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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