The Stacks project

Lemma 106.5.13. If $X$ is an irreducible, Jacobson, catenary, and locally Noetherian scheme of finite dimension, then $\dim U = \dim X$ for every non-empty open subset $U$ of $X$. Equivalently, $\dim _ x X$ is a constant function on $X$.

Proof. The equivalence of the two claims follows directly from the definitions. Suppose, then, that $U\subset X$ is a non-empty open subset. Certainly $\dim U \leq \dim X$, and we have to show that $\dim U \geq \dim X.$ Write $d = \dim X$, and choose a maximal strictly decreasing chain of irreducible closed subsets of $X$, say

\[ X = Z_0 \supset Z_1 \supset \ldots \supset Z_ d. \]

Since $X$ is Jacobson, the minimal irreducible closed subset $Z_ d$ is equal to $\{ x\} $ for some closed point $x$.

If $x \in U,$ then

\[ U = U \cap Z_0 \supset U\cap Z_1 \supset \ldots \supset U\cap Z_ d = \{ x\} \]

is a strictly decreasing chain of irreducible closed subsets of $U$, and so we conclude that $\dim U \geq d$, as required. Thus we may suppose that $x \not\in U.$

Consider the flat morphism $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x} \to X$. The non-empty (and hence dense) open subset $U$ of $X$ pulls back to an open subset $V \subset \mathop{\mathrm{Spec}}\mathcal{O}_{X,x}$. Replacing $U$ by a non-empty quasi-compact, and hence Noetherian, open subset, we may assume that the inclusion $U \to X$ is a quasi-compact morphism. Since the formation of scheme-theoretic images of quasi-compact morphisms commutes with flat base-change Morphisms, Lemma 29.25.16 we see that $V$ is dense in $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x}$, and so in particular non-empty, and of course $x \not\in V.$ (Here we use $x$ also to denote the closed point of $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x}$, since its image is equal to the given point $x \in X$.) Now $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x} \setminus \{ x\} $ is Jacobson Properties, Lemma 28.6.4 and hence $V$ contains a closed point $z$ of $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x} \setminus \{ x\} $. The closure in $X$ of the image of $z$ is then an irreducible closed subset $Z$ of $X$ containing $x$, whose intersection with $U$ is non-empty, and for which there is no irreducible closed subset properly contained in $Z$ and properly containing $\{ x\} $ (because pull-back to $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x}$ induces a bijection between irreducible closed subsets of $X$ containing $x$ and irreducible closed subsets of $\mathop{\mathrm{Spec}}\mathcal{O}_{X,x}$). Since $U \cap Z$ is a non-empty closed subset of $U$, it contains a point $u$ that is closed in $X$ (since $X$ is Jacobson), and since $U\cap Z$ is a non-empty (and hence dense) open subset of the irreducible set $Z$ (which contains a point not lying in $U$, namely $x$), the inclusion $\{ u\} \subset U\cap Z$ is proper.

As $X$ is catenary, the chain

\[ X = Z_0 \supset Z \supset \{ x\} = Z_ d \]

can be refined to a chain of length $d+1$, which must then be of the form

\[ X = Z_0 \supset W_1 \supset \ldots \supset W_{d-1} = Z \supset \{ x\} = Z_ d. \]

Since $U\cap Z$ is non-empty, we then find that

\[ U = U \cap Z_0 \supset U \cap W_1\supset \ldots \supset U\cap W_{d-1} = U\cap Z \supset \{ u\} \]

is a strictly decreasing chain of irreducible closed subsets of $U$ of length $d+1$, showing that $\dim U \geq d$, as required. $\square$

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