§2. Homogeneous prime spectra

2.1. Generalities on graded rings and modules

Notation.

Given a positively graded ring $S$, we denote by $S_n$ the subset of $S$ consisting of homogeneous elements of degree $n$ ($n\ge 0$), and by $S_{+}$ the (direct) sum of the $S_n$ for $n>0$. We have $1\in S_0$, S_0 is a subring of $S$, $S_{+}$ is a graded ideal of $S$, and $S$ is the direct sum of S_0 and $S_{+}$. If $M$ is a graded module over $S$ (with positive or negative degrees), we similarly denote by $M_n$ the S_0-module formed by the homogeneous elements of degree $n$ of $M$ (with $n\in \mathbb{Z}$).

For every integer $d>0$, we denote by $S^{(d)}$ the direct sum of the $S_{nd}$; by considering the elements of $S_{nd}$ as homogeneous of degree $n$, the $S_{nd}$ define on $S^{(d)}$ a graded ring structure.

For every integer $k$ such that $0\le k\le d-1$, we denote by $M^{(d,k)}$ the direct sum

of the $M_{nd+k}$ ($n\in \mathbb{Z}$); this is a graded $S^{(d)}$-module when we consider the elements of $M_{nd+k}$ as homogeneous of degree $n$. We write $M^{(d)}$ in place of $M^{(d,0)}$.

With the above notation, for every integer $n$ (positive or negative), we denote by $M(n)$ the graded $S$-module defined by $(M(n){)}_k=M_{n+k}$ for every $k\in \mathbb{Z}$. In particular, $S(n)$ will be a graded $S$-module such that $(S(n){)}_k=S_{n+k}$, with the convention $S_n=0$ for $n<0$. We say that a graded $S$-module $M$ is free if it is isomorphic, as a graded module, to a direct sum of modules of the form $S(n)$; since $S(n)$ is a monogeneous $S$-module, generated by the element 1 of $S$ considered as an element of degree $-n$, this is equivalent to saying that $M$ admits a basis over $S$ consisting of homogeneous elements.

We say that a graded $S$-module $M$ admits a finite presentation if there exists an exact sequence $P\to Q\to M\to 0$, where $P$ and $Q$ are finite direct sums of modules of the form $S(n)$ and the homomorphisms are of degree 0 (cf. (2.1.2)).

(2.1.2)

Let $M$, $N$ be two graded $S$-modules; we define on $M{\otimes }_{S}N$ a graded $S$-module structure as follows. On the tensor product $M{\otimes }_{\mathbb{Z}}N$, we can define a graded $\mathbb{Z}$-module structure (where $\mathbb{Z}$ is graded by ${\mathbb{Z}}_0=\mathbb{Z}$, ${\mathbb{Z}}_n=0$ for $n\ne 0$) by setting $(M{\otimes }_{\mathbb{Z}}N{)}_q={\oplus }_{m+n=q}{M}_m{\otimes }_{\mathbb{Z}}{N}_n$ (since $M$ and $N$ are respectively direct sums of the $M_m$ and the $N_n$, we know that $M{\otimes }_{\mathbb{Z}}N$ can be canonically identified with the direct sum of all the $M_m{\otimes }_{\mathbb{Z}}{N}_n$). This being so, we have $M{\otimes }_{S}N=(M{\otimes }_{\mathbb{Z}}N)/P$, where $P$ is the $\mathbb{Z}$-submodule of $M{\otimes }_{\mathbb{Z}}N$ generated by the elements $(xs)\otimes y-x\otimes (sy)$ for $x\in M$, $y\in N$, $s\in S$; it is clear that $P$ is a graded $\mathbb{Z}$-submodule of $M{\otimes }_{\mathbb{Z}}N$, and we see immediately that passing to the quotient yields a graded $S$-module structure on $M{\otimes }_{S}N$.

For two graded $S$-modules $M$, $N$, recall that a homomorphism $u:M\to N$ of $S$-modules is said to be of degree $k$ if $u(M_j)\subset N_{j+k}$ for every $j\in \mathbb{Z}$. If $H_n$ denotes the set of all homomorphisms of degree $n$ from $M$ to $N$, we denote by ${\mathrm{Hom}}_S(M,N)$ the (direct) sum of the $H_n$ ($n\in \mathbb{Z}$) inside the $S$-module $H$ of all ($S$-module) homomorphisms from $M$ to $N$; in general ${\mathrm{Hom}}_S(M,N)$ is not equal to the latter. However, we do have $H={\mathrm{Hom}}_S(M,N)$ when $M$ is of finite type; for we may then suppose $M$ generated by a finite number of homogeneous elements $x_i$ ($1\le i\le n$), and every homomorphism $u\in H$ can be written in a unique way as ${\sum }_{k\in \mathbb{Z}}{u}_k$, where for each $k$, $u_k(x_i)$ is equal to the homogeneous component of degree $k+deg(x_i)$ of $u(x_i)$ ($1\le i\le n$), which forces $u_k=0$ except for finitely many indices; by definition $u_k\in H_k$, whence the conclusion.

We say that the elements of degree 0 of ${\mathrm{Hom}}_S(M,N)$ are the homomorphisms of graded $S$-modules. It is clear that $S_m{H}_n\subset H_{m+n}$, so the $H_n$ define on ${\mathrm{Hom}}_S(M,N)$ a graded $S$-module structure.

It follows immediately from these definitions that we have

  M(m) ⊗_S N(n) = (M ⊗_S N)(m + n),                                       (2.1.2.1)

  Hom_S(M(m), N(n)) = (Hom_S(M, N))(n − m),                               (2.1.2.2)

for two graded $S$-modules $M$, $N$.

Let $S$, $S^{\prime }$ be two graded rings; a homomorphism of graded rings $\varphi :S\to S^{\prime }$ is a ring homomorphism such that $\varphi (S_n)\subset S_n^{\prime }$ for every $n\in \mathbb{Z}$ (in other words, $\varphi$ must be a homomorphism of degree 0 of graded $\mathbb{Z}$-modules). The data of such a homomorphism endows $S^{\prime }$ with a graded $S$-module structure; equipped with this structure and with its graded ring structure, we say that $S^{\prime }$ is a graded $S$-algebra.

If $M$ is a graded $S$-module, the tensor product $M{\otimes }_S{S}^{\prime }$ of graded $S$-modules is in a natural way endowed with a graded $S^{\prime }$-module structure, the grading being defined as above.

Lemma.

Let $S$ be a positively graded ring. For a subset $E$ of $S_{+}$ consisting of homogeneous elements to generate $S_{+}$ as an $S$-module, it is necessary and sufficient that $E$ generate $S$ as an S_0-algebra.

Proof. The condition is evidently sufficient; we show that it is necessary. Let $E_n$ (resp. $E^n$) be the set of elements of $E$ of degree equal to $n$ (resp. $\le n$); it suffices to show, by induction on $n>0$, that $S_n$ is the S_0-module generated by the elements of degree $n$ that are products of elements of $E^n$. This is evident for $n=1$ by virtue of the hypothesis; the same hypothesis also shows that $S_n={\sum }_{p=0}^{n-1}{S}_p{E}_{n-p}$, and the induction is then immediate.

Corollary.

For $S_{+}$ to be an ideal of finite type, it is necessary and sufficient that $S$ be an S_0-algebra of finite type.

Proof. We may always assume that a finite system of generators of the S_0-algebra $S$ (resp. of the $S$-ideal $S_{+}$) consists of homogeneous elements, by replacing each generator considered by its homogeneous components.

Corollary.

For $S$ to be Noetherian, it is necessary and sufficient that S_0 be Noetherian and that $S$ be an S_0-algebra of finite type.

Proof. The condition is evidently sufficient; it is necessary, since S_0 is isomorphic to $S/S_{+}$ and $S_{+}$ must be an ideal of finite type (2.1.4).

Lemma.

Let $S$ be a positively graded ring that is an S_0-algebra of finite type. Let $M$ be a graded $S$-module of finite type. Then:

(i) The $M_n$ are S_0-modules of finite type, and there exists an integer $n_0$ such that $M_n=0$ for $n\le n_0$.

(ii) There exists an integer $n_1$ and an integer $h>0$ such that, for every integer $n\ge n_1$, $M_{n+h}=S_h{M}_n$.

(iii) For every pair of integers $(d,k)$ with $d>0$, $0\le k\le d-1$, $M^{(d,k)}$ is an $S^{(d)}$-module of finite type.

(iv) For every integer $d>0$, $S^{(d)}$ is an S_0-algebra of finite type.

(v) There exists an integer $h>0$ such that $S_{mh}=(S_h{)}^m$ for every $m>0$.

(vi) For every integer $n>0$, there exists an integer $m_0$ such that $S_m\subset S_{+}^n$ for every $m\ge m_0$.

Proof. We may assume $S$ generated (as an S_0-algebra) by homogeneous elements $f_i$ of degree $h_i$ ($1\le i\le r$), and $M$ generated (as an $S$-module) by homogeneous elements $x_j$ of degree $k_j$ ($1\le j\le s$). It is clear that $M_n$ consists of linear combinations,

with coefficients in S_0, of the elements $f_1^{{\alpha }_1}\cdots f_r^{{\alpha }_r}{x}_j$ such that the ${\alpha }_i$ are integers $\ge 0$ satisfying $k_j+{\sum }_i{\alpha }_i{h}_i=n$; for each $j$, there are only finitely many systems $({\alpha }_i)$ satisfying this equation, since the $h_i$ are > 0, whence the first assertion of (i); the second is evident. On the other hand, let $h$ be the l.c.m. of the $h_i$ and set $g_i=f_i^{h/h_i}$ ($1\le i\le r$), so that all the $g_i$ are of degree $h$; let $z_{\mu }$ be the elements of $M$ of the form $f_1^{{\alpha }_1}\cdots f_r^{{\alpha }_r}{x}_j$ with $0\le {\alpha }_i<h/h_i$ for $1\le i\le r$; these elements are finite in number, so let $n_1$ be the largest of their degrees. It is clear that for $n\ge n_1$, every element of $M_{n+h}$ is a linear combination of the $z_{\mu }$ whose coefficients are monomials of degree > 0 with respect to the $g_i$, whence $M_{n+h}=S_h{M}_n$, which establishes (ii). In the same way, we see (for every $d>0$) that an element of $M^{(d,k)}$ is a linear combination, with coefficients in S_0, of elements of the form $g^d{f}_1^{{\alpha }_1}\cdots f_r^{{\alpha }_r}{x}_j$ with $0\le {\alpha }_i<d$, $g$ being a homogeneous element of $S$; whence (iii); (iv) then follows from (iii) and from Lemma (2.1.3), by taking $M=S_{+}$, since $(S_{+}{)}^{(d)}=(S^{(d)}{)}_{+}$. The assertion of (v) is deduced from (ii) by taking $M=S$. Finally, for a given $n$, there are only finitely many systems $({\alpha }_i)$ such that ${\alpha }_i\ge 0$ and ${\sum }_i{\alpha }_i<n$, so if $m_0$ is the largest value of the sum ${\sum }_i{\alpha }_i{h}_i$ for these systems, we have $S_m\subset S_{+}^n$ for $m>m_0$, which proves (vi).

Corollary.

If $S$ is Noetherian, so is $S^{(d)}$ for every integer $d>0$.

Proof. This follows from (2.1.5) and (2.1.6, (iv)).

(2.1.8)

Let $\mathfrak{p}$ be a graded prime ideal of the graded ring $S$; $\mathfrak{p}$ is then a direct sum of the subgroups ${\mathfrak{p}}_n=\mathfrak{p}\cap S_n$. Suppose that $\mathfrak{p}$ does not contain $S_{+}$. Then if $f\in S_{+}$ is not in $\mathfrak{p}$, the relation $f^{n}x\in \mathfrak{p}$ is equivalent to $x\in \mathfrak{p}$; in particular, if $f\in S_d$ ($d>0$), for every $x\in S_{m-nd}$, the relation $f^{n}x\in {\mathfrak{p}}_m$ is equivalent to $x\in {\mathfrak{p}}_{m-nd}$.

Proposition.

Let $n_0$ be an integer > 0; for every $n\ge n_0$, let ${\mathfrak{p}}_n$ be a subgroup of $S_n$. For there to exist a graded prime ideal $\mathfrak{p}$ of $S$ not containing $S_{+}$ and such that $\mathfrak{p}\cap S_n={\mathfrak{p}}_n$ for every $n\ge n_0$, it is necessary and sufficient that the following conditions be satisfied:

1° $S_m{\mathfrak{p}}_n\subset {\mathfrak{p}}_{m+n}$ for every $m\ge 0$ and every $n\ge n_0$.

2° For $m\ge n_0$, $n\ge n_0$, $f\in S_m$, $g\in S_n$, the relation $fg\in {\mathfrak{p}}_{m+n}$ implies $f\in {\mathfrak{p}}_m$ or $g\in {\mathfrak{p}}_n$.

3° ${\mathfrak{p}}_n\ne S_n$ for at least one $n\ge n_0$.

Moreover, the graded prime ideal $\mathfrak{p}$ is then unique.

Proof. It is evident that conditions 1° and 2° are necessary. Moreover, if $\mathfrak{p}\not\supset S_{+}$, there exists at least one $k>0$ such that $\mathfrak{p}\cap S_k\ne S_k$; if $f\in S_k$ is not in $\mathfrak{p}$, the relation $\mathfrak{p}\cap S_n=S_n$ implies $\mathfrak{p}\cap S_{n-mk}=S_{n-mk}$ by (2.1.8); hence, if $\mathfrak{p}\cap S_n=S_n$ from some value of $n$ onwards, we would have $\mathfrak{p}\supset S_{+}$ contrary to the hypothesis, which proves that 3° is necessary. Conversely, suppose conditions 1°, 2°, and 3° are satisfied. Note that if for an integer $d\ge n_0$, $f\in S_d$ is not in ${\mathfrak{p}}_d$, then, if $\mathfrak{p}$ exists, ${\mathfrak{p}}_m$, for $m<n_0$, is necessarily equal to the set of $x\in S_m$ such that $f^{r}x\in {\mathfrak{p}}_{m+rd}$, except for finitely many values of $r$. This already proves that if $\mathfrak{p}$ exists, it is unique. It remains to show

that if we define the ${\mathfrak{p}}_m$ for $m<n_0$ by the previous condition, then $\mathfrak{p}={\sum }_{n=0}^{\infty }{\mathfrak{p}}_n$ is a prime ideal. Note first that by virtue of 2°, for $m\ge n_0$, ${\mathfrak{p}}_m$ is also defined as the set of $x\in S_m$ such that $f^{r}x\in {\mathfrak{p}}_{m+rd}$ except for finitely many values of $r$. This being so, if $g\in S_m$, $x\in {\mathfrak{p}}_n$, we have $f^{r}gx\in {\mathfrak{p}}_{m+n+rd}$ except for finitely many values of $r$, so $gx\in {\mathfrak{p}}_{m+n}$, which proves that $\mathfrak{p}$ is an ideal of $S$. To establish that this ideal is prime, in other words that the ring $S/\mathfrak{p}$, graded by the subgroups $S_n/{\mathfrak{p}}_n$, is an integral domain, it suffices (by considering the components of highest degree of two elements of $S/\mathfrak{p}$) to prove that if $x\in S_m$, $y\in S_n$ are such that $x\notin {\mathfrak{p}}_m$, $y\notin {\mathfrak{p}}_n$, then $xy\notin {\mathfrak{p}}_{m+n}$. Otherwise, for $r$ large enough, we would have $f^{2r}xy\in {\mathfrak{p}}_{m+n+2rd}$; but $f^{r}y\notin {\mathfrak{p}}_{n+rd}$ for every $r>0$; it follows from 2° that, except for finitely many values of $r$, we have $f^{r}x\in {\mathfrak{p}}_{m+rd}$, and we conclude that $x\in {\mathfrak{p}}_m$ contrary to the hypothesis.

(2.1.10)

We say that a subset $\mathfrak{J}$ of $S_{+}$ is an ideal of $S_{+}$ if it is an ideal of $S$, and that $\mathfrak{J}$ is a graded prime ideal of $S_{+}$ if it is the intersection of $S_{+}$ and a graded prime ideal of $S$ not containing $S_{+}$ (this prime ideal is also unique by (2.1.9)). If $\mathfrak{J}$ is an ideal of $S_{+}$, the radical of $\mathfrak{J}$ in $S_{+}$ is the set of elements of $S_{+}$ having a power in $\mathfrak{J}$, in other words the set ${\mathfrak{r}}_{+}(\mathfrak{J})=\mathfrak{r}(\mathfrak{J})\cap S_{+}$; in particular, the radical of 0 in $S_{+}$ is called the nilradical of $S_{+}$ and denoted ${\mathfrak{N}}_{+}$: it is the set of nilpotent elements of $S_{+}$. If $\mathfrak{J}$ is a graded ideal of $S_{+}$, its radical ${\mathfrak{r}}_{+}(\mathfrak{J})$ is a graded ideal: passing to the quotient ring $S/\mathfrak{J}$, we reduce to the case $\mathfrak{J}=0$, and it remains to see that if $x=x_h+x_{h+1}+\cdots +x_k$ is nilpotent, then so are the $x_i\in S_i$ ($h\le i\le k$); we may assume $x_k\ne 0$, and the component of highest degree of $x^n$ is then $x_k^n$, hence $x_k$ is nilpotent, and we argue by induction on $k$. We say that the graded ring $S$ is essentially reduced if ${\mathfrak{N}}_{+}=0$, in other words if $S_{+}$ contains no nilpotent elements $\ne 0$.

(2.1.11)

We note that if, in the graded ring $S$, an element $x$ is a zero-divisor, so is its component of highest degree. We say that a ring $S$ is essentially integral if the ring $S_{+}$ (without the unit element) contains no zero-divisor and is $\ne 0$; it suffices for this that a homogeneous element $\ne 0$ in $S_{+}$ not be a zero-divisor in that ring. It is clear that if $\mathfrak{p}$ is a graded prime ideal of $S_{+}$, then $S/\mathfrak{p}$ is essentially integral.

Let $S$ be an essentially integral graded ring, and let $x_0\in S_0$: if there exists a homogeneous element $f\ne 0$ of $S_{+}$ such that $x_{0}f=0$, then $x_0{S}_{+}=0$, since $(x_{0}g)f=(x_{0}f)g=0$ for every $g\in S_{+}$, and the hypothesis thus forces $x_{0}g=0$. For $S$ to be integral, it is therefore necessary and sufficient that S_0 be integral and that the annihilator of $S_{+}$ in S_0 be reduced to 0.

2.2. Rings of fractions of a graded ring

(2.2.1)

Let $S$ be a positively graded ring, $f$ a homogeneous element of $S$ of degree $d>0$; then the ring of fractions $S^{\prime }=S_f$ is graded, taking for $S_n^{\prime }$ the set of the $x/f^k$, where $x\in S_{n+kd}$ with $k\ge 0$ (here $n$ may take arbitrary negative values); we denote by $S_{(f)}$ the subring $S_0^{\prime }=(S_f{)}_0$ of $S^{\prime }$ consisting of elements of degree 0.

If $f\in S_d$, then the monomials $(f/1{)}^h$ in $S_f$ ($h$ a positive or negative integer) form a free system over the ring $S_{(f)}$, and the set of their linear combinations is none other than

the ring $(S^{(d)}{)}_f$, which is therefore isomorphic to $S_{(f)}[T,T^{-1}]=S_{(f)}{\otimes }_{\mathbb{Z}}\mathbb{Z}[T,T^{-1}]$ (where $T$ is an indeterminate). Indeed, if we have a relation ${\sum }_{h=-a}^b{z}_h(f/1{)}^h=0$ with $z_h=x_h/f^m$, where the $x_h$ are in $S_{md}$, then this relation is by definition equivalent to the existence of a $k>-a$ such that ${\sum }_{h=-a}^b{f}^{h+k}{x}_h=0$, and since the degrees of the terms of this sum are distinct, we have $f^{h+k}{x}_h=0$ for every $h$, whence $z_h=0$ for every $h$.

If $M$ is a graded $S$-module, then $M^{\prime }=M_f$ is a graded $S_f$-module, $M_n^{\prime }$ being the set of $z/f^k$ with $z\in M_{n+kd}$ ($k\ge 0$); we denote by $M_{(f)}$ the set of homogeneous elements of degree 0 of $M^{\prime }$; it is immediate that $M_{(f)}$ is an $S_{(f)}$-module and that we have $(M^{(d)}{)}_f=M_{(f)}{\otimes }_{S_{(f)}}(S^{(d)}{)}_f$.

Lemma.

Let $d$, $e$ be integers > 0, $f\in S_d$, $g\in S_e$. There exists a canonical ring isomorphism

$$S_{(fg)}\cong (S_{(f)}{)}_{g^d/f^e};$$

if we canonically identify these two rings, there exists a canonical module isomorphism

$$M_{(fg)}\cong (M_{(f)}{)}_{g^d/f^e}.$$

Proof. Indeed, fg divides $f^e{g}^d$, and this latter element divides $(fg{)}^{de}$, so the graded rings $S_{fg}$ and $S_{f^e{g}^d}$ are canonically identified; on the other hand, $S_{f^e{g}^d}$ also identifies with $(S_{f^e}{)}_{g^d/1}$ (0, 1.4.6), and since $f^e/1$ is invertible in $S_{f^e}$, $S_{f^e{g}^d}$ also identifies with $(S_{f^e}{)}_{g^d/f^e}$. Now the element $g^d/f^e$ is of degree 0 in $S_{f^e}$; we conclude immediately that the subring of $(S_{f^e}{)}_{g^d/f^e}$ consisting of elements of degree 0 is $(S_{(f^e)}{)}_{g^d/f^e}$, and since we evidently have $S_{(f^e)}=S_{(f)}$, this proves the first part of the proposition; the second is established in the same way.

(2.2.3)

Under the hypotheses of (2.2.2), it is clear that the canonical homomorphism $S_f\to S_{fg}$ (0, 1.4.1), which sends $x/f^k$ to $g^{k}x/(fg{)}^k$, is of degree 0, and so by restriction yields a canonical homomorphism $S_{(f)}\to S_{(fg)}$, such that the diagram

            S_{(f)}
           /       \
          ↓         ↓
   S_{(fg)} ──→ (S_{(f)})_{g^d/f^e}

commutes. We similarly define a canonical homomorphism $M_{(f)}\to M_{(fg)}$.

Lemma.

If $f$, $g$ are two homogeneous elements of $S_{+}$, the ring $S_{(fg)}$ is generated by the union of the canonical images of $S_{(f)}$ and $S_{(g)}$.

Proof. By virtue of (2.2.2), it suffices to see that $1/(g^d/f^e)=f^{d+e}/(fg{)}^d$ belongs to the canonical image of $S_{(g)}$ in $S_{(fg)}$, which is evident by definition.

Proposition.

Let $d$ be an integer > 0 and let $f\in S_d$. Then there exists a canonical ring isomorphism $S_{(f)}\cong S^{(d)}/(f-1)S^{(d)}$; if we identify these two rings by this isomorphism, there exists a canonical module isomorphism $M_{(f)}\cong M^{(d)}/(f-1)M^{(d)}$.

Proof. The first of these isomorphisms is defined by sending $x/f^n$, where $x\in S_{nd}$, to the element $\bar{x}$, the class of $xmod(f-1)S^{(d)}$; this map is well-defined, since we have the congruence $f^{h}x\equiv x(mod(f-1)S^{(d)})$ for every $x\in S^{(d)}$, so if $f^{h}x=0$ for some $h>0$,

then we have $\bar{x}=0$. On the other hand, if $x\in S_{nd}$ is such that $x=(f-1)y$ with $y=y_{hd}+y_{(h+1)d}+\cdots +y_{kd}$ with $y_{jd}\in S_{jd}$ and $y_{hd}\ne 0$, then we necessarily have $h=n$ and $x=-y_{hd}$, along with the relations $y_{(j+1)d}=f{y}_{jd}$ for $h\le j\le k-1$, $f{y}_{kd}=0$, which finally gives $f^{k-n}x=0$; to every class $\bar{x}mod(f-1)S^{(d)}$ of an element $x\in S_{nd}$, we can therefore associate the element $x/f^n$ of $S_{(f)}$, since the remark above shows that this map is well-defined. It is immediate that the two maps so defined are ring homomorphisms, each the inverse of the other. We proceed in exactly the same way for $M$.

Corollary.

If $S$ is Noetherian, so is $S_{(f)}$ for every $f$ homogeneous of degree > 0.

Proof. This follows immediately from (2.1.7) and (2.2.5).

(2.2.7)

Let $T$ be a multiplicative subset of $S_{+}$ consisting of homogeneous elements; $T_0=T\cup 1$ is then a multiplicative subset of $S$; since the elements of T_0 are homogeneous, the ring $T_0^{-1}S$ is still graded in the evident way; we denote by $S_{(T)}$ the subring of $T_0^{-1}S$ consisting of elements of order 0, that is, of elements of the form $x/h$, where $h\in T$ and $x$ is homogeneous of degree equal to that of $h$. We know (0, 1.4.5) that $T_0^{-1}S$ is canonically identified with the inductive limit of the rings $S_f$, where $f$ runs over $T$ (for the canonical homomorphisms $S_f\to S_{fg}$); since this identification respects degrees, it identifies $S_{(T)}$ with the inductive limit of the $S_{(f)}$ for $f\in T$. For every graded $S$-module $M$, we similarly define the module $M_{(T)}$ (over the ring $S_{(T)}$) consisting of the elements of degree 0 of $T_0^{-1}M$, and we see that this module is the inductive limit of the $M_{(f)}$ for $f\in T$.

If $\mathfrak{p}$ is a graded prime ideal of $S_{+}$, we denote by $S_{(\mathfrak{p})}$ and $M_{(\mathfrak{p})}$ the ring $S_{(T)}$ and the module $M_{(T)}$ respectively, where $T$ is the set of homogeneous elements of $S_{+}$ not belonging to $\mathfrak{p}$.

2.3. Homogeneous prime spectrum of a graded ring

(2.3.1)

Given a positively graded ring $S$, we call the homogeneous prime spectrum of $S$, and denote by $\mathrm{Proj}(S)$, the set of graded prime ideals of $S_{+}$ (2.1.10), or equivalently the set of graded prime ideals of $S$ not containing $S_{+}$; we will define a scheme structure having $\mathrm{Proj}(S)$ as underlying set.

(2.3.2)

For every subset $E$ of $S$, let $V_{+}(E)$ be the set of graded prime ideals of $S$ containing $E$ and not containing $S_{+}$; this is thus the subset $V(E)\cap \mathrm{Proj}(S)$ of $\mathrm{Spec}(S)$. From (I, 1.1.2) we deduce:

  V_+(0) = Proj(S), V_+(S) = V_+(S_+) = ∅                                 (2.3.2.1)

  V_+(⋃_λ E_λ) = ⋂_λ V_+(E_λ)                                             (2.3.2.2)

$$V_{+}(E{E}^{\prime })=V_{+}(E)\cup V_{+}(E^{\prime })(\mathrm{2.3.2.3})$$

We do not change $V_{+}(E)$ by replacing $E$ with the graded ideal generated by $E$; moreover, if $\mathfrak{J}$ is a graded ideal of $S$, we have

  V_+(𝔍) = V_+(⋃_{q ≥ n} (𝔍 ∩ S_q))                                       (2.3.2.4)

for every $n>0$: indeed, if $\mathfrak{p}\in \mathrm{Proj}(S)$ contains the homogeneous elements of $\mathfrak{J}$ of degree $\ge n$, then since by hypothesis there exists a homogeneous element $f\in S_d$ not contained in $\mathfrak{p}$, for every $m\ge 0$ and every $x\in S_m\cap \mathfrak{J}$, we have $f^{r}x\in \mathfrak{J}\cap S_{m+rd}$ for all but finitely many values of $r$, so $f^{r}x\in \mathfrak{p}\cap S_{m+rd}$, which implies $x\in \mathfrak{p}\cap S_m$ (2.1.9).

Finally, we have, for every graded ideal $\mathfrak{J}$ of $S$,

$$V_{+}(\mathfrak{J})=V_{+}({\mathfrak{r}}_{+}(\mathfrak{J})).(\mathrm{2.3.2.5})$$

(2.3.3)

By definition, the $V_{+}(E)$ are the closed subsets of $X=\mathrm{Proj}(S)$ for the topology induced by the spectral topology of $\mathrm{Spec}(S)$, which we will also call the spectral topology on $X$. For every $f\in S$, we set

  D_+(f) = D(f) ∩ Proj(S) = Proj(S) ∖ V_+(f)                              (2.3.3.1)

and consequently, for two elements $f$, $g$ of $S$ (I, 1.1.9.1),

$$D_{+}(fg)=D_{+}(f)\cap D_{+}(g).(\mathrm{2.3.3.2})$$

Proposition.

As $f$ runs over the set of homogeneous elements of $S_{+}$, the $D_{+}(f)$ form a base for the topology of $X=\mathrm{Proj}(S)$.

Proof. It follows from (2.3.2.2) and (2.3.2.4) that every closed subset of $X$ is the intersection of sets of the form $V_{+}(f)$, where $f$ is homogeneous of degree > 0.

(2.3.5)

Let $f$ be a homogeneous element of $S_{+}$, of degree $d>0$; for every graded prime ideal $\mathfrak{p}$ of $S$ not containing $f$, we know that the set of $x/f^n$, where $x\in \mathfrak{p}$ and $n\ge 0$, is a prime ideal of the ring of fractions $S_f$ (0, 1.2.6); its intersection with $S_{(f)}$ is therefore a prime ideal of that ring, which we denote by ${\psi }_f(\mathfrak{p})$: it is the set of $x/f^n$ for $n\ge 0$, $x\in \mathfrak{p}\cap S_{nd}$. We have thus defined a map

$${\psi }_f:D_{+}(f)\to \mathrm{Spec}(S_{(f)});$$

moreover, if $g\in S_e$ is another homogeneous element of $S_{+}$, we have a commutative diagram

   D_+(f)  ──ψ_f──→  Spec(S_{(f)})                                        (2.3.5.1)
     ↑                    ↑
   D_+(fg) ──ψ_{fg}──→ Spec(S_{(fg)})

where the left vertical arrow is the inclusion and the right one is the map ${}^a{\omega }_{fg,f}$ deduced from the canonical homomorphism $\omega ={\omega }_{fg,f}:S_{(f)}\to S_{(fg)}$ (I, 1.2.1). Indeed, if $x/f^n\in {\omega }^{-1}({\psi }_{fg}(\mathfrak{p}))$, where $fg\notin \mathfrak{p}$, then by definition $g^{n}x/(fg{)}^n\in {\psi }_{fg}(\mathfrak{p})$, so $g^{n}x\in \mathfrak{p}$, and hence $x\in \mathfrak{p}$; the converse is evident.

Proposition.

The map ${\psi }_f$ is a homeomorphism from $D_{+}(f)$ to $\mathrm{Spec}(S_{(f)})$.

Proof. First, ${\psi }_f$ is continuous; for if $h\in S_{nd}$ is such that $h/f^n\in {\psi }_f(\mathfrak{p})$, then by definition $h\in \mathfrak{p}$ and conversely, so ${\psi }_f^{-1}(D(h/f^n))=D_{+}(hf)$, and our claim follows from (2.3.3.2). Furthermore, the $D_{+}(hf)$, where $h$ runs over the sets $S_{nd}$, form a base for the topology of $D_{+}(f)$ by (2.3.4) and (2.3.3.2); the

above thus proves, taking into account the (T₀) axiom valid in $D_{+}(f)$ and in $\mathrm{Spec}(S_{(f)})$, that ${\psi }_f$ is injective and that the inverse map ${\psi }_f(D_{+}(f))\to D_{+}(f)$ is continuous. Finally, to see that ${\psi }_f$ is surjective, we remark that if ${\mathfrak{q}}_0$ is a prime ideal of $S_{(f)}$, and if, for every $n>0$, we denote by ${\mathfrak{p}}_n$ the set of $x\in S_n$ such that $x^d/f^n\in {\mathfrak{q}}_0$, then the ${\mathfrak{p}}_n$ satisfy the conditions of (2.1.9): indeed, if $x\in S_n$, $y\in S_n$ are such that $x^d/f^n\in {\mathfrak{q}}_0$ and $y^d/f^n\in {\mathfrak{q}}_0$, then $(x+y{)}^{2d}/f^{2n}\in {\mathfrak{q}}_0$, whence $(x+y{)}^d/f^n\in {\mathfrak{q}}_0$ since ${\mathfrak{q}}_0$ is prime; this proves that the ${\mathfrak{p}}_n$ are subgroups of the $S_n$, and the verification of the other conditions of (2.1.9) is immediate, taking into account that ${\mathfrak{q}}_0$ is prime. If $\mathfrak{p}$ is the graded prime ideal of $S$ thus defined, then indeed ${\psi }_f(\mathfrak{p})={\mathfrak{q}}_0$, since for $x\in S_{nd}$, the relations $x/f^n\in {\mathfrak{q}}_0$ and $x^d/f^{nd}\in {\mathfrak{q}}_0$ are equivalent, ${\mathfrak{q}}_0$ being prime.

Corollary.

For $D_{+}(f)=\varnothing$, it is necessary and sufficient that $f$ be nilpotent.

Proof. For $\mathrm{Spec}(S_{(f)})=\varnothing$, it is necessary and sufficient that $S_{(f)}=0$, or again that 1 = 0 in $S_f$, which means by definition that $f$ is nilpotent.

Corollary.

Let $E$ be a subset of $S_{+}$. The following conditions are equivalent:

(a) $V_{+}(E)=X=\mathrm{Proj}(S)$.

(b) Every element of $E$ is nilpotent.

(c) The homogeneous components of every element of $E$ are nilpotent.

Proof. It is clear that (c) implies (b) and that (b) implies (a). If $\mathfrak{J}$ is the graded ideal of $S$ generated by $E$, condition (a) is equivalent to $V_{+}(\mathfrak{J})=X$; a fortiori, (a) implies that every homogeneous element $f\in \mathfrak{J}$ is such that $V_{+}(f)=X$, hence $f$ is nilpotent by (2.3.7).

Corollary.

If $\mathfrak{J}$ is a graded ideal of $S_{+}$, ${\mathfrak{r}}_{+}(\mathfrak{J})$ is the intersection of the graded prime ideals of $S_{+}$ containing $\mathfrak{J}$.

Proof. Considering the graded ring $S/\mathfrak{J}$, we reduce to the case $\mathfrak{J}=0$. We must prove that if $f\in S_{+}$ is not nilpotent, there is a graded prime ideal of $S$ not containing $f$; but at least one of the homogeneous components of $f$ is not nilpotent, so we may suppose $f$ homogeneous; the assertion then follows from (2.3.7).

(2.3.10)

For every subset $Y$ of $X=\mathrm{Proj}(S)$, let ${\mathfrak{j}}_{+}(Y)$ be the set of $f\in S_{+}$ such that $Y\subset V_{+}(f)$; equivalently, ${\mathfrak{j}}_{+}(Y)=\mathfrak{j}(Y)\cap S_{+}$; then ${\mathfrak{j}}_{+}(Y)$ is an ideal of $S_{+}$, equal to its radical in $S_{+}$.

Proposition.

(i) For every subset $E$ of $S_{+}$, ${\mathfrak{j}}_{+}(V_{+}(E))$ is the radical in $S_{+}$ of the graded ideal of $S_{+}$ generated by $E$.

(ii) For every subset $Y$ of $X$, $V_{+}({\mathfrak{j}}_{+}(Y))=\bar{Y}$, the closure of $Y$ in $X$.

Proof.

(i) If $\mathfrak{J}$ is the graded ideal of $S_{+}$ generated by $E$, then $V_{+}(E)=V_{+}(\mathfrak{J})$, and the assertion follows from (2.3.9).

(ii) Since $V_{+}(\mathfrak{J})={\bigcap }_{f\in \mathfrak{J}}{V}_{+}(f)$, the relation $Y\subset V_{+}(\mathfrak{J})$ implies $Y\subset V_{+}(f)$ for every $f\in \mathfrak{J}$, and hence ${\mathfrak{j}}_{+}(Y)\supset \mathfrak{J}$, whence $V_{+}({\mathfrak{j}}_{+}(Y))\subset V_{+}(\mathfrak{J})$, which proves (ii) by definition of the closed subsets.

Corollary.

The closed subsets $Y$ of $X=\mathrm{Proj}(S)$ and the graded ideals of $S_{+}$ equal to their radical in $S_{+}$ correspond bijectively via the inclusion-reversing maps $Y\mapsto {\mathfrak{j}}_{+}(Y)$, $\mathfrak{J}\mapsto V_{+}(\mathfrak{J})$; to the union $Y_1\cup Y_2$ of two closed subsets of $X$ corresponds

${\mathfrak{j}}_{+}(Y_1)\cap {\mathfrak{j}}_{+}(Y_2)$, and to the intersection of an arbitrary family $(Y_{\lambda })$ of closed subsets corresponds the radical in $S_{+}$ of the sum of the ${\mathfrak{j}}_{+}(Y_{\lambda })$.

Corollary.

Let $\mathfrak{J}$ be a graded ideal of $S_{+}$; for $V_{+}(\mathfrak{J})=\varnothing$, it is necessary and sufficient that every element of $S_{+}$ have a power in $\mathfrak{J}$.

This last corollary can also be expressed in one of the equivalent forms:

Corollary.

Let $(f_{\alpha })$ be a family of homogeneous elements of $S_{+}$. For the $D_{+}(f_{\alpha })$ to form a cover of $X=\mathrm{Proj}(S)$, it is necessary and sufficient that every element of $S_{+}$ have a power in the ideal generated by the $f_{\alpha }$.

Corollary.

Let $(f_{\alpha })$ be a family of homogeneous elements of $S_{+}$ and $f$ an element of $S_{+}$. The following are equivalent: (a) $D_{+}(f)\subset {\bigcup }_{\alpha }{D}_{+}(f_{\alpha })$; (b) $V_{+}(f)\supset {\bigcap }_{\alpha }{V}_{+}(f_{\alpha })$; (c) a power of $f$ belongs to the ideal generated by the $f_{\alpha }$.

Corollary.

For $X=\mathrm{Proj}(S)$ to be empty, it is necessary and sufficient that every element of $S_{+}$ be nilpotent.

Corollary.

In the bijective correspondence described in (2.3.12), the irreducible closed subsets of $X$ correspond to the graded prime ideals of $S_{+}$.

Proof. Indeed, if $Y=Y_1\cup Y_2$, where Y_1 and Y_2 are closed and distinct from $Y$, then

$${\mathfrak{j}}_{+}(Y)={\mathfrak{j}}_{+}(Y_1)\cap {\mathfrak{j}}_{+}(Y_2)$$

with the ideals ${\mathfrak{j}}_{+}(Y_1)$ and ${\mathfrak{j}}_{+}(Y_2)$ distinct from ${\mathfrak{j}}_{+}(Y)$, so ${\mathfrak{j}}_{+}(Y)$ is not prime. Conversely, if $\mathfrak{J}$ is a graded ideal of $S_{+}$ that is not prime, there exist two elements $f$, $g$ of $S_{+}$ such that $f\notin \mathfrak{J}$, $g\notin \mathfrak{J}$, $fg\in \mathfrak{J}$; then $V_{+}(f)\supset V_{+}(\mathfrak{J})$, $V_{+}(g)\supset V_{+}(\mathfrak{J})$, but $V_{+}(\mathfrak{J})\subset V_{+}(f)\cup V_{+}(g)$ by (2.3.2.3); we conclude that $V_{+}(\mathfrak{J})$ is the union of the closed subsets $V_{+}(f)\cap V_{+}(\mathfrak{J})$ and $V_{+}(g)\cap V_{+}(\mathfrak{J})$, which are distinct from $V_{+}(\mathfrak{J})$.

2.4. The scheme structure on $\mathrm{Proj}(S)$

(2.4.1)

Let $f$, $g$ be two homogeneous elements of $S_{+}$; consider the affine schemes $Y_f=\mathrm{Spec}(S_{(f)})$, $Y_g=\mathrm{Spec}(S_{(g)})$, and $Y_{fg}=\mathrm{Spec}(S_{(fg)})$. By (2.2.2), the morphism $w_{fg,f}=({}^a{\omega }_{fg,f},{\tilde{\omega }}_{fg,f})$ from $Y_{fg}$ to $Y_f$, corresponding to the canonical homomorphism ${\omega }_{fg,f}:S_{(f)}\to S_{(fg)}$, is an open immersion (I, 1.3.6). By means of the inverse homeomorphism of ${\psi }_f:D_{+}(f)\to Y_f$ (2.3.6), we can transport the affine scheme structure of $Y_f$ to $D_{+}(f)$; by the commutativity of diagram (2.3.5.1), the affine scheme $D_{+}(fg)$ is thus identified with the scheme induced on the open subset $D_{+}(fg)$ of the underlying space of the affine scheme $D_{+}(f)$. It is then clear (taking (2.3.4) into account) that $X=\mathrm{Proj}(S)$ is endowed with a unique prescheme structure whose restriction to each $D_{+}(f)$ is the affine scheme just defined. Moreover:

Proposition.

The prescheme $\mathrm{Proj}(S)$ is a scheme.

Proof. It suffices (I, 5.5.6) to show that, for any homogeneous $f$, $g$ in $S_{+}$, $D_{+}(f)\cap D_{+}(g)=D_{+}(fg)$ is affine and that its ring is generated by the canonical images of the rings of $D_{+}(f)$ and $D_{+}(g)$; the first point is evident by definition, and the second follows from (2.2.4).

Whenever we speak of the homogeneous prime spectrum $\mathrm{Proj}(S)$ as a scheme, it will always be with respect to the structure just defined.

Example.

Let $S=K[T_1,T_2]$, where $K$ is a field and T_1, T_2 are indeterminates, with $S$ graded by total degree. It follows from (2.3.14) that $\mathrm{Proj}(S)$ is the union of $D_{+}(T_1)$ and $D_{+}(T_2)$; we see immediately that these affine schemes are canonically isomorphic to K[T], and that $\mathrm{Proj}(S)$ is obtained by the gluing of these two affine schemes described in (I, 2.3.2) (cf. (7.4.14)).

Proposition.

Let $S$ be a positively graded ring, and let $X$ be the scheme $\mathrm{Proj}(S)$.

(i) If ${\mathfrak{N}}_{+}$ is the nilradical of $S_{+}$ (2.1.10), the scheme $X_{red}$ is canonically isomorphic to $\mathrm{Proj}(S/{\mathfrak{N}}_{+})$; in particular, if $S$ is essentially reduced, $\mathrm{Proj}(S)$ is reduced.

(ii) Suppose $S$ essentially reduced. Then for $X$ to be integral, it is necessary and sufficient that $S$ be essentially integral.

Proof.

(i) Let $\bar{S}$ be the graded ring $S/{\mathfrak{N}}_{+}$, and denote by $x\mapsto \bar{x}$ the canonical homomorphism $S\to \bar{S}$, of degree 0. For every $f\in S_d$ ($d>0$), the canonical homomorphism $S_f\to {\bar{S}}_{\bar{f}}$ (0, 1.5.1) is surjective and of degree 0, so by restriction yields a surjective homomorphism $S_{(f)}\to {\bar{S}}_{(\bar{f})}$; supposing $f\notin {\mathfrak{N}}_{+}$, we verify immediately that ${\bar{S}}_{(\bar{f})}$ is reduced and that the kernel of the previous homomorphism is the nilradical of $S_{(f)}$, in other words ${\bar{S}}_{(\bar{f})}=(S_{(f)}{)}_{red}$. To this homomorphism therefore corresponds a closed immersion $D_{+}(\bar{f})\to D_{+}(f)$ that identifies $D_{+}(\bar{f})$ with $(D_{+}(f){)}_{red}$ (I, 5.1.2), and which is in particular a homeomorphism of the underlying spaces of these two affine schemes. Furthermore, if $g\notin {\mathfrak{N}}_{+}$ is another homogeneous element of $S_{+}$, the diagram

   D_+(f̄)  ──→  D_+(f)
     ↑           ↑
   D_+(f̄ḡ) ──→  D_+(fg)

commutes; since moreover the $D_{+}(f)$, for $f$ homogeneous of degree > 0 and $f\notin {\mathfrak{N}}_{+}$, form a cover of $X=\mathrm{Proj}(S)$ (2.3.7), we see that the morphisms $D_{+}(\bar{f})\to D_{+}(f)$ are the restrictions of a closed immersion $\mathrm{Proj}(\bar{S})\to \mathrm{Proj}(S)$ which is a homeomorphism of the underlying spaces; whence the conclusion (I, 5.1.2).

(ii) Suppose $S$ essentially integral, in other words (0) is a graded prime ideal of $S_{+}$ distinct from $S_{+}$; then $X$ is reduced by (i) and irreducible by (2.3.17). Conversely, suppose $S$ essentially reduced and $X$ integral; then for $f\ne 0$ homogeneous in $S_{+}$, we have $D_{+}(f)\ne \varnothing$ (2.3.7); the hypothesis that $X$ is irreducible implies $D_{+}(f)\cap D_{+}(g)\ne \varnothing$ for $f$, $g$ homogeneous and $\ne 0$ in $S_{+}$; hence $fg\ne 0$ by (2.3.3.2), and we conclude that $S_{+}$ has no zero-divisors, whence the first assertion.

(2.4.5)

Given a commutative ring $A$, recall that a graded ring $S$ is called a graded $A$-algebra if it is endowed with an $A$-algebra structure such that each of the subgroups $S_n$ is an $A$-module; it suffices for this that S_0 be

an $A$-algebra, in other words we define on $S$ a graded $A$-algebra structure by defining an $A$-algebra structure on S_0 and setting, for $\alpha \in A$, $x\in S_n$, $\alpha \cdot x=(\alpha \cdot 1)x$.

Proposition.

Suppose $S$ is a graded $A$-algebra. Then on $X=\mathrm{Proj}(S)$, the structure sheaf ${\mathcal{O}}_X$ is an $A$-algebra (where $A$ is considered as a simple sheaf on $X$); in other words, $X$ is a scheme over $\mathrm{Spec}(A)$.

Proof. It suffices to note that for every $f$ homogeneous in $S_{+}$, $S_{(f)}$ is an algebra over $A$, and that the diagram

   S_{(f)} ──────→ S_{(fg)}
        ↖        ↗
           A

commutes, for $f$, $g$ homogeneous in $S_{+}$.

Proposition.

Let $S$ be a positively graded ring.

(i) For every integer $d>0$, there exists a canonical isomorphism of the scheme $\mathrm{Proj}(S)$ onto the scheme $\mathrm{Proj}(S^{(d)})$.

(ii) Let $S^{\prime }$ be the graded ring such that $S_0^{\prime }=\mathbb{Z}$, $S_n^{\prime }=S_n$ (considered as a $\mathbb{Z}$-module) for $n>0$. There exists a canonical isomorphism of the scheme $\mathrm{Proj}(S)$ onto the scheme $\mathrm{Proj}(S^{\prime })$.

Proof.

(i) We first show that the map $\mathfrak{p}\mapsto \mathfrak{p}\cap S^{(d)}$ is a bijection from the set $\mathrm{Proj}(S)$ to $\mathrm{Proj}(S^{(d)})$. Indeed, suppose given a graded prime ideal ${\mathfrak{p}}^{\prime }\in \mathrm{Proj}(S^{(d)})$, and set ${\mathfrak{p}}_{nd}={\mathfrak{p}}^{\prime }\cap S_{nd}$ ($n\ge 0$). For every $n>0$ not a multiple of $d$, define ${\mathfrak{p}}_n$ as the set of $x\in S_n$ such that $x^d\in {\mathfrak{p}}_{nd}$; if $x\in {\mathfrak{p}}_n$, $y\in {\mathfrak{p}}_n$, then $(x+y{)}^{2d}\in {\mathfrak{p}}_{2nd}$, so $(x+y{)}^d\in {\mathfrak{p}}_{nd}$ since ${\mathfrak{p}}^{\prime }$ is prime; it is immediate that the ${\mathfrak{p}}_n$ so defined for every $n\ge 0$ satisfy the conditions of (2.1.9), hence there exists a unique prime ideal $\mathfrak{p}\in \mathrm{Proj}(S)$ such that $\mathfrak{p}\cap S^{(d)}={\mathfrak{p}}^{\prime }$. Since for every $f$ homogeneous in $S_{+}$, $V_{+}(f)=V_{+}(f^d)$ (2.3.2.3), we see that the previous bijection is a homeomorphism of topological spaces. Finally, with the same notation, $S_{(f)}$ and $S_{(f^d)}$ are canonically identified (2.2.2), hence $\mathrm{Proj}(S)$ and $\mathrm{Proj}(S^{(d)})$ are canonically identified as schemes.

(ii) If, to every $\mathfrak{p}\in \mathrm{Proj}(S)$, we associate the unique prime ideal ${\mathfrak{p}}^{\prime }\in \mathrm{Proj}(S^{\prime })$ such that ${\mathfrak{p}}^{\prime }\cap S_n=\mathfrak{p}\cap S_n$ for every $n>0$ (2.1.9), it is clear that we define a canonical homeomorphism $\mathrm{Proj}(S)\cong \mathrm{Proj}(S^{\prime })$ of the underlying spaces, since $V_{+}(f)$ is the same set for $S$ and for $S^{\prime }$ when $f$ is a homogeneous element of $S_{+}$. Since moreover $S_{(f)}=S_{(f)}^{\prime }$, $\mathrm{Proj}(S)$ and $\mathrm{Proj}(S^{\prime })$ identify as schemes.

Corollary.

If $S$ is a graded $A$-algebra, $S_A^{\prime }$ the graded $A$-algebra such that $(S_A^{\prime }{)}_0=A$, $(S_A^{\prime }{)}_n=S_n$ for $n>0$, then there exists a canonical isomorphism of $\mathrm{Proj}(S)$ onto $\mathrm{Proj}(S_A^{\prime })$.

Proof. Indeed, these two schemes are both canonically isomorphic to $\mathrm{Proj}(S^{\prime })$, with the notation of (2.4.7, (ii)).

2.5. The sheaf associated to a graded module

(2.5.1)

Let $M$ be a graded $S$-module. For every $f$ homogeneous in $S_{+}$, $M_{(f)}$ is an $S_{(f)}$-module, and to it corresponds an associated quasi-coherent sheaf $\widetilde{M_{(f)}}$ on the affine scheme $\mathrm{Spec}(S_{(f)})$, identified with $D_{+}(f)$ (I, 1.3.4).

Proposition.

There exists on $X=\mathrm{Proj}(S)$ one and only one quasi-coherent ${\mathcal{O}}_X$-module $M$ such that for every $f$ homogeneous in $S_{+}$, we have $\Gamma (D_{+}(f),M)=M_{(f)}$, the restriction homomorphism $\Gamma (D_{+}(f),M)\to \Gamma (D_{+}(fg),M)$, for $f$, $g$ homogeneous in $S_{+}$, being the canonical homomorphism $M_{(f)}\to M_{(fg)}$ (2.2.3).