Lemma 10.22.2. Let $R$ be a ring. A connected component of $\mathop{\mathrm{Spec}}(R)$ is of the form $V(I)$, where $I$ is an ideal generated by idempotents such that every idempotent of $R$ either maps to $0$ or $1$ in $R/I$.

Proof. Let $\mathfrak p$ be a prime of $R$. By Lemma 10.17.9 we have see that the hypotheses of Topology, Lemma 5.12.10 are satisfied for the topological space $\mathop{\mathrm{Spec}}(R)$. Hence the connected component of $\mathfrak p$ in $\mathop{\mathrm{Spec}}(R)$ is the intersection of open and closed subsets of $\mathop{\mathrm{Spec}}(R)$ containing $\mathfrak p$. Hence it equals $V(I)$ where $I$ is generated by the idempotents $e \in R$ such that $e$ maps to $0$ in $\kappa (\mathfrak p)$, see Lemma 10.21.3. Any idempotent $e$ which is not in this collection clearly maps to $1$ in $R/I$. $\square$

Comment #8515 by on

In the third sentence, shouldn't one say "hence, by the proof of Lemma 10.22.1, it equals $V(I)$ where $I$ is generated by the idempotents..."?

Comment #8532 by on

Hmm, I think it is fine. I mean that ideal really does cut out the connected component of $\mathfrak p$ by the previous sentence, Lemma 10.21.3, the fact that $V(\{f_i\}) = \bigcap V(f_i)$, and $D(1 - e) = V(e)$ for an idempotent $e$. So I think the proof is fine as is.

I guess an alternative proof would be to first say that the connected component of $\mathfrak p$ is $V(I)$ as in Lemma 10.22.1 and then explain which idempotents are in $I$. Do people prefer this?

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