Is there a ring whose spectrum is homeomorphic to $\Bbb Z_p$?

Question

$\newcommand{\Z}{\mathbb{Z}}$

Can you provide a concrete example of a commutative ring $R$ such that $\mathrm{Spec}(R)$ (endowed with Zariski topology) is homeomorphic to $\Bbb Z_p$, the ring of $p$-adic integers (endowed with the $p$-adic absolute value)?

I think that such a ring $R$ should exist, because $X = (\Z_p, | \cdot |_p)$ is a spectral space :

• $X$ is sober, because it is Hausdorff.

• $X$ is compact, being a complete and totally bounded metric space.

• If $U,V\subseteq X$ are compact open sets, then $U\cap V$ is also compact and open ($X$ is Hausdorff).

• The compact open subsets of $X$ form a basis for the topology of $X$, since already the sets $p^n \Z_p$ provide a basis for the topology on $X$.

(I've seen that these conditions are equivalent to require that $X$ is an inverse limit of $T_0$ spaces, which is the case for $X = \Z_p$).

I don't really know what a ring $R$ with $\mathrm{Spec}(R) \cong \Z_p$ could look like. According to this, we should have $\dim(R) = 0$, and since $\Z_p$ is totally disconnected, I guess that $R$ could be decomposed as products of rings. I don't really know how $\mathrm{Spec}$ behaves with respect to (co)limits. Following Hochster's paper mentioned here seems to be difficult.

Any comment will be appreciated. Thank you!

Answer 1

Take any field $k$ and any profinite group $G$, and let $R = C^\infty(G, k)$ be the $k$-algebra of locally constant functions on $G$. Then essentially by definition, $R$ is the increasing union (to be more precise, a colimit) of the algebras $k^{G/N}$ of $k$-valued functions on $G/N$, as $N$ ranges over the open normal subgroups of $G$. The $Spec$ functor sends colimits of rings into limits of locally ringed topological spaces. In particular, we get a set-theoretic bijection

$ |Spec(R)| \cong \lim\limits_{\leftarrow} |Spec(k^{G/N})| \cong \lim\limits_{\leftarrow} G/N = G $.

The Zariski topology on $|Spec(R)|$ is the same as the inverse limit topology on the second term in the above equation. But because $k^{G/N}$ is a product of $|G/N|$-copies of the field $k$ as a ring, the Zariski topology on $|Spec(k^{G/N})|$ is just the discrete topology on the finite set $G/N$. So, the isomorphism above is in fact a homeomorphism.