The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 27.10.2. Let $X$ be a scheme. The following are equal

  1. The dimension of $X$.

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

  3. 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 25.13.2. 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 $X$. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 04MU. Beware of the difference between the letter 'O' and the digit '0'.