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Tag 00E8

Chapter 10: Commutative Algebra > Section 10.16: The spectrum of a ring

Lemma 10.16.10. Let $R$ be a ring. The space $\mathop{\rm Spec}(R)$ is quasi-compact.

Proof. It suffices to prove that any covering of $\mathop{\rm Spec}(R)$ by standard opens can be refined by a finite covering. Thus suppose that $\mathop{\rm Spec}(R) = \cup D(f_i)$ for a set of elements $\{f_i\}_{i\in I}$ of $R$. This means that $\cap V(f_i) = \emptyset$. According to Lemma 10.16.2 this means that $V(\{f_i \}) = \emptyset$. According to the same lemma this means that the ideal generated by the $f_i$ is the unit ideal of $R$. This means that we can write $1$ as a finite sum: $1 = \sum_{i \in J} r_i f_i$ with $J \subset I$ finite. And then it follows that $\mathop{\rm Spec}(R) = \cup_{i \in J} D(f_i)$. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 3143–3146 (see updates for more information).

    \begin{lemma}
    \label{lemma-quasi-compact}
    Let $R$ be a ring. The space $\Spec(R)$ is quasi-compact.
    \end{lemma}
    
    \begin{proof}
    It suffices to prove that any covering of $\Spec(R)$
    by standard opens can be refined by a finite covering.
    Thus suppose that $\Spec(R) = \cup D(f_i)$
    for a set of elements $\{f_i\}_{i\in I}$ of $R$. This means that
    $\cap V(f_i) = \emptyset$. According to Lemma
    \ref{lemma-Zariski-topology} this means that
    $V(\{f_i \}) = \emptyset$. According to the
    same lemma this means that the ideal generated
    by the $f_i$ is the unit ideal of $R$. This means
    that we can write $1$ as a {\it finite} sum:
    $1 = \sum_{i \in J} r_i f_i$ with $J \subset I$ finite.
    And then it follows that $\Spec(R)
    = \cup_{i \in J} D(f_i)$.
    \end{proof}

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