**Proof.**
By Nakayama's Lemma 10.20.1 there exists an element $f = 1 + i$, $i \in I$ such that $fI = 0$. Then $f^2 = f + fi = f$ is an idempotent. Consider the idempotent $e = 1 - f = -i \in I$. For $j \in I$ we have $ej = j - fj = j$ hence $I = (e)$. This proves (1).

Parts (2) and (3) follow from (1). Namely, we have $V(I) = V(e) = \mathop{\mathrm{Spec}}(R) \setminus D(e)$ which is open and closed by either Lemma 10.21.1 or Lemma 10.21.3. This proves (3). For (2) observe that the map $R \to R_{e'}$ is surjective since $x/(e')^ n = x/e' = xe'/(e')^2 = xe'/e' = x/1$ in $R_{e'}$. The kernel of the map $R \to R_{e'}$ is the set of elements of $R$ annihilated by a positive power of $e'$. Since $e'$ is idempotent this is the ideal of elements annihilated by $e'$ which is the ideal $I = (e)$ as $e + e' = 1$ is a pair of orthognal idempotents. This proves (2).
$\square$

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