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Tag 00JP

Chapter 10: Commutative Algebra > Section 10.56: Proj of a graded ring

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. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 12823–12865 (see updates for more information).

    \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}

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