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Tag 00F6

Chapter 10: Commutative Algebra > Section 10.28: Images of ring maps of finite presentation

Lemma 10.28.1. Let $U \subset \mathop{\rm Spec}(R)$ be open. The following are equivalent:

  1. $U$ is retrocompact in $\mathop{\rm Spec}(R)$,
  2. $U$ is quasi-compact,
  3. $U$ is a finite union of standard opens, and
  4. there exists a finitely generated ideal $I \subset R$ such that $X \setminus V(I) = U$.

Proof. We have (1) $\Rightarrow$ (2) because $\mathop{\rm Spec}(R)$ is quasi-compact, see Lemma 10.16.10. We have (2) $\Rightarrow$ (3) because standard opens form a basis for the topology. Proof of (3) $\Rightarrow$ (1). Let $U = \bigcup_{i = 1\ldots n} D(f_i)$. To show that $U$ is retrocompact in $\mathop{\rm Spec}(R)$ it suffices to show that $U \cap V$ is quasi-compact for any quasi-compact open $V$ of $\mathop{\rm Spec}(R)$. Write $V = \bigcup_{j = 1\ldots m} D(g_j)$ which is possible by (2) $\Rightarrow$ (3). Each standard open is homeomorphic to the spectrum of a ring and hence quasi-compact, see Lemmas 10.16.6 and 10.16.10. Thus $U \cap V = (\bigcup_{i = 1\ldots n} D(f_i)) \cap (\bigcup_{j = 1\ldots m} D(g_j)) = \bigcup_{i, j} D(f_i g_j)$ is a finite union of quasi-compact opens hence quasi-compact. To finish the proof note that (4) is equivalent to (3) by Lemma 10.16.2. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 5077–5088 (see updates for more information).

    \begin{lemma}
    \label{lemma-qc-open}
    Let $U \subset \Spec(R)$ be open. The following
    are equivalent:
    \begin{enumerate}
    \item $U$ is retrocompact in $\Spec(R)$,
    \item $U$ is quasi-compact,
    \item $U$ is a finite union of standard opens, and
    \item there exists a finitely generated ideal $I \subset R$ such
    that $X \setminus V(I) = U$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    We have (1) $\Rightarrow$ (2) because $\Spec(R)$ is quasi-compact, see
    Lemma \ref{lemma-quasi-compact}. We have (2) $\Rightarrow$ (3) because
    standard opens form a basis for the topology. Proof of (3) $\Rightarrow$ (1).
    Let $U = \bigcup_{i = 1\ldots n} D(f_i)$. To show that $U$ is retrocompact
    in $\Spec(R)$ it suffices to show that $U \cap V$ is quasi-compact for any
    quasi-compact open $V$ of $\Spec(R)$. Write
    $V = \bigcup_{j = 1\ldots m} D(g_j)$ which is possible by (2) $\Rightarrow$
    (3). Each standard open is homeomorphic to the spectrum of a ring and hence
    quasi-compact, see Lemmas \ref{lemma-standard-open} and
    \ref{lemma-quasi-compact}. Thus
    $U \cap V =
    (\bigcup_{i = 1\ldots n} D(f_i)) \cap (\bigcup_{j = 1\ldots m} D(g_j))
    = \bigcup_{i, j} D(f_i g_j)$ is a finite union of quasi-compact opens
    hence quasi-compact. To finish the proof note
    that (4) is equivalent to (3) by 
    Lemma \ref{lemma-Zariski-topology}.
    \end{proof}

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