Lemma 28.10.2. Let $X$ be a scheme. The following are equal

The dimension of $X$.

The supremum of the dimensions of the local rings of $X$.

The supremum of $\dim _ x(X)$ for $x \in X$.

Lemma 28.10.2. Let $X$ be a scheme. The following are equal

The dimension of $X$.

The supremum of the dimensions of the local rings of $X$.

The supremum of $\dim _ x(X)$ for $x \in X$.

**Proof.**
Note that given a chain of specializations

\[ \xi _ n \leadsto \xi _{n - 1} \leadsto \ldots \leadsto \xi _0 \]

of points of $X$ all of the points $\xi _ i$ correspond to prime ideals of the local ring of $X$ at $\xi _0$ by Schemes, Lemma 26.13.2. Hence we see that the dimension of $X$ is the supremum of the dimensions of its local rings. In particular $\dim _ x(X) \geq \dim (\mathcal{O}_{X, x})$ as $\dim _ x(X)$ is the minimum of the dimensions of open neighbourhoods of $x$. Thus $\sup _{x \in X} \dim _ x(X) \geq \dim (X)$. On the other hand, it is clear that $\sup _{x \in X} \dim _ x(X) \leq \dim (X)$ as $\dim (U) \leq \dim (X)$ for any open subset of $X$. $\square$

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