In Görtz/Wedhorn I read that for a scheme $S$ we define $\mathbb{A}^n_S:=\mathbb{A}^n_{\mathbb{Z}}\times_{\mathbb{Z}}S,$ and for a morphism $S'\rightarrow S$ of schemes we have $$\mathbb{A}^n_S\times_SS'= \mathbb{A}^n\times_{\mathbb{Z}}S\times_SS'\cong\mathbb{A}^n\times_{\mathbb{Z}}S'=\mathbb{A}^n_{S'}.$$ So we can take actually $\textit{any}$ morphism $S'\rightarrow S$ and we get always the same thing? In some sense this seems natural to me, since after composing with $S\rightarrow \operatorname{Spec}\mathbb{Z}$ all morphisms $S'\rightarrow S$ are equal anyway. But on the other hand it seems quite surprising/unnatural to me that we can have a scheme whose base change to another scheme is completely irrelevant of the actual morphism we make the base change with.
Question 1: should this surprise me?
Question 2: I guess I'm lacking the mathematical knowedledge to use the correct words, but is this in some sense an indicator that affine spaces $\mathbb{A}^n$ have some intrinsic property that they all share, irrelevant of the base, that make them "special"?
Best Answer
This isn't anything special about affine space. For example, the same statement is true if you replace affine space by projective space over the given schemes. More generally, given any scheme $B$, any $B$-scheme $X$ and any morphism of $B$-schemes $S' \to S$, we have $$X \times_B S' \cong X \times_B (S \times_S S') \cong (X \times_B S) \times_S S'.$$ Actually, this isn't even specific to schemes, the same is true of fiber products (also known as pullbacks) in any category—it follows directly from the universal property of fiber products.
What's special here is that, since $\operatorname{Spec} \mathbb{Z}$ is the terminal object in the category of schemes, there's a unique morphism from any scheme to $\operatorname{Spec} \mathbb{Z}$, so this leads to what you're observing about the choice of morphism $S' \to S$ not mattering. But this is also true in the relative setting, since for any scheme $B$, a $B$-scheme is just a morphism of schemes $X \to B$, and morphisms of $B$-schemes are required to be compatible with the morphism to $B$, so $B$ itself (with the identity morphism) is the terminal object in the category of $B$-schemes, so we have the same situation for fiber products in the category of $B$-schemes.