Lemma 10.116.4. Let $k$ be a field. Let $S' \to S$ be a surjection of finite type $k$ algebras. Let $\mathfrak p \subset S$ be a prime ideal, and let $\mathfrak p'$ be the corresponding prime ideal of $S'$. Let $X = \mathop{\mathrm{Spec}}(S)$, resp. $X' = \mathop{\mathrm{Spec}}(S')$, and let $x \in X$, resp. $x'\in X'$ be the point corresponding to $\mathfrak p$, resp. $\mathfrak p'$. Then

$\dim _{x'} X' - \dim _ x X = \text{height}(\mathfrak p') - \text{height}(\mathfrak p).$

Proof. Immediate from Lemma 10.116.3. $\square$

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