Lemma 10.21.1. Let $R$ be a ring. Let $T \subset \mathop{\mathrm{Spec}}(R)$ be a subset of the spectrum. The following are equivalent

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

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

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

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

**Proof.**
By Lemma 10.16.11 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 10.20.3. 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 10.16.2 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.
$\square$

## Comments (0)

There are also: