Lemma 10.116.2. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q \subset \mathfrak q' \subset S$ be distinct prime ideals. Then $\text{trdeg}_ k\ \kappa (\mathfrak q') < \text{trdeg}_ k\ \kappa (\mathfrak q)$.

Proof. By Lemma 10.116.1 we have $\dim V(\mathfrak q) = \text{trdeg}_ k\ \kappa (\mathfrak q)$ and similarly for $\mathfrak q'$. Hence the result follows as the strict inclusion $V(\mathfrak q') \subset V(\mathfrak q)$ implies a strict inequality of dimensions. $\square$

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