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Vakil 5.1.B, correspondence of points with irreducible closed subsets on general schemes

代数几何 Math StackExchange 2 票 1 回答 42 浏览 提问者: Jo Cantelmi 2026-06-29 19:56
algebraic-geometry schemes

问题内容

I'm working through Vakil and encountered the following exercise.

5.1.B. EXERCISE. Exercise 3.7.F showed that there is a bijection between irreducible closed subsets and points for affine schemes (the map sending a point p to the closed subset $\overline{\{p\}}$ is a bijection). Show that this is true of schemes in general.

Showing that the map of points is injective is not so bad invoking Exercise 3.7.F

However, stuggling with surjectivity has caused a lot of my intuition about the Zariski topology to break down on general schemes. (Edit: I see now that irreducible is the right notion to describe schemes on which every open set is dense)

As for surjectivity, I've thought about breaking down the scheme into irreducible components, but am unsure as to whether this argument works for general schemes as it does affine schemes.

Thank you so much!

回答 (1)

TY Mathers 4 票 已采纳 2026-06-29 20:47 原文

If you have an irreducible closed subset $Z$, choose some affine open $U$ such that $Z\cap U\neq\emptyset$. Then $Z\cap U$ is an irreducible closed subset of the affine scheme $U$, so you can invoke the previous exercise covering the affine case to show that $Z\cap U=\operatorname{cl}_U(\{p\})$ (this notation meaning the closure of $\{p\}$ in $U$) for some $p\in Z\cap U$.

Now show that for any other affine open $V$ which intersects $Z$, one also has $\operatorname{cl}_V(\{p\})=Z\cap V$, and use the fact that the closure can be computed over an (affine) cover $\mathcal U$ to deduce that the closure of $\{p\}$ in $X$ is equal to $Z$:

$$\operatorname{cl}_X(\{p\})=\bigcup_{V\in\mathcal U}\operatorname{cl}_V(V\cap\{p\}).$$