This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
In §Schemes, ⁋Example 10 we glued two affine lines \(\mathbb{A}^1=\Spec \mathbb{K}[\x]\) together in a suitable way to construct projective space \(\mathbb{P}^1\). In this post we generalize this to define projective schemes.
Projective Space
Generalizing §Schemes, ⁋Example 10 directly, it is not difficult to define \(\mathbb{P}^n\) as a scheme. However, in order to generalize this to define projective schemes, it helps to understand \(\mathbb{P}^n\) intuitively, so let us unpack it more carefully.
First we briefly review the projective space defined in topology. To construct the topological space \(\mathbb{P}^n\), we considered the topological space \(\mathbb{R}^{n+1}\setminus \{0\}\). Then on this we define the following equivalence relation
\[(x_0,\ldots, x_n)\sim (y_0,\ldots, y_n)\iff\text{$x_i=\lambda y_i$ for some $\lambda\neq 0$, for all $i$}\]projective space \(\mathbb{P}^n\) is defined as the quotient space \((\mathbb{R}^{n+1}\setminus \{0\})/{\sim}\), and the equivalence class containing \((x_0,\ldots, x_n)\) is denoted by \([x_0:x_1:\cdots:x_n]\) for notational convenience.
Now consider the canonical projection \(\pi:\mathbb{R}^{n+1}\setminus\{0\}\rightarrow \mathbb{P}^n\). Then the fiber over a point \([x_0:x_1:\cdots:x_n]\) in \(\mathbb{P}^n\) is by definition
\[\{(y_0,\ldots, y_n)\mid\text{$x_i=\lambda y_i$ for some $\lambda\neq 0$, for all $i$}\}\]that is, the set of points on the line passing through the origin and \((x_0,\ldots, x_n)\), excluding the origin. For this reason, \(\mathbb{P}^n\) is often thought of as the space of lines in \(\mathbb{R}^{n+1}\).
Meanwhile, in \(\mathbb{R}^{n+1}\), any plane \(P\) and any line not parallel to \(P\) must meet at a single point. Thus, defining the plane \(P_i\) as
\[P_i=\{\x_i=1\}=\{(x_0,\ldots, x_n)\mid x_i=1\}\]then, excluding the points in \(\mathbb{P}^n\) corresponding to lines perpendicular to the \(\x_i\)-axis, all remaining points are in one-to-one correspondence with points of \(P_i\), the remaining points are lines in the \(\x_0\x_1\cdots\x_{i-1}\x_{i+1}\cdots\x_n\)-plane, i.e., lines in \(\mathbb{R}^n\), so we obtain the decomposition
\[\mathbb{P}^n=\mathbb{R}^n\coprod \mathbb{P}^{n-1}\]This process is illustrated in the following figure for the case \(n=2\).
Writing this as a formula, for a point \([x_0:\cdots:x_n]\) in \(\mathbb{P}^n\), if \(x_i\neq 0\) then within the equivalence class of \([x_0:\cdots:x_n]\) we can (uniquely) find the point whose \(i\)-th coordinate is \(1\); regarding this point as a point of \(P_i\), we can identify the subset
\[U_i=\{[x_0:\cdots:x_n]\in \mathbb{P}^n\mid x_i\neq 0\}\]with \(P_i\cong \mathbb{R}^n\). On the other hand, the points in the complement of \(U_i\) are exactly those with \(x_i=0\), so by simply omitting the \(i\)-th coordinate we can understand them as points of \(\mathbb{P}^{n-1}\).
Explicitly, the above identification \(U_i\cong P_i\) is expressed by the formula
\[[x_0:\cdots:x_n]\text{ in $U_i\subseteq \mathbb{P}^n$}\leftrightarrow\left(\frac{x_0}{x_i},\ldots, \frac{x_{i-1}}{x_i},1,\frac{x_{i+1}}{x_i},\ldots, \frac{x_n}{x_i}\right)\text{ in $P_i\subseteq \mathbb{R}^{n+1}$}\]Meanwhile, the procedure of §Schemes, ⁋Example 10 reverses this process. That is, we are first given \(n+1\) copies of \(n\)-dimensional planes \(P_0,\ldots, P_n\) and we glue them via isomorphisms satisfying the cocycle condition. Then how the cocycle condition should be written is obtained by examining how a point of \(\mathbb{P}^n\) is written in different \(P_i\) and \(P_j\) under the above identification. Let us examine this. First, arbitrary points of \(P_i\) and \(P_j\) can be written in the form
\[(x_{0/i},\ldots, x_{(i-1)/i}, 1, x_{(i+1)/i}, \ldots, x_{n/i})\in P_i,\qquad (x_{0/j},\ldots, x_{(j-1)/j}, 1, x_{(j+1)/j}, \ldots, x_{n/j})\in P_j\]Now, if we assume that these points come from some point of \(\mathbb{P}^n\), then that point must lie in \(U_i\cap U_j\), and in this set we must have \(x_i,x_j\neq 0\), hence \(x_{j/i}, x_{i/j}\neq 0\). For notational convenience assume \(j>i\); using this fact,
\[[x_{0/i}:\ldots: x_{(i-1)/i}: 1: x_{(i+1)/i}: \ldots: x_{j/i}:\ldots, x_{n/i}]=\left[\frac{x_{0/i}}{x_{j/i}}:\ldots: \frac{x_{(i-1)/i}}{x_{j/i}}: \frac{1}{x_{j/i}}: \frac{x_{(i+1)/i}}{x_{j/i}}: \ldots: 1:\ldots, \frac{x_{n/i}}{x_{j/i}}\right]\]Therefore, for the point on the right-hand side to equal
\[[x_{0/j}:\ldots: x_{(j-1)/j}: 1: x_{(j+1)/j}: \ldots: x_{n/j}]\]the following formula
\[x_{k/i}/x_{j/i}=x_{k/j}\quad\text{for all $k\neq i,j$},\qquad\text{and}\qquad x_{i/j}=1/x_{j/i}\]must hold. Similarly, matching a point of \(P_j\) to a point of \(P_i\) would also yield formulas like \(x_{k/j}/x_{i/j}=x_{k/i}\), but because of \(x_{i/j}=1/x_{j/i}\) this is not a new formula.
Now let us generalize §Schemes, ⁋Example 10 based on this computation. First, consider \(n+1\) affine \(n\)-spaces
\[P_i=\Spec \mathbb{K}[\x_{0/i},\ldots, \x_{n/i}]/(x_{i/i}-1)=\Spec A^i\]Then for the open subschemes \(P_{ij}=D(\x_{j/i})=(A^i)_{\x_{j/i}}\) of \(P_i\) and the following ring homomorphism
\[(A^i)_{\x_{j/i}} \rightarrow (A^j)_{\x_{i/j}};\qquad \x_{k/i}\mapsto \x_{k/j}/\x_{i/j}\quad\text{for all $k\neq i,j$},\qquad\text{and}\qquad \x_{j/i}\mapsto 1/\x_{i/j}\]the isomorphisms \(\varphi_{ij}:P_{ij} \rightarrow P_{ji}\) defined through it almost trivially satisfy the cocycle condition of §Schemes, ⁋Lemma 9, and therefore a unique scheme \(\mathbb{P}^n\) is defined, and if we write elements of \(\mathbb{P}^n\) in the form \([x_0:\ldots:x_n]\), then \(U_i\) is exactly the set of points satisfying \(x_i\neq 0\).
Projective Schemes
As it stands, the above explanation has some incomplete parts. For example, that the \(U_i\) are open subschemes of \(\mathbb{P}^n\) is a consequence of §Schemes, ⁋Lemma 9, but by its very definition it seems that it should be an open set because it is the set where the function \(\x_i\) is nonzero. However, the problem is that \(\x_i\) is not a function on \(\mathbb{P}^n\). Indeed, even in the case \(n=1\) we have checked that \(\mathscr{O}_{\mathbb{P}^1}(\mathbb{P}^1)\cong \mathbb{K}\). This can also be verified purely by the topological construction: the function \(\x_i: \mathbb{R}^{n+1}\setminus\{0\} \rightarrow \mathbb{R}\) that takes a point \((x_0,\ldots, x_n)\) of \(\mathbb{R}^{n+1}\setminus \{0\}\) and returns \(x_i\) is not compatible with \(\sim\), and therefore does not define a function on \(\mathbb{P}^n\). As another example, if a function \(f: \mathbb{R}^2\setminus\{0\} \rightarrow \mathbb{R}\) defined on \(\mathbb{R}^2\setminus\{0\}\) is given by the formula
\[f(x_0,x_1)=x_0^2-x_1\]then
\[f(\lambda x_0,\lambda x_1)=\lambda^2x_0^2-\lambda x_1\neq f(x_0,x_1)\]so \(f\) is not well defined. Instead, if we take \(f\) to be a homogeneous polynomial, then even though \(f\) itself is not well defined as a function, its zero locus \(Z(f)\) is well defined. This is because of the formula
\[f(\lambda x_0,\ldots, \lambda x_n)=\lambda^{\deg f} f(x_0,\ldots, x_n),\qquad \lambda\neq 0\]In other words, to describe \(\mathbb{P}^n\) in a manner similar to the spectrum, we should not view \(\mathbb{A}^{n+1}\) simply as the spectrum of the ring \(\mathbb{K}[\x_0,\ldots, \x_n]\), but add degree information to regard it as a graded ring, and look at the zero loci of homogeneous elements rather than arbitrary elements. Then, recalling [Algebraic Structures] §Graded Rings, ⁋Proposition 6, our interest should also be in homogeneous ideals.
In the remainder of this post we follow the process of taking \(\Proj\) of a graded ring to obtain a projective scheme. To this end we fix some notation.
Unless stated otherwise, a graded ring is always assumed to be \(\mathbb{N}_{\geq0}\)-graded. That is, the ring of interest is always of the form
\[A_\bullet=\bigoplus_{i=0}^\infty A_i=A_0\oplus A_1\oplus\cdots\]Here, since \(A_0\) is itself a ring, \(A_\bullet\) can be regarded as a graded \(A_0\)-algebra, and for this reason we call \(A_0\) the base ring. Also, when we occasionally forget the grading structure on \(A_\bullet\) and regard it as an ordinary ring, we shall simply write \(A\).
Let a graded ring \(A_\bullet\) be given. Then the subset
\[A_+=\bigoplus_{i=1}^\infty A_i=A_1\oplus A_2\oplus\cdots\]is trivially a homogeneous ideal of \(A_\bullet\). However, if we consider the case \(A_\bullet=\mathbb{K}[\x_0,\ldots, \x_n]\), the point at which the function values of all elements of \(A_+\) vanish, that is, the point that is identically zero for all polynomials, is only the origin. Since the origin is the point omitted in constructing \(\mathbb{P}^n\), it is appropriate to exclude from our discussion any ideal containing the ideal \(A_+\). From this viewpoint we call \(A_+\) the irrelevant ideal.
Now, as a set, \(\Proj A_\bullet\) is defined as follows.
Definition 1 For a graded ring \(A_\bullet\), \(\Proj A_\bullet\) is the set
\[\Proj A_\bullet =\{\mathfrak{p}\in \Spec A\mid\text{$\mathfrak{p}$ is homogeneous and $A_+\not\subset \mathfrak{p}$}\}\]By definition \(\Proj A_\bullet\) is a subset of \(\Spec A\). That is, the points of \(\Proj A_\bullet\) are also points of \(\Spec A\). This would be somewhat awkward if we had used \(\mSpec A\) instead of \(\Spec A\), but \(\Spec A\) contains points corresponding to prime ideals in addition to traditional points. For example, considering the ideal \((\x_1-\x_2)\) of \(A=\mathbb{K}[\x_1,\x_2]\), we have \(\mathbb{K}[\x_1,\x_2]/(\x_1-\x_2)\cong \mathbb{K}[\x_1]\), so this ideal is a prime ideal. Moreover, when we regard \(\mathbb{K}[\x_1,\x_2]\) as a graded ring \(A_\bullet\), this ideal is a homogeneous prime ideal not containing \(A_+\), so it is also a point of \(\Proj A_\bullet\).
So far \(\Proj A_\bullet\) is only a set. To give it a topological structure we must use the zero locus of functions, and as observed above, we must use the zero locus of homogeneous polynomials.
Definition 2 Let a graded ring \(A_\bullet\) be given. For a homogeneous ideal \(\mathfrak{a}\) of \(A_\bullet\), define
\[Z_+(\mathfrak{a})=\{\mathfrak{p}\in\Proj A_\bullet\mid \mathfrak{a}\subseteq \mathfrak{p}\}\]Then, using the third result of [Commutative Algebra] §Localization of Graded Rings, ⁋Lemma 2, we can prove the following lemma, similar to §Spectra, ⁋Lemma 6 and §Spectra, ⁋Proposition 5.
Lemma 3 For a graded ring \(A_\bullet\), the following hold.
- For any homogeneous ideals \(\mathfrak{a},\mathfrak{b}\), we have \(Z_+(\mathfrak{a}\mathfrak{b})=Z_+(\mathfrak{a})\cup Z_+(\mathfrak{b})\).
- For any family \(\{\mathfrak{a}_i\}\) of homogeneous ideals, we have \(Z_+(\sum \mathfrak{a}_i)=\bigcap Z_+(\mathfrak{a}_i)\).
- For any homogeneous ideal \(\mathfrak{a}\), we have \(Z_+(\sqrt{\mathfrak{a}})=Z_+(\mathfrak{a})\).
- For any homogeneous ideal \(\mathfrak{a}\), we have \(Z_+(\mathfrak{a})=Z_+(\mathfrak{a}\cap A_+)\).
Of course, it is obvious that \(\mathfrak{a}\mathfrak{b}\), \(\sqrt{\mathfrak{a}}\), and \(\sum \mathfrak{a}_i\) appearing in the above lemma are homogeneous. The first through third results are already observed in the case of spectra; only the fourth is new.
Proof of Lemma 3
- It is obvious that a homogeneous prime ideal \(\mathfrak{p}\) containing \(\mathfrak{a}\) or \(\mathfrak{b}\) also contains the smaller homogeneous ideal \(\mathfrak{ab}\), so it suffices to show the reverse inclusion. Assume \(\mathfrak{p}\supset \mathfrak{ab}\). If \(\mathfrak{p}\not\supseteq \mathfrak{b}\), then we can find an element \(b\in\mathfrak{b}\) with \(b\not\in \mathfrak{p}\). Since \(\mathfrak{b}\) is homogeneous, we can decompose it into a sum of homogeneous elements
Meanwhile, for any homogeneous element \(a\in \mathfrak{a}\), we have \(ab\in \mathfrak{ab}\subseteq \mathfrak{p}\). Consider the element
\[ab=ab_1+\cdots+ab_n\]of \(\mathfrak{ab}\subseteq \mathfrak{p}\); since \(\mathfrak{p}\) is homogeneous, all the \(ab_i\) are elements of \(\mathfrak{p}\). On the other hand, by the previous assumption \(b\not\in \mathfrak{p}\), so there exists an \(i\) satisfying \(b_i\not\in \mathfrak{p}\), and then \(ab_i\) is a homogeneous element belonging to \(\mathfrak{p}\) with \(b_i\not\in \mathfrak{p}\), so by [Commutative Algebra] §Localization of Graded Rings, ⁋Lemma 2 we have \(a\in \mathfrak{p}\). Therefore \(\mathfrak{a}\subseteq \mathfrak{p}\).
- This is obvious because \(\sum \mathfrak{a}_i\) is defined as the smallest ideal containing all the ideals \(\mathfrak{a}_i\).
- [Commutative Algebra] §Properties of Localization, ⁋Corollary 8.
- By definition \(Z_+(\mathfrak{a})\subseteq Z_+(\mathfrak{a}\cap A_+)\) is obvious, so it suffices to show the reverse direction. That is, suppose \(\mathfrak{p}\) is a prime ideal containing all homogeneous elements of \(\mathfrak{a}\) of positive degree but not containing \(A_+\) as a whole; let us show that \(\mathfrak{a}\subseteq \mathfrak{p}\). To do this, it suffices to show that for any \(a\in \mathfrak{a}\cap A_0\), the above assumption implies that \(a\) also belongs to \(\mathfrak{p}\).
Now, since \(A_+\not\subset\mathfrak{p}\), there exists a homogeneous element \(f\) not belonging to \(\mathfrak{p}\). Then \(af\in \mathfrak{a}\cap A_+\subseteq \mathfrak{p}\), and since \(f\not\in \mathfrak{p}\), we have \(a\in \mathfrak{p}\).
From the results of this lemma, the first and second results allow us to define the following.
Definition 4 Let a graded ring \(A_\bullet\) be given. The unique topology on \(\Proj A_\bullet\) having the sets of the form \(Z_+(\mathfrak{a})\) as closed sets is called the Zariski topology.
Also, by the fourth result of this lemma, we know that in defining \(\Proj A_\bullet\) it suffices to consider only homogeneous ideals contained in \(A_+\). This is intuitively obvious: if we set \(A=\mathbb{K}[\x_0,\ldots, \x_n]\), the elements in \(A_0\) are constant functions anyway.
Now we define the following.
Definition 5 For any homogeneous element \(f\) of a graded ring \(A_\bullet\), we write \(D_+(f)\) for the complement of \(Z_+(f)\) in \(\Proj A_\bullet\).
The following corollary follows immediately from the first result of Lemma 3.
Corollary 6 We have \(D_+(f)\cap D_+(g)=D_+(fg)\).
Moreover, the following holds.
Corollary 7 The collection of \(D_+(f)\) forms a base for \(\Proj A_\bullet\).
Proof
Writing an arbitrary homogeneous ideal \(\mathfrak{a}\) of \(A\) using homogeneous generators as \(\mathfrak{a}=\sum_{i\in I} (f_i)\),
\[Z_+(\mathfrak{a})=\bigcap_{i\in I} Z_+((f_i))\]and therefore
\[D_+(\mathfrak{a})=\bigcup_{i\in I} D_+(f_i)\]Meanwhile, we have seen that in the spectrum \(\Spec A\) of a ring \(A\), for any element \(f\in A\), the set \(D(f)\) is isomorphic (as a scheme) to \(\Spec A_f\). A similar result holds for \(D_+(f)\).
Lemma 8 For a graded ring \(A_\bullet\) and any homogeneous element \(f\in A_\bullet\), define a function \(D_+(f) \rightarrow \Spec A_{(f)}\) by the formula
\[\mathfrak{p}\mapsto \mathfrak{p}A_f\cap A_{(f)}\]then this function is a homeomorphism. ([Commutative Algebra] §Localization of Graded Rings, ⁋Definition 5)
Proof
First, since \(f\not\in \mathfrak{p}\), via the localization \(A \rightarrow A_f\), \(\mathfrak{p}\) is carried to the prime ideal \(\mathfrak{p}A_f\) of \(A_f\). ([Commutative Algebra] §Localization, ⁋Proposition 8) Now the right-hand side of the claim is the preimage of \(\mathfrak{p}A_f\) under the inclusion \(i: A_{(f)} \rightarrow A_f\), so it becomes a prime ideal of \(A_{(f)}\).
Now let us define the inverse function \(\Spec A_{(f)} \rightarrow D_+(f)\) of this correspondence. Given an arbitrary prime ideal \(\mathfrak{q}\in\Spec A_{(f)}\), consider the homogeneous ideal \(\mathfrak{p}\) of \(A\) generated by those homogeneous elements \(x\) of \(A\) satisfying the condition
\[\frac{x^{\deg f}}{f^{\deg x}}\in \mathfrak{q}\]Then for any homogeneous elements \(x,y\in \mathfrak{p}\),
\[xy\in \mathfrak{p}\iff \frac{x^{\deg f}}{f^{\deg x}}\frac{y^{\deg f}}{f^{\deg y}}\in \mathfrak{q}\]so from the fact that \(\mathfrak{q}\) is a prime ideal, we know that \(\mathfrak{p}\) is a prime ideal. Now one can easily check that the correspondences \(\mathfrak{p}\mapsto \mathfrak{p}A_f\cap A_{(f)}\) and \(\mathfrak{q}\mapsto \mathfrak{p}\) are inverse to each other, and for any homogeneous ideal \(\mathfrak{a}\) of \(A_\bullet\), the closed set \(Z_+(\mathfrak{a})\cap D_+(f)\) of \(D_+(f)\) is carried by this function to the closed set \(Z(\mathfrak{a}A_f\cap A_{(f)})\) of \(\Spec A_{(f)}\), so we see that this is a homeomorphism.
Then the way to give a scheme structure to \(\Proj A_\bullet\) is now obvious. The proof of the following lemma is almost identical to Lemma 8.
Lemma 9 For a graded ring \(A_\bullet\) and nonzero homogeneous elements \(f,g\), there exists an isomorphism
\[\Spec A_{(fg)}\cong D(g^{\deg f}/f^{\deg g})\subseteq \Spec A_{(f)}\]Therefore, there exists an isomorphism between the principal open set \(D(f^{\deg g}/g^{\deg f})\subseteq \Spec A_{(f)}\) of \(\Spec A_{(g)}\) and the principal open set \(\Spec A_{(fg)}\cong D(g^{\deg f}/f^{\deg g})\) of \(\Spec A_{(f)}\). Now the following theorem is a simple computation.
Theorem 10 The \(\Spec A_{(f)}\) defined above, the open subschemes \(D(g^{\deg f}/f^{\deg g})\), and the isomorphisms
\[D(f^{\deg g}/g^{\deg f})\cong \Spec A_{(fg)}\cong D(g^{\deg f}/f^{\deg g})\]all satisfy the conditions of §Schemes, ⁋Lemma 9, and therefore give a unique scheme structure on \(\Proj A_\bullet\).
In particular, since \(\Proj A_\bullet\) is a locally ringed space, for any \(\mathfrak{p}\in \Proj A_\bullet\) the stalk \(\mathscr{O}_{\Proj A_\bullet,\mathfrak{p}}\) is a local ring. But anyway, since \(\mathfrak{p}\) can be placed in a suitable affine open neighborhood, the following can be shown by essentially the same procedure as §Affine Schemes, ⁋Lemma 8.
Lemma 11 For a graded ring \(A_\bullet\) and any \(\mathfrak{p}\in \Proj A_\bullet\), there exists an isomorphism
\[\mathscr{O}_{\Proj A_\bullet,\mathfrak{p}}\cong A_{(\mathfrak{p})}\]One slightly tricky point is that, unlike \(\Spec\), \(\Proj\) does not define a functor from \(\bgr_{\mathbb{N}_{\geq 0}}\cRing^\op\) to \(\LRS\). This is because even if a graded ring homomorphism \(\phi_\bullet:A_\bullet \rightarrow B_\bullet\) and an arbitrary homogeneous ideal \(\mathfrak{q}\) of \(B\) do not contain \(B_+\), their inverse image \(\phi^{-1}(\mathfrak{q})\) may contain \(A_+\).
Finally, we translate the projective space we examined at the very beginning for motivation into the language of algebraic geometry (almost) completely.
Example 12 In algebraic geometry, \(\mathbb{P}^n_\mathbb{K}\) is defined by the formula
\[\mathbb{P}^n_\mathbb{K}=\Proj \mathbb{K}[\x_0,\ldots, \x_n]\]Here the polynomial algebra \(\mathbb{K}[\x_0,\ldots, \x_n]\) is of course a graded ring with the grading given by degree.
Then the \(n+1\) open cover of projective space can be thought of in this language as
\[D_+(\x_i)\cong \Spec \mathbb{K}[\x_{0},\ldots, \x_{n}]_{(\x_{i})}\]by [Commutative Algebra] §Localization of Graded Rings, ⁋Proposition 6
\[\mathbb{K}[\x_{0},\ldots, \x_{n}]_{(\x_{i})}\cong \mathbb{K}[\x_{0/i},\ldots, \x_{n/i}]/(\x_{i/i}-1)\]explicitly this isomorphism is obtained by applying the first isomorphism theorem to the ring homomorphism
\[\mathbb{K}[\x_{0/i}, \ldots, \x_{n/i}]\rightarrow \mathbb{K}[\x_0,\ldots, \x_n]_{(\x_i)};\qquad \x_{k/i}\mapsto \frac{\x_k}{\x_i}\]Now any \(\mathfrak{p}\in \mathbb{P}^n_\mathbb{K}\) is contained in some \(D_+(\x_i)\). Suppose that via the above isomorphism the point \(\mathfrak{p}\) of \(D_+(\x_i)\) is carried to a point \(\mathfrak{q}\) of \(U_i=\Spec \mathbb{K}[\x_{0/i}, \ldots, \x_{n/i}]/(\x_{i/i}-1)\). Then in this case we naturally expect the isomorphism
\[\mathscr{O}_{\mathbb{P}^n_\mathbb{K},\mathfrak{p}}\cong \mathscr{O}_{U_i, \mathfrak{q}}\]And of course this holds. ([Commutative Algebra] §Localization of Graded Rings, ⁋Proposition 8)
References
[Har] R. Hartshorne, Algebraic geometry. Graduate texts in mathematics. Springer, 1977.
[Vak] R. Vakil, The rising sea: Foundation of algebraic geometry. Available online.
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