Lemma 10.20.5. Let $R$ be a ring. Let $I$ be a finitely generated ideal. Assume that $I = I^2$. Then $V(I)$ is open and closed in $\mathop{\mathrm{Spec}}(R)$, and $R/I \cong R_ e$ for some idempotent $e \in R$.

Proof. By Nakayama's Lemma 10.19.1 there exists an element $f = 1 + i$, $i \in I$ in $R$ such that $fI = 0$. It follows that $V(I) = D(f)$ by a simple argument. Also, $0 = fi = i + i^2$, and hence $f^2 = 1 + i + i + i^2 = 1 + i = f$, so $f$ is an idempotent. Consider the canonical map $R \to R_ f$. It is surjective since $x/f^ n = x/f = xf/f^2 = xf/f = x/1$ in $R_ f$. Any element of $I$ is in the kernel since $fI = 0$. If $x \mapsto 0$ in $R_ f$, then $f^ nx = 0$ for some $n > 0$ and hence $(1 + i)x = 0$ hence $x \in I$. $\square$

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).