Express $\mathbb{Z}$ as a direct limit of some rings with finite prime ideals

Question

By Hochster's theorem, $\mathrm{Spec}(\mathbb{Z})$ can be expressed as an inverse limit of some finite $T_0$-spaces, hence the ring $\mathbb{Z}$ should be able to be expressed as a direct limit of some rings with finite prime ideals. What is the expression? I did not figure out.

Answer 1

It is not true that $\mathbb{Z}$ is a direct limit of rings with finitely many prime ideals. Indeed, if $f:A\to\mathbb{Z}$ is any homomorphism, the unique homomorphism $i:\mathbb{Z}\to A$ satisfies $fi=1_\mathbb{Z}$, so $i^*f^*:\operatorname{Spec}(\mathbb{Z})\to \operatorname{Spec}(\mathbb{Z})$ is the identity. In particular, the image of $i^*f^*$ is infinite, so the image of $f^*$ is infinite, so $\operatorname{Spec}(A)$ must have infinitely many points.

What Hochster proved is that there exists an inverse system of finite topological spaces $X_i$ whose inverse limit is the space $\operatorname{Spec}(\mathbb{Z})$ and such that $X_i=\operatorname{Spec}(A_i)$ for some rings $A_i$. However, this does not mean that the rings $A_i$ form a direct system with colimit $\mathbb{Z}$, for several reasons. For one thing, the continuous maps $X_i\to X_j$ need not be the underlying maps of morphisms of schemes, so you don't necessarily get maps $A_j\to A_i$ of rings. For another thing, even if you did get a direct system of rings, there is no reason its colimit would have to be $\mathbb{Z}$ (all you can say is that Spec of the colimit is has a canonical continuous map to $\operatorname{Spec}(\mathbb{Z})$, which need not be a homeomorphism, let along the underlying map of an isomorphism of schemes).