Points of Proj of a graded ring

Let $S={⨁}_{d\ge 0}{S}_d$ be a commutative nonnegatively graded ring (Nonnegatively graded rings and modules, homogeneous elements, and twists) and let $S_{+}={⨁}_{d>0}{S}_d$ be its irrelevant ideal. Indeed $S_{+}$ is an ideal: it is an additive subgroup, and for $s_i\in S_i$ with $i>0$ and $s_j\in S_j$ with $j\ge 0$ one has $s_i{s}_j\in S_{i+j}$ with $i+j>0$.

An ideal $\mathfrak{p}\subseteq S$ is homogeneous when it is generated by homogeneous elements. Equivalently, $\mathfrak{p}$ is homogeneous when for every $a\in \mathfrak{p}$ each homogeneous component of $a$ lies in $\mathfrak{p}$: if $\mathfrak{p}=(h_{\lambda })$ with $h_{\lambda }$ homogeneous and $a={\sum }_d{a}_d\in \mathfrak{p}$, then each $a_d$ is an $S$-linear combination of the generators and so lies in $\mathfrak{p}$. A homogeneous ideal is a homogeneous prime when it is prime as an ideal.

The projective spectrum $\mathrm{Proj}S$ is the set of homogeneous prime ideals $\mathfrak{p}\subseteq S$ with $S_{+}\subseteq ̸\mathfrak{p}\mathrm{.}$ For a homogeneous ideal $I\subseteq S$ put $V_{+}(I)=\{\mathfrak{p}\in \mathrm{Proj}S:I\subseteq \mathfrak{p}\},$ and declare these the closed sets of the Zariski topology on $\mathrm{Proj}S$; a subset is open when its complement is closed. The family $\{V_{+}(I)\}$ is closed under the topological operations (see the Remarks), so the declaration defines a topology, which is the trace on $\mathrm{Proj}S$ of the Zariski topology of $\mathrm{Spec}S$ (The underlying space of an affine spectrum) restricted to the homogeneous primes that do not contain $S_{+}$, presented by homogeneous generators.

The zero ring is allowed: for $S=0$ the ring has no prime ideals at all, so $\mathrm{Proj}0=\mathrm{\varnothing }$ with its unique topology and $V_{+}(I)=\mathrm{\varnothing }$ for every homogeneous ideal $I\subseteq 0$. Consequently $\mathrm{Proj}S$ may be empty for nonzero $S$ as well, and all statements below are meant to include the empty case.

• The closed sets are the topological ones. $V_{+}(0)=\mathrm{Proj}S$, since every $\mathfrak{p}$ contains $0$, and $V_{+}(S)=\mathrm{\varnothing }$, since no prime ideal $\mathfrak{p}\subseteq S$ contains the unit ideal $S$. For a family $(I_{\lambda })$ of homogeneous ideals, ${\bigcap }_{\lambda }{V}_{+}(I_{\lambda })=V_{+}({\sum }_{\lambda }{I}_{\lambda })$ because $\mathfrak{p}$ contains every $I_{\lambda }$ if and only if it contains their sum, and ${\sum }_{\lambda }{I}_{\lambda }$ is homogeneous; the union of $V_{+}(I)$ and $V_{+}(J)$ is $V_{+}(IJ)$ because a prime containing the product $IJ$ contains $I$ or $J$, while a prime containing $I$ contains $IJ$; the product of homogeneous ideals is homogeneous. Intersections of the form ${\bigcap }_{\lambda }{V}_{+}(I_{\lambda })$ include arbitrary, possibly infinite, index families; for the empty family the intersection is $\mathrm{Proj}S=V_{+}(0)$.
• Uniqueness of the presentation is not claimed. Many homogeneous ideals present the same closed set; the relevant saturation statement for polynomial rings is proved later.
• The condition $S_{+}\subseteq ̸\mathfrak{p}$ excludes exactly the homogeneous primes above the irrelevant ideal; a homogeneous prime $\mathfrak{p}$ with $S_{+}\subseteq \mathfrak{p}$ is not a point of $\mathrm{Proj}S$.