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Tag 00JM

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

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 $G \subset \mathop{\rm Spec}(S)$ of $\mathbf{Z}$-graded primes of $S$ (with induced topology) maps homeomorphically to $\mathop{\rm Spec}(S_0)$.

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 $$ D_{+}(f) \leftarrow \{\mathbf{Z}\text{-graded primes of }S_f\} \to \mathop{\rm Spec}(S_{(f)}). $$
  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 $D(g) \subset \mathop{\rm Spec}(S)$ form a basis for the topology on $\mathop{\rm Spec}(S)$. 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 $D(f) = \mathop{\rm Spec}(S_f)$ 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 $\mathop{\rm Spec}(S_{(f)})$. 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 $\text{Proj}(S) \to \mathop{\rm Spec}(R)$ 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 $\mathop{\rm Spec}(S_{(f)})$ 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 $$ N = \mathop{\rm Coker}\Big( \bigoplus\nolimits_{j \in J} S(-d_j) \xrightarrow{(K_{ij})} S^{\oplus r} \Big) $$ which works. Some details omitted. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 12736–13152 (see updates for more information).

    \section{Proj of a graded ring}
    \label{section-proj}
    
    \noindent
    Let $S$ be a graded ring.
    A {\it 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
    $b \in \mathfrak p$.
    
    \begin{definition}
    \label{definition-proj}
    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 $\Spec(S)$
    and we endow it with the induced topology.
    The topological space $\text{Proj}(S)$ is called the
    {\it homogeneous spectrum} of the graded ring $S$.
    \end{definition}
    
    \noindent
    Note that by construction there is a continuous map
    $$
    \text{Proj}(S) \longrightarrow \Spec(S_0)
    $$
    
    \medskip\noindent
    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 $nd$.
    
    \begin{lemma}
    \label{lemma-Z-graded}
    Let $S$ be a $\mathbf{Z}$-graded ring containing a homogeneous
    invertible element of positive degree. Then the set
    $G \subset \Spec(S)$ of $\mathbf{Z}$-graded primes of $S$
    (with induced topology) maps homeomorphically to $\Spec(S_0)$.
    \end{lemma}
    
    \begin{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 $\mathfrak p_0$.
    
    \medskip\noindent
    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.
    \end{proof}
    
    \noindent
    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 $E \subset S$.
    
    \begin{lemma}[Topology on Proj]
    \label{lemma-topology-proj}
    Let $S = \oplus_{d \geq 0} S_d$ be a graded ring.
    \begin{enumerate}
    \item The sets $D_{+}(f)$ are open in $\text{Proj}(S)$.
    \item We have $D_{+}(ff') = D_{+}(f) \cap D_{+}(f')$.
    \item Let $g = g_0 + \ldots + g_m$ be an element
    of $S$ with $g_i \in S_i$. Then
    $$
    D(g) \cap \text{Proj}(S) =
    (D(g_0) \cap \text{Proj}(S))
    \cup
    \bigcup\nolimits_{i \geq 1} D_{+}(g_i).
    $$
    \item
    Let $g_0\in S_0$ be a homogeneous element of degree $0$. Then
    $$
    D(g_0) \cap \text{Proj}(S)
    =
    \bigcup\nolimits_{f \in S_d, \ d\geq 1} D_{+}(g_0 f).
    $$
    \item The open sets $D_{+}(f)$ form a
    basis for the topology of $\text{Proj}(S)$.
    \item 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
    $$
    D_{+}(f)
    \leftarrow
    \{\mathbf{Z}\text{-graded primes of }S_f\}
    \to
    \Spec(S_{(f)}).
    $$
    \item There exists an $S$ such that $\text{Proj}(S)$ is not
    quasi-compact.
    \item The sets $V_{+}(I)$ are closed.
    \item Any closed subset $T \subset \text{Proj}(S)$ is of
    the form $V_{+}(I)$ for some homogeneous ideal $I \subset S$.
    \item For any graded ideal $I \subset S$ we have
    $V_{+}(I) = \emptyset$ if and only if $S_{+} \subset \sqrt{I}$.
    \end{enumerate}
    \end{lemma}
    
    \begin{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.
    
    \medskip\noindent
    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 $T = V_{+}(I)$.
    
    \medskip\noindent
    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.
    
    \medskip\noindent
    The collection of opens $D(g) \cap \text{Proj}(S)$
    forms a basis for the topology since the standard opens
    $D(g) \subset \Spec(S)$ form a basis for the topology on
    $\Spec(S)$. 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.
    
    \medskip\noindent
    First we note that $D_{+}(f)$ may be identified
    with a subset (with induced topology) of $D(f) = \Spec(S_f)$
    via Lemma \ref{lemma-standard-open}. 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 $\Spec(S_{(f)})$. This follows from Lemma \ref{lemma-Z-graded}.
    
    \medskip\noindent
    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.
    
    \medskip\noindent
    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.
    \end{proof}
    
    \begin{example}
    \label{example-proj-polynomial-ring-1-variable}
    Let $R$ be a ring. If $S = R[X]$ with $\deg(X) = 1$, then the natural map
    $\text{Proj}(S) \to \Spec(R)$ 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 $\mathfrak p_0 = \mathfrak p \cap R$.
    \end{example}
    
    \noindent
    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 $f''(xf' - x'f) = 0$.
    
    \begin{lemma}
    \label{lemma-proj-prime}
    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
    $\Spec(S_{(f)})$ corresponding to $\mathfrak p$ as in
    Lemma \ref{lemma-topology-proj}. Then
    $S_{(\mathfrak p)} = (S_{(f)})_{\mathfrak p'}$
    and compatibly
    $M_{(\mathfrak p)} = (M_{(f)})_{\mathfrak p'}$.
    \end{lemma}
    
    \begin{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
    $(xf^m)/(g f^n)$.
    \end{proof}
    
    \noindent
    Here is a graded variant of Lemma \ref{lemma-silly}.
    
    \begin{lemma}
    \label{lemma-graded-silly}
    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 $i$.
    \end{lemma}
    
    \begin{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.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-smear-out}
    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 $S$.
    \end{lemma}
    
    \begin{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.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-graded-ring-minimal-prime}
    Let $S$ be a graded ring.
    \begin{enumerate}
    \item Any minimal prime of $S$ is a homogeneous ideal of $S$.
    \item Given a homogeneous ideal $I \subset S$ any minimal
    prime over $I$ is homogeneous.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    The first assertion holds because the prime $\mathfrak q$ constructed in
    Lemma \ref{lemma-smear-out} satisfies $\mathfrak q \subset \mathfrak p$.
    The second because we may consider $S/I$ and apply the first part.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-dehomogenize-finite-type}
    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
    \begin{enumerate}
    \item the ring $S_{(f)}$ is of finite type over $R$, and
    \item for any finite graded $S$-module $M$ the module $M_{(f)}$
    is a finite $S_{(f)}$-module.
    \end{enumerate}
    \end{lemma}
    
    \begin{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.
    
    \medskip\noindent
    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
    $e_i < \deg(f)$.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-homogenize}
    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
    \begin{enumerate}
    \item $R' \cong S_{(f)}$ and $M \cong N_{(f)}$ (as modules),
    \item $S_0 = R$ and $S$ is generated by finitely many elements
    of degree $1$ over $R$, and
    \item $N$ is a finite $S$-module.
    \end{enumerate}
    \end{lemma}
    
    \begin{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
    $$
    N = \Coker\Big(
    \bigoplus\nolimits_{j \in J} S(-d_j) \xrightarrow{(K_{ij})} S^{\oplus r}
    \Big)
    $$
    which works. Some details omitted.
    \end{proof}

    Comments (5)

    Comment #1373 by Keenan Kidwell on March 26, 2015 a 4:50 pm UTC

    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 March 26, 2015 a 5:08 pm UTC

    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 Johan (site) on April 1, 2015 a 7:49 pm UTC

    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 00JL 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 September 12, 2017 a 3:51 am UTC

    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 September 12, 2017 a 3:53 am UTC

    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!)

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