Lemma 10.114.6. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $X = \mathop{\mathrm{Spec}}(S)$. Let $\mathfrak m \subset S$ be a maximal ideal and let $x \in X$ be the associated closed point. Then $\dim _ x(X) = \dim (S_{\mathfrak m})$.

**Proof.**
This is a special case of Lemma 10.114.5.
$\square$

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