Lemma 29.29.6. Let $f : X \to Y$ be a morphism of locally Noetherian schemes which is flat, locally of finite type and of relative dimension $d$. For every point $x$ in $X$ with image $y$ in $Y$ we have $\dim _ x(X) = \dim _ y(Y) + d$.

Proof. After shrinking $X$ and $Y$ to open neighborhoods of $x$ and $y$, we can assume that $\dim (X) = \dim _ x(X)$ and $\dim (Y) = \dim _ y(Y)$, by definition of the dimension of a scheme at a point (Properties, Definition 28.10.1). The morphism $f$ is open by Lemmas 29.21.9 and 29.25.10. Hence we can shrink $Y$ to arrange that $f$ is surjective. It remains to show that $\dim (X) = \dim (Y) + d$.

Let $a$ be a point in $X$ with image $b$ in $Y$. By Algebra, Lemma 10.112.7,

$\dim (\mathcal{O}_{X,a}) = \dim (\mathcal{O}_{Y,b}) + \dim (\mathcal{O}_{X_ b, a}).$

Taking the supremum over all points $a$ in $X$, it follows that $\dim (X) = \dim (Y) + d$, as we want, see Properties, Lemma 28.10.2. $\square$

Comment #2305 by Daniel on

Typo in the 3rd line from the bottom: The last local ring in the equation.

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