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:

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 $$ \mathscr{O}_{{\rm Proj}G}(D_+(f)) = G_{(f)}$$ and prove that the map $G_{(f)} \to G_{(fg)}$ defined by $ a / f^n \mapsto a g^n / (fg)^n$ is localization at $ g^{\deg f} / f^{\deg g}$.

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.

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.

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.

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    $\begingroup$ This is more about $\rm Proj$ than about its structure sheaf, but I much prefer the definition ${\rm Proj}\ R = ({\rm Spec}\ R \setminus {\rm Spec} R_0)/$scaling to the gluing construction. (And in fact, I essentially never glue schemes together along open subschemes.) $\endgroup$ Jun 4, 2013 at 21:28
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    $\begingroup$ Hmm this sounds interesting as it is much closer to the "classical" definition of projective space. Two thinks I am a little unsure about though: How is $ {\rm spec} R_0 $ closed subset of $ {\rm spec } R $? How do you mod out by scaling? $\endgroup$ Jun 4, 2013 at 21:37
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    $\begingroup$ A ring $R$ is $\mathbb Z$-graded iff it carries an action of the multiplicative group (proof: define $R_i$ to be the subspace where $z\cdot$ acts by $z^i$). Then, that group will also act on ${\rm Spec}\ R$. Then for the first question, you need to look past the obvious inclusion $R_0 \to R$ to the less obvious quotient $R \to R_0$. This only works because $R$ is $\mathbb N$-graded, not just $\mathbb Z$-graded. $\endgroup$ Jun 5, 2013 at 0:57
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    $\begingroup$ See the answers to mathoverflow.net/questions/41624/… $\endgroup$ Jun 5, 2013 at 1:26

2 Answers 2


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?


If you look at the ordinary $\mathbb P^n$ then it has $n+1$ standard open patches of the form $C[X_0/X_i,X_1/X_i,\ldots,X_n/X_i]$. The $i$-th patch looks similar to a localization with respect to $X_i$ in the affine case, but not quite, since now all variables are degree less due to scaling. Now they want to do a similar thing to any graded ring $G$ and want to "localize" w.r.t. any homogeneous function $f$. To do that, just make $f$ invertible , and then take the subring of degree $0$ elements of $G[1/f]$ (due to scaling ). After we have that, we can easily go back and forth between affine and projective. For example, to explain the last statement in your algebra box :

Assume we have localized the projective variety $\operatorname{Proj}(G)$ to $\operatorname{Spec}(G_{(f)})$ for some homogeneous element $f$. Now we want to localize it further using $g$. $g$ is not an elemment of $G_{(f)}$, because it has degree, so instead we need to use $g^{\deg(f)}/f^{\deg(g)}$ for localization. What about the mapping $a/f^n \rightarrow ag^n/(fg)^n$ ? This maps does nothing, which it should, cause localization map does nothing. They only rewrite it because $ag^n/(fg)^n$ is easily recognized as a member of $G_{(fg)}$ while the former is not.


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