Lemma 10.22.1. Let $R$ be a ring. Let $T \subset \mathop{\mathrm{Spec}}(R)$ be a subset of the spectrum. The following are equivalent

1. $T$ is closed and is a union of connected components of $\mathop{\mathrm{Spec}}(R)$,

2. $T$ is an intersection of open and closed subsets of $\mathop{\mathrm{Spec}}(R)$, and

3. $T = V(I)$ where $I \subset R$ is an ideal generated by idempotents.

Moreover, the ideal in (3) if it exists is unique.

Proof. By Lemma 10.17.11 and Topology, Lemma 5.12.12 we see that (1) and (2) are equivalent. Assume (2) and write $T = \bigcap U_\alpha$ with $U_\alpha \subset \mathop{\mathrm{Spec}}(R)$ open and closed. Then $U_\alpha = D(e_\alpha )$ for some idempotent $e_\alpha \in R$ by Lemma 10.21.3. Then setting $I = (1 - e_\alpha )$ we see that $T = V(I)$, i.e., (3) holds. Finally, assume (3). Write $T = V(I)$ and $I = (e_\alpha )$ for some collection of idempotents $e_\alpha$. Then it is clear that $T = \bigcap V(e_\alpha ) = \bigcap D(1 - e_\alpha )$.

Suppose that $I$ is an ideal generated by idempotents. Let $e \in R$ be an idempotent such that $V(I) \subset V(e)$. Then by Lemma 10.17.2 we see that $e^ n \in I$ for some $n \geq 1$. As $e$ is an idempotent this means that $e \in I$. Hence we see that $I$ is generated by exactly those idempotents $e$ such that $T \subset V(e)$. In other words, the ideal $I$ is completely determined by the closed subset $T$ which proves uniqueness. $\square$

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