The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

10.21 Connected components of spectra

Connected components of spectra are not as easy to understand as one may think at first. This is because we are used to the topology of locally connected spaces, but the spectrum of a ring is in general not locally connected.

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

  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 10.16.11 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. $\square$

Lemma 10.21.2. 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 10.16.11 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 10.20.3. Any idempotent $e$ which is not in this collection clearly maps to $1$ in $R/I$. $\square$


Comments (2)

Comment #1146 by OlgaD on

Maybe instead hould be written: for every


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