$\mathbb{Z}$ is universally catenary?

Question

I'am reading Gortz, Wedhorn, Algebraic Geometry I, p.468

I'am trying to understand the underlined statement.

Since $A$ is finitely generated $\mathbb{Z}$-algebra, the structure morphism $\operatorname{Spec}(A) \to \operatorname{Spec}(\mathbb{Z})$ is locally of finite type (maybe?). So, if $\mathbb{Z}$ is universally catenary, then by the Remark 14.101-(4), $\operatorname{Spec}(\mathbb{Z})$ is universally catenary and by the Remark 14.101 - (5), $Y := \operatorname{Spec}(A)$ is universally catenary.

Is this argument may work? $\mathbb{Z}$ is universally catenary? If not, is there any other method to prove that $Y$ is universally catenary?

Answer 1

$A$ finitely generated as a $\Bbb Z$-algebra exactly means the structure morphism $\operatorname{Spec} A\to\operatorname{Spec}\Bbb Z$ is of finite type by definition.

Since $\Bbb Z$ is regular, proposition 14.103 applies. To check this, simply verify that every localization of $\Bbb Z$ at a prime ideal is a regular local ring: you just have to check $\Bbb Z_p$ and $\Bbb Q$, both of which should be quite direct.