
## 10.56 Proj of a graded ring

Let $S$ be a graded ring. A homogeneous ideal is simply an ideal $I \subset S$ which is also a graded submodule of $S$. Equivalently, it is an ideal generated by homogeneous elements. Equivalently, if $f \in I$ and

$f = f_0 + f_1 + \ldots + f_ n$

is the decomposition of $f$ into homogeneous parts in $S$ then $f_ i \in I$ for each $i$. To check that a homogeneous ideal $\mathfrak p$ is prime it suffices to check that if $ab \in \mathfrak p$ with $a, b$ homogeneous then either $a \in \mathfrak p$ or $b \in \mathfrak p$.

Definition 10.56.1. Let $S$ be a graded ring. We define $\text{Proj}(S)$ to be the set of homogeneous prime ideals $\mathfrak p$ of $S$ such that $S_{+} \not\subset \mathfrak p$. The set $\text{Proj}(S)$ is a subset of $\mathop{\mathrm{Spec}}(S)$ and we endow it with the induced topology. The topological space $\text{Proj}(S)$ is called the homogeneous spectrum of the graded ring $S$.

Note that by construction there is a continuous map

$\text{Proj}(S) \longrightarrow \mathop{\mathrm{Spec}}(S_0)$

Let $S = \oplus _{d \geq 0} S_ d$ be a graded ring. Let $f\in S_ d$ and assume that $d \geq 1$. We define $S_{(f)}$ to be the subring of $S_ f$ consisting of elements of the form $r/f^ n$ with $r$ homogeneous and $\deg (r) = nd$. If $M$ is a graded $S$-module, then we define the $S_{(f)}$-module $M_{(f)}$ as the sub module of $M_ f$ consisting of elements of the form $x/f^ n$ with $x$ homogeneous of degree $nd$.

Lemma 10.56.2. Let $S$ be a $\mathbf{Z}$-graded ring containing a homogeneous invertible element of positive degree. Then the set $G \subset \mathop{\mathrm{Spec}}(S)$ of $\mathbf{Z}$-graded primes of $S$ (with induced topology) maps homeomorphically to $\mathop{\mathrm{Spec}}(S_0)$.

Proof. First we show that the map is a bijection by constructing an inverse. Let $f \in S_ d$, $d > 0$ be invertible in $S$. If $\mathfrak p_0$ is a prime of $S_0$, then $\mathfrak p_0S$ is a $\mathbf{Z}$-graded ideal of $S$ such that $\mathfrak p_0S \cap S_0 = \mathfrak p_0$. And if $ab \in \mathfrak p_0S$ with $a$, $b$ homogeneous, then $a^ db^ d/f^{\deg (a) + \deg (b)} \in \mathfrak p_0$. Thus either $a^ d/f^{\deg (a)} \in \mathfrak p_0$ or $b^ d/f^{\deg (b)} \in \mathfrak p_0$, in other words either $a^ d \in \mathfrak p_0S$ or $b^ d \in \mathfrak p_0S$. It follows that $\sqrt{\mathfrak p_0S}$ is a $\mathbf{Z}$-graded prime ideal of $S$ whose intersection with $S_0$ is $\mathfrak p_0$.

To show that the map is a homeomorphism we show that the image of $G \cap D(g)$ is open. If $g = \sum g_ i$ with $g_ i \in S_ i$, then by the above $G \cap D(g)$ maps onto the set $\bigcup D(g_ i^ d/f^ i)$ which is open. $\square$

For $f \in S$ homogeneous of degree $> 0$ we define

$D_{+}(f) = \{ \mathfrak p \in \text{Proj}(S) \mid f \not\in \mathfrak p \} .$

Finally, for a homogeneous ideal $I \subset S$ we define

$V_{+}(I) = \{ \mathfrak p \in \text{Proj}(S) \mid I \subset \mathfrak p \} .$

We will use more generally the notation $V_{+}(E)$ for any set $E$ of homogeneous elements $E \subset S$.

Lemma 10.56.3 (Topology on Proj). Let $S = \oplus _{d \geq 0} S_ d$ be a graded ring.

1. The sets $D_{+}(f)$ are open in $\text{Proj}(S)$.

2. We have $D_{+}(ff') = D_{+}(f) \cap D_{+}(f')$.

3. Let $g = g_0 + \ldots + g_ m$ be an element of $S$ with $g_ i \in S_ i$. Then

$D(g) \cap \text{Proj}(S) = (D(g_0) \cap \text{Proj}(S)) \cup \bigcup \nolimits _{i \geq 1} D_{+}(g_ i).$
4. Let $g_0\in S_0$ be a homogeneous element of degree $0$. Then

$D(g_0) \cap \text{Proj}(S) = \bigcup \nolimits _{f \in S_ d, \ d\geq 1} D_{+}(g_0 f).$
5. The open sets $D_{+}(f)$ form a basis for the topology of $\text{Proj}(S)$.

6. Let $f \in S$ be homogeneous of positive degree. The ring $S_ f$ has a natural $\mathbf{Z}$-grading. The ring maps $S \to S_ f \leftarrow S_{(f)}$ induce homeomorphisms

$D_{+}(f) \leftarrow \{ \mathbf{Z}\text{-graded primes of }S_ f\} \to \mathop{\mathrm{Spec}}(S_{(f)}).$
7. There exists an $S$ such that $\text{Proj}(S)$ is not quasi-compact.

8. The sets $V_{+}(I)$ are closed.

9. Any closed subset $T \subset \text{Proj}(S)$ is of the form $V_{+}(I)$ for some homogeneous ideal $I \subset S$.

10. For any graded ideal $I \subset S$ we have $V_{+}(I) = \emptyset$ if and only if $S_{+} \subset \sqrt{I}$.

Proof. Since $D_{+}(f) = \text{Proj}(S) \cap D(f)$, these sets are open. Similarly the sets $V_{+}(I) = \text{Proj}(S) \cap V(E)$ are closed.

Suppose that $T \subset \text{Proj}(S)$ is closed. Then we can write $T = \text{Proj}(S) \cap V(J)$ for some ideal $J \subset S$. By definition of a homogeneous ideal if $g \in J$, $g = g_0 + \ldots + g_ m$ with $g_ d \in S_ d$ then $g_ d \in \mathfrak p$ for all $\mathfrak p \in T$. Thus, letting $I \subset S$ be the ideal generated by the homogeneous parts of the elements of $J$ we have $T = V_{+}(I)$.

The formula for $\text{Proj}(S) \cap D(g)$, with $g \in S$ is direct from the definitions. Consider the formula for $\text{Proj}(S) \cap D(g_0)$. The inclusion of the right hand side in the left hand side is obvious. For the other inclusion, suppose $g_0 \not\in \mathfrak p$ with $\mathfrak p \in \text{Proj}(S)$. If all $g_0f \in \mathfrak p$ for all homogeneous $f$ of positive degree, then we see that $S_{+} \subset \mathfrak p$ which is a contradiction. This gives the other inclusion.

The collection of opens $D(g) \cap \text{Proj}(S)$ forms a basis for the topology since the standard opens $D(g) \subset \mathop{\mathrm{Spec}}(S)$ form a basis for the topology on $\mathop{\mathrm{Spec}}(S)$. By the formulas above we can express $D(g) \cap \text{Proj}(S)$ as a union of opens $D_{+}(f)$. Hence the collection of opens $D_{+}(f)$ forms a basis for the topology also.

First we note that $D_{+}(f)$ may be identified with a subset (with induced topology) of $D(f) = \mathop{\mathrm{Spec}}(S_ f)$ via Lemma 10.16.6. Note that the ring $S_ f$ has a $\mathbf{Z}$-grading. The homogeneous elements are of the form $r/f^ n$ with $r \in S$ homogeneous and have degree $\deg (r/f^ n) = \deg (r) - n\deg (f)$. The subset $D_{+}(f)$ corresponds exactly to those prime ideals $\mathfrak p \subset S_ f$ which are $\mathbf{Z}$-graded ideals (i.e., generated by homogeneous elements). Hence we have to show that the set of $\mathbf{Z}$-graded prime ideals of $S_ f$ maps homeomorphically to $\mathop{\mathrm{Spec}}(S_{(f)})$. This follows from Lemma 10.56.2.

Let $S = \mathbf{Z}[X_1, X_2, X_3, \ldots ]$ with grading such that each $X_ i$ has degree $1$. Then it is easy to see that

$\text{Proj}(S) = \bigcup \nolimits _{i = 1}^\infty D_{+}(X_ i)$

does not have a finite refinement.

Let $I \subset S$ be a graded ideal. If $\sqrt{I} \supset S_{+}$ then $V_{+}(I) = \emptyset$ since every prime $\mathfrak p \in \text{Proj}(S)$ does not contain $S_{+}$ by definition. Conversely, suppose that $S_{+} \not\subset \sqrt{I}$. Then we can find an element $f \in S_{+}$ such that $f$ is not nilpotent modulo $I$. Clearly this means that one of the homogeneous parts of $f$ is not nilpotent modulo $I$, in other words we may (and do) assume that $f$ is homogeneous. This implies that $I S_ f \not= 0$, in other words that $(S/I)_ f$ is not zero. Hence $(S/I)_{(f)} \not= 0$ since it is a ring which maps into $(S/I)_ f$. Pick a prime $\mathfrak q \subset (S/I)_{(f)}$. This corresponds to a graded prime of $S/I$, not containing the irrelevant ideal $(S/I)_{+}$. And this in turn corresponds to a graded prime ideal $\mathfrak p$ of $S$, containing $I$ but not containing $S_{+}$ as desired. $\square$

Example 10.56.4. Let $R$ be a ring. If $S = R[X]$ with $\deg (X) = 1$, then the natural map $\text{Proj}(S) \to \mathop{\mathrm{Spec}}(R)$ is a bijection and in fact a homeomorphism. Namely, suppose $\mathfrak p \in \text{Proj}(S)$. Since $S_{+} \not\subset \mathfrak p$ we see that $X \not\in \mathfrak p$. Thus if $aX^ n \in \mathfrak p$ with $a \in R$ and $n > 0$, then $a \in \mathfrak p$. It follows that $\mathfrak p = \mathfrak p_0S$ with $\mathfrak p_0 = \mathfrak p \cap R$.

If $\mathfrak p \in \text{Proj}(S)$, then we define $S_{(\mathfrak p)}$ to be the ring whose elements are fractions $r/f$ where $r, f \in S$ are homogeneous elements of the same degree such that $f \not\in \mathfrak p$. As usual we say $r/f = r'/f'$ if and only if there exists some $f'' \in S$ homogeneous, $f'' \not\in \mathfrak p$ such that $f''(rf' - r'f) = 0$. Given a graded $S$-module $M$ we let $M_{(\mathfrak p)}$ be the $S_{(\mathfrak p)}$-module whose elements are fractions $x/f$ with $x \in M$ and $f \in S$ homogeneous of the same degree such that $f \not\in \mathfrak p$. We say $x/f = x'/f'$ if and only if there exists some $f'' \in S$ homogeneous, $f'' \not\in \mathfrak p$ such that $f''(xf' - x'f) = 0$.

Lemma 10.56.5. Let $S$ be a graded ring. Let $M$ be a graded $S$-module. Let $\mathfrak p$ be an element of $\text{Proj}(S)$. Let $f \in S$ be a homogeneous element of positive degree such that $f \not\in \mathfrak p$, i.e., $\mathfrak p \in D_{+}(f)$. Let $\mathfrak p' \subset S_{(f)}$ be the element of $\mathop{\mathrm{Spec}}(S_{(f)})$ corresponding to $\mathfrak p$ as in Lemma 10.56.3. Then $S_{(\mathfrak p)} = (S_{(f)})_{\mathfrak p'}$ and compatibly $M_{(\mathfrak p)} = (M_{(f)})_{\mathfrak p'}$.

Proof. We define a map $\psi : M_{(\mathfrak p)} \to (M_{(f)})_{\mathfrak p'}$. Let $x/g \in M_{(\mathfrak p)}$. We set

$\psi (x/g) = (x g^{\deg (f) - 1}/f^{\deg (x)})/(g^{\deg (f)}/f^{\deg (g)}).$

This makes sense since $\deg (x) = \deg (g)$ and since $g^{\deg (f)}/f^{\deg (g)} \not\in \mathfrak p'$. We omit the verification that $\psi$ is well defined, a module map and an isomorphism. Hint: the inverse sends $(x/f^ n)/(g/f^ m)$ to $(xf^ m)/(g f^ n)$. $\square$

Here is a graded variant of Lemma 10.14.2.

Lemma 10.56.6. Suppose $S$ is a graded ring, $\mathfrak p_ i$, $i = 1, \ldots , r$ homogeneous prime ideals and $I \subset S_{+}$ a graded ideal. Assume $I \not\subset \mathfrak p_ i$ for all $i$. Then there exists a homogeneous element $x\in I$ of positive degree such that $x\not\in \mathfrak p_ i$ for all $i$.

Proof. We may assume there are no inclusions among the $\mathfrak p_ i$. The result is true for $r = 1$. Suppose the result holds for $r - 1$. Pick $x \in I$ homogeneous of positive degree such that $x \not\in \mathfrak p_ i$ for all $i = 1, \ldots , r - 1$. If $x \not\in \mathfrak p_ r$ we are done. So assume $x \in \mathfrak p_ r$. If $I \mathfrak p_1 \ldots \mathfrak p_{r-1} \subset \mathfrak p_ r$ then $I \subset \mathfrak p_ r$ a contradiction. Pick $y \in I\mathfrak p_1 \ldots \mathfrak p_{r-1}$ homogeneous and $y \not\in \mathfrak p_ r$. Then $x^{\deg (y)} + y^{\deg (x)}$ works. $\square$

Lemma 10.56.7. Let $S$ be a graded ring. Let $\mathfrak p \subset S$ be a prime. Let $\mathfrak q$ be the homogeneous ideal of $S$ generated by the homogeneous elements of $\mathfrak p$. Then $\mathfrak q$ is a prime ideal of $S$.

Proof. Suppose $f, g \in S$ are such that $fg \in \mathfrak q$. Let $f_ d$ (resp. $g_ e$) be the homogeneous part of $f$ (resp. $g$) of degree $d$ (resp. $e$). Assume $d, e$ are maxima such that $f_ d \not= 0$ and $g_ e \not= 0$. By assumption we can write $fg = \sum a_ i f_ i$ with $f_ i \in \mathfrak p$ homogeneous. Say $\deg (f_ i) = d_ i$. Then $f_ d g_ e = \sum a_ i' f_ i$ with $a_ i'$ to homogeneous par of degree $d + e - d_ i$ of $a_ i$ (or $0$ if $d + e -d_ i < 0$). Hence $f_ d \in \mathfrak p$ or $g_ e \in \mathfrak p$. Hence $f_ d \in \mathfrak q$ or $g_ e \in \mathfrak q$. In the first case replace $f$ by $f - f_ d$, in the second case replace $g$ by $g - g_ e$. Then still $fg \in \mathfrak q$ but the discrete invariant $d + e$ has been decreased. Thus we may continue in this fashion until either $f$ or $g$ is zero. This clearly shows that $fg \in \mathfrak q$ implies either $f \in \mathfrak q$ or $g \in \mathfrak q$ as desired. $\square$

Lemma 10.56.8. Let $S$ be a graded ring.

1. Any minimal prime of $S$ is a homogeneous ideal of $S$.

2. Given a homogeneous ideal $I \subset S$ any minimal prime over $I$ is homogeneous.

Proof. The first assertion holds because the prime $\mathfrak q$ constructed in Lemma 10.56.7 satisfies $\mathfrak q \subset \mathfrak p$. The second because we may consider $S/I$ and apply the first part. $\square$

Lemma 10.56.9. Let $R$ be a ring. Let $S$ be a graded $R$-algebra. Let $f \in S_{+}$ be homogeneous. Assume that $S$ is of finite type over $R$. Then

1. the ring $S_{(f)}$ is of finite type over $R$, and

2. for any finite graded $S$-module $M$ the module $M_{(f)}$ is a finite $S_{(f)}$-module.

Proof. Choose $f_1, \ldots , f_ n \in S$ which generate $S$ as an $R$-algebra. We may assume that each $f_ i$ is homogeneous (by decomposing each $f_ i$ into its homogeneous components). An element of $S_{(f)}$ is a sum of the form

$\sum \nolimits _{e\deg (f) = \sum e_ i\deg (f_ i)} \lambda _{e_1 \ldots e_ n} f_1^{e_1} \ldots f_ n^{e_ n}/f^ e$

with $\lambda _{e_1 \ldots e_ n} \in R$. Thus $S_{(f)}$ is generated as an $R$-algebra by the $f_1^{e_1} \ldots f_ n^{e_ n} /f^ e$ with the property that $e\deg (f) = \sum e_ i\deg (f_ i)$. If $e_ i \geq \deg (f)$ then we can write this as

$f_1^{e_1} \ldots f_ n^{e_ n}/f^ e = f_ i^{\deg (f)}/f^{\deg (f_ i)} \cdot f_1^{e_1} \ldots f_ i^{e_ i - \deg (f)} \ldots f_ n^{e_ n}/f^{e - \deg (f_ i)}$

Thus we only need the elements $f_ i^{\deg (f)}/f^{\deg (f_ i)}$ as well as the elements $f_1^{e_1} \ldots f_ n^{e_ n} /f^ e$ with $e \deg (f) = \sum e_ i \deg (f_ i)$ and $e_ i < \deg (f)$. This is a finite list and we see that (1) is true.

To see (2) suppose that $M$ is generated by homogeneous elements $x_1, \ldots , x_ m$. Then arguing as above we find that $M_{(f)}$ is generated as an $S_{(f)}$-module by the finite list of elements of the form $f_1^{e_1} \ldots f_ n^{e_ n} x_ j /f^ e$ with $e \deg (f) = \sum e_ i \deg (f_ i) + \deg (x_ j)$ and $e_ i < \deg (f)$. $\square$

Lemma 10.56.10. Let $R$ be a ring. Let $R'$ be a finite type $R$-algebra, and let $M$ be a finite $R'$-module. There exists a graded $R$-algebra $S$, a graded $S$-module $N$ and an element $f \in S$ homogeneous of degree $1$ such that

1. $R' \cong S_{(f)}$ and $M \cong N_{(f)}$ (as modules),

2. $S_0 = R$ and $S$ is generated by finitely many elements of degree $1$ over $R$, and

3. $N$ is a finite $S$-module.

Proof. We may write $R' = R[x_1, \ldots , x_ n]/I$ for some ideal $I$. For an element $g \in R[x_1, \ldots , x_ n]$ denote $\tilde g \in R[X_0, \ldots , X_ n]$ the element homogeneous of minimal degree such that $g = \tilde g(1, x_1, \ldots , x_ n)$. Let $\tilde I \subset R[X_0, \ldots , X_ n]$ generated by all elements $\tilde g$, $g \in I$. Set $S = R[X_0, \ldots , X_ n]/\tilde I$ and denote $f$ the image of $X_0$ in $S$. By construction we have an isomorphism

$S_{(f)} \longrightarrow R', \quad X_ i/X_0 \longmapsto x_ i.$

To do the same thing with the module $M$ we choose a presentation

$M = (R')^{\oplus r}/\sum \nolimits _{j \in J} R'k_ j$

with $k_ j = (k_{1j}, \ldots , k_{rj})$. Let $d_{ij} = \deg (\tilde k_{ij})$. Set $d_ j = \max \{ d_{ij}\}$. Set $K_{ij} = X_0^{d_ j - d_{ij}}\tilde k_{ij}$ which is homogeneous of degree $d_ j$. With this notation we set

$N = \mathop{\mathrm{Coker}}\Big( \bigoplus \nolimits _{j \in J} S(-d_ j) \xrightarrow {(K_{ij})} S^{\oplus r} \Big)$

which works. Some details omitted. $\square$

Comment #1373 by Keenan Kidwell on

In the previous section (Tag 00JL), the convention is made that a graded ring is a $\mathbf{Z}_{\geq 0}$-graded ring. But Tag 00JO makes reference to $\mathbf{Z}$-graded rings, the definition of which is obvious, but slightly clashes with the aforementioned convention, right? The remark at the beginning of this section that primality of a homogeneous ideal in a graded ring can be checked on homogeneous elements" in the natural sense is used implicitly for $\mathbf{Z}$-graded rings in the proof of 00JO, as is the fact that the radical of a homogeneous ideal (in a $\mathbf{Z}$-graded ring!) is itself homogeneous. The latter fact follows from the former via Tag 00JT, which shows that, while a priori the radical of an ideal $I$ is the intersection of the primes which contain it, if $I$ is homogeneous, the radical is already the intersection of the homogeneous primes which contain it. I'm just pointing a bunch of stuff out and not offering any constructive suggestions. I guess what I would propose is the following. Introduce the notion of a $\mathbf{Z}$-graded ring (to be distinguished from a graded ring, which by convention is $\mathbf{Z}_{\geq 0}$-graded). Give the all-important example of $S_f$ for $S$ graded and $f$ of positive degree (which is what we care about anyway). Perhaps the result about checking primality of homogeneous ideals (which holds in $\mathbf{Z}$-graded rings) should get its own tag at the beginning of this section. One can then deduce 00JT from this result immediately (the proof of 00JT is basically the proof of the primality-checking thing). Now insert the corollary that radicals of homogeneous ideals (in $\mathbf{Z}$-graded rings) are homogeneous. And cite this in the proof of 00JO. (The argument there proves that if $a$ and $b$ are homogeneous elements with $ab$ in the radical of $\mathfrak{p}_0S$, then one of $a,b$ is in the radical of $\mathfrak{p}_0S$, so there is a direct but implicit application of the primality-checking result.) If this seems worthwhile, I could do it and submit it. I wrote this stuff up for myself at some point already anyway. But maybe I'm just being pedantic.

Comment #1374 by Keenan Kidwell on

Also, I think one should argue that $(\mathfrak{p}\cap S_0)S=\mathfrak{p}$ for a homogeneous prime $\mathfrak{p}$ of $S$ (or it's obvious?). The forward containment is obvious. If $s\in\mathfrak{p}$ is homogeneous, then $f^{-\deg(s)}s\in\mathfrak{p}\cap S_0$, so $s=f^{\deg(s)}(f^{-\deg(s)}s)\in(\mathfrak{p}\cap S_0)S$. This is enough to conclude the reverse containment since $\mathfrak{p}$ is generated by its homogeneous elements.

Comment #1386 by on

You are right. In fact, we should introduce $\Gamma$-graded rings, where $\Gamma$ is a commutative(?) monoid. Then we should introduce $\Gamma$-graded modules. Except then there is the funny business with our graded modules over a graded ring not following this pattern. So I guess then we should introduce, given a homomorphism $\Gamma \to \Gamma'$ of monoids, $\Gamma'$-graded modules over a $\Gamma$-graded ring. Sigh!

Yeah, some of this stuff has to go into Section 10.55 and note how that section does not have formal definitions. The initial write-up made the, reasonable I feel, assumption that the reader may already have an idea of what a $\mathbf{Z}$-graded ring is. I've decided to leave it as is for now, but feel free to edit a bit and send in the result.

Comment #2803 by Saurav on

For part 10, lemma 10.56.3 : $V_+(I) = V(I) \cap Proj(S)$. One can then use the fact that if $\mathfrak{p}^h = \oplus \mathfrak{p} \cap S_d$, the homogeneous part of a prime ideal is also a prime. Therefore $V_+(I) = \emptyset$ if and only if $\mathfrak{p}^h \supset S_+$ for every $\mathfrak{p}^h\in V(I)$ which implies $S_+ \subset \mathfrak{p}^h$. (Shorter argument!)

Comment #2804 by Saurav on

For part 10, lemma 10.56.3 : $V_+(I) = V(I) \cap Proj(S)$. One can then use the fact that $\mathfrak{p}^h = \oplus \mathfrak{p} \cap S_d$, the homogeneous part of a prime ideal is also a prime. Therefore $V_+(I) = \emptyset$ if and only if $\mathfrak{p}^h \supset S_+$ for every $\mathfrak{p}\in V(I)$ which implies $S_+ \subset \mathfrak{p}$. (Shorter argument!)

Comment #2906 by on

@#2803 and #2804. Yes, I sort of agree. But not enough to replace the current one.

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