Lemma 10.21.3. 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 10.17.10, 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 10.17.2. 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 10.17.2. 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 10.17.2. It is easy to see this gives the desired prime. $\square$

Comment #2242 by on

Let the ring be $\Pi_{n \in \mathbb{Z}} \mathbb{F_2}$. Then the idempotents do not correspond to clopen subsets of $\operatorname{Spec} R$,since the idempotent $(1,0,....)$ doesn't correspond to anything.

Should the bijection be with connected components of $\operatorname{Spec} R$?

Comment #2278 by on

Dear Hari Rau-Murthy, hmm, why do you say what you say? The idempotent $(1, 0, 0, \ldots)$ corresponds to the open point $\Spec(\mathbf{F}_2)$ corresponding to the first factor $\mathbf{F}_2$ of the infinite product.

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