Lemma 10.21.1. Let $R$ be a ring. Let $e \in R$ be an idempotent. In this case

## 10.21 Open and closed subsets of spectra

It turns out that open and closed subsets of a spectrum correspond to idempotents of the ring.

**Proof.**
Note that an idempotent $e$ of a domain is either $1$ or $0$. Hence we see that

Similarly we have

Since the image of $e$ in any residue field is either $1$ or $0$ we deduce that $D(e)$ and $D(1-e)$ cover all of $\mathop{\mathrm{Spec}}(R)$. $\square$

Lemma 10.21.2. Let $R_1$ and $R_2$ be rings. Let $R = R_1 \times R_2$. The maps $R \to R_1$, $(x, y) \mapsto x$ and $R \to R_2$, $(x, y) \mapsto y$ induce continuous maps $\mathop{\mathrm{Spec}}(R_1) \to \mathop{\mathrm{Spec}}(R)$ and $\mathop{\mathrm{Spec}}(R_2) \to \mathop{\mathrm{Spec}}(R)$. The induced map

is a homeomorphism. In other words, the spectrum of $R = R_1\times R_2$ is the disjoint union of the spectrum of $R_1$ and the spectrum of $R_2$.

**Proof.**
Write $1 = e_1 + e_2$ with $e_1 = (1, 0)$ and $e_2 = (0, 1)$. Note that $e_1$ and $e_2 = 1 - e_1$ are idempotents. We leave it to the reader to show that $R_1 = R_{e_1}$ is the localization of $R$ at $e_1$. Similarly for $e_2$. Thus the statement of the lemma follows from Lemma 10.21.1 combined with Lemma 10.17.6.
$\square$

We reprove the following lemma later after introducing a glueing lemma for functions. See Section 10.24.

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. $\square$

Lemma 10.21.4. Let $R$ be a nonzero ring. Then $\mathop{\mathrm{Spec}}(R)$ is connected if and only if $R$ has no nontrivial idempotents.

**Proof.**
Obvious from Lemma 10.21.3 and the definition of a connected topological space.
$\square$

Lemma 10.21.5. Let $R$ be a ring. Let $I$ be a finitely generated ideal. Assume that $I = I^2$. Then $V(I)$ is open and closed in $\mathop{\mathrm{Spec}}(R)$, and $R/I \cong R_ e$ for some idempotent $e \in R$.

**Proof.**
By Nakayama's Lemma 10.20.1 there exists an element $f = 1 + i$, $i \in I$ in $R$ such that $fI = 0$. It follows that $V(I) = D(f)$ by a simple argument. Also, $0 = fi = i + i^2$, and hence $f^2 = 1 + i + i + i^2 = 1 + i = f$, so $f$ is an idempotent. Consider the canonical map $R \to R_ f$. It is surjective since $x/f^ n = x/f = xf/f^2 = xf/f = x/1$ in $R_ f$. Any element of $I$ is in the kernel since $fI = 0$. If $x \mapsto 0$ in $R_ f$, then $f^ nx = 0$ for some $n > 0$ and hence $(1 + i)x = 0$ hence $x \in I$.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)