Lemma 10.21.5. Let $I \subset R$ be a finitely generated ideal of a ring $R$ such that $I = I^2$. Then

1. there exists an idempotent $e \in R$ such that $I = (e)$,

2. $R/I \cong R_{e'}$ for the idempotent $e' = 1 - e \in R$, and

3. $V(I)$ is open and closed in $\mathop{\mathrm{Spec}}(R)$.

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 orthogonal idempotents. This proves (2). $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).