Hi Mathoverflow. This question is about building intuition for the Proj construction. When I first started learning about schemes, I found the construction of the structure sheaves on Spec and Proj very confusing. However, after enough time had passed I began to understand the construction for Spec: It is sort of an algebraic partition of unity argument. The construction of the structure sheaf on Proj is still very mysterious to me. For completeness it goes something like this:



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> > Let $G$ be a graded ring. Take $ f \in G $ homogeneous of degree $ d \geq 1 $. We have a homeomorphism $D\_+(f) \cong{\rm spec} \, G\_{(f)} $ which sends $D\_+(fg)$ to $D(g^{\deg f} / f^{\deg g})$. It first expands the prime to $G\_f$ and then contracts the result to $G\_{(f)}$. Motivated by this, we can define $$ \mathscr{O}\_{{\rm Proj}G}(D\_+(f)) = G\_{(f)}$$
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I can fill in the details but they are messy and I have very little idea what the details actually mean. Thinking about $ \mathbb{P}^n $ as a variety, I understand why $D\_+(X\_0)$ should be $ {\rm spec} \mathbb{C} [X\_1/X\_0,X\_2/X\_0, \dots, X\_n / X\_0] $ but I don't really understand how this intuition translates into the commutative algebra which is boxed above. 


Also, for homogeneous elements of degree higher than $1$ I have no idea what is going on. I understand that geometrically the veronese map should be involved but I don't understand how that intuition translates into the messy proof which I am able to write down. 



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> >  **Question** : Is anyone able to explain the idea behind this construction? Note that this explanation could lie in either the realm of algebraic geometry or commutative algebra.
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It is very possible that there is not a nice way to think about this construction which would make me sad because it is so fundamental. I hope that there are some good responses which help me fix my ignorance!


EDIT: I just went through the proof again and wrote up the details.


Let $ G$ be a graded ring. The first order of business is to explain why we have a homeomorphism $ D\_+(f) \cong {\rm Spec} G\_{(f)} $ when $ f \in G $ is homogeneous with degree $d \geq 1 $. Fix $ \mathfrak{p} \in D\_+(f)$. Then $ \mathfrak{p}$ is the generic point of some irreducible closed subset $Z$ in $ {\rm Spec} G$. Since $ f \not\in \mathfrak{p}$ it follows that $ G\_f \mathfrak{p}$ is the generic point of the irreducible closed subset $ Z \cap {\rm Spec} G\_f$ in $ {\rm spec} G\_f$. Assume that $g \in G $ is homogeneous. Then $ g $ vanishes at $ \mathfrak{p} $ iff $ g / 1 $ vanishes at $ G\_f \mathfrak{p}$ iff $g^{\deg f} / f^{\deg g}$ vanishes at $ G\_f \mathfrak{p}$ iff $g^{\deg f} / f^{\deg g} \in G\_f \mathfrak{p} \cap G\_{(f)} $. This proves that the map $D\_+(f) \to {\rm Spec} G\_{(f)}$ is injective. We need to prove that the map is surjective. Let $ \mathfrak{q}$ be a prime ideal in $G\_{(f)}$. Motivated by the above argument, we define
$$ \mathfrak{p}\_n = \{ g \in G\_n : g^{\deg f} / f^{\deg g} \in \mathfrak{q} \} $$
Then $ \mathfrak{p} = \oplus\_{n \geq 0} \mathfrak{p}\_n \in D\_+(f)$ maps to $ \mathfrak{q} $. Since $ g $ vanishes at $ \mathfrak{p} $ iff $g^{\deg f} / f^{\deg g} \in G\_f \mathfrak{p} \cap G\_{(f)} $ the map is a homeomorphism which sends $D\_+(fg)$ to $D(g^{\deg f} / f^{\deg g})$. All that remains is to prove that the map $ G\_{(f)} \to G\_{(fg)}$ defined by $ a / f^n \mapsto a g^n / (fg)^n$ is the localization map. From the affine case we know that $ G\_{f} \to G\_{fg}$ is the localization map at $ g / 1 $, therefore the only non units which $ G\_{(f)} \to G\_{(fg)}$ sends to units are powers of $ g^{\deg f} / f^{\deg g}$. I suspect that this last bit can be made rigorous.



Intuition for coordinate patches on Proj of a graded ring.
You in fact already found out the most natural way to think about the Proj construction, namely in terms of the projective space. Consider $X := \mathbb{P}^n$ with homogeneous coordinates $[x\_0: \cdots : x\_n]$. Then what are the basic open sets in $\mathbb{P}^n$? These are simply the complement of zero sets $V(f)$ of homogeneous polynomials $f \in S:= k[x\_0, \ldots, x\_n]$. I suggest that you try to see for yourself that $U\_f := X \setminus V(f)$ is indeed isomorphic to $Spec~ S\_{(f)}$. (Hint: at first check this for linear polynomials - which follows almost by definition. Then for an arbitrary $f$, write $U\_f$ as the union of $U\_f \setminus V(x\_i)$ and figure out what is the coordinate ring of $U\_f \setminus V(x\_i)$. $U\_f \setminus V(x\_i)$ is also the complement in $U\_{x\_i}$ of the zero set of an element in the coordinate ring of $U\_{x\_i}$ - what is this element?)


For me at least it was illuminating to get the hands dirty: start with the definition of projective space as the space of lines in $k^{n+1}$ (with the natural topology given by identification of the lines not passing through a hyperplane with $k^n$) - and check explicitly what the notions in the definition of the Proj construction mean. 


To get flavours of more general situations, I would again go through examples. E.g. for the scenarios that $S\_0$ may not be a field, consider $X := \mathbb{C}^2 \times \mathbb{P}^2(\mathbb{C})$ and see why this is isomorphic to $Proj~ A[z\_0, z\_1, z\_2]$, where $A := \mathbb{C}[x,y]$. For a bit more involved example, consider the subvariety $V$ of $X$ defined by the zero set of polynomials $f \in A$ and a homogeneous polynomial $g \in \mathbb{C}[z\_0, z\_1, z\_2]$, and check that $V$ is isomorphic to the subscheme of $Proj ~ A[z\_0, z\_1, z\_2]$ defined by the ideal generated by $f$ and $g$. Can you see how to interpret the zero degree elements of the graded ring?


