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Tag 00EC

Chapter 10: Commutative Algebra > Section 10.20: Open and closed subsets of spectra

Lemma 10.20.1. Let $R$ be a ring. Let $e \in R$ be an idempotent. In this case $$ \mathop{\rm Spec}(R) = D(e) \amalg D(1-e). $$

Proof. Note that an idempotent $e$ of a domain is either $1$ or $0$. Hence we see that \begin{eqnarray*} D(e) & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e \not\in \mathfrak p \} \\ & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e \not = 0\text{ in }\kappa(\mathfrak p) \} \\ & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e = 1\text{ in }\kappa(\mathfrak p) \} \end{eqnarray*} Similarly we have \begin{eqnarray*} D(1-e) & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid 1 - e \not\in \mathfrak p \} \\ & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e \not = 1\text{ in }\kappa(\mathfrak p) \} \\ & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e = 0\text{ in }\kappa(\mathfrak p) \} \end{eqnarray*} Since the image of $e$ in any residue field is either $1$ or $0$ we deduce that $D(e)$ and $D(1-e)$ cover all of $\mathop{\rm Spec}(R)$. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 3447–3454 (see updates for more information).

    \begin{lemma}
    \label{lemma-idempotent-spec}
    Let $R$ be a ring. Let $e \in R$ be an idempotent.
    In this case
    $$
    \Spec(R) = D(e) \amalg D(1-e).
    $$
    \end{lemma}
    
    \begin{proof}
    Note that an idempotent $e$ of a domain is either $1$ or $0$.
    Hence we see that
    \begin{eqnarray*}
    D(e)
    & = &
    \{ \mathfrak p \in \Spec(R)
    \mid
    e \not\in \mathfrak p \} \\
    & = &
    \{ \mathfrak p \in \Spec(R)
    \mid
    e \not = 0\text{ in }\kappa(\mathfrak p) \} \\
    & = &
    \{ \mathfrak p \in \Spec(R)
    \mid
    e = 1\text{ in }\kappa(\mathfrak p) \}
    \end{eqnarray*}
    Similarly we have
    \begin{eqnarray*}
    D(1-e)
    & = &
    \{ \mathfrak p \in \Spec(R)
    \mid
    1 - e \not\in \mathfrak p \} \\
    & = &
    \{ \mathfrak p \in \Spec(R)
    \mid
    e \not = 1\text{ in }\kappa(\mathfrak p) \} \\
    & = &
    \{ \mathfrak p \in \Spec(R)
    \mid
    e = 0\text{ in }\kappa(\mathfrak p) \}
    \end{eqnarray*}
    Since the image of $e$ in any residue field is either $1$ or $0$
    we deduce that $D(e)$ and $D(1-e)$ cover all of $\Spec(R)$.
    \end{proof}

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