Is the set of affine morphisms the smallest set in this specific sense?

Question

One motivation for affine morphisms I have seen is that:

• $\operatorname{Spec} A \to \operatorname{Spec} \Bbb Z$ should be an affine morphism for any ring $A$.
• The set of affine morphisms should be closed under composition and base change.

Is every affine morphisms realized by a finite application of the above two rules? Or since the pullback of a pullback is a pullback, is every affine morphism $f:X\to Y$ the pullback of some morphism of the form $\operatorname{Spec} A \to \operatorname{Spec} \Bbb Z$?

Just restricting attention to affine schemes, this question becomes:

For every morphism $f: B\to C$, is there a ring $A$ such that $C = B\otimes_\Bbb Z A$ and $f$ is the morphisms $B\to A$ associated to this tensor product?

Answer 1

Let $B=\mathbb{Z}[X]$. If the condition holds, there is some $A$ such that $C\cong \mathbb{Z}[X]\otimes_\mathbb{Z} A \cong A[X]$.

But lots of rings are not polynomial rings over anything, e.g. finite rings or fields.

Even if $C\cong A[X]$, there are plenty of morphisms $\mathbb{Z}[X]\to A[X]$ that are not the base change of $\mathbb{Z}\to A$.