# An example of a local integral domain with special spectrum

I am looking for a local integral domain $$(D, m)$$ with $$Spec(D)=\{0,m\}\cup\{ P_i\}_i$$ such that $$P_i's$$ are incomparable (that is, $$P_i\not\subseteq P_j$$ and $$P_j\not\subseteq P_i$$ for $$i\not= j$$) and $$\cap_i P_i=0$$ and $$\cup_i P_i\not=m$$, where $$Spec(D)$$ is the set of all prime ideals of $$D$$ ($$D$$ is not Noetherian in general).

• Do you insist that $\mathrm{Spec}(D)$ should be countable? Commented Jul 20, 2022 at 13:20
• @Laurent Moret-Bailly: I edited the question. No $Spec(D)$ need not be countable. Commented Jul 21, 2022 at 19:30

Let $$k$$ be a field. Put $$R=k[[x,y]]$$ and let $$D\subset R$$ be the subring $$k+xR$$. It consists of power series without any term $$ay^n$$ ($$a\in k^\times$$, $$n>0$$), or (equivalently) series $$f(x,y)$$ such that $$f(0,y)\in k$$. It is easy to see that $$D$$ is local with maximal ideal $$m=\{f\in D\mid f(0,0)=0\}$$.

I claim that $$D$$ is as required.

First, $$D[x^{-1}]\to R[x^{-1}]$$ is an isomorphism: it is clearly injective by localization, and surjective because for $$g\in R$$ and $$m\in\mathbb{N}$$, we have $$x^{-m}g=x^{-m-1}(xg)$$ and $$xg\in D$$. Thus, the nonzero primes $$P_i$$ of $$D$$ not containing $$x$$ correspond bijectively to the primes of height one in $$R$$ distinct from $$xR$$, and there are infinitely many of these. On the other hand, we have $$m^2\subset xD$$ (check!), so the only prime of $$D$$ containing $$x$$ is $$m$$. Since $$x$$ is not in any of the $$P_i$$'s, their union is not $$m$$.

Here is a geometric explanation: we can view $$D$$ as the fibre product ring $$R\times_{R/xR}k$$. It follows that the natural diagram $$\begin{array}{rcl} \mathrm{Spec}(R/xR) & \longrightarrow & \mathrm{Spec}(R) \\ \downarrow\;&&\;\downarrow\\ \mathrm{Spec}(k) & \longrightarrow & \mathrm{Spec}(D) \end{array}$$ is a pushout of ringed spaces; this is Theorem 5.1 in this paper by Ferrand. In other words, $$\mathrm{Spec}(D)$$ is obtained from $$\mathrm{Spec}(R)$$ by crushing'' the closed subset $$\mathrm{Spec}(R/xR)\cong \mathrm{Spec}(k[[y]])$$ to the point $$\mathrm{Spec}(k)$$.

Put $$D_0 = \mathbb{Q}[T,X_1,X_2,\cdots]/I_0$$, where $$I_0$$ is generated by the elements $$X_i^2-TX_{i+1}$$, for every $$i \in \mathbb{N}$$. This is a domain. Let $$\mathfrak{m}_0$$ $$=$$ $$(T,X_1,X_2,\cdots)$$, and take $$D = (D_0)_{\mathfrak{m}_0}$$, a local ring with maximal ideal $$\mathfrak{m} = \mathfrak{m}_0 D$$.

We have $$D_0 = \varinjlim D_n$$, where $$D_n = \mathbb{Q}[T,X_1,X_2,\cdots,X_n]/I_n$$, with $$I_n$$ generated by the $$X_i^2-TX_{i+1}$$ for $$i < n$$. Since the $$D_n$$ have Krull dimension 2, It follows that $$\mathrm{dim}(D_0) = 2$$, and hence $$\mathrm{dim}(D) = 2$$ as well.

The only prime ideal of $$D$$ that contains $$T$$ (or rather, its image in $$D$$) is $$\mathfrak{m}$$, which is of height 2, so $$\mathfrak{m}$$ is not the union of the height 1 primes of $$D$$.

Let $$R = \mathbb{Q}[T,X_1]$$. It is a subring of $$D_0$$. If $$\mathfrak{p}$$ and $$\mathfrak{q}$$ are different height 1 prime ideals of $$D_0$$ contained in $$\mathfrak{m}_0$$ with $$\mathfrak{p} \cap R$$ = $$\mathfrak{q} \cap R$$, there is an $$a$$ with $$a \in \mathfrak{p} - \mathfrak{q}$$, say. As $$TX_2=X_1^2$$, $$TX_3=X_2^2$$, etc., in $$D_0$$, there exists an $$n$$ such that $$T^na$$ $$\in$$ $$R$$. But then $$T^na$$ $$\in$$ $$\mathfrak{q} \cap R$$, contradiction. This shows that $$\mathrm{spec}(D_0)$$ is at most countable.

Note that for $$z \in \mathbb{Q}$$, the ideal $$\mathfrak{p}_z := \sum_{i \in \mathbb{N}} (X_i - z^{2^{i-1}}T)\cdot D_0$$ is a height 1 prime ideal of $$D_0$$. If $$z_1 \ne z_2$$ and $$\mathfrak{p}_{z_1}$$ = $$\mathfrak{p}_{z_2}$$, then $$X_1-z_1T$$ and $$X_1-z_2T$$ are in this ideal, hence so is $$T$$, contradiction. Hence $$\mathrm{spec}(D_0)$$, and therefore also $$\mathrm{spec}(D)$$, is indeed a countably infinite set.

And if $$a \ne 0$$ is in all height 1 primes of $$D_0$$ contained in $$\mathfrak{m}$$, then, for a suitable $$n \in \mathbb{N}$$, the non-zero element $$T^na$$ is in $$R$$ and in infinitely many prime ideals of $$R$$, another contradiction.