
Lemma 10.28.1. Let $U \subset \mathop{\mathrm{Spec}}(R)$ be open. The following are equivalent:

1. $U$ is retrocompact in $\mathop{\mathrm{Spec}}(R)$,

2. $U$ is quasi-compact,

3. $U$ is a finite union of standard opens, and

4. there exists a finitely generated ideal $I \subset R$ such that $X \setminus V(I) = U$.

Proof. We have (1) $\Rightarrow$ (2) because $\mathop{\mathrm{Spec}}(R)$ is quasi-compact, see Lemma 10.16.10. We have (2) $\Rightarrow$ (3) because standard opens form a basis for the topology. Proof of (3) $\Rightarrow$ (1). Let $U = \bigcup _{i = 1\ldots n} D(f_ i)$. To show that $U$ is retrocompact in $\mathop{\mathrm{Spec}}(R)$ it suffices to show that $U \cap V$ is quasi-compact for any quasi-compact open $V$ of $\mathop{\mathrm{Spec}}(R)$. Write $V = \bigcup _{j = 1\ldots m} D(g_ j)$ which is possible by (2) $\Rightarrow$ (3). Each standard open is homeomorphic to the spectrum of a ring and hence quasi-compact, see Lemmas 10.16.6 and 10.16.10. Thus $U \cap V = (\bigcup _{i = 1\ldots n} D(f_ i)) \cap (\bigcup _{j = 1\ldots m} D(g_ j)) = \bigcup _{i, j} D(f_ i g_ j)$ is a finite union of quasi-compact opens hence quasi-compact. To finish the proof note that (4) is equivalent to (3) by Lemma 10.16.2. $\square$

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