Lemma 10.21.1. Let $R$ be a ring. Let $e \in R$ be an idempotent. In this case

$\mathop{\mathrm{Spec}}(R) = D(e) \amalg D(1-e).$

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{\mathrm{Spec}}(R) \mid e \not\in \mathfrak p \} \\ & = & \{ \mathfrak p \in \mathop{\mathrm{Spec}}(R) \mid e \not= 0\text{ in }\kappa (\mathfrak p) \} \\ & = & \{ \mathfrak p \in \mathop{\mathrm{Spec}}(R) \mid e = 1\text{ in }\kappa (\mathfrak p) \} \end{eqnarray*}

Similarly we have

\begin{eqnarray*} D(1-e) & = & \{ \mathfrak p \in \mathop{\mathrm{Spec}}(R) \mid 1 - e \not\in \mathfrak p \} \\ & = & \{ \mathfrak p \in \mathop{\mathrm{Spec}}(R) \mid e \not= 1\text{ in }\kappa (\mathfrak p) \} \\ & = & \{ \mathfrak p \in \mathop{\mathrm{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 $\mathop{\mathrm{Spec}}(R)$. $\square$

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