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Hartshorne, Algebraic Geometry, Exercise II.3.20, reads (in part):

Let $X$ be an integral scheme of finite type over a field $k$.

(a) [Prove:] For any closed point $P \in X$, $\dim X = \dim \mathcal{O}_P$, where for rings we always mean the Krull dimension.

[...]

I'm stuck on a point which is completely tangential to the actual business about dimension. Here's my proof:

Let $\emptyset \neq X_0 \subset X_1 \subset \cdots \subset X_n \subseteq X$ be an increasing chain of irreducible closed subsets of $X$ of maximum length, so that $n = \dim X$. (In fact $X_n = X$ since $X$ is irreducible.) Let $U \subseteq X$ be an open affine subset which meets $X_0$, and write $A := \mathcal{O}_X(U)$. As $X$ is of finite type over $k$, $A$ is a finitely-generated $k$-algebra by Exercise II.3.3(b).

It follows from general topology that $\dim U = \dim X$. [etc.]

We've now done enough to reduce the problem to the affine case. To see this, consider a closed point $P \in X$. $P$ is contained in some open affine set $V$, which intersects $U$ since $X$ is irreducible. Let $Q$ be a closed point in $U \cap V$. Assuming the affine case, we have $$\dim \mathcal{O}_{X, P} = \dim V = \dim \mathcal{O}_{X, Q} = \dim U = \dim X$$ which establishes the desired result.

So suppose that $X = \operatorname{Spec} A$ for some finitely-generated $k$-algebra $A$. [etc.]

I'm concerned with the bolded sentence, "Let $Q$ be a closed point of $U \cap V$." In particular, how do I ensure that $U \cap V$ contains a closed point? As I understand it, the hypothesis that $X$ is integral of finite type over a field is vital here -- if I take two arbitrary open affines and glue them together then I can easily create a situation where $U \cap V$ doesn't contain a closed point.

However, I'm having trouble seeing how to get anything out of this hypothesis. For instance, $U \cap V$ isn't necessarily affine, and though it contains some affine set a point could be closed in that affine but not in $U \cap V$.

Am I just being stupid here? How do you do this?

Daniel McLaury
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1 Answers1

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You are completely on the right track. $U \cap V$ contains an affine $W$, and $W$ will have a closed point $x$ (closed in $W$). But since $X$ is of finite type over a field, $x$ is in fact closed in $X$ (and hence in particular closed in $U \cap V$) as well. This is an immediate consequence of the following result.

Let $X$ be a scheme, which is locally of finite type over a field $k$. A point $x \in X$ is closed, if and only if the residue field $\kappa(x)$ is a finite algebraic extension of $k$.

Cf. for example Görtz-Wedhorn, Proposition 3.33. Indeed, using this result one can even show that on any scheme $X$ as above the set of closed points is dense in $X$.

Nils Matthes
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