## Tag `00EC`

Chapter 10: Commutative Algebra > Section 10.20: Open and closed subsets of spectra

Lemma 10.20.1. Let $R$ be a ring. Let $e \in R$ be an idempotent. In this case $$ \mathop{\rm Spec}(R) = D(e) \amalg D(1-e). $$

Proof.Note that an idempotent $e$ of a domain is either $1$ or $0$. Hence we see that \begin{eqnarray*} D(e) & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e \not\in \mathfrak p \} \\ & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e \not = 0\text{ in }\kappa(\mathfrak p) \} \\ & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e = 1\text{ in }\kappa(\mathfrak p) \} \end{eqnarray*} Similarly we have \begin{eqnarray*} D(1-e) & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid 1 - e \not\in \mathfrak p \} \\ & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e \not = 1\text{ in }\kappa(\mathfrak p) \} \\ & = & \{ \mathfrak p \in \mathop{\rm Spec}(R) \mid e = 0\text{ in }\kappa(\mathfrak p) \} \end{eqnarray*} 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 $\mathop{\rm Spec}(R)$. $\square$

The code snippet corresponding to this tag is a part of the file `algebra.tex` and is located in lines 3477–3484 (see updates for more information).

```
\begin{lemma}
\label{lemma-idempotent-spec}
Let $R$ be a ring. Let $e \in R$ be an idempotent.
In this case
$$
\Spec(R) = D(e) \amalg D(1-e).
$$
\end{lemma}
\begin{proof}
Note that an idempotent $e$ of a domain is either $1$ or $0$.
Hence we see that
\begin{eqnarray*}
D(e)
& = &
\{ \mathfrak p \in \Spec(R)
\mid
e \not\in \mathfrak p \} \\
& = &
\{ \mathfrak p \in \Spec(R)
\mid
e \not = 0\text{ in }\kappa(\mathfrak p) \} \\
& = &
\{ \mathfrak p \in \Spec(R)
\mid
e = 1\text{ in }\kappa(\mathfrak p) \}
\end{eqnarray*}
Similarly we have
\begin{eqnarray*}
D(1-e)
& = &
\{ \mathfrak p \in \Spec(R)
\mid
1 - e \not\in \mathfrak p \} \\
& = &
\{ \mathfrak p \in \Spec(R)
\mid
e \not = 1\text{ in }\kappa(\mathfrak p) \} \\
& = &
\{ \mathfrak p \in \Spec(R)
\mid
e = 0\text{ in }\kappa(\mathfrak p) \}
\end{eqnarray*}
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 $\Spec(R)$.
\end{proof}
```

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