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*}

Lemma 10.20.3. Let $R$ be a ring. For each $U \subset \mathop{\mathrm{Spec}}(R)$ which is open and closed there exists a unique idempotent $e \in R$ such that $U = D(e)$. This induces a 1-1 correspondence between open and closed subsets $U \subset \mathop{\mathrm{Spec}}(R)$ and idempotents $e \in R$.

Proof. Let $U \subset \mathop{\mathrm{Spec}}(R)$ be open and closed. Since $U$ is closed it is quasi-compact by Lemma 10.16.10, and similarly for its complement. Write $U = \bigcup _{i = 1}^ n D(f_ i)$ as a finite union of standard opens. Similarly, write $\mathop{\mathrm{Spec}}(R) \setminus U = \bigcup _{j = 1}^ m D(g_ j)$ as a finite union of standard opens. Since $\emptyset = D(f_ i) \cap D(g_ j) = D(f_ i g_ j)$ we see that $f_ i g_ j$ is nilpotent by Lemma 10.16.2. Let $I = (f_1, \ldots , f_ n) \subset R$ and let $J = (g_1, \ldots , g_ m) \subset R$. Note that $V(J)$ equals $U$, that $V(I)$ equals the complement of $U$, so $\mathop{\mathrm{Spec}}(R) = V(I) \amalg V(J)$. By the remark on nilpotency above, we see that $(IJ)^ N = (0)$ for some sufficiently large integer $N$. Since $\bigcup D(f_ i) \cup \bigcup D(g_ j) = \mathop{\mathrm{Spec}}(R)$ we see that $I + J = R$, see Lemma 10.16.2. By raising this equation to the $2N$th power we conclude that $I^ N + J^ N = R$. Write $1 = x + y$ with $x \in I^ N$ and $y \in J^ N$. Then $0 = xy = x(1 - x)$ as $I^ N J^ N = (0)$. Thus $x = x^2$ is idempotent and contained in $I^ N \subset I$. The idempotent $y = 1 - x$ is contained in $J^ N \subset J$. This shows that the idempotent $x$ maps to $1$ in every residue field $\kappa (\mathfrak p)$ for $\mathfrak p \in V(J)$ and that $x$ maps to $0$ in $\kappa (\mathfrak p)$ for every $\mathfrak p \in V(I)$.

To see uniqueness suppose that $e_1, e_2$ are distinct idempotents in $R$. We have to show there exists a prime $\mathfrak p$ such that $e_1 \in \mathfrak p$ and $e_2 \not\in \mathfrak p$, or conversely. Write $e_ i' = 1 - e_ i$. If $e_1 \not= e_2$, then $0 \not= e_1 - e_2 = e_1(e_2 + e_2') - (e_1 + e_1')e_2 = e_1 e_2' - e_1' e_2$. Hence either the idempotent $e_1 e_2' \not= 0$ or $e_1' e_2 \not= 0$. An idempotent is not nilpotent, and hence we find a prime $\mathfrak p$ such that either $e_1e_2' \not\in \mathfrak p$ or $e_1'e_2 \not\in \mathfrak p$, by Lemma 10.16.2. It is easy to see this gives the desired prime. $\square$


Comments (2)

Comment #2242 by on

Let the ring be . Then the idempotents do not correspond to clopen subsets of ,since the idempotent doesn't correspond to anything.

Should the bijection be with connected components of ?

Comment #2278 by on

Dear Hari Rau-Murthy, hmm, why do you say what you say? The idempotent corresponds to the open point corresponding to the first factor of the infinite product.


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