Lemma 33.19.3. Let $f : X \to Y$ be locally of finite type. Let $x \in X$ be a point with image $y \in Y$ such that $\mathcal{O}_{Y, y}$ is Noetherian. Let $d \geq 0$ be an integer such that for every generic point $\eta$ of an irreducible component of $X$ which contains $x$, we have $f(\eta ) \not= y$ and $\dim _\eta (X_{f(\eta )}) = d$. Then $\dim _ x(X_ y) \leq d + \dim (\mathcal{O}_{Y, y}) - 1$.

Proof. Exactly as in the proof of Lemma 33.19.1 we reduce to the case $X = \mathop{\mathrm{Spec}}(A)$ with $A$ a domain and $Y = \mathop{\mathrm{Spec}}(B)$ where $B$ is a Noetherian local ring whose maximal ideal corresponds to $y$. After replacing $B$ by $B/\mathop{\mathrm{Ker}}(B \to A)$ we may assume that $B$ is a domain and that $B \subset A$. Then we use the dimension formula (Morphisms, Lemma 29.52.1) to get

$\dim (\mathcal{O}_{X, x}) + \text{trdeg}_{\kappa (y)} \kappa (x) \leq \dim (B) + \text{trdeg}_ B(A)$

We have $\text{trdeg}_ B(A) = d$ by our assumption that $\dim _\eta (X_\xi ) = d$, see Morphisms, Lemma 29.28.1. Since $\mathcal{O}_{X, x} \to \mathcal{O}_{X_ y, x}$ has a kernel (as $f(\eta ) \not= y$) and since $\mathcal{O}_{X, x}$ is a Noetherian domain we see that $\dim (\mathcal{O}_{X, x}) > \dim (\mathcal{O}_{X_ y, x})$. We conclude that

$\dim _ x(X_ y) = \dim (\mathcal{O}_{X_ y, x}) + \text{trdeg}_{\kappa (y)} \kappa (x) < \dim (B) + d$

(equality by Morphisms, Lemma 29.28.1) which proves what we want. $\square$

Comment #6843 by 羽山籍真 on

In the last formula of the proof, $X_s$ should be $X_y$. Also, two rows above, $\mathcal{O}_{X_s,x}$ should be $\mathcal{O}_{X_y,x}$.

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