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 10.115.3. Let $k$ be a field. Let $S$ be a finite type $k$ algebra. Let $X = \mathop{\mathrm{Spec}}(S)$. Let $\mathfrak p \subset S$ be a prime ideal, and let $x \in X$ be the corresponding point. Then we have

\[ \dim _ x(X) = \dim (S_{\mathfrak p}) + \text{trdeg}_ k\ \kappa (\mathfrak p). \]

Proof. By Lemma 10.115.1 we know that $r = \text{trdeg}_ k\ \kappa (\mathfrak p)$ is equal to the dimension of $V(\mathfrak p)$. Pick any maximal chain of primes $\mathfrak p \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ r$ starting with $\mathfrak p$ in $S$. This has length $r$ by Lemma 10.113.4. Let $\mathfrak q_ j$, $j \in J$ be the minimal primes of $S$ which are contained in $\mathfrak p$. These correspond $1-1$ to minimal primes in $S_{\mathfrak p}$ via the rule $\mathfrak q_ j \mapsto \mathfrak q_ jS_{\mathfrak p}$. By Lemma 10.113.5 we know that $\dim _ x(X)$ is equal to the maximum of the dimensions of the rings $S/\mathfrak q_ j$. For each $j$ pick a maximal chain of primes $\mathfrak q_ j \subset \mathfrak p'_1 \subset \ldots \subset \mathfrak p'_{s(j)} = \mathfrak p$. Then $\dim (S_{\mathfrak p}) = \max _{j \in J} s(j)$. Now, each chain

\[ \mathfrak q_ i \subset \mathfrak p'_1 \subset \ldots \subset \mathfrak p'_{s(j)} = \mathfrak p \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ r \]

is a maximal chain in $S/\mathfrak q_ j$, and by what was said before we have $\dim _ x(X) = \max _{j \in J} r + s(j)$. The lemma follows. $\square$


Comments (1)

Comment #715 by Keenan Kidwell on

The notation is being used for two different sets of primes. Maybe put a prime on the second set?


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