The Stacks project

Proof. Recall that $\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$ is the set of points of $Y$ specializing to $y$, see Schemes, Lemma 26.13.2. Thus we may replace $Y$ by $\mathop{\mathrm{Spec}}(\mathcal{O}_{Y, y})$ and assume $Y = \mathop{\mathrm{Spec}}(B)$ where $B$ is a Noetherian local ring of dimension $\leq 1$ and $y$ is the closed point. We may also replace $X$ by an affine neighbourhood of $x$.

Let $X = \bigcup X_ i$ be the irreducible components of $X$ viewed as reduced closed subschemes. If we can show each fibre $X_{i, y}$ has dimension $d$, then $X_ y = \bigcup X_{i, y}$ has dimension $d$ as well. Thus we may assume $X$ is an integral scheme.

If $X \to Y$ maps the generic point $\eta $ of $X$ to $y$, then $X$ is a scheme over $\kappa (y)$ and the result is true by assumption. Assume that $X$ maps $\eta $ to a point $\xi \in Y$ corresponding to a minimal prime $\mathfrak q$ of $B$ different from $\mathfrak m_ B$. We obtain a factorization $X \to \mathop{\mathrm{Spec}}(B/\mathfrak q) \to \mathop{\mathrm{Spec}}(B)$. By the dimension formula (Morphisms, Lemma 29.52.1) we have

We have $\dim (B/\mathfrak q) = 1$. We have $\text{trdeg}_{\kappa (\mathfrak q)}(R(X)) = d$ by our assumption that $\dim _\eta (X_\xi ) = d$, see Morphisms, Lemma 29.28.1. Since $\mathcal{O}_{X, x} \to \mathcal{O}_{X_ s, x}$ has a kernel (as $\eta \mapsto \xi \not= y$) 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

(Morphisms, Lemma 29.28.1). On the other hand, we have $\dim _ x(X_ s) \geq \dim _\eta (X_{f(\eta )}) = d$ by Morphisms, Lemma 29.28.4. $\square$