**Proof.**
Denote $R = S_{\mathfrak q}$ and denote its maximal by $\mathfrak m$ and its residue field $\kappa $. By Lemma 10.140.4 and 10.131.9 we see that there is a short exact sequence

\[ 0 \to \mathfrak m/\mathfrak m^2 \to \Omega _{R/k} \otimes _ R \kappa \to \Omega _{\kappa /k} \to 0 \]

Note that $\Omega _{R/k} = \Omega _{S/k, \mathfrak q}$, see Lemma 10.131.8. Moreover, since $\kappa $ is separable over $k$ we have $\dim _{\kappa } \Omega _{\kappa /k} = \text{trdeg}_ k(\kappa )$. Hence we get

\[ \dim _{\kappa } \Omega _{R/k} \otimes _ R \kappa = \dim _\kappa \mathfrak m/\mathfrak m^2 + \text{trdeg}_ k (\kappa ) \geq \dim R + \text{trdeg}_ k (\kappa ) = \dim _{\mathfrak q} S \]

(see Lemma 10.116.3 for the last equality) with equality if and only if $R$ is regular. Thus we win by applying Lemma 10.140.3.
$\square$

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