

Let $I \subset S$ be a graded ideal. If $\sqrt{I} \supset S_{+}$ then $V_{+}(I) = \emptyset$ since every prime $\mathfrak p \in \text{Proj}(S)$ does not contain $S_{+}$ by definition. Conversely, suppose that $S_{+} \not\subset \sqrt{I}$. Then we can find an element $f \in S_{+}$ such that $f$ is not nilpotent modulo $I$. Clearly this means that one of the homogeneous parts of $f$ is not nilpotent modulo $I$, in other words we may (and do) assume that $f$ is homogeneous. This implies that $I S_ f \not= 0$, in other words that $(S/I)_ f$ is not zero. Hence $(S/I)_{(f)} \not= 0$ since it is a ring which maps into $(S/I)_ f$. Pick a prime $\mathfrak q \subset (S/I)_{(f)}$. This corresponds to a graded prime of $S/I$, not containing the irrelevant ideal $(S/I)_{+}$. And this in turn corresponds to a graded prime ideal $\mathfrak p$ of $S$, containing $I$ but not containing $S_{+}$ as desired. $\square$

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