The Proj Construction

Let $S = S_0 \oplus S_1 \oplus \cdots$ be a $\mathbb{Z}_{\geq 0}$-graded ring. Call $S_{+} = S_1 \oplus S_2 \oplus \cdots$ the irrevelant ideal.  We call a homogeneous prime ideal $\mathfrak{p}$ of $S$  relevant if it does not contain $S_+$.

We will define $\text{Proj} S$.

As a Set

As a set, let $\text{Proj} S$ be the set of relevant homogeneous prime ideals of $S$. 

As a Topological Space

As a subset of $\text{Spec} S$, $\text{Proj}(S)$ inherits the Zarisky topology. In particular, its closed subsets are

$$\mathbb{V}_+(E) = \mathbb{V}(E) \cap \text{Proj} (S)$$ . It has a basis of distinguished opens $D_+(f) = D(f) \cap \text{Proj} (S).$

As a Scheme

Clearly the $D_+(f)$'s (where $f$ runs over the homogeneous elements of $S$) form an open cover of $\text{Proj} S$. We will identify each $D_+(f)$ with an affine scheme in a compatible way. Thus gluing these schemes give a scheme structure for $\text{Proj} S$.

Affine structures on Distinguished Opens

Claim: $$D_+(f) \cong \text{Spec} S_{(f)}$$

where $\text{Spec} S_{(f)}$ is the $0$-degree component of the graded ring $S_f$. 

Proof. Define a map $\psi_f: D_+(f) \to \text{Spec} S_{(f)}$ by

$$\mathfrak{p} \mapsto \mathfrak{p} S_f \cap S_{(f)}.$$

(Note that the definition makes sense since if $\mathfrak{p} \not \ni f$ then $\mathfrak{p} S_f$ is a prime ideal of $S_f$ so its contraction to $S_{(f)}$ is a prime ideal of the latter.)

We want to show that this is a homeomorphism of topological spaces.

First, we show that $\psi_f$ is injective.

Indeed, suppose $\psi(\mathfrak{p}) = \psi(\mathfrak{q})$, i.e. the elements of degree $0$ in $\mathfrak{p}S_f$ are the same as the $0$-degree elements in $\mathfrak{q}S_f$. 

Let $d = \deg(f)$ and $h \in \mathfrak{p}$ be of degree $m$. Then $h^d/ f^m$ is of degree $0$ in $\mathfrak{p}S_f$ so must be an element of  $\mathfrak{q}S_f$. This implies $h^d/1 \in \mathfrak{q}S_f$ and thus $h \in \mathfrak{q}$. 

Here we used the obsevation that if $h^m/f^n \in \mathfrak{p}S_f$ for some $m, n$ then $h \in \mathfrak{p}$; and that we can turn every element of $\mathfrak{p}$ into an element of degree $0$ in $\mathfrak{p}S_f$. 

Now we show that $\psi_f$ is surjective. 

From the above discussion, we notice that $h \in \mathfrak{p}_m$ iff $\frac{h^d}{f^m} \in \mathfrak{p} S_f \cap S_{(f)}$. We use this as a characterization of the graded components of $\mathfrak{p}$.

Let $P$ be a prime ideal of $S_{(f)}$. For each $m$, define

$$\mathfrak{p}_m = \left\{h \in S_m \mid \frac{h^d}{f^m} \in P \right\}.$$

We need to check that the $\mathfrak{p}_m$ are subgroups of $S_m$, that they satisfy the grading criterion, so that $\mathfrak{p}:= \oplus \mathfrak{p}_m$ is a graded ideal of $S$. Then we need to check that $\mathfrak{p}$ is a relevant prime ideal.

It's easy to see that $\psi_f$ is a homeomorphism.